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L for all n E N or
(ii) cp (xn) - L # < (xo) and cp (xn) < L for all n E N.
In both cases, since u is continuous at xo E Idiff and {xn} C Idiff, we have (6.46)
u*
(xn)) = t (x,) - tc (xo) =
u* (cP (xo))
6.3. Derivative of u*
211
By Proposition 6.3 we have that u* is decreasing and right continuous. Hence, (u*)_ (L) > u* (L) = (u*)+ (L). In case (i) we have that p (x,,) -,. L+ and so by (6.46) and the fact that u* is right continuous, it follows that u* (L) = u* (tp (xa)). On the other hand, in case (ii) we have that co (x,,) -p L-, and so by (6.46), (u`)_ (L) = u* (cp (xo)). Since L ip (xo), in both cases we have that cp (xo) lies in or at the end of an interval of constancy of u*. Since (u*)' exists at cp (xo), then, necessarily, (u*)' (gyp (xo)) = 0,
which is a contradiction. Next we consider the case in which API rd,ff is continuous at xo but u' (xo) _ 0. Let {x,,} C Iduff be such that xn -+ xo and cp (x,,) # c (xo) for all n E N
(note that since xa is a point of density one for Idiff and (p is measurepreserving, it cannot map sets of positive measure onto a single point, and so there are plenty of such sequences). Then u* (rp (xn))
- u* (W (xo)) a (xn) - tp (xo) _ u
u (xa)
xn-x0
xn-x0
SP(xn)-cp(xo) If (u* )' (rp (xo)) 74 0, then (6.47)
lim `p (xn) - W (xo) = 0.
n-+oo
xn - xo
On the other hand, if (p (x,,) = p(xo), then (6.47) still holds. This shows that (x) -' (.To) = 0T lim
x-x0
which contradicts the previous step.
Step 4: Let xo E Idiff be such that (u*)' ((p (xo)) 0 0. Then by Step 3, u' (xo) 0 0 and p : Id;ff --> R is continuous at xo. By Step 1, for all x E Id;ff sufficiently close to x0 we may write u* (cp (x)) - u* (_P (xo)) V (x) - _ (x)
cp(x)-cp(xo)
x-xo
u (x) - u (.To)
x - xo
Letting x - xo, we obtain that `plydiff is differentiable at xa E Id;ff with (6.48)
(u*), (T (xo)) (Ild,ft)'(xo) = u' (xo)
.
By Step 2 we have that (wIJdIff) (xo) > 1. Hence, if El :_ {x E Iduff : (u*)' (ip (x)) 0 0 }
,
6. Decreasing Rearrangement
212
then by (6.48),
I
Jj.
(x)) (I1dIff)xr dx
(u*)' (cp (x))I P dx < fE, I
I
= I Iu' (x)Ip
dx <-
f Iu' (x)Ip
dx.
Since (u*)' (tp (x)) = 0 in Iduff \ El and £' (I> \ 'duff) = 0, we have that
j
(x)) IP dx < f Iu' (x) Ip dx.
I (u«)'
,
Finally, using the fact that cp : I> - F> is measure-preserving (see Corollary 6.24), we obtain that
IF> I(u*)'(t)IP dt=
f
I(u*)'(V(x))Ip dx <- f Iu'(x)Ip dx,
which completes the proof. The next exercise shows that decreasing rearrangement does not preserve higher-order regularity.
Exercise 6.29. Consider the function )u2: [0, 2]
u(x):=3+(x-2
[0, oo) defined by
x2)4
Prove that (u*)' is discontinuous.
We refer to the papers of Cianchi [35] and Dahlberg [43] for more information on the regularity of (u*)'.
Corollary 6.30 (Singular functions). Let I C R be an interval and let u : I -> (0, oo) be a function vanishing at infinity and differentiable f- 1 -a. e. in I. Then u is a singular function if and only if u* : (I*)* [0, oo) is a singular function.
Proof. If u is singular, then u' (x) = 0 for G1-a.e. x E I, and so by (6.41) we have that . I(u*)' (t) I dt s f Iu' (x) I dx = 0,
f
which implies that (u*)' (t) = 0 for £1-a.e. t E I*. Conversely, assume that u* : (I*)O -> 10, oo) is singular. Then by Theorem 3.72, there exists a Lebesgue measurable set Fo C (I*)* such that C1((I*)' \ Fo) = 0 and ,C' (u* (Fo)) = 0.
Let I> and F> be the sets defined in (6.38) (with I in place of E). By Corollary 6.24 the function W : I> --+ F> is measure-preserving and u* (go (x)) = u (x) for ,C'-a.e. X E I>.
6.3. Derivative of u*
213
Define Go :_ (I \ I>) U cp-1 (Fo n F>). We claim that C (I \ Go) = 0 and .C1(u (Go)) = 0. To see this, note that since u = 0 in I \ I>, we have that G1 (u (I \ I>)) = 0. Moreover, £1 (u (cp-1(Fo n F>))) = C1(u` (Fo n F>)) = 0. Thus, G1 (u (Go)) = 0. On the other hand,
1(I \ Go) = c1 (I> \ cp-1(Fo n F>)) = C1 (co 1(F>) \ cP-1(Fo n F>)) ='C (,p-' (F> \Fo)) = C' (F> \Fo) = 0.
This proves that the claim holds. We are now in a position to apply Theorem 3.72 to conclude that u is singular. 0
Chapter 7
Functions of Bounded Variation and Sobolev Functions Prospective Grad Students, III: "Will your qualifying exams procedure utterly destroy my dignity and sense of self-respect?" - Jorge Chain, www.plidcomics.com
In Chapter 5 we have seen that to every function u of bounded pointwise variation we may associate a signed measure A, for which the integration by parts formula (5.53) holds. From the distributional point of view (see Chapter 9), this implies that the distributional or weak derivative of u is the signed measure A,,. In the following section we will see that this leads us to an "alternative" characterization of the space of functions of bounded pointwise variation, which has the advantage of carrying over to functions defined in RN
7.1. BV (Q) Versus BPV (Q) Definition 7.1. Given an open set Q C R, the space of functions of bounded variation BV (1k) is defined as the space of all functions u E Ll (Sl) for which there exists a finite signed Radon measure A : B (I)) - 1[8 such that
ud dx = - J ip dA
in
for all W E C' (1k). The measure A is called the weak or distributional derivative of u and is denoted Du. The next result establishes the relationship between functions of bounded variation and functions of bounded pointwise variation. 215
7. Functions of Bounded Variation and Sobolev Functions
216
Theorem 7.2. Let 0 C IR be an open set. If u : £1-> R is integrable and if it belongs to BPV (Q), then u E BV (S1) and IDul (S2) < Varu.
Conversely, if u E BV (Q), then u admits a right continuous representative
la in BPV (0) such that Var u = IDul (fl) .
We begin with an auxiliary result.
Lemma 7.3. Given an open interval I C R, let u E L10 (I) be such that
I
ucp'dx=0
for all cp E Cr' (I). Then there exists a constant c E R such that u (x) = c for L1-a.e. X E I.
Proof. Step 1: The proof is quite simple in the case in which u E L' (I) and I = (a, b), and thus we begin with this case. A density argument shows
that b
( 7.1)
urp' dx = 0
L
for all 9 E C1 ([a, b]) with cp (a) = v (b) = 0. We claim that b
A for all w E C ([a, b]) with fQ w dt = 0. To we this, fix any such w and define the function := dt, x E I.
jw(t)
Since cp (a) = 0 and rp (b) = fb w (t) dt = 0, we have that cp is admissible in (7.1), and so the claim holds. Let now w E C ([a, b]). Taking w - 4 fQ w dt in (7.2), we get
bu(w
1. which can be written as
- blaJbwdt) do;=0, a
I wlu-b-a Ja udt dx=0. b
((
rb
1
Since this is true for all w E C ([a, b)), again by density we obtain that
!abXE(u_baLit) 1
dx=0
217
7.1. BV (S2) Versus BPV (S2)
for all Lebesgue measurable sets E C (a, b). This implies that (why?) fb
u(x)-b1a J udt=0 a
for G1-a.e. x E I. Step 2: In the general case, fI u dx may not be defined. To circumvent this problem, we fix a function 0 E CC (I) such that LO dx = 1. We will prove that
U (X) - j (u &) dt = 0 for G1-a.e. x E I. For every function w E CC (I) we can find a function cp E CC (I) such
that gyp' (x) = w (x)
- (if w ds) % (x) , x E I.
/
Indeed, find an interval [a, b] C I such that supp w U supp'r& C [a, b] and define the function W (x) .- T I Lw (t) -
(fwds) .0 (t)1
dt,
x E I.
a
Since +p (a) = 0 and W' (x) = 0 in (inf I, a], we have that p = 0 in (inf I, a]. Similarly, since V (b) =
Ja
b
(f) (tJ dt
1W (t) -
(
= J [W (t) -
w ds)
+1i
(t)] dt = 0
and cp' (x) = 0 in [b, sup I), we have that W = 0 in [b, sup I). Thus rp E C,1 (I) and so, by hypothesis,
j[_ (jwds)1] dx=0, or, equivalently,
J to [u JJJI
JI
(utP) ds] dx = 0
for all to E CC (I). Using a density rargument, it follows that
f XE IU - J (ut) I
dig]
dx = 0
I
for all Lebesgue measurable sets E C I with finite measure. As before, this implies that
u (x) - 1(W) ds = for G1-a.e. x E I.
218
7. Functions of Bounded Variation and Sobolev Functions
Proof of Theorem 7.2. By working on each connected component of fl, we may assume that f = I, with I C ]R an open interval. Assume that u : I -> R is integrable and that it belongs to BPV (I). Let A,, be defined as in (5.19). It follows by Theorem 5.13 that 1At,l is finite with 1,\. I (I) < Var u.
Moreover, by Corollary 5.41 we have that
1 for all cpEQ1 (I). Hence, n E BV (I). Conversely, if u E BV (I), let A : B (I) -+ R be the finite signed Radon measure given in Definition 7.1. Then ucp' dx =
- J pd,\
for all cp E C- (I). Define ua as in (5.17), where a E I is a Lebesgue point of u and -y = u (a). Then by Remark 5.12, the function ua is right continuous and has bounded pointwise variation and Var u.\ = JA I (I). Moreover, by Corollary 5.41, 1,
uacp'dx = -J cpdA J
for all VECc' (I). It follows that
J.
(u-ua)cp'dx = 0
for all cp E C' (I). By the previous lemma, this implies that u-u,\ is constant
for G1-a.e. x E I. Since a E I is a Lebesgue point of u and ua (a) = u (a), we have that u (x) = ua (x) for £'-a.e. x E I, and so ua is a representative of U.
0
Remark 7.4. Since a function of bounded pointwise variation is bounded (see Corollary 2.23), if I is bounded, then u is integrable, and so in the first part of the statement of Theorem 7.2 the hypothesis that u is integrable is redundant. However, when I is unbounded, the function u - 1 has bounded pointwise variation, but it is not integrable and thus does not belong to BV (I). An alternative proof of the first part of the previous theorem is given by the next exercise.
Exercise 7.5. Let U E BVP ((a, b)).
7.1. BV (S2) Versus BPV (fl)
219
(i) Prove that u is Riemann integrable and that for every cp E C' (a, b) the Riemann integral Jab u(p' dx
is well-defined.
(ii) Prove that for every .p E C' (a, b) and for every e > 0 there exists a partition of [a, b],
xo:=a<xl <...<xn:=b, such that
jb
<
u (xi) co' (xi) (xi - xi-1) + S. i-1 (iii) Prove that the partition in part (ii) may be chosen so that n
n
u (xi) w' (xi) (xi - xi-1) !5 i=1
u (xi) [W (xi) - V (xi-1)I + C. i=1
(iv) Prove that for every cp E CC (a, b), t' dx S sup 191 Var(a,b) u. a
[a,bJ
(v) Prove that the functional --+ R 11W'dx d
is linear and that it can be extended uniquely to a linear continuous functional L : Co (a, b) - R, with IILII(Co(a,b)) < Var(a,b) U.
Exercise 7.6. Let St C R be an open set. Prove that the space BV (92) is a Banach space with the norm IIUIIBV(n) := IIUIIL'(n) + IDuI (SZ)
Next we introduce a notion of pointwise variation that does not change if we change the representative of a function u E L' (I).
Definition 7.7. Let f l C 1 be an open set and let u : f2 -+ 1 be a locally integrable function. The essential variation of u is defined as essVar u := inf {Var v : v : 0 -, R, v (x) = u (x) for ,C'-a.e. x E fZ } .
220
7. Functions of Bounded Variation and Sobolev Functions
It follows from the definition that if S2 C R is an open set and u : 92 R : 92 --> R that coincides with u except on a set of Lebesgue measure zero,
is a locally integrable function, then for every function v essVar u = essVar v.
Theorem 7.8. Let Il c R be an open set and let u E L1(Il) . Then essVar u < oo if and only if u belongs to BV (S2). Moreover, I Dul (12) = essVar u.
Proof. Assume that essVar u < oo. Then for every e > 0 there exists a representative v of u such that
Varv <essVaru+s < 00. By Theorem 7.2, it follows that v (and hence u) belongs to BV (S2) and I Dul (1) < Var v < essVar u + e.
Given the arbitrariness of E > 0, we conclude that I Dul (fl) < essVar u.
Conversely, assume that u E BV (Il). Then by Theorem 7.2, u admits a right continuous representative U, which has bounded pointwise variation and VarU = IDul (0). In particular, essVar u <_ Var V = I Dul (St)
.
0
This concludes the proof.
Exercise 7.9. Let I C R be an open interval and let u E BV (I). Prove that n
essVarr u = sup E Iu (x;) - u (xi-1)1 i=1
where the supremum is taken over all partitions P :_ {xo,... , xn} of I such that each xz is a Lebesgue point of u, i = 1, ... , n, n E N. In the remainder of this section we extend some of the previous results to functions that have only locally bounded variation. The space of functions of locally bounded variation BV., (S2) is defined as the space of all functions u E I.i C (fl) such that u E BV (n') for all open sets Cl' Cc Cl. Note that for a function u E BVbc, (Cl), the distributional derivative Du may not be a finite signed measure, but it may be defined as the difference of two nonnegative Radon measures.
7.1. BV (12) Versus BPV (St)
221
Proposition 7.10. Let St C R be an open set and Let u E BV]", (St). Then there exist two Radon measures µ : ,B (ft) - [0, co] and v : B (ft) - [0, co] such that
j
f
cp d1.t - f cp dv
for all cp E Q' (Q).
Proof. Let ftn be an increasing sequence of open sets such that 0, cc ft and
00
UOn=f1. n=1
Since u E BV (ftn), there exists a finite signed Radon measure A(n) 8 (Stn) - R such that
L
ucp' dx =
- Ion
cp d
A()
for all cp E C' (1l, ). We claim that the restriction of A(7h+1) to B (On) coincides with A("). Indeed, since every cp E C' (SZn) can be extended to for every cp E zero outside On so that it belongs to C' we have
-r
(p
dA(n+1)
= -J
nn
'pdA(n+1) = fnn4-1 ucp'dx
nn+1
=f ucp' dx = n
and so
f
Jnn
cp dA(n),
- A(n)) = 0.
cod
1J
Since A(':+1) and A(n) are finite signed measures, a density argument yields
that ,pd
(A('+') -A(n)) = 0
ff2.
for all cp E CC (Stn), which implies that A(n+1) restricted to B (Stn) coincides with A(n). Thus, if cp E Cc (fl), letting n be so large that supp cp C Stn, we may define L (cp) :_
fscdA().
The functional L is well-defined, linear, and locally bounded. By the Riesz representation theorem (see Theorem B.115) there exist two Radon measures /i (St) -' [0, oo] and v : 5 (SZ) --+jcodu_/codv [0, oo] such that L (cp) =
7. Functions of Bounded Variation and Sobolev Functions
222
for all cp E CC (2). Moreover, for every cp E CC (12),
fwP'dx= - f cp dl- J cp dy. n
l
st
0 In view of Corollary 5.41 we have the following theorem.
Theorem 7.11. Let 12 C R be an open set. If u : fl -+ R has locally bounded pointwise variation, then u has locally bounded variation. Conversely, if u E L10C (0) has locally bounded variation, then u admits a right continuous representative that has locally bounded pointwise variation.
Proof. The proof is very similar to the one of Theorem 7.2 and is left as an exercise.
7.2. Sobolev Functions Versus Absolutely Continuous Functions In this section we introduce the notion of Sobolev functions and study their relation to absolutely continuous functions.
Definition 7.12. Given an open set fl C R and 1 < p < oo, the Sobolev space W 1-P (S2) is the space of all functions u r= L' (0) for which there exists a function v E LP (S2) such that
L
ucp' dx = - J vcp dx n
for all cp E C' (S2). The function v is called the weak or distributional derivative of u and is denoted u'. The space IV, "P (S2) is defined as the space of all functions u E Ll 42) such that u e W 1,P (S2') for all open sets 12' cc S2.
Note that W1"1(1) C BV (S2). Indeed, if u E W1,1 (fl), then in the definition of BV (0), we can take the signed measure
A(E):=u'dx, EC8(12). We will show that there is a very close relation between Sobolev functions and absolutely continuous functions. We begin by studying the case 1 < p < 00.
Theorem 7.13. Let S2 C R be an open set, let u : S2 - R, and let 1 <_ p < oo. Then u E W" (12) if and only if it admits an absolutely continuous representative u : f2 -+ R such that u and its classical derivative u' belong to LP (S2). Moreover, if p > 1, then z is Holder continuous of exponent 1/p'.
7.2. Sobolev Functions Versus Absolutely Continuous Functions
223
Proof. By working on each connected component of fl, we may assume that SZ = I, with I C R an open interval. If fA : I - R is absolutely continuous, then by Corollary 3.37,
juip'dx=
- J u dx
for all p E C1 (I). Since R E L" (I) and 2b' E 1]' (I), it follows that u' is the weak derivative of u, and so u E W 1,P (1) -
Conversely, assume that u E W 1,P (I) and let v be its weak derivative. Fix a Lebesgue point xo E I of u and define (7.3)
u (x) := u (xo) +
dt,
x E I.
o
Since v E I' (fZ), we have that v is locally integrable, and thus, by applying Lemma 3.31 to closed intervals [a, b] C fI, we have that -a is locally absolutely continuous, with u' (x) = v (x) for G1-a.e. x E I, and, in turn, by Corollary 3.26 we have that u is actually absolutely continuous in I. Using Corollary 3.37, we get juc'dx= -
u'cp dx = - J vV dx
J
for all cp E CC (I). Hence, we have that
(u-u)rp'dx=0 for all cp E C' (I). By Lemma 7.3 it follows that u - u is constant L1-a.e. in I, and since xo E I is a Lebesgue point of u and u (xo) = u (xo), we have that u = u G1-a.e. in I. This shows that u has an absolutely continuous representative. To prove that u. is Holder continuous of exponent 1/p', let x, y r: I, with x < y. By Theorem 3.30,
u (x) - u (y) =
v (t) dt,
and so, by Holder's inequality, x
(7.4)
Iu (x) - u (y)1 <-
Iv (t)I dt < (x - y)1IP,
v
< (x -
lIp
r rx
`J v
Iv (t) IP dt)
11P y)1/P'
Jo Iv (t)1" dt)
which shows that u is Holder continuous of exponent 1/p'.
0
Corollary 7.14. Let fl C R be an open bounded set and let u : Il -> R. Then u E W1,1 (f) if and only if it admits an absolutely continuous representative
7. Functions of Bounded Variation and Sobolev Functions
224
Proof. If u : SZ - R is absolutely continuous, then by Corollary 3.9, u is bounded, and so integrable, and u' is integrable. We are now in a position to apply the previous theorem.
Corollary 7.15. Let tZ C R be an open bounded set, let u : 0 -' R, and let 1 < p < oo. Then u e W'.11(0) if and only if it admits an absolutely continuous representative z : S2 -> R such that its classical derivative u` belongs to LP (Q).
Proof. Since for sets of finite measure I? (f') C L' (0), we have that W 1,P (52) C W1,1 (Sl). Thus, the result follows from Corollary 7.14.
The next exercise gives an example of an absolutely continuous function in L' (R) \ W1'1 (R).
Exercise 7.16. Let 1 if xE n n if xE and let
(n-1+2k ,n-1+2-j
mEN,0
nEN,0
411-1n-1+2
fog (t) dt if x > O, U(X) '_ I u (-x) if X < 0.
(i) Prove that u is absolutely continuous. (ii) Prove that u E L' (R). (iii) Prove that u' E ne>o L1+-- (R) \ L' (R). Finally, we consider the case p = oo.
Theorem 7.17. Let I C R be an open interval. Then u E Wi,00 (I) if and only if it admits a bounded, Lipschitz continuous representative u : I - R. Proof. If u : I -, R is Lipschitz continuous, then it is absolutely continuous, and so, as in the proof of Theorem 7.13, we conclude that u E W1"1 (I). Conversely, assume that u E W1,°° (I) and define the function u as in (7.3). Since v E L00 (I), we have that v is Lipschitz continuous and locally integrable, and thus we may proceed as in the proof of Theorem 7.13 to conclude that u = u G1-a.e. in 1.
Exercise 7.18. Let St C R be an open set and let 1 < p < oo. Prove that the space W' ' (St) is a Banach space with the norm IIUIIwl.Ptn)
IIUIILP() +
IIT'IILP(n)
Hint: Use the Ascoli-Arzela theorem.
Exercise 7.19. Let St C R be an open set and let 1 < p < oo. Prove that the space W 1,P (SZ) is a separable space.
7.2. Sobolev Functions Versus Absolutely Continuous Functions
225
Exercise 7.20. Consider the function
u(x):=xlog!, xE(0,1). Prove that for every 0 < a < 1 the function u is Holder continuous of exponent 0 < a < 1 but not Lipschitz. Is u absolutely continuous?
Theorem 7.21 (Poincare's inequality). Let I = (a, b) and let 1 < p < oo. Then
fb
J
(7.5)
2
Iu (x) - uj Ip dx <
(b - a) P
r J
b lu' (x) Ip dx
for all is r= W1,P ((a, b)), where b
1
ur' b-aJ u(x)dx. a Proof. Fix is E W ',P ((a, b) ). By Corollaries 7.14 and 7.15 there exists an absolutely continuous representative of is, u : (a, b) - R, such that u' belong to LP ((a, b)). By Exercise 3.7 we may extend fs to [a, b] in such a way that the extension is still absolutely continuous. Since u is continuous, by the mean value theorem there exists xp E (a, b) such that ra
uj= b 1 aJ
ii (x)
dx=u(xo).
a
By Theorem 3.30, for all x E (a, b),
u (x) - ul =
u' (t) dt, xo
and so as in (7.4), 1/p
b
I u (x) - nil < [x -X01 l/p `f I u' (t) I' dt)
j6/'
Raising both sides to exponent p and integrating in x gives
jb
dx
<
(b - a)p P
where we have used the fact that
rs
J
dx
J
lu' (t) Ip da
lu'(t)I' dt,
0
= p - 1.
Exercise 7.22. Let is E AC ((a, b)). (i) Prove that for all x E (a, b), 1 f l (s - a) u' (s) ds - f b (b - s) u' (a) ds] R (X) - ui = b 1 a f LLL
Ja
JJJ
7. Functions of Bounded Variation and Sobolev Functions
226
(ii) Prove that
m a a ($-a) (b-8) =
4a)x
(b
(iii) Prove that
f b Iu (x) - uiI dx < b - a f b It/ (x) I dz. 2
a
a
(iv) Prove that the constant b-a is sharp. We conclude this section with a weighted Poincare's inequality, which will be used to prove Poincar6's inequality in convex domains (see Theorem 12.30).
Proposition 7.23 (Weighted Poincar4's inequality). Let g : [0, d] - [0, oo), g 0, be increasing in [0, c] and decreasing in [c, d] for some 0 < c < d and let 1 < p < oo. Then there exists a constant Cp > 0 depending only on p such that (7.6)
fiu (x) -:s(o,d) p g (x) dx < CPd& J d
for all u E W 1-P ((0, d)), where u(o'd)
fog (x) dx
f
a
I u' (x) Ip g (x) dx
u (x) g (x) dx.
Proof. If g = 0 on some interval [0, b] (respectively, [b, d]), then all the integrals involved reduce to integrals over [b, d] (respectively, (0, b)). Thus, we
can assume that g is strictly positive in (0, d). Also, by a scaling argument, it is enough to prove the result for d = 1. Finally, by dividing the inequality (7.6) by fag (x) dx, we may assume that I
A By Corollaries 7.14 and 7.15, without loss of generality we may assume that u is absolutely continuous. By' Theorem 3.30 and (7.7),
Lf
u (x) - u(o,l) = u (x) - J 1u (t) g (t) dt = fo 1 [u (x) - u (t)] g (t) dt
f Jf= 0
u' (a) g (t) dsdt
=
JJu'(8)9(t) d8dt - J
= fo u' (s)1 g (t) dtds o
f
x
f
d8 dt
x
u' (s) J g (t) dtds, e
7.2. Sobolev Functions Versus Absolutely Continuous Functions
227
where in the last equality we have used F ubini's theorem. Hence,
Iu (x) - u(o.l) I <
f
s
I u' (s) I
J
1
g (t) dtds +
1
J lu' (s) I
! g (t) dtds.
Raising both sides to the power p, multiplying by g (x), and integrating over (0,1] gives
Iu (x) - ZL(0,1)
Jo
IP
9 (x) d x < f 1 ( f z Iu' (s) I 0g (t) dtds P
+ f 1 Jul (s) I f g (t) dtds j x 8
(x) dx
(JZ
< 2P-1 fl + 2P-1 f
g
Jul (s) I
1 a
l J
\z
1
8
J
j
Jul (s) I
g (t) dtds I P g (x) dx 1
g (t) dtds ]Pg (x) dx
S
=:Z+2Z, where in the second inequality we have used the inequality n
n
P
< nP-1: ap,
sail
i=1
i=1
where a; E R, i = 1, ... , n, which follows from the convexity of the function f (t) :_ ItIP, t E R. We estimate 1. If 0 C t < s < c, then by hypothesis g (t) < g (s), so for
0<x
/ rx I
Jo
I u' (s) I
fs u' (s) I [9 (s)] p J
pa `P / g (t) dtds I <
< <
,
a
1
P
[g (t)] P dtds
(jau(8)g(8)]*
ds IP
f
'
Iu'(s)IPg(s) ds,
where we have used Holder's inequality twice and (7.7). On the other hand,
if c < s < x < 1, then by hypothesis g (s) > g (x), and so for c < x < 1 we
7. Functions of Bounded Variation and Sobolev Functions
228
have that
(LX Iu' (s) I f9(t)
_
J c I u' (s) I
dtds)
r g (t) dtds + J y I u' (8) I
<
2p-1
(J C Its' (s) I
<
2p-1
(J 1
<
2p-1 r
!
g (x)
f
2p-1 (fx Iu' (8)1 ds)
f " g (t) dtd8) P g (x) +
ju' (s) Ip
I,u' (8) I p
g (s) ds)
g(x) +
2p-1
g (8) d8 J g (x) + 2"1
g (t) dtds) p g (x)
(j (j
Iu' (s)I
g (x) ds)P
Ig (8))'
Iu' (8)lpg (8)
d8
again by Holder's inequality and (7.7). Thus, for all0 < x < 1, Joz Iu' (s) I f e g (t) dsds) g (x) 2p-1
(f
I,u'
(s) I p g (s)
ds) (g (x) + 1),
and, in turn, using (7.7) once more,
1 < C (p) 1 1 Iu (s)Ipg (s) ds Jol (g (x) + 1) dx r1
=
2C (p)
l
I1' (8) I' g (8) ds. 0
A similar estimate holds for 22, thus giving the desired result for p > 1. The proof in the case p = 1 is simpler, since there is no need to use Holder's inequality.
Part 2
Functions of Several Variables
Chapter 8
Absolutely Continuous Functions and Change of Variables Prospective Grad Students, IV. "Can you really live comfortably in this major metropolitan area with that stipend, or will I find myself living out of a closet working part time as a shoe sales-
man?" - Jorge Cham, www.phdcomics.com
In this chapter we extend some of the concepts and results of Chapter 3 to functions of several variables, including the notion of absolute continuity, the Lusin (N) property, and some change of variables formulas.
8.1. The Euclidean Space RI In the remainder of this book the implicit space is the Euclidean space RN. The elements of RN are N-tuples of real numbers x = (Si, ... , xN).1 Given
x = (x1, ... , XN) E RN and y = (yl, ... , YN) E RN, the Euclidean inner product of x and y is the real number defined by N
x'y
xiq
ili7
i=1
iWhen there is no possibility of confusion, we will also use xl,z2, etc., to denote different points of RN. Thus, depending on the context, z; is either a point of RN or the ith coordinate of the point z E RN. 231
S. Absolutely Continuous Functions and Change of Variables
232
while the Euclidean norm of x is the number N
txz.
IxI:=
When working with Euclidean spaces of different dimension, we will sometime use the notation IxI N for Ixl. We will often use the Cauchy's inequality: Ix - V1 <_ ixl lyI
for all x, y E RN. The unit vectors
el := (1,0,...,0), e2:= (0,1,0,...,0) eN := (0,.. ., 0, 1)
form the standard, or canonical, orthonormal basis of RN. For each x E RN we have
N
X = (xl, ... , XN)
xiei.
W e will refer to x = (x1, ... , rN) as background coordinates. We will call local coordinates y = (yl, ..., yN) at a point x = xo new coordinates related to background coordinates by a rigid motion, that is, an affine transformation of the form
y := Lxo (x) = Rxo (x - xo) , (L0)1 (y) = xa + (R0)-1 y, where R,,. is an orthogonal N x N matrix. Given x = (X 1, ... , xN) E RN, for every i = 1, ... , N we denote by xs the (N - 1)-dimensional vector obtained by removing the ith component from x and with an abuse of notation we write (8.1)
x = (xs, xi) E RN-1 X R.
(8.2)
When i = N, we will also use the simpler notation
x = (x', x v) E R'-' x R.
(8.3)
Given x E RN and r > 0, the open ball of center x and radius r is the set
B(x,r):= {yERN: Ix-9!I
Q (x, r) := x + l- 2' )
i
= {y E
RN
:
iyi - xi, <
i for all i = 1,...,N Y
8.1. The Euclidean Space RN
233
The unit sphere SN-1 is the boundary of the unit ball B (0, 1); that is, S^'-1 ={y ERN: Iyj=1}. Given a set E C RN, a transformation 19 : E -> IBM is the differentiable
at some interior point xo E E° if there exists a linear transformation L RN - RM depending on xo such that IP (x) - 'P (xo) - L (x - xo)Int = 0. lim x-tx° Ix-xp1N The linear transformation L is called the differential of 4< at xo and is denoted d4' (xe). It may be shown that if i is differentiable at xo, then (i) qP is continuous at xo,
(ii) 41 admits all partial derivatives
4< {xo + to } -
091 (8.4)
ax,
(xo)
(xo) := m
at xo, (iii) 4< admits all directional derivatives
Q (xo + ty) t av (xo) :- l ta at xo, (iv) for all v = (vl, ... , vN) E SN-, 01Q
(MO)
v E SN-1,
N
av
(xo) = E ax i=1
(x0) vi,
(v) for all v = (v1, ... , vN) E SN-1, alp
dW(xo)(v). 8v (fro) =
Exercise 8.1. Construct a transformation P : RIV -p IBM that satisfies properties (i)-(iv) at some xo r= RN but which is not differentiable at xo. Compare the previous exercise with Exercise 11.48. Given a set E C RN and a transformation T : E -> IBM, the gradient of 4< = ('P 1, . . . ,'L)
at some interior point xo E E°, whenever it exists, is the
M x N matrix Vi I (xo) VIP (xo)
VTM (xo) where V%Pi (xo) .=
(1(xo),...,.(xo)),i=1,...,M. When M = N,
VW (xo) is an N x N square matrix and its determinant is called the Jacobian
234
8. Absolutely Continuous Functions and Change of Variables
of'P at xo and is denoted by JT (xo). Thus,
`
JW(xo):=detVT(xo)=detI O!'(xo)I exi
Another common notation for the Jacobian of
at xo is
TN) (xo).
('Q I,
8 (xl, , XN) For more information on differentiability of functions of several variables, we refer the reader to the books of Bartle 1151 and Fleming [621.
8.2. Absolutely Continuous Functions of Several Variables Before introducing the notion of absolute continuity for functions of several variables, we begin by proving some results on differentiable transformations. In what follows, 1 C RN is an open set, possibly unbounded.
The first result of this section, which is the analog of Corollary 3.14, implies, in particular, that every differentiable transformation IF : tZ -p RN has the Lusin (N) property; that is, it maps sets of 'CN-measure zero into sets of £N-measure zero. Proposition 8.2. Let 92 C RN be an open set and let T : fI - RN. Assume that there exists a set E C S2 of Lebesgue measure zero such that 9 is differentiable for all x E E. Then GN ('P (E)) = 0.
Proof. For every n, k E N let .,,,k be the set of all points x E E for which dirt (x, 00) > and IT (x) -'f` (v)I < k 12., - VI
for all y E fl such that [x - yI < n. Since %F is differentiable on E, it follows
that
00
E = U En,k, n,k=1
and so it suffices to show that £N ('P (En,k)) = 0. To see this, fix n, k E N and e > 0. Since .CN (En,k) = 0, there exists a sequence {Q (xi, ri)} of cubes with center xi and side length ri < 2 1Nn such that
En,k C UQ(Xi,ri) i and
rv <6.
8.2. Absolutely Continuous Functions of Several Variables
235
If x E En,k n Q (x47 ri), then Ix - xil < riv 1Y <- 1 and, since dist (x, 8.Q) > ,, it follows that xi c 12. By the definition of E7z,k we have that JtW (x) - W (xi)I S k Ix - xil < krivlY
for all x E E,,,k nQ (xi, ri). Thus, P (E,,,k n Q (xi, ri)) is contained in a cube of center I& (x4) and side length kriV, and so Lo (IF (E,,,k n Q (xi, ri))) <_ kNrNN
Summing over i, we get Go (W (E,,,k)) <
L
(W (E,s,k n Q (xi, ri)))
i
< kNN 2 > rN < kNN h E.
i
Given the arbitrariness of E > 0, we obtain that GN (' (Ef,k)) = 0.
Corollary 8.3. Let i C RN be any open set and let W : ft -, RN be a differentiable transformation. Then 'I' has the Lusin (N) property; that is, GN (IF (E)) = 0 for every Lebesgue measurable set E C 5t with tCN (E) = 0. Using Proposition 8.2, we can prove the following result, which extends Lemma 3.13.
Theorem 8.4. Let it C RN be any open set and let T : SZ -f RN. Assume that there exist a set E C 1 (not necessarily measurable) and M >_ 0 such that IF is differentiable for all x E E and I JW (x) I < M for all x E E. Then
Go (IP (E)) < MLO (E).
To prove the theorem, we need some auxiliary results. The first is a special case of the change of variables formula for multiple integrals.
Lemma 8.5. Let L : RN -> RN be an invertible linear transformation. Then the change of variables formula (8.5)
r
u (y) dy = Ides LI J N u (L (x)) dx
holds for every Lebesgue integrable function u : RN
R.
S. Absolutely Continuous Functions and Change of Variables
236
Proof. Any invertible linear transformation can be written as a composition of linear invertible transformations of three basic types: to (x)
(8x1, x2, ... , XN),
a(x) 8ij (x) = Sij (XI,...,xi,...)xj,...,XN) := (x1,...,xj,...,xi,...,xN) for x = (x1, x2, ... , xN) E RN and where s e R \ {0}. Since the determinant
of a composition of two invertible linear transformations is the product of their determinants, it suffices to verify the result for these three special types of transformations. By Fubini's theorem and the one-dimensional change of variables v (yl) dyl = 181
v (8x1) dx 1,
JR
which holds for all Lebesgue integrable functions v : R -i R, we have (t8 (x)) dx
Idet tsl I
RN RI
f
JU
RN-1JR
= ISI J
RN.1
=
(Sx1) x2, ... ,
XN) dxldx2 ... d IN
Ju(Y1x2i...xN)dY1dx2...dxN
u (y) dy.
AN
Similarly, again by Fubini's theorem and the one-dimensional change of variables JR
v (y1) dyl = f v (xi + s) dx1, R
which holds for all Lebesgue integrable functions v : R -4R and all s E R, we have
Ju(a(x)) dx = J N_1 .1R 2(x1+x2,x2,...,XN)
Ju(Y1x2...xN)dv1dx2...dxN RN -1 -1
=
f
RN
u (y) dy.
8.2. Absolutely Continuous Functions of Several Variables
237
Finally,
f
N
n (sij (x)) dx
= fR ... fR u(xlxj)
xi,
xN)
dxl...dxj...dxq...dxN
= fR - - - fR U (xl, ... xi, ... , xj.... , xN) dxl ... dxi ... dxj ... dxN
=
J
u (y) dy, fN
0
again by Fubini's theorem.
Remark 8.6. By taking u = XE in (8.5), where E C RN is a Lebesgue measurable set with finite measure, we get LN (L (E)) = Idet LI LN (E)
.
In particular, if L is a rigid motion, then Idet LI = 1, and so LN (L (E)) _ LN (E). In the following lemma for every set F C RN and every e > 0 we use the notation Z. :_ {x E RN : dist (x, F) < e}.
Lemma 8.7. Let L : RN RN be a linear transformation and let F L (Q(xo,r)), where xo E RN and r > 0. Then for every e > 0, LN (Fir) <_ LN \Q (xo, r)) (Idet LI -I- C (N) (IILII +
E)N-1 E)
Proof. Step 1: Let E C RN be a Lebesgue measurable set contained in the intersection of a hyperplane H with a closed ball B (yo, r), where yo E E
and r > 0. We claim that for every e > 0, LN(Es) !5
2N(r+e)N-1e.
In view of the previous remark, by applying a rigid motion, we may assume
that yo = 0 and that H = {xN = 0}. It follows that Ee is contained in the rectangular parallelepiped R (-e - r, E + r)N-1 x (-e, e), and so LN (Ea) C LN (E) = 2Ne (e +r) N-1
This proves the claim.
Step 2: To prove the lemma, assume first that det L = 0. In this case F is contained in a hyperplane H. Taking yo := L (xo), for all x E Q (xo, r) we have
IL (x) -yoI = IL(x - xo)I S IILII Ix - xol <
2 IILIIr,
8. Absolutely Continuous Functions and Change of Variables
238
and so we may apply the previous step (with s and r replaced by sr and 'C IILII r, respectively) to conclude that N-i E)N-i < C (N) rN (IILII + LN (F&r) < 2Ner 2 IILII r + Er Suppose next that det L j 0. Applying the previous lemma with u XF, we have that
LN (F) = L"" (L (Q(xo,r))) = Idet LI LN (Q(xo,r)) = Idet LI rN, and so it remains to show that the open set FEr\F has measure not exceeding
C (N) rN (IILII + E)N-i E. Since F is compact, for each y E FFr \ F there exists x E OF such that Iy - xI = dist (y, F). Since OF is the image by L of the boundary of Q (xo, r), it is enough to prove that if F' is a face of Q (xo, r) and G := L (F'), then +e)N-iE.
LN(GEr)
To prove this, observe that L (F') is contained in a hyperplane. Choosing yo to be the image of the center xo of F', for all x E F' we have IL (x) - yoI = IL (x - xo)I <- IILII Ix
- xol <- 2-1 IILII r,
and so we may apply the previous step (with e and r replaced by sr and
'
IILII r, respectively) to conclude that N-i
LN (GEr) < 2Nr (er +
2
II LII r
< C (N) LN (Q (x-o, r)) (IILII +
s)N-i
E.
0
This completes the proof.
We turn to the proof of Theorem 8.4.
Proof of Theorem 8.4. Without loss of generality, we may assume that ,Co (E) < oo. Fix e > 0 and choose an open subset A C f such that E C A and (8.6)
Vv (A) < Lo (E) + e.
We claim that for every xo E E there exists r 0 > 0 such that Q (xo, r) C A and (8.7)
Lo (41 (Q(xo,r))) < (M + e) LN (Q(xor))
8.2. Absolutely Continuous Functions of Several Variables
239
for all 0 < r < r,,,. Since 'I' is differentiable at xo, for every 8 > 0 there exists rya > 0 such that for all 0 < r < rx0 we have that Q (so, rxo) C A and
I'p(x)-'p(xo)-V'I'(xo)(x-xo)I <6Ix-xo1
from the point V t (xo) x. It follows that this translate of t (Q (xo, r)) is
contained in the set Fa#r, where F := L OTOI r)) and L : RN -> RN is the linear transformation L (x) := VIP (xo) x,
x E ]RN.
Hence, by Lemma 8.7, for all 0 < r < rxo,
LQ ('p (Q(xo,r))) = Go (-'I' (xo) + V p (X,)) X,, +'I' (Q(xo,r))) < GN (F,,.....)
(8.8)
< LN (Q (xo, r)) (Idet LI + C (N) (IILII + 6)rr-1 b) Since, by hypothesis, IdetLI = Idet V'I' (xo)I = IJ'I' (xo)I < M and IILII
= IV'I' (xo)I, taking b > 0 so small that C (N) (I VT (xo)I +
6)N-1
b < s,
we have proved the claim. Let F be the family of all closed cubes contained in A that are centered
at x E E and satisfy (8.7). By the Vitali-Besicovitch covering theorem (see Theorem B.118) there exists a countable family {Q,,} C F of pairwise disjoint cubes such that
Lrv E\UQ,,
0.
Let F:=E\U,LQ,,. Then ,Co (10 (E))
(irs (UQn)) +Lo (ID (F)). >L
S. Absolutely Continuous Functions and Change of Variables
240
By Proposition 8.2 we have that ,rCN ('W (F)) = 0, while by Lemma 8.7, the fact that the cubes are disjoint and contained in A, and (8.6),
£o
(u)) C E Qn
n
CN
(w
.
(Q ))
: 5 (M + E)
E'LN
(en)
n
<(M+s)RCN(A)<(M+e)('Ca The result now follows from the arbitrariness of e > 0.
Remark 8.8. In Chapter 3 we used Lemma 3.13 to prove Corollary 3.14. Here we do the opposite: We first prove Proposition 8.2 and then use it to prove Theorem 8.4. Also, the proof of Theorem 8.4 is significantly more involved than in the one-dimensional case. Note that by mimicking the onedimensional proof of Lemma 3.13, one can prove only the weaker conclusion
that if T is differentiable for all x E E and
IVT(x)I <M for all xEE, then
Go (W (E)) < MLQ (E). In what follows, it is of importance to control the Jacobian rather than the gradient of IF.
Remark 8.9. By taking M = 0 in Theorem 8.4, it follows that if fl C P.N is an open set, T : SZ - RN and if E C 1 is a Lebesgue measurable set on which 11 is differentiable and JW = 0, then RCN (11 (E)) = 0. This result is closely related to a theorem of Said [149], where a similar result is proved for smooth functions 'P : SZ -, R1. As a consequence of the previous theorem, we can prove that Lemma 3.16 continues to hold in RN. Theorem 8.10 (Varberg). Let ft C RN be any open set, let 11 : S1 RN, and let E C f2 be a Lebesgue measurable set on which T is differentiable. Then T (E) is Lebesgue measurable and GN
('I' (E)) <_ fE IJ'P (x) I dx.
Proof. Step 1: Assume first that E has finite measure. We first prove that F (E) is Lebesgue measurable. Write E = E. U Eo, where E. is an F, set and Eo is a set of Lebesgue measure zero. Then T (E) _'P (E00) U' (Eo). By Proposition 8.2 the set IQ (Eo) has Lebesgue measure zero, and so it is Lebesgue measurable by the completeness of the Lebesgue measure, while by the continuity of IQ the set T (EOO) is Lebesgue measurable. Fix e > 0 and for every k E N write
Ek:={xEE; (k-1)e
8.2. Absolutely Continuous Functions of Several Variables
241
Then Ek is Lebesgue measurable (why?), and so is ' (E1). By Theorem 8.4,
j(k
keN (Ek)
£N ('p (E)) < [-` jN (T (.))
,CN (E'-)
- 1) eGN (E") + e k
k
JE* I J1/ (x) I dx + sGN (E) =
f
J
IJk (x)I dx + eCN (E)
and it suffices to let s - 0+. Step 2: If E has infinite measure, for every n r= N write
E := EnB(0,n). By the previous step applied to E. we have that IF (En) is Lebesgue measurable, and so P (E) is Lebesgue measurable, since it is a countable union of Lebesgue measurable sets. Moreover, again by the previous step, ,CN (1, (En)) < fs. IJW (x)l dx < JE IJW (x)I dx. S ince
ID (En) C 'k (E,,+1), letting n lim GN (T (En)) = GN
n-4oo
oo, it follows by Proposition B.9 that
U T (En)
_ RCN ('p (E)) ,
n
and so the proof is complete. As in the one-dimensional case, the previous theorem allows us to prove several important results for absolutely continuous transformations.
Definition 8.11. Let St C R' be an open set. A bounded continuous function ' : SZ -, RN is said to be absolutely continuous if for every e > 0 there exists 6 > 0 such that if Q1 i ... , Qe are cubes, with pairwise disjoint interior contained in f? and such that GN (Qn) < d, then I
E Co (W (Qn)) < e. n=1
The next result is a partial analog of Theorem 3.12. Theorem 8.12. Let SZ C RN be an open set and let W : fl --4 RN. Assume the following.
S. Absolutely Continuous Functions and Change of Variables
242
(i) T is bounded and continuous on fl. (ii) ' is differentiable CN-a.e. in fl and JT is Lebesgue integrable. (iii) T has the (N) property; that is, it maps sets of Lebesgue measure zero into sets of Lebesgue measure zero. Then IQ is absolutely continuous.
Proof. Let Q1i ... , Qr be cubes with pairwise disjoint interior contained in fl and f o r each n = 1, ... , e let En be the subset of Qn n fl where T is differentiable. Then the set En is Lebesgue measurable, and, by Theorem 8.10, so is the set it (E.n). Moreover, using properties (ii) and (iii),
t E,CN ('p (Q, , )) ,
e
Q
_ >2JCN (,p (Q , i )) .
n=1
n=1
_
f
< EJ
nU Q+. IJQ (x) I dx
n=1 fQ
1JW (x)I dx
n=1 En
Jk (x) I dx.
n-,
Since JW is Lebesgue integrable, the right-hand side can be made arbitrarily e
small provided L CN (Qn) is sufficiently small. n=1
Exercise 8.13. Let fl C RN be an open set and let 'P : fl - RN be bounded and continuous on fl and differentiable for all but a countable subset of 12 and such that J'P is Lebesgue integrable. Prove that T is absolutely continuous.
Remark 8.14. We will see later on that if the open set fl is sufficiently regular, then every function if : fl -+ RN in the Sobolev space W'P (fl; RN), with p > N, has an absolutely continuous representative (respectively, locally absolutely continuous if fl C RN is an arbitrary open set). Another important class of absolutely continuous functions is given by Lipschitz con-
tinuous functions' : 1-> RN.
8.3. Change of Variables for Multiple Integrals In this section we prove a change of variables formula for multiple integrals. The proof will make use of Brouwer's fixed point theorem [261.
Theorem 8.15 (Brouwer's fixed point theorem). Let K C RN be a nonempty compact convex set and let IT : K -> K be a continuous transformation. Then there exists x E K such that IQ (x) = x. The proof that we present here is due to Lax (1041 and makes use of an interesting change of variables formula (see Theorem 8.17 below).
8.3. Change of Variables for Multiple Integrals
243
Exercise 8.16. Let W : RN RN be of class C2 and let Ml,..., MN be the cofactors of the first row of the Jacobian matrix J9.2
(i) Prove that E
can be written symbolically as
i=1
V V",2 det
OWN (ii) Let N = 2. Using part (i), prove that 0M1 8x1
0.412
+ ax2 = o
(iii) Let N > 2 and prove that
det
V'P2
N
=
V'P2
E det k=2
VWN
OWN
where the subscript k means that the differential operator V in the first row acts only on the kth row. (iv) Prove that V 0'Q2 = 0
det
WN k for all k = 2, ... , N and conclude that N 8n'it o.
Theorem 8.17 (Lax). Let W E C2 (RN; RN) be such that W is the identity outside some sphere, say, the unit sphere:
+l; (x) = x for Jxj > 1. 2If A = (a;f)j
9=j... N
is an N x N matrix, the cofactor of the entry a{j is defined by Mr1:_(-1):+?detCj,
i,7=1,...,N,
where Cj is the (N - 1) x (N - 1) matrix obtained from A by removing the ith row and the jtb column.
244
8. Absolutely Continuous Functions and Change of Variables
Then the following change of variables formula holds for every function u E C'I (EN): (8.9)
jen u (y) dy = LN u (W (x)) J'P (x) dx.
Proof. We use notation (8.2). Define (8.10)
v (y) = v (y'i, yi)
y E RN.
J-700 u (yi, t) dt,
Then v is differentiable and by the fundamental theorem of calculus, We claim that
Z
= U.
V(VOW) (8.11)
VW2
(u o T) JT = det
V'N By the chain rule, for all x E RIV, IV
v (v o 1Y) (x)
= i-1 F,
Ov i (W (x)) VWi (x)
.
Thus, the vector V (v o 41) can be written as a linear combination of the vectors v W 1, ... , V'F jv. The last N - 1 of these vectors are the last N - 1 rows of the matrix on the right-hand side of (8.11), and therefore these can be subtracted from V (v o') without altering the value of the determinant (8.11). This leaves us with
2 (V0W)VW;
b
let
v'P2
= let
I
O W)
\\
0q'2
VW1
VWN
= (i-oW) J'k
1 (
noI)JW,
where in the second identity we have factored out the scalar 7g o W. Hence, (8.11) holds. Since u has compact support, there exists r > 0 such that u - 0 outside
the cube Q (0, r). Without loss of generality, we may take r > 2, so that is the identity outside the unit Q (0, r) contains the unit ball. Since sphere, it suffices to restrict the integrals in (8.9) to Q (0, r). By expanding
8.3. Change of Variables for Multiple Integrals
245
the determinant on the right-hand side of (8.11) according to the first row, we get
(uo')J = Mla(v°T) + axl
(8.12)
+MNa(yo11),
axN where All, ... , MN are the cofactors of the first row of the Jacobian matrix
JW. By Fubini's theorem and a one-dimensional integration by parts, for all i = 2, ... , N, 8 {v o
Q(OMi
,r)
(x)
}
(x) dx =
axi
J( -a,2) rr
N-`
j r lbli (x)
a(voty) axe
,
{ e} dxidx;
2
(8.13)
(v o') (x) aMi (x) dx (9xi
Q(O,r)
+
rr
(
ri (xi, xi) (v o ) (xi
N_1
dx4
2'2)
Since W is the identity on 8Q (0, r) and u - 0 outside the cube Q (0, r), for
all i=2,...,Nwe have (see 8.10)
v(W(x,± )) -v\xi'±2/ =0 N-1.
for all x; E (- 2 ,
Hence, (8.13) reduces to
2)
(8.14)
(x)
4%
a(u 0 T) 0-Ti
Q(o,r) )
(x) dx = -
(v o W) (x) JQ(O,r)
aM 8xi
(x) dx
for alli=2,...,N. On the other hand, for i = 1 we have
v \ \xl' 2I11I = v r\x21} rr
f-00 u (xi, t) dt,
while
r
1l1l
v(11 (x''-2/I =v-2/ _J0o u(xi,t) dt=0, (XI,
where we have used the fact that u 0 outside Q (0, r). Since Ml (x) = 1 when IF (x) = x, again by Fubini's theorem we have that l
r r
\-a 2 )
r%
r
(
(X' X1) (v O `Y) (xl>xl) I "2 ,
22
J oo u (xi, t) dt dx1 = JQ(O,r) U (X) dx,
8. Absolutely Continuous Functions and Change of Variables
246
and so (8.13) for i = 1 becomes
I
1bi18(yoT)
dx =
8x1
Q(O,r)
j - f rQ(O,r) (voW) 8Ml dx+ J 8x1
udx.
Q(O,r)
Summing this identity together with the N - 1 ones in (8.14) and using the previous exercise and (8.12) yields N
Mya(y0T)dx
i=1
Q(O,r)
8xi
(v o T) E
J
(O,r)
i=1
Ldx + J 8xi
4
Q(O,r)
u dx.
This proves the result.
Exercise 8.18. Using mollifiers (see Appendix C) prove that the previous theorem continues to hold if E C1 (RN; RN) is the identity outside some sphere and u E CC (RN).
Although the transformation T in the previous theorem (and more generally in the previous exercise) is not assumed to be one-to-one or onto, it actually turns out that 12 is onto. Corollary 8.19. Let 11 E C' (]RN; RN) be the identity outside the unit ball. Then 1F is onto.
Proof. Assume that there exists a point yo E RN \ %p (RN). Since 19 is the identity outside B (0, 1), it follows that yo E B (0, 1). By the continuity of 'P we have that the set 91 (B (0,1)) is closed, and thus we may find a ball B (yo, ro) C B (0, 1) that does not intersect IF (B (0, 1)) (and, in turn, T (RN) either). Let 0 < r < ro and define U(Y)
to
if y E B (yo, r) , otherwise.
Note that supp u C B (yo, r) , and so it does not intersect 'P (RN). Thus u o 91 = 0. On the other hand, fRN a (y) dy > 0 and this contradicts the previous theorem.
We now turn to the proof of Brouwer's fixed point theorem.
Proof of Theorem 8.15. Step 1: Assume that K = B (0, 1) and let IF E C (B (0,1); RN) be the identity on the unit sphere. We claim that
8.3. Change of Variables for Multiple Integrals
247
T (.(o 1)) D B (0, 1). Extend W to be the identity outside B (0, 1). Using standard mollifiers, we may construct a sequence of transformations and {'P.} C C1 (RN; RN) such that Pn is the identity outside B (0, 1 + Wn -4 'P uniformly on compact sets. By the previous corollary (which continues to hold when the unit ball is replaced by any ball) we have that each 'Pn is onto, and so B (0,1) C'Pn (B (0,1 + n)} . Hence, for each y E B (0,1)
(x,) = y. Let {xnk} be a subsethere is x E B (0, 1 + such that quence of {x} such that xnk - x E B (0, 1). By uniform convergence in B (0, 2) we have that v = T.k (xTk } 1 'f` (x) as k --> oo. It follows that 19 (x) = y. Step 2: Let W : B (0, 1) -t B (0,1) be a continuous transformation. Assume
x for all x E B (0, 1). Define 4P : B (0,1) -> SN-1 as follows. For each x E B (0,1) let 41> (x) be the intersection with the sphere SN-1 of the ray from ' (x) to
by contradiction that T has no fixed point, that is, that ' (x) x, precisely3,
-b (x):=x+F(x)(x-*(x)), where
x
(x) -
(x) - I X12 +
Ix1212 -I- Ix -'P (x)12
IT -'P(x)I Then fi E C (B (0,1); RN),
TR (x) := R9 (Rx) ,
x E B (0, 1).
Step 4: Let K C RN be a nonempty compact convex set and let' : K -> K be a continuous transformation. Find R > 0 such that K C B (0, R) and 3To obtain F, we consider the line t1Y (x) + (1 - t) z,
t E R,
through the distinct points II' (x) and x and then find t E R such that
+(1-f)n12 = tt Ix - T (x)l2 + 2t (z %Y (x) - 1x12) + 1x12 It suffices to solve for t.
.
248
S. Absolutely Continuous Functions and Change of Variables
for each x E B (0, R) consider the continuous transformation
(x) :_ ' (II (x)) x E B (0, R), where I1 : RN -+ K is the projection onto the convex set K. Note that -P (K) C K C B (0, R), and so by the continuity of II we have that I : B (0, R) -i B (0, R) is continuous. By the previous step there exists x E B (0, R) such that
x=c(x)=W(II (x)).
On the other hand, since 11) (K) C K, we have that x E K, and so II (x) = x. Thus, the previous identity reduces to x = %P (x) and the proof is completed. 17
As a corollary of Brouwer's fixed point theorem we have the following result, which will be needed in the proof of the change of variables formula below.
Corollary 8.20. Let T E C (B (0,1); RN) be such that 14, (x) - xI < e
for all x E SN-1, where 0 < e < 1. Then IF (B (0,1)) J B (0,1- e). Proof. Assume by contradiction that there exists a point yo E B (0, 1 - e) 1 IF (B (0,1)). By hypothesis IW (x) I > 1 - E if x E SN-1. Thus, yo T (SN-1), and therefore W (x) # yo for all x E B (0, 1). Define : B (0, 1) SN-1 by
yo-%F(x) Iyo - W (x)I' Then W is continuous. To reach a contradiction in view of Brouwer's fixed point theorem, it remains to show that has no fixed points. Since
(B
(0,1)) C
sN-1,
the only possible fixed points lie on the unit sphere. On the other hand, if SN-1, then xE x 4 (x) < 0. Hence, c cannot have fixed points in which is a contradiction and completes the proof.
SN-1,
We are now ready to prove the change of variables formula for multiple integrals.
Theorem 8.21 (Change of variables for multiple integrals). Let n C 1R' be an open set and let W : n -> RN be continuous. Assume that there exist two Lebesgue measurable sets F, G C n on which W is differentiable and one-to-one, respectively. If
8.3. Change of Variables for Multiple Integrals
(i) GN (SZ
(ii)
£N
249
F) = 0,
\ (W (IZ 1 F)) = 0,
(iii) LN (,p (S \ G)) = 0, then the change of variables formula
I
(E)
u (y) dy = fE u (I (x)) I J (x) I dx
holds for every Lebesgue measurable set E C f2 and for every Lebesgue measurable function u : ID (E) - [-oo, oo], which is either Lebesgue integrable or has a sign.
Proof. Step 1: We begin by showing that IV maps Lebesgue measurable subsets of F into Lebesgue measurable sets. Let E C S l be a Lebesgue measurable set. In view of property (ii) and the completeness of the Lebesgue
measure, it suffices to show that E n F is Lebesgue measurable. But this follows from Theorem 8.10.
Step 2: For every Lebesgue measurable set E C RN define
µ(E):=CN(IF (EnFnG)). Since CN is countably additive and IV is one-to-one on G, it follows that µ is a Borel measure. Moreover, since IQ is continuous, it maps compact sets into compact sets, and so µ is finite on compact sets. Thus, is is a Radon measure. Moreover, p. is absolutely continuous with respect to the Lebesgue measure. Indeed, if E C RN is such that CN (E) = 0, then by Proposition 8.2 the set T (E n F n G) has Lebesgue measure zero. Thus, by the Radon-Nilcodym theorem (see Theorem B.65) there exists a nonnegative locally integrable function
:
dGN
A (E)
RN
10, ooj such that
= fE dLN
(x) dx
for all Lebesgue measurable sets E C IR'. In particular, if E C SZ is a Lebesgue measurable set, then
T (E) =' (EnFnG) U IF (E\ F)u'P (E\G). By (ii) and (iii) we get that (8.15)
CN(,p (E))=,CN('(EnFnG)) =
(E) = JE dCN (x)
for every Lebesgue measurable set E C S1.
dx
8. Absolutely Continuous Functions and Change of Variables
250
Let now H be a Borel set in 19 (fl) with finite measure. Since ' is continuous, the set E := W-1 (H) is Lebesgue measurable, and so by (8.15), (8.16)
(q, (E))
401) XH (H) dy = GN (H) = GN
= Jb) dx = fy XH ('I' (x))
(x) dx,
where we have used the fact that XE (x) = XH (I' (x)) for all x E 1. On the other hand, if H is only a Lebesgue measurable set in W (12) with finite measure, then '-1 (H) may not be Lebesgue measurable (see Exercise 3.56). In this case find two Borel subsets B1 and B2 of (St) such that B1 C H C B2 and GN (Bi) = GN (H) = GN (B2) < oo (see Proposition C.3). Then by (8.16),
dx = j
XB2
OF (x)) fly (x) dx < 001
and since
XB, ('IP (x)) N (x) <_ XH (`I' (x)) dG dG N (x) 5 XB, ('I' (x)) dGN (x) , it follows that the three functions coincide Vv-a.e. In particular, the middle function is Lebesgue measurable and we have that (8.17)
40) XH (v) dy = fn XH ('P (x)) d N (x) dx.
Note that we do not assert here that the function XH ('P (x)) is Lebesgue measurable, since this would imply that W-1 (H) is Lebesgue measurable. In view of (8.17), for every Lebesgue measurable simple function s ' (fl) -, [0, oo) vanishing outside a set of finite measure, we have fl,(ft) s (y) dy =
fn a
(T
(x)) dGN (x) dx, and Corollary B.37, together with the Lebesgue monotone convergence theorem, allows us to conclude that fir(n) u
(Y) dy = fn u (q, (x)) dCN (x) dx
for every Lebesgue measurable function u : 'P (0) -+ 10, oo]. The latter identity implies in particular that if u is integrable over T (fl), then so is (u o F) dLN over Cl. Using this fact, we conclude that (8-18)
fl M
u (I!) dy = / u ('P (x))
(x) dx
8.3. Change of Variables for Multiple Integrals
251
for every integrable function u : T (Q) -- R. To replace Sl with a Lebesgue measurable set E C fl in (8.18), we first prove that (8.19)
W) dGN (x) = XE (x) dGN (x)
for ,CN-a.e. x E RN. To see this, let Mo be the set of all x E RN for which the previous equality fails. Since 'I' is one-to-one in G, we have that ' (Mo) C >y (fl \ G) . It follows by (iii) that LN (q, (A,If)) = 0. By (8.15) we have that _WA (x) = 0 for ,CN-a.e. z E Mo. Hence, (8.19) holds, and so, replacing u (y) with u (y) XW(E) (y), (8.18) becomes
u (y) dy = L(E)
J
U OF {x)) Xww(E) ('t (x))
JN (x) dx
jE u (IF (x)) dCN (x) d
for every integrable function u : J (E) - R. Finally, to conclude the proof, it remains to show that dGN (x) = I JT (x)I
for LN-a.e. x E fl. By the Besicovitch derivation theorem (see Theorem
B.119) there exists a Borel set M C tZ, with LN (M) = 0, such that for every xo E f \ M, (8.20)
d1l
dLN
i (Q (xo, r)) _ LN ('_ (Q (xo, r))) E R, Um r-.o+ LN (Q (xo, r)) r---)O+ £N (Q (xo, r))
(x0) = lira
-
where in the second equality we have used (8.15). Later on in the proof we will use the fact that we can also consider balls B (xo, r) in place of cubes Q (xo, r) in (8.20). By Lemma 8.7 for all 8 > 0 we may find r,,,, > 0 such that RCN ('I' (Q (xo, r)))
< £N (Q (xo, r)) (i'i' (xo)I + C (N) (IV9 (xo)I +
6)N-1 6)
for all 0 < r < rxo. Dividing by rN and letting r -> 0+, it follows that dIA
< IJW (xo)I + C (N) (I VIP (xo)I + dLN (xo)
Given the arbitrariness of d, we obtain that
0 < GN (.TO) < IJl (xo)I
5)N-1
J.
S. Absolutely Continuous Functions and Change of Variables
252
To prove the converse inequality, it suffices to consider the case J%P (xo) # 0.
Let r,,o > 0 be so small that B (xo, rx°) C 92 and for V E B (0, rxo) define
`(y) := (VW (--o))-' ('I' (y + .TO) -'I' (xo)) . Note that c (0) = 0. Moreover, since %P is differentiable at xo, we have that 4) is differentiable at 0 with
V (0) = (oW (xO))-n (V (xa)) = IN, where IN is the identity matrix. Hence, for every 0 < e < 1 there exists 0 < 8 < r,o such that I4D (ii) -yl <elyl for all 0 < ly) < 8. Let 0 < r < 8. Then by Corollary 8.20 applied to B (0, r) in place of B (0, 1), we obtain that (B (0, r)) 3 B (0, (1 - E) r). Hence, (1 - e)N CN (B (xo, r)) = (1 - e)N JN (B (0, r)) < LN (4' (B (0, r))) = IJW (xo)I-i £N (`I' (B (xo, r))) , where we have used the translation invariance of the Lebesgue measure and Lemma 8.5. This implies that
(1-e)
Nr
1JT (xo)I < GNN12 (B (xo,r))) ,C (B (xo, r))
for all 0 < r < 8. Letting r - 0+ (and using (8.20) with balls in place of cubes), we obtain that
(1 - e)N IJ1 (xo)I <
(XO) d,CN /4 and given the arbitrariness of e, we conclude the proof.
Remark 8.22. (i) The hypothesis that f is open may be replaced by the hypothesis that fI is any set whose boundary has Lebesgue measure zero. Indeed, in this case we have that LN (n \ St°) = 0, while by (8.15) we have LN (W (11 \ fZ°)) = 0. It is now enough to apply the
theorem to W. (ii) Note that under the hypotheses of the previous theorem we obtain, as a by-product, that I JW I is a locally integrable function such that
JEJJT(x),dx
8.3. Change of Variables for Multiple Integrals
253
(iii) Brouwer's fixed point theorem was used in the previous proof to guarantee the validity of Corollary 8.20. Its use can be avoided if W : fl -r RN satisfies additional regularity conditions (see, e.g., [15]).
As a consequence of the previous theorem and of Proposition 8.2 we obtain some classical changes of variables.
Corollary 8.23. Let fZ C RN be an open set and let q : SZ -' kN be differentiable. Assume that there exists a Lebesgue measurable set F C St, with,CN (ti \ F) = 0, on which' is one-to-one. Then the change of variables formula
f(E) u (y) dy = AY
f
E
u (11 (x)) IJP (x)I dx
holds for every Lebesgue measurable set E C IZ and for every Lebesgue mea-
surable function u : 19 (E) -> [-oo, oo], which is either integrable or has a sign.
We will see below that, under suitable regularity hypotheses on the open set SZ, an important class of transformations satisfying the hypotheses of Theorem 8.21 is given by transformations IF : f - RN belonging to the Sobolev space W 1 ' (11; RN), p > N, and that are one-to-one except at
most on a set of Lebesgue measure zero. Another extremely useful class of transformations is given by (locally) Lipschitz continuous functions T : S2 - RN, which are again one-to-one except at most on a set of Lebesgue measure zero.
Perhaps one of the most important applications of Theorem 8.21 and Remark 8.22 is given by spherical coordinates in RN. We proceed by induc-
tion. For N = 2 we consider the standard polar coordinate system. Given a point x = (x1, ... , xy.) E RN, let r Ixl, let 01 be the angle from the positive x1-axis to x, precisely, 81 := cos 1 (21 }
r
,
0<01 < ir,
and let (p, 02, ... , ON-1) be the spherical coordinates for xi = (r21 ... ) xN) E RN-1, where p := IxIIN-1 = rsin01. The coordinates of x are
xl=rcosOi, x2 = r sin 01 cos 02,
xN_1 = r sin 01 sin 02
sin ON-2 COS 8N-1
xN = r sin 01 sin 02
Sin ON-2 Sin 8N-1
254
8. Absolutely Continuous Functions and Change of Variables
This defines a coordinate system (r, 81i ... , 8N_1), which is invertible except on a set of GN-measure zero. The Jacobian is sinN-3 02.. -sin ON-2 81 J'P (r, 81, ... , 8N-1) = rN-i sinN-2
Example 8.24. Let u (x) = g (Ixl), where g : [r1, r2] --+ R is continuous and
0 < r1 < r2. Then r2
f
u (x) dx = AN i
rl
9 (r) I.N-1 dr
To find fN, take g = 1. Then Q. 1v (r2 - rN) = LN (B (0, r2))
- LN (B (0, ri)) = fN
N where aN = GN (B (0, 1)), and so ON = NaN. It turns out that ON is the N - 1 surface measure of SN-1 = 8B (0,1).
Chapter 9
Distributions Prospective Grad Students, V. "Are your health-care plans affordable, or will I end up going to a dentist that operates out of a tnziler?" -Jorge Chant, www.phdcomics.com
Throughout this chapter the implicit space is the Euclidean space RN and 1 C RN is an open set, not necessarily bounded. For a multi-index a = (al, ... , aN) E (No)N we set a° TT
aikI
:= axi1 ... axN
Ia l :- a1 + ... + aN,
where x = (xl, ... , xN). Given two multi-indexes a = (al,... , aN) and ON), we write a <,3 if ai < $ for all i = 1,...,N. Q= For every nonnegative integer m E No we denote by Cm (fl) the vector space of all functions that are continuous together with their partial deriva00
tives up to order m. We set C00 (fl) := nC' (St) and we define C,," (fl) m=0
and C,,° (n) as the subspaces of Cm (fl) and C°° (Cl), respectively, consisting of all functions that have compact support.
9.1. The Spaces DK (Cl), V (0), and V (Q) In this section, given an open set Cl C RN, we construct a topology on the space Cr (ft). We begin by considering the space of all functions with support in a given compact set K C Q. In this subspace the topology should be consistent with the natural notion of convergence, which is uniform convergence of the functions and of all their partial derivatives of any order. Consider an open set Cl C RN and fix a compact set K C Cl. Let DK (fl) be the set of all functions in CC° (Cl) whose support is contained in K, that 255
9. Distributions
256
is,
V K (Sl) := {q E C C ° (Sl) : supp 0 C K}.
For each j E No define the norm II - IIKj on DK (Sl) by
I
I
J'
where the supremum is taken over all x E K and all multi-indices a with j al < j. By Theorem A.25, the family of norms { II-IIK, }3 turns DK (fl) into a locally convex space and a base for the topology rK is given by all sets of the form
EDK(S1):IIOIIKj,< 1,..., II0IIKjk(0)<ek}
where ji, .. , jk E No and el,... , ek E N, k E N. Taking
j:= max{j1,...,jk}, t:= max{t1,...,ek}, it follows that (9.1)
r E DK (S2) :
VK,;,t
c
E DK (C) ;
II4IIKj < 11011 Kj, <
1, ... ,
II OII K,;k
(0) < ek
}
,
and so it suffices to consider as a local base for the topology rK the family of sets VK,J,1, where j E No and £ E N.
Exercise 9.1. Let 0 C RN be an open set and let K C Sl be a compact set. Define (9.2)
dK (0,0) := m
2a 1
lIKa.,
0,
E DK (l) .
(i) Prove that dK is a metric. (ii) Prove that the topology TK is determined by the metric dK. (iii) Prove that DK (El) is complete. To construct a topology on CC° (El), let Bo be the collection of all convex,
balanced' sets U C C°° (fl) such that (9.3)
U n DK (Sl) E TK
for every compact set K C E. 'We recall that a subset E of a vector space X is balanced if tx E E for all x E E and t E [-1,1].
9.1. The Spaces DK (f2), D (0), and D' (fl)
257
Theorem 9.2. Let 0 C RN be an open set. The family
B:={4)+V: is a base for a locally convex Hattsdorf topology r on C,,' (f2) that turns C°° (ft) into a topological vector space.
Proof. Step 1: We first prove that Bo is nonempty. Indeed, for every j E No and .£ E N define the norms (9.4)
I1
11
sup
la
(x) 11
where the supremum is taken over all x E fZ and all multi-indices a with Ial < j, and
Vj.t0ECf (1k):110II3<
(9.5)
ll
Then (9.6)
Vj,t fl DK (12)
E VK (n)IIOIIK,j <
Q1
= VKj,I E TK
by (9.1), and so Vjt belongs to Bo.
Step 2: To prove that B is a base for a topology, it suffices to verify the following two conditions:
(i) for every 0 E C°° (f2) there exists U E B such that 0 E U; (ii) for every U1, U2 E B, with Ui fl U2 # 0, and for every 0 E U1 fl U2 there exists U3 E B such that 0 E U3 and u3 C U1 fl u2. To prove (i), let 0 E C,,° (f2). Fix j E No and I E N and consider the set Vj,1 defined in the previous step. Then Vjt E Bo, and so U :_ 0 + Vj,j E B. To verify property (ii), let 01, 02 E C:°° (fl) and V1, V2 E Bp be such that
(01+Vi)n(¢2+V2)0 0 and fix 0 E (4)i + Vi) n (4)2 + V2). Since the supports of 01, 4)2i and 0 are compact sets contained in 52, we may find a compact set K C 12 such that K D supp ¢1 U supp 02 U supp 4).
Note that if (¢ - Oj) (x) 0 for some x E 12 and for some i E {1,2}, then necessarily x E supp Oi U supp ¢, and so ¢ - ¢i E DK (f2), i = 1, 2. Since 0 - 4j E vi fl DK (fl) E 1 K, i = 1, 2, using the continuity of scalar multiplication in VK (f2), we may find 0 E (0, 1) such that
4)-4) E (1-B)(V=fVK(ft))C(1-8)Vj for i = 1, 2. By the convexity of the sets Vi we have that
0-Oi+9ViC(1-0)V +Ovi=Vi
9. Distributions
258
for i = 1, 2, so that
o+B(V1f1V2) C (o1+vi)n(02+V2) Thus, B is a base for a topology r given by all unions of members of B. Step 3: Next we show that (C'° (S2) , T) is a topological vector space.
To prove the continuity of scalar multiplication at a point (to, 00) E R x C,,° (S2), consider an (open) neighborhood U E T of to0o. Since B is a
base for the topology, we may find V E Bo such that to0o + V C U. Let K := supp 00. Then 00 E DK (1) and since V fl DK (II) E TK, by the continuity of scalar multiplication in VK (il) we may find b > 0 so small that 600 E 2 (V nVK (ft)) C 2V. Let
8:=
2(Itol+8)-
Then for every It - tol < b and ¢ E 00 + sV we have that
tai-tooo=t(0-0o)+(t-to)0oEtsV+2VC2V+2V=V, where we have used the fact that V is convex and balanced. Hence, to r= toOo + V C U for every It - t01 < b and every ¢ E Oo + sV, which proves continuity of scalar multiplication at the point (to, Oo). To prove the continuity of addition at a point (101, 02) E (SZ) X CC° (St), consider a neighborhood U E T of 01 + 02. Since B is a base for the topology, we may find V E B0 such that 01 + 02 + V C U. The convexity of V E Bo implies that
`i1 + 2V) + (02 + 2V) = (¢1 +02)+ V. By (9.3) for every compact set K C St we have that V fl DK (SZ) E TK, and since the topology TK turns DK (f?) into a topological vector space, it also
follows that 71V fl DK (f) E TK. Thus, 1V E Bo and, in turn, fil + IV,
02+zVEB. Step 4: Finally, to prove that (C,,'° (11) , T) is a Hausdorff topological vector space, it suffices to show that singletons are closed. Let 01, 02 E Cc° (St) be two distinct elements and define V
E 0C° ()) : sup 10 (X) I < sup 101 (x) - 02 (X) I I XEf
xEn
.
In view of (9.5) the set V belongs to Bo and 01 0 ¢2 + V. Hence, toll is closed and the proof is completed.
0
9.1. The Spaces DK (S2), D (0), and D' (fl)
259
The space C,,' (f') endowed with the topology T is denoted V (a) and its elements are called testing functions. A natural question is why we define T in such a convoluted way, instead of directly considering the locally convex topology generated by the family of norms (11-i11}
)EN0
defined in (9.4). The problem is that this topology
would not be complete.
Exercise 9.3. Take N = 1, SZ = R, and consider a function ¢ E C°° (R) with support in [0, 1] such that 0 > 0 in (0, 1). Prove that the sequence
(x)(x-1)+10(x-2)+---+1O(x-n),
XER,
is a Cauchy sequence in the topology generated by the family of norms {II-ii;}
jENdefined in (9.4), but its limit does not have compact support, 0
and so it does not belong to C°° (R).
Exercise 9.4. Let S2 C RN be an open set. Let {K,b},SEN. C fl be an increasing sequence of compact sets, with KO := 0, such that
dist (K., 8Kn+1) > 0 and 00
UKn=SZ. n=1
Given two sequences m := {?7tn}nE1qO C No and a :_ with Mn - oo and an -+ 0, for every 0 E V (St) define pm,a (01 :=
u
Esup
n
xs
Kn
C (0, oo),
1 E I E(x)
an
(i) Prove that pm,a is a seminorm. (ii) Prove that the family of seminorms {pn,,a}me, where m and a vary among all sequences as above, generates the topology r defined in Theorem 9.2. We now show that the topology r, when restricted to DK (S2), for some compact set K C fl, does not produce more open sets than the ones in TK.
Theorem 9.5. Let S1 C RK be an open set. Then for every compact set K C f the topology TK coincides with the relative topology of DK (St) as a subset of D (S2).
Proof. Fix a compact set K C Sl and let U E r. We claim that u fl DK (SZ) belongs to TK. To see this, it suffices to consider the case in which Uf1DK (fl)
9. Distributions
260
is nonempty. Let 0 E UfVK (fl). Since B is a base for T, there exists V E Bo such that 0 + V C U. Hence,
o+ (v n VK (Sl)) c U n VK (n). Since 0 E DK (Q) and VnVK (fl) E rK, we have that #+(V n DK (a)) E TK. This shows that every point of U n VK (11) is an interior point with respect to rK, and so U n DK (12) E TK. Conversely, let U E rK. We claim that U = V n DK (f2)
(9.7)
for some V E T. Since the family of sets VK,j,r, where j E No and £ E N, is a local base for the topology TIC (see (9.1)), for every ¢ E U we may find ji E No and to E N such that 0 + VK,am.I C U.
Let V3,,,[,, be defined as in (9.5). By Step 1 of the proof of Theorem 9.2,
(o +
n DK (11) = q + VKi,,e,,
CU
and 0 + Vj.,,ld E B. In turn, the set
V := U (0+Vioto) OEu
0
belongs to r and (9.7) holds.
Exercise 9.6. Let Il c RN be an open set. Prove that for every xo E fl, for every compact set K C fl, and for every r > 0, the set U := {0 E DK (S2)
(xo) I < r}
is open with respect to TK. Next we study topologically bounded sets in D (fl) (see Definition A.18). The following result will be used to study convergence with respect to the topology r.
Theorem 9.7. Let fl c RN be an open set and let W C V (fl) be a topologically bounded set. Then there exists a compact set K C fl such that W C VK (11). Moreover, for every j E No there exists a constant Mj > 0 such that II0IIj<_M,
forall0EW.
9.1. The Spaces VK (Sl), V (fl), and D' (fl)
261
Proof. Assume by contradiction that W is not contained in VK (fl) for any compact set K C Q. Let {Kn} C fl be an increasing sequence of compact sets such that dist (K,,, 3K,+1) > 0 and 00
UKn=SI. n=1
Then for each n E N we may find a function On E W and a point xn E ..+l \ Kn such(( that On (x,) # 0. Define
ED (Q): I0(xn)I < 1 Ion(xn)I for all nEN }. n
U:_ ll
Since each compact set K C fl contains only finitely many x,,, by the previous exercise we have that U n DK (fl) E TK, and so U E T. Using the fact that the set W is topologically bounded, we may find t > 0 such that W C W. Consider an integer n > t. Then 0. (xn) 0 and
t Ian(xn)I n which implies that On V U, or, equivalently, that lon(xn)I,
W, which is a
i that W C DK (Sl) for some compact set K C Sl. contradiction. This shows To prove the final part of the statement, note that by Theorem 9.5 the set W = W n DK (fl) is bounded with respect to the topology TK, and so, by Corollary A.26, for each j E No the set
{0E W} is bounded in R. We are now ready to characterize convergence with respect to the topology T and to prove the completeness of D (Cl) (recall Exercise 9.3).
Theorem 9.8. Let Cl C R IV be an open set. Then the space D(fl) is complete. Moreover, a sequence {4,j C V (Cl) convetges to E D (Cl) with respect to r if and only if (i) there exists a compact set K C Cl such that the support of every 0. is contained in K, a
0"
'. 8xa
(ii) lim
19 0 uniformly on K for every multi-index a.
= 8X=
Proof. Let Jon I C V (Cl) be a Cauchy sequence. By Proposition A.20, the set {4 : n E N} is topologically bounded. Hence, we may apply Theorem 9.7 to find a compact set K C Sl such that {0n : n E N} C DK (Cl). In turn,
9. Distributions
262
by Theorem 9.5, we have that {din} is a Cauchy sequence in DK (fl). Hence, for all integers j E No and Q E N there exists an integer tT E N such that 1
on - O,, E VK,7,1 =
10
E DK (f2) :
e
for all k, n > n; that is, for every multi-index a, with Ial < j, we have that
sup .EK
ax (x) - axa (x)
<
a
This implies that
j axn
is a Cauchy sequence in the space of continuous
bounded functions, and so it converges uniformly in fl to a function iba with support in K. In particular, taking a = 0, we have that (bn converges uniformly in fZ to a function 9yo with support in K and 8x = lka for every multi-index co, with lal < j (why?). Given the arbitrariness of j e No, we conclude that ''o E DK (fZ) and that the sequence {fin} converges to '/'o E V (f2) with respect to r. Thus, D (f1) is complete.
Exercise 9.9. Let fl C RN be an open set. (i) Prove that for every compact K C ft, the space DK (f2) is closed and has empty interior in D (f2). (ii) Prove that 1) (fl) is not metrizable.
In the previous exercise we have proved that V (fl) is not a metrizable space. Despite this fact, we can still prove that linear functionals defined on D (f2) are continuous if and only if they are sequentially continuous. Precisely, the following result holds.
Theorem 9.10. Let ft C RN be an open set and let T : D (f2) --+ R be linear. Then the following properties are equivalent:
(i) T is continuous. (ii) T is bounded. (iii) If {On} C D (fl) converges to 0 E V (fl) with respect to r, then
T to DK (fl) is continuous for every compact set KCS2. (v) For every compact set K C S2 there exist an integer j E No and a constant CK > 0 such that (9.8)
IT (0) 1
CK IIOIIKJ
9.1. The Spaces DK (St), V (11), and D' (ft)
263
foraU0EVK(f?). Proof. (i)=(ii) If T is continuous, then T is bounded by Theorem A.29. Assume that T is bounded and let {q,,} C V (f') converge to E V (12) with respect to r. Since D (St) is a topological vector space, by replacing {¢n} with {0n - 0}, without loss of generality, we may assume that {on} converges to 0 with respect to r. By Proposition A.20, the set {0n : n E N} is topologically bounded. Since T is bounded, it follows that the set IT (On) : n E N} is bounded. By Theorem 9.8 there is a compact set K C Cl such that dDK (¢n, 0) - 0 as n - oo. Since VK (Cl) is metrizable, it follows by Theorem A.29 that T (on) -- 0. (iii)=(iv) Fix a compact set K C Cl and assume that {0n} c DK (Cl) is such that dDK (0n, 0) -* 0 as n - oo. By Theorem 9.8 we have that {0n} converges to 0 with respect to T. Hence, by property (iii), lim
n-.oo
T(4 )=0.
Using Theorem A.29 once more, we get that the restriction of T to DK (Cl) is continuous. For every e > 0 and for every compact set K C Cl the restriction of T to VK (Cl) is continuous at zero, and so
T-1((-e, e)) n VK (fl) E TK. Hence, T'1 ((-6, e)) E r, which shows that T is continuous at zero and, by linearity, everywhere.
(iv)q(v) Assume that (iv) holds and fix a compact set K C Cl. Since T restricted to VK (Cl) is continuous at the origin, given E = 1 there exist j E Na and I E N such that VKj,e C T-1 ((-1, 1)), that is,
IT(O)I<1 for all 0 E DK (Cl) with II#IIKj < I If 0 E VK (Cl) and 0 54 0, then II 0II K.j
0 and 1
2e II0IIKj Kj
1
71
By the linearity of T it follows that IT(O)I <2eII0IIA, which gives (iv). Conversely, if (v) holds, then by taking a sufficiently large, we have that IT(O)I < e for all E VKJ,I, which shows continuity of T restricted to DK (Cl).
9. Distributions
264
The dual of V (S2) is denoted D' (S2) and its elements are called distribu-
tions. We often use the duality notation (T, 0) to denote T (¢). The space D' (S2) is given the weak star topology, so that a sequence {T.} C D' (S2) converges to T E D' (1) if T,s (0) -. T (0) for every 0 E D (f2). In this case we say that {T.j converges to T in the sense of distributions. Exercise 9.11. Let SZ C RN be an open set and let T E V'(12). (i) Prove that if 0,,0 E C00 (12), then for every multi-index fl the Leibnitz formula
8a (f) = 8x01
e1-Q0 err c°QOxa- Oxa
holds for some cap E R. (ii) Prove that if ii E C°° (s2), then the linear functional OT : V (S2) IR, defined by (OT) (0) := T (,y4)) ,
0 E D (11) ,
is a distribution.
9.2. Order of a Distribution In this section we define the order of a distribution. We begin by observing that the integer j E No in (9.8) may change with the compact set K C S2.
If the same integer will do for all compact sets K C 0, then the smallest integer j E No for which (9.8) holds for all compact sets K C S2 is called the
order of the distribution T. If no such integer exists, then the distribution T is said to have infinite order.
Example 9.12. Let 12 C RN be an open set. (i) Let A be a signed Radon measure on ft. The functional Ta (0) :=
f4)dA
is a distribution of order zero.
(ii) Fix xo E 0. The functional &, defined by 6. (4)) = 4) (xo), 4 E V (0), is a distribution and is called the delta Dirac with mass at x0. The distribution 8ro has order zero. (iii) Let u E Li (52). The functional T. (0) :=
f
4) (x) u (x) dx,
is a distribution of order zero.
0 E D (1l) ,
9.2. Order of a Distribution
265
Actually, it turns out that all distributions of order zero may be identified with measures.
Theorem 9.13. Let 11 C RN be an open set and let T E D' (& ).
(i) If T is positive, that is, if T (0) > 0 for all nonnegative functions 0 E V (fl), then there exists a unique Radon measure µ : 13 (f2) -> [0, oo] such that
r
T (0) = J 0dµ for all 0 E D (1l) . s2
(ii) If T has order zero, then there exist two Radon measure 14,111 B (1) - [0, ool such that
T(0) =
Jif
0dµ1 -
r
Jsa
0 dµ2 for all q E D (Q)
.
Proof. (i) We claim that T has order zero. Fix a compact set K C f1 and find an open set 01 such that
KC01CC11. Construct a smooth cut-off function c E C°O (f) such that co - 1 on K, supp cp C SZl, and 0 < Sp < 1 (see Exercise C.22). In particular, Sp E D (SZ). Since 9 1 on K and rp > 0, for every 0 E DK (Il) we have that 10 W1 <- 11011 K,0 (P (x)
for all x E fl, and so r(IIOIIK,oV-0) >>_0,
T(0+II0IIK,oSp)
?0;
that is, by the linearity of T, IT(0)I <- II0IIx,oT(W)
for all 0 E DK(a),
which shows that T has order zero. Let (Oil C fi be an increasing sequence of bounded open sets such that
ni CC tli+i and
o°
U SZ, = SZ. i=1
We start by showing that for every fixed i E N the distribution T can be extended in a unique way as a linear continuous map on CJ (Q1). Since K1 := SZ, is a compact set contained in n and T has order zero, by what we just proved there exists a constant C; > 0 such that (9.9)
IT (0)1 5 CGII0IIKi,o
for all 0 E DK; (0). If 0 E C, (Sts), then dist (supp 0, c( 1) > 0, and thus if we consider 0n := gyp, *0, where cp are standard mollifiers (with a := 1) and n
n
9. Distributions
266
n < dist (supp gyp, 8Sti), we have that {4,,a} C DK, (f2) and
uniformly
on Ki. It follows by (9.9) that IT(4,n.-41)1 !5 Cz110. -01IIK,,a--+ o
as 1, n -* oo. Hence, IT (4,,a)} is a Cauchy sequence, and therefore it converges to a limit that we denote by T, (0). Moreover, if 4, > 0, then 0, > 0 also, and so Ti (0) > 0. Note that, again by (9.9), Ti (0) is independent of the choice of the approximating sequence {4,,a}.
By the linearity of T it follows that Ti : CC (a,) - R is linear and positive, while by (9.9) we have that IT (0)I 5 Ci II4I1c,A)
for all 0 E C, (St,). Since Il, C St,+i, it follows that T,+i (4') = T, (¢)
for all 4' E CC (Sti)
.
Thus {T,} defines a unique linear positive extension of T to the union of all CC (SZi), which coincides with CC (0). The result now follows from the Riesz representation theorem in CC (C') for positive linear functionals (see Theorem B.115). (ii) The second part of the proof of (i) continues to hold in this case. By (9.9) and the fact that {(l,} covers St, we have that the extended functional is locally bounded, and so the result now follows from the Riesz representation theorem in Cc (12) for locally bounded linear functionals (see Theorem B.115). 0
9.3. Derivatives of Distributions and Distributions as Derivatives We now define the notion of a derivative of a distribution.
Definition 9.14. Let St C RN be an open set and let T E D' (S2). Given a multi-index a, we define the ath derivative of T as a 8z:a
(4,) = (-1)IaIT
0 E D(1Z).
For j E N the symbol D'T stands for the collection of all ath distributional derivatives of T with I a I = j.
Remark 9.15. It can be verified that is still a distribution. Indeed, let K C 12 be a compact set. By Theorem 9.10 there exist an integer j E No and a constant CK > 0 such that IT (0)1<-CKI10IIK,j
9.3. Derivatives of Distributions and Distributions as Derivatives
for all
E VK (1Z). It follows that
la (O)I= for all
267
IT(210-)I :SCKII
E DK (fr), which shows that s
`CKIIOIIK'j+j.j
810'-IIKj
E 1Y (St), again by Theorem 9.10.
In particular, if u E L1oc (Il) and a is a multi-index, then the ath weak,
or di8tributional, derivative of u is the distribution j''.
Example 9.16. Let SZ C RN be an open set and let u E Lj (0). The Laplacian of u in the sense of distribution is the distribution
N a2Tu
TAu :=
a2x. 4
that is, 2
N
T
axi 2
s`
z
d=1
for all
i=1
axs-
i=1
N
=
9} l
2 s.
dx
uaxx, dx -
E D (St). So a function u E L10C (Q) is subharm.onic if "Au > 0",
that is,
Tau(0)=J uI4dx>0 for all 0 E V (Cl) with 0 > 0. By Theorem 9.13 there exists a unique (positive) Radon measure µ : B (Cl) - [0, oo] such that
TDu (0) = in uA4 dx =
Jn
0 dµ for all
E D (S2)
Thus, with an abuse of notation, we may write Au = ia. Similarly, for U E Lic (Cl; RN) one can define the divergence of u in the sense of distributions.
Definition 9.17. Let Cl C R v be an open set, let u E Li (Cl), and let a be a multi-index. If there exists a function va E L je (Cl) such that a
a u(0)
for all 0ED(Cl),
then vQ is called the ath weak, or distributional, derivative of Tu. We write 8
.= Va.
9. Distributions
268
Thus, a function va E Li°C (Q) is the nth weak derivative of u r= Ll (12)
if F
(9.10)
in
Ovcdx = (-1)I"I In uLO- dx
for all 0 E C'° (SZ).
Exercise 9.18. Let u : R -i R be defined as follows:
if -7r <x <0,
cos x
-
u(x):=
if 0 <x <1r,
7r
otherwise.
0
(i) Calculate the derivative of u in the sense of distributions and find its order. (ii) Let v be the restriction of u in the interval (-ir, 7r). Calculate the first and second derivative of v in the sense of distributions in (-7r, 7r) and find their orders. (iii) Let w : l[8 -p R be differentiable in R \ {a} and assume that there exist lim w (x) E R, Jim w (x) E R.
z"a-
x-"a+
Calculate the derivative of w in the sense of distributions. What can you say about its order? Under what conditions can you conclude
that it has order zero? Exercise 9.19. Assume that u E L' ((-oo, 6) U (6, oo)) for every 6 > 0 and define the principal value integral 00
PV foo u (x) dx
J
lim
00
u (x) dx +
u (x) dxa-4o+
s
whenever it exists. For 0 E D (R) define 00
T (0) :=
¢ (x) log IxI dx. -CO
Prove that
T'
PV
r
J
0 0 (x) - dx,
00
p
T" (4) = - PV J
0 0 (x)
00
Z
0 (0) dx.
Exercise 9.20. Let fl C RN be an open set and let {Tn.} C D' (Sl) be such that the limit slim T. T (4) #CO
exists in R for every 0 E D (fl).
(i) Prove that T E D' (Il).
9.3. Derivatives of Distributions and Distributions as Derivatives
269
E V (a),
(ii) Prove that for every multi-index a and for every
In the next few theorems we characterize distributions as weak derivatives of continuous functions.
Theorem 9.21. Let Q C RN be an open set and let T E D' (n). Then for every compact set K C St theme exist a continuous function u : SZ - R and a multi-index a such that
j uaLO dx
T
8xa
lyy
forall0EDK(Q). Proof. In what follows, we use the notation (E.2) in Appendix E. Without loss of generality we may assume that K C Q 10,1]N. By the mean value theorem, for every zfi E DQ (RN), for all i = 1, ..., N, and for all x E Q we have
Ivy (x)l = lip (x;, xi) -'(°)I
(9.11)
= 8Lo (x;, t) (xi - 0) Q 18 I
.
x
For x r= Q set
Q(x).={yEQ:0<_y4<_xi,i=1,...,N}. Then for vy E DQ (RN) and x E Q we have X1 alp 10
X2
0
r JJJJffoxl
8x1
(t11, x2, ..., xN) dy1
(Yl, x2, ... , XN) dy1
a2
y2
f
r
xi (yl, 0, x3, ... ) XN) + !
JO
Jo
JXIL 0
8x1
8x18x2
(y1) X12, x3, ... , X N) dy2dy1
/fp
J
Xa
0
where fl := (1, (9.12)
O2
8x18x2
...
(y1, y2, x3, ... , XN) dy2dyl Q(X)
1), and so a
a
(x) = 4(m)
for allxEQ.
8xR' (y) dy
8x (y) dy,
9. Distributions
270
Fix an integer j E No (to be determined later) and let a be a multi-index with Jal < j. By repeated applications of (9.11) we obtain that ( 9.13 )
<max maxi axa II
Q
gxjp
<
a
I- f
TP to+i lp1)A
dx{1+
Q
dy,
where j)3 = (j, .. , j) and where in the last inequality we have used (9.12). By (9.12) the linear operator
D-6: DK P)
DK (M
H aim
ax
is one-to-one, and hence so is the linear operator L := D'1O : DK (1k) -t DK (f2)
aU+i),s0
8x(j+i)0 since
Dip= DIO
j times
Let Y := L (VK (fl)) and define the linear functional Tl : Y -> R as follows. Given 0 E Y, there exists a unique 0 E DK (fl) such that 8u+i),so Define
Ti('5):=T(O) Since T E D' (11), by Theorem 9.10 there exist an integer j E No and a constant CK > 0 such that for all 0 E VK (S2), IT (0)I !5 CK II0IIK,j <_ CK fK
exU+1)P
dx,
where in the last inequality we have used (9.13). It follows from the definition
of Tl that IT1(b)I C CK L I-0I dxx
for all /i E Y, and thus we may apply the Hahn-Banach theorem (see Theorem A.30) to extend Tl as a continuous linear functional defined in L' (K). By the Riesz representation theorem in L' (K) (see Theorem B.95) there exists a function v E L00 (K) such that Ti (10 = jK uo dx
9.3. Derivatives of Distributions and Distributions as Derivatives
271
for all -i E L' (K). In particular, if 0 E DK (12), then aU+000 dx. T Tl (OX0+00) = JK V axU+1)O Extend v by zero outside K and define
l ... 1 ON v (11) dyN ... dyl
u (x)
J 00 f o0 Then u is continuous and by integrating by parts N times we have that for all 0 E DK (a), a(j+2)/3 N
(-1)
T
f
.1 K
u gX(i+2)9
dx.
To complete the proof, we may define a := (j + 2)13 and, if needed, replace
u with -u. If T has compact support, then the previous local result becomes global.
Definition 9.22 (Support of a distribution). Let 12 C RN be an open set and let T E D' (12). If 12' C 0 is open, then we write that T = 0 in 12' if T (0) = 0 for all 0 E D (S2'). The support of T is the complement of U relative to fl, where U is the union of all open subsets 12' C S2 in which
T=O. The support of T will be written as supp T.
Exercise 9.23. Let 12 C RN be an open set and let T E D' (12). Show that a =0 T = 0 in the (possibly empty) open set 12 \ supp T. Prove also that TJF in 12 \ suppT for every multi-index a. We now prove a global version of Theorem 9.21 for distributions with compact support.
Theorem 9.24. Let 12 C RN be an open set and let T E D' (12) be such that supp T is a compact set of 12. Then (i) there exist an integer j E No and a constant 0 > 0 such that IT(0)1
for all 0 E V (12) (in particular, T has finite order t:5 j), (ii) if U is an open set, with suppT C U C 12, then for each multiindex a, with a < 3 := (,Q + 2, ... , 2 + 2), there exists a function vQ E C (11), with supp ve C U, such that
T (4) _
a
iT axe
for all 0 E D (12).
(0) =
a
Lo dx (-1) 10" i r 'aaxe n
9. Distributions
272
Proof. (i) Consider an open set U, with supp T C U CC S2, and (see Exercise C.22) construct a function 0 E D(St) such that 0 = 1 on U. Let K := supp . We claim that ?PT = T. Indeed, if E D (cl), then since = 0 on U, and so
z/1 = 1 on U, we have that 0 -
supp (o - ''O) fl supp T = 0, which implies that T(0-tPO)=0,
or, equivalently, T (4)) = T ( 4)) _ (')T) (0) .
Hence, the claim holds. Since T E D` (cl), by Theorem 9.10 there exist an integer j E No and a constant CK > 0 such that IT (0)15 CK II0IIKj
for all 0 E DK (Il). On the other hand, if 0 E D (Il), then z'¢ E DK A, and so, since OT = T, IT (0)1 = IT(OO)I < CK IIt4)IIK,J
By the Leibnitz formula (see Exercise 9.11), II7P0II K j 5 C.'O II4II KJ) and so IT (4))I <- CKC, II0IIKJ
which shows part (i). (ii) Consider an open set A, with supp T C A CC U. By Theorem 9.21 with A in place of K there exists a continuous function u : f2 - R such that
T1)IPI j44dT ( t+2 , . . . , 1+2 ) . Consider an open set V, with supp T C V CC A, and as in (i) construct a function ifi E V (S2) such that iG = 1 on V and K := supp,0 C A. Then, by (i), for all 4)E D (1l) we have that 00 E DA (0) and f o r all 0 E D A (n), where Q
(-1)101
T (4)) = T
_
ca,s
f u(9ON)
dx
uaQ-a0 8-
fn
8X$-a axa
dx
'
and so it suffices to take va
This concludes the proof.
caQ
(-1)119-aI
ax.a-a
O
9.3. Derivatives of Distributions and Distributions as Derivatives
273
Finally, using partitions of unity, we have a similar representation for every distribution. Theorem 9.25. Let fl C 1[SN be an open set and let T E Di (S2). Then for each multi-index a there exists a function va E C (S2) such that
(i) each compact set K C f2 intersects the supports of finitely many Va,
(ii) for all 0 E V (f2) 0
,,, / 1N
T
11\
a
Y IIQI
1
vQ Q
a
axe
ax.
If T has finite order, then only finitely many vQ different from zero are needed.
Proof. (i) Let S be a countable dense set in S2, e.g., S :_ {x e QN fl 0), and consider the countable family F of open cubes
.F:={Q(x,r): rEQ,0
By the density of
00
S2 =
U
Q(xn,r2n).
N ote
that every compact set K C Q intersects only finitely many cubes Q (xn, rn). For each n E N construct a function ¢n E D (1) such that (b = 1 on Q (x.,1.) and supp On C Q (xn, Use this family to construct a partition of unity {rJin} C D (fl) with supp 7yn C Q (x., rn) for each n E N (see the proof of Theorem C.21). For each n E N the distribution ipnT has support contained in Q (x,,, rn), and so it has finite order in. Hence, by the previous theorem we may find finitely many functions {vE} C C (0), with
a < fl.
+ 2, ... , Qn + 2), with supp vin) C Q (xn, rn), such that 'qXa
for all , E V (ft). For every multi-index a for which a < pn fails define v,(,n)
:- 0. Thus, 1) 1`1
a
forall0ED(fl).
( van) L o- dx
J
9. Distributions
274
Hence, for every multi-index a we may define V
00
.=
n=1
Since each compact set intersects the support of only finitely many v&n), the function va is continuous and (i) holds. To prove (ii), let 0 E D (SZ). Since is a partition of unity, we have
that
00
=T, V)n0, n-I where the sum is actually finite since 0 has compact support. Since the sum is finite and T is linear, we have that 00
T (cb) _ >2 T ( n=1
00
(-1)1°I
n'1')
J fan)8xa dx
n=1 a
n=1
(.
E
LoJ 00 a 8xa 8xo a a :a=1 This shows (ii). Finally, if T has finite order $, then -O.T has finite order R. < Q, and so it suffices to consider only multi-indices a < ,l3 := (f + 2, ... , e + 2). Exercise 9.26. Let f2 C RN be an open set. Prove that every continuous linear functional on C00 (1l) is of the form f H T (f), where T is a distribution with compact support.
Exercise 9.27. Let 1 = (a1, bl) x ... x (aN, bN). (i) Prove that 0 E V (0) is such that
1 O (x) dx = 0 if and only if N
80s
Eax; i=1 for some 01, ... , ON E V (SZ). Hint: Use induction on N and look at Step 1 of the proof of Lemma 7.3.
(ii) Prove that if T E D' (12) is such that = 0 for all i = 1, ... , N, then there exists a constant cin R such that T (O) = c
0 ( x) dx
for all 0 E V (f2), i.e., T is constant.
9.4. Convolutions
275
(iii) Prove that if Q C RN is an open connected set and T E D' (fI) is such that = 0 for all i = 1, ... , N, then T is constant.
9.4. Convolutions In this section we work mostly with c = RN. We recall that given two functions 0, cp E C°° (RN), one of which with compact support, their convolution is the function 0 * cp defined by
(sp * 0) (x) = j
(9.14)
11R N
co (x - y) 0 (y) dy,
x E RN.
Motivated by this formula, given T E D' (RN) and ¢ E V (RN), we define the convolution of T and 0 by (T * 0) (x) := T (9.15) x E RN,
(x - y), y E RN. Note that T * ¢ is a function. In the special case in which T = Tp for some rp E CO° (RN), the two formulas (9.14) and (9.15) agree. where 4,X (y)
Exercise 9.28. Let 0, cp E V (RN). For h > 0 define uh (x) ._ $N
tp (x - hy) c (hy) ,
x E RN.
YIEZN
Prove that {uh} converges uniformly to cp * . Hint: Use Riemann sums.
Theorem 9.29. Let T E D' (RN) and -0, ip E V (RN). Then (i) T * 0 E Coo (RN), (ii) supp (T * 0) C supp T + supp 4), (iii) for every multi-index a,
a (T*4))=T* (iv)
Q10) = L
/
Q *4)
(T*gyp)*4)=T*((p*O).
Proof. If xf,, --+ x in RN, then for every y E RN, 4)
(y) = 4(xn.-y) - 4'(x-y) = 9x(y)
and conditions (i) and (ii) of Theorem 9.8 are satisfied. Hence, {pxn} converges to 9 with respect to r, and so by Theorem 9.10,
(9)
(T * 0) (xn) = T -* T (9) = (T * 4') (x) which proves that T * 0 is a continuous function. To prove (ii), note that if x E RN is such that supp 4'X fl supp T = 01
9. Distributions
276
then (T * 4)) (x) = 0. Thus, supp (T * 4i) C (x E RN supp 4x fl supp T # 0) = supp T + supp 0.
Next we prove (iii). Let e; be an element of the canonical basis of RN and for every x E RN and h # 0 consider the function Oa,h,i (y) :_
4) (x + he; - y) - 0 (x - y),
yE
h
1[$N.
# (x - y) for all y E RN and conditions (i) and (ii) of Theorem 9.8 are satisfied (why?). Hence, {4) converges to (A. ) x with respect to r as h -p 0, and so, by the linearity of T and As h -* 0, we have that 0x.h,i (y)
Theorem 9.10, (T * 4)) (x + he;) - (T * 4)) (x) h
T
T
(8" \ Ox:
as h -p 0, which proves that z = T *. Moreover, since for all x, y E RN,
(ax )z (y) = for all x
8x (x - y) =
-a (x - y)
av (y),
RN we have a
* 4)/ (x) =
a
(4)x)
=TI _(
_ -T
&Yi
)= (T*-)
N-i which, together with an inducti\on ar\gument, gi ves (iii).
Finally, to prove (iv), we approximate the Riemann sum
(
x),
function cp * 0 with the
uh (x) := hN E Sp (x - hy) 0 (hy) ,
x E RN,
YEZN
where h > 0. Note that for every multi-index a, a
8xh (x) = hN 1/E$N axa Since {
} converges uniformly to
8xa
(x. - hy)
(hy) ,
x E IlBN.
* 0 by the previous exercise and
*0_ &60*0) 0x°i
9.4. Convolutions
277
(why?), by Theorem 9.8 we have that {uh) converges to V*4) with respect to r as h -> 0. It follows that for every x E RN, {(nh)'} converges to (cp * Of with respect to r as h --+ 0. By the linearity of T and by Theorem 9.10 we have that
(T * (tP * )) (x) = T ((co * Of) = lim T ((uhf) = lim (T * Uh) (x) h-,0
= lim hN h-,0
h"O
(T * gyp) (x - hy) 4) (hy) YEZN
_ ((T * 9) * 0) (x), where we have used the previous exercise. This completes the proof.
0
As a consequence of the previous theorem, we can approximate convolutions with C°° functions.
Theorem 9.30. Let T E V1(RN) and let {rpE} f, e > 0, be a family of standard mollifiers2. Then {T * cp} converges to T in the sense of distributions as E -> 0+; that is,
sh 0 for every 0 E D (RN).
I
N
(T * coe) (x) 0 (x) dx = T (0)
Proof. By Theorems C.19(i) and C.20 and Theorem 9.8 we have that for every 0 r =D (RN) the sequence {ape * 4i} converges to 0 with respect to r as
For every 0 E D (RN) define fi (x)
4) (-x) ,
xE
RAr.
Then by Theorem 9.10, T (0) = (T * ) (0) = T
Jim T ((APE * )c)
= Urn (T * (cpE * )) (0) = Eli m ((T * we) * ) (0)
= lim f
e-.O+ f1,N
(T * cpe) (y)
(0 - y) dy
= lim f (TN * (pe) (y) 0 (y) dy, e 4O+
where we have used Theorem 9.29(iv).
Exercise 9.31. Let Il C RN be an open set and let T E D` (fl). 2See Appendix C.
O
278
9. Distributions
(i) Prove that there exists a sequence {T,,} C V (ft) such that each T, has support compactly contained in f and {T.} converges to T in the sense of distributions. (ii) Prove that CLOD (S2) is dense in D' (S2) with respect to the weak star
topology of V (ci).
Chapter 10
Sobolev Spaces Newton's first law of gradrwtion: A grad student in procrastination tends to stay in procrastination unless an external force is applied to it. -Jorge Cham, www.phdcomics.com
In this chapter we define Sobolev functions on domains of RN, N > 1, and begin to study their properties.
10.1. Definition and Main Properties Definition 10.1. Let ft C RN be an open set and let 1 < p < oo. The Sobolev space W" (S2) is the space of all functions u E LP (Q) whose dis-
tributional first-order partial derivatives belong to V (0); that is, for all i = 1, ... , N there exists a function gi E 1/ (S2) such that (10.1)
u-Lo-
in exi
dx = - in gi4dx
for all 0 E CC° (S2). The function gi is called the weak, or distributional, partial derivative of u with respect to xi and is denoted O. In terms of distributions the previous definition means that for all i = 1, . . . , N (see Definition 9.17).
= Ti
Remark 10.2 (Important). Following the literature, we use the same notation to indicate the weak and the classical (in the sense of (8.4)) partial derivatives of a function. Unfortunately, this results in endless confusion for students. In the remainder of this book, when u E W4(92), unless otherwise specified, is the weak partial derivative of u. 279
10. Sobolev Spaces
280
For u E W1P (f2) we set
Vu :=
au
au.
X_I'
XN)
As usual, we define W 1,' (n; Rd) as the space of all functions u = (ui,... , ud) such that u; E W 1°P (ff) for all i = 1, ... , d. Also, 1 (SZ) :_ {u E L, a (11): u E Wl,P (l?) for all open sets S)' CC fl).
Exercise 10.3. Let Il = B (0,1). Show that the function u : 0
R,
defined by
u (x) = u (XI, ... , xrv):=
1
ifxN>0,
0
if xN < 0,
does not belong to W1,P(1Z) for any 1
Exercise 10.4. Let fZ C RN be an open bounded set, let xo E SZ, and let 1 < p < oo. Prove that if u c C A n C1 (S2, {xo}) is such that the (classical) gradient Vu belongs to LP (ft; RN), then u E W '-P (f2).
We now show that W1.P (ti) is a Banach space.
Theorem 10.5. Let ti C RN be an open set and let 1 < P:5 oo. Then (i) the space W 1,p (ti) is a Banach space with the norm IIUIIW1.p(n)
IItIIL,P(f) + IIVUllLP(n;RN)
(ii) the space H1 (Q) := W1,2 (f2) is a Hilbert space with the inner product N p
n1 n
(u, v)Hi (n) := J uv dx +
J
au av
a7adx.
Proof. We only prove part (i). Let {u,=} C W1.P (ft) be a Cauchy sequence;
that is,
O= urn
Ilud -
lion (lIui
t,n-> 00
- unIIL'(n) + IIVui - Vu,,,IILP(n,RtN))
.
Since LP (f2) is a Banach space, there exist u, v1i ... , vN E LP (ft) such that Jim Thin - ulIL,(n) = 0,
aax ,l
a
- Vi LP(a) = 0
for all i = 1, ... , N. Fix i = 1, ... , N. We claim that let 0 E Cc' (fl) and note that ,!
1P LU, axi aEx = - Jn 28n a "
clx.
= v,. To see this,
10.1. Definition and Main Properties
281
Letting n -> oo in the previous equality yields
in
bv:dx = -
u
8o
J8x;
dx
(S2), which proves the claim. Thus, u E W' (n). This
for all 0 E
0
completes the proof.
Remark 10.6. The previous theorem continues to hold if we consider the equivalent norms n IIuIIWI.P()
(IIuII() +
-P
au P
OOLF))
or
IIUIIWI.p(S2) :=
IIUIIp() +
E-1
I8xiILP(cl)
for1
au axi
Lo(O)
,
au "11 8xN
forp=oo. Exercise 10.7. Let SZ C RN be an open set and let 1 < p < oo. (i) Prove that a subset of a separable metric space is separable. (ii) Prove that W IM (S2) is separable. Hint: Consider the mapping W1,P (n) - IF (St) X LP (!a; RIV)
u H (u, VU).
Exercise 10.8. Let 11 C RN be an open set. Prove that W"O° (Cl) is not separable.
Exercise 10.9. Let Cl C RN be an open set, let k E N, with k >- 2, and let 1 < p < oo. Define by induction the Sobolev space WkP (Cl) as Wk,P (S2) := {u E L" (Sa) : Vu E
Wk-"P (Cl; RN)}
.
Prove that W k,P (St) is a Banach space and that u E LP (fl) belongs to Wk,P (Cl) if and only if for every multi-index a there exists a function gg E LP (SZ) such that
for all 0 E
(SZ).
f 9,,0dx
10. Sobolev Spaces
282
Let Q C RN be an open set and let 1 < p < oo. The space Wa'1' (S2) is defined as the closure of the space CC° (Q) in W 1'Y' (f2) (with respect to the topology of W1'P (12)).
Remark 10.10. It is important to observe that since uniform convergence 1 preserves continuity, functions in Wa'0° (fl) are necessarily of class C' (S2). In particular, piecewise affine functions do not belong to W0'0O (S2). This is the reason why some authors refer to Wa'0O (11) as the closure of the space CC° (92) in W 1'°° (11) with respect to the weak star topology of W 1'°° (f2),
rather than the strong topology. These two definitions are not equivalent. Thus, caution is needed when applying results for Wo'OO (S2).
We conclude this section by observing that in several applications in partial differential equations and in the calculus of variations, when one works with domains S2 of infinite measure, the Sobolev space W1'1' (0), 1 < p < oo, may not be the right space to work with. Indeed, as the following exercise shows, there exist solutions of the Dirichlet problem for the Laplacian with the property that u r= Lj ° (0), Vu E LP (SZ; RN), but u does not belong to W i,p (S2).
Exercise 10.11. Let 0 := {x E RN : IxI > 1) and let
u(x):= J 1- x12-N if N > 3 and x E fl, logIxI
ifN=2andxE0.
(i) Prove that u is a classical solution of the Dirichlet problem
©u=0 in fl, u=0
on 852.
(ii) Prove that u E Li (St) for all 1
q < oo, but u
LQ (0) for any
1
(iv) Prove that
E LP (0) for all 1 < p < oo.
To provide a functional setting for Dirichlet problems as in the previous exercise, we introduce the space L1" (i).
Definition 10.12. Let 12 C RN be an open set and let 1 < p < oo. The Sobolev space L1" (c) is the space of all functions u E Li o (S2) whose distributional gradient Vu belongs to LP (S2; R"'). When 0 is connected, to define a norm in L1MT' (S2), fix a nonempty open
set 12' CC 0 and define IItIILI.n(n) := II'IILI(n') + IIVuIILP(n;RN) .
10.2. Density of Smooth Functions
283
Exercise 10.13. Let 1 C RN be an open connected set and let 1 < p < oo. Prove that L1,P (I) is a Banach space. Note that the inclusion
W1"P(1)CLl,n(1) holds. We will see later on that as a consequence of Poincare's inequality for sufficiently regular domains of finite measure the spaces L1' (S2) and W 1,p (f) actually coincide. However, for domains of infinite measure (see Exercise 10.11) or irregular domains with finite measure (see Exercise 11.7) this is not the case.
10.2. Density of Smooth Functions The next exercise shows that Sobolev functions can behave quite badly.
Exercise 10.14. Let SZ := B (0,1) C RN. (i) Consider the function 'L (x)
Ila I
x00,
where a > 0. (a) Prove that the (classical) gradient Vu belongs to L jC (St; RN)
if and only if a+1
p > N). (ii) Let
c fZ be a countable dense set and define
u(X):_E21x-1xnla n=1
00
xE01U{a:n}, n=1
where 0 < a < N - 1. Prove that u E W' ' (S2) if and only if (a + 1) p < N, but u is unbounded in each open subset of Q. Thus, functions with a weak derivative can be quite pathological. Despite the previous example, we can prove the following result.
Theorem 10.15 (Meyers-Serrin). Let 12 C RN be an open set and let 1 < p < oo. Then the space 000 (SZ) fl W l,p (S2) is dense in W 1,P (n).
We begin with an auxiliary result. We use standard mollifiers (see Appendix C).
10. Sobolev Spaces
284
Lemma 10.16. Let ft C RN be an open set, let 1 < p < oo, and let u G W 1,P (0). For every e > 0 define uE := V, * u in f2e, where cps is a standard mollifier and Us :_ {x E f2: dist (x, aft) > e} .
(10.2) Then
lim CL juE - uIPdxMA+(IVu_- -Vul''dx)p=0. In particular, if SZ' C S2, with dist (f2', 8f2) > 0, then
Ilue -
0
as e - 0+.
Proof. By Theorem C.20 we have that ue E C°° (f2e) and for x E f2e and
1,...,N, (x)
8xi
f g
=
(x - y) u (y) dy = 8xi
Jsa
8yi
ape (x - Y) avi (y) dy = (We *
(x - y) u (v) dl!
axi) (x)'
where we have used (10.1) and the fact that for each x E f2e the function cp,, (x - ) belongs to C.° (1), since supp cps (x - -) C B (x, e) C f2. The , result now follows from Theorem C.19 applied to the functions u and
i = 1,...,N. Remark 10.17. Note that if f2 = RN, then Sle = RN. Hence, u,, -> u in W 1.P (RN).
Exercise 10.18. Let 0 C RN be an open set and let 1 < p < oo. Prove that if u r= W 1,P (0) and cp E C' ° (RN), then 'pu E W1,P (l).
We now turn to the proof of the Meyers-Serrin theorem.
Proof of Theorem 10.15. Let f2z CC f2i+1 be such that 00
O= u i i=1
and consider a smooth partition of unity F subordinated to the open cover {f2i+1 \ f i-11, where f2_1 = f?.c := 0.1 For each i E N let /i be the sum of all the finitely many 1i E T such that supp -0 C Hi+1 \ ?2i_ 1 and such 1Note that we do not work simply with {its+1 \? p, since, otherwise, O1 would not be covered.
10.2. Density of Smooth Functions
285
that they have not already been selected at previous steps j < i. Then
Wi E C,° (c +1 \7) and 00
E bi = 1 i=1
(10.3)
in S2.
Fix 7? > 0. For each i E N we have that (10.4)
supp (Zit) C I +1 \ ni-1,
and so, by the previous lemma, we may find ei > 0 so small that (10.5)
supp (vliiu)£{ C ni+1 \ Ii-1
and
-
77
II(pit)f, -1AU11wt.P(n) < , where we have used the previous exercise. Note that in view of (10.5), for every f1' c c fZ only finitely many ni+1
ni-1 cover flr, and so the function CO
v
(Oiu)£{ L=1
belongs to C0O (11). In particular, v E W1 v (Z). For X E f2ei by (10.3), (10.4), and (10.5), (10.6)
(V)iu),(x)
u (x) = 1 (Piu) (x) , v (x) =
.
i=1
i=1
Hence, e
re
(10.7)
IIu - VII W,.P(nj)
C!
111piu)E,
2i < 7
-1
i.1
.
L.1
Letting £ - oo, it follows from the Lebesgue monotone convergence theorem that Iju - vIIw,.,(n) < tl. This also implies that u-v (and, in turn, v) belongs to the space W1,r (f1).
Remark 10.19. We will see later on that if u e L1,r (ft) (see Definition 10.12), then u E Lpj,, (f2). Hence, we can adapt the proof of the MeyersSerrin theorem to show that if u E L1'31 (fl), then for every e > 0 there exists a function v E C°° (fl) n L',P (Cl) such that IIu - VIIwI.n(n) < F^,
despite the fact that neither u nor v need belong to W1' (f1).
Exercise 10.20. Prove that the function u (x) := IxI, x E (-1, 1), belongs to W1i0O (-1, 1) but not to the closure of COO (-1, 1) n W1'00 (-1, 1).
10. Sobolev Spaces
286
The previous exercise shows that the Meyers-Serrin theorem is false for p = oo. This is intuitively clear, since if n C RN is an open set and c C00 (5)) fl W1'0° (5)) is such that Ilun - ullw,,,,(n) -' 0, then u E Cl (5)) (why?). However, the following weaker version of the Meyers-Serrin theorem holds.
Exercise 10.21. Let 5Z C RN be an open set with finite measure and let u E W l,00 (52). Modify the proof of the Meyers-Serrin theorem to show that there exists a sequence {utz} C COO (5)) fl W 10O (5t) such that as n -, oo,
Ilun - ulltm) - 0,
IIVuuIIL.(n;RN) - IIVUIIL-(n;RN)
and Vu. (x) - Vu (x) for GN-a.e. x E 52. We recall that given an open set fl C RN, the space COO () is defined as the space of all functions in CO° (0) that can be extended to functions in COO (RN) (see Appendix C).
Exercise 10.22. Let Sl = B (0, 1) \ {x E RN : XN = 0}. Show that the function u : 5Z - R, defined by
u(x) = u(xi,...,xN)
1
ifxN>O,
0 if xN < 0,
belongs to W 1,P (5Z) for all 1 < p < oo, but it cannot be approximated by functions in C°O (a) . The previous exercise shows that in the Meyers-Serrin theorem for general open sets 0 we may not replace C°° (5)) with C°O A. The problem is that the domain lies on both sides of its boundary. This motivates the next definition.
Definition 10.23. An open set St C RN satisfies the segment property if for every x0 E 85Z there exist r > 0 and a vector v E RN \ {0} such that if x E S2 fl B (xo, r), then x + Ov E 5 for all 0< 0 < 1. It turns out that the segment property is related to the regularity of the domain. Indeed, we have the following result. Theorem 10.24. An open set SZ C RN satisfies the segment property if and only if
(i) 85)=8(), (ii) for each point x0 E 851 there exist a neighborhood A of xo, local coordinates y = (y', tN) E RN-1 x R, with y = 0 at x = xo, and a function f r= C (QN_1 (0,r)), r > 0, such that
00 fl A = { (y', f (y')) : y' E QN-1(0, r) } .
10.2. Density of Smooth Functions
287
'Ib highlight the dependence of f on the point xO in (ii), we will some-
times write f = f
xo).
The boundary 01) of an open set fl C RN satisfying properties (i) and (ii) is called of class C. Before proving the theorem, some remarks are in order. Without loss of generality, in the definition of segment property one can replace the ball B (xo, r) with any small (open) neighborhood of xo and, similarly, in part (ii) the (N - 1)-dimensional cube QN-1 (0, r) can be replaced by any small (open) neighborhood of 0 in RN-1. We will use this fact without further notice.
We remark that in part (ii) background coordinates x = (x1.... , xN), in terms of which ) is defined, and local coordinates y = (yl, ... , yN) at a point x = xo are related by an affine transformation of the form yr y := Lxo (x) = R.. (x - xo) , (L..)-' (y) = xO + where R,,,, is an orthogonal N x N matrix. Given xO E 851, t > 0, and a function f = f xo) as in (ii), we define the sets (10.9) V (xo, r, t):= {(I/, yN) : &I E QN-1 (00'), -t < YN - f (it) < t),
(10.8)
(10.10) U (xo, r, t) :=
(V (xo, r, t)).
Remark 10.25. Condition (i) ensures that the domain 1 does not lie on both side of its boundary. Sets like
RN\{xERN: xN=O} are excluded.
Exercise 10.26. Let n C R2 be a bounded open set satisfying condition (ii) of the previous theorem. Prove that Oil = 8 (M. Hint: Prove that if 81) # 8 (T), then there exists a component r of 81) that is a closed curve and such that R2 \ I' is connected. Prove that this violates the Jordan curve theorem (see Theorem 4.56). By the previous remark, by eventually changing the sign of yN, we can always assume that
0 n A = {(y',YN)EflnA: y'EQN-1(0,r),1/N>f&)}
(10.11)
We now turn to the proof of Theorem 10.24.
Proof of Theorem 10.24. Assume that 81) is of class C. Then for every xo E 8St there exist a neighborhood A of xO, local coordinates y = (y', VN) E RN-1 x R, with y = 0 at x = xo, and a function f E C (QN_1 (0, r)), r > 0,
such that
S)nA= {(y',yN) E0nA: y'EQN-1(0,r), yN> f (y')}.
10. Sobolev Spaces
288
By taking r > 0 smaller, if necessary, and t > 0 sufficiently small, we have that the open set U (xo, r, t) defined in (10.10) is contained in A. Then the neighborhood U (xo, r, ) of xo and the vector v :_ (L y0)-1 ((0, ... , 0, ) ) satisfy the definition of segment property. Conversely, assume that f2 C RN satisfies the segment property. Fix xo E 8f2 and find a neighborhood W of xo and a vector v E RN \ {0} such that if x e S2 fl W, then x + 9v E fl for all 0 < 9 < 1. Take local coordinates
y = (y', yN) E RN-l x R. with y = 0 at x = xo and with y = eN at x = xo +'. For x E U n W, let y := Ly0 (x). The segment
Leo({x+9v: 0<9<1})=y+{(0,0IvI): 0<9<1} will be called the arrow from V. From now on, for simplicity, we omit mention of Lx0, and thus we write 0 E 8f2, for example, in place of 0 E (8f2). Consider the closed segment
S = {o} x [-77, q]
where 0 <,q < IvI is chosen so small that S C W. Note that the points of S with yN > 0 are in 52 because they are on the arrow from 0 E 852, while the points of S with yN < 0 are not in SZ because the arrows from them contain 0 that is not in 52. Let Sl
{0} x L-77, -1771
S.
{0} x
,
Sm :_ {0} X I
1
271, 177]
L
277,71
(here the subscripts stand for lower, middle, upper). Since SS, Sm, and S. are compact sets, we may find finite open covers of cubes centered in S and contained in (RN \ SZ) n w, w, and 52 n W, respectively. By taking r > 0 sufficiently small, we obtain that the union of all these cubes contains the open set V := QN-1 (0, r) X
that 1 := QN-1 (0,r) x (-77, -7177) C RN \?!, and that V,,!= QN-1(0, r) x (271,71} C Q. Define f E QN-1 (0,r) -} R by
f
inf {yN : (y', YN) E U n V } .
Note that -277 < f < 277 because V C RN \? and V. C f2, and the infimum is attained because 52 is closed. Moreover, for every y = (y', yN) E V, if yN > f (y'), then y E 12, since y is on the arrow from (y', f (y')) E 52, while if yN < f (y'), then y E RN \ SZ by definition of f (y'). Thus, for every y' E QN_1(0, r) the point (y', f (y')) belongs to 852 and is the only such point in V.
10.2. Density of Smooth Functions
289
It remains to prove that f is continuous. Let y' E QN_ 1 (0, r) and let 0 < e < 4r?. Define (y', f (y') + e) , yBy what we just proved and the bound If I
(y', f (y') -5) .
y+
<-
2
q, we have that y+ E t2 f1 V
and y- E (RN \ ?7) fl V. Since these two sets are open, there exist cubes QN (y+, d) C n fl v and QN (y-, d) C (RN 112) fl v. We claim that
If(y')-f(z')I
Example 10.27. Let N = 2. The set
B(0,1)\{x=(x1ix2) ER2: 0<x1 <1,x2=0} fails to satisfy both conditions (i) and (ii) of the previous theorem. In particular, property (11) fails at the boundary points (0, 0) and (1, 0).
The following example shows that domains with continuous boundary may have exponential cusps.
Exercise 10.28. Prove that the domain
<exp` f x= (xl,x2)ER2: 0<xl <1,-exp<x2 X1 X1 / is of class C and that the function f in (ii) of the previous theorem cannot be taken of class C°'01 for any a > 0. We now prove the density of functions C°° (RN) in W',P (fl) for domains with continuous boundary.
Theorem 10.29. Let 1 C RN be an open set whose boundary is of class C. Then the restriction to fl of functions in C°° (RN) is dense in W" (1Z) for
1
Zb prove the theorem, we need an auxiliary result.
Lemma 10.30. Let 0 C RN be an open set, let 1 <_ p < oo, and let u E L" (11). Extend u by zero outside fl. Then for every E > 0 there exists 8 > 0 such that Iu (x + h) - u (x) Ip dx < E is, for ail h E R,Nwith IhI <- S.
10. Sobolev Spaces
290
Proof. Exercise. We now turn to the proof of Theorem 10.29.
Proof. By the Meyers-Serrin theorem, without loss of generality, we may assume that u r= co O(S2) n W1,P (S2). For every xo E 80 there exist a neighborhood A,,a of xo, local coordinates y = (y', yN) E RN-1 x R, with y = 0 at x = xo, and a function f E C (QN_i (0, r)) , r > 0, such that 80 n Ayo = { (y', f (y')) = y' E QN-1 (0, 01, and (see (10.11))
(10.12) c n A
(y ' , yN) E S2 n A ,,
/1
: y' E QN-1 (0, r), yN > f (y') }
.
If the set 11 \ Uxeo Ax is nonempty, for every xo E SZ \ U.,Easa A, let A.,. be any open ball centered at xp and contained in f2. The family {A.':}.Eye is an be a smooth partition of unity subordinated to open cover of S2. Let {A2 }.E-0 .
Fix n r= N and define i := uip E W1,P (Il) (see Exercise 10.18), where we extend un to be zero outside supp'+bn. There are two cases.
If supp 0 is contained in n, then we set vn :_ Onu E C:° (RN). If supp, i , is not contained in S2, then fix any x E 80 such that supp lpn C A.,,. For t > 0, using local coordinates in Ay,,, define the function un,t R by tn,t (y', yN) := U. (y', yN + t) , where
At = { (y', YN) E RN : y' E QN-1 (0, r) , yN >
f
Note that un,t is well-defined for t > 0 sufficiently small and that A D
n A.,,. Fix 91 > 0. By the previous lemma there exists to > 0 so small that supp'trn + B (0, tn) C A,,, and (10.13)
2n.
Ilun,tn
Construct a function 0,s E Coo (RN) such that 4n (y', yN) = 1 if y` E and 0m (y', yN) = 0 if y' E QN-1 (0, r) QN-1(0, r) and yN > f (y') and yN < f (y') - 3+ and define V.:= O.Un,t,,
Then v E C.° (RN) (provided we define un,t,. to be zero, whenever 0n is zero) and supp v C A%. Note that v, = un,t,, in S1, and so (10.13) may be
10.2. Density of Smooth Functions
291
rewritten as (10.14)
2n
Define the function
00
V:_Evn. n=1
Note that
00
supp v C U := U At. n=1
Since for every open bounded set U' C U only finitely many A! cover U', it follows that v E C°° (RN). In particular, v E W1 ' (St). Moreover, as in the last part of the proof of the Meyers-Serrin theorem, for every open bounded set fl' C fl we may find an e such that for every x E fl' we have e
U (x) _ >2 (1I u) (x) , n=1
t v (x) _ E vn (x) n=1
Hence, It
e
1k' - VlIw1.pp) <-
IIIPnu -
<_ 17.
Vnllw1.p(n) <_ E
4=1
2
Letting Ill / f2, it follows from the Lebesgue dominated convergence theorem that ilu - vlIyy,,p(n) < 71. This also implies that u - v (and in turn v) belongs to the space W1,P (fl). Finally it suffices to approximate v with a function in C:° (RN).
Remark 10.31. Note that in view of Theorem 10.29 it follows that for all
1
W' (RN) = Wi,P (RN)
.
In Exercise 10.39 below we will show that there exists an open bounded domain SZ C R2, with 812 = 8 (a), such that W 1,P (11) n C (i is not dense in W 1,P (Sl), 1 < p < oo.
As a corollary of Theorem 10.29, we can prove that piecewise affine functions are dense in W 1,P (12), whenever fI C 18N satisfies the segment property.
Definition 10.32. An N-simplex 0 is the convex hull of N + 1 points xi E RN (the vertices of A) that are not contained in a hyperplane, namely, N+1
XER
X
N+1 ,
j=1
E93=1
1=1
,
10. Sobolev Spaces
292
and foralliE{1,...,N+1}, {x3-xi:jG{1,...,N+1}\{i}} is a basis of RN. Let PA be the family of all continuous functions u : RN - R for which there exists a finite number of N-simplexes A 1, ... , Al with pairwise disjoint interiors such that the restriction of u to each Ai is affine and u = 0 outside e
UAi.
i=1
Theorem 10.33. Let S1 C RN be an open set whose boundary is of class C. Then the restriction to Q of functions in PA is dense in W ',P (St) for
1
out loss of generality, that u E C°° (RN). Let K := suppu and consider B (0, R) D K. Fix a > 0 and construct N-simplexes 1, ... , At with pairwise disjoint interiors such that
t
KCUAiCB(0,R) i=1 and
diem Ai > 4'i
for all i = 1, ... , t, where ri is the radius of the inscribed ball in Ai. Let v be the continuous piecewise affine function that coincides with u in all the vertices of the simplexes. Then by Taylor's formula sup Iu (x) - v (x) I C Ch2 sup I V 2u (x) I < Ch2 sup I V2u (x) I , xEA,
WEA;
xERN
sup IVu(x) - Vv(x)I < Ch sup IV2u(x)I
xEA:
for all i =
xERN
where
h := max diam ii. 1
Thus, t
f
f
f Ivu - 0vip da < f .ffRN f
IQtb - vvIP dx =
SZ
f
1041 - 9JIp (fix
(
f
{
G111 (Di)
A similar estimate holds for fa Iu - vir dx. By choosing h sufficiently small (and so increasing the number t of N-simplexes), we have the desired approximation.
10.3. Absolute Continuity on Lines
293
Remark 10.34. If f2 C PEN is an open set and u E W' -P (Sl) for 1 < p < oo, then for any Sl' cc SI we can construct a sequence of fu,,) C PA such that un -r u in W 1P (Cl'). Indeed, by the Meyers-Serrin theorem we may find v E C°° (Sl) fl W 1,y (S1) such that v,, -> u in W 1,p (St). Construct a cut-off function iI' E C° (Cl) such that ip = 1 in W. Since -Ov. E CC° (Cl), we may extend iftfz to be zero outside Cl and then apply the previous theorem to
find u, E PA such that Il-Ovn - U-11W..p(sz) <- n
Using the fact that ip = 1 in Cl', it follows that IIu-unhIW1,p(n,)
<-
Thin-umIIWI.p(n')
<_ Ilu -
(Q) + 110n - ufhI Wl.P(SZ) -
It suffices to let n -> oo.
10.3. Absolute Continuity on Lines The next theorem relates weak partial derivatives with the (classical) partial derivatives. This result is the analog of Theorem 7.13. In what follows, we use the notation (E.2) in Appendix E. Given x; E RN-1 and a set E C IAN, we write (10.15)
E ._ (xi E R :
E E).
Moreover, if v : Cl --+ R is Lebesgue integrable, with a slight abuse of nota-
tion, if fl, is empty, we set
I
v (x;, xi) dxi := 0,
sax,
{
so that by Fubini's theorem (10.16)
Jv(cc)dx=J N-1 fa v (x, xi) dxjdx. ,
Theorem 10.35 (Absolute continuity on lines). Let Cl C RN be an open set and let 1 < p < oo. A function u E LP (Cl) belongs to the space W 1,P (Cl) if and only if it has a representative i that is absolutely continuous on LN-1a.e. line segments of Cl that are parallel to the coordinate axes and whose first-order (classical) partial derivatives belong to LP (Cl). Moreover the (classical) partial derivatives of u agree £N-a.e. with the weak derivatives of u.
Proof. Step 1: Assume that u E W 1°1' (Cl). Consider a sequence of standard mollifiers {cp£}E>o and for every E > 0 define uE := ii * cp£ in SZ£. By
10. Sobolev Spaces
294
Lemma 10.16,
Slim fad I vu, (x) - Vu (x) I" dx = 0.
Using (10.16), it follows by Fubini's theorem that for all i = 1, ... , N, lim
I Du£ (x;, x,)
- Vu (xs, x,) I' dxi dxi = 0,
and so we may find a subsequence {Fm} such that for all i = 1,., -, N and for GN-1-a.e. xi E RN-1, (10.17)
lim f vu, (xi, xi) - Vu (x;, xi) n-4oo (n }¢,
dxi = 0.
I
Set u,, := ue2 and
E:= {x E fZ : lim un (x) exists in 11I'}
n+*
.
Note that for every x E 0 we have that x E 12,. for all n sufficiently large (depending on x), and thus ia,, (x) is well-defined for all n sufficiently large. In turn, the set E is well-defined. Define
(x) -
lim un (x) if x E E, n-oo otherwise.
0
Since {u,} converges pointwise to u at every Lebesgue point of u by Theorem
C.19, we have that E contains every Lebesgue point of u. Moreover, since by Corollary B.122 the complement in ft of the set of Lebesgue points has Lebesgue measure zero, it follows that LN (ft \ E) = 0. This shows that the function I is a representative of u. It remains to prove that Ui has the desired properties. By Fubini's theorem for every i = 1, ... , N we have that IVu (x;,xi)I'* dxi) dx; <00
RN-i and
(Ixi E ftx,
G1
:
(xi)xi) 0 E}} dxi = 0,
and so we may find a set Ni C RN-1, with GN-1(Ni) = 0, such that for all x; E RN-1 \ Ni for which ge is nonempty, we have that (10.18)
/
IVu
dxi < oo,
s1z,
(10.17) holds for all i = 1, ... , N, and
x,) E E for £1-a.e. x, E !Z
.
10.3. Absolute Continuity on Lines
295
Consider a closed rectangle R:= [ai, b1] X ... X [aN, bN] C 11,
with ai and b1 rationals for all i = 1, ..., N. Since dist (R, 3) > 0, we have that R C 1Z for e > 0 sufficiently small, and so by (10.17) for all i = 1, ... , N and for all xs E R; \ Ni, (10.19)
iim 1
I V u£n (4 rj) - Vu (xi, xi) I" dxi = 0,
a
where, using the previous notation, R = Ri x jai, bi]. Define
v, (t) u,t (xi, t) , t E [ai, bi] . Let to r: [ai, bi) be such that (x;, to) E E. Then v. (to) -> u (x;, to) E R. Moreover, since v,s E C°° ([ai, bi]), we have that
t v.. (t) = v.. (to) +
v', (s) ds
J
for all t E [ai, bi], and thus by (10.19), for all t E [ai, bi] we have that there exists the limit lim v" (t) - ny+oo { V. (to) + ! vrt (a) dot to
(x;,
ftt C7Xi (xi, s) ds =: v (t) o
Note that by the definitions of E and u, this implies, in particular, that (10.20) {xz} x [ai,bi] C E and that (10.21)
u (x;, t) = v (t) = za (x;, to) +
rt
s) do
for all t E [ai, bi]. Hence, by Lemma 3.31, the function u is absolutely continuous in [ai, bi] and 1 . (x;, t) t) for 41-a.e. t E [ai, bi].
Now if R = [al, bl] x . . . X [NN] is another rectangle contained in fl and such that [ai, bi] rl [a3, bi] # 0, taking an x that is admissible for both R and R and to E [a.i, bi] fl
it follows from (10.20) and
(10.21) that v is absolutely continuous in jai, bi) U 6d. Since Sl can be written as a countable union of closed rectangles of this type and since the union of countably many sets of,CN_1-measure zero still has 0_1-measure zero, by (10.18), (10.21), and Corollary 3.26 we have that for RCN-1-a.e. [ai,
xi E RN-1 \ Ni for which n,,, is nonempty, the function v is absolutely continuous on any connected component of Q.,.
10. Sobolev Spaces
296
Step 2: Assume that u admits a representative u that is absolutely continuous on .C1-1-a.e. line segments of S2 that are parallel to the coordinate axes and whose first-order (classical) partial derivatives belong to IF (S)). Fix i = 1, ... , N and let x; E RN-1 be such that u is absolutely continuous on the open set Slx' . Then for every function e C°° (0), by the integration by parts formula for absolutely continuous functions, we have
s,
t)
e' (x;, t) dt =
axi
c`>-
- Js, 8xi (x=, t)' (xq7 t) dt.
Since this holds for LN-1-a.e. x E RN-1 for which S2x is nonempty, integrating over RN-1 and using Fubini's theorem yields
sl u (x) 8-0 (x) dT_ - J
a
(x) do (x) dx,
which implies that -- E L" (SZ) is the weak partial derivative of u with
0
respect to xi . This shows that u E W1,n (S2).
Remark 10.36. If instead of assuming that u c W 1-P (S2) we only assume that u E L1' (1Z), then Step 1 of the previous proof still proves that u has a representative u that is absolutely continuous on RCN-1-a.e. line segments of f2 that are parallel to the coordinate axes and whose first-order (classical) partial derivatives belong to L" (12). As a consequence of Theorem 10.35 and of the properties of absolutely continuous functions we have the following results.
Exercise 10.37. Let 12 C RN be an open set and let 1 < p < oo.
(i) (Chain rule) Let f : R -+ R be Lipschitz and let u E 147 1 (Q). Assume that f (0) = 0 if St has infinite measure. Prove that f o u E W l,P (S1) and that for all i = 1, ... , N and for LN-a.e. x E 12, U)
0 (fexi0
(x)
= f (u (x)) aU (T),
8u where f (u (x)) ax (x) is interpreted to be zero whenever (x) 8xi 0. What can you say about the case p = oo? (ii) (Product rule) Let u, v E Wl,p (12) fl L°O (St). Prove that uv E W l,p (0) n L°° (0) for all i = 1, ... , N and that for CN-a.e. x E 0,
a v) (x) = v (x) ax (x) + u (x) What can you say about the case p = oo?
(x)
.
10.3. Absolute Continuity on Lines
297
(iii) (Reflection) Let 11 = R+ := {(x',XN) E and let u E W 1,P
RN-1
x R : XN > 0}
(RN). Prove that the function
t
v (x) .
u(x)
ifxN>0,
IL W, -XN) if xN < 0
belongs to W 1,P (RN) and that for all i = 1, ... , N and for CN-a.e.
xEIItN
ifxN>0,
8 (x)
8v (x) axt
(-1)61N 8u
8xi (x', -XN)
ifxN<0,
where biN is the Kronecker delta, that is, SiN = 1 if i = N and aiN = 0 otherwise.
(1), and let u (iv) Let E C R be such that G1 (E) = 0, let u E be its precise representative given in Theorem 10.35. Prove that Vu (x) = 0 for GN-a.e. x E (u)-1(E). (v) Prove Exercise 10.14 using Theorem 10.35.
(vi) Prove analogous versions of parts (i)-(v) in the case in which the functions u and v are in L1,P (f2).
Exercise 10.38. Let Q C RN be an open set and let 1 < p < oo. Let f : Rd -' R be a Lipschitz function such that the set EJ :_ f u E Rd : f is not differentiable at u} is purely 7.11-unrectifiable. Assume that f (0) = 0 if fZ has infinite measure.
Prove that f o u E W1,P (SZ) for all u E W1.r (11; Rd) and that for all i = 1, ... , N and for GN-a.e. x E 0,
8 You) /
i
where
d
(x) = L..: of (u (x)) a ss (x) , j=1 loui
; (u (x)) a j (x) is interpreted to be zero whenever
i (x) = 0.
The next exercise shows that there exists an open bounded domain Il C R2, with 8S2 = 8 (St), such that W 1,p (il) n C (St) is not dense in W 1,p (11), 1 < p < 00 (see Theorem 10.29).
Exercise 10.39. (i) Construct a sequence of open balls {B ((t,,, 0), r,)} C Ilt2
10. Sobolev Spaces
298
with 0
Er,z < Z, n=1
in such a way that Un° 1 (tn - rn, to + rn.) is dense in (0,1) and
as2=a(N). (ii) Let 00
B((1,0),1), U B((ts,0),rn)
52
n=1
and for (x1, x2) E i) define
for0<xl
1
2L (xl, x2)
for 0 < x1 < '1, x2 < 0, anything convenient otherwise. 0
Prove that u E W11P (SZ) for all 1 < p < oo.
(iii) Let r be the set of x 1 E (4 , 2) with (x 1, 0) E M. Assume by contradiction that there exists a sequence {un} C W1.P (cl) f1C (SZ)
that converges to u in W 1,P (52) and prove that for all n E N and for all x2 E (0, 1),
f lu, (xl, x2) - un (xl, -xa)I dxl
Jr
f
"(xl,s) dxIds.
X(-x2,z2)
ax 2
(iv) Using Fubini's theorem, prove that there is a subsequence {unk } of {u,} such that for ,C1-a.e. X2 E (0, ),
k r JI1nk(x1x2)_v1zk(x1_x2)I dr1 = J
r
lu (1x2) - u (xl-x2)I dx1
and show that this is in contradiction to part (iii).
10.4. Duals and Weak Convergence In this section we study the dual spaces of W1,P (fI) and Wo'P (52). We begin with the case 1 < p < oo. The idea is to consider W1,P (Cl) as a closed subspace of LP (Cl; RN+1) In this way any element of L E (W 1,p (Cl))' can be extended by the Hahn-Banach theorem to an element of (LP (Cl; The Riesz representation theorem in LP (Cl; RN+1) will give a representation for the extension of L and, in turn, for L. RN+1))'.
10.4. Duals and Weak Convergence
299
In this section it will be convenient to consider in W 1,P (SZ) the equivalent norm N
.ate
(10.22) II
D
IILD(0)
axi
lip
We begin by observing that if p' is the Holder conjugate exponent of p and fo, ... , fN E L/' (St), then the functional (10.23)
L (u) :=
I
N
fo (x) u (x) +
U E W" (n),
&c,
A (x) 8 i (x) i=1
belongs to (W1,P (Il))'. Indeed, by Holder's inequality (for functions and for finite sums) we have that N
P
IIBaxUi
IL(u)I
IIfoIILP'(n) IIuIILP(n) +
i-0
Ilfillp"
)
IIfiIILP`(0)
IILD(c)
IIuIILp(Q) + E Il i=1
p
aU axi
r
II LD(S2}
for all u E W1,n (SZ), and thus by (10.22), (10.24)
IILII(WI,P(n))'
IL
up
uEwsp
sa
N
5
IlfiliLD,(0)
a IIuIIWU)Ip
Note that in general we do not have equality in the previous inequality (see Remark 10.42 below). In view of the previous inequality, in the space LP (St; RN+1) we use the equivalent norm IV
IIg.IILP(c)
II9IILD(C RM+1) i=0
where 9 = (go, ... , gN). Similarly, in LI" (n.'; IAN}1) we take N
II!IILn'(n;RN+1)
E i=o
P,
Exercise 10.40. Let 1 < p < oo and let LP (1;1RN+l) and LPG (0; RN+1) be equipped with the norms just defined. Using the Riesz representation theorem in L' (0), prove that for every L E (LP (S2; RN+1))' there exists a
10. Sobolev Spaces
300
unique f = (fo, ... , fN) E LP' (a; RN+l) such that N
L (g) = f E fi (x) 9i (x) dx for all g = (go, .. , 9N) E LP (ft; RN+1) . 1=0
Prove also that N
llLII(LP(cy;RN+i))' = E i=o
Theorem 10.41 (Riesz's representation theorem in W 1,P). Let 1 < p < 00 and let p, be its Holder conjugate exponent. Then for every bounded linear functional L E (WI,P (ti))' there exist fo, ... , fN E 17' (ft) such that (10.25)
L (u) =
f J
cN
fo (x) u (x) + L, fi (x) 8 (x) dx i=1
for all u E W 1,P (ft) and N
(10.26)
IILIl(wl.P(fI))' = E llfillLP'(S2} i=0
Proof. Consider the application (10.27)
T : W I,P ()) - LP (ft; RN+1)
au
8u
u'8x ,...,axN 1
The operator T is one-to-one and continuous and it preserves the norm; that is, IIT (u)IILP(SZ;DtI}1) = IIullW'.P(st)
for all u E W"P (ft). Hence, also by Theorem 10.5, the subspace Y T (WI,P (11)) is closed in LP (ft; functional
LI : Y
RN+1).
Given L E (W1°P (ft))', define the
R
g = (go,... , 9N) ' - L
(T-1((9o,
... , 9N))) .
Since T preserves the norm, we have that L1 is linear and continuous, with IIL1IIY- = IILJJ(WI.P(st))'
By the Hahn-Banach theorem (see Theorem A.30) we may extend L1 as a continuous linear operator L1 : LP (fl; V*1+1) -> P.
10.4. Duals and Weak Convergence
301
in such a way that IILIII
(Lp(0AN+1)}-
= IIL1IlY, = IILII(W=,P(n))'
Note that since Y is closed, it is not dense in LP (12; RN+1), and so the extension will not be unique. By the previous exercise there exist unique functions fo,... , fN E 1/ (S2) such that N
L1 (9)
dx
= fn fo (x) 9o (x) + > fi (x) 9i (x) i=1
for all g E LP (S2; RN+1) and N
IILII(wl.P(n)}+ = IIL1II(Ln(.;eN+'))!
_
j7 Ilfzll
'(n}
It follows that
L (u) =
r
Jsz
N
(foX)X)+f(X)X)) dx i=1
0
for all u E W1'P (S2).
Remark 10.42. Note that the previous theorem does not imply that the dual of W 1,P (0) is LP' (S2; Indeed, if f = Yo,..., fN) belongs to RN+1).
LP' (c'; RN+1), then we have shown that the functional L defined in (10.23) belongs to (W' P (9))' and that (10.24) holds. On the other hand, by Theorem 10.41 there exists h = (ho, ... , hN) in LP' (S2; RN+1) (possibly different
from f) such that T
N
IIhiII .,
IILII(wl.n(n)}'
cz
i-0
Thus, we have that N
IILII(wl.P(ca)' = min E IIhiiiLq-(
h E L" (s2; RN+1)
i=0 such that (10.25) holds
To explore the possible lack of uniqueness, assume that h = (h4, ... , hN) E iY (S2; RN+1)
.
10. Sobolev Spaces
302
is such that N
au (fou+fi.)ax
dx
J
" au hou+Fh,a
ax
for all u E W1' (Sl). In particular, we obtain that
J (fo+fi:) dx = J
rtx
i=1
for all we get
i=1
E V (]ESN). If we rewrite this identity in the sense of distributions, N
Tfoxn (0) -
a`y (10) = Thoxn ( i=1
)i=1
8T
8
;
n
(ib)
for all ¢ E D (RN), or, equivalently,
foxn - div (FXn) = hoXa - div (Hxn)
for all 0 E D (RN), where F := (fl,..., fN) and H := (hl,..., hN). Hence, we have shown that given f = (fo,..., fN) E LP' (fl; RN+'-), any solution h = (ho,..., hN) E Z' (Q; RN+1) (in the sense of distributions) of the partial differential equation div (HXn) - hoX& = div (Fxn) - foxsa in RN will give rise to a different representation in (10.25).
In view of Theorem 10.41, we can characterize weak convergence in
W''()).
Exercise 10.43. Let Il C RN be an open set, let 1 < p < oo, and let {u, } C W 1,P (1k). Prove that u,a -s u in W 1.P (f) if and only if u, -s u in LP (S2) and Vu,, - Vu in LP (fl; RN). If SZ has additional properties, in the previous exercise one can replace
un - u in LP (1) with u, - u in LP (1) (see Exercise 11.14). Theorem 10.44 (Compactness). Let 11 C RN be an open set and let 1 < p < oo. Assume that {u,s} C W1"(fl) is bounded. Then there exist a subsequence {ufk } of {u,,} and u E W 1.P (Cl) such that u, - u in W 1,P (11).
Proof. Since NO and {Vun} are bounded in the reflexive Banach spaces L" (Cl) and LP (Cl; RN), respectively, we may select the subsequence {u., }
such that u,,,, -i u in LP (fl) and
s vi in LP (1Z) for all t = 1, ... , N
10-4. Duals and Weak Convergence
303
and for some functions u, v1, ... , vN E LP (il). It remains to show that u e W 1,P (SZ). For every ¢ E (0), i = 1, ... , N, and k E N we have
Jin unx
a
20dx =-
J a k 0 dx. i
Letting k -> oo in the previous equality yields
Ju±-dx = ax; which shows that
n
vidx,
0
= vi. Hence, U E W1,P (a).
Remark 10.45. The previous result fails for p = 1. Indeed, in this case, L' (SZ) is not reflexive (so we do not have weak sequential compactness of bounded sequences) and it is not the dual of a separable space (so we do not have weak star sequential compactness of bounded sequences). To recover some compactness, we use the embedding L1(SZ) -> Mb (1l)
U -' A. (E) where
Au (E) := JE (x) dx,
E E B (il)
.
Note that (why?)
r L Jul dx = IIAUIIMb(n) Thus, given a bounded sequence {u.} C L1 (Q), one can only conclude that A in there exist a subsequence {i } and A E Mb (f) such that W1,1 (11), Mb (SZ). In particular, given a bounded sequence one will C only recover some compactness in the space BV (a). See Theorem 13.35.
Exercise 10.46. Let SZ = B (0,1). Construct a bounded sequence {un} C W1,1 (il) converging strongly in L1 (1Z) to the function given in Exercise 10.3.
Every functional L E (W 1'P (0))' is the extension of a distribution T E More precisely, if L has the form (10.25), then its restriction to D (it) is given by D' (11).
(10.28)
T (0) := Tfo (0)
for all 0 E D (SZ) . e=1
Conversely, a distribution T E D' (SZ) of the form (10.28), where fo,... , fN E
LP' (fl), can be extended to an element of (W1,P (1k))', but this extension may not be unique, unless W 1,p (SZ) = Wo'1' (SZ).
Definition 10.47. For 1 < p:5 oo let p' be its Holder conjugate exponent. The dual space of TV0'r (SZ) is denoted by W-1,p (n).
10. Sobolev Spaces
304
Exercise 10.48. Let fZ C RN be an open set, let 1 < p < oo, and let T E D' (fl) be of the form (10.28), where fo, ... , fN E L/ (1). Prove that T may be uniquely extended to a functional L E W-1,p (a). In view of the previous exercise we have the following characterization of the dual of W3"P (S2).
Corollary 10.49 (Riesz's representation theorem in Let fl C RN be an open set and let 1 < p < oo. Then W-1,P(fl) can be identified with the subspace of distributions T of the form N
T=Tfo_1: fir`, i=1
where fo, ... , fN E LP' (fl).
Next we study the case p = oo. We endow W1,1 (1k) with the equivalent norm IIuIIW1,00(n) := max
...
IIUIIL-(sz)' II ax,
II aXN
By Theorem B.96 the dual of L°O (i]) can be identified with the space of all bounded finitely additive signed measures that are absolutely continuous with respect to the Lebesgue measure restricted to fl. Thus, if Ao, ... , AN are any such measures, then the functional N
r
udAo+> J
L(u):=,
dAi,
uEW1'OO(fl),
belongs to (W1" (S2))'. Indeed, since each Ao E (L°° (Il))', we have that N
IL (u)j< IIAOII(L-(n))' IIUIIL-(n) + Max 0
II
II
(
II
L°O( n ))
IlaOU
IIuIIL-(n)'
IL(fl)
for all u E W '°° (i), and thus, IILII(wl
-(n))' =
sup uEW1.°O(iZ)\{0}
IL
IIUIIwu)(n) < O
Reasoning exactly as in the proof of Theorem 10.41, we have the following result.
10.5. A Characterization of W 1,P (1)
305
Theorem 10.50 (Riesz's representation theorem in W1,1). For every linear continuous functional L E (W1"0° (fl))' there exist bounded finitely additive signed measures AO,.. -, AN that are absolutely continuous with respect to the
Lebesgue measure GN restricted to fl such that
L(u) =
(10.29)
f udAo+FJaxi _ dAj
,152
i=1
n
for all u E W1,0O (fl).
Exercise 10.51. Let fl C RN be an open set. Prove that un W1,P (fl) if and only if un
-
u in
u in LOO (1) and Vu,, - Vu in L°O (fl; RN).
The proofs of the next results are left as an exercise.
Corollary 10.52 (Riesz's representation theorem in W1'OO). The space W-1.1 (fl)
can be identified with the space of all distributions T of the form N
T(u)=TA0(u)-EA (u), axi i=1
where A0, ... , AN are bounded finitely additive measures that are absolutely
continuous with respect to the Lebesgue measure GN restricted to D.
Theorem 10.53 (Compactness). Let fl C RN be an open set. Assume that {un} C W1'O° (fl) is bounded. Then there exist a subsequence {unk} of {un} and u E W1"O° (fl) such that unk - u in W1"1 (fl).
10.5. A Characterization of Wl.n (fl) In this section we give a characterization of W" (fl) in terms of difference quotients. An analogous result has been given in Corollaries 2.17 and 2.43 for functions of bounded pointwise variation in one variable. As we already mentioned, results of this type are often useful in the regularity theory for partial differential equations (see, e.g., [24] and [531). Moreover, they provide characterizations that do not involve derivatives and thus they can be used to extend the definition of Sobolev spaces to more abstract settings (see, e.g., [11] and [81]). Let SZ C RN be an open set and for every i =1, ... , N and h > 0, let
flh,i:={xEfl: x+teiEfl forall0
(a+b)'' < a''+bp for all a, b > 0.
10. Sobolev Spaces
306
(ii) Prove that if p > 1 and r > 0, then there exists a constant C = C (E, p) > 0 such that
(a+b)p <(I+ E)aP+CbP for all a, b > 0. Hint: Use the convexity of the function g (t) = ItJP,
tGR. Theorem 10.55. Let 1 C
RN be an open set and let u E W ',P (Il). 1 <-
p < oo. Then for every i = 1, ... , N and h > 0, Iu (x + hei) - u (x)I P dx < fah'i
Iu (x + hei) - u (x)IP Jszh ,
18u (x)IP dx ax,
J
hP
"
dx
=
hP
(J
18u (x) Ip dx) si
axi
A
< o0
Conversely, if u re L" (0), 1 < p < oo, is such that (10.32)
lim inf
Iu (x + he - u (x)IP
(foh'i
dx
for every i = 1, ... , N, then u E W '.P (cl).
Proof. Step 1: Assume that u E c, (11) n W 1'p (Q) and fix a compact set KCSZ. Let d := min {dirt (K, 00) ,1 } . By Taylor's formula there exists a constant CK > 0 (depending on u) such 2
that
u (x + hei) - u (x) - a (x) hI < CKh2 for all x E K and 0 < h < d. Hence,
su
(x)Ih <_ Iu(x+he;)-u(x)I +CKh2 axi for all x E K and 0 < h < d. By the previous exercise, for every 8 > 0 we have
au (x) hP < (1 + 6) Iu (x + hei) - u (x)IP + CK,rh2P axi for all x E K and 0 < h < d. Dividing by hP and integrating over K yields I
I
axi
(x) I P dx < (1 + 8) fK
<(1+8)J
ju (x + hei) - u (x) I P
dx + CK,dZN (K) hP
hei) -u(x)IPdx+CK,aGN(K)hP.
10.5. A Characterization of W 1,P (SZ)
307
By the previous exercise, P
8u
dx
(x)
UK 18xi
I
P
)
Iu (x + hei) - u (x) I P
< (1 + b) a
hP
Oh.;
P
dx
+ (CK,oGN (K) hP) P .
Letting h - 0+, we obtain 8u
U 8xi
(x)I
P
P
r
1
dx) < (1 + 8)P l h'O f
Iu(x+hei) - u(x)IP
J.h,i
P
dx
If b - 0+ and K / SZ, by the Lebesgue monotone convergence theorem we get
\a dx J< 1im o f 18xi (x) Un 8u
P
I
f
hi
Iu (x +hei) - u (x) I" hP
P
dx
Step 2: If u E W1.P (1), let SZ' cc fl and for 0 < e < dist (SZ', 8SZ) define
uE :_ APE * u, where W, is a standard mollifier. By Corollary B.83, for 0 < h < dist (SZ', OIl) - e, IuE (x + hei) - uE (x)IP dx P
< (f(")h,i I f <_
f
(y)Iu(x+hei-y)-u(x-y)I
(h+i f
'P, (Y)
-y)-u(x-y)Idx
dx) 1 P
dy
< (fnh,i Iu (z + hei) - u (z) IP dz where we have made the change of variables z = x - y and used the fact that fRN APE dy = 1. Letting h - 0+ and using the previous step applied to uE gives (10.33)
Cf I
I
`(x)Pdx P
(ki
Iu(x+hei) -u(x)IPdx
P
10. Sobolev Spaces
308
Since a -,
in LP (1') ass --+ 0+ by Lemma 10.16, we obtain
(10.34)
au I
axi
(x)
P I
(
11 P
dx) < 1h- O nf
f
lu (x + hei) -u(x)I" lip
h,
dx
By letting 11' / f2 and using the Lebesgue monotone convergence theorem, we obtain
8u
(10.35)
IOxi
P
(x)
\P
<
dx 1
sth.:
hmo
lu(x+hei)-u(x)lP dx
P
hl'
Step 3: To prove the converse inequality, we use the notation introduced in Theorem 10.35. Let u be the representative of u given in Theorem 10.35. For all h > 0 and for GN-1-a.e. x E RN-1 for which f2x{ is nonempty, by the fundamental theorem of calculus we have 1
Iu (x;, xi + h) - u (xi, xi) I = I f
th)) dtl dt (u (x:, x1 + "U
_<
l hl fo, I axi (xi, xi + th) I dt.
Raising to the power p and integrating over 11h,i, by Holder's inequality we get
I7L (x;, xi + h) - u (xi, xi) I P dx
<
(j1
lhlP J1 h.i
I
a
+ th) I dt )P dx
(xi, xi
(x1, xi +
< l hI P Lh.Ii 1 I ate:
th) 1P dtdx P
lhI
<- FLIP
+ th) I dxdt
fo, Jna,jiOxi (x" xz P
I
L
I
axi (J)
I
dy,
where we have used Tonelli's theorem and the change of variables y _ (x;, xi + th). Hence, lim sup
In (x + hei) - u (x) l P
P
dx
<
rIu
(x)
IP)
hP (foh,i st axi which, together with the previous step, completes the proof of (10.31). Step 4: To prove the final statement of the theorem, let u E L" (Cl), 1 < p < oo, be such that (10.32) holds for every i = 1, ... , N. We claim that h-.O
10.5. A Characterization of W 1,P (11)
309
u E W1,P (S2). To see this, let S2' CC S2. Then reasoning as in Step 2, by (10.33) we get
sup
f (' lJ
I(x)
O<e
iA
IP
dx
< 00
for every i = I,-, N. Since u£ - u in LP (W) by Theorem C.19, it follows from Theorem 10.44 that u E W1,' (0') and that (10.34) holds for every i = 1, ... , N. Given the arbitrariness of 0', taking a sequence 123 CC SZ with S1,
92, we conclude that u E M ,111.f (S2). Since (10.34) holds for all 52j,
letting j -ti oo, we obtain (10.35), which shows that u E W 1,P (S2).
Exercise 10.56. Prove that for p = 1 the last part of the statement of the theorem is false. Hint: It is enough to construct an example for N = 1.
Exercise 10.57. State and prove the analog of the previous result for L1'P (S2).
Chapter 11
Sobolev Spaces: Embeddings Newton's second law of graduation: The age, a, of a doctoral process is directly proportional to the flexibility, f, given by the advisor and inversely proportional to the student's motivation, M.
-Jorge Cham, www.phdoomics.com
In this chapter we are interested in finding an exponent q such that (11.1)
IIUIIL4(RN) <_ c
IIVUIIL"(RNAN)
for all u in an appropriate subfamily of W1'p (RN) or L1' (RN). Assume for simplicity that u E Q1 (RN) and for r > 0 define the resealed function tar (x) := u (rx) ,
x E RN.
If (11.1) holds for ur, we get Q
(LN
Iu(rx)IQ dx J
= (.LN CC
Iur (x)IQ
dx) Q
U. Ivur (x) 1P dx
=C rP
f.."
Ivu(rx)Ipdx
,
d)
311
11. Sobolev Spaces: Embeddings
312
that is.
J
Q
N
lu(y)14 dy
dy) a
erl-p+Q
<
\J IVu(i,)Ip
.
If 1- L + 4 > 0, let r -* 0+ to conclude that u = 0, while if 1- N + N < 0, let r -i oo to conclude again that u - 0. Hence, the only possible case is when
N
N
q
p
- 1.
So in order for q to be positive, we need p < N, in which case, q=p*._
Np
N-p. The number p* is called Sobolev critical exponent.
11.1. Embeddings: 1 < p < N In this section we prove inequality (11.1).
Definition 11.1. Let E C RN be a Lebesgue measurable set and let u : E - R be a Lebesgue measurable function. The function u is said to vanish at infinity if for every t > 0, (11.2)
GN({xEE: Iu(x)I>t})
Note that the previous definition is automatically satisfied if E has finite measure, while if E has infinite measure and u does not satisfy (11.2), then it cannot belong to any LQ (E) for 1 < q < oo.
Theorem 11.2 (Sobolev-Gagliardo-Nirenberg's embedding theorem). Let
1 < p < N. Then there exists a constant C = C (N, p) > 0 such that for every function u E L1,n (RN) vanishing at infinity,
_ (11.3)
1
Ivu(x)IP
LM 1u (x)I* dx)
dx)
C (JM In particular, W 1,P (RN) is continuously embedded in LQ (RN) for all p < q < p*. The proof makes use of the following result.
Exercise 11.3. Prove that if u E LP (R) for some 1 < p < 00, then for every representative v of u, lim i of Iv (x)I = 0,
lim
Iv (x)I = 0
and prove that in general one cannot replace the limit inferiors with actual limits.
11.1. Embeddings: 1 < p < N
313
In what follows, we use the notation (E.2) in Appendix E.
Lemma 11.4. Let N _> 2 and let ui E Lt"-1 (RN-1), i = 1,.. . , N. Then the function u1 (XI) u2 (x2) ... uN (x'N),
u (x)
x E RN,
belongs to L1 (RN) and N IIUIIL1(RN)
< rj
IIuiIILN-1(RN-1)
i=1
Proof. The proof is by induction on N. If N = 2, then
u(x)
Ui (x2) u2 (x1) ,
x = (X i, x2) E R2.
Integrating both sides with respect to x and using Tonelli's theorem, we get I Iu (x)I dx = j 1U1 (X2)1 dx2
J
Iu2 (X01 dx1.
Assume next that the result is true for N and let us prove it for N + 1. Let
x E RN+1,
u (x) := u1 (x1) U2 (x2) ... uN+1 (xN+1),
where ui E LN (RN), i = 1, ... , N + 1. Fix xN+i E R. Integrating both sides of the previous identity with respect to x1i ... , xN and using Holder's inequality, we get (11.4)
JaN Iu (x) I dx1... dxN N-1 N
< II uN+1II LN(RN)
dx1... dxN
(LNlU(xaI
.
For every i = 1, ... , N we denote by x;' the (N - 1)-dimensional vector obtained by removing the last component from x; and with an abuse of notation we write x'j = (xa',xN+1) E ISBN-1 X R. Since XN+1 is fixed, by the induction hypothesis applied to the functions N
Vi (xi)
IN (xi, xN+1) I N-1
a
xi E
RN-1,
i = 1, ... , N, we obtain that N
N
JRNil lul (4)l i=1
NN_1
dx1 ... dXN < rj IIviIILN-1(RN-1) i=1
In
RN-1
N
IN (xi , xN+1) J 'V
n 7
11. Sobolev Spaces: Embeddings
314
and so from (11.4),
f Iu (x)I dx1 "dXN RN
N
< IIuN+1IILN(RN)
i=1
1
(JN1 I i (x,xN+1)I N dx.
Integrating both sides with respect to xN+1 and using the extended Holder inequality (see the Exercise B.80(i)), with 1
+...+
1
N
and Tonelli's theorem, we get N+1
f
V+I
lu (x)I dx1... dxN+1 <_ n IIuiIILN(RN) i=1
0
which concludes the proof.
We now turn to the proof of the Sobolev-Gagliardo-Nirenberg embedding theorem.
Proof of Theorem 11.2. Step 1: We prove (11.3) in the case p = 1 and under the additional hypothesis that it. E L1 (RN) n Cl (RN) with Vi E L1 (RN; RN). Fix i = 1,.. . , N. By Fubini's theorem for Chr-1-a.e. xa E RN-1 the function v (t) := u t), t E R, belongs to L (R)flC' (R) and yr E L' (R). Fix any such x E R^'-1. By Exercise 11.3, lim inf Iv (t) I = 0,
and so we may find a sequence tn -> -oo such that v On) - 0. Hence, for every t E R we have that t v(t) =v(t,,.)+ ft. v' (s) do. In turn, IV MI <- Iv (tn)I + fR Iv' (s) I ds
for all t E R and u E N. Letting n - oo, we conclude that for each xi E JR we have 1
Iu(xivxi)I
< f Ix' (xi,yi)l dyi.
11.1. Embeddings: 1 < p < N
315
Hence, we have shown that for all i = 1, ... , N and for GN-a.e. x E RN we have Iu (x) I
S
f I axi (xi, yi)
I
dFli.
Multiplying these N inequalities and raising to the power N1 1, we get N7 Iu (x)I Nf
l\
1 < 11(JR 18 i=1
N
1
=: II wi (xi)
(xi, Ii) dpi) I
'
i=1
//
for £N-a.e. x E RN. We now apply the previous lemma to the function N
w(x):=flwi(xi), xERN, i=1 to obtain that
fRN lu(x)I
N dx < JRN Iw (x)I dx < H IIwiIILN-1(RN-1) i
N(
J
=1
(x)I dx J
N
N
NI 1< \JRN I Vu(x) I
dx)
N1
where we have used Tonelli's theorem. This gives the desired inequality for
p=1.
Step 2: Assume next that 1 < p < N and that u E LP* (RN) fl C1 (RN) with Vu E LP (RN; RN). Define
v := IuI4,
q:=
p (N - 1
N-p
Note that since q > 1, we have that v E C' (RN). Applying Step 1 to the function v, we get N-1
N-1
N
(LN
N
N )/I IvI = fRN
:5f
N
dxj
Ivvl dx < q
1J
f
Iul4-1 IVul dx
RN 1
< q (fRN Iul(q-i)d
A
YP
(LN I VulP dx,
where in the last inequality we have used Holder's inequality.
11. Sobolev Spaces: Embeddings
316
Since (q - 1) p' = p", if u
0, we obtain
\\;j (f.N IUI da]
dx\\
f,tN
p
)
Iul
1
q
f NIvulpdx)P
which proves the result. Step 3: Next consider the case u r= LP* (RN) with distributional gradient Vu E LP (RN; RN) for 1 < p < N. Define uF := u*cp£, where V, is a standard mollifier. Then by Theorems C.19 and C.20 we have that u£ E LP* (RN) f1 C°° (R'V ), while, reasoning as in Lemma 10.16 and using Theorem C.19 once more, we obtain that V ue E LP (RN; RN). Hence, by the previous two steps }N
(IRN IueI "-Y [
NA < q
(1r
IVUElp dx
)
Letting s - 0+ and using Theorem C.19 and Lemma 10.16 one more time, we obtain the same inequality for u. Step 4: Finally we prove (11.3) in the general case. Assume that u E L1,' (RN) vanishes at infinity. For n E N and x E RN define
0
< lu(x)l < n, if Iu(x)I < ,s,
n-n
if JU(X)l >n1.
lu (x) I - n if
vn (x) :=
By the chain rule (see Exercise 10.37(1) and (vi)) for LN-a.e. T E RN, Ivv,z(x)I =
pvu(x)I
therwlsex)I < n,
and so Vvn E LP (RN; RN), while
I
Iv Idx = fi2i>} Ivldx { N fr
n-In
£N({xE RN: Iu(x)I>-}}
11.1. Embeddings: 1 < p < N
317
since u is vanishing at infinity. Hence, by the previous step applied to vn,
j Iu (z) I
n
`
f
< q
P
N N
1 J
f
Np
N-D
"
dx
f N [71,1 "Np R
(JRN
l
f
dx
(f1-L1n
IQulp dx
IVulp
(,fix) Np
1 n
1
IVvJ' dx) " = q
N-p
P.
R.N
First letting n -' oo and using Fatou's lemma, we obtain the desired result. Step 5: To prove the last part of the theorem, assume that u E W 11P (RN) .
Then by the previous steps we know that u E if (RN), and so we can now use Exercise B.80(ii) to conclude that u E E4 (RN) for all p < q < p* Indeed, assume that p < q < p"` and write 1
0
q
p
1-0 +
p*
for some 0 < 8 < 1. Then IIILIIL0(Rx) <-
(IIUIILP(RN))° (IIuIILP(RN) + IIUIILP`(RN)
where we have used Young's inequality (see (B.17)) with exponents
and
Hence, IIUIIL-(RN) < IIUIILP(RN) + IIUIILn*(RN) < 1IuI6(RN) +CIIVuIILP(RN;RN)
<- (1 + C) IIlII W1.P(RN) ,
which shows that the immersion (or embedding) W1" (RN) -> E4 (RN)
UIU is a continuous linear operator.
Remark 11.5. (i) If 1 < p < N, then the best constant in (11.3) is given by 1
1
p-1
Ira NP
N-p
1-p
r{1+ 2}r(N)
1r (p)rl1+N- P)
and equality holds in (11.3) if U (X) _
a+
11
»,
x E 18N,
blxlP-1/
where a, b are positive constants (see [13J, [1651).
11. Sobolev Spaces: Embeddings
318
If p = 1 < N, then the best constant in (11.3) is given by
}lN
f/
rr
2/ C= zN{r11+ ``sharp,
}
.
inequality)))
To see that this
it suffices to take if IxI < 1,
is 1
un(x):=
1+n-nlxl if1
if IxI > 1 + n, which is a sequence of Lipschitz functions that converges to the characteristic function of the unit ball. It is possible to show that 0
equality does not hold in (11.3) for functions in W1,1 (RN) but that
(11.3) still holds in BV (RN) (see Chapter 13), with the norm of Vu in L' replaced by the total variation of the vectorial Radon measure Du, and that equality holds for the characteristic function of the unit ball. (ii) In Steps 1 and 2 of the previous proof we could have used Theorem 10.35, and so avoid Step 3.
Exercise 11.6. Let k E N and 1 < p < oo be such that k > 2 and kp < N. Prove that (i) Wk+jp (RN) is continuously embedded in W""4 (RN) for all j E N
and forallp-5 q<
'
(ii) Wk,p (RN) is continuously embedded in LQ (RN) for all p < q <
NP
IV-kP.
Next we discuss the validity of the Sobolev-Gagliardo-Nirenberg embedding theorem for arbitrary domains. Exercise 11.7 (Rooms and passages). Let {hn} and {52n} be two sequences of positive numbers such that 00
I: hn=I
h+1 < 1,
0 < const. <
0 < b2n < h2n+1,
n
n=1
and for n E N let
n
Cn =
hi. i=1
Define Sl C R2 to be the union of all sets of the form Rj P1+1
(c, - hj,cj) x (-2h3, [cj, cj +' hj+i] x
2,
2h;m
J
,
(_i+i5i+i) 1
1
,
11.1. Embeddings: 1 < p < N
319
R1
R5 t
Figure 1. Rooms and passages.
for j = 1, 3,5.... (see Figure 1). (i) Prove that 8fl is the range of a rectifiable curve but that it is not of class C.
(ii) Let
hn =
1
31
62n= 3,
nl
n3
and for j = 1, 3, 5, ... define 3
log 2j
in R,,
Kj
K,,+(K,+2-K,)
in Pj+1.
hj+1
Prove that it E W1,2 (s)) but u 0 LQ (i)) for any q > 2. (iii) Let p > 1, q > 1 (2p - 1), 1
h2n-1 = h2n := -,
52n
3pn2q+p,
and for n E N define u (x1, x2) :=
1
1 in R2n-1
and
Vu (x1, x2) :=
(n + 1)Q - n4 1
0
in Pen.
nP
Prove that Vu E L2 (Q; R2'c2) but it 0 L2 (1). The previous exercise shows that Theorem 11.2 fails for general domains. The problem is the regularity of the boundary.
11. Sobolev Spaces: Embeddings
320
Definition 11.8. Given 1 < p < oo, an open set St C RN is called an extension domain for the Sobolev space W 1,P (S1) if there exists a continuous
linear operator E : W 1,P (Q) - W1,P (RN)
with the property that for all u E W1,P (a), C (u) (x) = u (x) for CN-a.e.
xE1.
Note that the extension operator 6 strongly depends on p. In the next chapter we will show that "nice" Lipschitz domains are extension domains for all 1 < p < oo.
Corollary 11.9. Let 1 < p < N and let Il C RN be an extension domain for W1,P (Sl). Then there exists a constant C = C (p, N, 12) > 0 such that IIUIILQ(n) <_ C II UII W1,P(n)
for all p
< C IIuII W 1.P(n)
for all u E W 1,P (S2). On the other hand, by Theorem 11.2, for all p < q < p* we have that IIuIIL4(n) = IIE (u)IILQ(n) _< IIe (U)IILQ(RN)
C1 IIe (u)IIW1.P(RN)
< C1C IIuIIw1.P(n) ,
where C1 = C1 (N, p) and we have used the fact that S (u) (x) = u (x) for CN-a.e. x E SZ.
Next we show that if p < q < p* and Il is an extension domain for W1,P (S2) with finite measure, then the embedding W1,P (cZ) -' Lq (12)
u'-* u is actually compact.
Theorem 11.10 (Rellich-Kondrachov). Let 1 < p < N and let S2 C RN be an extension domain for W" (Il) with finite measure. Let {un} C W1,P (SZ) be a bounded sequence. Then there exist a subsequence {un,, } of {un} and a function u E LP' (St) such that un,r - u in Lq (S2) for all 1 < q < p*.
The proof makes use of the following auxiliary results.
11.1. Embeddings: 1 < p < N
321
Lemma 11.11. Let 1 < p < oo and let u E W1.p (RN). Then for all
hERN\{0},
f
v
Iu (x + h) - ti (x)I" dx < IhIpf IVu (x) I'' dx. N
Proof. Assume that u E W4 (RN) n C- (RN). For x E RN and h E RN \ {0} by the fundamental theorem of calculus we have that Iu (x + h) - u (x) l =
I
d (u (x + th)) dtl dt fo 1
i
<- IhIJo IVu(x+th)I dt. Raising to the power p and integrating over RN, by Holder's inequality we get
JAN
I u (x + h) - u (x)Idx < <
dt) dx
fIN f'°ivux +th )Ip dtdx 1
IVu(x+th)I" dxdt
= IhIPfa
= IhIpJJN IVu(y)Ip dy, where we have used Tonelli's theorem and the change of variables y = x+th. To remove the additional hypothesis that u E C°O (RN), it suffices to apply the previous inequality to ue := cpe*u, where We is a standard mollifier, and let e --+ 0+ (see Theorem C.19 and Lemma 10.16). 0
Lemma 11.12. Let 1 < p < oo and let u E W1,p (RN) . For k E N consider standard mollilers of the form cQk (x) = kNcp (kx) ,
x E RN,
where W is defined in (C.8). Then
f I(u * (0k) (x) - u(x)Il dx < C(N 'P)
1 N
IVu(x)Ip dx.
Proof. By Holder's inequality and (C.5) we have (11.5)
I(u*'Pk)(x)-u(x)I'5 f
(x-y)Iu(y)-u(x)1pdy
< C (N) kN
JB(o,k)
Iu (x +h) - u (x) Ip A.
11. Sobolev Spaces: Embeddings
322
Hence,
JN (u. * Wk) (x) - u (x) I" dx Iu (x + h) - u (x) IP dxdh. < C (N) kN 1B (0,) J
(11.6)
N
In turn, by the previous lemma and Tonelli's theorem we getf
J I(u * Wk) (x) - u(x)IP dx < C (N) kN J IVu(x)I' dx J N
N
IhI'° dh
B(04)
C (N, p) / N I V (x) I" dx. O
We now turn to the proof of the Rellich-Kondrachov theorem.
Proof of Theorem 11.10. Since St is an extension domain for W1 (SZ), we can extend each un to a function un E W'"P (RN) is such a way that the sequence {un} remains bounded in W1,p (RN). It follows by Theorem 11.2 that the sequence {u'} is bounded in LP (RN). Since p* > 1, by the LP*
(RN) (see Theorem B.91 and Corollary A.60) we may find a subsequence {uk} of {un} such that reflexivity of
unk-u in LP*(W"). We claim that u-k -> u in LP (0). For simplicity, for every v E LP (RN) we Set v(k) := u * Wk. By the previous lemma and the fact that {ua} is bounded in W1.P (RN), we get slip J
nEN RN
dx < C (N, p) sup f
(nn)(k) -
dx <
nEN RN
p k
and so (11.7)
inn sup
f p
k -too nEN RN
I (un){k)
- u- lp dx = 0.
By Minkowski's inequality h un - UIILP(n) < kUn)(k) - U"IILP (
+
)
u(k)II j,r(n)
+ 11 U(*) - uII
jr(St)
Fix e > 0. By (11.7) and Theorem C.19 there exists k depending only on e such that for all k > k and all n E N the first and last terms in the previous
11.1. Embeddings: 1 < p < N
323
inequality are both bounded by c, and so (11.8)
II U. - U11 L-(0) < I(un)(k) - u(k) 11
ILP()
+ 2E
for all k > k and all n E N. Hence, to complete the proof, it suffices to show
that (11.9)
Since un
un)kukn II(
IILP(SZ)
u in LP* (RN), it follows that for all x E RN, (un)(k) (x) =
vi (x - y) un (y) dy (x - y) v (y) dy =
u(L*) (x)
as n -+ oo. Moreover, by (11.5) (with u replaced by un) and the fact that {u,} is bounded in 1'P (RN), we get (u)(')
(x)-u(k)(x)I'<ekN JJfB(0,1/k) f Iu.(x+h)-u(x+h)Ipdh < ckN
for all x E RN and all n E N. Since fl has finite measure, we are in a position to apply the Lebesgue dominated convergence theorem to conclude that (11.9) holds. Hence, we have shown that u in U (Sl). Since {un.,, } is bounded in LP (RN), by Vitali's convergence theorem (see Theorem B.101) this implies that u,,,` - u in LQ (f2) for all 1 < q
Remark 11.13. Note that when p > 1 the function u belongs to W1'P(fl). This follows from Theorem 10.44. On the other hand, when p = 1, we can only conclude that u E BV (Il) (see Theorem 13.35).
Exercise 11.14. Prove that if n is an extension domain for W" (a), 1 < p < N, with finite measure, then u, - u in W1 (S2) if and only if un u in f' (fZ) and Vu - Vu in LP (); ISSN). The following exercises show that compactness fails for q = ps even for nice domains and that for general domains even the embedding W" (SZ) U
may fail to be compact.
LP (S2)
U
324
11. Sobolev Spaces: Embeddings
Exercise 11.15. Let n = B (0,1) C RN, let 1 < p < N, and consider the sequence of functions
fI --+ R defined by
un (x) _
nN"
A
(1 - n Ixl) if IxI < n,
if IxI > 1. Prove that {u n} is bounded in W 1,P (fl) but that it does not admit any 0
subsequence strongly convergent in LP' (fl).
Exercise 11.16. In Exercise 11.7 take (hem)a
62.
,
for some a > 3, and for j = 1, 3, 5, ..., consider functions uj : 0 --+ R such
that
1hj
uj (X1, x2) =
0
in Rj,
in 0\(Pj_1URjUPj+1)
and 1
Vuj (X1, r2) _
in P?_,,
(hll_l,0) 1 (-hjhj+l,0
J
in Pj+1
Prove that the sequence {uj} is bounded in W1,2 (B (0, 1)) but that it does not admit any subsequence strongly convergent in L2 (n). The previous exercise shows that the embedding u E W1,P (fI) - LP (n) fails to be compact for arbitrary open bounded sets. The following exercise gives the only possible type of compactness that remains.
Exercise 11.17. Prove that if f1 C RN is an open set with finite measure and 1 < p < oo, then the embedding W 1" (Sl) --> L4 (tI)
uHu is compact for all 1 < q < p.
Exercise 11.18. Let fl C RN be an open set and let 1 < p < oo. Prove that the embedding W "P (l1) - LP (SI)
UI-'u is compact if and only if sup
fn\nn Jul" dx :
<-1 = 0,
where SI := {x E 0 : disc (x.,81) > n and IxI < n}.
11.1. Embeddings: 1 < p < N
325
The Rellich-Kondrachov theorem and its variations hold for special domains with finite measure. The next two exercises show that if we restrict our attention to the class of radial functions, then we have compactness in the entire space RN for all N < q < p", but not for q = N or q = p".
Exercise 11.10 (Radial functions, I). Let N > 2, let f E C1(10, oo)), and let
u(x)
(11.10)
f (IzI), -TERN . (i) Let 1 < p < oo. Find necessary and sufficient conditions on f for u to be in W' ,P (RN).
(ii) Let a > 0. Prove that for r > 0, (r2a f2 (r)),
< [(raf
(r))12
+ (1,f
(r))2
=r2a [(fi (r))2 + f2 (r)] + a (r2a-1 f2 (r))' - a (a - 1)
r2a-2 f2 (r).
(iii) Prove that for every N E N with N > 3 and for all r > N - 1, rN-1 f2 (r) < 2
J0
r tN-1 [(f' (t))2 + (f (t))2] dt.
(iv) Prove that for all r > 1, 00
r f 2 (r) < 2
Jr
t [(f' (t))2 + (f (t))2] dt.
(v) Prove that if the function u defined in (11.10) belongs to W1,2 (RN),
N > 2, then C
It(x)I: iN) I
IIUIIw1.2(RN)
a
forallxERN,with IxI>N-1. C W 1,2 (RN) n C' (RN)
Exercise 11.20 (Radial functions, II). Let be a sequence of radial functions with sup IItIIw'.2(itN) < 00n
(i) Prove that 21
urn sup
R-oo a
dx = 0.
RN\B(o,R)
(ii) Prove that there exist a subsequence {u,yk} of and a function u E W1.2 (RN) such that u in Lq (RN) for all 2 < q < N-2' 2N
11. Sobolev Spaces: Embeddings
326
(iii) Let V E C' (R) be such that supp cp C 10, 11, cp # 0, and define u,3 (x) = an.cp (IxI - n),
x E RN.
Find an in such a way that {u.} is bounded in W1"2 (RN) but does not converge in L2 (RN). (iv) Let cp be as in part (iii) and define
um (x)=a W(2'LIxI-1),
xERN.
Find an in such a way that {un} is bounded in W1.2 (RN) but does not converge in L2* (RN).
We conclude this section by showing that the Rellich-Kondrachov theorem continues to hold for bounded domains with boundary of class C.
Theorem 11.21. Let 1 < p < oo and let n C RN be a bounded domain whose boundary is of class C. Then the embedding W 1,P (0) -- LP (1)
is compact.
Proof. For every xo E M there exist a neighborhood Ax0 of xo, local coordinates y = (y', yN) E RN-1 x R, with y = 0 at x = xo, and a function f E C (QN_1 (0, r)), r > 0, such that n fl Ayo = { (y', YN) E f n A.. : y' E QN-1 (0, r) , 1N > f (y') )
By taking r > 0 smaller, if necessary, and t > 0 sufficiently small, we have that the open set U (xo, r, t) defined in (10.10) is contained in A.,a, so that, by eventually replacing A,,, with U (xo, r, t), without loss of generality, we may assume that
(11.11) nn Axa
= { (Y1', Y!N) : 1/ E QN-1 (0, r) , f (Y1') < YIN
Assume that u E CC ° (RN) . Fix y' E QN_ 1 (0, r) and let
f (y) < Y!N < f (y') + t,
f (y') + 2 <'r < f (y') + t.
By the fundamental theorem of calculus we have that
/
u (y', Y!N) = u (y', r)
T
JyN y!
(y', s) ds,
11.1. Embeddings: 1 < p < N
327
and so by the convexity of g (t) = ItIP, t E R, and Holder's inequality, P
f&)+t
Iu(y,yN)IP<2p-1Iu(y',r)IP+2P-1
dS
ayN(y,'s)I
LN
I
f(y')+t 8u
2P-1 j u (y', r) 1P + 2P-1tP
P
OYN
YN
(y', s) I ds.
By averaging in r over (f (y') + 2, f (y') + t), we obtain J U (y', yN) I P <
t
+
f(y')+1
f
2P-1 t
Iu (y', r) IP dr f (y ')+t
JyN
au JN
I
P
(Y" s)
I
ds.
Now we integrate in (y, yN) over QN-1 (0, r) x (f (y') , f (y') + b), where
0<6
j
f (y')+s lu
I(y')
(y,YN) 1P dyNdy
f(y' )+ t
2P5
Iu(y',r)IP drdy'
t jQN-1(0,r) Jf(y')+2
+ 21t
'QN -1(Q,r)
f
f(y ')+t
P I
N
ayN (y', s)
I
dsdy'.
Since an c RN is compact, we can find xii ... , xM E an such that M
an C U U (xi, ri, ti) i=1
and (11.11) holds for each
i = 1, ... , M. For every e > 0 let
SZf:={xE1Z: dist(x,81l)>e}. By taking 0 < 6 < min1
n n ( ri, 5), \ E, C U VrT (xi, i=1
where U (xi, ri, 6) is defined in (10.10). Hence, by (11.12) we get that
r
cd ju (x) IP dx <
'M' f Iu (x) IP dx + 2P-1Mt b
J
where t_ := mine
you (x) IP dx,
11. Sobolev Spaces: Embeddings
328
Since the restriction to Sl of functions in Cr (RN) is dense in W",P (fl) by Theorem 10.29, the previous inequality continues to hold for all u E W" P (St). In particular, sup Ilullt,1l,p(n)«
+ 2p-' Mt 6 -> 0 l(x)l' dx < 2PM45 tl\f,a
f
as S -> 0, and so the result follows from Exercise 11.18.
0
11.2. Embeddings: p = N The heuristic argument at the beginning of the chapter shows that when p > N, we cannot expect an inequality of the form IIuIILs(RN) <_ C IIVtIILP(RN;RN).
However, we could still have embeddings of the type W 1,P (RN) -> Lq (RN)
that is, inequalities of the type IIUIILQ(IlCN) 5
CIIUIIwl.p(gN).
We now show that this is the case when p = N. We begin by observing
that when p / N, then p* / oo, and so one would be tempted to say W1,N (RN), then u E LOO (RN). For N = 1 this is true since if that if u E u E W1,1 (R), then a representative u is absolutely continuous in R so that
z
-C(x) = U (O) +
jii(s) ds,
and since u = u' E Ll (R), we have that I is bounded and continuous. For N > 1 this is not the case, as the next exercise shows.
Exercise 11.22. Let n = B (0,1) C RN, N > 1. Prove that the function U (X) := log (log (1 + ICI )} '
x E B (O,1) \ {0} ,
belongs to W1,N (B (0, 1)) but not to L00 (B (0, 1)). However, we have the following result.
Theorem 11.23. The space W1,N (RN) is continuously embedded in the space Lq (RN) for all N < q < oo.
11.2. Embeddings: p = N
329
Proof. Let u E W1,N (RN). Define v := lult, where t > 1 will be determined so that v E W1,1 (RN) By Theorem 11.2 with p = 1 and Exercise 10.37(i), N-1
(
fRN
July dx
N-1
_1 !
N
N
r dx
IvI'
<
J lovl dxt N
< t (j
lult-1IVul dx
j
RN
(t-1)N' dxl
N` l ul
\JRN I V UI N
dx)
'
where in the last inequality we have used Holder's inequality. Hence, N-1 NE
!
<
C( U.. lul' dx} < C Us (11.13)
` N-1
lul(t_1)
dx)
`f
fR
N-1
lul(t-1)N1
dx) N t=1 /
1
tN
IVuIN dx ]
RN
N
/// 1
+ (JN R IVUIN
dx) N J
where we have used Young's inequality (see (B.17)) with exponents t and tt1' Taking t = N yields (LN lul N-1 dx)
=IV
< C [(J N IuIN dx) + (f N louIN dx)
so that u E LT (RN) with continuous embedding. Reasoning as in Step 5 of the proof of Theorem 11.2, we conclude that IIUIIM(RIV)
for allN
.
Taking t = N + 1 < N(N+1)
N-1
C IIuIIw1.N(RN)
IV2
in (11.13) and using what we just proved gives
N-1
dx N N+1)
(JRNlul hr4
1
=Nr
A IVt`IN
+ (fRN
IIuIIw1,N(RN),
and so the embedding W1'a (RN) -4 L4 (RN) u I- u
dx)
11. Sobolev Spaces: Embeddings
330
is continuous for all N < q < N Nil . We proceed in this fashion taking 0 t = N + 2, N + 3, etc. Exercise 11.24. Let k E N and 1 < p < oo be such that k > 2 and kp = N. Prove that (i)
Wk+,j,P (RN) is continuously embedded in W3"9 (RN) for all j E N
and for all p
Exercise 11.25. Prove that for every function u E WN,1 (RN), aNu IIUIILOQ(RN) 5
0X1 ...OXN IILN(RN)
Exercise 11.26. Let Q C RN be an extension domain for W1,N (S2)
(i) Prove that there is a constant C = C (N, n) > 0 such that IIUIILq(n) !5 C IIUIIW1.n(n)
for allN
Note that in general, if u E W1,N (RN), we cannot conclude that u E LQ (RN) for 1 < q < N. However, since u E Lq (RN) for all q > N, the function u must decays faster than algebraically at infinity. Indeed, we will show that it must have exponential decay. For every m E N consider the function m-1
00
(11.14)
expm (s) :=
c 1n.s" = exp (s) - E 1n.sn,
n=m
1
s E R,
1
n=0
and let (11.15)
7N
NQN
,
where, we recall, QN is the surface area of the unit sphere, that is, ON
HN-1 (SN-1) =
N
2w 2
r(2)
11.2. Embeddings: p = N
331
The next theorem, which is due to Adachi and Tanaka [4], gives an embedding of the space W1,N (RN) into the Orlicz space generated by the function expN_1 (sN/(N-1)) We introduce the notion of Orkcz space.
Definition 11.27. Let E C RN be a Lebesgue measurable set and let [0, oo) -i [0, oo] be a convex, lower semicontinuous function such that 4) (0) = 0 and 4) is not identically zero or infinity. The Orlicz space L" (E) generated by the Orlicz function 4) is the space of all Lebesgue measurable 4)
:
functions u : E - R such that u (x)I l dx < oo \ s /l `I
1
for some s > 0 (depending on u).
Exercise 11.28. Let E and 4) be as in the previous definition. (i) Prove that Lt (E) is(ra normed space with the norm
11u04,:=inf{s>0:
(11.16)
f
dx<111
(Iu(x)P)
E
JJ
(ii) Prove that L` (E) is a Banach space. The following theorem is the main theorem of this section. To the author's knowledge the first result of this kind in bounded domains is due to Trudinger [168] (see also [130] and see [107] for some recent results).
Theorem 11.29. Suppose N > 2. Then for every ry E (0, -tN) there exists a constant C = C (N, -y) > 0 such that N
expN_1
Iu
(X)IN' x N)I
yIIVUIIN' (
dx xN )
<
Cy
N
for all u E W1,N (RN) \ {0}. In particular, if 4 (s) := expN_1 s > 0, then (11.18)
IIotLIILN(RN;RNxN)
(ir'),
IItII4-, < II 'tIILN(RN;RNxN)
for all u E W 1,N (RN)
Remark 11.30. Note that the inequality (11.17) is scale invariant; that is, if for r > 0 we define the rescaled function ur (x) := u (rx) ,
x E RN,
then
IIurIILN(RN)
rN IIuIILN(RN)'
IIVtrIILN(RN;RNxN) =
IIVUIILN(RN;RNxN)
11. Sobolev Spaces: Embeddings
332
and
R
'?
eXpN-1
IIV
Iu r (x)IN,
dx
rI) j,N (RN.RNxN) N'
=
expN-1
TN
N,
?
RN
dx.
IIVuIILN(RN;RNXN)
To prove Theorem 11.29, we use the notion of symmetric rearrangement (see Chapter 16).
Proof of Theorem 11.29. Fix u E W 1,N (RN) \ {0} and define v (x) :=
lu (x)I
xERN. IIVuIILN(RN;RNxN)'
Then
IIVVIILN(RN;RNxN) = 1, and (11.17) reduces to
(11.19)
jN
eXpN-1
(-,v'" (x)) dx <
C.1 IIvII N(RN) .
Let v* be the spherically symmetric rearrangement of v. Then v* (x) _ w (IxI), where w is nonnegative, decreasing, and locally absolutely continuous by Theorem 16.17. Hence, Vv* (x) = w (IxI) IxI
for GN-a.e. x E RN. Using spherical coordinates and Theorem 16.17, it follows that 00
(11.20)
Iw'(r)INrN-idr
/3N 0
= Ilov*II2 (R"' tNxN) N(RN;RNxN) = 1.
<_ IIvVII
Define
ro:=inf{r>0: w (r) <_ 1}.
(11.21)
Since w (r) --+ 0 as r -# coo, we have that ro must be finite. Using spherical coordinates and Theorem 16.10, we have that
JN (11.22)
(v" (x)) dx = JRN eXpN-1
expN-1
(.Y
(v*)N' (x)) dx
= ON f eXpN-1 (1N' (r)) 9.N-1 dr a ,
To
expN-1`(ryw
= JON
'
(r)) rN-1 dr
o
+ I6N
eXpN-1
(yw" (r)) rN-1 dr =: Z + 11. J
11.2. Embeddings: p = N
333
To estimate Z, it is enough to consider the case that ro > 0, so that w (ro) = 1 by (11.21). Since w is locally absolutely continuous, by the fundamental theorem of calculus (see Theorem 3.30), Holder's inequality, (11.20), and (11.21), for 0 < r < ro we have that Iro
w(r) dT <1 +
w (r) = w (ro) <1+
f
TN, dr
ro
r
f
Iw, (T)IN TN-l
dr N (log r)
r
o
1 +'3 N
N (log).
By the convexity of the function sN' for every e > 0 we may find a constant C£ = C£ (N) > 0 such that
(1+s)N' < (1+e)sN'+C£ for all s > 0. Hence,
r
wN' (r)
(11.23)
1 + e) QN N- 1 log o + C£
for all 0 < r < ro. Since -y < yN, we may take e so small that
y(1+e)
NN-' < N. Hence, also by (11.23), Pro
Z < QNeycr J
' log ro rN-1 dr exp y (1 + e) QNR 1
r
o
(11.24)
QNeroNjro rN-1-y(1+3N
=
&0c`
' dr
ra =: C1 (N, y) r0
N-y(l+e)QN"--1 On the other hand, by the Lebesgue monotone convergence theorem and the fact that w (r) < 1 for all r > ro by (11.21), we have that or,
ZZ=QN n=N-1 W!
" Jr.
00
< 13N
n=N-1
yL.
v'
= ON (expN-1 y)
J°° o
f
oo
o
wN (r)
rN-1
dr
wN (r) rN-1 dr.
11. Sobolev Spaces: Embeddings
334
Combining this estimate with (11.22) and (11.24), we get (11.25)
JRN eXpN-1 (7vN' (x)) dx
0 + (expN-17) ON
Jrp
wN (r)
rN-1
dr.
Using spherical coordinates and Theorem 16.10, we have that 00
ON J
wN (r)
rN-1
dr < J (v*)N (x) dx = RN
ro
JRN vN (x) dx.
Thus, to obtain (11.19), it remains to estimate ro in the case that ro > 0. By (11.21) and the fact that w is decreasing, we have that w (r) > w (ro) = 1 if and only if 0 < r < ro. Hence,
{xERN: v'(x)> 1} ={xERN: w(Ixl)> 1}=B(0,ro). By Proposition 16.6, weI have that i1 aNrp = GN ({x E RN : v* (x) > 1})
=RCN({xER": v(x)>1}) <J
vN(x)dx,
{v>1}
and so we have proved (11.19) and, in turn, (11.17). By (11.17), the number s = IIVUIILN(RN;RNxN) is admissible in the defi0 nition of IIufl*, (see (11.16)), and so (11.18) follows.
The previous theorem is complemented by the following exercise, which shows that (11.17) fails for all 7 > 7N.
Exercise 11.31. Suppose N > 2 and let r > 0. (i) Construct a sequence C W1,N (RN) of nonnegative radial functions such that supp u, C B (0, r), IlovnllLN(RN;RNXN) = 1, and
fRN expN-1 7N< (x) dx
-00.
IIvnIILN(RN)
(ii) Prove that for every open set 1 C RN and for every C > 0, the inequality
eXpN-1 7N
Iu (x) I
N'
IIVt IILN(Q;RN)
IIUIILN(n)
dx < C (IIIILN(fl;RN))
fails for some u E Wo'N (St) \ {0}.
N
11.3. Embeddings: p > N
335
Remark 11.32. The analog of this result for bounded domain can be found in Exercise 12.20. See also Remark 12.21 for more references on this topics.
11.3. Embeddings: p > N We recall that, given an open set SZ C RN, a function u : SZ -* R is Holder continuous with exponent a > 0 if there exists a constant C > 0 such that
lu(r)-u(y)I
Exercise 11.33. Let 1 C RN be an open set and let a > 0. (i) Prove that if a > 1 and if fZ is connected, then any function that is Holder continuous with exponent a is constant.
(ii) Prove that the space C°'a R, 0 < a < 1, is a Banach space with the norm IIUIICO.a(
}
sup Iu (x) I + su z, ysup#7J
lu ( I
-
yl(01
Note that if SZ is bounded, then every function u : SZ -+ R that is Holder
continuous with exponent a > 0 is uniformly continuous and thus it can be uniquely extended to a bounded continuous function on RN. Thus, in the definition of CO,' (SZ) one can drop the requirement that the functions are bounded. The next theorem shows that if p > N, a function u E Wln (RN) has a representative in the space C0,1 P (RN).
Theorem 11.34 (Morrey). Let N < p < oo. Then the space W1,p (RN) is continuously embedded in C°"- a (RN). Moreover, if u E 1v'.P (RN) and u is its representative in C°'1 p (RN), then lim u (x) = 0. jxI-'oo
Proof. Let u E W4 (RN) n C°° (RN) and let Qr be any cube with sides of length r parallel to the axes. Fix x, y E Q,, and let
g(t):=u(tx+(1-t)y), 0
-u(y) = g(1) -g(0) =
g'(t) dt fo
= J Vu(tx+(1-t)y).(x-y) dt. 0
11. Sobolev Spaces: Embeddings
336
Averaging in the x variable over Q, yields
uQr - u (y) =
Vu (tx + (1 - t) y) (x - y) dtdx,
rN JQr J
where uQ,. is the integral average of u over Q, that is, uQr :=
1
N r
u (x) dx. Qr
Hence,
J J lla (tx+(1-t)y)IIx1-ysi dtdx
IuQr-u(y)I
Qr 0
s=1
N
-
jjj
N
r1
fo
r
ax (tx + (1 - t) y)I dx dt :
taxti N
(1-t)y+Q,
(z)I dzdt,
where we have used the fact that I x; - y, I < r in Qr, Tonelli's theorem, and the change of variables z = tx + (1 - t) y (so that dz = tNdx). By Holder's inequality and the fact that (1 - t) y + Qt C Qr, we now have N
IuQr - u (y)l
(11.26)
I
rN-1 Jp
N
(tNa
(1(1 -) t + Qr= Oxi rrN-1 N- P tN- n < N IIVUIILP(Qr;RN) tN dt Jo
= Np rl- p _N
(z)p dz
p
dt
I
IIouIILp (QriR) N.
Since this is true for all y E Qr, if x, y E Qr, then lu (x) - u (y)I <- Iu (x) - uQrl + IU (y)
<
p-2NpN rl
tQrl
N p IIouIILp(Qr;RN)
Now if x, y E RN, consider a cube Qr containing x and y and of side length r := 2 Ix - yl. Then the previous inequality yields Np
(11.27)
lu (x) - u (y)I <- C (N, p) Ix - yl' < C (N, p) Is
- yl1
IloullLp(Qr;RN) p II7UIILp(RN;RN)
Hence, u is Holder continuous of exponent 1 - N . To prove that u E N C0.1- (IEBN), it remains to show that u is bounded. Let x E RN and
11.3. Embeddings: p > N
337
consider a cube Q1 containing x and of side length one. By (11.26) we get (11.28)
lu (x)I s ItQI I + Iu (x) - tQ' 1:5 L1' (x) dx I + C (N, p) L
S IItIIL,(Q,) + C (N,p) IIVuIIl (RN;RN) s C (N,p) IIuIIW,.p(RN) ,
where we have used Holder's inequality. Next we remove the extra hypothesis that u E C' (RN). Given any u r= W 1'P (R'V ), let x, y E RN be two Lebesgue points of u and let u£ := u * rp,, where V£ is a standard mollifier. By (11.27) we have that
Iu£(x)-uE(y)I
Iu (x) - u (y)I <_ C (N, p) Ix - yl' P IIVuIILp(RN;RN)
for all Lebesgue points x, y E RN of u. This implies that
u : {Lebesgue points of u} - R can be uniquely extended to RN as a Holder continuous function u of exponent 1 - E in such a way that (11.29) holds for all x,y r= RN. With a similar argument from (11.28) we conclude that Iu (x)I < C (N, p)
(11.30)
IIuIIw1.p(RN)
for all x E RN. Hence, IIull
CO
,1_ p
ly+
)
sup Ila (x)I +
xERN
sup x,YERN,xOyr
N I (x) - ft (Y) N
Ix - VIl p
<- C (N, p) Ilu1I W1.p(RN)
Finally, we prove that ii (z) - 0 as IxI -> oo. Let {un} C C°° (RN) be any sequence that converges to u in W1,P (RN). The inequality (11.30) implies, in particular, that u E L°O (RN), with IIUIIL-(RN) <_ C (N,p) Ilullwl.p(RN) .
Replacing u with u - u,, gives IIu - UPIIL-(RN) < C (N,p) IIu - u.Ik v .p(RN) ,
and so IIu - unIIL°°(RN) -, 0 as n -+ oo. Fix e > 0 and find n E N such that IIu - unIIL-(RN) << E
11. Sobolev Spaces: Embeddings
338
for all n > n. Since tin E C°° (RN), there exists Rn > 0 such that u,-, (x) = 0 for all IxI > Ra. Hence, for LN-a.e. X E RN with IxI > Ra we get
lu(x)I = I" (x) - ua(x)I < Ilu-u,LIIL'(RN) <_ E, and, since u is continuous, we get that the previous inequality actually holds for all x e RN with IxI > Rn.
Remark 11.35. The first part of the proof actually shows that if u E L1"P (RN) for some N < p < oo, then a representative u of u is Holder continuous with exponent 1 -
N
and
Iu(x) - u (y)I <- C (N,p) Ix
- Yl'
P Ilo'IIL"(RN;RN)
for all x, y E RN. As a consequence of the previous theorem we obtain the following result.
Corollary 11.36. If u E W IM (RN), N < p < oo, and u is its representative in CO.1 P (RN), then u is differentiable at LN-a.e. x r= RN and the weak partial derivatives of u coincide with the (classical) partial derivatives of u LN-a.e. in RN.
Proof. Let xo E RN be a p-Lebesgue point for Vu (see Corollary B.123), that is, rl im
-4 fQ(xo,r) I VU (x) - Vu (xo) lP dx = 0,
where as usual Q (xo, r) := xo + (-,) N, and define v(x) := i (x) - u (xo) - Vu (xo) (x - xo) , x E RN. Reasoning as in the first part of the proof of Theorem 11.34, if x E RN and r := 2 Ix - xpl, we obtain that Iv (x) - v (xo)I <_ C (N,p) rl IIVVIILP(Q(xo,T)N) or, equivalently, Iu(x)-fl (xo) -Vu(xo) (x - xo)I Ix - xol
C (N,p) N (j1q(z,r) I Vu (v) - Vu (xo)IP dy
r7
P
)'L i
P
=C(N,p) rN f (zo,r) IVu(y)-Vu(xo)I'dy as x
xo. This completes the proof.
-0 0
11.3. Embeddings: p > N
339
Exercise 11.37. Let fZ C RN be an open set and let p > N. Prove that if IQ is an extension domain for W 1'' (0), then the space W l,-' (f') can be N
continuously embedded in C0,1- P (S2) and that if u E W i.p (1) and u is its N representative in CO,1- v (), then ii is differentiable at GN-a.e. X E fl and, if Sl is unbounded, u (x) = 0. lim 00
ZEU, IsH
Exercise 11.38. Let St C RN be an open set and let p > N. Prove that if 11 is a bounded extension domain for W 1p (Ii), then the RellichKondrachov theorem (Theorem 11.10) continues to hold for p > N, that is, for all 0 < a < 1 - n the embedding
W' (0) -' Co,a P) is compact.
Exercise 11.39. Let Ii C RN be an open set and let p > N. Prove that if Il is an extension domain for W 1,p (Ii) with finite measure, then the embedding
WI.P (a) -+ II' (ci) is compact.
The following exercise shows that if u E W1,N (f2), where i C RN is an open set, then one cannot have differentiability C,v-a.e. in i (if u E W 1P (S2), with p > N, then this is true by Corollary 11.36).
Exercise 11.40 (Indian mystic's bed of nails). Let Q :_ (0,1)2 be the unit cube in R2, subdivide it into 4" subcubes each having side length 2-k, and {x(1),... , x(4k) } be th e set of centers of the subcubes. Define let Ck e-2k
gk
(t)
in
t+ 1+e-2k ,
e-2 k I
n
+
e-2k
if
Ix
- t<1
0<<
,
and
fk (x) _ {
gk Ek
0
-
('n)
otherwise,
ek for some x(n) E Ck ,
11. Sobolev Spaces: Embeddings
340
where 0 < ek < 4-k are chosen sufficiently small. Define 00
u(x):=F 2 fk(x), xEQ. k=1
(i) Prove that u E W1,2 (Q) n C (Q). (ii) Let X E Q be such that either fk (x) = 0 for all k or fk (x) = 0 for all k sufficiently large. Prove that u is not differentiable at x. (iii) Prove that if Ek is chosen sufficiently small, then u is not differentiable on a set of positive measure. The following result is another important consequence of Theorem 11.34.
Corollary 11.41. Let IF E W1,P (RN; RN), N < p < oo, and let iY be its representative in C°'1- p (RN; RN) Then there exists a constant C = C (N, p) > 0 such that for every Lebesgue measurable set E C RN the set T (E) is Lebesgue measurable and GN
(E)) < C (GN (E))- A
(E;RNxN).
In particular, i has the (N) property; that is, it maps sets of LN-measure zero into sets of £N-measure zero.
Proof. In the proof of Theorem 11.34, we have seen that for every cube Q with sides parallel to the axes and for every x, y E Q, NP J
(x) - `I`4 (y) 1 < C (N,p) (diamQ)1
IIV'&=IILP(Q,;RN)
foralli=l,...,N. Hence, (11.31)
CN (1 (Q)) < 2N (diam 1 < C (N, p)
(Q))N
(diamQ)(1- p)N
IIoTIIL
(Q;RNxN)
Next, if E is an arbitrary Lebesgue measurable set, for every e > 0 find a countable family {Q,,} of pairwise disjoint cubes such that
ECUQ,, =:E0 n and
GN (Qn) <_ GN (E) + E. n
11.4. Lipschitz Fbnctions
341
Then by (11.31) and Holder's inequality (which one?) we have
GQ (T (E)) c > GN (p (Q.)) n
C C(N,p)
(diamQ,) 1
n N IIQWIILP(Qn;kNxN)
n N
<_ C (N, p)
(diann)N
P
n
< C (N, p)
(N(Qn)
N
}
A (IIvII9P(Q,,;RNxN)
n
linT V
IL
(Eo;W
')
Ti
< C (N, p) (L'V (E) + s)1
Letting e -> 0+ yields ( (E))
G;
P
I I"T II t (Eo;IRN x N )
< C (GN (E))1- P IIo'PIIL
.
(E;RNxN)
This inequality implies, in particular, that has the (N) property. It remains to show that f (E) is Lebesgue measurable. This follows as in the proof of Theorem 8.10. Assume first that E has finite measure and
write E = E U Eo, where E is an FF set and Eo is a set of Lebesgue measure zero. Then T (E) = T (E.,) U'P (Eo). By the (N) property the set I' (Eo) has Lebesgue measure zero, and so it is Lebesgue measurable by the completeness of the Lebesgue measure, while by the continuity of T the set T is Lebesgue measurable. If E is not bounded, we may write it as the countable union of bounded Lebesgue measurable sets, and so by what we just proved, the set ' (E) is Lebesgue measurable, since it may be written as a countable union of Lebesgue measurable sets. Remark 11.42. The previous corollary still holds if 'P E L1,P (P.5; PRN) for
some N
Exercise 11.43. Let 11 C RN be an open set and let 11 : fL -- RN be oneto-one on a set whose complement in S"t has Lebesgue measure zero. Prove that if 'P E W "P (f'; RN), N < p < oo, then it has a representative lY for which all the hypotheses of Theorem 8.21 are satisfied.
11.4. Lipschitz Functions In this section we discuss some properties of Lipschitz functions and their relation with W1T0O. The next exercise shows that a Lipschitz function can
11. Sobolev Spaces: Embedding's
342
always be extended to RN. We recall that if E C RN and u : E -+ R, then u is Lipschitz continuous if UI
Lip (u; E) =
sup
-
lu
(y)l < 00.
IX
We usually write Lip u in place of Lip (u; E), whenever the underlying set is clear.
Exercise 11.44. Let E C RN and let u : E - R be a Lipschitz function. Define
v(x) := inf {u(y) + Lip (u; E) Ix - yl : y E E} ,
x E RN.
(i) Prove that v is Lipschitz with Lipschitz constant Lip (u; E). (ii) Prove that v (x) = u (x) for all x E E. (iii) Prove that there exists a function w : 1[8N --+ R with Lipschitz constant at most Lip (u; E) such that w (x) = u (x) for all x E E and
inf w=in u, supw=sup u. ]RN B
]RN
The next exercise shows that there exist functions that are in W1'00 (fl) but are not Lipschitz continuous.
Exercise 11.45. Let SZ C JR2 be the open set defined in polar coordinates by -7r < B < it and r > 1. Prove that the function u = 8 belongs to W 1°O0 (0) but that it is not Lipschitz continuous. The relation between Lipschitz functions and functions in W 1>°O is discussed in the next exercise.
Exercise 11.46 (Lipschitz functions and W1"0°). (i) Given a function u r=
Li 0 (RN) prove that u has a representative is that is Lipschitz continuous if and only if its distributional gradient Vu belongs to L0° (RN; RN) .
(ii) Prove that a function u E L10 (RN) has a representative ft that is bounded and Lipschitz continuous if and only if u E W1"1 (RN). (iii) Let fl c RN be an open set and let u : fl -> R be bounded and Lipschitz continuous. Prove that u E W"O° (11). (iv) Let SZ C RN be an extension domain for W1"°° (Sl) and let u E W1,00 (n). Prove that u has a representative u that is bounded and Lipschitz continuous. In view of the previous exercise and Corollary 11.36 we have the following
important result.
11.4. Lipschitz Functions
343
Exercise 11.47 (Rademacher). Let 11 C RN be an open set. Prove that every Lipschitz function u : f2 -> R is differentiable at GN-a.e. x E Q.
Next we present a proof of Rademacher's theorem that does not make use of Corollary 11.36 and of the theory of Sobolev spaces. We begin with an auxiliary result.
Exercise 11.48. Let u : RN - R be a Lipschitz continuous function. Let D C SN-1 be dense in SN-1 and let xo E RN be such that there exist all partial derivatives 5OFu (xo), i = 1, ... , N, all directional derivatives 6u (xo) for v E D, and N
a (xo) Vi
(xo) _ {=1
for all v = (v1, ... , vN) E D. Prove that u is differentiable at xo. Compare the previous exercise with Exercise 8.1.
Theorem 11.49 (Rademacher). Let u : RN R be a Lipschitz continuous function. Then u is differentiable at LN-a.e. x E RN. Proof. If N = 1, then u is absolutely continuous and the result follows from Lebesgue's theorem (Theorem 1.21). Thus, assume that N > 2. For every v E SN-1 let
E
x E RN : there exists - (x)
and let H,, be the hyperplane orthogonal to v. For every xo E H,,, the function
g(t) := u (xo + tv) ,
t E R,
is Lipschitz continuous and thus, by the result for N = 1, g is differentiable for Gl-a.e. t E R. Hence, VO;u (xo + tv) exists for Gl-a.e. t E R. This shows
that xo + tv E E for Gl-a.e. t E R. Since the set E is a Borel set (why?), it follows by Tonelli's theorem that
Next consider a countable set D C SN-1 dense in SN-1. By what we just (xo) for proved, for LN-a.e. x E RN there exist (x) for all v E D and
alli=1,...,N.
We claim that N
(11.32)
a (xo) = E a- (xo) vi i.=1
11. Sobolev Spaces: Embeddings
344
for ,CN-a.e. x E RN and for all v E D. To see this, fix u E D and let co E Q' (RN). Since u is Lipschitz, by the Lebesgue dominated convergence theorem, Urn fRN T6 t->o+ r(
tll
(x + t -
(X) (
U
(
u (s) W (x - tvt - W l
)
JAN [
-I
)
(x)
N 49V
x) dx,
(x) a (x) dx.
By using the change of variables y = x + tv on the left-hand side of the first equality, we see that the two left-hand sides coincide. Hence,
I
Similarly,
I
N
au (x) P (x) dx =
au IV
N (9-Ti
-J
(xo) vzv (x) dx _ -
N
av (x) u (x) dx.
f
aW
N
N axe (xo) utu (x) dx.
Since the two right-hand sides coincide, we obtain that N
JRN au
(x)
(x) dx = JRN E aX (xo)
(x) dx
s_1
for all cp E Cc', (RN) . Given the arbitrariness of 'P E CG (RN), it follows that
(11.32) holds for GN-a.e. X E RN.
We are now in a position to apply Exercise 11.48 to obtain the desired result. As a corollary of Rademacher's theorem we can show the following result.
Theorem 11.50 (Stepanoff ). Every function u : RN --> R ie differentiable at £ N-a.e. x in the (possibly empty) set Eu
x E RN . lim_sup Iu (ly)
I(x) < oo}
_
-
Proof. Assume that E,, is nonempty. Let {B (x., rn)}n be the family of all balls with xn E QN and r1 > 0, r. E Q, such that u : B (x, , rn) -* R is bounded. It follows from the definition of Eu that C E,y
00
U B (xn., rn)
.
n=1
Set Bn := B (xn, N). For every x E Bn define vn (x) := sup {v (x) : v : B,, -> R is Lipschitz,
Lip v < n, and v < uIB. }
11.4. Lipschitz Fbnctions
345
and
W. (x) := inf {w (x) : w : Bn -t R is Lipschitz, Lip w < n, and to > ulBn } .
Then (exercise) Lip v < n, Lip wf, < n, and w,, < ul B. < vn. By Exercise 11.44 and Rademacher's theorem, v,, and wn are differentiable at GN-a.e. x E B. Define Fn :_ {x E B,z : vn and wn are differentiable at x)
.
Then LN (Bn \ Fn) = 0. By Exercise 10.37(iv) applied to v - wn, we have that Vvn (x) = Vwn (x) for LN-a.e. x E Bn such that v,, (x) = wn (x). Set C. := {x E F. : vn (x) = wn (x) and Vvn (x) 96 Vwn (x)} . Then RCN (Gn) = 0. Hence, the set w
E:= U n=m
has Lebesgue measure zero.
We claim that u is differentiable at all x e R. \ E. To see this, fix x E Et, \ E. By definition of Eu, there exist r > 0 and L > 0 such that ju (y) - u (x)I < L ly - xI
(11.33)
for all y E B (x, r). Let n > L be such that B C B (x, r) and x E B,,. Since x 0 E, it follows that x E F. By definition of vn and wn and by (11.33) we have that
u(x) - nly -xj <
bu(y) :5 wn(y) tu(x)+nly-xj
for all y E B,,. In particular, for y = x we obtain v, (x) = u (x) = w (x). Moreover, since x 0 Gn, we have that Vvn (x) = Vw (x) =: A. Hence, using the previous inequality, together with the facts that vn and wn are differentiable at x and v (x) = u (x) = w,s (x), it follows that
(y) - u (x) - A (y - x) inf u 0 = lim vn (y) - u (x) - A (y - x) < lim y--s X IV-X1 I
limn sup
u (y) - u(x) - A (y - x) 1Y-X1
< Jim v-
ly - xl
= 0,
which shows that u is differentiable at x. This concludes the proof. The following change of variables formula for Sobolev functions is another consequence of Rademacher's theorem.
11. Sobolev Spaces: Embeddings
346
Theorem 11.51 (Change of variables). Let Q, a' C RN be open sets, let T : S2' - St be invertible, with IF and 19-1 Lipschitz functions, and let u E W 11P (SZ), 1 < p < oo. Then u o 'P E Wi.p (SY) and for all i. = 1, ... , N and for GN-a.e. y E St',
8 (a ) (y) = N_ j= I
a (IF (y)) 8yz
(y) .
Proof. Let Stn CC Qn+1 be such that 00
Van.
n=1
Fix n E N and for 0 < e < dist (Stn, 00) define u, := u * AE in Stn. Let SZn :_ W(S2n) and for y E Stn set (11.34)
V. (y) := U. ('y (y)) .
By Rademacher's theorem the Lipschitz continuous function IF is differentiable CN-a.e. in Sin. Since uE E C°° (Q,,), we conclude that for all i = 1, ... , N and for RCN-a.e. V E f1n, N avE(y)=au£(`I'(y))a'(y).
(11.35)
p=1
ON
Hence,
1:5
axj
ayyt
N
(y) 15 E Lip'P.i l
and so or
IVve (y) I' dy <_ C
=c
f
I
,
IVue (IF (y)) I' dy
Jr-1(c,,)
f
op (y))
IouE ('P (y)) I p I JIF (x) I dy
IouE (x)Ip dx,
where we have used the fact that I JW I is bounded from below by a positive
constant (since IF-1 is Lipschitz) and Theorem 8.21. Taking a sequence e; - 0+, the previous inequality implies that
Jsr
'
ve. (y) - %t (&) Ip dal
C J0 IVue3 (y) - Vn (y) lp dx
for all j, d E N. Similarly,
f I vet (y) - ve, (y) Ip dy < c f
an
Iue, (y) - uec (y) l p dx
11.4. Lipschitz l
ctions
347
for all j, l E N. Since u,, -' u in W ',P (Stn), it follows that {v£j } is a Cauchy sequence in W1,P (SZn), and so it converges to a function vin E W' ,P (Stn).
By extracting a subsequence, we can assume that v,, and Vv,, converge, respectively, to v(n) and Vv(n) pointwise GN-a.e. in 1 ,. Since T-1 has the
(N) property and uff -> u and Vu,, -' Vu pointwise GN-a.e. in Stn, it follows that uf, o 41 - u o IQ and (Vu) o
%F pointwise GN-a.e.
in SYn. In view of (11.34) and (11.35), we conclude that v(n) (y) = u ('Y (y)) for GN-a.e. y E Stn and that for all i = 1, ... , N and for G1v-a.e. y E P 'n,
8 (u o )
(11.36)
ON
N
8u
j=1
C7xj
(y) _
(T (y)) -,
;-
ON
(y)
Since u o T = v(") in SZn, it follows that u o %P E W1,P (Q') for all n E N. Reasoning as in the first part of the proof, it follows from (11.36) and Theorem 8.21 that
f
,
f -IA =Cf
IV (u o 41) (y)IP dy < C
IVu (IF (y)) I" I J' (x) I dy
IVu (x)IP dx,
and similarly
f (uoW)(y)" dy
Cf lu(x)Ipdx.
Since it o 4r E W1,P (ft;n) for all it E N, applying either Theorem 10.35 or the definition of weak derivatives yields u o T E W1,P (St'). 0
Exercise 11.52. Let f : (0, R) -p R, where 0 < R < oo, and let 1 < p < oo. Find necessary and sufficient conditions on f for the radial function u(x) := f (IxI) , x E B (0, R) , to belong to W1,P (B (0, R)).
Chapter 1 2
Sobolev Spaces:
Farther Properties Newton's third law of graduation: For every action towards graduation there is an equal and opposite distraction.
-Jorge Cham, www.phdcomics.com
12.1. Extension Domains As we have seen in the previous chapter, several embeddings that are valid for the entire space RN continue to hold for extension domains. The next
exercise shows that in general an arbitrary open set is not an extension domain for W1'p. Thus, an important problem is to characterize the class of domains that are extension domains for Wl-. To the author's knowledge this problem is still open, with the exception of the case N = 2 and p = 2 (see the paper of Jones [921).
Exercise 12.1. Let 1 < p < oo, let fl = {x = (xl, x2) E R2 : 0 < x1 < 1, 0 < x2 < (xl)a}, where a > 1, and let u : fl -> R be defined by
u(x)
(x1)1-l3
where
1
,
x E fl,
a+1 P
Prove that u E W 1,P (f) and that if p is sufficiently close to ap , then u cannot be extended to a function in W1' (R2). Hint: Consider the three
cases1
Next we show that extension domains strongly depend on p. 349
12. Sobolev Spaces: Further Properties
350
Exercise 12.2. Let Sl be as in the previous exercise, and using polar coordinates (r, 9), let u /: R2 \ S2 - R be defined as u (x1, x2) := r1-,6vl' (9) fo (r cos 9, r sin 9) ,
where # > 0, 0 E CO0 ([0, 2ir]), with 0 = 1 for 9 small and 9 E [7r, 2-7r], and W E
= 0 for
(R2) with cp = 1 for 0 < r < 1.
(i) Let 1 < p < oo and find for which values of 6 the function u belongs to WI'P (R2 \?j) .
(ii) Prove that if 0 <# <
is sufficiently close to 2, then u cannot be extended to a function in W 1,P (iii) For every function v E W1,1 (R2 with v = 0 for xi > Z, define (R2).
ifx0f,
I v(x)
v (x ) +
E4 (v) (x)
( v+
( x)
- v - (x ))
if x E fl
where v- (x) := v (x1, -x2) and v+ (x) := v (x1, 2 (x 1)' - x2). Prove that Ea (v) E W 1,1 (R2) and that I14 (v)IIw1.11R2) 5 C
lW1"1
(iv) Prove R2 \ SZ is an extension domain for Hint: Use cutoff functions to reduce to part (iii) and away from the origin use Theorem 12.15 below.
The previous exercise shows that R2 \N is an extension domain for W 1,I but not for W1' with p > 1. We now prove that if 8 is sufficiently regular,
then it is possible to construct an extension operator C that works for all Sobolev spaces W1,P (1). We begin with the important special case in which 11 := `(x', XN) E
e-1 x R : x.Nr > f (x') } ,
where f : RN-1 -> R is a Lipschitz function.
Theorem 12.3. Let f : RN-1 - R be a Lipschitz function and let (12.1) 0:= l(x', xN) E RN-1 x R : xN > f (xr)} . Then for all 1 < p < oo there exists a continuous linear operator
e : W 1,P (Sl) -> W 1P (RN) such that for all u E W 1'' (a), E (u) (x) = u (x) and (12.2) (12.3)
for EN-a.e. x r= St
IIC(u)IIr,.(RN) = 2IlullLVfnl /
IIVC(u)Ilp'(mN;OtNxN) S (2+Lip
f)IIVUlIr'(O;RNxN).
12.1. Extension Domains
351
Proof. The idea of the proof is to first flatten the boundary to reduce to the case in which SZ = RN and then use a reflection argument (see Exercise 10.37). We only prove the case 1 < p < 00 and leave the easier case p = 00 as an exercise. Consider the transformation
IF :RN,RN (z', ZN) '-' (z', zN + f (z')) Note that W is invertible, with inverse given by
.
tI,-1:RNRN (X', xN)
(x', xN
- f (x')) .
Moreover, for all y, z E RN,
(J)
(z)
(V,YN + f (Y))
(z',zN + f (z'))
= I (F!' - z', f (J') - f (z') - YIN + Z.) I <
Iy'-z'I2+((Lipf)Iy'-z'I+IYN-zn.I)2
«Iy - zI, which shows that' (and similarly T-1) is Lipschitz continuous. Since f is Lipschitz, by Rademacher's theorem (see Theorem 11.49) it is differentiable for LN'1-a.e. z' E RN-1, and so for any such z' E RN-1 and for all ZN E R we have VIQ (z) =
L(z') ... which implies that det V ' (z) = 1. Note that t (RN) = Sl. Given a function u E W1 ,p (St), 1 < p < oo, define the function
w(z):=u('P(z))=u(z',zN+f (z')), zER
.
By Theorem 11.51 the function w belongs to W 1,P (RN) and the usual chain rule formula for the partial derivatives holds. By Exercise 10.37 the function w : RN R, defined by V (x) :=
I
w (z)
w (z', -ZN)
if ZN > 0, if ZN < 0,
belongs to W 1,P (RN) and the usual chain rule formula for the partial derivatives holds. Define the function v : RN R by (12.4)
v (x) :_ (w o T -1) (x) _ 1
(2)
u
if 2f (x') - XN) if XN < f (x') .
12. Sobolev Spaces: Further Properties
352
Again by Theorem 11.51, we have that v E W 1,P (RN) and the usual chain rule formula for the partial derivatives holds. Using Theorem 8.21 and the fact that det V W = det V' -1 = 1, we have
that P fRIV\-0 Iv (x)l" dx = fRN\-a Iu (x', 2f (x) - XN) I dx
= jIiiQai)IP
dy.
t
Since for all i = 1, ... , N - 1 and for GN-a.e. x E RN (12.5)
(9V
,8x;
(x)
8
u (x', 2f (x') - XN) + 8xN (x', 2f (x') - X N) ax (x')
again by Theorem 8.21 we have that P av dx JmN\s1 8x (x}
r
P
I
I
(fRN\?J exi
(.T '
I 2f (x) - xN)
P I
P
dx 1
+Lipf fR N\ I-aXN
(x',2f (x') dy)P
(1 + Lipf) (j IVu(y)I' Similarly, using the fact that for LN-a.e. X E RN 1
N
8V
8xN (x)
we obtain
a
1V
(x', 2f (x')
- XN)
P
,
- X N) '
xN) I p dx N n a N (x', 2f (x') -
(x) IP dx =
I
flI
j
Eltb
IOXN (y)
p dtJ
Hence, the linear extension operator
u E W1'P (fl) '- 9 (u) := v E W1'P (R") is continuous and satisfies (12.2) and (12.3).
0
Remark 12.4. Note that the operator E defined in the previous theorem does not depend on p. However, it has the disadvantage that it cannot be used for higher-order Sobolev spaces, unless one assumes that f is more regular (see (12.5)).
12.1. Extension Domains
353
In order to construct an extension operator that can be used for higherorder Sobolev spaces, we need to introduce the notion of regularized distance.
Exercise 12.5 (Regularized distance). Let Q C RN be an open set, let {Q (xn, r-)} be a Whitney decomposition of 11 (see Theorem C.26), and let 0 < E < 4. Consider a nonnegative function cp E Ca' (RN) such that cp (x) = 1 for all x E Q (0,1) and V (x) = 0 for all x E RN \ Q (0,1 + e) and for each n set 9n (x) := V
x - xn1
l
r,,
,
X E RN
Define the regularized distance diam Q (x, , rn)cpn (x) ,
d,.eg (x) :_
x E RN.
n
(1) Prove that C1 dist (x, RN \n) < dteg (x) < 02 dist (x, RN \ Q) for all x E f and for some constants C1, C2 > 0 depending only on N. (ii) Prove that drag E C°° (0). (iii) Prove that for every multi-index a, OadQg
(x) < Ca (dist (x,RN \ ))1-lal
for all x E fl and for some constant C, > 0 depending only on N and a. Exercise 12.6. Let fl C RN be as in (12.1) and let d1eg be its regularized distance. Prove that there exists a constant C = C (Lip f) > 0 such that d1eg (x) ? C (f (2) - X N)
for all xERN \?I. The next exercise gives another extension operator (see (1600.
Exercise 12.7 (Stein). Let n be as in (12.1) and let !i : [1, oo) - R be a continuous function such that for every integer m E N, lim 0 (t) = 0
t,oo and
f(t) d=1,
tin
jtb(t) dt=0.
12. Sobolev Spaces: Further Properties
354
For every u E W"P (fl), 1 < p < oo, define
I u (x)
if x E 52,
El (u) (x)
where d
dt if x E R N \ 11,
is the regularized distance.
(i) Prove that if u E W'M (St) n Cc° (RN), then E1 (u) is continuous. (ii) Prove that if u E W"' (52) n C,1 (RN), then E1 (u) is differentiable for GN-a.e. x E RN.
(iii) Prove that if u E W'M (0) n C°° (RN), then E1 (u) E W1" (RN) with IIEI (U)IIw1.n(RN) <_ C (N, Lip f) IIuIIwl.P(n).
(iv) Prove that E1 can be extended to a continuous linear operator from
W' (St) to W" (RN). Remark 12.8. By constructing o with the additional property that 00
A
tmib (t) dt = 0
for all m E N, one can actually show that if u E Wm,P (11) for some k E N, then E1 (u) E Wm'P (RN) and the operator El is continuous. We refer to the book of Stein [160] for more details. Next we extend Theorem 12.3 to more general domains.
Definition 12.9. The boundary 812 of an open set 52 C RN is locally Lipschitz if for each point xo E 8St there exist a neighborhood A of xo, local
coordinates y = (y', yN) E RN-1 x R, with y = 0 at x = xO, a Lipschitz function f : RN-1 - R, and r > 0, such that St n A = { (y', yN) E Q n A : y' E QN-1(0, r) , yN > .f (y') }l
If 812 is bounded, then we refer to locally Lipschitz boundaries simply as Lipschitz.
Definition 12.10. The boundary Oft of an open set 52 C RN is uniformly Lipschitz if there exist e, L > 0, M E N, and a locally finite countable open cover {52n} of 8St such that
(i) if x E 852, then B (x, e) C Stn for some n E N, (ii) no point of RN is contained in more than M of the Stn's,
(iii) for each n there exist local coordinates y = (y',yN) E RN-1 X R and a Lipschitz function f : RN-1 - R (both depending on n), with Lip f < L, such that ERN-1 xR: yN> f(y')}. nn nfl =52nn{(y',yN)
12.1. Extension Domains
355
Exercise 12.11. Let 12 C RN be an open set such that 812 is bounded. Prove that 8S2 is uniformly Lipschitz if and only if it is Lipschitz.
Exercise 12.12. Let 11 C R be such that
Q = U In' n
where the In are open intervals such that length I,, > S for all n E N and
dist(In,Ik)>Sfor all n,kENwith n
k and for some 6 > 0.
(i) Prove that 812 is uniformly Lipschitz. (ii) Prove that the condition length In > S for all n E N is necessary to have an extension operator from W'-' (12) to W 'm (R). (iii) Prove that the condition dist (In, Ik) > S for all n, k e N is necessary to have an extension operator from W1'O° (12) to W1'O° (R).
In the next exercise, by a finite cone having vertex v E RN, length e, and vertex angle 2a E (0, ir) we mean a set of the form Ki,,R := v + RK, where R is an orthogonal N x N matrix and
K:_ {(x',XN)
ERN-1
x R: xN>x1cota}flB(0,Q).
Exercise 12.13. Let S2 C RN be an open set whose boundary 812 is uniformly Lipschitz.
(i) Prove that for every x E 811 there exists a finite cone KK,R, with vertex x, length e, and vertex angle depending on L, such that KX,R \ {x} C S2.
(ii) Prove that if 12 is unbounded, then GN (ft) = oo. The next theorem shows that domains with uniformly Lipschitz boundary are extension domains for W1.P (S2) for all 1 < p < oo. In the proof we will use the following exercise.
Exercise 12.14. Let wn : RN -' [0, oa] be a sequence of Lebesgue measurable functions and let
w(x):=Ewn(x), xERN. n=1
Assume that there exists an integer M E N such that for every x E most M terms wn (x) are nonzero. Prove that for every 1 <_p<00, p
00
IIVIILp(RN) < Ma
IIwnhIPP(RN)
n=1
while IIWIIW(RN) < MSUP IIwnIIL-(RN) n
,
RN at
12. Sobolev Spaces: Further Properties
356
Theorem 12.15. Let Il C RN be an open set with uniformly Lipschitz boundary. Then for all 1 < p< oc there exists a continuous linear operator
e : W"' (il) . W 1p (RN ) such that for all u E W1"" (0),
for GN-a.e. x E St
E (u) (x) = u (x) and (12.6) (12.7)
116 (u)IILD(nRN) < (1 + 2M) IIullLp(Q),
We (u)IIL6(RN;RNxN) <- C (1 + M (1 + L))
x
+
e
IIouIIwl,P(n;,NxN)
1
for some constant C = C (N) > 0. Proof. We only prove the case 1 < p < oo and leave the easier case p = 00 as an exercise. For every set E C RN and every r > 0, we define
E' .={xERN: B(x,r)CE}. We observe that ET C E and that condition (i) in Definition 12.10 reads
U nn.
an C
n
Define the regularized functions (12.8)
;
On :=
Xfln /4,
where cps is a standard mollifier. Then
suppOnC0n,
(12.9)
On
1 in n,12
By Theorem C.20 we have that D¢,, = V`p: * Xu/4, and so for all x E Q. we have that (12.10)
Io
n (x)I
I
N
(y)I dy
I V O+ (y - x)I IIL1(RN;RN)
£
Next consider the three open sets no (12.11)
S2+
{x E RN : dist (x, fZ) < } 4
,
(x E RN : dist (x, On) < 4
St- :_ {x E 0 : dist (x, on) > 4 }
,
12.1. Extension Domains
357
and define the regularized functions (12.12)
/o
{o
* Xszo,
xn f.
0±
Then Oo = 1 in N, 0+ (x) = 1 if x E RN and dist (x, M) < 2, and 0- (x) = 1 if x E SZ and dist (x, 8(i) > 2 . Moreover, the supports of Quo, 0+, and 0_ are contained, respectively, in an neighborhood of fl, in an s neighborhood of Bfl, and in f2. Finally, reasoning as in (12.10), we have that (12.13)
IIVOoll., IIVOtll. < e
Note that (12.14)
supp0oC{xERN: 0+(x)+0_(x)>1}.
Thus, with a slight abuse of notation, we may define (12.15)
14+ = 00
+
0-
00
0++0-'
where we interpret the right-hand sides to be zero whenever 00 = 0. Again by (12.14) we have that all the derivatives of ifif are bounded by !2. Also, ?p+ -}- (i_ = 1 in SZ and 0+ = _ = 0 outside an' neighborhood of Il. We are finally ready to construct the linear extension operator. Given a function u E W1P (n), 1 < p < oo, since supp (On.u) C SZ for each n by (12.9), by condition (i) in Definition 12.10 and Theorem 12.3 we can extend to a function vn E W1.P (RN) in such a way that (12.16)
IIvnIILP(RN) = 2IIOnuIILP(n. IlovniIf,,(RN;BtNxN)
),
(2 + L) IIv
Again with a slight abuse of notation we define E On (x) vn (x) (12.17)
E (u) (x)
(x) k
+ t/'_ (x) u (x) ,
02k (x)
x E RN.
Note that if x E RN is such that dist (x, 80) < 2, then there exists an n such that y E StnJ2, and so 0n (x) = 1 by (12.9). In particular, since all the functions 0n are nonnegative, it follows that (12.18)
0. (x) > 1,
if x E supper+, then n
and thus the first term on the right-hand side of (12.17) is well-defined, provided we interpret it to be zero whenever 0+ = 0. Similarly, since supp,0_ C supp O_ C 91, the term ifr_ (x) u (x) is well-defined, provided we set it to be zero outside 1?.
12. Sobolev Spaces: Fbrther Properties
358
It remains to show that the linear operator .6 (u) is an extension operator and that it is bounded. For the former, it suffices to observe that if x r= St, then vn (x) = 0,b (x) u (x), and so
E (u) (x) = b+ (x) u (x) + ib- (x) u (x) = u (x) by (12.15). To obtain the latter, we observe that by (12.17), condition (ii) in 1, Definition 12.10, the previous exercise, (12.18), the fact that 0 < and (12.16), in this order, we have
A
lie (u)II LD(.N) < M n
dx) f Ivnlp n
+ IIUIILD(SI) i
A
f LJuIPEi
<2M7
iPdx+IIUIILP
1)
(1 + 2M)
where in the last inequality we have used the fact that E I0ni < M. NxN), we use the fact that since {Stn} is locally finite, any bounded neighborhood of every point x E RN intersects only finitely many Stn's. Hence, we can calculate To estimate II VC
(u)IILA(RN.
E K V0. + On0un)
E Onvn
VC (u) _ k
k
V. + I'+
n
Ek 02
r{ E unOnJ 1 (l E, iv4 J \ \ 2 // - 2'P+
+ u0IP- + 5-Vu.
k
Using (12.16)-(12.18), condition (ii) in Definition 12.10, the previous exercise, and the facts that 0
(12.19)
Ii
ME,
EI0nlp<_M,
n
LIv4nIp<_M
n
12.2. Poincare Inequalities
359
we obtain 1
P
j,,
Iloe (U) II LP(RN;RNxN) < CM '
+
C IIUIILP(n) + C
< CM7 (1 + L) > n
+
IvnIP + ivv, r dx
C s
IIVUIIW1.P(n;RNXN)
1
r
J
IIlILP(n) + C
P
IOnuiP + IV (Onu)I' dx IIouIIW1,P(n;RNXN)
< C (1 + M (1 + L)) 1 IIUIILP(n) +
IIVUIIW1.P(n;RNXN)).
To conclude the proof, it suffices to observe that since {Sln} is locally finite, in a neighborhood of every point the infinite sum in (12.17) is finite. Hence, we can now invoke Theorem 10.35 and the previous estimates to conclude 0 that E (u) E W1"P (RN).
Remark 12.16. Using Poincare's inequality (see the next section), in the case of bounded Lipschitz domains one can show that in the inequality (12.7) it is possible to drop the term E IIuIILP(n). This was first proved by Tartar (see [401 and Lemma 2.6 in [11).
12.2. Poincare Inequalities The next result provides an equivalent norm in Wa'P (fZ) for a large class of domains St.
Theorem 12.17 (Poincare's inequality in W)'' (fl)). Assume that the open set fl C RN has finite width, that is, it lies between two parallel hyperplanes, and let 1 < p < oo. Then for all u E Wo'P ([1),
jIu(xW' dx <_
p
jIVu(x)IP dx,
where d is the distance between the two hyperplanes.
Proof. Without loss of generality, up to a rotation and translation, we may assume that St lies between the two parallel hyperplanes xN = 0 and xN = d > 0. For u E C°° (S1), by the fundamental theorem of calculus and
12. Sobolev Spaces: Further Properties
360
Holder's inequality, we have fxN
IU (4 xN)I = Iu (x',xN) - u (x', 0)I =
r xN l J
I
j
aN
(2, t) dtl
1
ON
(x t) lp dt)
Raising to the power p and integrating over RN-1 x [0, d], by Tbnefi's theorem we get
lEN-i x [p,d]
I0 xN (x', xN) I p dx < / u
Id
p
ON
fd = S2 I a N
_
(Plpp
(11) Ip dy
ICI a
(x', t)
I
dtdxNdx
xN l dxN
(1)Ipdy.
0
This concludes the proof.
Remark 12.18. Thus if the open set Sl C RN has finite width, then the seminorm IIVul1Lp(nDtN) is actually an equivalent norm in the space Wo,r
(n).
Exercise 12.19. Show that if the open set SZ C RN contains a sequence of where X. E RN and r,,, -* oo, then the previous Poincare's balls B (xn, inequality fails.
Exercise 12.20. Let ) C
RN be an open bounded set and let yN be the
constant defined in (11.15).
(1) Prove that for every -y E (0, 7N) there exists a constant C., > 0 such that for every bounded open set N1 (12.20)
fn exp
lu (x)I
dx
C"CN (a)
IIVuIILN(n;ItN)
for all uE Wo'N(St) \{0}. (ii) Prove that the previous inequality fails if 7 > yN. Remark 12.21. The inequality (12.20) was first established by Trudinger in [168]. In (130], Moser proved that for bounded domains the inequality (12.20) also holds for y = 7N. The proof is not immediate and we refer to [130] for more details. See also the recent paper of Li and Ruf [107] and the bibliography contained therein for some recent results on unbounded domains.
12.2. Poincare Inequalities
361
Exercise 12.22. Assume that the open set fZ C RN has finite width and let 1 <- p < oo. Prove that for every functional L E W-1"1 (f') there exist fl, ... , fN E L1'' (fl) such that L (u)
f
N
a-1
A (x) (9xi (x) dx
for all u E Wo'p (S2).
Another form of Poincare's inequality is
1 Iu (x) - uE I1 dx < C J
Ivul' dx, S2
where E C SZ is a Lebesgue measurable set with finite positive measure and
uE := GN (E)
(12.21)
[ u(x) dx.
Theorem 12.23 (Poincar6's inequality in W1.n (SZ)). Let 1:5 p < oo and let 1 C RN be a connected extension domain for W 1'P (SZ) with finite measure. Let E C SZ be a Lebesgue measurable set with positive measure. Then there exists a constant C = C (p,1, E) > 0 such that for all u E W 1,P (SZ),
fo
Iu(x) -uEIp dx < Cjnivu(x)I" dx.
Proof. Assume by contradiction that the result is false. Then we may find a sequence {u,,} C W 1'p (SZ) such that
in
lun (x)
- (u )EIP dx > n I IVun (x)I" dx > 0.
Define
vn:_
Un - (un)E Thin - (u+z)EIILP(n)
Then v,, E W1'P (91) and IIVnIILp(n) =1,
(un)E = 01
J In
Ivtinlp dx < 1.
n
By the Rellich-Kondrachov theorem (Theorem 11.10) and Exercises 11.26 and 11.37 there exists a subsequence {..k} } such that vnk -> v in LP (fl) for some function v E Lp (S2) with IIVIILP(f) = 1,
uE = 0.
12. Sobolev Spaces: Further Properties
362
f
Moreover, f o r every
f
Jn
v
axi
E C 1 (fZ) and i = 1, ... , N, by Holder's inequality ,0
a dx. = lim
dx = lim
loco fn
Oxi
li o Un
IVv,,kI
r
sunk dx 8x{
k-400
\ Jn jiff dxA
dx J p y fl)
0,
and so v r= W 1,' (f2) with Vv = 0. Since f2 is connected, this implies that v is constant, but since vE = 0, then, necessarily, v = 0. This contradicts the fact that IIyIILp(,) = 1 and completes the proof.
Exercise 12.24. Let 1 < p < N and let S2 C RN be a connected extension domain for W1' (f2) with finite measure. Let E C 0 be a Lebesgue measurable set with positive measure. Prove that there exists a constant C = C (p, f2, E) > 0 such that for all u E W 1,P (l),
(fIux) - uEI"" dxl
n
<
(jIvu(x)I" dx)"
.
C Exercise 12.25. Prove that if 0 C RN is an open connected set with finite measure, 1 < q < p < oo, and E C 0 is a Lebesgue measurable set with positive measure, then there exists a constant C = C (p, q, f2, E) > 0 such that for all u E W l,p (ft),
(L1u(x) -uEIq dx °
(jIvu(x)l"
dx)D
.
Exercise 12.26. Let f2 c RN bean open connected set with finite measure and let 1 < p < oo. Assume that the embedding W1,p (ft) -> l" (f?)
U'+U is compact and prove that there exists a constant C = C (p, ft) > 0 such that for all u E W 1,P (ft),
/ lu (x) - un Ip dx < C
f
I Vu (x) Ip dx.
Exercise 12.27. Let f2 C R3 be a connected bounded extension domain for W 1,2 (f2) and consider the subset of W 1,2 (s2) given by
s:= {uEwh12(c): fu3(x)dx=0}. (i) Prove there exists a constant C = C (f2) > 0 such that for all u E S, 2
(L1'I2 &)'i `C `fn ivu(x) 12 dx}
2
.
12.2. Poincare Inequalities
363
(ii) What are the properties of the subset S that you used in the proof?
Corollary 12.28 (Poincare's inequality for continuous domains). Let 1 <
p < oo and let fl c RN be a connected bounded domain whose boundary is of class C. Then there exists a constant C = C (p, fl) > 0 such that for all u E W1"P (52),
Iu (x) - ur I' dx
J
I Vu (x) I" dx.
Proof. In view of Theorem 11.21 we are in a position to apply Exercise
0
12.26.
In many applications it is of interest to have an estimate of the constant C (p, S2, E). In the remainder of the chapter we study important special classes of domains for which there are explicit bounds on the Poincare constant C (p, 0, E). We begin by considering rectangles.
Proposition 12.29 (Poincare's inequality for rectangles). Let 1 < p < 00 and let R = (0, al) x ... x (0, aN) C RN. Then for all u E W1"' (R),
lu(x) - uRI" dx < N" (max{al.... ,aN})pJ IVu(x)I" dx. R
JR
Proof. Without loss of generality, by Theorem 10.29 we may assume that u E Coo (R), and, by eventually replacing u with v (y) := u (Ay), V E aR for an appropriate A > 0, we may assume that
max{a,,...,aN} = 1.
(12.22)
By the fundamental theorem of calculus, for all x, y E R,
Iu(x) - u(y)I <- Iu(x) -u(xl,...,xN-1,yN)I +... + Iu(x1,y2,...,YN) - u(y)I N
<
.; I
C7x '
(xl,...,xy-l, t,yi+1,...,yN)I
dt.
12. Sobolev Spaces: Fl irther Properties
364
By Holder's inequality, the convexity of the function g (t) = ItIP, and (12.22), N
P
f' IDu(xl,...,xi-1, t,yi+l,...,YN)I dt
lu(x) - u(y) IV < i=1
N (f,j
Iou(xl,...,xi-1, t,yi+1,...,YN)IP dt
< N
< NP-1 E i=1
rat
J0
i Y
P
IVu (xl,... , xi-1, t, yi+1, ... , yN) IP dt.
Hence, again by Holder's inequality,
1 Iu (x) - URIP dx < ()CN (R))P JR <
<
=
(/ N
1
(R))P-p
\P
\UR
lu (x) - u (y)I dy J dx
RR
lu (x) - u (y) IV dydx
Np-1 fJfJa*
Vu (xl,...,xi-l,t,yi+1,...,yN)IP dtdydx
ai L IDu (z)IP dz < NP J IVu (z)IP dz,
NP-1
i=1
where we have used the fact that p - Pr = 1, F1lbini's theorem, and (12.22). This concludes the proof. Next we study convex domains. In the literature there are simpler proofs (see, e.g., [73]). The present one gives a sharper constant and the approach is quite interesting.
Theorem 12.30 (Poincare's inequality for convex sets). Let 1 < p < oo and let 0 C RN be an open bounded convex set. Then there exists a constant
C = C (N, p) > 0 such that (12.23)
1 Iu (x) - uQI P dx < C (diam S2)P J
IDu (x) IV dx
for all u E Wl,P (S2).
We begin with some preliminary results.
Exercise 12.31. Let D C RN be a convex set. (i) Prove that the interior and the closure of D are convex.
12.2. Poincare Inequalities
365
(ii) Let u : D - R be a continuous function such that
u(X1 +x21 >
u(xl)+u(x2)
2
2
for all x1, x2 E D. Prove that u is concave.
Exercise 12.32. Given a set E C RN, prove that the distance function d (x) = dirt (x, E) is Lipschitz continuous with Lipschitz constant at most one.
Lemma 12.33. Let SZ C ][8N be an open bounded convex set. Then for every
small 6 > 0, the set fla := {x E f2: diet (x, 8f2) > 6} is a convex set with SZa C Q.
Proof. We begin by showing that the function d (x) := dist (x, 811) is concave in Q. To see this, consider x1i x2 E SZ and let dl := d(x1) and d2 := d (x2). Fix any direction v c SN'l and consider the trapezoid with vertices x1, x2,1!1, y2, where y1 := xl+d1v and y2 := x2+d2v. Since the open balls B (xi, di), i = 1, 2, are contained in fI by the definition of di, it follows that Vi E art, i = 1, 2, and so the trapezoid also lies in N by the convexity of
(see Exercise 12.31). In particular, if x := ZLIa and do := 4, then the point y :_ 24a = x + dov lies on the edge of the trapezoid and therefore I
also in U. As the vector v varies in SN-1, the corresponding point y fills the surface of the ball B (x, do). Hence, B (x, do) C ?Z, and so d (x) >- do, that is,
d (x1 + x21) 2
> d (xl) + d (x2) 2
Since d is Lipschitz continuous by the previous exercise, it follows from Exercise 12.31 that d is concave. Next we show that 11. is convex. Let X1, X2 E 1t and let 6 E (0, 1). By the concavity of d,
d(9x1+(1 -9)x2) >- 8d(xi)+(1-8)d(x2) > S,
0
andso8x1+(1-8)x2Ef2o.
Lemma 12.34. Let fZ C RN be an open bounded convex set and let u E L,
(SZ) be such that
ju(x) dx=0
.
12. Sobolev Spaces: Further Properties
366
Then for every 6 > 0 there exists a decomposition of fl into a finite number of pairwise disjoint convex domains fl,,, n = 1,...,4 such that
_ it = U fn) e
r&=1
1 u (x) dx = 0,
n=1 .... , $,
and each Sl,a is thin in all but one direction, that is, in an appropriate coordinate system y = (y', yN) E RN-1 x R, Sl C [0, b[ x ... x [0, b] x [0, diam ll] .
Proof. For each a E [0, 27r] there is a unique hyperplane H,, C RN with normal (cos a, sin a, 0,. .. , 0) that divides it into two convex sets fl,, and f1 of equal volume. Let
I (a) :=
u (x) dx. Ire.
S ince
I (a) = -1 (a + zr), by continuity there exists ao such that I (ao) = 0.
Since
f u (x) dx = f &0
1.,00
u (x) dx = 0,
we now apply the same reasoning to each of the sets cl"Pto and ft,1,1,0. Continuing in this way, we can subdivide Cl into convex sets (L with the property that each of these sets is contained between two hyperplanes with normal of the form (cos p,,, sin Q,,, 0, ... , 0) and at distance 6 and the average of u vanishes on each of these sets.
Fix any
By a rotation we can assume that the normal of the two
hyperplanes is (1, 0, ... , 0). In these new coordinates we apply the same reasoning using hyperplanes with normals of the form (0, cos a, sin a, 0, ... , 0). We continue in this way to obtain the desired conclusion.
Proof of Theorem 12.30. Step 1: We begin by showing that we may assume u E C1 R. By Lemma 12.33 for every 6 > 0, the set f2i is convex and
dist( ,8Sl)=S>0. Consider a sequence of standard mollifiers {cps}E>O. For 0 < e < d define us := u * cps in fl.. Then us E C00 (17a) by Theorem C.20, and so, if (12.23) holds for the pair u£, SZa, we get
f,l,u,(x)-(,u,).,I'"dx
I Vu , (x) I" dx.
12.2. Poincare Inequalities
367
Letting e - 0+ and using Theorems C.19 and C.20 rwe obtain
J
Ip dx < C (N, p) diam clo J Iou (x)Ip dx
I u (x) -
rna
a
< C (N, p) diam St in J Vu (x) I P dx.
It now suffices to let 8 -' 0+.
Step 2: By the previous step and by replacing u with u - un, we may assume that u E C' (St) and that
u (x) dx=0. By uniform continuity, given 0 < e < 1, there exists 8 > 0 such that (12.24)
IIu (xl)Ip - Iu (x2)lpl + Ilou (xl)I" - IVu (x2)IpI <- e,
(12.25)
Iu(x1)-u(x2)I+IVu(x1)-Vu(x2)I <<e
for all xl, x2 E St with Ix1 - x21 < 8. By Lemma 12.34 we can decompose St into a finite number of pairwise disjoint convex domains Stn, n = 1, ... , Q, such that P
fu(x)dx=On = 1, ... , t,
St = U Stn,
(12.26)
tn
n=1
and for every n there is an appropriate coordinate system y = (y', yN) E RN-1 x R such that the projection of On into the yN-axis is (0, dn), where do < diam St, and (12.27)
On C [o°
N - 1]
x ... X [0,
N - 1]
x [0' dn]
In particular, (12.28)
(y', YN) - (0, yN) 15 8
for all (y', yN) E Stn.
Fix one such Stn. By (12.24), (12.25), (12.26), and (12.28), (12.29)
in u (y) dy n
(12.30)
(12.31)
if
f
u (0, YN) dy1 < erN (Stn)
,
Stn
Iu(y)I1 dy- fn Iu(O,yN)Ip dyl <.GN(Stn), n
IfnlftN(y)Ipdy-fnlN
yN)lpdyl <eGN()
Let 9 (yN)
fRN-1 Xn,, (Y, YN) dy'.
12. Sobolev Spaces: Further Properties
368
By Fubini's theorem,
jrl u(0,YN) dy=1
(12.32)
dn
u(O,YN)9(yN) dyN,
o
d f Iu(0,yN)IP dy = fo Iu(O,yN)I"9(!N) dyN,
(12.33) (12.34)
in.
i
YN)IP
(O,
n.
I
N
dy = f
d . 1TYN &U (0, YN) IP 9 (YN) dYN
By Minkowski's inequality with respect to the measure g (ypr) dyN we have
that 1 a4,
(10
dYN)p
Iu(O,YN)IP9(YN)
/
rf ..nlu (0, 9N) -'l(o.a.) I p 9 (yN) dYN } p 77
(12.35)
<_ l
1
\\JJ
`p
ci 1P
+ (Jo
I
9 (ON) dYN)
1+ 11,
where
fo u(O,yN)g(yN) d11N d04 g (yN) dyN
Since the function gT is concave by Remark C.8, we are in a position to apply Proposition 7.23 to conclude that l1
P9(YN)
Z
= Cpd
If
Cp diam i
Jp
d&N J/
YN)IP
I
'9U (O,
dye
N
P
u
(1 kYA
(?!) I
411 J p
/
-I' FCN
where we have used (12.34), (12.31), the inequality (12.37)
(a + b)1/P < (a + b)1"P + (a + b)1"P
and the fact that d,, < diam 1, in this order.
for a,b > O,
12.2. Poincare Inequalities
369
To estimate 11, we distinguish two cases. If then by (12.26) and (12.29),
" g (yN) dyN > ELN (ft,,.),
fd
fUJ (0,YN)9(VN) 1
dyNI
d
(
9(yN) dyN0
-1+1a
do
u (0, YN) 9 (YN) dyN -
< sLN (nn)
Jst., u (y) dy \ Jo l
g (yN) d11N 1
1
//
`+P
g (yN) dyN 1
(10
(SZn))
/I
while if f d" g (yN) dyN < C,CN (Stn), then
fd- u(0,yN)9(yN) dyN
ZZ
p
\Jo
J0"9(1!N) dyN
<MUI
/
9(yN) dyNI
1
< M (ELN (Stn)) P,
9 (YN) dyN)
` 0
where
M :=
a
Jul+maxlouI+1
?F
since u E C' (12). Combining both cases, we have that ZZ < (1 + M) (ELN
(12.38)
It now follows from (12.35)-(12.38) that
1
U
an l u (0, YN) I p 9 (YN) dyN 11
< Cp diam 1
un
+(Cpdiamft+1+M) Set C:= max {Cp diam n + 1 + M, Cp}. We will change C from line to line. By (12.30), (12.37), and the previous inequality, i (Lv,IuIp
dy)p
<_
(f A Iu(0,yN)1'9(yN) dyN)P + (eCN(an.))
<_ C diam St
Un.
I au Ip dyl P + C (eCN 8yN
//1
Using a discrete Minkowski inequality in the form 1
E
P
(an+bn)p n=1
E
p
< 'aP + L n=1
1
l
n-1
p bP_)
(SIn))i
12. Sobolev Spaces: Further Properties
370
where an, bn > 0, it follows by summing over all n that 1
(LutaY
n=1
fnlulp dy
p
p
P
diam 0 fEn I e
< C diem Sl < C (ciiam 0
d y J P+ (efN (S2n)) P 16
NI
r
I au l p dy
JN n=1 J \J
P
I Vulp dy) P
1
1
P
P
+ EP
,CN (SZn)
n=1
+ CEa (,CN (0)) P
Since C does not depend on c, it suffices to let e -+ 0+.
0
Remark 12.35. It can be shown that for p = 1 and p = 2 one can take the constant C in (12.23) to be 1 and 1 , respectively (see [16] and [2], respectively), and that there exist convex domains for which these constants are optimal.
Next we consider star-shaped sets. We recall that a set E C RN is starshaped with respect to a point xo E E if Ox + (1 - 0) xo E E for all 0 E (0,1) and for all x E E.
Theorem 12.36 (Poincare's inequality for star-shaped sets). Let 1 < p < 0o and let SZ C RN be an open set star-shaped with respect to xo E St and such that Q (xo, 40 C fl C B (xo, R)
(12.39)
for some r, R > 0. Then there exists a constant C = C (N) > 0 such that for all u E W1"p (S2),
I
R ) N-1
Iu (x) - of I" dx < CRP (
r R
in lVu (x) Ip dx.
Proof. By the Meyers-Serrin theorem (Theorem 10.15), without loss of generality, we may assume that u E COO (52) n W IM (Sl) and that xo = 0. Hence, we may write yESN-1,OSp
where f : SN-1 - [0, oo) is the radial function of C), that is,
f(y) :=sup{p> 0: pyE 11}.
12.2. Poincare Inequalities
371
Note that for all y E SN-1 we have that f (y) y E Otl and 2r < f (y) < R by (12.39).
Step 1: Assume that u vanishes in B (0, r). Since u (ry) = 0 for all y E SN-1, using polar coordinates and the fundamental theorem of calculus, for
all r < p < f (y) we have that P OP
U (P?I) = u (Py) - u (ry) =
(ty) dt,
J
and so by Holder's inequality, Iu (Py) r=
fu r
(Jr,!±(y) I d t ) < p
1(ty) lp dt
P
llI
P
Ipu (ty) I" dt.
r
Hence, by F ubini's theorem f(3!) 0 0
I u (py)I''
PN-1 dp
f('y)
< fo
P
pP,+N-1
I Vu (ty)I? dtdp
Ir. R
ff (v)
(i pr+N-2dp) `O
RP+N-1
p+N-1 < R"
(R
dt
1(y)
f jf(y)
Ipu {ty}Ip dt
N-1
V u(ty)I"tN1 dt.
Integrating with respect to y over SN-1 and using spherical coordinates yields
f
(R)N_l Iu(x)I" dx < R"
r
IVu(x)I ate.
n Step 2: Assume next that u r= COO (1k) n W1.P (1k) is such that uq(o,4r) = 0. Then by Proposition 12.29, n
(12.40)
r
Iu (x)I" dx < CP(N) rP J
IVu(x)I1 dx.
JJQ(0,4r)
Consider the function 0
x -' 1
if x E B (0, r) , if x E B (0, 2r) \ B (O, r) ,
if x B (0, 2r)
and write
u=(1-p)u+qu=:ul+u2.
12. Sobolev Spaces: Further Properties
372
Since u2 vanishes in B (0, r), by the previous step i
R
dx ] D
(in 'u
(R}
N-1
Ivu21p
D
d2)
N-1
1 D
IuVV + cpvuIP dx
in N-1
< R (r)
(12.41)
1
lx + (f Ivlp dx
f
RD
rP Q(0,4r)
<
RiR
r
1
D 1
//
12
N-1
1
f ivulp dx) A,
C (N)
where we have used Minkowski's inequality, the fact that Vco = 0 outside B (0, 2r), and (12.40). On the other hand, since ul is zero outside Q (0, 4r). 1
1
Jul Ip 12
}
dx/
11
((O4r) lul lp f Iulpdx
JQ(0,4r)
j
± u2Ip
+
)'L
IVulp dx 1
NN-i
D C(N)( N-1
A
)
D
IU21"dx
(fQ(0,4r)
<
+R(r)
dx
Jul
Q(0,4r)
l
1
fn IVulp
dx) D
,
where we have used (12.40) and (12.41). Hence, by Minkowski's and Holder's inequalities
r/' (L Iu - un Ip dx )
A
< 2 ( fn luip dx)
-P
< 2 f Iullp dx)
+2
(p Iu2IP dx) A SE
SZ
NI-1
<
1
(Rl D C (N) R
(L IvuIP
dx) D
.
12.2. Poincare Inequalities
373
Step 3: Finally, if u E C°O (n) n W (S2) is such that 'ttuQ(0,1r) # 0, apply the previous step to the function u - uQ(0,4r) to conclude that
lu - unip dr)
D
(fa
lu - uQ(0.4r) - (u - UQ(0,4r))IlI N-1
P dx) P
!r
)
1
C(N)R(R) P (J IVu(x)IPdx)p. 0 For p > 1 the constant RR (R) N-1 can be significantly improved using Muckenhoupt's weighted norm inequality for the maximal function.
Theorem 12.37 (Muckenhoupt). Let J C R be an interval, let 1 < p < oo, and let U : J -> [0, oo) be a Lebesgue measurable function. Then there is a constant C > 0 such that for all Lebesgue measurable functions u : J -> R, (12.42)
f((Miu)(x))PU(z) dx < C
j
Iu (x) I-' U (x) dx
if and only if there is a constant K > 0 such that for all intervals I C J,
P1
JU(x)dxIf
KJII.
dx
1
(U (x)) P-1
Furthermore, if C and K are the least constants for which the previous inequalities are true, then C< K < 2"K. In the previous theorem, (12.43)
sup 1 fvlu(t)ldt, yEJ\(v) y - x
(M1 u) (x) :=
x E J.
The proof of this result is rather involved and goes beyond the purposes of this book. We refer to [131] for more details.
Exercise 12.38. Let 0 < r < 2 and let (12.44)
U(t) := min {(r +
1t1)N-1
,
RN-lJl
,
t E R.
Prove that for all 1 < p < oo, (12.45)
K (r, R, p, N) :=
sup I interval
1 JU(t)dt1" I
ifl
(R)N-P
<
xr/ logN-1 (1 1
1
I (U (t))p=1
+
if p = N,
ifp>N.
12. Sobolev Spaces: Further Properties
374
Corollary 12.39. Let 1 < p < oo and let SZ C RN be an open set star-shaped with respect to xo E S2 and such that Q (xo, 4r) C SZ C B (xo, R)
for some r, R > 0. Then there exists a constant C = C (N) > 0 such that for all u E Wl,P (Sl),
fn
Iu (x) - unIP dx < CR1'K (r, R, p, N) fn IVu (x)1' dx,
when K is defined in (12.45). Proof. The only change is in Step 1 of the proof of Theorem 12.36, where instead of using Holder's inequality, we write
f
a
Iu (PY)I" =
I
a
(t3!) I
dt)
1P I < R1' (p_rL
a(ty) I dt)P
< RP ((M1 g) (P))P,
where g (P)
Op
(PY)I X(r,f(1l)) (P) ,
P > Of
and M1 g is defined in (12.43). In turn, since u = 0 in B (0, r),
f
f (P) Iu (PY)W P
PN-1 dp =
0
f (Y)
Iu (PY) PP
Jr
PN-1 dp
< RP f f(y) ((M1 g) <
PPPN-1
dp
RP rf(P) ((M1
r
g) (p))p U (P) dp rf{1r)
su
r
aP(py)
P
rAV) I au
P
U(P)dp PN-1
dp = RPK (r, R, p, N) ,f 8P (PY) r where U and K (r, R, p, N) are defined in (12.44) and (12.45), respectively, and we have used (12.42). Integrating with respect to y over SN-1 yields I
fiu (x) IP dx <- RPK (r, R, p, N)
J
I Vu (x) IP dx.
0 Exercise 12.40. Let 1 < p < 00, let f : QN-1(0, r) -' [0, oo) be a lower semicontinuous function such that 0 < r < f (x') : R for all x' E QN-1 (0, r), and let
n:= B (0, 2r) U {(x', xN) E QN-1 (0, r) x R : -a < xN < f (X')) .
12.2. Poincare Inequalities
375
Prove that for all u r= W 1,p (Q),
J
Iu(x) -univ dx
Hint: Use rectangular coordinates in place of polar coordinates to obtain the weight U = 1.
Chapter 13
Functions of Bounded Variation Living with P.Q.S, I. What is PQS? "Post-Quals-Slump" or PQS, affects 99% of all grad students. It is the #1 cause of delayed graduation dates and contributes to the A.D. degree drop rate.
-Jorge Cham, www.phdoomics.com
In this chapter we introduce the space of functions of bounded variation in domains of RN and study some of their basic properties. What is covered
here is just the tip of the iceberg. We refer the interested reader to the monographs [10], [54], [58], [74], and [182] for more information on the subject.
13.1. Definition and Main Properties Definition 13.1. Let Q C RN be an open set. We define the space of functions of bounded variation BV (f2) as the space of all functions u E L' (1) whose distributional first-order partial derivatives are finite signed Radon measures; that is, for all i = 1,.. . , N there exists a finite signed measure Ai : B (fl) -> R such that (13-1)
dx.bdA1 fn
for all 0 E C°° (Il). The measure A; is called the weak, or distributional, partial derivative of u with respect to xi and is denoted Diu. We define
BY0 (D) := {u r= L (Il) : u E BV (i') for all open sets Il' cc S2}
.
377
13. Functions of Bounded Variation
378
For u E BV (92) we set
Du := (D1u,... , DNu). Thus, if u E BV (S2), then Du E Mb (fl; RN), and so the total variation measure of Du, defined by 00
IDuI (E) := sup EIDu (En)I
(13.2)
E E B(S2)
,
,
n=1
where the supremum is taken over all partitions {En} C B (f2) of E, is a finite Radon measure (see Proposition B.75). Moreover, since Mb (S2; RN) may be identified with the dual of Co (SI; RN) (see Theorem B.114), we have
that IDuI (n) = IIDuIIMbARtN) N
f
J idD:u :
= sup i=1
4- E Co (fl; RN)
RN)
,
c
< 00.
Definition 13.2. Let Sl C RN be an open set and let u r= Ll (S2). The variation of u in f is defined by V (u, 12) := sup
t
-u dx :
4=1 J axi
lb E CC° (s2; RN) ,
111111
co(n;mN)
<1
Exercise 13.3. Let 12 C RN be an open set and let u E L'10C (St). Prove the following.
(i) If the distributional gradient Du of u belongs to Mb (St; RN), then IDul (11) = V (u, St) .
(ii) If V (u, 12) < oo, then the distributional gradient Du of u belongs to Mb (SZ; RN). In particular, if u E L1 (St), then u belongs to BV (SZ)
if and only if V (u, SI) < oo. Hint: Use the Riesz representation theorem in Co (SZ; RN)
(iii) If {un} C Ll (fl) is a sequence of functions converging to u in
Li
C
(S2), then
V (u, fl) < lim inf V (un, f2).
Exercise 13.4. Let SZ C RN be an open set. Prove that the space BV (SZ) is a Banach space with the norm IIuIIBV(fl)
IIUIILI(u) + IDuI (12) .
13.1. Definition and Main Properties
379
It follows from the definition of BV (1Z) that W1°1(SZ) C BV (fl) .
We have already seen that for N = 1 the inclusion is strict. Exercise 13.5 shows that this is also the case in higher dimensions.
Exercise 13.5. Let fl C RN be an open set and let E C RN, N > 2, be a bounded set with C2 boundary.
(i) Prove that XE 0 W1" (Q)(H) Prove that I D (xE) I (a) <_
NN-1 (aE
n Sl) .
Hint: Use Exercise 13.3(i).
(iii) Prove that the normal v to t9E may be extended to a function C1 (RN. RAT )
(iv) Prove that ID (XE) I (0)= NN-1 (aE f1 1l) .
(13.3)
The previous example shows that characteristic functions of smooth sets belong to BV (a). More generally, we have the following.
Definition 13.6. Let E C RN be a Lebesgue measurable set and let SZ C RN be an open set. The perimeter of E in fl, denoted P (E, Sl), is the variation of XE in 12, that is, P (E, f2) == V (XE, H)
=Sup
2 i Cax : t e C 0; RN) , IID IIGo(n;R,v) < 1
s_1 E axi
The set E is said to have finite perimeter in fl if P (E, fl) < oo. If SZ = RN, we write
P (E) := P (E, RN) .
Remark 13.7. In view of Exercise 13.3, if fl c RN is an open set and E C RN is a Lebesgue measurable set with GN (E n fl) < oo, then XE belongs to BV (1) if and only if P (E, 1) < oo. Note that Exercise 13.5 shows that for smooth sets E the perimeter of E in 11 is simply NN-1 (&E n S2). The next exercise shows that this fact no longer holds if the boundary of the set E is not (sufficiently) smooth.
13. Functions of Bounded Variation
380
Exercise 13.8. Let QN = {xfl}FEN and let A C RN be the open set
1).
00
A:= UB(xni2n
Jn=t
(1) Prove that &'(A) < oo but C' (8A) = oo. (ii) Prove that fN'1 (8A) = co. (iii) Prove that A has finite perimeter. Hint: Define »:
1
Am := U B
n, 2 }
,
Um := XAm
///
n=1
and use Exercise 13.3(iii).
13.2. Approximation by Smooth Functions Since every function u r: COO (0) n BV (tt) belongs to W 1.1(12) (why?) and
IDuI (0) = in IVul dx,
the closure of coo (St) fl BV (S2) in BV (t2) is W" (tZ). Thus, we cannot expect the Meyers-Serrin theorem (see Theorem 10.15) to hold in BV (ti). However, the following weaker version holds.
Theorem 13.9. Let SZ C RN be an open set and let u E BV (ti). Then there exists a sequence {u,,} C C°O (St) n W1.1 (Il) such that u.,, -> u in L1 (ti) and lim
fn IVunl dx = IDuI (fl).
We begin with an auxiliary result.
Lemma 13.10. Let tZ C RN be an open set and let u E BV (ti). For every e > 0 define ue := cpe * u in 1ls, where We is a standard mollifaer and tl is defined in (C.7). Then (13.4)
6llim
+0+
/
IVUEl dx = IDuI (0)
.
e
Moreover for every open set Cl' C Cl, with dist (SZ', O.Q) > 0 and IDul (8Si') _ 0,
(13.5)
lim L IvuEI dx = IDul
e-' 0+
(ci') .
13.2. Approximation by Smooth Functions
381
Proof. Step 1: For every i = 1, ... , N, define ai := Diu E Mb (1). By Theorem C.20, for every x E fZE we have
r
2u-£
(13.6)
(x - y) u (y) dy 8xi (x) = f naVE axi
n hi
(x - p) u (J) dy
= fnWe(x - y) dAi(y), where we have used (13.1) and the fact that the function We (x - ) belongs to C°° (1), since supp We (x - ) C B (x, c). Let - E C°° (cle; RN) be such that II4IICo(n,;R^') : 1 and extend 4) to be zero outside SZE. By (13.6) and an integration by parts, we have
f
,
' (x) `Fi (x) dx
(x) dx
U. (x)
axi
f, f 4)i(x)we(x-y) dai(y)dx
a
n
_-
f
n
(4)i)e (y) dAi (y) = J a (:i)e (y) u (v) dy,
i2
if
ON
where we have used Fubini's theorem and (13.1). Hence,
JUEdiV4)dX=JUd1V4)edX.
(13.7)
Moreover, for each x E f2E, by Jensen's inequality and by the facts that
It 1S1and fo(x -y) dy = 1, we have
E (fn We (x - y) IN (y) dy M
((4) )E)2 (x)
2
N
(13.8)
<
=
W. (x - y) lb? (y) dy
fnWe (x-y)It (Y)I2dy:5f We(x-y)dy=1,
and (since supp 41 C f1E) (13.9)
supp 4 ' e C {x E 1 : dist (x, supp -6) < e} C Q.
Thus, by (13.7) and Exercise 13.3,
fuediv4)dx=fudiv4)edx< IDuI (f2) . Taking the supremum over all admissible 4) and using Exercise 13.3 once more, this time applied to f2E, yields (13.10)
1 DueI (fee) <_ 1 Dul (1) .
13. Functions of Bounded Variation
382
In particular, ug E BV (SZg) fl coo (nr). Letting e - 0+, we conclude that lim sup IDuEI (Or) < IDuI (St) .
(13.11)
E-*o+
To prove the converse inequality, fix an open set 0' C SZ with the property that dist (S2', 81k) > 0. Since uE -r u in L1 (Il') as a -> 0+ (see Theorem C.19), by Exercise 13.3, IDuI (cl') < limiinf IDuEI
(13.12)
Since 12E D 1' for all 0 < e < dist (Q', 8Q), it follows that IDuI (SZ') < lim inf IDuEI (f2E)
.
Letting 11' / SZ and using Proposition B.9 gives (13.13)
IDuI (SZ) < lim inf IDuEI (QE)
,
E--,O+
which, together with (13.11), gives (13.4). Step 2: Fix an open set fl' C fl, with dist (fZ', On) > 0 and IDuI (8St') = 0. In view of (13.12), it remains to show that lim sup I Du, I (ii') < IDuI o+
We proceed as in the previous step, with the only change that we now consider E C,° (SZ'; RN) such that II-,D IICo(nr;mN) 5 1. If 0 < e < dist (SZ', 8t ), then by (13.9), supp,pF C (0')E := {x E SZ : dirt (x, St') < E } C SZ,
and so (13.10) should be replaced by IDuEI (f') <_ IDuI ((12')E) .
Letting e -+ 0+ and using Proposition B.9, we conclude that lim sup IDuE I (ci') C IDuI (?) = IDuI Z__+04
since by hypothesis IDuI (80') = 0.
Remark 13.11. If SZ = RN, then Z = RN, and so u£ -r u in L1 (RN) and by (13.5), (13.14)
lim J IDuEI dx = IDul
E-4o+
(RN).
N
Note that the previous proof continues to hold if u E L'10C (RN) and its distributional gradient Du belongs to Mb (RN; RN). The only difference is that now u£ -> u in Li° (RN).
13.2. Approximation by Smooth Functions
383
We observe that if u = XE, then (13.14) gives lim
a_.0+
IV (XE)EI dx = P (E).
N
This identity can be used to define the perimeter of a set. This approach was taken by De Giorgi [46], [45], who considered a somewhat different family of mollifiers.
Exercise 13.12. Let f1 C RN be an open set, let u E BV (0), and let {u,,} C BV (11) be a sequence of functions converging in L L (Ii) to u and such that IDul (fl)
n I Dun.I (s1) .
Prove that D2u j Diu in Mb (0) for all i = 1, ... , N and that IDul (K') = n-oo lien IDu,,l (n') for every open set 11' C Il, with I DuI (as1' fl n) = 0.
Exercise 13.13. Let S1 C RN be an open set and let u E BV (fl). Fix
1
(i) Prove that IDauI (SZ) = sup {j OdDiu : 0 E C,, (0), II0IICo(a) <_ 1 y . J
(ii) Prove that
J
IsueIdx
for all e > 0, that lim e-'.O
faxi I dx = I D=uI (11) 1!2u-'-e
and that for every open set fl' C S1, with dist (S1', as1) > 0 and ID:uI (an') = 0, lin+ o fn'
I axe I
dx = l
(iii) Prove that if u E Lt1'0 (RN) and its distributional gradient Du belongs to Mb (RN; RN), then aue
lim RN f axi e-0+ I
I
dx = IDsuI (RN).
13. Functions of Bounded Variation
384
Exercise 13.14. Let Q C RN be an open set. Prove that if u E BV (f2)
and¢ECO°(f),then ¢uEBV(f)and ID (Ou) I (n) <- IIfiIICo(c) IDul (a') + IIVOIIc0(c;RN) IIUIILI(01)
where n' C f2 is any open set such that supp 0 C n'. We now turn to the proof of Theorem 13.9.
Proof of Theorem 13.9. Since the measure IDul is finite,
lim IDul n {x e n : disc (x, ffl) >
j+00
1,
IxI <j
} = 0. 111
///
Fix vi > 0 and let jo E N be so large that (13.15)
IDuI [ SZ \ E`
for all j > jo. For i
ni
{x E Q : dist(x,on) > , Ixl <j }} < n
N define 1
{x E n : dist (x, 8 )) > ?0 + ., IxI < jo + i}
.
As in the proof of the Meyers-Serrin theorem, consider a smooth partition of unity F subordinated to the open cover {ni+1 where n-1 = no :_ 0. For each i E N let iii be the sum of all the finitely many 1; E F such that supp O C S2i+1 \ iZi_ 1 and such that they have not already been selected at previous steps j < i. For every i E N find ci > 0 so small that supp (Aria),,{ C ni+1 \ ni-1 and (13.16)
Jo
I('Oiu)e;
- 4Piul dx <
Define
2
,
jI(tV4/Ji)e. -u0(;I dx <
17
2i'
00
vn
(omeD i=1
where we have extended each (i/iiu),, to be zero outside its support. As in the Meyers-Serrin theorem we have that v,7 E C00 (St) and (see the proof of (10.7))
in IT` -
vrl dx
Thus, v,t -+ u in L1(1?) as ri Next we prove that (13.17)
lim sup IDunl (n) < IDul (n) n- 00
13.2. Approximation by Smooth Functions
385
Let ' E CC° (fl; RN) be such that II'(DIIco(n;RN) < 1. Since by the previous exercise Oju E BV (St) for every i E N, by (13.7) with ib u in place of u we have in (,O u)E, div 4) dx =
J in
Oju div E; dx,
and so, using the facts that the support of is compact and that the partition of unity is locally finit e, we have that (13.18)
jvidiv1Idx =
di v 4) dx =
,
00
r
Oc'
J Pju div fiE, dx
00
i=1
J
u[div(
s1
Since II1GiE; Ilco(s1;RN) < 1 by (13.8), it follows from Exercise 13.3 applied
to 1 and to SZi+1524_1 \judiv(tPiei) (see also (10.5)) thatjudiv(tPie) 11 =
dx +
dx i=2
00
< IDuI (52) + E IDuI (SZi+1 \ ii-1
-
i=2
Using telescopic series, the right-hand side of the previous inequality may be bounded by IDuI (52) + 3 IDuI (11 \ 01) < IDuI (fl) + 3,q
by (13.15), and so 11 < IDuI (1) + 371.
(13.19)
To estimate 12, note that by Fubini's theorem 00
(uVbi)E, . 41 dx = - > ( [(uV j)E; - uVt/, ]
12
i=1 J
i=1 JJ
dx,
00
where in the second identity we have used the fact that E VO, = 0 by i=1
(10.3) and the local finiteness of the partition of unity. We now use (13.16) and the fact that II"I IIco(s1;RN) : 1 to conclude that 12 < ri, which, together with (13.18) and (13.19), yields
'ci
vi div
dx < IDuI (5Z) + 3ri.
Taking the supremum over all admissible 0 and using Exercise 13.3 gives IDv,7I (n) <- I DuI (sZ) + 3n.
It now suffices to let ri -> 0+ to obtain (13.17).
13. Rmctions of Bounded Variation
386
To prove the converse inequality, note that, since vn 17 - 0+, by Exercise 13.3, I Dul (Sl) < lim inf I Dv,71(a)
u in L' (Sl) as
.
0
This concludes the proof.
Remark 13.15. Note that the previous proof continues to hold if u E Lip (Q) and its distributional gradient Du belongs to Mb (fl; RN). The only difference is that now {un} C Coo (Sl)nWW
(St) and ue - u in L oc (fl).
Exercise 13.16. Prove that the sequence constructed in Theorem 13.9 has the property that lim e-.o+
Oun
fn 8xi
dx =
ID2uI (11)
for every i = 1, ... , N.
13.3. Bounded Pointwise Variation on Lines In Theorem 10.35 we have seen that if a function u belongs to W1,P (Cl), then
it admits a representative that is absolutely continuous on GN-1-a.e. line segments of Cl that are parallel to the coordinate axes and whose first-order (classical) partial derivatives agree LN-a.e. with the weak derivatives of u. In this section we prove a similar result for a function in BV (Cl). In what follows, we use the notation (E.2) in Appendix E. Given an open rectangle R C RN, i.e., R = (ai, bi) x ... x (aN, bN),
following (E.2), we consider the rectangle R to be the Cartesian product of a rectangle RR C RN-1 of the variable x; and of an interval Ri C R in the xi variable and we write
R=R,xRj. Given x E RN-1 and a set E C RN, we write (13.20)
Exis ._ {xi E R : (xii, xi) E E}
.
Moreover, if v is a real-valued function defined GN-a.e. on an open set fl C RN-1 for which Cli. is nonempty we let essVarnx, v (x'j, ) RN, for all xi E be the essential pointwise variation of the function xi E Sle, '- v (xz, ) (see Definition 7.7 and Exercise 7.9). If v is Lebesgue integrable in Cl, with a slight abuse of notation, if is empty, we set
v (xi, xi) dxi := 0, f2x1
13.3. Bounded Pointwise Variation on Lines
387
so that by Fubini's theorem
L Finally, for every set F in some Euclidean space R" and for every e > 0 we set
FF :_ {y E F : dist (y, 8F) > E}
(13.21)
.
Let S2 C RN be an open set and let u E BV (1). Setting A := Du E Mb (S2; RN), by Theorem B.77, the vectorial measure A can be expressed in the form A (E) = f P (x) dx + \s (E)
(13.22)
E
for every Lebesgue measurable set E C RN, where T E Li (t1; R')measure. and A8 is a vector-valued measure singular with respect to the Lebesgue Here IF is the Radon-Nikodym derivative dD u of the vectorial measure Du with respect to the Lebesgue measure GN restricted to Q. The following result shows the relation between the weak derivatives of a function in BV (f2) and the (classical) partial derivatives of its representative (when they exist).
Theorem 13.17. Let 11 C RN be an open set, let u E BV (11), and let 1 < i < N. If v is any function equivalent to u for which the (classical) partial derivative
(x) exists in R for GN-a.e. x E S2, then ay
axi (x)
(x)
for GN-a.e. x E fl, where I is the vectorial function given in (13.22). Proof. Consider a convex function f : R -, 10, oo) such that f (e + t) :5 f (s) + ItI
(13.23)
for all a, t E R. We claim that for every rectangle R whose closure is contained in fl, (13.24)
JR
f (i!-) xi
Oadx. ,p+ JR, I\ axi
dx < lim inf
13. Functions of Bounded Variation
388
To begin with, for every function 0 E CO0 (R), for x = all It > 0 sufficiently small one has
f (0 (x;, xi + h) - p (x;, xi))JJ f
(X!" R)
f
h
xi) E R and for
del
8x1
i
xi+h
ff
f ! ax. (x
L.
`
,
s)JJ ds
s
by Jensen's inequality. Integrating this inequality along the interval (Rj)h (see (13.21)) and using Tonelli's theorem yields
f
(xi, xi + h) - 0 (214 -Ti) \
dxi
h
1
<_
Ia
f
f(R,)h fxj
s) ] dsdxi
(x;,
fa
8x1
///
fs
=
Ja-h
h
R.;
JR,
f (a8x1 (x
axi,
axi\ ,
s))
duds
8)
ds.
Integrating over (RR)6 and using Tonelli's theorem, one then obtains
x+ h) -
fRh' ( Setting letting a
x)1 JJ
h
(xj,
<-
J
. 8x1
R
ue in the previous inequality, replacing R with R, and then 0+, we deduce from Fatou's lemma that
(!())
r f (iz(xx+ h) - u (x, xi))
dx < lim inf J f dx. e- oJRe 8x1 Since v is a representative of u, the previous inequality may be written as !L,
h
Rh
J
h
f
h
J
M dx < limanf
J
+
Rs
a46s dx. f ((x))
JJJ
If we now let h - 0+ and use Fatou's lemma, we obtain (13.24). Next we prove that (13.25)
f (1) dx < Rwhere
jf
(I') dx + i.i i (R),
is the total variation measure of A!. By (13.6) and (13.22) for IA xERe we have
axi (x) =
jc(xv)'Iui(Y) dy + f Pe (x - y) deli (y) .
13.3. Bounded Pointwise Variation on Lines
389
Therefore by virtue of (13.23) and Jensen's inequality
f \ a i (x)/ < f \JS coe (x - y) wi (Y) dY) + I f w.. (x -1l) dAl (Y) 5 f coe(x - y)f (Wi(y))
dIA I (y)
Integrating over RE and applying Tonelli's theorem yields (13.25). Combining (13.24) and (13.25), we obtain (13.26)
ff (p-) dx <
dx + JA (R)
.
By Besicovitch's derivation theorem (see Theorem B.119) lim Iii I (Q (xo, r)) = 0
rN
r-+0+
for GN-a.e. xo E Q. Let xo be any such point and assume that xo is also a Lebesgue point for f (n-) and f (W2). By replacing R with Q (xo,r) in (13.26), dividing by rN, and letting r - 0+, we obtain (13.27)
f (8xi (xo)) 5 f (Ti (xo))
Now choose f to be the function
f1(t).
_ f et
if t < 0,
t+1 ift>0.
Since f1 is monotone increasing, it follows from (13.27) that - (x) < Ti (x) for GN-a.e. x E Q. Similarly, setting f (t) := f1(-t), we find that (x) > Wi (x) for GN-a.e. x E St.
Next we prove that Theorem 10.35 continues to hold for BV (St).
Definition 13.18. Let f2 C RN be an open set. A function u : St -- R is called of bounded pointwise variation in the sense of Cesari if there exist an equivalent function v defined GN-a.e. in 12 and N nonnegative functions V1, , VN E Ll (RN-1) such that (13.28)
essVarQs, v (x:, ) < V (x=) i
-
for all xi E RN-1 for which Sty' is nonempty and for all 1 < i < N.
Theorem 13.19 (Serrin). Let ft C RN be an open set and let u E L' (St). Then u E BV (St) if and only if u is of bounded pointwise variation in the sense of Cesari. Moreover, if u E BV (ft), then u has a representative ii that admits (classical) partial derivatives GN-a.e. in f2 and Vu (x) = IY (x) for GN-a.e. x E St, where T is the vectorial function given in (13.22).
13. Functions of Bounded Variation
390
We begin with some preliminary results, which are of interest in themselves.
Lemma 13.20. Let R C RN be an open rectangle, let i E {1, ... , N}, and let u : R -r R be a Lebesgue measurable function that is monotone on
RCN-1-
a. e. line parallel to the xi axis. Then the (classical) partial derivative UX-i exists and is finite LN-a.e. in R. Moreover, is Lebesgue measurable.
Proof. For x =
xi) E R and n E N we define the upper right (respec-
tively, left) sequential derivative
D u (x) := lim sup
t(x;,xzfn)-u(x;,xz) n
and the lower right (respectively, left) sequential derivative
Dfu (x) .= lim inf
u(x'- xi± n1) -u(xz,xi)
n-'ao
''
n
Then the four functions D±u and D±u are Lebesgue measurable. Moreover, by Lebesgue's theorem (Theorem 1.21), for GN-1-a.e. x E R; we have that a (xi, xi) exists and is finite for ,C1-a.e. xi E R. By Tonelli's theorem D±u and Dfu are finite sCN-a.e. in R and
J I D±u (x) - Dtu (x) I dx
=
JR;
I Dt. (x;, xi) - D±u (xi, xi) I dx;dxi = 0. Rf
It follows that both limits lim n-'oo
u(xi'xi+n)-u(xs,xi) ,
limn
u( 'x:
n)-u(Xi'xi) -n
n
exist in R and are equal LN-a.e. in R. Let v (x) be the common limit where it exists; otherwise put v (x) := oo. Let E be the set of points x = (x'j, xi) E R such that v (x) is finite and u (xy, ) is monotone. Note that Vv (R \ E) = 0.
If x = (x;, xi) E E, with, say, u (x=, ) increasing, and n1 < h < n, then
u x" xi + n+m ) - u (xi, xi) < u (x{, xi + h) - u (xi, xi) n+1
h
<
u(4xi+n) -u(xi,x,) 1
n+1
13.3. Bounded Pointwise Variation on Lines
391
Since both the right- and left-hand sides of the above inequality tend to v (x) as h - 0+, we conclude that u (x, xi + h) - u (x;, xi) = v (x) . lim h
h-.0+
Similarly, we have that
u (xy, xi + h) - u (x;, xi) = v (x) h h-.oand so (x) = v (x), which shows that exists and is finite CN-a.e. in R. Moreover, 4 is Lebesgue measurable. lim
Lemma 13.21. Let R C RN be an open rectangle and let u be a Lebesgue integrable real-valued function defined LN-a.e. in R and such that for some variable xi, 1 < i < N, and for all x'z- E R;, (13.29) essVarR{ u (x4, ) < V (zO for some nonnegative Lebesgue integrable function Vi defined in R. Then there exist two Lebesgue integrable functions ul and u2 on R, each increasing in the variable xi, such that
= ul (x) - u2 (x) for £N-a.e. x E R. In particular, the (classical) partial derivative 11
u (x1
exists
and is finite RCN-a.e. in R.
Proof. We construct a function v equivalent to u but with fewer discontinuities. Let S' be the set of points x; E R' such that u (x'j, -) is defined for G1-a.e. xi E R1 and Vi oo. By hypothesis GN-1(Rz \ S') = 0. For every x E S', essVarR{ u (2<, ) < 00
by (13.29) and the definition of S'. Hence, by Theorems 7.2 and 7.8 we may find a right continuous function w such that
w (x;, xi) = u (x;, xi)
(13.30)
for G1-a.e. xi E Ri
and VarR; w (x', -) = essVarR, u (x4, ) < 00For x = (x;, xi) E R define
v (x) :=
(13 . 31)
\ S') = 0, it follows by (13.30) and by Tonelli's theorem that v (x) = u (x) for GN-a.e. x E R. By the completeness of the Lebesgue measure, the function v is Lebesgue measurable. Since
£N-1 (Rt,
w (x;, xi) if x; E S', ot herwise . 0
13. Functions of Bounded Variation
392
Now let a E R, be such that
f,Iv(xa)Idx
xi) if xi ? a,
Var[a,x,j v .
- V&r[xi,aj v (4, xi)
if xi < a.
We claim that V is Lebesgue measurable. To see this, denote by Pk the partition of the interval R, by k + 1 equally spaced points
tO
and for x = (xi, xi) E R define Vk (x', xi) := Iv (xi, tj) - v (xi, tj-1) I sgn (tj - a), where the sum is extended over all partition points tj which lie in the interval of endpoints a and xi. Since v (x;, ) is right continuous, it follows by Exercise 2.49 that
V (xt, xi) = lira Vk (xi, xi) for all x = (x;, xi) E R. Thus, to prove the measurability of V, it remains to show that each Vk is Lebesgue measurable. This follows from the fact that, for each x; fixed, the function Vk (x;, ) is a step function. More precisely, for xi between any two consecutive points of Pk we have that Vk (x'i,xi) = finite sum of Lebesgue measurable functions of 2. Hence, we have shown that V is Lebesgue measurable. We can now define (13.32)
ul (x) := 2 [V (x) + v W),
u2 (x) := 2 [V (x) - v (x)]
for x = (x;, xi) E R. The functions ul and u2 are Lebesgue measurable, increasing in the variable xi, and u = ul - u2 GN-a.e. in R. Furthermore, Iu1(x) I + Iul (x) I < (varRui
+ (VarJ, u2 (xi, .) +
2 Iv (xi, a)
= VarR, v (xi, ) + I v (xi, a) I < Vi (xi) + Iv (x2, a) I, and so ul and u2 are Lebesgue integrable.
The last part of the statement follows by applying the previous lemma to the functions ul and u2. 11
13.3. Bounded Pointwise Variation on Lines
393
Remark 13.22. Note that by Exercise 2.20 for every E R,, for every xi E Ri the functions u1 (x;, ) and u2 (e, -) defined in (13.32) cannot both be discontinuous at xi. We now turn to the proof of Theorem 13.19.
Proof of Theorem 13.19. Step 1: We prove that if u is of locally bounded pointwise variation in the sense of Cesari, then u E BVi0 (SZ). Let v, V1,. .. , VN be as in Definition 13.18. For E Cc' (SZ) and for every
fixed i, 1 < i < N, set L (0) :=
ju±dx. axi
The functional L is linear. We claim that it is continuous with respect to the topology of CO (fl). To see this, let us first observe that, since essVara , v (x;, ) < Vi (XD
for all xi E RN-1 for which 1L is nonempty, for LN_ 1-a.e. x E RN-1 we have that Vi (x4) (and, in turn, essVarR{ v (x=, )) is finite. Thus, by Theorem 7.8 for GN-1-a.e. xi E RN-1 for which f e is nonempty the function v (xi, )
belongs to BV (f
that
), and so there exists a measure a..,,, E Mt,
(f) such
_ .1., and I
I (i ,)
V (x,) ,
= essVarcx, v
again by Theorem 7.8. It follows that L (0)
- II u8xi dx = m"_, f u.- dxsdx` RN-1
,
dA dx,
and so L ([/7) I < max 1,01
n
Vi dx;.
JRN1 Since L is continuous on Co (SZ), by the Hahn-Banach theorem (see Theorem
A.30) it may be extended to a continuous linear functional on Co (f2). But then by the Riesz representation theorem in Co (f') (see Theorem B.114) there exists a finite Radon measure Ai E Mb (ti) such that
L (O) = - fn O dai
fn
and jail (SI) = I1LlI(co(n))' <_ f N_, Vi (xt) dx=.
13. Fbnctions of Bounded Variation
394
This shows that u E BV (fl). Step 2: Assume that u E BV., (fl). We prove that u is of bounded pointwise variation in the sense of Cesari. Let {rpe}E>O be a sequence of standard mollifiers and define uE := u * VE in StE. Set
E
(13.33)
x E fl : lim uE (x) exists in R
I
and
if x E E,
lim uE (x)
v (x) :=
(13.34)
E- O+
otherwise.
0
Since {uE} converges pointwise to u at every Lebesgue point of u by Theorem
C.19, we have that E contains every Lebesgue point of u. Moreover, since by Corollary B.122 the complement in 1 of the set of Lebesgue points has Lebesgue measure zero, it follows that GN (St \ E) = 0. This shows that the function v is a representative of u. Let i be a fixed integer, 1 < i < N. By Exercise 13.13 we get (13.35)
J
6
dx < I Daul (fl) .
I
We now take e := n and for simplicity we set un := ui,,, (1j ) On
(c)
,
and so on.
En
For x; E RN-1 define (13.36)
." axi
Vi (xi) := lim f J n-oo (n
xi) dxi I
if Sl., is nonempty and Vi (xt) := 0 otherwise. The function Vi is Lebesgue measurable on RN-1. By Fatou's lemma,f(n.,). Fubini's theorem, and (13.35), I V (x2) dx'i < lim inf n-oo RN-1 RN-i
J
=1nm fJ
l
Stn
(x , xi)
I
aunldx
Hence, (13.37)
J N-i
Vi
dxZ < IDiul (Sl)
Since LN (11 \ E) = 0, by Fubini's theorem (13.38)
G1(cu,\E1)=0
.
I
dxidx;
13.3. Bounded Pointwise Variation on Lines
for
395
LN-1-a.e. x E RN-1 for which f2x' is nonempty. Fix any such x E
RN-1
and let Ri C R be any maximal open interval such that {x } x Ri C fl; e. We claim that essVarj, v (.,xi) < V (xi) . (13.39)
By (13.38), (xi, xi) E E for G1-a.e. xi E Ri. Thus, to prove (13.39), it is enough to consider partitions contained in EE' (see Definition 7.7 and Exercise 7.9). Let
inf Ri < to < t1 <
< tk < sup Ri,
where tj E Ex, for all j = 0, ... , k. Then [to, tk] C (Ri),t for all n sufficiently large, and so, by the fundamental theorem of calculus,
t-au
In (x, tj) - un (x, tj-1) I <
J'' _
j=1
J
a (x, xi) dxj=1
1
Oxi
Letting n -' oo, by (13.33), (13.34), and (13.36) we get k
j=1
IV (xi, ti) - v (xi, ti-1) I : V (xi)
.
Taking the supremum over all admissible partitions yields (13.39). By eventually redefining Vi to be infinite on a set of LN-1-measure zero, we have proved that (13.28) holds. This shows that u is of bounded pointwise variation in the sense of Cesari. Step 3: We prove the last part of the theorem. Since the function v defined in (13.34) satisfies the hypotheses of Lemma 13.21, we may find two Lebesgue integrable functions vl and V2 on R, each increasing in the variable xi, such
that v (x) = v1 (x) - v2 (x) for CJ-a.e. x E R. Let S' be(( the set of points x; E R; such that v1 (xi, xi) = EllM (vl )E (xi, xi)
,
v2 (xi, xi) = Ely0+ (v2), (xi, xi)
for G'-a.e. xi E Ri. Note that GN-1 (Ri \ S') = 0. Fix (xa, xi) E S' x R1. We assert that if the limit lim (vi), (x;, xi) does E0+ not exist, then the limit lim (v2)E (x;, xi) does exist. Indeed, since v1 (xi, ) E-0+ and (v1)E (x;, ) are increasing, if the limit lim (vi), (x;,xi) does not exist, E0+ then (why?)
-
vj (xi, xi) < limEyO+ inf (vI )E (x;, xi) < lim sup (v1)E (X" xi) < vl (xi, x:) C-O+
13. Functions of Bounded Variation
396
Thus, v1 (x;, ) is discontinuous at xi. But then by Remark 13.22, v2 (xs, ) must be continuous at xi, and hence 412 (xz1 xi) = eliO (V2) e (xi, xi)
We now define the function u (x) := lim sup i (x) E R. The function ti is a precise representative of u and is independent of i. We also introduce the functions ul and t12 by means of lim sup (vi (x;, xi) if x E S', a-o+ u1 (x) otherwise, V (x)
l'im iuf (v2)e (x'i, xi) if xi E S',
U2 (x)
otherwise.
0
Then ill and 4i2 are equivalent to v1 and v2, respectively, and so they are increasing on GN-1-a.e. lines parallel to the xi axis. Furthermore from the identity ue = (VI - V2)e = (vi),,. - (v2)e
and the assertion of the preceding paragraph it follows thatl
u (x) = ul (x) - u2 (x)
for all x E R. By Lemma 13.20 the partial derivatives s and a exist
0
and are finite £N-a.e. in R. The same is true for
Exercise 13.23. Let u E WO (R+) and v E
W1,1 (RN). Prove that the
function uw . RN -> R, defined by w
(x)
U (x) { V (x)
if xpr > 0, if xN < 0,
belongs to BV (RN).
Exercise 13.24. Let 0, 0' C RN be open sets, let' :1Z' --+ 0 be invertible, with fi and (D-1 Lipschitz functions, and let u E BV (1k). Prove that uofi E BV (V). 1The key point here is that given two sequences of real numbers {a_} and {b, ), in general one can only conclude that liim sup (a. + bn) < lim sup nn + Jim sup b. ,
n-oo n-00 ft-00 with the strict inequality possible. However, if one of the two sequences converges, say {an}, then lim sup (a., + 84,) = lim an + lint sup b,,. n-eo n-*o n-cc
13.4. Coarea Formula for BV Functions
397
13.4. Coarea Formula for BV Functions In this section we prove the coarea formula for a function u E BV (a). This formula relates the total variation measure IDul with the perimeter of its superlevel sets {x E 12 : u (x) > t}.
Theorem 13.25 (Coarea formula). Let fI C RN be an open set and let u E L10C (Sl). Then
V(u,1l)=J P({xE 1 : u(x)>t}, 12) dt. R
In particular, if u E BV (Q), then the set {x E 12 : u (x) > t} has finite perimeter in 12 for G1-a.e. t E R and (13.40)
IDul (1l) = J P ({x E 12 : u (x) > t}, 11) dt. R
We begin by proving the theorem for linear functions.
Lemma 13.26. Let u : RN - R be an affine function of the form
xERN, where a E R and b E ]RN. Then for every Lebesgue measurable set E C RN the function
t E R - RN-1 (E n u 1({t})) is Lebesgue measurable and (13.41)
IbI GN (E) =
J
RN-1 (E n u-1({t})) dt.
Proof. If b = 0, then there is nothing to prove, since both sides of (13.41) are zero. Thus, assume that b 54 0. Hence, u-1 ({t}) = {x E RN : a + b . x = t} is the translate of a hyperplane in RN, and so the measure xN-1 restricted to u-1 ({t}) coincides with the measure RCN-1. It follows by Tonelli's theorem
that the function
t E R f-, ?LN-1 (E n ul ({t})) is Lebesgue measurable. Since GN and 7-1N-1 are both rotation invariant, without loss of generality we may assume that n = e1. By the change of
13. Functions of Bounded Variation
398
variables t = JbI a + a we obtain
(En u-1 ({t})) dt = IbI = IbI
futxN-1 (En u-1({jbI s + a})) ds
J %N-1 ({x E E : xl = s}) ds R
= IbI J LN-1({x E E : xl = s}) de R
_ JbI LN(E), where in the last inequality we have used Tonelli's theorem.
Remark 13.27. Note that Vu - b. Exercise 13.28. Let fl C RN be an open set and let u E LloC (Sl). Prove that the function
u(x)>t},Q) is Lebesgue measurable.
We are now ready to prove Theorem 13.25. In what follows, for t E R we set
SZt:={xEft:u(x)>t}. Proof of Theorem 13.25. Step 1: We claim that if {u,,} C LbC (St) converges to u in Li C (f2), then there exists a subsequence (u,,,} of {u,,} such that P (flt, fZ) : lim inf P (S2t k, n) k-*oo
f o r G1-a.e. t E R and JR
P ((Zt,11) dt <- lim inf J P (SZt k, f2) dt, k-boo
R
where Stt := {x E ft : un (x) > t}. Indeed, for every x E SZ and u E N,
f
IXn, (x) - Xn, (x) I dt
= Jmin{u,.(s),u(x)}
dt = Iun (x) -,u (x) I .
Hence, by Fubini's theorem, for every ft' CC f2,
jf
,
Ixn, (x) - xn, (--)I dxdt = fol IUn (x) - u (x)I dx -r 0.
By considering an increasing sequence fj' / ft, with fZj CC fl and using a diagonal argument, we may find a subsequence of {u,, } such that Xni k
Xns in
L'10i
(11) for
1-a.e. t E R. It follows by Exercise 13.3 that
p (ftt, fZ} < lim inf p (SZi k, ft) k-+oo
13.4. Coarea Formula for BV Functions
399
for G1-a.e. t E R. In turn, by Fatou's lemma, P (nt, Cl) dt < lim inf k-1oo
JR
f
P (f k, Z) dt.
R
Step 2: We prove that if u is piecewise affine in the sense of Definition 10.32, then
I. ID (xn,)I (f) dt = f Ivul dx. To see this, let
be N-simplexes with pairwise disjoint interiors e
such that the restriction of u to each Ai is affine and u = 0 outside U a;, 1=1
O;, where asERand bi ERN. Then by the previous lemma, applied to the affine function ui (x) = ai + bi x, x E RN,
f
IDul dx =
Ibil £N (CZ n o;) _
1-IN-1
(fZn, 1nu-1 Qt})) dt.
a_1 fIR
i_1
Since by Exercise 13.5 (note that in Cl n a; the function u coincides with the smooth function v (x) := a; + bi x, x E RN) NN-1 (a n Ai n u-1 ({t})) , ID (xct)I (l n Di) = we have
E jnN_1 () nt1nu 1({t})) dt=
jlD(x)l(ni) dt i=1
= L1D(x)1(c1) dt. Step 3: We prove that if u E coo (CZ) n W,.',' (CZ), then
fit
P A, fZ) dt < f lVul dx.
It suffices to assume that fn Ioul dx < oo. Fix an open set Sl' CC Q. By Remark 10.34 we may find a sequence C W1,1 (CZ') of piecewise affine functions such that u,, -> u in W 1,1 (CZ'). By the previous two steps (with f? replaced by CZ') we may find a subsequence (not relabeled) of {u,,} such
that JR D 7rsa,) I (CZ') dt < lim oof
j ID (% (n% } I {St') dt
_ lim. f IVu,,l dx = Sz,
f
IVul dx <
f
IVul dx.
13. Functions of Bounded Variation
400
Since Xn= = Xn= in 12', we have that
ID
(X.,)
JID(xc)I (1') dt <
f
I (S2')
= ID (Xn,)I (12'). Hence,
iVul dx.
CC fl, in the
By considering an increasing sequence fll J` 12, with 1
previous inequality and using the Lebesgue monotone convergence theorem, we obtain the result.
Step 4: We prove that for u E Ll10 (Sl), P (Slt, f2) dt < V (u, fl)
.
JR
It suffices to assume that V (u, Sl) < oo. Hence, by Exercise 13.3 we have that the distributional gradient Du of u belongs to Mb (S2; RN). By Theorem 13.9 and Remark 13.15 there exists a sequence C COO (12) f1 W1 (Sl) such that u,, --+ u in Li0 (S2) and lim
f IVu,,I dx = IDuI (0).
,n-+00 Q
By Step 1 and Step 3 applied to each function u,,,
f I D (xn,)I (0) dt < lim:°uf f ID (xn-) I (S1) dt < nim
Step 5: We claim that for every
I
f
I VunI dx = IDuI (fl).
si
E C, ° (12;RN) with II,(PIICo(n;RJv) < 1,
f
u div -Z dx = J
JR n
div 15 dxdt.
To see this, assume first that u > 0. Then for CN-a.e. x E 0, 00
Xsa, (x) dt,
U (X) = 0
and so, by F ubini's theorem 00
u (x) div
(x)
dx =
J°xa= (x) div r 00
=J
f
(x)
Xc, (x) div 415 (x) dxdt
f fn 00
=
div ' (x) dxdt.
o
Similarly, if u < 0, then for £ "-a.e. x E Sl, U (X) =
f
to 00
dtdx
(xn, (x) -1) dt,
13.5. Embeddings and Isoperimetric Inequalities
401
and so
Ju(x)div4'(x) dx =
f f°(x(x) -1) div 4' (x) dt dx Z
l J
=
(X., (x) -1) div 4 (x) dxdt 0
=J
div 4 (x) dxdt, 0o
Izt
where we have used the fact that f. div 4' dx = 0 by the divergence theorem
and the fact that - E q° (fi; RN). In the general case, write u = u+ - u-. Note that for t > 0,
Sit={xEf2: u(x)>t}={xE0: u+ (x)>t}, while for t < 0,
fit={xEfi: u(x)>t}= {xEfl: -u (x)>t}. Hence,
If,
u (x) div 4' (x) dx = / (u+ (x) - u (x)) div 4' (x) dx r
= JR fl
div 4) (x) dxdt,
which proves the claim.
Step 6: We show that for u E Lj (fi), V (u, fi) <
jP(fitfi) dt.
Let t r: C°° (0; RN) with II`'IIcc(n;RN) : 1. By the previous step and the definition of the variation of the perimeter of fit in fl,
J
u div iDdx = SZ
JR
j div 4' dxdt < j P (fit, f2) dt. ae
R
This concludes the proof.
O
Remark 13.29. We refer to [54] for a different proof of Step 3, which does not make use of piecewise affine functions.
13.5. Embeddings and Isoperimetric Inequalities In this section we prove that the Sobolev-Gagliardo-Nirenberg theorem continues to hold for functions of bounded variation. We recall that for N > 2, N 1* = N-1'
13. Functions of Bounded Variation
402
Theorem 13.30. Let u E LloC (RN), N > 2, be a function vanishing at 1
infinity such that its distributional gradient Du belongs to Mb (RN; RN) Then them exists a constant C = C (N) > 0 such that
(LNIU(x)I' dx}
<_ C IDUI (RN) .
In particular, BV (RN) is continuously embedded in Lq (RN) for all 1 <
q<1`.
Proof. Consider a sequence of standard mollifiers {ipJe>U and define uE u * cpE. By Lemma 13.10 and Remark 13.11, uE -> u in L'10C (RN) and lira J
e0+ RN
IVu6j dx = IDuI (RN).
Applying the Sobolev-Gagliardo-Nirenberg embedding theorem to each u,, we obtain Jf
IuE (x) 1'*
dx' 1
f
IpuE (x)I dx. < c JRN RN
It now suffices to,flet s - 0+ and to use Fatou's lemma.
Definition 13.31. An open set fZ C RN is called an extension domain for the space BV (11) if there exists a continuous linear operator
e:BV(fZ)- BV(RN) such that (1) for all u E BV (fl), E (u) (x) = u (x) for Vv-a.e. x E fl, (ii) E (u) E 111.1 (RN) whenever u E W "1 (fl). Next we prove that the Rellich-Kondrachov theorem continues to hold in BV (0).
Theorem 13.32 (Rellich-Kondrachov). Let St C RN be an extension domain for BV (11) with finite measure. Let {un} C BV (fl) be a bounded sequence. Then there exist a subsequence {u,,k } of {u,,} and a function u E BV (fl) fl L1' (1) such that u,,, --> u in LQ (f1) for all 1 < q < 1'. The proof is very similar to that of Theorem 11.10. The following result is the analog of Lemma 11.11.
Lemma 13.33. Let u E BV (RN). Then for all h E RN \ {0}, JRN Iu (x + h) - u (x)I dx < IhI IDuI (RN).
13.5. Embeddings and Isoperimetric Inequalities
403
Proof. Assume first that u E W1'1 (RN)f1C0o (RN). Then by Lemma 11.11
we have that forrallhCRN\{0}, JRN
f lu(x+h) -u(x)I dx < lhl J N IVu(x)Idx.
To remove the additional hypothesis that u E CO° (RN), it suffices to apply the previous inequality to uE := cp,*u, where pE is a standard mollifier, and to let e --+ 0 (see Theorem C.19 and Lemma 13.10). 0
The following result is the analog of Lemma 11.12.
Lemma 13.34. Let u e BV (RN). For k E N consider standard mollsf&evs of the form cpk (x) := kNcp (kx) , x E R"', where cp is defined in (C.8). Then JRN KU * cPk) (x) - u (x)l dx < C (k ) IDul (RN)
Proof. By Lemma 11.12 (see (11.6)) we have N
IN * Wk) (x) - u (x) l dx
< C (N) kN B((ofRN ju (x + h) - u (x)l dxdh. ,k) I n turn, by the previous lemma we get
fRS I(u * cok) (x) - u(x)I dx < C (N) kN IDul (RN) Br(0,) Ihl dh
=
C (N ) k
IDul (RN)
0 We now turn to the proof of the Rellich-Kondrachov theorem.
Proof of Theorem 13.32. The proof of the first statement is the same as that for Theorem 11.10 with the only difference that we use the previous lemma in place of Lemma 11.12. It remains to prove that u belongs to BV (St). For all i = 1, ... , N and k E N define Ax') Dfunk. Then sup I
Akb l(n)<00
13. Functions of Bounded Variation
404
for all i = 1, ... , N. Since Mb (fl) is the dual of Co (fl) and Co (ft) is separable, we may select a further subsequence, not relabeled, such that for each i = 1, ... , N, Ank - Ai in Mb (H) for some finite signed Radon measures )q,.. . , AN E Mb (fl). For every
f
dx = - a =k O dx.
unk
8xi n Letting k -+ oo in the previous equality yields
fuiLdx
- f 4 dAi,
which shows that - = Ai. Hence, u E BV (fl) and the proof is complete. As a consequence of the previous theorem, we can prove a compactness result in BV (n) (see Remark 10.45).
Theorem 13.35 (Compactness). Let Q C RN be an open set. Assume that {un} C BV (fl) is bounded. Then there exist a subsequence {unk} of {un} and u E BV (Sl) such that unk -+ u in Lioc (fl) and Diunk ` Diu in Mb (ft) for all i = 1, ... , N.
Proof. Fix an open set fll cc fl. Construct a cut-off function 0 E C°° (ft) such that 0 - 1 in fl1 and for every n E N define := r 0 (x) ui (x) if x E St, vn (x)
otherwise.
Sl 0
Since supp vn C supp 0 and the latter set is compact, we are in a position to apply the Rellich-Kondrachov theorem to find a subsequence {vn,1} of
and a function u0) E BV (fl1) such that vn,1 -+ uM in L1 (a). Using the fact that 0 - 1 in f11, it follows that un,1 -+ u(') in L' (fl1). By considering an increasing sequence of open sets flj cc flj+l cc fl such that 00
S2= UHj j=1
and using a diagonal argument (see, e.g., the proof of the Helly theorem), we may find a subsequence {unk } and a function u E BV,,, (Sl) such that unk u in L' (flj) as k -+ oo for every j E N. By selecting a further subsequence, not relabeled, we may assume that unk (x) u (x) for GN-a.e. x E I. Hence, by Fatou's lemma
J
S2
Iu (x) I dx < lim inf j IUnk (x)p dx < oo, k-'OO
13.5. Embeddings and Isoperimetric Inequalities
405
u in Q (S1), it follows by
which implies that u E L' (S2). Since Exercise 13.3 that V (u, 0) < lim inf V (u,,,,, f2) = lim inf
I (f2) < oo,
kyoo
k-too
which implies that u E BV (s1).
D
Remark 13.36. In particular, if n C RN is an open set and {u,a} C W 1,1(f2) is bounded, then it follows from the previous theorem that there exist a subsequence {ufzk} of {uf,} and u E BV (fl) such that u,,, --+ u in Li (S1) and n k 1 Dt-u in Mb (fl) for all i = 1, ... , N, where C
7 k dx, A n y (E) := J 2! axi
E E 13 (SZ) .
E
Exercise 13.37. Let S2 C RN be an open set with uniformly Lipschitz boundary. Prove that f2 C RN is an extension domain for BV (S2). Hint: Use Theorems 12.15 and 13.9.
Exercise 13.38 (Poincare's inequality). Let f2 C RN be a connected extension domain for BV (1k) with finite measure. Let E C Sl be a Lebesgue measurable set with positive measure. Prove that there exists a constant C = C (S2, E) > 0 such that for all u E BV (f2),
/r 1 J ju (x) - uEl1* dx)
< C IDul (Sl)
,
where, we recall,
uE
LN (E)
JE u (x) dr..
Exercise 13.39 (Poincare's inequality for balls). Let Q = B (xo, R) C RN. Prove that there exists a constant C = C (N) > 0 (independent of R and xo) such that for all u E BV (S1),
(flux) - u&iI'* dx l 3
< C IDul (0).
As a corollary of Theorem 13.30 we can prove the following isoperimetric inequality (see also Theorem C.13).
Theorem 13.40 (Isoperimetric inequality). Lei E C RN, N > 2, be a set of finite perimeter. Then either E or RN \ E has finite Lebesgue measure and (13.42)
min{LN(E),LN (Rv \E)}
for some constant C = C (N) > 0.
13. Functions of Bounded Variation
406
Proof. If LN (E) < oo, then XE E BV (RN), and so (13.42) follows from Theorem 13.30 applied to xE. It remains to show that if E has finite perimeter, then either E or RN \ E has finite Lebesgue measure. Consider a ball BR := B (0, R) and set u - XE- We claim that min {LN (BR n E) , LN (BR \ E)} ? <- C (N) ID (XE)I (BR).
To see this, note that UBR
_
LN (BR n E) , LN (BR)
1
LN
(BR) JBR
and so
= 1-
Iu - ugR I1* BR
1CN
BN
(BR(Bn R)E) 11*
LN (BR n E)
/RCN (BR n E)\ 1* N
+ I\ LN (BR)
11
(BR
\
E)
If LN (BR n E) < LN (BR \ E), then i
Uft
Iu - UBRI1* dx 1
>
L LN (BR)
) (LN (BR n E))
lz-
> l (CN (BR n E)) 1Z
= 2 min{LN(BRnE),LN(BR \E)}T". The other case is analogous. By applying Poincare's inequality for balls (see the previous exercise), we get that the left-hand side of the previous inequality is bounded from above by C (N) ID (XE)I (Bp), and so
2min{LN(BRnE),LN(BR\E)}1 <-C(N)ID(XE)I (BR) <- C (N) ID (XE) I (RN) .
Hence, the claim is proved. By letting R -> oo in the previous inequality and using Proposition B.9, 0 it follows that either E or RN \ E has finite Lebesgue measure.
Thus, we have shown that the Sobolev-Gagliardo-Nirenberg theorem implies the isoperimetric inequality. Next we show that the opposite is also true.
13.5. Embeddings and Isoperimetric Inequalities
407
Theorem 13.41. Assume that the isoperimetric inequality (13.42) holds for all sets with finite perimeter. Then there exists a constant C = C (N) > 0 such that 1W
(LNIU(x)11 ,
dx)
for all u E BV (RN).
Proof. Assume first that u > 0 and that u E Coo (RN) n W""l (RN). For t E R, define
At:={xERN: u(x)>t}. Then by the coarea formula (13.40) and the isoperimetric inequality (13.42), 00
IVul dx =
(13.43) JRN
J
(GN (At))' T dt.
P (At) dt >
For every t > 0, consider the function ut := min {u, t} and let
f (t)
UN ( ut
(x))1* dx
1
.
Note that f is finite, since
J (u(x))dx < tl.-1 N
f
u (x) dx + t1GN (A) < oo.
{u
Moreover, f is increasing in [0, oo) and by the Lebesgue monotone convergence theorem, (13.44)
tlim f
(t) = (JRN (u (x))' dx
while by the fact that 0 < (ut (x))" < (ul (x))' for all t E [0, 1], it follows by the Lebesgue dominated convergence theorem that (13.45)
slim f (t) = 0 = f (0).
Furthermore, for h, t > 0, by Minkowski's inequality,
0< f(t+h)-f(t)<- (¼!
-ut(x)l1.
Iut+h(xW
dx)
N
where in the last inequality we have used the facts that ut+h (x) = u (x) _ ut (x) if u (x) < t, while
lut+h(x)-ut(x)l =min{u(x),t+h}-t
13. F znctions of Bounded Variation
408
for £1-a.e. t > 0. It follows by the fundamental theorem of calculus, (13.45), (13.43), and (13.44) that
(JRN
(u (x))dxJ
= im f (t) =
i
00
r 00
Jo
f' (s) ds
(GN (A8))' ds < C f
dx.
N RN
The general case follows as in the proof of Theorem 13.30.
0
13.6. Density of Smooth Sets In this section we prove that sets of finite perimeter can be approximated by smooth sets. The proof makes use of a theorem of Sard (see [124], [1481).
Theorem 13.42 (Sard). Let S2 C RN be an open set, let u E C°° (S2), and let
E:={xES2: Vu(x)=0}. Then £' (u (E)) = 0. The proof uses the following covering result (see Theorem 1.31).
Lemma 13.43 (Vitali). Let E C RN be the union of a finite number of balls B (xi, ri), i = 1, ... , e. Then there exists a subset Z C {1, ... ,1} such that the balls B (xi, ri) with i E I are pairuise disjoint and
E c U B (xi, 3ri). iEZ
Proof. Without loss of generality we may assume that
rl>r2>...>rp. Put it := 1 and discard all the balls that intersect B (xl, rl). Let i2 be the first integer, if it exists, such that B (xi2, riz) does not intersect B (xi, ri). If i2 does not exist, then set I :_ {i1}, while if i2 exists, discard all the balls B (xi, ri), with i > i2, that intersect B (xiz, ri2). Let i3 > i2 be the first integer, if it exists, such that B (xi3, ri3) does not intersect B (xiz, riz). If i3 does not exist, then set I :_ {i1, i2}; if i3 exists, continue the process. Since there are only a finite number of balls, the process stops after a finite number of steps and we obtain the desired set I C {1, ... , e}. By construction, the balls B (xi, ri) with i E I are pairwise disjoint. To prove the last statement, let x E E. Then there exists i = 1, ... , e such that x E B (xi, ri). If i E 1, then there is nothing to prove. If i V Z, then B (xi, ri) is one of the balls that has been discarded and thus there exists j E Z such
13.6. Density of Smooth Sets
409
that rj >- ri and B (x3, r3) n B (xi, ri) # Ql. Let z E B (x3, rj) fl B (xi, ri). Then
Ix - xjI
We are now ready to prove Theorem 13.42. We use the notation (E.3) in Appendix E.
Proof of Theorem 13.42. Step 1: Assume that I is bounded and let EN :_ {x E n : Vu (x) = 0, V2u (x) = 0'..., V NU (x) = 0} . We claim that ,C' (U (EN)) = 0. To see this, fix e > 0 and let xo E EN. By Taylor's formula centered at xc we have N
u(x)= E 1aJ=O
Ira
(xo)(x-xo)"+0(Ix-x0 I N) `
= u (x0) '+' O (Ix - XOINI
as x -+ xO, and thus there exists rx0 > 0 such that <eIx-
lu(x)-u(xo)I
xoIN
for all x E B (x0, rx0) C fl. In particular, for all 0 < r < rxo,
osc u:5 2erN.
(13.46)
B(xo,r)
Fix a compact set K C EN. Since {B (x, is an open cover for K, we may find a finite subcover. By the previous lemma there exists a disjoint finite subfamily { B (xi, ) ) 1, such that n
K C U B (xi, i=1
Hence,
L' (u (K)) <
Gl (u (K fl B (xi, ri))) i=1 C2e7'rNC62.3N
aN(3)N
<62aNN,CN(n),
i=1
i=1
where we have used (13.46) and the fact that {B (xi,))
is a disjoint family. Given the arbitrariness of e > 0, we conclude that C' (u (K)) = 0. 1
13. Functions of Bounded Variation
410
Let now {K,z} be an increasing sequence of compact sets such that 00
12= UKn. n=1
Then EN fl Kn is a compact set (since u is of class C°O (i))), and so, by what we just proved, C1 (u (EN fl Ku)) = 0
for every n E N. Thus, G1 (u (EN)) = 0. Step 2: We now prove the theorem by induction on N. The previous step
shows that the result is true for N = 1. Thus we assume that the result is true for dimension N - 1 and we prove it for dimension N. Again by the previous step it remains to show that 41 (u (E \ EN)) = 0. If T E E \ EN, then there exist a multi-index a, with l al < N, and i E { 1, ... , N} such that 0
a
(13.47)
a (x)
0,
a ( 8,U (x) # 0. a
8xi
Let Ea,i be the set of points x E E \ EN for which (13.47) holds. Since
E\EN =
U
Ea,i,
a multi-index, IaI
to prove that G1 (u (E, EN)) = 0, it is enough to show that C1 (u (Ea,i)) = 0. Without loss of generality, we may assume that i = N. Fix xo r= Ea,N
and set v := 8 Q . Then v (xo) = 0 and
TXT
(xo) # 0 and thus, by the
implicit function theorem, we may find a cube Q
r r x T) = QN-1 (20, r) X (zoN-,xoN+ Cn (o, 2
and a function ip E C00 (QN-1 (x0, r)) such that Ea,N n Q (.To, r) = {x E Q (xo, r) : v (x) = 0}
C {x' E QN-1 (xo, r) : xN = 'Pp (x') } .
The union of all these cubes as xo varies in Ea,N gives an open cover for Ea,N. Since from this open cover we can always select a countable subcover (why?), it suffices to prove that C1 (u (Ea,N n Q (xo, r))) = 0.
For every x' E QN-1 (xQ, r) define the function
f (x') := u (x',ip (x'))
13.6. Density of Smooth Sets
and let E'
411
1(x')) E Ea,N fl Q (xo, r)}.
{x' E QN_t (xo, r)
Since
Vu = 0 in Ea,N, it follows by the chain rule and the definition of f that px, f (x') = 0 for all x' E E. By the induction hypothesis
£l(f(E'))=0. But
f (E') = { u (x', ? (x'))
:
(x', -0 (x')) E Ea,N fl Q (xo, r) }
= u (Ea,N f1 Q (xo, r))
and so £' (u (E,,,,= fl Q (xo, r))) = 0 and the proof is complete.
Exercise 13.44. What is the minimal regularity on u that is necessary to carry out the previous proof? The following result is an important consequence of the previous theorem:
Corollary 13.45 (Sard). Let fi C RN be an open set and let u E C°° (St). Then for £'-a.e. t E R the sets {x E St : u (x) = t} are C'-manifolds.
Proof. Let
E:={xESt: Vu(x)=0}. By the previous theorem G1 (u (E)) = 0. Hence if t E R \ u (E), then by the implicit function theorem {x E fl : u (x) = t} is a C°°-manifold.
We now present the density result mentioned at the beginning of this section.
Theorem 13.46 (Density of smooth sets). Let E C RN, N > 2, be a set of finite perimeter. Then there exists a sequence of sets {En} with smooth boundary such that GN (EnOE) -> 0
and P (En) - P (E).
Proof. By the isoperimetric inequality, either E or its complement have finite Lebesgue measure. Since P (E) = P (RN \ E) (why?), without loss of generality, we may assume that LN (E) < oo. Hence, XE E BV (RN). For every e > 0 define uE := cpE * XE, where c°E is a standard mollifier. By
Lemma 13.10, ue -' XE in L' (RN) and lim
IVuel dx = ID (XE)I (R"') =
P (E).
e.0+ N Note that since 0 < XE < 1, then 0 < ui,, < 1. By Step 1 of the proof of Theorem 13.25 there exists a sequence en - 0+ such that liminfP(A=) > P(At) n-00
13. Functions of Bounded Variation
412
for Gl-a.e. t E R, where for t E R,
At:={xERN: XE(x)>t}, At:={xERN: ufn(x)>t}. Since
At _
E ift<1, 0
if t > 1,
it follows that
lim inf P (At) > P (E) -oo
(13.48)
for £'-a.e. t E [0,1]. On the other hand, by the coarea formula, Fatou's lemma, and the fact that 0 < uEn < 1,
P (E) = lim
I VuEn I dx = lim
n-oo J fo Together with (13.48), this implies that
1
P (At) dt >
r
J
lim inf P (At) dt.
lim inf P (A') = P (E) n-oo for Gl-a.e. t E [0, 1]. Since uEn E C°° (RN), it follows by Sard's theorem that the sets
{xERN: U",(x)=t} are C°O-manifolds for Gi-a.e. t E [0, 1J. Hence, we may find t E (0,1) such
that lim inf P (At) = P (E) and the sets {x E RN : uEn (x) = t} are C°O-manifolds for all n E N. Define En := Ant. It remains to show that GN (EnLE) - 0. To see this,
note that
UEn(x)-XE(x)>t for X E En \ E, while
XE(x)-uen(x)>1-t for x EE \ En. Hence, f/ N
IUEn - XEI dx >
JEn\E
IUEn - XEI
dx+ fE\En
IuEn - XEI dX
2: tGN(En\E)+(1-t)GN(E\En), which implies that CN (EnOE) -> 0, since 0 < t < 1. This concludes the proof.
0
Exercise 13.47. Prove that in general we cannot approximate E with smooth sets contained outside E (or inside). Hint: Consider the set A given in Exercise 13.8.
13.7. A Characterization of BV (12)
413
13.7. A Characterization of BV (f2) In this section we give a characterization of BV ((2) in terms of difference quotients. This is the extension of Corollaries 2.17 and 2.43 in higher dimension. A similar characterization has been given in Section 10.5 for the Sobolev space W1" (S2), 1 < p < oo. Let fl c RN be an open set and for every i = 1, ... , N and h > 0, let
f1h,i:={xEf2: x+heiEf2}. Theorem 13.48. Let Sl C RN be an open set and let u E BV (12). Then for every i = N and h > 0, lu (x + hei) - u (x)I dx < h IDiuI (R)
J and
(13.49)
u (x + h h) - u (x)I dx = ID:uI
hli lim
(11).
nh,t
Conversely, if u E L' (S2) is such that
r
(13.50)
liminf h-Of Jflh i
Iu(x+hei)-u(x)I h
dx < 00
for every i = 1, ... , N, then u E BV (S2). Proof. The proof is very similar to the one of Theorem 10.55 and we only indicate the main changes. Step 1 of that proof remains unchanged. In Step 2 we assume that 0' satisfies the additional hypothesis IDuI (Oil') = 0 and we proceed as before to obtain (10.33), which now reads (13.51)
r 8u£ 8xi
(x) dx < limo f
Iu (x + hei) - u (x)
dx.
Inh's
We are now in a position to apply Lemma 13.10 to obtain (13.52)
I Diul (St') <
j u (x + h ) - u (x) l
h,
hm inf
dx.
By letting 0' / S2 along a sequence and using Proposition B.9, we have that (13.53)
IDiuI (St) < liminf J
h,s
Iu (x + h i) - u (x)I h
dx,
which completes Step 2.
Step 3: To prove the converse inequality, we follow the proof of Lemma 13.33 with the only exception that the vector h there should now be replaced
13. Functions of Bounded Variation
414
by the vector he;, where h > 0, and we use Exercise 13.16. Hence, we have
that
xi + h) - u
Iu
xi) I dx < h ID2uI (0) .
htlh,i
In turn,
r
lim sup h-0+ JIZh,;
lu (x + h
- u (x) I
h)
dx < IDtuI (fl),
which, together with the previous step, completes the proof of (13.49). Step 4: To prove the final statement of the theorem, let u E Ll (f2) be such that (13.50) is true for every i = 1, ... , N. We claim that u E BV (0). To see this, let 11' C :C 0. Then reasoning as in Step 2, by (13.51) we get sup 0<e
I
t' 8xi
(x)) dx < oo
for every i = 1, ... , N. Since of -' u in Ll (f") by Theorem C.19, it follows
from Theorem 13.35 that u E BV (S2') and that (13.52) holds for every i = 1, ... , N. Given the arbitrariness of 1?, taking a sequence Std CC 11 with Q, / 1, we conclude that u E BV,, (f'). Since (13.52) holds for all f1j, letting j -> oo, we obtain (13.53), which shows that u E BV (12).
0
Exercise 13.49. Let St C R'' be an open set, let u E BV (fl), and let f : R - R be a Lipschitz function with f (0) = 0 if GN (fl) = oo. Prove that f o u E BV (el) and that ID (f o u)I (92):5 (Lip f) IDuI (11)
Remark 13.50. One can actually show that a chain rule formula holds in BV (11). We refer to the paper of Ambrosio and Dal Maso [9] (see also [106] ).
Chapter .L4
Besov Spaces Living with P.Q.S, II: What are the symptoms of PQS? Symptoms include: Inability to maintain interest in research area; Advisor Avoidance; Increased web-surfing; Healthy sleepingpat-tern; Cynicism towards Academia.
-Jorge Cham, www.phdoomics.com
In this chapter we introduce the Besov spaces BS'p'0. Besides their intrinsic interest, these spaces will play a crucial role in characterizing the traces of Sobolev functions of several variables (see Chapter 15). In the literature there are several different approaches to Besov spaces. One of the most successful is by interpolation. We refer to [7], [166], and [167]. Here we define B',P,e by integration properties and in the proofs we use only the inequalities of Hardy, Holder, and Young, together with some integral identities. In what follows, we use the notation (E.2) in Appendix E.
14.1. Besov Spaces BB,P,e, 0 < s < 1 In this section we define the Besov spaces B',",9 (RN). Given a function u : RN _ 1P, for every h E R, i = 1, ... , N, and x E RN, we define (14.1)
(x) := u (x + hei) - u (x) = u
xa + h) - u (x;, xi) ,
where ei is the ith vector of the canonical basis in
RN.
If N = 1, we write
Ohu :_ flu.
Definition 14.1. Let 1 < p, 0 < oo and 0 < a < 1. A function u E Ll (RN) belongs to the Besov space B",AB (R') if C
IIUIIB..r.e(RN) := IItLIILP(KN) +'UIB,.r.a(RN) < 00, 415
14. Besov Spaces
416
N IuIBa.P.e(RN)
E (JO B
A
s
11A
` 1
i
tIILI(RN) hl+s9
N
IuIB.P.00(RN) :=
sup h>O
, III ulI
BQ,P (RN) := B"AP"P (R'v)
.
In the sequel, we will often use the notation (14.2)
ipj (h)
II,4UIILp(RN,'
where i = 1,... , N and h > 0. Remark 14.2. (i) Given a function u E L10C (R"), to see that the seminorm IUI B8.D.e(RN) is well-defined, consider a representative u of u that is
Borel measurable and for every i = 1, ... , N let wi : RN x (0, cc) -> R be the function defined by
w' (x, h) := u (x + hes).
Then w' is a Borel function, since it is the composition of i with the continuous function 9i : RN x (0, oo) - RN given by g` (x, h) x + he;,
(ii) Observe that if 0 = p = oo, then a function u E Lj (RN) belongs to B8,00,00 (RN) if and only if it has a representative in C0" (RN).
(iii) In the sequel we will often use the fact that if u E LP (RN), 1 < p <_ oo, then for every h > 0, i = I,-, N, I
'&.a ull(RN) < 2 IIIILP(RN) . L-P
This follows from Minkowski's inequality and the change of variables y = x + hei.
Proposition 14.3 (Completeness). Let 1 < p, 8 < oo and 0 < s < 1. Then the Besov space B",P.e (RN) is a Banach space.
Proof. We prove the result for 8 < oo. Let {ua} C B".P.e (RN) be a Cauchy sequence. Then {un} is a Cauchy sequence of LP (RN) and for
14.1. Besov Spaces B',pe, 0 < s < 1
417
every i = 1, ... , N the sequence of functions vn R' x (0, oo) - I8, defined by
$
v (x, h)
a
t un W,
is a Cauchy sequence in LP (RN x (0, co)), where p :_ (p, ... , p, 0) E II$N+1 (see Appendix Q. By Theorem C.38 we may find u ELT' (RN) and v' E LP (RN x (0, co)), i= 1,...,N,such that un - uinLP(RN) and v,.- v' in LP (RN X (0, 00)). Extract a subsequence {u,,,; } of {u7,} such that u1,j, (x) -+ u (x) for LN-a.e.
x E RN and v;t (x, h) -> v' (x, h) for LN-a.e. x e RN and L1-a.e. h > 0, i = 1, ... , N. It follows that for ,CN-a.e. x r= W' and LI-a.e. h > 0 the function v' (x, h) coincides with the function w1. (x, h) :=
0, u (x)
1
.
hB+s
Hence, u E Bs,P,e (RN) and IIt
- ullBe.n.e(RN) -+ 0.
Exercise 14.4. Prove that Bs"°° (RN) is a Banach space. Next we discuss the density of smooth functions in
Bs,r,e (RN).
Proposition 14.5. Let 0 < s < 1, 1 < p < oo, and 1 < 0 < oo. For any u E Ll (RN), let uE :_ SPa * u, where (pr is a standard mollifier. Then (14.3)
IueIB8.D.e(RN)
<- IuI B..y.e(RN)
for all e > 0 and (14.4)
olio IuEIBe.v..(RN) = IuIBs,D.e(RN)
Moreover, if p < oo, 0 < oo, and u E gs,n,e (RN), then (14.5)
lim o+IuE - ulB,.P.e(RN) = 0.
In particular, if p < oo and 0 < oo, then Coo (R1') fl Bs,P,O (RN) is dense in Bs,P,e (RN).
Proof. Since 4z uE _ E * / u, by Theorem C.19 we have that (14.6)
IIOs
IILP(RIN)
<
for all h>Oande>Oand (14.7)
ue IIL-V(R.V)
11A- u 11LV(RN)
as a -> 0+ for all h > 0. It follows from (14.6) that (14.3) holds. In turn, limSUP IuEIBs.v.B(RN) 6--,Q+
IulBs.v.9(RN)
14. Besov Spaces
418
To prove the opposite inequality, assume first that 0 < oo. For h > 0 and e > 0 we define fE (h)
II2nuEIILP(RN)'
h1+490
f (h)
h1+s8 FA'uIILP(RN)
Since fe (h) -> f (h) for every h > 0 by (14.7), by Fatou's lemma we have
that 00
00
10
f (h) dh = fo
00
lim fe (h) dh < lim inf
t -«0+
fe (h) dh. a
Thus, (14.4) holds.
for all h > 0 by
If 0 = oo, then since (14.7), we have that h$
III
eEm h8
LP(RN}
lim inf sup s Ilk cell N 4F-p0+ h>0 h LP(R )
for all h > 0. It follows that sup
a
a
< lim inf sup LP(RN) - e--'0+ h>0 h
I
h>o h
11'&
IIL-I(RN)
and so (14.4) is satisfied.
Assume next that p < oo, 0 < oo, and u E B',P,e (RN). By Theorem C.19(iv), (14.8)
IIA4 ne - At uII
LP(RN)
-' 0.
For h > 0 and e > 0 define the functions 1
ge (h) := h1+ 0 lohue
- a,
g (h) == h1+ae
uIIe
L P(RN)
'
I.
By hypothesis g E L' ((0, oo)), and by (14.6), Minkowski's inequality, and the convexity of the function Iyle we have that ge (h) 5 leg (h)
for all h>0. SincegE(h)->0forallh>0by(14.8),weareinapositionto apply the Lebesgue dominated convergence theorem to conclude that (14.5) holds.
14.2. Dependence of B',P,e on s
419
(RN) Exercise 14.6. Let f : R - R be a Lipschitz function and let u r= Ll 10
be such that IUIB,,p,a(N) < oc for some 0 < 8 < 1, 1 < p < oo, and 1 < 0 < oo. Prove that If o UIB5.P.e(RN) :5 (Lip f) I t6I B..p,o(RN)
14.2. Dependence of B',P,e on s In this section we prove that for 0 < t < s < 1, W1,P (RN) C Ba,P,O (RN) C Bt,P,e (1N)
Theorem 14.7. Let 0 < t < 8 <
< p < oo, and 1 < 0 < oo. Then
1,11
there exists a constant C = C (t, 0) > 0 such that IUIBa,P,B(RN) + C IIUIIL.(RN)
IUIBt.P.e(RN)
for all u E B',P,B (RN). In particular, Bs,P,e (RN) C Bt,P,e (RN).
Proof. Assume first that 1 < 0 < oo. Then by Exercise 10.54, the fact that hs < ht for 0 < h < 1, and Remark 14.2, we have r poo
hdh \ a
r`J
x h
(14.9) I
l+te 1
dh uJJO
LP(RN)
h1+ts) +
(/
h
oo
00
6
hl+ )
< (J0
dh h1+te}
g
+ 2 IIIIILP(RN)
U
1
h1+te
dh)
This proves the desired inequality when 1 < 0 < oo. When 0 = cc, it is enough to replace the integrals by suprema, precisely,
o
0 ht
sup
0
Il'ahullLnlRN)
+ >p ht
11'&#U11 LP(RN)
+2
IIUIILP(RN) .
0
This concludes the proof.
The proof of the embedding
W1,P (RN) C Bs,P,9 (RN) is very similar.
Theorem 14.8. Let 0 < s < 1, 1 < p:5 oo, and 1 < 0 < oo. Then there exists a constant C = C (a, 0) > 0 such that IkIB+.P.a(RN) !5 C IIUIIW1.P(RN)
for all u E W 1,P (RN). In particular, W 1,P (RN) C B',P,e (RN).
1 4. Besov Spaces
420
Proof. If 1 < p < oo, then by Theorem 10.55, h
(14.10)
I1Ai UIILP(RN) < h
11
axi LP(RN) .
The same inequality holds for p = oo, since in this case u has a representative u that is Lipschitz continuous by Exercise 11.46.
Assume first that 1 < 0 < oo. Reasoning as in (14.9), by (14.10) and Remark 14.2, we have (14.11) dh 0
II
a
1
IILN(RN)
00
(8
(L
+ 2 11U16(RN)
1)9)
hl
\ Ji
g
h1+8° dh1
1
+2
"U = ((113) 0) 1 II ax
( TO1
IIUIILP(RN)
IILP(RN)
while if 0 = oo, sup h8 h>0
( RN )
< sup hl-' Il Ou.
axi
0
.9u 11
axi
IILP(RN)
+sup h>1 sh11AhU l Lp(R N )
+2 IIUIILP(RN)
11
LP(RN)
0
This concludes the proof.
For p = 1 we actually have the stronger result:
Theorem 14.9. Let 0 < s < 1 and 15 8 < oo. Then there exists a constant C = C (s, 9) > 0 such that IUIB$.i.a(RN)
<_ C II uII BU(RN)
for all u E BV (RN). In particular, BV (RN) C
B8,l,e (RN)
Proof. The proof is exactly the same as that of the previous theorem, with the only difference that we use Theorem 13.48 to conclude that (14.12)
I1Ahu11Ll(RN)
< h IDiuI (RN)
.
0
1-
14.3. The Limit of Be'p'0 ass -* 0+ and s
421
14.3. The Limit of BS'"'e as s - 0+ and s -+ 1In the proof of Theorem 14.8 (see (14.11)) we have shown that if u E
W1P(RN), 1
(s(1 -s}8}B
dh
u11°
ie
LP(RN) h1+'6
0
u
< 87
aX
+ 2(1 - s) I IIUIILP(RN) . LP(RN)
Hence, lim sup 0+(s (1 - 8) 0)
2 11u16(RN)
a a MI' uIILP(RN)
while oo
lim sup (8 (1 - 8) 8) a f
\0
JJt
tIIN d+-
1
au
a
Lp(R )
IIOXiIILp(gN)
In this section we prove that the opposite inequalities hold, so that the norm in LP (RN) and the seminorm in W1'P (IAN) can be regarded as limits of seminorms in Be'p'e (RN) as 8 - 0+ and s -t, 1-, respectively.
Theorem 14.10. Let u E L110C (RN) be a function vanishing at infinity such that oo for some 0 < so < 1 and 1 < p, 0 < oo. Then u E LP (RN) if and only if lgmonf (s (1 - s) 0) A I uI B,,P,e(RN) < 00.
Moreover, if u E LP (RN), then elli (s (1 - 8) 0) 9 IUIBa.P.e(RN) = 2N IIuIILP(RN) .
Let 0 < 8 < oo and s E (0,1) and for every nonnegative Lebesgue measurable function g : (0, oo) - [0, oo) define
r
r00 9e (h)
II9IIa,8 := ` s (l - s) 8 J a
h1+&9
if0
if0=00.
Sh) e
a
dh)
14. Besov Spaces
422
Lemma 14.11. Let 0 < 0 < oo and let g : [0, oo) - [0, oo) be a Lebesgue measurable function such that 11911,0,8 < oo for some so E (0, 1). Then (14.13)
lim inf g (h) < lim inf II9II,,6 < lim sup 11911,,6 < lim sup g (h) h-.oo s_.0+ s_.o+ h-oo
and
(14.14)
Jim inf g
h-0+
(h) 11911 _< lim inf < lm sup g (h) , s6 < lim sup 11911,6 h h s-18-1-
Proof. We prove only part (14.13)1 and leave (14.13)2 and (14.14) as exer-
cises. Let f := lim infh.w g (h). Note that f E [0, oo]. If f = 0, then there is nothing to prove. Thus, assume that f > 0 and fix any 0 < 11 < f. Then there exists Al > 0 such that
g(h)>fl
(14.15)
for all h > M. Write Al
119116,8=s(1-s)e
e i+shedh+s(1-s)6
oo
hei+.hedh=:2+22.
J1
fo
for all 0 < s < so, 6
Z < s (1 - s)
and so letting s
g1+06 dh
9M(SO-,)9 f
< sos
_ (1-so)OM1sO-s)611911,0,6'
0+, we (obtain
lim Z=0. On the other hand, by (14.15), 00 1 I Z2> s(1-s)94 Im h1+sedh=(1-s)Mse el.
Hence,
liminf22 > fa. Letting f, / f gives the desired result.
0
Exercise 14.12. Prove (14.13)2 and (14.14). Exercise 14.13. Let g : [0, oo) - [0, oo) be a Lebesgue measurable function such that IIgII,,,,. < oc for some so E (0, 1).
(i) Prove that if g is decreasing, then there exists eli
o IIgLI,,L = him g (h).
What happens if we remove the hypothesis that g is decreasing?
14.3. The Limit of Ba.P,e as s - 0+ and s - 1-
423
(ii) Prove that if EfL is increasing, then there exists lim II9Ila,oo
lim g (h) = h-o+ h
What happens if we remove the hypothesis that 2V is increasing?
Lemma 14.14. If u E LP (RN), 1 < p < oo, then for every i = 1,...,N, dx)P
(Ahu(x)IP
hllmoo Ull
(f
=2
RN
lu(x)IP dx)P
.
Proof. Since u E LP (RN), by Theorem C.23 for every e > 0 we may find a function v E CC (RN) such that
(LN lu (x ) - v (x)Idx)
(14.16)
Hence by Minkowski's inequality, IQhu(x)I P \P <
dx) (LN ltP (x) Ip
dx )
(IRN
_< e.
P
1
+ (I /r +`J
(14.17)
lu(xi,xi+h) -v(x2,xi+h)IP dx J
P
N lu(x)-v(x)IPdx)v
l RN
<(LN I Ahu(x)IPdx)P+2e, 1
and, similarly, (14.18)
(f
A; u (x) P dxl P <
(
N
Iphv (x)IP dx) P + 2e.
RIV
Next we write JRN
Dv
(x)IPdx
_v f (x)
P dx +
x;+hI?IxdI
=
=:I+11.
I Ohv (x) I P dx
f J
Ov (x) I P dx
I/xj+hl
+J
I Aby (y) I P dy
Iy,I
Since v E CC (RN), there exists R > 0 such that v (x) = 0 for all x = (x;,xi) E RN-I x R such that IxzI + IxiI > R. Note that if Ixi+hi ? IxiI
14. Besov Spaces
424
and IxLI + kxiI > R, then IxsI + Ixi + hi > R, and so v (x;, xi + h) = 0. Hence we may write
=J
1iv(x)" dx.
{Ixr+hI> k.I}n{ I,I+Ix,I
Since the set {Ixzl + Ixtl < R} has finite measure and Iisv (x)I < 2 IIvll0, we may apply Lebesgue's dominated convergence theorem to conclude that
lim . = lim /
h-+oo J{Ixi+hI>Ix,I}n{Ix
I AA v (x) I'
dx
I+Ix,I
1v(x)I-dx=
f
RN
4Jx'jl+jxjj
Iv(x)I"dx,
where we have used the facts that
{xERN.
Ixil}/RN
and v (xi, xi + h) = 0 for all h sufficiently large (depending on x) due to the fact that v E CC Similarly, (RN).
fim T2= JRN Iv (x) 11 d, and so 1
lim (I IZOv (x) Ip
l dx /
Using (14.16)-(14.18), we obtain
the//
h-000
D
=2
N
Iv (x) Ip dx
1 lD .
desired result.
Exercise 14.15. Let u E Lj (RN) be a function vanishing at infinity and C
let 1 < p < oo. Prove that hm inf
JRx
IhM' dx - f
lu (x) I," dx;. ,Y
Hint: Assume first that u E C, (RN), then that u E LP (RN), and then do the general case.
We turn to the proof of Theorem 14.10.
Proof of Theorem 14.10. If u E L9 (RN), then by Lemma 14.11 (with g (h) := II,&huIIL,(RN)) and Lemma 14.14,
lim (S (1 - 8) p6),
2N IIUIILP(RN)
On the other hand, if lim inf (a (1 - s) pO) I 8->0+
e(RN) < 00,
.
14.4. Dependence of Bs,p,e on 0
425
then by Lemma 14.11 (with g (h) := II AhuII LP(RN)), Exercise 14.15, and the
fact that the limit inferior of a sum is greater than the sum of the limit inferiors, N
E lim
lim inf (s (1 - s) pO) 8
f
i=1
(p dx) P
(fRN I Ahu (x)
ff
>N
J
P
JNIu(x)IPdx
.
R
Thus, U E LP (RN).
Theorem 14.16. Let u E L1oc (RN) and let 1 < 9 < oo and 1 < p < oo. (i) If p > 1, then the distributional gradient Vu belongs to LP (RNxN) if and only if lim inf (s (1 - s) 9) e I uI BS,P,e(RN) < 00. s-1-
Moreover, if the distributional gradient Vu belongs to LP (RNXN), then N
hm (s (1 - 3) 9) 6
I uI Ba.p.9(RN)
(f.
N
/ \ I
ax=
(x)
1
P I
dx I P
/
(ii) If p = 1, then the distributional gradient Du belongs to Mb (RN) if and only if liminf (s (1 - s) 9)1 IUIB8,1,8(RN) < 00"-1-
Moreover, if the distributional gradient Du belongs to Mb (RN), then N
lim (s (1 - s) 9)8
s- 1-
IUIBs,I,a(RN)
_
IDiuI
(RN).
Proof. We proceed exactly as in the proof of Theorem 14.10, with the only difference that in place of Lemma 14.14 and Exercise 14.15 we use Theorems 10.55 and 13.48, respectively.
14.4. Dependence of B8,P0 B on 0 We study the relation between different Besov spaces Bs,P,e (RN) as 9 varies.
Theorem 14.17. Let 0 < s < 1, 1 < p < oo, and 1 < 91 < 92 < oo. Then there exists a constant C = C (N, p, s, 01, 92) > 0 such that I uI BS.P.82(RN) :5 C
(RN)
for all u E L10C (RN). In particular, B8,AOI (RN) C B8,p,e2 (RN).
14. Besov Spaces
426
We begin with an auxiliary result.
Lemma 14.18. There exists a constant C > 0 such that for every u E L" (IR), 1 < p< oo, and for all h > 0, C IIAhUIILP(iC) < h
h
II''uIILF(R) dry.
Proof. Consider a function cp E C9° (R) such that supp cp C [0, 1J and fR cp (y) dy = 1. For x E R and h > Owe write
u(x) = jra(y)u(x+hy) dy+JRco(y)[u(x+hy) -u(x)] dy (14.19)
= h f cp (z h Ilt
') u (z) dz + fo
cP
Ahyu (x) dy
U (x, h) + V (x, h).
By Theorem C.20,
l
rz -x u(z) dz
-1
aU(x,h) ax
hJ
h
=
JR cpl
1
h2
(z h x) [u (z) - u (x)] dz
fR cP' (y) [u (x + hy) - u (x)] dy
=--h
f'W (y)ahyu(x) dy,
where we have used the fact that fR cp' ( ) dz = 0. By Theorem C.20, the fundamental theorem of calculus, and Fubini's theorem, (14.20)
U (x + q, h) - U (x, h) = f
_
-
(x + s, h) d.9
-1 f j o
' (y)
Vu (x + s) dsdy.
By (14.19) and (14.20) we obtain
!'u(x) = U(x+q,h) - U (x, h) + V (x + 17, h) -V(x,h) 1
=
-1 J J ho 0
11
gyp' (y) &4'u (x + s) dsdy
Dux dy, 0
14.4. Dependence of B''P'0 on 0
427
and so, taking r) = h, we get IAhU (x) I <
jh IAhyu (x
j1
+ s) dsdy
h
+ C Jf 1 IOh&,Ahu (x) I dy.
We now take the norm in L" (R) on both sides and use Corollary B.83 to get
II
C LP(R)
Uh
o1
dsdy
s)II
h
LP(R)
+ CJ
1
LP(R)
I
dy =:I+IT.
First making the change of variables z = x + s and then 77 = by gives
I
h
=
f1
h
dsdy = C
I'
1 Il,AhyuIIL9(R) dy
IlAhYullL.(R)
C rh IID'UIIL,(R) d7j. h
J0
On the other hand, by Minkowski's inequality and a change of variables, we have that I
M)
(14.21)
11 Al (u ( + h) - u (-)) IILP(R)
I( 2
+ h)IILa(R) +
IIAhYull(,,)
IIAhYHIILa(R)'
and so
ZZ < C J0' Idy = 2
jh II'IIL'{R) d77,
where we have made the change of variables1 = hy. This concludes the proof.
Corollary 14.19. There exists a constant C > 0 such that for every u E LP (RN), 1 < p < oo, and for all i = 1, ... , N and h > 0, h
11'&lUIL
(RN)
h
f Ila,UlI
P(RN)
Proof. We only prove the case 1 < p < oo. By Fubini's theorem we have that for RCN-1-a.e. x' E RN'' the function u (xz, -) belongs to LP (I8). Fix any such x= E RN-i. Applying Lemma 14.18 to u(xi, ) gives h
C Ilahu (4, ) IILa(R) < h
p
4
II' &7U (x, -)IILp(R) d77.
14. Besov Spaces
428
Taking the norm in LP (RN-1) on both sides in the variable x; and using Corollary B.83 and Tonelli's theorem yields 1Jp
C II
II
LP
C W (JRN-i
c
dldxi
Il'&i u (xi, .) IILP(R)
p
P
0
Jh
W
IIazUIILP(RN) dal.
Exercise 14.20. Prove the case p = oo. Hint: Use the definition of essential supremum and Fubini's theorem. Exercise 14.21. Let 1 < 91 < 82 < oo.
of nonnegative numbers, prove that
(i) Given a sequence 00
oo
Ean n=1
Eany (?z=I
(ii) Given an increasing function g : [0, r) 10, oo), r > 0, and q > 1, prove that there exists a constant C > 0 independent of g, q, r, 81, and 02 such that
(fr
62
\hq}
7
dh
h
J0
dh h
\hq)
(iii) Using part (ii), prove that sup 9 (h) < C
o
r
9 (h) \ B` dh hq J h
0
e,
Proof of Theorem 14.17. Step 1: Assume first that 02 < co. For all i = 1, ... , N, let iJii be the function defined in (14.2). By Corollary 14.19, h
(14.22)
7yi (h) <_
h
0Oi (11) dy
Fix r > 0. Using (14.22) and Exercise 14.21(ii), where the function g is the increasing function h
9(h):=J
i(11)dy,
hE[0,r],
14.5. Dependence of B''P'0 on s and p
429
and the change of variables y = hz, we obtain that
(10
ih'+se2
(h))e' dh}
02
r
lea
(10 i {?/) dF/
(10
h
1
r
(jr
el
h
1
V)s (y) dy)
hl+(i+s)01 `10
hi+e (J1
1
ei
ipj (hz) dz
l
By Corollary B.83 and the change of variables = hz we get
11 (ff {
h 1+B1
{hz}}ei dh/
1+3e1
Jo
dz
1
i
x(jzri
f1
=C
i
4}
(04
1
dz
m
< By letting r -> coo, we obtain
([0
``1 hi+sez
('i (h))e2 dh 1
(r00
'r2
1
1ael (pi
4)
'r, .
Step 2: If 02 = oo, we proceed exactly as in the previous step, using Exercise 14.21(iii) in place of Exercise 14.21(ii), to conclude that sup
0
1
s
h
i (h) < sup hl+s f ii (y) dy
Since the estimate for I is the same, we conclude as before.
14.5. Dependence of Bs,p°e on s and p The next theorem shows that by lowering the regularity parameter s of functions in Besov spaces, one can increase their integrability parameter p.
Theorem 14.22. Let u E Li L (R1) be such that IUIB,,P,e{RN) < oo for some 0 < s < 1, 1 < p < oo, and l < 9 < oo. Then for every 0 < t < s there exists a constant C = C (N, p, s, t, 0) > 0 such that I1IB9.4.8(RN) !5 C IUIBa,P.e(RN) ,
430
14. Besov Spaces
where
+N=s+N. P
q
We divide the proof into a few lemmas.
Lemma 14.23. There exists a constant C > 0 such that for all u E Coo (R),
C fh
Iu (xI
(14.23)
+C
ju (x + y)j dy
J hJ£J
_
3
IA'?u (x + y)I dtldyde
forallxERandallh>0. Proof. Consider a function cp E C'° (P.) such that supp cp C [0,11 and fR cp (y) dy = 1. For x E R and h > 0 define the function
U(x,h) :=
rh
Ja
u(x+y)co (h)
dy.
Since
ds=1,
hJoh'Plhl by the continuity of u we obtain
liin U (x, h) =
h
fh m
u (x + y) cP
hdy = u x) .
By the fundamental theorem of calculus we get Ph
u (x) = U (x, h) -
(14.24)
We now compute au x'{
.
Since 9
have
\/
au (X, )
dC.
= cp (1) = 0, by Theorem B.53 we
f u(x+y) w'\ } dy
aUa
_ - 3 J u(x+y) [e,() +yco' (f)J dy _-g
f
u (x + y)
ay (4Yc
\
)) dy.
14.5. Dependence of B',P,O on s and p
431
Integrating by parts and using Fubini's theorem and again the fact that cp (1) = 0, we obtain OU(( 2
f f u'(x+Y)mo ( ) dy
f'u,(X+Y)w (
) Jo
dr dy
where
T:= {(y,77) ER2: 0
ff I1
I- JC-Zu(x+z+n)cP'(z") dq -u(x+z)cp (z)] dz f
f-Z
JO JO
[u (x + z + 17) - u (x + z)] gyp' (z
\
± ")
d17dz,
where we have used the fundamental theorem of calculus. Hence, by (14.24),
u (x) = h f u (x + y) P (k) dy h
0
f f-Z 1 h +Jo Jo Jo
z+
\
3
C
")
l
dndzd,
which gives the desired estimate.
Remark 14.24. (i) From (14.23) we get h
lit (x) I
lu (x + y) l dy
h
h3 +C
J0
c fh
d
j
h rh-v 0
IDu (x + )I dijdy h h-" C2
ff
ju (x + y)l dy+ _ dndy for allxEJRandallh>0. Since 0<,
T2
1
+ y)2, (rl
0
14. Besov Spaces
432
and so (14.25)
Iu(x)I <_
c h JohI u(x+v)I
dy+C 1
h
Jy IA(17+Y)2 )I dr1dy
(ii) Note that in the proof of Lemma 14.23 we have only used the fact that y (1) = 0. Hence, the function
sv (x) .= 2 (1- x) could have been used.
Lemma 14.25. Let 1 <- p < q
oo and let u E C°O (R). Then for all
h>0,
IILIILQ(a) <
C
h--
IIuIIrfl(m) + C10
II'&"UIILa(e) dr),
1+n
where C=C(p,q)>0. Proof. By Lemma 14.23, for all x E R and h > 0 we have
Iu (x)I < -
jh
Iu (x + y)I dy
+Cf
( 14.26)
0
j j jWu 0
(x + y) I drjdydc
0
f (x) + g (x)
.
To estimate the norm of the function f in LQ, for all x E R we write C
f (x) =
fI x[o,h[ (y) Iu (x + y) I dy =
fI X(o,h) (z - x) Iu (z) I dz,
where we have used the change of variableskz = x + y. Applying the general form of Young's inequality (see Theorem C.16) with 1
r
:= 1
-fl
p
q -1)
(note that the roles of q and r are exchanged here), we obtain (14.27)
h
Ilf IILQ(m}
f
m
lu (x)Ip dx) p
hpC qf1R
Iu (x)I
fR IX[°'h) (x)
I.
dx )
n dx) p
Similarly, by Tonelli's theorem and the change of variables z = x + y, h
g (x) = C f
ff £
3
X[°,EJ (z - x) Ia"14 (z) I
dzdrjd.
14.5. Dependence of B*,P,e on s and p
433
Using Corollary B.83 twice and reasoning as in (14.27), we get
ff
f1
h
II9IIL4(R) <_ C
/h f < CJ
- 11f
drjdd R)
1-(I
C
IIANILP(R) drjdl;
f{
=Cfh
X" (z - .) IA"u (z)I dzl)La
1
2+1-4 IItgUIILP(R) drjde
By Tonelli's theorem we have that
T=
(14.28)
h
/'h
CJo
IIi
UIILP(R) f
2+1-q
dij
C1 h q1+p-4 IIL'UIILP(R) dn. If we now combine (14.26), (14.27), and (14.28), we obtain the desired result.
Lemma 14.26. Assume that the vectors p = (pi.... 9PN), 4 = (q1, . . , qN) and the real numbers 8, s, and t satisfy the relations 1 < 8 < oo, 1 < pi <
qi
µ:=1-11 1Pi - i>0. s qi /
(14.29)
i=1
Then for every u E C°° (RN) and j = 1, ... , N, C
f
00
h1+te IIAJUIILQ(RN) dh1$ N
< CE
\f
h1+8e VtuIILP(RN)
dh)
if 0 < oo, while N SUP SUP h >O
ht OIILq(RN) <
Ck k=1 h0 h8
IUIILP(RN)
if 9 = oo.
Proof. We recall that we are using the notation introduced in (E.2) in Appendix E.
14. Besov Spaces
434
Step 1: Assume that q differs from p by only one component qk > pk. Let v E C00 (RN) . We claim that C
_
IIVIIL9(RN)
IIVIILP(mN)
h Pk
Qk
+Cf0
h 17
1+Pk1 - Qk1
IIAkVIILP(RN)
dt.
To see this, assume that 1 < k < N (the cases k = 1 and k = N are simpler) and write x= (yrxk, z) E RA;-1 X R X RN-k and
PI _ (P1, ...,pk-1) , P, = (PN-k, ,PN) Note that p = (p', pk, p"), while q = (p', qk, p"). Fix z E the previous lemma to the function u (xk) := IIv (', xk, z)IILP'(Rk-1) ,
RN-k
and apply
xk E R,
to obtain
C
IIv (', ., z)IIL(P'.Qk)(Rk) <_
hPx
IIv
1
(-'.'Z) IIL(P-.11)(Rk
Qk
h
+C 0
7
,+ 1 _ 1 IIAkv pk
4k
z) IIL(P'.Pk)(Rk) d`f'
We now take the norm in LI" (RN- k) in z on both sides and use Corollary B.83 to obtain the claim.
Step 2: Assume that q differs from p by only one component qk > pk. Take v := A4 u in the previous step. Then (14.30) I
1C
Lq(RN)
h. Pk
L II&hUII LP(RN)
Qk
oh
1
C
h nl+Pk_k
ll7f.
Reasoning as in (14.21), we have (Ajhu)
2 IIAkuIILP(RI) IILP(IE') IIIk and so the previous inequality reduces to (14.31)
II'&3u1I
La(RN)
Ci
<
h
j
j -L-C
+
0
IIu11 LP(RN) h
1
1+ 1 _ 1 II©kUIILP(RN) di7. Pk
Qk
14.5. Dependence of B'P'0 on s and p
435
We now distinguish two cases. If 0 < oo, by the change of variables 11 = hC we have
t p q I4uIILp(R-)
II°3uIIL9(nv)
C
L1
+
h Pk
01
qk
dC-
t1+ PkL_ 9kL I°uIILP(IRN)
S
By (14.29) and the fact that q differs from p by only one component qk > pr,, we have that 1 1 t=has= s--+-, Pk qk
1+E and taking the norm in
and so multiplying the previous inequality by >a
L9 ((0, oo)) in 0 on both sides, we get
JOh.l+w (°°
ILq(EN)
II
dh) 1
dhJIh1+88
< C (j°° +C
J
O
e
B
00 1
h1+80
j1
1
O
1
CPk-k
°h
I
LP(V')
d
=:I+II.
dh
By Corollary B.83 we have that
r
7.1
pl
J1+(J°° 1
0
S
h1+88
II
lv
1+
e
uIILp(IRN) dh) I
8
fjp
JO
1
4k
1+88 JO 1 A
d
where in the first identity we have made the change of variables h =
and
used the facts that dh = -14 dry and that s - PA; -L + ek > 0 by (14.29). This
completes the proof in this case. Assume next that 8 = oo and let A := sup,,>0 ,, II°xUIILP(RN). Multi-
plying both sides of (14.31) by , and using the fact that t + pk - - = s by
14. Besov Spaces
436
(14.29), we have that 1t
T
II°,u I
La(RN}
c
` hC
LP(k ) + h C IHLP(RN) + N
< C II
UiILv (RN )
=1
=
77 k - kfa
0
h 0
Ak1
II°kUIILp (RN )
d
Ld
+ C sup ,>07s u°kuIILP(RN) .
Hence,
sup
V81114(R-1 :5-Sup 4>0 h II°UII LP(RN)
h>a t
+ Csup
1 II°kUIILP
n>0 7
RN ) .
Step 3: If q and p now differ by more than one component, we introduce the intermediate vectors
p() :_ (gi,...,gj-1,PjtPj+1,...)pN), qu)
(ql,
, qj-1, gj,Pj+1,
PN)
for j = 1,...,N and 8(1)
s(j+l) := 8(i)
-
1
Pi
+
1
qj
for j = 1, ... , N - 1 and apply the previous step several times.
O
We now turn to the proof of Theorem 14.22.
Proof of Theorem 14.22. Step 1: Assume that u E COD (IAN). Since
µ:=1-N+q8>0, ps we are in a position to apply Lemma 14.26, with p :_ (p) ... j p)7 q (q, ... , q), to obtain that I1IBl,q.B(RN)
!5 C I
Step 2: The additional assumption that u E C°O (RI) can be removed by mollification. To see this, let ue := cpe * u, where SPe is a standard mollifier. By the previous step, I-EI$=.4.8(RN)
C IueIBs.P.e(RN) .
It is now enough to apply Proposition 14.5 (see (14.4) and (14.3)).
14.6. Embedding of B,P,e into Lq
437
Remark 14.27. Under the hypotheses of the previous theorem we also have
C
IUIBt,9(RN) <
where
IUIB=.P(1! N)
N N t+-=s+-. p q
To see this, note that in view of Theorem 14.17 and the previous theorem, C
IUIBI,9.P(RN)
As a consequence of the previous theorem and of Theorem 14.17 we obtain an extension of Morrey's embedding theorem to Besov spaces. Corollary 14.28 (Morrey). Let u E L IC (RN) be such that I uI B..P,e(RN) <
oo for 0 < a < 1, p > s , and 1 < 0 < oo. Then a representative is of u is Holder continuous with exponent s (14.32)
and
Iii (x)-ii (y)I
IkIB,.P.a(RN)
for all x, y E RN.
Proof. By Theorem 14.17, IUIB-,,,-SRN) <- C
IUIB-.P.B(RN) < 00,
and by Theorem 14.22 with q = oo and t = s - p , IuIBs- P
C IUIB..P,-(RN) < 00.
By the definition of the seminorm 1.1
B° P'
,
it follows that u has a
(RN)
representative as satisfying (14.32). This concludes the proof.
14.6. Embedding of B',p'e into Lq Next we extend the Sobolev-Gagliardo-Nirenberg theorem (Theorem 11.2) to Besov spaces. Theorem 14.29. Let u E L10C (RN) be a function vanishing at infinity such that I uI Bs,F.e( N) < oo
for some 0 < e < 1, 1 < p < $, and 1 < 0 < N
Then there exists a constant C = C (N, p, s, 0) > 0 such that N-s
U..
Iu (x) I N-SP dx)
"i'
-< C I
uIB..P.s(RN)
In particular, Bd' '8 (W') is continuously embedded in Lq (RN) for all p <
q < N ep We divide the proof into a few lemmas.
14. Besov Spaces
438
Lemma 14.30. Let 00
00
f (x) := 0
x E R,
(z+1;) 2
o
: [0, oo) x [0, oo) - [0, oo) is a Lebesgue measurable function. Then
where
00 (14.33)
IIf IIL(R)
(LI(z,r dz
C' x
p
f
1
d4l
4
for all 1 < p < q < oo and ail l < 9 < q. Proof. Step 1: To estimate IIfIIL,(R), we use Proposition B.81. Thus, let 9 E L' (R) with g > 0, where q' is the Holder conjugate exponent of q. Define
a:= +--p 1
(14.34)
1
1
9
q
and
IF (y, 0 :=
g (x)
T-00
(y-x+t)
2
dx,
y E R, e > 0.
By Tonelli's theorem, the change of variables y = x + z (so that dy = dz), and Holder's inequality applied twice, we have that
f
00
f (x) 9 (x) dx = <_
Jo a III (',
f f P (y, e) W (y, ) dydC )IILn(R) °` III (', 0II&(R) dC
< IIk10 IID (', )IIII(R)IILB((o,00))
where with an abuse of notation the norms and are taken with respect to the variable 1;. Since aO = 1 + 0 Proposition B.81, to prove (14.33), it suffices to show that (14.35)
1:= IIIT
(.1
p
-
1),
by
0I6'(R) IILa'((0,oo)) ` c II911L4'(R)
where p' and 9' are the Holder conjugate exponents of p and 9, respectively.
439
14.6. Embedding of BB,P,O into LQ
Step 2: Assume that p > 1 and 9' > p'. Hence, if 0 > 1, by Corollary B.83 (where the exponent p there is replaced here with 0' > 1) we have that P
Z=
dS
[r° UR Iba41 (y,c)IP' dy
f
ff
<
)71;7 ,
00 ,fiSt
dy
I caT (y, )I B' d
J \ Jo
Since for 0 < x < y we have that ca < (y - x + c)*, it follows that c°` IW (Y'01:5
9 (x) 00 (y - x
+)2-a
dx.
Using Corollary B.83 once more (with the exponent p there replaced here with 0') and (14.34), we obtain that
f
,
Z< JR
I
07
00
Y
oo
9 (x)
1
(y - x + O2-°", ,
P
=C R
-oo
P'
9 x
(y -
x)1-'-p+Q
W
dx
dy
when 0 > 1. If 0 = 1, then 0' = oo, and it is enough to replace the LB integral norm in c with esssupg>0. Thus, in both cases we obtain
Z
(JY
P
9W
oo Ix -
dx
dy
.
yIl-p+q
Since p > 1, we are in a position to apply Proposition C.31 (with N = 1 and where the numbers a, p in the proposition are replaced here with p - 4 and q') to conclude that the right-hand side of the previous inequality is bounded by C II9II Lv' (R)
Step 3: If p = 1, then p' = oo, and so for every fixed t; > 0, II`I'
esssup yER
f
g (x) by (x) dx,
R
where 1
by (x)
(x)
(y - x + b)2'
x E R.
14. Besov Spaces
440
Let g* : [0, oo) - (0, oo) and h; : [0, oo) -> (0, oo) be the decreasing arrangement of g and hy, respectively. Then
h; (z) = (z
re-
z > 0,
and so, by the Hardy-Littlewood inequality (see Theorem 6.13),
f
00
g (x) by (x) dx <
R
dz.
Jo
(z + W
Hence,
7<
(14.36)
[a f1
cg+
}2 I dz11 Le, W),00))
< (z +we have that
If 8 > 1, then, since for z, 6 > 0,
([0
T
e'
9* (z)
JOO
= C II9IIL¢(R) , where we have used (14.34), Proposition C.31 in the second inequality, and Theorem 6.15 in the last equality. If 8 = 1, then by (114.36), C II9RII1,4'((0,00))
2 < esssup l 2- 9 £>o
l
9# (z) )2 dff
o
(z +
] dzJJ/
1
< (>O
II9'IIq((0,Q)) (L°° (z
(
C II9'IILQ'((0,00)) = C II9IILQ'(R)
)
dz Y
where we have used Holder's inequality and Theorem 6.15.
Step 4: Finally, if p > 1 and 8' < p', then 00
<
Li0esssup
f
(y,
)I
r
00
esssup VER
{3r,
)I
(y)I" d)
(JR oo
dy 1
7
d
where in the second inequality we have used Holder's inequality with exponent . and the fact that ()
14.6. Embedding of B",»,e into Lq
441
The first term on the right-hand side of the previous inequality can be estimated as in the case p = 1 (see Step 3) while the second term can be estimated as in the case p' = 0' (see Step 2). Thus, we get (14.35) even in this case.
D
Combining Lemmas 14.23, 14.30, and 14.26, we obtain the SobolevGagliardo-Nirenberg theorem.
Proof of Theorem 14.29. Step 1: Assume that N = 1 and that u E COO (R) n L sP (R). By Holder's inequality h
1
lu (x + y)I dy <
J
1
" IIUII L
i-( h
(R)
Hence, letting h -> oo in (14.25), we obtain (14.37)
lu (x) I
-L
.
(x + 71)2
0
to obtain that
We may now apply Lemma 14.30 with q = i 00
(14.38)
1
Ilull Lr-
C
(R)
JO
+se
UR
fp
IO u (z)
I
1
B
dz
p
dC
Step 2: Assume next that N > 1 and that u E COO (RN) n LN ep (RN).
Fr
Reasoning as in Step 1, we have
00 I ONU (x', x
l u (x) I<
(z
Jo
Jo
z) I
+ + h)2
dhdz,
where we are using the notation introduced in (E.3) in Appendix E. Let q := 7 and let q' be its Holder's conjugate exponent. By Corollary B.83, IIu (',xN)II
L
(RN1)
ff 0
0
(z +1 h) 2
U
xN + x) II
I
N
' (RN-1)
dhdz.
We now take the norm in Lq (R) in XN on both sides and apply Lemma 14.30 to obtain o°
(LN
Iu (x) I
dx)
1
< C (10 h1+pe
II
where
Np
Np
( N - sp' 'N-spP
NUII LQ(RN)
dh)
,
14. Besov Spaces
442
We are now in a position to apply Lemma 14.26 to the right-hand side of the previous inequality. Indeed, letting
we have that
(N-1)
R:=1-
- N - sp1 1
= N'
Np
s
and so 1e
{JO B hl+AO
IIANUIILq(RN) dh M
< CE =t
(
1
/
8
dh) hl+so Ik*LII L P(Rx)
Step 3: To remove the hypothesis that u E C°O (RN), we apply Proposition 14.5. The case in which u E LjO0 (RN) is a function vanishing at infinity as in the statement of the theorem is left as an exercise (see Exercise 14.6 and Step 4 of the proof of Theorem 11.2). As a corollary of Theorems 14.17, 14.22, and 14.29 we obtain the following result.
Corollary 14.31. Let0<s<1 and1
If0
,
then Bt,q (RN) is continuously embedded in Be,p.q (RN) (and in B-,p (RN))
14.7. Embedding of W1' into We now study the relation between Besov and Sobolev spaces.
Theorem 14.32. Let u E L'10o (RN) be such that its distributional gradient Vu belongs to Lp (RN; RN) for some 1 < p < oo. If N > 2, p < q < oo,
0
t+N=1+ Q,
(14.39)
then there exists a constant C = C (N, p, q, 0) > 0 such that (14.40)
IUIBc.o.e(RN)
C (hN IVu (x)I' dx 1J ' ie
.
///
In particular, W1'P (RN) is continuously embedded in Bt"Q (RN).
We divide the proof into a few lemmas.
14.7. Embedding of W ',P into Bt,q
443
andlet1
Lemma 14.33.
and1<0
1-1--+->0, P r 1
(14.41)
1
ao:=
t-1-1+1 E (1t-1 i r 1-l-p+
\\
1l
1-1 /
1
Then
(14.42)
Q(R) .
IUIB7,q.e(R) < C IIU IIL,P(R)
For 8 = oo the value ao = 1-1 1-7 is allowed.
Proof. We only prove the lemma in the case 0 < oo. In view of Theorem 14.17, to prove (14.42), it suffices to show that (14.43)
IuIBt,4,1(R)
C Ilu IILP(R)
By (14.31) and (14.10) for all h > 0 we have
II°hU ll
L°(R) <-
ChpIi
< Ch'
1a
I+C LP(R)
P+Q Ilu IILP(R) + C
Ih o
1+1-1 p q
d77
17
IIU'IILP(R)
f
h
1
1
1
17P
q
= Ch' P+Q Ilu IILP(R) Hence,
(14.44)
II,&hUIILo
hl+t
R O
< Ch-t P+Q IIhu
Let
(14.45)
a-
1
1
q
r
1
1
p
r
IILP(R)
dri
14. Besov Spaces
444
Then = n + 1 T°`, and so by Exercise B.80, (14.44), and (14.10) once more, for every e A > 0,
100 dh IIAhuII Lq(R) hI+t
f
A
IIAhUIILq(R)
TI -+t + I
A
II& t
hl+t
IIL4(nt)
h-t-? dh
C Ilu'IILP(R) j
IIAh
IlAhUllLIP
IILI(R)
hl+t 00
C1 AIt
1 + 1 beIILP(R) + P
IIu'IILP(R)
q
JA
IIAhU 1-0) h1+tL
a
Note that 1 - t - 1P+ 19> 0, since a0 <1 by (14.41)2. Using the fact that ao > i-i by (14.41)2, it follows that t-l-a (1 - 1) > 0, and thus we obtain that 1-a 1 h l-a dh dh
. III
uIIL.(R)
hl+t-a -
foo
h
hi
hl+t-I-a(l-I)
IuI o0
1-a
BI-r-°°(R) Z
t - I - a (1 - l)
1 hl+t-I-all-t) dh
IkIBla°°(R)
Hence,
foo IIAhUIILQ(R) h H
G,A1-t+!9-P
ilu'IIL-(R) +
CA'-t+9-'PA
At-I all-I)
+ At-I all-O
Taking
Ilu'IILP(R) IUIBJ'aW(R)
AaBI-a.
-a
A :_ (BA)9P
and using (14.45) and (14.41)2, we obtain (14.43). This concludes the proof.
0 Exercise 14.34. Prove the case 8 = oo. We turn to the proof of Theorem 14.32
14.7. Embedding of W ',P into Bt.q
445
Proof of Theorem 14.32. In view of Theorem 14.17 it suffices to prove (14.40) for 8 = q.
Step 1: We claim that there exist I E (0, t) and r > q such that (14.41) holds and p - apq > 0, or, equivalently, t-1
11=I
(14.46)
1 - I - 1 + 1 > o,
ql
p
-t
11=I
P
<
t-1 P
ql
r <
P
q
To see this, note that by (14.39) we have that t = 1 - v + q > 1 - E.. Hence, we may choose I such that max{1-p N,0JJJJ}
(14.47)
t.
l
Since 1 - A < I, using (14.39), we have that 1-1
(14.48)
qi
_ t=I tP < min
1-1
ql
-
_
lq tt
"
- L- > 0. We claim that
1
1P 1q q
P
1
r:
Indeed, by (14.39) and the facts that N > 2 and p < q we have that (14.49)
1-t-(P
(N-1)(p-9)>0,
q)
and so 1-1
_
t-1
1-1
P
q
q
(14.50)
1-t
<
_
p
t.-t
P q
On the other hand, since I < t by (14.47) and p < q, we have that t_t
1-1
(14.51)
- tP < Q
q1
Hence, (14.48) holds.
Next, we claim that 1- I - + > 0. First, observe that 1-1- + 1 > 0 by (14.49) and the fact that 1p< t. Next, by (14.39) and the fact that p p
1+1-1>1-N+N=t, q p p q which implies that 1-1 q,
P
and so the claim holds.
t-1 I
P
q
>I-1 +-, p 1
14. Besov Spaces
446
To prove (14.46), it now suffices to choose a number r > r. > q so close
to r. that 1- l - p + ,1-, > O and such that 11-1
q
_
P <-<-.
t=t
1-t
1
1
r
r.
Step 2: Here we use the notation introduced in (E.3) in Appendix E. We recall that B = q. We begin by estimating ONu. By the previous lemma, o0 1 hl+tq
J
I DNu (x', XN) Iq dxNdh
J
coq
N
11
1-ao
o0
(x', IILP(R) f
IILr(R) ha+g)
.
'
Integrating both sides in x' over IltN-1 and using Tonelli's theorem and Holder's inequality with exponents -- and P we get ,
f
hl+ II NuII Lq(RN) dh
r
J
coq
(x ')
N-1
Lp(R)
(
oo
1-ao
h
i
(x ' )IIL*(R) h
CII a,NIILP(RN,
(1-a
(1-ap)P
(1
P-aoq
dh) P-aoq
hl+tq
IIGNU (x''
)IILr(R)
By Corollary B.83 (with p replaced with (1- ao)
q), we obtain that
Z
(1-uo)P9
1
-aoq
q
X
J0
hl+lq
LN_1 II,&NU (x', .) IILr(R)
= C IIVuIIaL°P(RN) (
,
1-ao
o0
\Jo
dh)
h1+lq
where r is the vector of components
r1:=
r
1-ao P-aoq
ifi=N, ifi54N,
=: Z.
447
14.7. Embedding of W 1,P into B1,q
i= 11... , N. By Young's inequality (see (B.17) in Appendix B), we have 00
(14.52)
Oh1+ts < CE
IIANU 11L-1(RtN) dh
jjoujj,(P-N)
+s
J
h1+tQ
III''
IIL*(IN) dh.
Using (14.39) and (14.41), we have that N
=
1
t
tt
1- 1) Q ri
N
+(p-aoq
qN
[p+p-tq-2 ]+1(q-p) t- ++(N-1) r pq(1-t)-(q-p)
1
t
1
q
t - N + -1 + p-tq+(d-t )(q-p)
1
q-p
q
t
t>0.
t-tq+Z+p
q-p
q
///
Since the vectors q := (q... .. , q) and r and the real numbers i and t satisfy
the relations 1
00
f 0
h1+lq
I
dh
II°hUIILe(RN)
dh,
i=1
and so, also from (14.52), 00 h1+=Q
1I$1U11q(N) dh
N 00 CE IIVuIILp(RN) + EC E j hi+ i=1
IIAhnIILa(RN)
dh.
0
A similar inequality holds with Oku in place of ONU (to see this, one can either change the order of the variables or proceed somewhat as in Step 1 of the proof of Lemma 14.26).
14. Besov Spaces
448
Summing over all k yields N
fm
hl'
+tq
II'&UIILQ(RN) dh
< CE I I V u I I j,,.
+ EC
tq i=1
Ills
UIILq(RA') A,
or, equivalently, N p
/t1+tq
(I2
n UIILQ(RN) dh <
1
q
SEC
I
where e > 0 is chosen so small that 1 - EC > 0.
Remark 14.35. The previous result continues to hold for N = 1 and p > 1 (see [191), but it fails if N = 1, p = 1. and 9 < oo.
Exercise 14.36. Construct a sequence
of piecewise affine functions
bounded in W1'1 (R) such that sup nEN
J
unI dx < oo,
IIC
but IunlB1/Q,q.s(R) -> 00
asn-sooforevery q>1and9
Du belongs to Mb (RN). Prove that if N > 2, 1 < q < oo, 0 < t < 1, and
t+N= 1+ N q
then there exists a constant C = C (N, q) > 0 such that IvIB,.q(RN) < C IDtal (RN) .
In particular, BV (RN) is continuously embedded in Bt,q (RN)
14.8. Besov Spaces and Fractional Sobolev Spaces In this section we study the relation between Besov spaces and fractional Sobolev spaces.
Definition 14.38. Let 1 < p < oo and 0 < s < 1. A function u E L" (RN) belongs to the fractional Sobolev space W8' (RN) if IIuIIL?(RN) + ItIwi.P(RN) < 00)
14.8. Besov Spaces and Fractional Sobolev Spaces
449
where IUI Wa.P(RN) := (LN JRN
IIX (x)- YIN - u (Y) (p dxdy
1/P
(y
The following proposition shows that in the range of exponents considered in this chapter, the fractional Sobolev space W''p (RN) coincides with the Besov space B''p (RN).
Exercise 14.39. Let 1 < p < oo and 0 < s < 1. (i) Prove that for every a > 0 and a > N21 f0
xN-2
f
1
(x2 + a2)a
dx < ti' (a, N) a2a-(N-1)
(ii) Prove that for every u E B',p (RN), IUIW..P(RN) <- C
IUIBS.P(RN) ,
where the constant C depends only on p and s. Hint: Write
u(x) - u(Y) =
[u(x1,x2,...,xN)
- u(ylax21...,XN)]
+...+Iu(Yl,...,yN-1,xN)-u(Y1,..,YN)] Proposition 14.40. Let 1 < p < oo and 0 < s < 1. Then the seminorms I'Iwa,P(RN) and I'IB9,P(RN) are equivalent.
Proof. We use the notation (E.2) in Appendix E. In view of the previous exercise it suffices to show that ILIB9.P(RN) <- CIuIWa.P(RN)
for all u E W'"P (RN). Let u E C0° (RN). For every xi, y'1 E RN-1, xl E R, and h > 0 we have
IDiu (x)
p
= lu (xi + h, xi) - u (xi, xi) Ip 2p-1 Iu (x1
<
+
+ h, xl) -u (xl + 2h, Yi) Ip
2p-1
Iu (xi + zh,y'1) - u (xi, x') 1P. Integrating the previous inequality in y'1 over the (N - 1)-dimensional ball centered at x1 and having radius 2h yields
Aiu(x)p
C hN-1
LN_l(X) lu (x +
- u (x1, x)
dy'1
14. Besov Spaces
450
Hence,
f
00
JN 00
O
P
(x)
JA
dxdh
J fBl(,*)
Iul+h,x')-u(xl+yh ,)I<
RN
00
+C 00
!R1JBNl(x)
Iu(xi+.,yi)-u(xl,x'i)I"dy`ldxdh hsp+N
C(1-f-1T). We estimate 1. By Tonelli's theorem and the change of variables zl = xl+h,
I may be written as lu (xl + h, xi) - u (xl + JRN-1 fOCO JBN-l(x1, ) JR
- fR_l.fo LNl(X)J _i> R
f LfRNl L RN4
0O
. yi) IP
I((2 h*P+N
Iu (zl)
,
dx 1 d yl
dzldyidhdxl 1I
hsP+N
xi) - u (zl
Ixi-VLl
U sP+ N
-
2, V11)
dhdx 1
1P
dhdyldzldxi
zl-Iii-y;I Iu(zl,xi) -u(yl,yi)IP dyldyldz dx 1 1 RN-1 -1 -00 (zi - y1) eP+N where in the last identity we have made the change of variables yl = zl - z. Since in the last integral zl - yl ? Ix'i - VII, we have that C
fjN
I(zijxi) - (yi,yi)I < /2- (zl
- yi),
and so
1
Cf
J R LN -1 R IIu (zl,(zi,x) x') - (yl, Y,) 1,,,+N
r
I u (x) - u (y) I P dxdy.
JR"rJRN Ix-yIN+sP The term 11 may be estimated in a similar way. Similar inequalities hold for 0; u, i = 2, ... , N. We omit the details. 0
Chapter 15
Sobolev Spaces: Traces Living with P.Q.S, III: How long does PQS last? PQS is incurable. With proper treatment, its main symptoms may last up to 5 years. -Jorge Cham, www.phdcomics.com
In this chapter we characterize the trace of a Sobolev function u E W 1,P (9),
namely, the "restriction" of u to the boundary 8SI. We begin with the simplest case in which Sl is the half-space
R+ = {x= (x',XN) ERN-i xI8: xN>0}. If U E Wi,P (iE) with p > N, using a reflection argument (see Exercise 10.37(iii)), we can extend u to W1,P (RN) and then apply Morrey's theorem
to conclude that u has a Holder continuous representative u. Thus, the value of u on the boundary of ft, namely, on the hyperplane xN = 0, is well-defined.
The situation is quite different when 1 < p : N (unless N = 1). Note that in dimension N = 2, by Theorem 10.35 we know that there is a representative a such that u x2) is absolutely continuous for C1-a.e. X2 E R, but we do not know apriori if x2 = 0 is an admissible value, so that in general the pointwise value of u 0) may not make sense. In what follows, we need to distinguish the cases p = 1 and p > 1. We begin with p = 1.
15.1. Traces of Functions in W 1,1 (Q) Theorem 15.1. Let N > 2 and let X be the family of all functions u E L1,1 (iE) vanishing at infinity. Then there exists a linear operator Tr:X-,L1(WN-1)
451
15. Sobolev Spaces: Traces
452
such that
(i) Tr (u) (x') = u (x', 0) for all x' E RN-1 and for all u E xnc (R+}, (ii) for all u E X,
(x')I dx' < jN
(15 1)
la8u
(x) dx,
RN-1
(iii) for all V; E CC (RN), u E X, and i = 1, ... , N, (15-2)
uOx dx = Je+ where v = -eN .
%b_
dx + f
ip Tr(u)vidx',
RN-1
Proof. Step 1: Assume first that u E
LA11 (RN) n C' (RN) with Vu E
L1 (R)V;RN). Reasoning as in the first step of the proof of Theorem 11.2, for every x' E we have that R-1v-1
Iu (x', 0) I <_
r I axN (x', xN)
I
dZN.
Integrate both sides with respect to x' and use Tonelli's theorem to conclude
that fRN-1 Iu (x', 0) I dx' <- RN f l a&U (x) dx. I
Define Tr (u) (x') := u (x', 0). Then Tr is a linear operator satisfying (ii). Moreover, classical integration by parts yields (15.2).
Step 2: To remove the additional assumption that u E LN' 1 (RN) n C1 (RN) with Vu E Ll (RN; RN), note that given a function u E X, using reflection (see Exercise 10.37), we can extend u to a function u E L1,1 (RN) vanishing at infinity. In turn, by Theorem 11.2, we have that u E L N-1 11 (RN). Let uE :_ cpE * it, where tpE is a standard mollifier. As in the proof of Lemma 10.16, we obtain that lim 11U, - u11 C--+O+
L7
(RN)
= 0,
0.
lim 11DuE - VU 11 L, RN.RN
Since, by Step 1,
J N-' Iu (x', 0} - ue' (x', 0) I dx' <
JRN
I axN
(x) - axN (x) dx I
for all e,,-' > 0, we may extend Tr uniquely as a linear operator TV : X -r L' (RN-1) satisfying properties
The function Tr (u) is called the trace of it on xN = 0.
0
15.1. 'Daces of Functions in IV',' (l)
453
Remark 15.2. In particular, it follows from the previous theorem that the linear operator
Tr : W1'1 (R+) - L'
(RN-1)
is continuous and satisfies (i)-(iii).
Exercise 15.3. Adapt the previous proof to conclude that there exists a linear operator a : Wlo1-1c (R+r) --, Lloc (RN-1) such that (i)-(iii) hold (with the obvious modifications).
Exercise 15.4. Let u, v E
W1,1 (RN). Prove that for
all i= 1'...,N
ua!dx=_ f va- dx+ fTr(u)Tr(v)v:dx', f +8xi U(
Ox{
RN-1
where v = -eN. Exercise 15.5. Let u E (W1,1 (RN) and v E W1,1 (RN), where RN
{ (x', XN) E
RN-1 X R : xN < 01.
(i) Prove that the function w : RN - R, defined by W (x) :_
!
if x E RN,
v (x)
belongs to BV (RN).
(ii) Prove that the function w belongs to W1,1 (RN) if and only if Tr (u) = Tr (v).
Next we prove that the operator Tr is onto.
Theorem 15.6 (Gagliardo). Let g E L1 (RN-1), N > 2. Then for every 0 < e < 1 there exists a function u E W 1'1 (R+) such that Tr (u) = g and
f+ Iu (x)I dx < e
f
IVu (x) I
RN-1 I g (x+) I dx',
dz < (1 + e)
&.,. RN-1 Ig (x') I
Proof. If g = 0, it suffices to take u = 0. Thus, assume that g ,-6 0. By Theorem C.23 there exists a sequence {gn} C C:° (RN-1) such that gf1 -p g in V (RN-1). For each k E N there exists N1. E N such that for all n >_ NA,, C
11 9- - 9 IIL1 (RN-1) 5
2I
II9IIL1(RN-1)
.
15. Sobolev Spaces: Traces
454
and ho := 0. The
Let nk := max{Nk,Nk_1 + 1} and define hk := sequence {hk} satisfies the inequalities (15.3)
IIhk+1-
hkllL1(RN-1)
IIhkIIL1(RN-1)
< 2k IIgIILI(RN_1)
< (1 + E)
for all k E N,
II9IILI(RN-1)
for all kEN0.
Construct a strictly decreasing sequence {ti.} C (0, 1), k E No, such that
tk->0and (15.4)
Itk+1 - tkI <
II9IILI(RN-1) k2 IIVx'hk+lIIL1 + IloilhkllL1 + 1
to
<e 4
For x E RN define
ifxN>to,
0
tk - xN hk+1 (x') + xN - tk+1 hk (x') if tk+1 C xN < tk.
(x)
tk - tk+1
tk - tk+1
Using Theorem 10.35, we show that u E W1,1 (1[$N) . Indeed, using the facts
that for tk+1 < xN < tk, Iu (x)I < Ihk+1 (x') I + Ihk (x') I (15.5)
I aa
(x) < I
I
b
L
8u 8xN
(x )
I
(x') + exk I
,
(xj) I
for all i = 1, ... , N - 1,
Z
< I hk+1 (x') - hk (x') I tk - tk+1
and that u = 0 for XN > to, by (15.3) and (15.4) we have that X00
tk
r JR+ Iul dx Ik0
fJJff
+l
JN-1 Iul dx+dN
00
C
Itk+1 - t,I (lIhk+1IlLtN_1) + IIhkIILI(RN-1)) k=0 00
<-
4II9IIL1(RN_1)
E Itk+1- tkl = 4t0 k=0
<_ E
II9IIL1(RN-1)
II9IIL1(RN-1)
15.1. Traces of Functions in W1,1 (n)
455
Similarly, by (15.3) and (15.5), 00
N
dx
+
-
tk
au
dx dxN
Jtk+1 JRN-1 axN 00
E < _0 L._1 Ih k+l
(s')
- hk
I dxr
k
00
<-
IIhk+1- hkIIL1(RN-1)
IIh1IIL1(RN-1) +
k=1
<- (1 + 2e) II9IIL1(RN-1)
while for i = 1, ... , N -1, 'ix
0,44
+
I
V-
I
00
=
>
0
11.
k=O 00
jN-1
E Itk+1 - tkI k=0
<E
ax=
dx'dXN ahk
ahk+1
(x') D dzr 8xi (x1) + axt
LAr-1
II9IIL1(RN-1)
by (15.4) and (15.5). Since u is locally absolutely continuous on GN-1-a.e. lines of R that are parallel to the coordinate axes (why?), by Theorem 10.35 we have that u E W1"1 (R+). It remains to show that Tr (u) = g. Reasoning as in the proof of Theorem 15.1, we have that J,aN1
Iu (x', ZN) - Tr (u) (x') I dx' <
J
su
xN (x) dx'dxN
xN LN_l
for G1-a.e. xN > 0. Let E be the set of xN > 0 for which the previous inequality holds. Then, lim
(15.6)
ZN_,O+ RN-1 xNEE
u(x',ZN) -Tr(u)(x')I dx'=0.
On the other hand, by the definition of u we have that for tk+I < xN < tk, Iu (x', ZN) - g (x') I
tk - xN P-i
l(hk+l I tk - tk+ 1
(XI)
-TN
- g (x')) + tk - j+11 (hk lx')
<- Ihk+1(x') - g (x') I + Ihk (x') - 9 (x') I ,
and so, since hk - g in V (RN-1),
lim J N-1
XN
O+
Iu (x'? ZN)
- 9 (x') I dx' = 0,
g(am))
15. Sobolev Spaces: Traces
456
which, together with (15.6), implies that Tr (u) (x') = g (x') for .£N-1-a.e.
p
x'ERN-1
Remark 15.7. (i) Note that from the proof we actually obtain a stronger estimate than the one in the statement of the theorem, precisely, r
aU I
JN axnr
dx<(1+e)
RN-1
(x') I dx',
l
andfori=1,...,N-1,
f
8u ax;
d x <_ e
l
lg (x')( dx'.
ieN-
Moreover, the function u is locally Lipschitz away from the boundary.
Peetre [1371 has proved that there does not exist a bounded linear operator L :L1 (RN-1) -p w1,1 (RN) 9'-a L (g)
with the property that Tr (L (g)) = g. Next we extend the previous theorem to special Lipschitz domains.
Theorem 15.8. Letl!f : RN-1 -> R be a Lipschitz function, N >- 2, and let RN-1 X
fl ;= `(x',xN) E
(15.7)
R : XN > f (x')}
.
There exists a continuous linear operator flN-1)
TY : W1,1 (n) -> L1 (an, such that
(i) Tr (u) = u on an for all u E W 1,1 (n) n c (ii) for all u E W1,1 (n),
dN-1 < i + (Lip f I2 (x) l dx, IT (u) Jill JS2 OXN (iii) for all hii E C'I (RN), u E W1,1 (n), and i = 1, ... , N, I
fu2±dx=_f!.dx+f
rJ
where v is the outward unit normal to an, that is, for RCN-1-a.e. x' E RN-1 (15.8)
v (x" f (x')) =
VIIf (x)
(VII-+
-1
(x'F V11-+ Ivx-f (x')I2
15.1. paces of Functions in W1,1 (n)
457
Proof. Set w(z) := u($(z))= u(z',zN+f (z')),
z r= R+,
where 'Y : RN -> RN is defined by
FY (z) := (z', zN + f (z'))
z E RN.
,
As in the proof of Theorem 12.3 we have that given by 'P
is invertible, with inverse
-1: RN SRN (x', XN) - (x', xN - f (x+) )
and that P and '-1 are Lipschitz, W (RN) = 0, and det V (z', zN) = 1 for GN-1-a.e. z' E RN-1 and for all ZN E R. Hence, by Exercise 10.37(iv), we have that w r= W1,1 (RN), with (15.9)
(z)
T
ax{ N(z)=
a
''N + .f (z')) + 8xN (2, zN + f (z')) 8f (z') ,
(z',ZN+f (z'))
It follows by Theorem 15.1 and Exercise 15.4 that Tr (w) E L1 (RN-1) (15.10)
LN....l Tr (w) (2)1 dz' <
f
N
I
azN (z)
dz,
and for all v E W"' (RN) and i = 1, ... , N, (15.11)
fa+
w
av aw v -dz+ -dz= 8zi f N-i Tr(w)Tr(v)v1dz', fut + 8 z;
where v = -eN. Note that u = w o'-1. Since P -1 (8SZ) = RN-1 x {0}, by identifying RN-1 x {0} with RN-1, with a slight abuse of notation, for x = (x', f (x')) E M we may define Tr (u) (x', f (x')) := Tr (w) o'P -1 (x) = Tr (w) (x') .
458
15. Sobolev Spaces: Traces
Then by (15.10),
.fm
ITr
dHN-1 (u)I
1 + IVz,f (x')I2 dx'
= JRN-1 I Tr (w) W)
'
1 + ILip/I2 .N-1 I Tr (w) (x') I dx'
<
f2 (z) 1R + 5ZN V1+ f r f 18u (x) I d x, ILiP 1
I
l I
OXN
where in the last equality we have used (15.9), Theorems 8.21 and 11.51, and the fact that det VIP = det VIP-' = 1. To prove part (ii), let tJ' E Cc' (RN) and define 0:= TlioW-1. By Theorem 11.51, ¢ E W1'1 (R+), and so (15.11) holds. Using (15.9) (for u and O) and Theorems 8.21 and 11.51, we have that
(15.12)
Rw N +
dz + 20 19ZN
0
19W
49ZN
fRN +
dz =
fn
u
OXN
dx +
Tau
ff, axN
dx,
while
IRN-1 Tr (w) (z') 0 (z', 0) dz'
(u) (x', f (x')) 0 (x', f (x')) dx'
RN-1
Ivx f
=-
fRll-1
'k
(u) (x', f (X,)) 0 (x', f (x,))
(x'}I2
1+
dx'
(x')I
= f8 S'2
l
where we have used (15.8). Together with (15.11) and (15.12), this shows
that (15.13)
1 u-
J
dx +
N
J
4'x0' dx =
J
¢ Tr (u) vN
dHN-1
15.1. Traces of Functions in W 1,1 (11)
459
On the other hand, if i = 1, ... , N -1, once more using (15.9) and Theorems 8.21 and 11.51 in (15.11), we get
o=f
v' (z) dz + fl,
w (z)
(z) az (z) dz
+
= fa u (x) 8x (x) dx + fn ¢ (x) ax (x) dx
f
+ n u (x) a N (x) a
(x) axN (x) of (x') dx.
(xf) dx + fn 0
Hence,
1uadx+ J axi
¢6audx in 8x;
l_ O
where
\.l
(,l u O sl
r
1
(&L
J
dx +
N
is a mollification of
f.
axN
af8xi} dx/
A in RN-1. Define '
oe (x)
0 (x) axi) (x')
,
x E RN.
a
Then ¢f E CC (W") and
80. ax N
L0 (x r axN (x} (OxIJE )
Hence, by (15.13),
JuP±dx+fdx a1 exi elo+ Un uaxN dx + fn El Tr (u) vN J
axN dHN-1 =
dx
-1
OTr (,u)
L
VN
N
To conclude the proof, it suffices to observe that by (15.8), for RCN-1-a.e. X' E
RN-1
of (x') vN (x', If (x')) = - 8xi
8x4
Vi
10x, f (x') I
x
= Ui (x" f (x'))
0 The previous theorem continues to hold with W 1,1(0) replaced with L1,1 (S2).
15. Sobolev Spaces: Traces
460
Exercise 15.9. Let Il C RN be as in (15.7) and let it E W 1,1(11) and £N-1-a
4) E Cc' (RN). Prove that for
e. x' E RN-1,
Tr (u.4)) (x', f (x')) = 0 (x', f (X')) Tr (u) (x', f (x')) .
Finally, we extend the previous result to open sets f C RN with uniformly Lipschitz boundary.
Theorem 15.10. Let 0 C RN, N > 2, be an open set whose boundary O1 is uniformly Lipschitz. There exists a continuous linear operator
Tr : W1'1(f) -, L'
(an,?-1N-1)
such that (i) Tr (u) = u on c7S2 for all u E W 1"1(1) fl C (Sa), (ii) for all ip r= C1 (RN), u E W 1-I (12), and i = 1,... , N,
r 8x1
dx _-
au Ip
dX
+f 0Tr(U)v'dhN-1, ,
8x1 n where v is the outward unit normal to OIL
Proof. Let e, L > 0, M E N, and
be given as in Definition 12.10.
We proceed exactly as in the proof of Theorem 12.15 and define 4)7L1 Ilo, Qt,
4)o, 4f, i/if as in (12.8), (12.11), (12.12), (12.15), respectively. Then for all x E Cl we may write (see (12.17) ) (15.14)
'4)+ (x) 4)n (x)
u (x) _
c 4)k (X)
u (x) + t/i_ (x) u (x)
u,z (x) + u_ (x) . m
By (12.15) the support of 0_ is contained in ft Hence, we define the function u_ satisfies (ii), that is
Tr (u_) ;= 0. We claim (15.15)
8x1
If,
E CI (RI"). To see this, construct a cut-off function 4) E C' (Il) E CC (RN) we have that 0 E C' (Cl), and so, by the definition of weak derivative, for all i = 1, ... , N for all
such that 0 = 1 on supp L. Then for every we have ia u_
OW) dx = ax{
n
¢
dx. 8x1
Since 4) = 1 on supp t/i_, we have that = 0 on the support of u_, and so the previous equality reduces to (15.15). This proves the claim.
15.1. 'Daces of Functions in IV1,1 (fl)
461
Next we study the functions un. Fix n. By property (iii) of Definition 12.10 there exist local coordinates y = (y', yN) E RN-1 x R and a Lipschitz function f : RN-1 -> R (both depending on n), with Lip f < L, such that
fto n ft = fin rl An, where
A.:= {(Y',yN) E
RN-1 X
R: YN > f (%()I Since by (12.9) the support of un is contained in fln, we may extend un to be zero in An \f2n. Thus, we are in a position to apply the previous theorem to obtain a function Tr (un) E L1 (&A., 9`(N-1) such that (i)-(iii) of Theorem 15.8 hold (with An and un in place of ft and u). Since -
k+ In
U.
02
it follows by Exercise 15.9 that Tr (u.n) =
,,2 u
01I8An 11
k
Since the support of On is contained in fLn C An, it follows that b,IOA,, = on OA n fl f2. Hence, the same holds for Tr (un). This shows that
f
(15.16)
n
dhN-1 <
1Tr (uva)I
1 + L2 J
IDunI dx
A9"
n"A n
=
1 + L2 j
I Vunl dx.
JJJ (2nS2n
Moreover, for all Vi E Cr' (RN) and i = 1, ... , N,
Ja
un Oxi dx +
ax
dx =
A.
u, e dx + fAn' - dx 0 Tr (un) vi
JOAWn
fannOA.,
dRN-1
1
R (un) vi
dHN-1
Hence, (15.17)
J in
un± dx = -J barn dx+ J axi axi S2n&A,,
0Tr(un)vid7{N-1
S2
Define
Tr (u)
'Ir (un)IannOAmn +'Ir (u_) =
Tr (un) Iad2naA.n .
0
15. Sobolev Spaces: Traces
462
Then summing (15.15) and (15.17) for all n and using (15.14), we get
JuP-±dx= >2 8xi
= - >2 n
J
_
u 8xiL dx + fin u_ ± dx 8xi
Jin
92!1 d x axi
8xi
d2,
+l
ast
j (
,11
sut
1I axi
(fix + >2 1"'n&A:.. n
,h Tr l(u)1 vi
'1
Tr (Yin) vz
dnN-1
dl.{N-1
for all' E C' (RN) and i = 1, ... , N, where we have used the fact that the support of 0 intersects only finitely many Stn, since {Stn} is locally finite. On the other hand, from (15.16),
JITr(u)I dnN1 n
n
<
-+L 2 >2
1
n
Jana, IVunl dx.
To estimate the right-hand side of the previous inequality, we use the fact that since {Stn} is locally finite, any bounded neighborhood of every point x E RN intersects only finitely many Stn's. Hence, in 0 fl1 On, by (15.14), 4 VYSn =
} (0 VV;+ + 2vy+0.Von)
-1;+.0n
s
'
(;:5)
u
2
+ '0+0n Vu. E 02 k
k
Since
1, using (12.10), (12.13), and (12.19), we get that in On Sin,,
Iounl <
CM
Iul + IVul .
Hence, ITr (un)I a-HN-
ITr(u)I dhN-1 <
fan CM n
Jan
< This completes the proof.
1
-+L-2 f
Jul
dx +
1 + L2
J
I Vul dx.
0
Remark 15.11. Note that by Exercise 12.13, if 1 C RN is an unbounded open set whose boundary 80 is uniformly Lipschitz, then n has infinite measure. On the other hand, if Si C RN is an unbounded open set with
15.1. 'Daces of Functions in W1,1 (fl)
463
finite measure, then 812 is unbounded. In this case the function 1 belongs to W1°1(12), but its trace does not belong to L' (812,%N-1). Therefore in this case, if, say, 811 is locally Lipschitz, then we can construct only a locally bounded linear operator Tr : W1'1 (C2) - L110
(811,?5tN-1)
Exercise 15.12 (Gagliardo). Let Cl C RN, N > 2, be an open set whose NN-1), then
boundary OC2 is uniformly Lipschitz. Prove that if g E Ll (On,
there exists a function ti E W1,1 42) and a constant C = C (Cl) > 0 such that Tr (u) = gjIu(x)I and
dx+IQIVu(x)I dx
in
dfN-1.
Exercise 15.13. Let f2 C RN, N > 2, be an open set whose boundary 812 W1,1 (RN 1
is uniformly Lipschitz and let u E W1,1 (11) and v r=
(i) Prove that the function zu : RN -r R, defined by u (x) if x E 12,
fw(x)
v(x) ifxERN\I,
belongs to BV (RN). W1,1 (RN) if
(ii) Prove that the function w belongs to Tr (u) = Tr (v). Exercise 15.14. Let Q,.= (0,
1)N-1
and only if
and Q = (0,1)N.
(i) Prove that if u E C' (RN) n W"1 (Q), then for all XN E (0, 1),
f
1u (x'+ 0) I
dx' <
f
Iu (x', xN) ( dx' +XN
+f J tIaN (Y,yN) dy'dyN. QXN (ii) Prove that if u E W1,1(Q), then for all e E (0, 1),
fq
I'I
(u) (x') I dx' < 11iu(y)I dy
+f
eu
I
dy. 8XN (y) I
C W1,1 (Q) converges weakly in Wl,l (Q) to a function u E W 1,1 (Q), then T5r converges to Tr (u) in Ll (Q'). Hint: Use part (ii) and the Rellich-Kondrachov theorem. (iv) Construct a sequence {u,} C W1'1 ((0, 1)) bounded in W1" ((0, 1))
(iii) Prove that if
such that u - 1 in L' ((0, 1)) and Tr (u,s) (0) -> 0. Why is this not in contradiction to the previous part?
15. Sobolev Spaces: 7kaces
464
15.2. Traces of Functions in BV (1Z) In this section we prove that every function in BV (fl) has a trace in Ll (Bp, WN-1), provided Sl is sufficiently regular. As usual we begin with
the case in which Sl = R.
Theorem 15.15. Let N > 2. Then there exists a linear operator Tr : BV (R+) -' Ll (RN-1) such that
(i) Tr (u) (x') = u (x', 0) for all x' E
RN-1 and for all
u E BV (R+) n
c (R+), (ii) for all u E BV (R+),
J
N-1
ITr (u) (x') I dx' <_ IDul (R+) ,
(iii) for all t/ E C (RN), u ErBV (RN), and i = 1,...,N,
Ju
(15.18)
where v = -eN. Proof. By Theorem 13.9 for every u E BV (RN) there exists a sequence {u,,} C Coo (R+) n W1'1 (R+)
such that u,, --+ u in Ll (R+), Vu, GN l R+
Du in the sense of measures
and
lim fRN noo RN
dx = IDuI (R+)
.
As in Exercise 15.14 for every E > 0 and n, m E N we have that
JNi
ITr (un - um) (x) I dx' <
+ E
+
N-1 X (O,e)
I (u, - um) (y) I dy
8 (uz
JRN-1 x (a,c)
'1
<-
J
N-1x(O,e)
- un) (y) dy 19xN I(un - u.) (y) I dy I
RN-1x(Q,E) 9 N
I O'XN (y) dy. (y) dy + fN-1x(O,E) R I
Letting n, m ->roo, we obtain ITr (un - urn) (x') I dx' < 2 IDul lim sup n,rn-oo RN-1
J
(RN-1 x (0,.
15.3. Traces of Functions in W 1,P (fl), p > 1
465
Letting a -> 0+ in the previous inequality, we have that {Tr (u a)} is a Cauchy sequence in L1 (RN-1) and thus it converges to a function Tr (u) in
L' (RN-1). Moreover, by (15.1) and (15.2) we have that for all n E N,
f
v_1
IT (U-) (x') I dx' < J
N
l
a
dx,
N (x)
and for all .0 E CC (RN) and it = 1, ... , N,
8xi J un.a0dxb+
R+
Oxi
f dx+ J RN-1 ipTY(un)v4dxl.
Letting n -+ oo gives (ii) and (iii). Since W1"' (RN) C BV (R+), it follows from Theorem 15.6 that
Tr : BV (R+) - Ll (RN-') is onto.
The analog of Theorem 15.10 is given by the following theorem. The proof is very similar to the one of Theorem 15.10 and is left as an exercise.
Theorem 15.16. Let Q C RN, N > 2, be an open set whose boundary 8f1 is uniformly Lipschitz. There exists a continuous linear operator Tr : BV (fZ) - L1 (8ft, xN-1) such that
(i) Tr (u) = u on t7I for all u E BV (C2) n c (St), (ii) for all V; E C' (RN), u E BV (0), and i = 1, ... , N,
-J ODiudx+
f
bTr
(u)VidfN-1
where v is the outward unit normal to O.Q.
15.3. Traces of Functions in
Wl,P (0),
p>1
In this section we study the trace of functions in bV'4r' (ft) for 1 < p < N. We will see that the situation is quite different from the case p = 1.
Theorem 15.17. Let 1 < p < N and let Xp be the family of all functions u E L''P (R+) vanishing at infinity. Then there exist a linear operator p(N-1)
Tr : X, -r L (RN-11 and a constant C = C (N, p) > 0 such that 1 (i) Tr (u) (x') = u (x', 0) for all u E X. fl C
R
),
466
15. Sobolev Spaces: Traces
(ii) for all u E Xp,
I(x')
N-y
I"('N
//p
-T,(.) d.1//f AIN-1' <
J
1J
P
I
Du()Ip x dx
RN
(iii) for all 'k E C' (RN), u E Xp, and i = 1, ... , N,
fE u! dx = _ f 8xi
g+
aU
dx +
fix{
f
RN-1
0a (u) vi dx
where v = -eN.
Proof. Step 1: Assume first that u E LP' (RN) fl Cl (RN) with Vu E LP (RN; RN). Reasoning as in the first step of the proof of Theorem 11.2, for every x' E RN-1 and for r > 1 we have that ju
(x', 0) I'
< rf
Iu (x', XN)
00
O
Ir-'
OXN
(x , xN) dxN.
Integrate both sides with respect to x' and use Tonelli's theorem to conclude
that Iu (x', 0) Ir dx' <- r fRN Iu (x)I'-1
(15.19)
1
< r (JR4N
It,(x)I(r-1)
dx
fN
dx
1
eu (x)P dx
axN
R
where we have used Holder's inequality. Taking
p(N - 1)
N-p I
we have that (r -1) p' = ps, and so, by Theorem 11.2,
(INR 1
_) N-p
P(
lu (x'? o) I
dx' < C
u (x)Idx
1
I
P
OXN
1i +1
+ 1 = p1N p
-
.
so that
PIN-1
rJJ/'
IVu(x)IPdxIP
N-D
NP
N-1
1
P I Vu (x)IP dx 1J
dx' C . (xO) u < (JR1 Step 2: To remove the additional assumption that u E LPG (RN) /nC1 (RN) with Vu E LP (RN; RN), we proceed as in the proof of Theorem 15.1 by I
15.3. Traces of Functions in W 1,P (l), p > 1
467
first extending a function u E X. to a function u E Ll.n (RN) vanishing at infinity and such that 1
l1
/' (JRN IVu(x)Ip dx
y
y
ou(x)IP dx
(fRRN +
and then using mollifiers. We omit the details.
U
Remark 15.18. In particular, it follows from the previous theorem that the linear operator 1
T r: W ' (RN)
r LN
(RN-1)
is continuous and satisfies (I)-(iii). Moreover, taking r = p in (15.19) yields rRN-1 lu (x', 0) Ip dx' < p (fR' Iu (x) Ip dx
N (x)
(fR48
< p fR+ Iu (x)Ip dx + p
`
(x)
dx
' dx,
where we have used Young's inequality (see (B.17) in Appendix B). Note that this inequality actually holds for every 1 < p < oo. Hence, reasoning as in the proof of Theorem 15.1, we conclude that
JtN-' ITr (u) (x+)
lp
d2, < pfR+ Iu
(x)IP dx + p fR+ 18XN (x)I dx,
for all u E W1,n (R+).
Exercise 15.19. Prove that if p = N, then for every N < q < oo there exists a continuous linear operator Tr: W 1'N (RN)
--> Lq
(RN-1)
such that (i)-(iii) of the previous theorem hold (with the obvious modifications).
Unlike the case p = 1, when 1 < p < oo, the trace operator Tr : W1,P (R+) Lp (RN-1) is not onto. We now show that if u E W1P (RN), 1 < p < oo, then its trace Tr (u) belongs to the Besov space B1 p'p (RN-1) (see Definition 14.1).
Theorem 15.20. Let 1 < p < oo and let N > 2. Then there exists a constant C = C (p, N) > 0 such that for all u E L1.' (I8+), (15.20)
IVu (x)Ip dx
Imo' (uIBh1(RN_t) <_ C f1R
.
15. Sobolev Spaces: Traces
468
Proof. Assume that u E L" (RN) n C°O (RN). For every x' E RN-1 and
xN>0Write u (x', 0) = u (x', XN) - (u (X', ,TN) - u (x', 0)) = u (x', XN) - ANu (x', 0) ,
where we have used the notation of (14.1). Integrating in xN over the interval (0, h), h > 0, yields Iu (x', 0) I
f h Iu (x', 2N) I dxN +
J f h IAN u (x', 0) I dxN. u, where i = 1, ... 11F-1,
We now replace the function u with the function 1
to obtain
Ih
1
1, u (x', 0), < <
Ahu (x', XN) dxN +
h. J 0
1
jh
IAN Au ( , o) I dxN
h Iu(2;'N)IN h(IAN
IAN u(x',0)I) dxN,
(x +
+If
0) E RN-1 x R. By the fundamental theorem of calculus and
where ej
Tonelli's theorem weff have that IXhu (x', 0)
h
hf
s
(x' +ddN
+ h J0 JO \ a N
rh fh
1
JJ 0
+
f
I
0
a
(x' + he" z) + I
I
N (x', z) I) dzdxN
(x` + e xN) I dt dxN
h l
o
IB-N (x'+hey, z)I+1019XUN
(x',z) l) dz
By Corollary B.83 we get o(., 0)
<
IILP(K^`-1) h
h f fa1(')II h
dddxN
l
+Cf h
II a N (. + hei, z) II LP(RN-1)
< Cf h
II8U
(-,xN)II
,
RN-1)
+IIaN ( x)
+ II O N
(',xN)IILP
dz II
L
dxN. ]RN-1)
15.3. Traces of Functions in W1,P (fl), p > 1
469
Fix e > 0 small and apply Holder's inequality to obtain
hP
-Ep f hMN
a (l
Hence,
f
00 1
V
o
D;u(,0) KPOW-1) A - C j x
(1
a
LP(RN-1)
p
WN
h
aN
Ir
LP(.N-1)dxNdh
P
xN AN
00
x!
hps-EP
J
+
(',XN)Ilp
00 F-9
00
xN)
axi
+ aXN LP{RN-'}
P
/_
\ II-'xN) IILP(RN_1))
1
dhdxN
it -rep
8u p
8u
Oxi LP(R+)
+ 1a
P IrNHLP(Rfl)
where we have used Tonelli's theorem.
To remove the extra assumption that u E C00 (RN), we first use a reflection argument to extend every function u E L1,P (RN) to a function u E L1,P (RN) and then consider a sequence of fuel, where uE := u * We and the We are standard mollifiers. Since ue --+ u in L10 0 (RN), Vu5 (RN-1), Vu in L (RN; RN), and ue 0) - Tr (u) in L1 by selecting a subsequence, we may assume that u. (x', 0) -+ Tr (u) (x') for GN-i-a.e. x' E RN-1. Hence, by Fatou's lemma,
ffR
ITr (u) (x' + he;) - Tr (u) (x') Ip N- 1
r°° < lim inf s- 0+
J0
f
dx dT
hP
RN
S C (N,p) Eli o+
dx'dh
hp Ius (x' + hei, 0) - uE (x', 0) I P
IVuE (x)IP dx = C (N,p) f fRRThis N
concludes the proof.
IVu (x)IP dx.
D
The previous result shows in particular that
Tr (W1,P (RN)) C B'-*` (RN-') l1
We now prove the opposite inclusion.
N
1!
15. Sobolev Spaces: Traces
470
Theorem 15.21. Let 1 < p < oo, let N > 2, and let g E B1-n'p (RN-1) Then there exists a function u E W 1,P (R+) such that Tr (u) = g and (15.21)
II 4w1,P([N< C "
11 Bl-p."(RN-1\
t°
1
7
where C=C(N,p)>0. Proof. Let cp E C' ° (RN-1) be such that supp V C BN_ 1(0,1) and cp (x') dx' = 1.
LN-I For x' G RN-1 and xN > 0 define
x ,_ xNpr
V (2, ZN)
1
xN-1 N
9 (p) dp'-
iP
JIRN_1
By Theorem C.19 (with xN in place of E) we have that for all xN > 0,
f
(15.22)
N-1
IV (x',xN)Ip do;' < J
Ig (x') I" dx'. N-1
With a slight abuse of notation, for every i = 1, ... , N - 1 we write
x' = (x'i, xi) E
RN-2
X It
By Theorem C.20 (where xN plays the role of E), for any i = 1, ... , N - 1 we have that
I
RN8cp
_, xai
X
x'
dpr
-
g (VI)
-IN
aIP('N11l
19
1
.
xN JRN-i
L (N) [g (x? - y') - g (xi' - y!', xi) ] dy',
where in the second equality we used the fact that
a R axi
x' xN
g
(Vill,
xi) dyi = g (t!"$ xi)
8cp x - 3 dz/i = 0 fJR axi xN I 1\
Since supp cp C BN_ 1 (0,1), we have that (15.23)
av axi (x) I
:5
c
N
xN
I g (xi' BN-1(o,xN)
Vi", x;
- fi) - 9 (Z - y"o xi) I
dgf'.
15.3. Traces of Functions in W1,P (n), p > 1
471
Raising both sides to the power p, integrating in x over R+, and using Holder's inequality, we get
f
P
r
(x)I dx
JR+ C7xi
C
g
Np
R+ xN
B
(x_ y') - g (xi' _
XN
J
LN_I(O,XN) I g
xP
f
dx
(x'
- y') - g (x- y', xi) IP dy'dx
kg (x' - ii) - (x- y'xi)Idydydx
,-1
f
P+N 'r'N
)xN
< C RN R+ C2,
P
I dy'
I
(N-1)(p-1)
y',xi)
where in the last inequality we used Tonelli's theorem. Again by Tonelli's theorem and the changes of variables z° = x'j' - y:" and zi = xi - yi we get
that
fxN
x
Z- 10"0 J BN-2(O,xN) O oo
N-2 x1+N-1 N
Jo
NJR fN-2
xN
1
dz'dxdytdy'dxN
O&'g (z') (P dz'dyidXN
xN JRN-
I
00
1 f f N-1 I Qv:g (z) I" dz'dyidXN =: CI'.
=C J N
By Tonelli's theorem once more,
2=
l
00
00
I QYig R!``-1
=C LM....1
xp+1 dz'dyi
(z) IP 7/.
I
N
W) 1P dz' dyi . yip
Hence, we have shown that
jN+ a (x) foralli=1,...,N-1. (15.24)
IP
I
To estimate
YXN ,
dx < C
N-1 I9 (x' jwr J J
g (x') I P
dx'dh
we write
g (ii') - g (x') N-1
(gh, W
_ i=1
... , lip xi+1, ... , xN-1) - 9 (y1, ... yi-1, xi) ... 9 XN-01
15. Sobolev Spaces: Traces
472
Since fRN-1 cp (x') dx' = 1, we have that
' -g
1
v (x', ZN)
= : LN_l 1
xN -1
xN -g(y1,...,yi-1)xi,...,xN-1)] dy'
x [g(yl,...,yi,xi+1,...) xN-1)
+g(x'). By Theorem B.53 we obtain that av
axN (x XN) =
a
N-1 4-1
ey'
1
N-1 N-i &N xN
fRE
xN
f
x [g(yl)...,yi,xi+,,...,xN-1)
-g(yl)...,yi-1,xi,...,xN-1)] d!!
In turn, ev , (x , xN) axN N-1 C,
N
i=1 xN BN-i(O,XN)
Ig(x1 -y1,...,xi-yi,xi+1,...,xN-1)
-g (xl - y1, ... , xi-1 - yi-1, xi, ... , xN-1) I dy
i
We can now continue as before (see (15.23)) to conclude that (15.25)
r f RN +
N-1
In I
a- (x)
41X < C
4-1 J
J N-1
I9 (x' + he' - 9 (z' ' dx'dh.
We are now ready to define the desired function u; precisely, for x = (x', xN) E 1R
we define 5A
u (x) := a-P V(x)By (15.22) and Tonelli's theorem we have
f+ I u (x)I n dx = <
Je00-xN / N-1 Iv (x', xN) I" dX,dXN Jf
j
=
e--N dxN
LN_l
JN.1
Ig (x')1i dx'
Ig (x')IP dx',
while for i = 1, ... , N - 1,
axi (x) I =
(
e
a (x) 1:5 1 a (x) I,
15.3. Traces of Functions in W1,P (fl), p > 1
473
and so, by (15.24), we obtain that the estimate (15.24) also holds for -. On the other hand, ou
8
2A
(x) = e
axN (x) -
-e 'v (x),
and so, again by (15.22) and (15.25), IP
t
I axN
(1001
dX)
<
fRN I vxN I p dX) 1
t
+
I9 (z' + he) - g (z')Ip dz'dh) N-1
/
hP
fRN I -e
t
+C
Pvlp &
(fRN-1
I9I
D
dx')
.
Since u E C1 (R+) , we have shown that u E W lp (R+) and that (15.21) holds. To conclude the proof, it remains to show that Tr (u) = g. By Proposition 14.5 we may find a sequence {gn} C C°° (RN-1)nBl-a'p (RN-1) such
that II 9 -
Let
R
N-1)- 0.
s
un (x) := e- P vn (x)
be the corresponding sequence. Reasoning as in the proof of Theorem C.19(i), we have that vn E Co (R+), with vn (x', 0) = gn (x'). By (15.21) we have that u,, - u in W 1,P (RN), and since Tr : W1,P (R+) -> LP (RN-1)
is a continuous operator, we obtain that Tr (u) = g.
0
Exercise 15.22. Under the hypotheses of the previous theorem and using the same notation, prove that v, E Co (R). The analog of Theorem 15.10 is given by the following theorem. The proof is very similar to the one of Theorem 15.10 (see also the proof of Theorem 12.15 and Exercise 12.14) and is left as an exercise.
Theorem 15.23. Let St C RN, N > 2, be an open set whose boundary 8Sz is uniformly Lipschitz and let 1 < p < oo. Then there exists a continuous linear operator Tr : W1,P (1) --+ LP (on, jjN-1) such that
(i) Tr (u) = u on 8fl for all u E W1,P (cl) n C (0),
15. Sobolev Spaces: Traces
474
(ii) for all 0 E C (RN), u E W1,P (n), and i = 1,... , N,
- fn
dx +
f
Tr' (u) vi
d'HN-1,
where v is the outward unit normal to 811. To characterize the traces of Sobolev functions in W 1,P (ft) when p > 1, we need to extend the definition of Besov spaces to on. To give an intrinsic definition that does not depend on local charts, it is convenient to use the equivalent norm introduced in Section 14.8.
Definition 15.24. Let 1 C RN, N > 2, be an open set such that Oil is locally Lipschitz and let 1 < p, 0 < oo and 0 < s < 1. A function g E LP (8ft, NN-1) belongs to the Besov space Bs,P,e (On, xN-1) if II9II B=.P.e(80,-{N-1)
II9IIl.P(e9,7{N-1) + I9IB8,P.e(00,IHN-1) < 00,
where I9I Ba.P.6(8n,fN-1) I
B
IgW-g(y)l P dW N-1 (x)
f (fan Ix - yI(N-1+88)P
P
B
d71 N-1 (Y)
For simplicity we write B",P,e (Oil) := BB,P,e (ail, xN-1) Moreover, if 8 = p, we set B$,P (on) := B8,RP (ail).
Exercise 15.25 (Gagliardo). Let fl C RN, N > 2, be an open set whose boundary &St is uniformly Lipschitz and let 1 < p < oo.
(i) Prove that if 1 < p < oo, then there exists a constant C = C (p, N, ft) > 0 such that III' (u)IIB1-p,P(8a) < C IItIIWI.P(n)
for all u E W1,P (f2).
(ii) Conversely, if 1 < p < oo, prove that for every g E Bl-p'P (aft) there exists a function u E W1,P (R+) such that Tr (u) = g and IIuII W1.P(n) -< C II9IIB1-P,P(aa) ,
where C = C (N, p, ft) > 0. As in Remark 15.11, if ft C RN is locally Lipschitz but aft is unbounded, then the results in the previous exercise continue to hold locally.
15.4. A Characterization of WW (SZ) in Terms of Traces
475
Exercise 15.26. Let SZ C RN be an open set whose boundary 8f2 is uniformly Lipschitz and let u r= W l,p (f) and v r: W 1,P (RN \ Sl), 1 < p < 00. Prove that the function f u (x) if x E fl, w (x) v (x) if x r= RN \ SZ belongs to W1,P (RN) if and only if Tr (u) = Tr (v).
Exercise 15.27. Let 1 C R2 and u E W 1,2 (0) be given as in Exercise 11.7. Prove that the trace of u on the line segment r :_ (0, t) x {0} is not in Lq (I') for any q > 1.
Exercise 15.28. Let Q':= (0,1)N-1, Q = (0,1)N, and 1 < p < oo. (i) Prove that if u E C1 (RN) fl W 1,p (Q), then for all xN E (0, 1),
fQ,lu(x',0)Ipdx'
+ C (p) xnT 1 0
'
f , axN au
(11', YN) lp
dy'dyN
(ii) Prove that if u E W1,P (Q), then for all e E (0,1), I'1
C (p) (u) (x') l p dx' <
JQ,
f
Iu (p) 1,1 dy ' x (0,e) Ep-1
+ C (p)
p
f
Ip I
8xN (y)
dy.
W',p (Q) converges weakly in W' (Q) to a (iii) Prove that if {u.,,} C function u E W'P (Q), then Tr (u,) converges to Tr (u) in LP (Q'). Hint: Use the Rellich-Kondrachov theorem.
15.4. A Characterization of Wo'p (0) in Terms of Traces In this section we show that if the domain 0 is sufficiently regular, then we may characterize W0'" (fl) as the subspace of functions in W1,P (SZ) with trace zero.
Theorem 15.29. Let f C RN, N > 2, be an open set whose boundary aft is uniformly Lipschitz, let 1 < p < oo, and let u E W1,P (SZ). Then Tr (u) = 0 if and only if u E I VW'p (S1) .
Proof. If u E CC° (0), then Tr (u) = 0, and so, since WO"' (1) is the closure
and the trace operator is of C°° (i)) with respect to the norm continuous with respect to strong convergence in W" (f?), it follows that Tr (u) = 0 for all u E Wa'p (SZ).
15. Sobolev Spaces: Traces
476
'Ib prove the converse implication, let u E W 1,P (f') be such that Tr (as) =
0. Using partitions of unity and flattening out 811, we may assume that SZ = RN and that u = 0 for all x E RN with I xI > R. For x E RN define _ u (x) if xN > 0, (a) V(X) if xN < 0. 0 By Theorems 15.1 and 15.17 and the fact that Tr (u) = 0, for all E Q' (I8N) and i = 1, ... , N we have that It N v Ox
=
JR
f+
u ax dx
lox
)
which shows that v E W1,P (RN), with
8v Ox,
(x) =
r
(x) if xN > 0,
t0
if xN < 0,
i = I,-, N. We now translate v upwards. More precisely, for t > 0 and x E RN define
Vt(x):=V(x,2N-t) Note that the support of vt is a compact set contained in RN-1 x (2, oo). For each 71 > 0, by Lemma 10.30 we may find t so small that IIu - VIII 4 l.°(RN) <71-
It now suffices to consider a sequence of mollifications {(vt)s}E>o of vt. Find
0 < e < a so small that II (vt)E - VIII 41.P(RN) <-11.
1RN
Since, for x E RN,
(vt)f(x) =
J
pe(x-v) vi (y) dy = f
_1
x (`oo)
p(x-y) vt() dy
and suppV, C B (0, e), it follows that if 0 < xN < e, then (vt)E (x) = 0. Thus ve E C'° (RN). This shows that u E W" (RN). O
Chapter 16
Sobolev Spaces: Symmetrization Living with P. Q.S, I V: How do 1 treat PQS? Experimental treat-
ment such as Advisor-Pressure (AP), Spousal-Income-Rustintion (SIF), Lack-of-Savings-Realization (LSR) and "cocktails" of these have known to cause a remission of the disease long enough to at least defend a thesis. More severe treatments are required to actually finish writing the thesis. - Jorge Cham, www.phdcomics.com
In the proof of the embedding theorem for p = N (see Theorem 11.29) we have used the notion of spherically symmetric rearrangement of a function. In this chapter we introduce and study this concept. For more information on this topics and on its applications to partial differential equations we refer the reader to the monographs of Kawohl 1971 and Kesavan [98].
16.1. Symmetrization in 17 Spaces As in Chapter 6, given a Lebesgue measurable set E C R' and a Lebesgue (0, oo], the distribution function of u is the measurable function u : E function gu : [0, oo) - [0, LN (E)], defined by (16.1)
qu(a):=CN({xEE: u(x)>s}), s>0.
Similarly, we define (16.2)
c (s) := RCN ({x E E : u (x) > s}) ,
s > 0.
We recall that u : E - [0, oo] vanishes at infinity if u is Lebesgue measurable and g,, (s) coo for s > 0 (gu (0) can be infinite). The proof of the following proposition is very similar to that of Proposition 6.1 and is left as an exercise. 477
16. Sobolev Spaces: Symmetrization
478
Proposition 16.1. Let E C RN be a Lebesgue measurable set and let u, v, u : E -, 10, oo], n G N, be Lebesgue measurable functions. Then the following properties hold:
(i) The function gu : [0, oo) -> [0, RCN (E)] is decreasing and right continuous, while cu : [0, oo) - [0,GN (E)] is decreasing. (ii) If u vanishes at infinity, then cu is left continuous, lim Su (s) = 0, Iim Qu (s) = S-»oo
a+oo
and Vu and c are continuous at s > 0 if and only if
GN({xEE: u(x)=s})=0. In particular, Su (s) = Lou (a) for all s > 0 except for at most a countable number.
(iii) If u (x) < v (x) for LN-a.e. x E E, then Lou < p,,. In particular, if u (x) = v (x) for GN-a.e. x E E, then pu = Q,,.
(iv) If u,b (x) / u (x) for LN-a.e. xEE, then p / Lou, Let E C RN be a Lebesgue measurable set and let u : E - [0, oo] be a Lebesgue measurable function. As in Chapter 6, the function u* : [0, oo) -[0, ool, defined by
u* (t) := inf {s E [0, oo) : Lou (8):5 t},
(16.3)
t > 0,
is the decreasing rearrangement of u.
Proposition 16.2. Let E C RN be a Lebesgue measurable set and let u : E -, [0, ool be a measurable function. Then the following properties hold:
(i) The function u* is decreasing and right continuous and (16.4)
u* (0) = esssup U. E
(ii) For all s, t > 0, u* (t) > s if and only if ou (s) > t. (iii) If u vanishes at infinity, then for all 0 < t < pu (0), t < su (u* (t)). (iv) If u vanishes at infinity, then for all t > 0, pu (u* (t)) < t, and if Lou (u* (ti)) < ti for some tl > 0, then u* is constant on [to, t1J, where (16.5)
to
pu (u* (tl))
Proof. Properties (i) and (ii) follow as in the proof of Proposition 6.3. (iii) If pu (0) > t > 0, then u* (t) > 0 by property (ii). Fix 0 < s < u* (t). Then pu (s) > t by part (ii), and since pu cu, we have S. (s) > t. Using the fact that Su is left continuous, letting s / u* (t), we get -;. (u* (t)) > t.
16.1. Symmetrization in LP Spaces
479
(iv) If u* (t) < s, then by part (ii), ou (s) < t. Taking s = u* (t) gives Lou (u* (t)) < t. Assume next that Lou (u* (t1)) < tl for some tl > 0 and let to be defined as in (16.5). For to < t < tl we have Lou (u* (t1)) = to < t and hence u* (t1)
is an admissible s in the definition of u* (t), and so u* (t) < u* (t1). But since u* is decreasing, this implies that u* (t) = u* (t1) in [to, ti].
0
Remark 16.3. If in the previous proposition we assume that u is also bounded, then by (16.4), u* (0) < oo. Hence, reasoning as in part (iv), we have that ou (u* (0)) = 0.
Exercise 16.4. State and prove the analogous statement of part (iv) of Proposition 16.2 for the function cu.
Definition 16.5. Let E C RN be a Lebesgue measurable set and let u : E [0, oo] be a Lebesgue measurable function. The function u# : RN -> [0, oo], defined by (16.6)
u# (y) := u* (aN I yI N) = inf Is E [0, 00) : ou (s) <_ aN I yI N }
for y E RN, is the spherically symmetric rearrangement of u. The function ul is also called the Schwarz symmetric rearrangement of u. Note that if u vanishes at infinity, then by Proposition 16.1(ii), ul (y) < 00
for all y 40. Given a Lebesgue measurable set F C RN, with 0 < GN (F) < oo, we define the spherically symmetric rearrangement of the set F to be the open ball centered at the origin and with the same measure of F, precisely F1 := B (0, r), where
r :=
(16.7)
LN(F)11p
aN/
where we recall that aN is the measure of the unit ball. If F has measure zero, then we take FO to be the empty set, while if F has infinite measure, then we take FO to be RN. Note that Fa is always open even if F is not. The next proposition collects some elementary properties of symmetrizar tion.
Proposition 16.6. Let E C RN be a Lebesgue measurable set and let u : E -> [0, oo] be a measurable function. Then the following properties hold:
(i) For all s > 0,
{yERhh1:u(Y)>8}={xEE:u(x)>s}I.
16. Sobolev Spaces: Symmetrization
480
(ii) For all s > 0, GN ({y E R'V : u# (y) > s}) = ,CN ({x E E : u(x) > s})
(16.8)
.
In particular, if u vanishes at infinity, then so does uo. Moreover,
LN ({y E E1t : ul (y) = 0}) < LN ({xEE : u (x) = 0}) with equality holding if and only if either
£N({xEE: u(x)>0})
,CN({xEE: u(x)>0})=oo and ,CN({xEE: u(x)=0})=0. (iii) The function ub is lower semicontinuous and (16.9)
up (0) = esssup u. E
(iv) If v : E -r [0, oo] is another Lebesgue measurable function with u < v ,CN-a.e. on E, then u# < v1 on RN. (v) If {un} is an increasing sequence of nonnegative Lebesgue measur-
able functions u : E -r [0, oo] such that u moo U,,, (x) = u (x)
for all x E E, then lim t4 (y) = u# (y)
n-+oo
for all y E RN.
Proof. (i) For s > 0,
IV R
: up (y) > s} =
{y
RN : u* (ON jY1N) > s}
By Proposition 16.2 we have that u* (ON I yI N) > a if and only if OU (s) > ON IYIN, and so
N:
p
YER N yERN : IyI <
aN (Ou(s)) CN({xEE: u(x.)>s}))N ON
={xEE: u(x)>s}p, where in the last equality we have used (16.7).
(ii) By part (i) the equality (16.8) holds for 0 a > 0. The second part follows as in the proof of Proposition 6.3(v).
16.1. Symmetrization in L" Spaces
481
(iii) To prove that ui is lower semicontinuous, it suffices to show that the set {y E RN : ul (y) > s} is open for all s E R. Ifs > 0, then by part (i) the set {y E RN : ua (y) > s} is an open ball or empty or RN, while if s<0,then {y E RN: uO(y)>s} = RN since uO > 0. Properties (iv) and (v) follow as in the proof of Proposition 6.10(1) and (ii), respectively.
Corollary 16.7. Let E C RN be a Lebesgue measurable set and let u E LOO (E) be nonnegative. Then ul belongs to L°° (En) and IIUIILOO(E)
Proof. By (16.6) and Propositions 16.2 and 16.6 we have that up (0) = if (0) = IIUIIL-(E) < oo. IIu'IIL-(EI) = 11
Exercise 16.8 (Uniqueness of Schwarz symmetrization). (i) Let fl, f2
:
(0, oo) - [0, oo] be two decreasing right continuous functions such
that
£N({xERN: fl (IxI)>s}) ='CN({xERN: f2 (IxI)>s}) for all s > 0. Prove that f, = f2. (ii) Let u : RN -- [0, oo] be a function vanishing at infinity such that
u (x) = f (IxI)
for LN-a.e. x E RN for some decreasing function f : (0, 00) -+ [0, 00] [0, oo). Prove that there exists a unique function v : RN vanishing at infinity such that (a) v (x) = f (IxI) for all x E RN \ {0} for some decreasing right continuous function f : (0, 00) - [0, 00), (b) v (0) = clim f (t), (c) u (x) = v (x) for CN-a.e. x E RN. Prove also that the function v is lower semicontinuous. [0, oo] be as in part (ii). Prove that u (x) = ul (x) (iii) Let u : RN for GN-a.e. x E RN. The proofs of the following results are very similar to the analogous ones in Chapter 6 and are left as an exercise.
16. Sobolev Spaces: Symmetrization
482
Theorem 16.9 (Hardy-Littlewood's inequality). Let E C RN be a Lebesgue measurable set and let u, v : E -> [0, oo) be two Lebesgue measurable functions. Then IE u (x) v (x) dx < f u# (v) v# (y) dy.
Theorem 16.10. Let E C RN be a Lebesgue measurable set, let u : E -> (0, oo) be a function vanishing at infinity, and let f : [0, oo) - [0, oo) be a Borel function. Then
ff
(16.10)
(uU
(y)) dy < fE f (u (x)) dx,
with equality holding if f (0) = 0 or LP' ({x E E : u (x) > 0}) < oo or
CN({xEE: u(x)>0})=oo andCN({xEE: u(x)=0})=0. In particular, for any p > 0,
L (UI (y)) P dy = L (u (x))" dx. Theorem 16.11. Let Q : R -> [0, oo) be a convex function such that 41 (0) _
0, let E C RN be a Lebesgue measurable set, and let u, v : E - [0, oo) be two functions vanishing at infinity. Then
f In particular,
T
f
(ul (l!) - vI (y)) dy < fE' (u (x) - v (x)) dx.
(y)I'
u# (y) - v n
dy <
JE lu(x) - v(x)I' dx
for all 1 < p < oo and the operator u i- u# is a continuous operator from 12 (E) into LP (E#).
16.2. Symmetrization of Lipschitz Functions In the next sections we prove that if u E W 1a' (RN) is nonnegative, 1 < p < oo, then u# E W 1,P (RN) and uvu III Lp(RN;RNXN) s IIVUIILP(RN;RNXN).
The proof is divided into several parts. The scheme is the following: Prove (16.11) for nonnegative functions in W 1,co (RN) vanishing at infinity.
Prove (16.11) for nonnegative piecewise afilne functions in W1" (RN). Prove the general case using density of piecewise affine functions. We begin by considering Lipschitz functions.
16.2. Symmetrization of Lipschitz Functions
483
Proposition 16.12. Let u E W 1,00 (RN) be a nonnegative function vanishing at infinity. Then u# : RN -p [0, oo) belongs to W1'00 (RN) and vanishes at infinity and IIu#IIL=(RN)
= IItIIL-(RN) ,
IIVUIILOO(RN;RNxN) .
II'Lm(jj1NANxN)
Proof. By Corollary 16.7 we have that
IIuI LOO(RN) -
IIuIIL-(RN) < 00.
Let yl,y2 E RN, with IFIII < Iy2I, and define L := claim that
IIVuIIL-(RN,RNXN).
We
0:5 u1 (yi)-ul(y2)
t):=up(y1)-u#(y2)>0. Without loss of generality, we may assume that n > 0, since otherwise there is nothing to prove. then, using the Lipschitz If xo E u 1 ([u# (yl) , oo)) and x E B (xo, continuity of u (see Exercise 11.46), we have L),
u# (yl) < u (xo) < u (x) + L Ix - xoI < u (x) + 91, which implies that u (x) > u# (y1) - ?l = u# (YO, that is, x E u 1 ((u# (y2), oo)). Hence,
\ u([u(yl) oo)) +B(0,) Cu ((u(y2)oo)) and so, by the Brunn-Minkowski inequality (see Theorem C.7),
( N (a-' ({u(yi) ,oo))))*
+ (CNN (B
(o!L)))*
5 (rN (_1 ([u#
(y1) , oo))
LL + B (0, L)))
} < (GN (a_-i ((U# (Y2),oo))))
or, equivalently, by (16.1) and (16.2),
0 < 1) (y1)
- u# (y2) = n < L
Qu
R
(u# (y2))
aN
)
l
_ (cuut(Y1)))'
By Proposition 16.2(iii) and (iv), Remark 16.3, and (16.6) we have that the right-hand side of the previous inequality is less than or equal to Iy2I - I.
16. Sobolev Spaces: Symmetrization
484
This proves the claim. Using Exercise 11.46 once more, it follows that ul E W',- (RN). Moreover, (why?) livu III LoO(RN.RNxN)
<
IIVUIILOO(RN;RNxN)
0
This completes the proof.
Remark 16.13. Note that the previous proof also shows that u* : [0, oo) [0, oo) is Lipschitz continuous.
16.3. Symmetrization of Piecewise Affine Functions We recall that PA is the family of all continuous functions u : RN --+ R for which there exist N-simplexes A1, ... , At with pairwise disjoint interiors e
such that the restriction of u to each Ai is affine and u = 0 outside U Ai. i=1
Theorem 16.14. Let u E W""p (RN) n PA, 1 < p < oo, be a nonnegative function. Then uO E W 1,P (RN) n W1,00 (RN) and (16.12)
JRN
N7hL ()r dy < (N IVu(x)Ip dx.
Note that if u E W""p (RN) n PA is nonnegative, then for each s > 0 the set
{xERN: u(x)>s} is the union of finitely many N-simplexes.
The proof of Theorem 16.14 will rely on the following version of the isoperimetric inequality (see Theorem C.13).
Exercise 16.15. Prove that if E C RN is given by the union of finitely many N-simplexes, then N-I
Gn
aN
<
HN-1 (OE) ON
Another important tool is given by the coarea formula (see Theorem 13.25 and Lemma 13.26).
Exercise 16.16. Let u E W1 ,P (]RN) n PA, let g : RN - [0, oo] be a Borel function, and let E C RN be a Lebesgue measurable set. Prove that (16.13)
JR
J
f1u-1({s})
Hint: Use Lemma 13.26.
g (y)
dNN-1
(y)
ds =
JE
g (x) IVu (x) I dx.
16.3. Symmetrization of Piecewise Afline Fbnctions
485
We now turn to the proof of Theorem 16.14. Without further notice, in the proof we will use the fact that the Lebesgue-Stieltjes signed measure A.- generated by u* is negative, since u* is decreasing (see Theorem 5.13). Proof of Theorem 16.14. Since u G W1-1 (RN), it follows from Proposition 16.12 that u# belongs to W1,°° (RN). Moreover, ua belongs to LP (RN) by Theorem 16.10. Thus, it remains to show that (16.12) holds. Let 0 = yo < y1 < < it be the finite set of values assumed by u on the vertices of the N-simplexes Al i ... , A . Since u# E W 1°O° (RN), it follows that u E W1 (][8N). Hence, we are in a position to apply Exercise 10.37(iv) to conclude that Vu# (x) = 0 for GN-a.e. x E (ub)-1({yi}). Hence, it suffices to show that IVu#
(y)1'
ua
dy <
f IDu(x)lP
dx,
u,
where
Up:_ {yERN: yi-1
Since ul (y) := u` (CiN IyIN), y E RN, using spherical coordinates and the fact that QN = NaN, we have that
Jb/aN
I(u*)' (aNrN) IPr(N-1)(P+1) dr
(y) I' dy = 3ni 1
U, IVU
=a
'
r
I
(u*)' (T) IP l T ) ' N
aN
a{
dT N-1
= -YN
N
(u*)/ (T) IP-1
JI
(u ")f (T) dr,
( a1v r)P
+t
where in the second equality we have used the change of variables T = aNrN. If p > 1, then by Remark 16.13 and Exercise 5.21 we have that
A. (B) = fB ( u*)' (t) dt for every Borel set B C 10, oo), and so by Theorem 5.42 and Remark 5.43, N-1
IVu# (y) lp dy = pN J
f (s)
IN-1
Lou S
N
ds
N
where (16.14)
f(S):=
P-1) N1 I(u')'(9u(8))I'
1 (-0,1(3))' C(N
sE(yi-lryi)
16. Sobolev Spaces: Symmetrization
486
Note that f is a Borel function by Exercise 1.41 and Proposition 16.1(i) and (ii). By Exercise 16.15 and Proposition 16.1 (ii), for s E (7s_1, -ti),
iN-1 (-1({}))
(Qu(8)'\l
(16.15)
aN and so by Exercise 16.16,
f
Uts
ON
fy
IVu# (y) I' dy < ON 1
1
7i_1
-ON 1
fui
?jN-1 (u 1
f (a)
l ds ({s}))
f (u (.T)) IVu(x)I dx-
We now use Holder's inequality and Theorem 16.10 to conclude that
J©u() Idy < f3N 1 (fU, IVu (x)Id)
{
\Ju{ If (u (x))l'
l
'
P
= aN 1 UU, Iou(x)IY
dx) ())P'
(fu: If
n
dy
We claim that 1
I"("
f (u# (y)) < #N
(16.16)
for GN-a.e. y E U;. By (16.6) and Proposition 16.2(iv) we have that Lou
(Ul (y)) = Au. (u* (aN IyIN)) <_ ON IyIN
for all y E U. We now distinguish two cases. If Au (u# (y)) = aN I YIN' then by (16.14), Au
f (u# ( y )) = I (u*)I ( Au ('a# ( y ))) I
aN
Ip-1
(u*)' (aNIVV")
Iy1(N-1)
_
l
aN 1 I Vup (x)
. I
1
If Au (u# (y)) < aN IyIN, then by Proposition 16.2(iv), u* is constant on
[9u (u# (y)) aN IyIN] If u* is differentiable at eu (u# (y)), then )
(a*)` (0- (u(y))) = 0, and so
f (UN (y)) = 0 < Thus, we have proved (16.16).
aN
1
I Vul (x)
Ip-1 .
16.4. Symmetrization in W '-P and BV
487
Hence,
f
(x)Ip
I Vu#
f dx < (J Iou (x)lP dx)A \\ U
f
J
U
P IVua (x) Ip dx
,
which gives the desired inequality. If p = 1, we proceed similarly to obtain bt
J Ua
Vu# (x) I dx
l Nl = -0, J.aN t/ J (us)' (T) dT r7i
={3NJ
NN -1
Au
(a)
ds
aN
xN-i u-1
s
ds =
I Vu (x) I dx, U,
where we have again used Proposition 16.12, Theorem 5.42, Remark 5.43, (16.15), and Exercise 16.16.
16.4. Symmetrization in W 1,P and BV In this section we prove that the operator u H ul maps nonnegative functions in Wl.p (RN), 1 < p < oo, into nonnegative functions in Wl,p (RN). We begin with the case 1 < p < oo.
Theorem 16.17. Let U E W1,p (RN), 1 < p < oo, be a nonnegative function. Then ul E W1 'P (RN) and (16.17)
N
Vu# (y)IP dy :5JRN IVu (x)Ip dx.
.
Proof. By Theorem 10.33 we can construct a sequence of piecewise affine
functions {un.} C W1,P (RN) with compact support such that u, - u in W1'P (RN). Then by Theorem 16.14, un E Wisp (RN), JRN lun 111) p dy = J1N Iun (x) Idx, a nd
IIVunhILp(RNANxN) . I
Thus, {um} is bounded in W1,P (RN). On the other hand, since U!" -+ u# in LP (RN) by Theorem 16.11, it follows by Theorem 10.44 and the lower
16. Sobolev Spaces: Symmetrization
488
semicontinuity of the norms that ua E W" (RN) with IILp(RN.RNxN < lnm of II II
DUIILpQtNANXN)
The previous proof in the case p = 1 only shows that u# E BV (RN). Indeed, we have the following result.
Theorem 16.18. Let u E BV (RN) be a nonnegative function. Then ua E BV (RN) and (RN) < IDuI (RN) . l
Proof. By Theorem 13.9 there exists a sequence C COO (RN) n W1'1 (RN)
such that un -> u in L' (RN) and DuI (RN)
lim fRMN IVun (x)I dx fl-'
By Theorem 10.33 for every n E N there exists a piecewise affine function v E W111 (RN) with compact support such that 1
IIV, - urallwl.l(RN) < n
Hence vn -), u in L1 (RN) and lim
n-4oo fRX
lava (x)I dx = IDul (18N) .
We now proceed as in the previous theorem. By Theorem 16.14, vn E w1,1 (RN),
hN I() I dy =
f
N
Ivn (x)I dx,
and
II'ILlNNxN) <
IIVVnhIL1(RN;]RNxN) .
Thus, {v} is bounded in W1'1 (RN). On the other hand, since von -> u# in L1 (RN) by Theorem 16.11, it follows by Theorem 13.35 and Exercise 13.3 that u# E BV (RN) with (Dual (HN) C nm of
I
(RN;R,vxx)
< inn f IIVvnhIL1(RNMNxN) = IDuI (RN).
16.4. Symmetrization in W '-P and BV
489
Since W"1 (RN) C BV (RN), the previous theorem shows that if it E W 1°1 (RN), then ui E BV (RN). Next we show that ui E W" (RN). Since ua (y) = u* (ow IVIN), V E RN, it is enough to show that u* is locally absolutely continuous (why?).
Theorem 16.19. Let it E W" (RN) be a nonnegative function. Then ug E W1'1 (RN) and JRN Vua (y) I dy <
J
IVu (x) I dx. N
Exercise 16.20. Let it E W1,1 (RN)
(i) Prove that if N = 1, then 0
u(x)=Jii(x -t )dt for Gl-a.e. x E R.
(ii) Prove that if N > 2, then for every z E SN-land for RCN-a.e. x E RN, U (x) =
JVlt(x_tz).zdt.
(iii) Prove that if N 2:2, then for GN-a.e. X E RN u (x) =
1
N
N i=1
! JgN 8xi (x
-
yz
y)
dy.
IYIN
We begin with a preliminary result that is of interest in and of itself.
-
Proposition 16.21. Let it E W1,1 (RN), N > 2. Then N-1
sup
>oSN({xERN:u(x)_s})] v
a
N
Proof. By the previous exercise and a change of variables, for RCN-a.e. x E RN
dy. Iu (x)I < NaN JRN Iou(y)I Ix - VIN-1
Let E8 := {x E RN : i(x) > s}. Integrating the previous inequality over E. and using Tonelli's theorem yields eGN (E'.):5
JE. u (x)I dx < NaN J N
IVu (Y)I
JEIx -SIN-1 dxdy.
16. Sobolev Spaces: Symmetrization
490
Let r > 0 be such that £N (B (0, r)) = £N (Es). By the change of variables z = y - x and the Hardy-Littlewood inequality (see Theorem 16.9),
JE. FX-
- f y+E.
YIN-1
Iz7N-1
dz
(_y+EE)' I
= Naevr = NaN 1IN (f N
zIN-1
dz
(Es))1'
which, together with the previous inequality, gives s,CN (Es) :5
1
1 (CN (Es))1I N fRN IVu /(FIJI dy.
IN N
Hence,
81 N (Ea)1 7
fIVu (v)I dy-
IN
0
It now suffices to take the supremum over all s > 0. We are now ready to prove Theorem 16.19.
Proof of Theorem 16.19. Step 1: By Theorem 16.18 we have that ua E BV (RN) with IDuo I (RN) <
N) .
I1ou1lL1(RN
We claim that if N > 2, then for all 0 < a < b we have (u.* (a) - u* (b)) a te - ' <
1
11N
J
IVu(x) I dx.
For 0 < 81 < s2 define 0
f (s)
if s<-81,
8-81 if 81 < s < $2, s2-s1 ifa>s2.
Since f is a Lipschitz function and If o ul < u, by the chain rule (see Exercise 10.37) the function v := f o u belongs to W 1,1 (RN) and for all i = 1, ... , N and for ,CN-a.e. x E RN,
if u (x) c 81,
0
8x,
8u
(x) = f' (u (x)) 49 (a) =
(x) if 81 < u (x) < 82i
if u (x) > s2.
0
Applying the previous proposition to v, we obtain sups ECN ({x E a>o
RN. v(x)>s})]
Ni
aNN {sl
IVul dx.
16.4. Symmetrization in W 1,P and BV
491
Since s
ups[CN({xERN: u(x)>s})]
N 1
>(s2-31)['CN ({xERN: u(x)>s2})] N we get that
(s2 -Si) [,C" ({x E RN : u (x) > 92})
x-i
1
<
11N
N
Jal
Vudx.
Let 0 < a < b and consider si := if (b), 82 := u* (a). Then by Proposition 16.2(111),
V
(u* (a) - u * (b)) a N
l1
N,
(16.18)
< (u* (a) p- u* (b)) [,CN ({x E RN : u (x) > u* (a)})1
< JIN 1 J
[ou (x)I dx
.
C'N
Step 2: We show that the function u* is absolutely continuous on closed subintervals of (O,oo). If N = 1, this follows from Corollary 6.26. Thus,
assume that N > 2. Let I = [a, b] c (0, oo) and fix e > 0. Since Vu E L1 (RN; ISNXN), we may find 6 > 0 such that (16.19)
1
IVu(x)I dx
I IN N-1 <ec a
for every Lebesgue measurable set E C RN with GN (E) < 6. Consider any finite number of nonoverlapping intervals (ak, bk), k = 1, ... , e, with [ak, bk] C [a, b] and (bk - ak) < 6.
k.1
By Proposition 16.2, P
£N U {x E RN : u* (bk) < u (x) < u* (ak) } (k=1 e
E'C,11
: u*(bk)
k=1 P
P
(bk - ak) < 6,
[Lou (u* (bk)) - c (u* (ak))] < k=1
k=1
16. Sobolev Spaces: Symmetrization
492
and so, by (16.18) and (16.19),
t 1 u* (ak)I < > Iu* (bk) 11N N-1 k=1
aN aN-
r U {u*(bk)
IDu (x)I dx < e.
k=1
This proves the absolute continuity of u*.
0
Appendix A
Functional Analysis When probed under pressure, grad students will either blurt out what they are doing (but won't know if it means anything), or they will blurt out what they plan to do (but won't know how to do it). -Jorge Cham, www.phdcomics.com
In this chapter we present without proofs all the results from functional analysis used in the text.
A.I. Metric Spaces Definition A.I. A metric on a set X is a map d : X x X -+ 10, oo) such that (i) d (x, y) < d (x, z) + d (x, y) for all x, If, x E X. (ii) d (x, y) = d (y, x) for all x, y E X,
(iii) d (x, y) = 0 if and only if x = y.
A metric space (X, d) is a set X endowed with a metric d. When there is no possibility of confusion, we abbreviate by saying that X is a metric space.
If r > 0, the (open) ball of center xo E X and radius r is the set
B (xo, r) :_ {x E X : d (xo, x) < r). If x E X and E C X, the distance of x from the set E is defined by dist (x, E) := inf {d (x, y) : y E E} ,
while the distance between two sets El, E2 C X is defined by
dist (El, E2) := inf {d (x, y) : x E El, y E E2} . 493
A. Functional Analysis
494
A sequence {x,,} C X converges (strongly) to x r= X if lira d (x,,, xe) = 0.
n-,oo
A Cauchy sequence in a metric space is a sequence {x,z} C X such that lim d (xn, xm) = 0.
n,m-*oo
A metric space X is said to be complete if every Cauchy sequence is convergent.
A.2. Topological Spaces Definition A.2. Let X be a nonempty set. A collection r C P (X) is a topology if the following hold.
(i) 0,XET.
(ii) IfUi Erfor i= 1,...,M,then U1fl...numEr. (iii) If {U«}«Ef is an arbitrary collection of elements of r, then U«Ef U« belongs to r.
Example A.3. Given a nonempty set X, the smallest topology consists of {0, X}, while the largest topology contains all subsets as open sets. The pair (X, 7) is called a topological apace and the elements of r are open sets. For simplicity, we often apply the term topological space only to
X. A set C C X is closed if its complement X \ C is open. The closure E of a set E C X is the smallest closed set that contains E. The anterior E° of a set E C X is the union of all its open subsets. A point xo E X is an accumulation point for a set E C X if for every open set U that contains x4 there exists x E E fl U, with x # xo. A subset E of a topological space X is said to be dense if its closure is the entire space, i.e., E = X. We say that a topological space is separable if it contains a countable dense subset. Given a point x E X, a neighborhood' of x is an open set U E T that contains x. Given a set E C X, a neighborhood of E is an open set U E T that contains E. A topological space is a Hausdorff space if for all x, y E X with x y there exist two disjoint neighborhoods of x and y. Given a topological space X and a sequence {x,a}, we say that {xn} converges to a point x E X if for every neighborhood U of x we have that x, E U for all n sufficiently large. Note that unless the space is Hausdorff, the limit may not be unique. 1The reader should be warned that in some texts (e.g., [1441) the definition of neighborhood is different.
A.2. Topological Spaces
495
A topological space is a normal space if for every pair of disjoint closed sets C1, C2 C X there exist two disjoint neighborhoods of C1 and C2.
Definition A.4. Let X, Y be two topological spaces and let f : X -> Y. We say that f is continuous if f -1(U) is open for every open set U C Y. The space of all continuous functions f : X -> Y is denoted by C (X; Y). If Y = i$, we write C (X) in place of C (X; R). The next theorem gives an important characterization of normal spaces.
Theorem A.5 (Urysohn). A topological space X is normal if and only if for all disjoint closed sets C1, C2 C X there exists a continuous function f : X -, 10,1] such that f - 1 in C1 and f - 0 in C2. We now introduce the notion of a base for a topology. Let (X, rr) be a topological space. A family (3 of open sets of X is a base for the topology z if every open set U E r may be written as the union of elements of Q. Given a point x E X, a family fax of neighborhoods of x is a local base at x if every neighborhood of x contains an element of 0..
Proposition A.6. Let X be a nonempty set and let f3 C P (X) be a family of sets such that (i) for every x E X there exists B E 0 such that x E B, (ii) for every BI, B2 E fl, with B1 fl B2 0, and for every x E B1 fl B2 there exists B3 E (3 such that x E B3 and B3 C Bl f1 B2.
Then the collection r C P (X) of arbitrary unions of members of f3 is a topology for which J3 is a base.
As we will see later on, the metrizability and the normability of a given topology depend on the properties of a base (see Theorems A.14 and A.36 below).
Definition A.7. Let (X, r) be a topological space. (i) The space X satisfies the first axiom of countability if every x E X admits a countable base of open sets. (ii) The space X satisfies the second axiom of countability if it has a countable base.
In the text we use several notions of compactness.
Definition A.S. Let X be a topological space. (i) A set K C X is compact if for every open cover of K, i.e., for every collection {Ua} of elements of r such that U,, U. D K, there exists a finite subcover (i.e., a finite subcollection of {U0} whose union still contains K).
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496
(ii) A set E C X is relatively compact (or precompact) if its closure E is compact. (iii) A set E C X is a-compact if it can be written as a countable union of compact sets. (iv) The topological space X is locally compact if every point x E X has a neighborhood whose closure is compact.
Remark A.9. A closed subset of a compact topological space is compact. On the other hand, a compact set of a Hausdorff space is closed. If X is a topological space, we denote by C,, (X) the space of all continuous functions u : X -+ R whose support is compact. The next theorem is used to construct cut-off functions and partitions of unity.
Theorem A.10. If X is a locally compact Hausdorff space and K C U C X,
with K compact and U open, then there exists W open such that W is
compact and KCWCWCU. Corollary A.11. If X is a locally compact Hausdorf space and K C U C X, with K compact and U open, then there exists a function cp E C, (X) such that 0 < W :S 1, cp - 1 on K, and Sp - 0 on X \ U. The function cp is usually referred to as a cut-off function. To introduce partitions of unity, we need the notion of a locally finite family.
Definition A.12. Let X be a topological space and let F be a collection of subsets of X. Then (i) F is locally finite if every x E X has a neighborhood meeting only finitely many U E F, (ii) F is a-locally finite if 00
F=U n=1
where each I, is a locally finite collection in X. We are now ready to introduce partitions of unity.
Definition A.13. If X is a topological space, a partition of unity on X is a family {cpi}4EA of continuous functions (pi X -t 10,1] such that
E (x) = 1 iEA
A.3. Topological Vector Spaces
497
for all x E X. A partition of unity is locally finite if for every x E X there exists a neighborhood U of x such that the set {i E A : U n supp cps 34 0} is finite. If {U,}3E y is an open cover of X, a partition of unity subordinated to the cover {U3}3E- is a partition of unity {cpq}jEA such that for every i E A, supp Vi C U3 for some j E
A metric space (X, d) can always be turned into a topological space (X, T) by taking as a base for the topology r the family of all open balls.
We then say that r is determined by d. Note that (X, r) is a Hausdorff normal space. A topological space X is metrizable if its topology can be determined by a metric. Theorem A.14. A topological space is metrizable if and only if the following properties hold:
(i) Singletons are closed. (ii) For every closed set C C X and for every x 0 C there exist disjoint open neighborhoods of C and x. (iii) X has a o--locally finite base.
A.3. Topological Vector Spaces Let X be a vector space over R and let E C X. The set E is said to be balanced if tx E E for all x E E and t E[-1,1]. We say that E C X is absorbing if for every x E X there exists t > 0 such that sx E E for all
0<s
Definition A.15. Given a vector space X over R endowed with a topology r, the pair (X, r) is called a topological vector space if the functions
XxX -'X, (x, y) i--> x + y
and
R xX - X, (t, x) H tx
are continuous with respect to r.
Remark A.M. (i) In a topological vector space a set U is open if and only if x + U is open for all x E X. Hence, to give a base, it is enough to give a local base at the origin.
(ii) Using the continuity of addition and scalar multiplication, it is possible to show that each neighborhood U of the origin is absorbing
and it contains a neighborhood W of zero such that W + W C U and W C U, as well as a balanced neighborhood of zero. As a corollary of Theorem A.14 we have the following:
A. Functional Analysis
498
Corollary A.17. A topological vector space X is metrixable if and only if (i) singletons are closed, (ii) X has a countable base.
Definition A.18. Let X be a topological vector space. A set E C X is said to be topologically bounded if for each neighborhood U of 0 there exists
t > 0 such that E C W. Note that when the topology r is generated by a metric d, sets bounded in the topological sense and in the metric sense may be different. To see this, it suffices to observe that the metric dl := i generates the same topology
as d, but since dl < 1, every set in X is bounded with respect to dl. We now define Cauchy sequences in a topological vector space.
Definition A.19. Let X be a topological vector space. A sequence {x,, } C X is called a Cauchy sequence if for every neighborhood U of the origin there exists an integer W E N such that
xn - x,n E U for all n, m > W. The space X is complete if every Cauchy sequence is convergent.
Note that Cauchy (and hence convergent) sequences are bounded in the topological sense.
Proposition A.20. Let X be a topological vector space and let {xn} C X be a Cauchy sequence. Then the set {xn : n E N} is topologically bounded.
Topologically bounded sets play an important role in the normability of locally convex topological vector spaces (see Theorem A.36 below).
Definition A.21. A topological vector space X is locally convex if every point x E X has a neighborhood that is convex. Proposition A.22. A locally convex topological vector space admits a local base at the origin consisting of balanced convex neighborhoods of zero.
Let X be a vector space over R and let E C X. The function pE : X - R defined by
pE (x) := inf {t > 0 : x E tE},
x E X,
is called the gauge or Minkowaki functional of E.
Definition A.23. Let X be a vector space over R. A function p : X -' R is called a seminorm if p (x + y) <_ p (x) + p (y)
for allx,yEXandp(tx)=Itip(x)forall xEXand tEPC
A.3. Topological Vector Spaces
499
Remark A.24. Let X be a vector space over R and let E C X. The gauge PE of E is a seminorm if and only if E is balanced, absorbing, and convex.
Theorem A.25. If F is a balanced, convex local base of 0 for a locally convex topological vector space X, then the family {pu : U E .F} is a family of continuous seminorms. Conversely, given a family P of seminorms on a vector space X, the collection of all finite intersections of sets of the form
V(p,n):={xEX: p(x)
We now give some necessary and sufficient conditions for the topology T given in the previous theorem to be Hausdorff and for a set to be topologically bounded.
Corollary A.26. Let P be a family of seminorms on a vector space X and let r be the locally convex topology generated by P. Then (i) T is Hausdorff if and only if p (x) = 0 for all p E P implies that x = 0, (ii) a set E C X is topologically bounded if and only if the set p (E) is
bounded in R forallpEP. We now introduce the notion of dual space.
Definition A.27. Let X and Y be two vector spaces. A map L : X - Y is called a linear operator if
(i) L (x + y) = L (x) + L (y) for all x, y E X, (ii) L (tx) = t L (x) for all xEX and t E R.
If X and Y are topological vector spaces, then the vector space of all continuous linear operators from X to Y is denoted by C (X; Y) . In the special case Y = R, the space C(X;R) is called the (topological) dual space of X and it is denoted by X'. The elements of X' are also called continuous linear functionals. The bilinear (i.e., linear in each variable) mapping (A.1)
X' x X
-a
(L, x) i- L (x) is called the duality pairing. The dual space L (X'; R) of X' is called the bidual space of X and it is denoted by X".
A. Functional Analysis
500
Definition A.28. Let X, Y be topological vector spaces. An operator L : X -i, Y is bounded if it sends topologically bounded sets of X into topologically bounded sets of Y.
Theorem A.29. Let X, Y be topological vector spaces and let L : X -, Y be a linear operator. Consider the following properties: (i) L is continuous. (ii) L is bounded.
(iii) If {x.} C X and x.
0, then {L (xm)} is a topologically bounded
set.
(iv) If {xn} C X and x!, -+ 0, then L (x,z) - 0. Then (i)
(iii).
. (ii)
Moreover, if X is metrizable, then (iv)
(iii)
(i)
so that all four properties are equivalent.
Theorem A.30 (Hahn-Banach's theorem, analytic form). Let X be a vector space, let Y be a subspace of X, and let p : X - R be a convex function. Then for every linear functional L : Y -+ R such that L (x) < p (x)
for all x E Y
there exists a linear functional Ll : X -* R such that
L1(x) = L (x)
for all x E Y
and
L(x)
L(x)>a forallxEE
and
L(x)
Corollary A.32. Let X be a topological vector space. Then the dual X' of X is not the null space if and only if X has a convex neighborhood of the origin strictly contained in X. In view of the previous corollary it is natural to restrict our attention to locally convex topological vector spaces.
A.4. Normed Spaces
501
Theorem A.33 (Hahn-Banach's theorem, second geometric form). Let X be a locally convex topological vector space and let C, K C X be nonempty disjoint convex sets, with C closed and K compact. Then there exist a continuous linear functional L : X --+ R and two numbers a E R and e > 0 such that L (x) < a- E for all x E C and L (x) > a -}- a for all x E K.
A.4. Normed Spaces Definition A.34. A norm on a vector space X is a map X -> [0,00)
such that (i) Iix + vII < iIxII + Ilyil for all x, y E X. (ii) litxil = Iti lixII for all x E X and t E R,
(iii) IIxil = 0 implies x = 0.
A nonmed space (X, 11 Il) is a vector space X endowed with a norm For simplicity, we often say that X is a normed space. If for every x, y E X we define
d(x,y) := IIx - yll, then (X, d) is a metric space. We say that a normed space X is a Banach space if it is complete as a metric space. Example A.35. Let X be a topological space and consider the space CC (X)
of all continuous functions u : X -y R whose support is compact. For u E CC (X) define iiulL0 :=
max
Ju (x)i
Then is a norm. In general C. (X) is not complete. We denote by Co (X) the completion of CC (X) relative to the supremum norm 11-1100. if X is compact, then
Co(X)=Cc (X)=C(X). If X is a locally compact Hausdorff space, then it can be shown that u E Co (X) if and only if u E C (X) and for every e > 0 there exists a compact set K C X such that
Ju(x)l <E forallxEX\K. We also define C,, (X; Rm) and Co (X; Rm) as the spaces of all functions u : X -> RI whose components belong to Cc, (X) and Co (X), respectively. A topological vector space is normable if its topology can be determined by a norm.
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502
Theorem A.36. A topological vector space X is normable if and only if it is locally convex and it has a topologically bounded neighborhood of 0. Two norms 1 1-1 11 and 11'112 are equivalent if there exists a positive constant
C > 0 such that
forallxEX.
CIIxII1 <_ 11x112<_CII2II1
Equivalent norms induce the same topology on X.
Proposition A.37. Let X and Y be normed spaces 4vith norms II-II. and II. "Y, respectively.
(i) A linear operator L : X -* Y is continuous if and only if IILII,c(x;Y) :=
sup
II L (x) IlY
< 00.
IIxIIx
(ii) The mapping L E L (X; Y) H II LII C(x;Y) is a norm.
(iii) if Y is a Banach space, then so is C (X; Y). Conversely, if X 34 {0) and L (X; Y) is a Banach space, then so is Y.
As a corollary of the Hahn-Banach theorem (Theorem A.30) one has the following result.
Corollary A.38. Let X be a normed space. Then for all x E X, IIxIIx -=LEX',ILllr,<1
IL(x)l IIxIIx
Definition A.39. Let X and Y be normed spaces with norms II IIx and ".IIY. A continuous linear operator L E L (X; Y) is said to be compact if it maps every bounded subset of X onto a relatively compact subset of Y. In particular, if L is compact, then from every bounded sequence {x"} C X we may extract a subsequence I such that {L (x,,,,)) converges in Y.
Definition A.40. We say that the normed space X is embedded in the normed space Y and we write
XyY if X is a vector subspace of Y and the immersion
i.X - Y is continuous.
A.5. Weak Topologies
503
Note that since the immersion is linear, in view of Proposition A.37 the continuity of i is equivalent to requiring the existence of a constant M > 0 such that ""xj'y SM""x'jx for all x E X. We say that X is compactly embedded in Y if the immersion i is a compact operator.
A.5. Weak Topologies Given a locally convex topological vector space X, for each L E X' the function PL : X - [0, oo) defined by (A.2) pL (x) :_ L (x)I , x E X, is a seminorm. In view of Theorem A.25, the family of seminorms U'L}LEx'
generates a locally convex topology a (X, X') on the space X, called the weak topology, such that each PL is continuous with respect to or (X, X'). In turn, this implies that every L E X' is a (X, X') continuous.
Theorem A.41. Let X be a locally convex topological vector space and let E C X. Then (i) E is bounded with respect to the (strong) topology if and only if it is weakly bounded,
(ii) if E is convex, then E is closed if and only if it is weakly closed.
Definition A.42. Given a locally convex topological vector space X, a sequence {x,} C X converges weakly to x E X if it converges to x with respect to the weak topology a (X, X').
We write xn - x. In view of Theorem A.25 and (A.2), we have the following result.
Proposition A.43. Let X be a locally convex topological vector space. A sequence {x,z} C X converges weakly to x E X if and only if lim L (x,,) = L (x)
n-Poo
for every LEX'. Similarly, given a locally convex topological vector space X, for each x E X the function p. : X' -* [0, oo) defined by (A.3)
pz (L) := IL (x) I
,
L E X',
is a seminorm. In view of Theorem A.25, the family of seminorms {px}.Ex
generates a locally convex topology a (X', X) on the space X', called the weak star topology, such that each px is continuous with respect to a (X', X).
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504
Definition A.44. Let X be a locally convex topological vector space. A sequence {Ln} C X' is weak star convergent to L in X' if it converges to L with respect to the weak star topology a(X',X).
We write L j L. In view of Theorem A.25 and (A.3), we have the following result.
Proposition A.45. Let X be a locally convex topological vector space. A sequence {L} C X' converges weak star to L E X' if and only if lim L,a (x) = L (x) a-loo
for every x E X. Theorem A.46 (Banach-Alaoglu). If U is a neighborhood of 0 in a locally convex topological vector space X, then
K :_ {L E X' : IL (x)l < 1 for every x E U} is weak star compact.
Corollary A.47. If X is a nonmed space, then the closed unit ball of X',
{LEX': JILIIx,<1}, is weak
star compact.
If X is separable, it actually turns out that weak star compact sets are metrizable, and thus one can work with the friendlier notion of sequential compactness. Theorem A.48. Let X be a separable locally convex topological vector space and let K C X' be weak star compact. Then (K, a (X', X)) is metrizable. Hence, also in view of the Banach-Alaoglu theorem, we have the following:
Corollary A.49. Let U be a neighborhood of 0 in a separable locally convex topological vector space X and let {L,z} C X' be such that ILn W1 C 1
for every x E U and for all n E N.
Then there exists a subsequence that is weak star convergent. In particular, if X is a separable Banach space and {Ln} C X' is any bounded sequence in X', then there exists a subsequence that is weak star convergent. For Banach spaces the converse of Theorem A.48 holds:
Theorem A.50. Let X be a Banach space. Then the unit ball B (0; 1) in X' endowed with the weak star topology is metrizable if and only if X is separable.
A.5. Weak Topologies
505
Proposition A.51. Let X be a Banach space. If a sequence {Ln} C X' converges weak star to L E X', then it is bounded and IILIIx, <_ lim inf
-oo
IILnIIx,.
Proposition A.52. Let X be a Banach space. If X' is separable, then so is X. The converse is false in general (take, for example, the separable space L' (RN) and its dual L°° (RN)). We now study analogous results for the weak topology. An infinitedimensional Banach space when endowed with the weak topology is never metrizable. However, we have the following:
Theorem A.53. Let X be a Banach space whose dual X' is separable. Then the unit ball B (0; 1) endowed with the weak topology is metrizable.
Definition A.54. Let X be a locally convex topological vector space. A set K C X is called sequentially weakly compact if every sequence {xn} C K has a subsequence converging weakly to a point in K.
Theorem A.55. Let X be a Banach space. If K C X is weakly compact, then it is weakly sequentially compact.
Using Banach-Alaoglu's theorem, one can prove the following theorem:
Theorem A.56 (Eberlein-gmulian). Let E be a subset of a Banach space X. Then the weak closure of E is weakly compact if and only if for every sequence {xn} C E there exists a subsequence weakly convergent to some element of X. As an immediate application of the Hahn-Banach theorem we have the following:
Proposition A.57. Let X be a normed space and consider the linear operator mapping J : X - X" defined by J (x) (L) := L (x) ,
L E X'.
Then IIJ (x)IIx,, = IIxIIx for all x E X. In particular, J is injective and continuous.
Definition A.58. A normed space X is reflexive if J (X) = V. In this case it is possible to identify X with its bidual X".
Theorem A.59 (Kakutani). A Banach space is reflexive if and only if the closed unit ball {x E X : IIxil S 1} is weakly compact.
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In view of the previous theorem and the Eberlein-Smulian theorem we have the following corollary:
Corollary A.60. Let X be a reflexive Banach space and let
C X be a
bounded sequence. Then there exists a subsequence that is weakly convergent.
Proposition A.61. A norm.ed space X is reflexive if and only if X' is reflexive.
The following proposition is used throughout the text, sometimes without mention.
Proposition A.62. Let X be a Banach space. If a sequence {x,z} C X converges weakly to x E X, then it is bounded and IIxII <_ limf IIxnII
A.6. Hilbert Spaces Definition A.63. An inner product on a vector space X is a map
X x X -i R such that (i) (x, y) = (y, x) for all x, y E X, (ii) (sx + ty, z) = s (x, z) + t (y, z) for all x, y, z E X and s, t E R, (iii) (x, x) > 0 for every x E X and (x, x) = 0 if and only if x = 0. If for every x E X we define IIxII
(xx),
then X becomes a normed space. We say that a normed space X is a Hilbert space if it is a Banach space.
Theorem A.64. Let (X, II'II) be a nonmed space. Then there exists an inner product X x X - R such that IIxII = (xx) for all x E X if and satisfies the parallelogram law only if II' II IIx + y112 + IIx - tll2 = 2 IIx112 + 2 IIyhl2
for all x,yGE X. Remark A.65. If 11.11 satisfies the parallelogram law, then the inner product in the previous theorem is defined as (x, y)
for all x,yEX.
4
[llx+vfl2 - IIx - III2]
Appendix B
Measures ANOVA: Analysis of Value. Is your research worth anything? Significance is determined by comparing one's research with the
Dull Hypothesis: Hp: /Al = p2? where Hp: the Dull Hypothesis, is,: significance of your research, p2: significance of a monkey typing randomly on a typewriter in a forest where no one can hear it. - Jorge Cham, www.phdcomic.com
In this chapter we present, without proofs, all the results in abstract measure
theory and Lebesgue integration that we used in the text. This chapter draws upon [65], to which we refer for the proofs of all the results that cannot be found in classical texts (such as [48], [54], [64], [1431). Here, we prove only a few results that are used in this text and that cannot be found in 1651 or in classical textbooks.
B.1. Outer Measures and Measures Definition B. 1. Let X be a nonempty set. A map ps : P (X) -1, [0, oo] is called an outer measure if (i) (ii)
(0) = 0,
(E) < p* (F) for all E C F C X, (iii) p* (UO0 1 E.) < E° 1 µs (E.) for every countable collection {E} of subsets of X.
Remark B.2. The reader should be warned that in several books (e.g., [54], 158]) outer measures are called measures.
To construct outer measures, we usually start with a family G of "elementary sets" (e.g., cubes in RN) for which we have an "elementary notion of measure" p : 9 -> [0, oo] (i.e., in 1ll;N the volume of a cube Q (x, r) is rN). 507
508
B. Measures
Proposition B.3. Let X be a nonempty set and let 9 C P (X) be such that 0 E 9 and there exists {Xn} C c with X = U°°_1 Xn. Let p : G --f [0, oo] be such that p (0) = 0. Then the map p.* : P (X) - [0, oo[, defined by
µ (E) := inf
p (E.) :
C Q, E C U E.
In=1 is an outer measure.
E C X,
n=1
Remark B.4. Note that if E E 9, then taking E1 = E, E,z := 0 for all n > 2, it follows from the definition of lc* that p* (E) < p (E), with the strict inequality possible. However, if p is countably subadditive, that is, if
p (E) < 1: p(En) n=1
for all ECU°D_1Enwith EE9and{E,,}C9,then µpons. The main problem with outer measures is that they are not additive on disjoint sets. To circumvent this problem, we restrict an outer measure lc* : P (X) -, [0, oo] to a smaller class of subsets for which additivity of disjoint sets holds. Definition B.5. Let X be a nonempty set and let µ* : P (X) - [0, oo] be an outer measure. A set E C X is said to be µ*-measurable if
µ*(F)=µ*(Ff1E)+AC*(F\E) for all sets F C X.
We will see below in Theorem B.13 that the restriction of A* to the class
9X* :_ {E C X : E is p*-measurable}
is additive, actually countably additive, and that the class U1* has some important properties; precisely, it is a o -algebra.
Definition B.6. Let X be a nonempty set. A collection 9)1 C P (X) is an algebra if the following hold. (i) 0 E 9}t.
(ii) IfEE9?t,then X\EEM. (iii) If E1, E2 E WI, then E1 U E2 E WI.
fit is said to be a v-algebra if it satisfies (i)-(ii) and (iii)l if {En} C 9)t, then U°n°_1 En E M.
B.I. Outer Measures and Measures
509
If 9A is a a-algebra, then the pair (X, WI) is called a measurable space.
Let X be a nonempty set. Given a subset F C P (X), the smallest (in the sense of inclusion) c-algebra that contains.F is given by the intersection
of all a-algebras on X that contain F. If X is a topological space, then the Borel a-algebra B (X) is the smallest u-algebra containing all open subsets of X.
Definition B.7. Let X be a nonempty set and let 9.)I C P (X) be an algebra. A map A. 9)1 --> 10, oo] is called a (positive) finitely additive measure if p. (El U E2) _ p (El) + A (E2)
p (0) = 0,
for all E1, E2EWIwith El fE2=0. Definition B.8. Let X be a nonempty set, let WI C P (X) be a a-algebra. A map p : 9)i -- (0, oo) is called a (positive) measure if 00
u
00
f U E. = E I (E.)
0,
n=1
n=1
for every countable collection {En} C fit of pairwise disjoint sets. The triple (X, fit, p) is said to be a measure space.
Given a measure space (X, fit, p), where X is a topological space, the measure p is called a Borel measure if fii contains B (X). One of the most important properties of measures is given in the next proposition.
Proposition B.9. Let (X, MI, p) be a measure space. W If {En} is an increasing sequence of subsets of 91, then 00
Fb U En = n=1
Ulim,
-400
p(En)
(ii) If {En} is a decreasing sequence of subsets of 911 and IA (El) < oo, then
00
p
I
I
n=1
En = lim p (E,L). n"oo
Definition B.10. Let (X, fit, p) be a measure space. (i) The measure p is said to be complete if for every E E W with p (E) = 0 it follows that every F C E belongs to fit.
(ii) A set E E fit has o-finite p-measure if it can be written as a countable union of measurable sets of finite measure; p is said to
be a-finite if X has o-finite p-measure; Is is said to be finite if p (X) < oo.
B. Measures
510
(iii) The measure µ : fit -> [0, oo] is said to have the finite subset property, or to be semifinite, if for every E E fit, with p (E) > 0, there
exists F E 9)t, with F C E, such that 0 < µ (F) < oo. (iv) A set E E fit of positive measure is said to be an atom if for every
F E fit, with F C E, either to (F) = 0 or to (F) = µ (E). The measure µ is said to be nonatomic if there are no atoms, that is, if for every set E E fit of positive measure there exists F E fit, with F C E, such that 0 < it (F) < µ (E). Analogous definitions can be given for finitely additive measures.
Remark B.11. Note that a o-finite measure has the finite subset property. The Hausdorff measure 9-le (see Appendix C) is an example of a non-a-finite measure with the finite subset property.
The following is a useful property of finite measures.
Proposition B.12. Let (X, fit, ti) be a measure space with p finite and let {Ej }JEA C fit be an arbitrary family of pairwise disjoint subsets of X. Then tc (Ej) = 0 for all but at most countably many j E A.
Theorem B.13 (Caratheodory). Let X be a nonempty set and let u* P (X) - [0, oo] be an outer measure. Then (B.1)
fit* := {E C X : E is µ*-measurable}
is a a-algebra and µ* : fit* - [0, oo] is a complete measure. FYom Proposition B.3 and Remark B.4 we have the following.
Corollary B.14. Let X be a nonempty set, let fit C P (X) be an algebra, and let µ : fit - [0, oo] be a finitely additive measure. Let a* be the outer measure defined in Proposition B.3 (with 9:= fit and p := µ). Then every element of fit is p*-measurable. Moreover, if it is countably additive, then µ* = p on fit. Remark B.15. Note that the previous result implies that every countably additive measure µ : fit - [0, oo] defined on an algebra fit may be extended as a measure to a a-algebra that contains M. It actually turns out that when p is a-finite, this extension is unique. Corollary B.16. Let I C R be an interval and let µ, v : B (I) -> [0, oo] be two measures such that A ((a, b]) = v ((a, b]) < 00
for all intervals (a, b] C I. Then p = v.
B.2. Measurable and Integrable Functions
511
Proof. Let fit := {E E B (I) : µ (E) = v (E)}.
Then fit is an algebra that contains all intervals (a, b] C I. Moreover, µ, v : fit - [0, oo] are a-finite. Hence, by the previous remark, µ and v coincide on the smallest a-algebra that contains all intervals (a, b] C I. This a-algebra is B (I). 0
Remark B.17. In the previous corollary, we could have used any other family (e.g., open, closed) of intervals that generates B (I).
Using Caratheodory's theorem, we have created a large class of complete measures. The next problem is to understand the class fit* of the µ*-measurable sets. To do this, we consider a special class of outer measures.
Definition B.18. Let X be a metric space and let u* : P (X) -, [0, oo] be an outer measure. Then p* is said to be a metric outer measure if
p*(EUF) _!s*(E)+p*(F) for all sets E, F C X, with dist (E, F) := inf {d (x, y) : x E E, y E F} > 0.
Proposition B.19. Let X be a metric space and let µ* : P (X) -* [0, oo] be a metric outer measure. Then every Borel set is µ*-measurable.
B.2. Measurable and Integrable Functions In this section we introduce the notions of measurable and integrable functions.
Definition B.20. Let X and Y be nonempty sets and let fit and 91 be algebras on X and Y, respectively. A function u : X -' Y is said to be measurable if u-' (F) E fit for every set F E 9!.
If X and Y are topological spaces, fit := B (X) and M := B (Y), then a measurable function u : X --> Y will be called a Borel function, or Borel measurable. If in a topological space (usually, RN, R, or [-oo, oo]) the aalgebra is not specified, it is understood that we take the Borel a-algebra B (X).
Proposition B.21. If fit is a a-algebra on a set X and 9! is the smallest aalgebra that contains a given family G of subsets of a set Y, then u : X - Y is measurable if and only if ul (F) E 9)! for every set F E G.
B. Measures
512
Remark B.22. In particular, if in the previous proposition Y is a topological space and 91= B (Y), then it suffices to verify that u-1 (A) E fit for every open set A C Y. Moreover, if Y = R (respectively, Y = [-oo, 001), then it suffices to check that u1 ((a, oo)) E fit (respectively, u-1 ((a, oo]) E fit) for every a E R.
Remark B.23. If X and Y are topological spaces, then, in view of the previous remark, every continuous function u : X --> Y is a Borel function.
Proposition B.24. Let (X, M), (Y, 91), (Z, Q) be measurable spaces and let u : X -> Y and v : Y --+ Z be two measurable functions. Then the function v o u = X -> Z is measurable.
Corollary B.25. Let (X, fit) be a measurable space and let u : X -> R (respectively, u : X - [-oo, oo]) be a measurable function. Then uz, lul,
u+, u , cu, where c E R, are measurable. Remark B.26. If c = 0 and u : X
[-oo, oo], the function cu is defined
to be identically equal to zero. Given two measurable spaces (X, MI) and (Y, 92), we denote by MI 091 C
P (X x Y) the smallest a-algebra that contains all sets of the form E x F, where E E fit, F E 7t. Then fit 091 is called the product a-algebra of fit and 91.
Proposition
B.27. Let (X, M?), (Y1, 911), ... , (Yn, 91..) be measurable spaces and consider (Yi x ... x Yt, 9T1 0 ... (9 9Tn) .
Then the vector-valued function u : X -+ Y1 x - x Yn is measurable if and only if its components uu : X -> Yi are measurable functions for all
i= 1,...,n. Corollary B.28. Let (X, fit) be a measurable space and let u : X -+ R and v : X -). R be two measurable functions. Then u + v, uv, min {u, v}, max {u, v} are measurable.
Remark B.29. The previous corollary continues to hold if R is replaced by [-oo, oo], provided u + v are well-defined, i.e., (u (x), v (x)) {± (oo, -oo)} for all x E X. Concerning uv, we define (uv) (x) := 0 whenever u (x) or v (x) is zero.
Proposition B.30. Let (X, WI) be a measurable space and let un : X [-oo, oo], n E N, be measurable functions. Then the functions sup un, inf un, lim inf u,t, lim sup u,i n
are measurable.
n
n-yoo
B.2. Measurable and Integrable Functions
513
Remark B.31. The previous proposition uses in a crucial way the fact that fit is a v-algebra. Let (X, fit, µ) be a measure space. We will see later on that the Lebesgue integration does not "see" sets of measure zero. Also, several properties (existence of pointwise limit of a sequence of functions, convergence of a series of functions) do not hold at every point x E X. Thus, it is important to work with functions u that are defined only on X \ E with µ (E) = 0. For this reason, we extend Definition B.20 to read as follows.
Definition B.32. Let (X, fit) and (Y, 91) be two measurable spaces and let
p : fit - [0, oo] be a measure. Given a function u : X \ E - Y where p (E) = 0, u is said to be measurable over X if u-1 (F) E fit for every set
FE91. In general, measurability of u on X \ E does not entail the measurability of an arbitrary extension of u to X unless p is complete. However, if we define W (x)
U(X) ifxEX\E,
if x E E, yi where yl E Y, then w is measurable. Hence, in general there are extensions
of u that are measurable and others that are not. The next result shows that if the measure p is complete, then this cannot happen. In what follows, if p is a measure, we write that a property holds µ-a.e.
on a measurable set E if there exists a measurable set F C E such that p (F) = 0 and the property holds everywhere on the set E \ F. Proposition B.33. Let (X, fit) and (Y, 91) be two measurable spaces and let u : X -> Y be a measurable function. Let p : fit -> [0, oo] be a complete measure.
If v : X - Y is a function such that u (x) = v (x) for p-a.e.
x E X, then v is measurable.
Going back to the setting in which u : X \ E -+ Y with p (E) = 0, since Lebesgue integration does not take into account sets of measure zero, we will see that integration of u depends mostly on its measurability on X \ E.
Corollary B.34. Let (X, fit) be a measurable space and let u : X [-oo, oo], n E N, be measurable functions. Let p : fit -> [0, oo] be a complete measure. If there exists limn_oo u (x) for p-a.e. x E X, then limn_oo u,, is measurable.
We are now in a position to introduce the notion of integral.
Definition B.35. Let X be a nonempty set and let fit be an algebra on X. A simple function is a measurable function s : X -- R whose range consists of finitely many points.
B. Measures
514
If cl, ... , ee are the distinct values of s, then we write
t
s=EenXE"' n=1
where 71 CE" is the characteristic function of the set
En:={xEX:8(x)=cn}, if x E En, otherwise. If µ is a finitely additive (positive) measure on X and s > 0, then for every measurable set E E X12 we define the Lebesgue integral of s over E as I 1
XE " (x)
0
ad/L>cnp(EnnE),
(B.2)
n=1
where if c = 0 and it (En n E) = oo, then we use the convention c.A (E,, n E) := 0.
Theorem B.36. Let X be a nonempty set, let WI be an algebra on X, and let u : X - [0, oo] be a measurable function. Then there exists a sequence {sn} of simple functions such that
081(x) <s2(x) < C s,,(a')-*u(x) for every x E X. The convergence is uniform on every set on which u is bounded from above.
Corollary B.37. Let (X, 0, p) be a measure space and let u : X - [0, oo] be a measurable function. If the set {x E X : u (x) > 01 has u-finite measure and u is finite µ-a.e., then there exists a sequence of simple functions, each of them bounded and vanishing outside a set of finite measure (depending on n), such that 0 'C 81 (x) < 82 (x) 'C ... < sn (x) -' u (x) for every x C= X.
In the remainder of this section, WI is a c-algebra and U a (countably additive) measure. In view of the previous theorem, if u : X -. [0, oo] is a
IJJJ
measurable function, then we define its (Lebesgue) integral over a measurable
set E as
IJE udu:=sup{ fE sdp: ssimple,0<s
B.2. Measurable and Integrable Functions
515
Proposition B.38. Let (X, fit, p) be a measure space and let u, v : X [-oo, oo] be two measurable functions.
(i) If 0 < u < v, then fE u dµ < fE v dp for every measurable set E. (ii) If c E [0, oo], then fE cu dµ = c fE u dµ (here we set Ooo 0). (iii) If E E fit and u (x) = 0 for µ-a. e. x E E, then fE u dµ = 0, even if µ (E) = oo. (iv) If E E fit and µ (E) = 0, then fE u dµ = 0, even if u =_ oo in E. (v) fE u d1A = fx XEu dµ for every measurable set E. The next results are central to the theory of integration of nonnegative functions.
Theorem B.39 (Lebesgue's monotone convergence theorem). Let (X, WI, p) be a measure space and let un : X -, [0, oo] be a sequence of measurable functions such that
0<'u1(x):u2(x) :51 ...
lim
n-.oo
x
Remark B.40. The previous theorem continues to hold if we assume that un (x) - u (x) for µ-a.e. x E X. Indeed, in view of Proposition B.38(iv), it suffices to redefine un and u to be zero in the set of measure zero in which there is no pointwise convergence.
Corollary B.41. Let (X, fit, µ) be a measure space and let u, v : X - [0, oo) be two measurable functions. Then
udµ+fxvdµ.
Jx(u+v) dµ = JX
Corollary B.42. Let (X, fit, µ) be a measure space and let un : X - [0, oo] be a sequence of measurable functions. Then
n=1
fx undp= f
>undp.
x n=1
Example B.43. Given a doubly indexed sequence {ank}, with ank > 0 for all n, k E N, we have 00
00
00
00
E E ank = > E ank'
n=1k=1
k=1n=1
B. Measures
516
To see this, it suffices to consider X = N with the counting measurer and to define un : N - [0, oo] by un (k) := ank. Then
JX
00 ank, un dµ = E
k=1
and the result now follows from the previous corollary.
Lemma B.44 (Fatou's lemma). Let (X, fit, µ) be a measure space. (i) If un X - [[0, oo] is a sequence of measurable functions, then lim inf un dµ < lim inf
JX n- oo
n-too
X
u,1 dµ.
(ii) If un X - [-oo, oo] is a sequence of measurable functions such that
un
Corollary B.45. Let (X, fit, it) be a measure space and let u : X -' [0, oo] be a measurable function. Then
Ix udtt=0 if and only if u (x) = 0 for µ-a. e. X E X X.
In order to extend the notion of integral to functions of arbitrary sign,
consider u : X - [-oo, oo]. Note that u = u+ - u- and Jul = u+ + u- and that u is measurable if and only if u+ and u are measurable. Also, if u is bounded, then so are u+ and u-, and in view of Theorem B.36, u is then the uniform limit of a sequence of simple functions.
Definition B.46. Let (X, fit, Et) be a measure space and let u : X [-oo, oo] be a measurable function. Given a measurable set E E fit, if at least one of the two integrals fE u+ dp and fE u- dµ is finite, then we define the (Lebesgue) integral of u over the measurable set E by
JE
:=
fEU+dit
-
fEu
dµ.
'Given a set X, the counting measure u : 7, (X) - 10, ool is defined by
A(E):= for every E C X.
the number of elements of E 00
if E is a finite set, otherwise
B.2. Measurable and Integrable Functions
517
If both fE u+ dµ and fE u- dµ are finite, then u is said to be (Lebesgue) integrable over the measurable set E.
A measurable function u : X -+ [-oo, oo] is Lebesgue integrable over the measurable set E if and only if
1 Jul dµ
Proposition B.48. Let (X, 9A,µ) be a measure space and let u, v : X [-oo, oo] be two integrable functions.
(i) If a,,8 E R, then au +,Qv is integrable andr
J(+
dµ = a
Jx udµ + f3 xJ v d.
(ii) If u (x) = v (x) for µ--a.e. x E X, then
fud/L=JvdIL(iii)
(B.3)
I fx udµl <_ fx Jul dµ.
Part (ii) of the previous proposition motivates the next definition. Let (X, 97i, µ) be a measure space. If F E 911 is such that µ (F) = 0 and
u : X \ F - [-oo, oo] is a measurable function in the sense of Definition B.32, then we define the (Lebesgue) integral of u over the measurable set E as the Lebesgue integral of the function V (x)
_
u(x) ifxEX\F, 0
otherwise,
provided fE v dµ is well-defined. Note that in this case
JEvdµ = JEVdµ, where v (x)
_
u(x) ifxEX\F,
1 w (x) otherwise and w is an arbitrary measurable function defined on F. If the measure µ is complete, then fE v dµ is well-defined if and only if fE\F u dµ is well-defined.
B. Measures
518
For functions of arbitrary sign we have the following convergence result.
Theorem B.49 (Lebesgue's dominated convergence theorem). Let (X, fit, p) be a measure space and let un : X -p [-oo, oo] be a sequence of measurable functions such that mwu,b(xW =u(x)
for p-a.e. x E X. If there exists a Lebesgue integrable function v such that
lun(x)I
j (Un - ul du = 0. Jx
In particular, lim fX un dp =
n-+o0
Jx
udµ.
Corollary B.50. Let (X, fit, p) be a measure space and let un : X [-oo, oo] be a sequence of measurable functions. If ao
EJ IuIdt
U (x) :_ E u.,t. (x) n=1
defined for µ-a. e. x E X, is integrable, and
fxundy-Jx>u n=1 r=i
dit.
Theorem B.51 (Jensen's inequality). Let V be a Banach space and let f :
V -1 be bounded from above in a neighborhood of a point. Then f is convex if and only if for every measure space (X, 0, p), where X has at least two distinct elements and u (X) = 1, and every function g E Li ((X, fit, p) ; V), then (B.4)
f (f g
dy) :5 f
f o g dµ.
B.3. Integrals Depending on a Parameter
519
B.3. Integrals Depending on a Parameter In this section we study the continuity and the differentiability of functions of the type
F(y)=ff(xv) dA (x), yEY. We begin by studying continuity.
Theorem B.52. Let (X, 91I, p.) be a measure space, let Y be a metric space, and let f : X x Y -' R be a function. Assume that for each fixed y E Y the function x E X-+ f (x, y) is measurable and that there exists yo E Y such that lim f (x, y) = f (x, yo) y-'No
for every x E X. Assume also that there exists an integrable function g X -> [0, oo) such that If (x,y)1<_9(x) for p-a.e. x E X and for all y E Y. Then the function F : Y -' R, defined by
F(y)=Jxf(x,y)dA(x), yEY, is well-defined and is continuous at yo.
Next we study the differentiability of F.
Theorem B.53. Let Y be an interval of R and assume that for each fixed x E X the function y E Y p-' f (x, y) is differentiable and that for each
fixed y E Y the functions x E X '-, f (x, y) and x E X '- X (x, y) are measurable. Assume also that for some yo E Y the function x E X -f (x, yo) is integrable and that there exists an integrable function h : X -> [0, oo) such that
of (x,y)I < h(x) for 14-a. e. x E X and for ally E Y. Then the function F : Y - R, defined by
F (y) = Ix f
(x,
y) dA (x) ,
yEY,
is well-defined and differentiable, with F' (y)
= Ix d'+J
(x,
y) dµ (x) .
The next example shows that, without the integrability of g, Theorem B.52 fails.
520
B. Measures
Exercise B.54. Consider the function IILY4
f (x, y) _
0
if IVI < IxI , if IyI ? IxI
Prove that the function
F (y) =
ff(xu) dx,
Y E R,
is well-defined and is not continuous at y = 0.
B.4. Product Spaces We recall that, given two measurable spaces (X, 911) and (Y, 91), we denote by 931 0 9Z C P (X x Y) the smallest a-algebra that contains all sets of the
form E x F, where E E 931, F E 97. Then 931 0 91 is called the product or-algebra of 9R and 91.
Exercise B.55. Let X and Y be topological spaces and let B (X) and B (Y) be their respective Borel a-algebras. Prove that
B(X)0B(Y) CB(X xY). Show also that if X and Y are separable metric spaces, then
B(X)®B(Y)=B(X xY). In particular, B (RN) = B (R) 0... ®B (R). Let (X, 931, jc) and (Y, 91, v) be two measure spaces. For every E E X x Y define 00
(B.5)
(p x v)* (E) := inf
p (F.) v (Gn) : {Fn} C 931, {G,,} C 91, n=1 00
ECU (Fn X Gn)
.
n=1
By Proposition B.3, (p x v)` : P (X) - [0, oo] is an outer measure, called the product outer measure of ,a and v. By Caratheodory's theorem, the restriction of (p x v)* to the a-algebra 911 x 91 of (p x v)*-measurable sets is a complete measure, denoted by p x v and called the product measure of It and v. Note that 911 x 91 is, in general, larger than the product v-algebra 9'11®9.
Theorem B.56. Let (X, fit, p) and (Y, 97, v) be two measure spaces.
(i) If F E M and G E 91, then F x G is (,u x v)`-measurable and (B.6)
(pxv)(FxG)=p(F)v(G).
B.4. Product Spaces
521
(ii) If u and v are complete and E has a-finite (p x v)-measure, then for u-a.e. x r: X the section
E,:_ {y e Y : (x, y) E E} belongs to the cr-algebra 9 and for v-a.e. y E Y the section
Ey := {x E X : (x, y) E E} belongs to the o-algebra fit. Moreover, the functions y u-r us (Ev) v (E.,) are measurable and and x
(p x v) (E) =
f(E) dv (y) _ Jv(Ex) dµ (x)
.
The previous result is a particular case of Tonelli's theorem in the case
that u = XE Theorem B.57 (Tonelli). Let (X, fit, u) and (Y, 97, v) be two measure spaces. Assume that p. and v are complete and or-finite and let u : X x Y [0, ooj be an (9A x 91) -measurable function. Then for p-a. e. x E X the function it (x, ) is measurable and the function fj, u y) dv (y) is measurable. Similarly, for v-a.e. y E Y the function it y) is measurable and the function fX u (x, ) dA (x) is measurable. Moreover,
IXXY
u (x, y) d (u x v) (x, y)
fX =
\JY u (x, y) dv (y))
f (f' ) dµ
du (x)
(x)) dv (y).
Throughout this book, for simplicity, we write
XY
u (x7'll) dv (y) du (x)
fX fy u
(x, y) dv (y)dp (x)
Exercise B.58. Prove that in the case that it : X x Y -+ [0, oo] is (91 0 T)measurable, then Tonelli's theorem still holds even if the measures p and v are not complete, and the statements are satisfied for every x E X and
y E Y (as opposed to for p-a.e. xEXand for va.e. yEY). Theorem B.59 (Fubini). Let (X, Wt, A) and (Y, 01, v) be two measure spa-
ces. Assume that u and v are complete and let u : X x Y -* [-oo, oo] be (p x v)-integrable. Then for p-a.e. x E X the function is vintegrable and the function fy u
y) dv (y) is ,a-integrable.
522
B. Measures
Similarly, for v-a.e. y E Y the function u y) is µ-integrable and the function fX u (x, ) dµ (x) is v-integrable. Moreover,
J xY u (x, y) d (µ x v) (x, y) = J X JY u (x,1!) dv (y) dA (x) J u (x, y) dp (x) dv (y)
f
Exercise B.60. Prove that in the case that u : X x Y -' [-oo, oo] is (9R (9 9t)-measurable, then Fubini's theorem still holds even if the measures µ and v are not complete. The following result is a simple consequence of Tonelli's theorem.
Theorem B.61. Let (X, 9Jt, µ) be a measure space and let u : X -' [0, oo] be a measurable function. Then 00
fud=j({xEX:u(x)>t})dt.
(B.7)
B.5. Radon-Nikodym's and Lebesgue's Decomposition Theorems Definition B.62. Let (X, sJJ) be a measurable space and let µ, v : 97t [0, oo] be two measures. The measure v is said to be absolutely continuous with respect to µ, and we write v << µ, if for every E E fit with µ (E) = 0
we have v(E)=0. Proposition B.63. Let (X, 9)t) be a measurable space and let µ, v : Wi -+ [0, oo] be two measures with v finite. Then v is absolutely continuous with respect to µ if and only if for every e > 0 there exists 6 > 0 such that v (E) < e
(B.8)
for every measurable set E C X with µ (E) < 5.
Proposition B.64. Let (X, 9R) be a measurable space and let µ, v : Wi [0, oo] be two measures. For every E E 91t define (B.9)
v(
(E) := sup { J u dµ : u : X -' [0, oo] is measurable,
I
E
fE' udµ < v(E) for allE'CE,E'E9Jt}. 1JJ
Then vac is a measure, with vas << µ, and for each E E fit the supremum in the definition of vac is actually attained by a function u admissible for va., (E). Moreover, if va, is a-finite, then u may be chosen independently of the set E.
B.6. Signed Measures
523
Theorem B.65 (Radon-Nikodym). Let (X, fit) be a measurable space and let it, v : fit - [0, oo] be two measures, with. is or-finite and v absolutely continuous with respect to A. Then there exists a measurable function u : X - [0, oo] such that
v(E)=JEudµ for every E E fit. The function u is unique up to a set of u-measure zero. The function u is called the Radon-Nikodym derivative of v with respect
to ic, and we write u = k . Definition B.66. Let (X, fit) be a measurable space. Two measures it, v : fit - [0, oo] are said to be mutually singular, and we write v 1 ,a, if there exist two disjoint sets X., X,, E fit such that X = Xu U X and for every
EE9Awehave
µ(E) =it (EnXµ), v(E) Proposition B.67. Let (X, fit) be a measurable space and let it, v : fit [0, oo] be two measures. For every E E fit define (B.10)
v, (E) := sup {v (F) : F C E, F E fit, µ (F) = 01.
Then v, is a measure and for each E E fit the supremum in the definition of v, is actually attained by a measurable set. Moreover, if v, is o -finite, then v, 1 µ.
Theorem B.68 (Lebesgue's decomposition theorem). Let (X, fit) be a measurable space and let it, v : fit -+ [0, oo] be two measures, with u o -finite. Then (B.11)
V = Vac + vs,
where va, and v, are defined in (B.9) and (B.10), respectively, and vac << It.
Moreover, if v is o -finite, then v, 1 µ and the decomposition (B.11) is unique, that is, if v = vac + I/a,
for some measures vac, v,, with vac << ib and v, 1 µ, then
vac =v.c
and v,=v,.
B.6. Signed Measures Definition B.69. Let X be a nonempty set and letW be an algebra on X. A finitely additive signed measure is a function A : fit [-oo, oo] such that (i) A (0) = 0,
B. Measures
524
(ii) A takes at most one of the two values oo and -oo, that is, either A : W - (-oo, co] or A : W- [-oo, oo), (iii) if El, E2 E W are disjoint sets, then A(E1UEI) = A(E1)+A(E2). Definition B.70. Let (X, fit) be a measurable space. A signed measure is a function A : fit -* [-oo, oo] such that
(i) A(0)=0, (ii) A takes at most one of the two values oo and -oo, that is, either
A: %1-4 (-oo,oo]or A:W->[-oo,oo), (iii) for every countable collection {E2} C fit of pairwise disjoint sets, 00
A
00
A(E.). UEn =I: n=
» =1
1
Proposition B.71. Let (X,0)1) be a measurable space and let A : fit -> [-co, oo] be a signed measure. For every E E fit define (B.12) A+ (E) : = sup {A (F) : F C E, F E DYR}, (B.13)
A-(E)
-inf{A(F): FC E, FE fit}
=sup{-A(F): FCE, FEfit}. Then A+ and A- are measures. Moreover, if A : 9)1 -p I-oo, oo), then for every E E WI we have (B.14)
A+(E)=sup{A(F): FCE,FEfit,A-(F)=0},
A+ is finite, and A = A+ - A- .
Theorem B.72 (Jordan's decomposition theorem). Let (X, fit) be a measurable space and let A : fit -+ [-oo, oo] be a signed measure. Then there exists a unique pair (A+, A-) of mutually singular (nonnegative) measures, one of which is finite, such that A = A+ - A-. The measures A+, A- are called, respectively, the upper and lower variation of A, while the measure
1,\I:=,\+ +xis called the total variation of A A. We say that A is bounded, or finite, if the measure J A I is finite.
Proposition B.73. Let (X, fit) be a measurable space and let A : fit -> [-oo, oo] be a signed measure. Then for every E E 9)1, CO
IAI(E)=sup EIA(E.)I n=1
B.6. Signed Measures
525
where the supremum is taken over all partitions { E } C 911 of E.
Definition B.74. Let (X, 9931) be a measurable space. A set function A = (Al, ... , Am) : fit
RI is a vectorial measure if each component A, : 9912 - R
is a signed measure, i = 1, . . . , m. Proposition B.75. Let (X, 9931) be a measurable space and let A : fit -> R'" be a signed measure. Then the function IAA :9N -p [0, oo], defined by W
11\1(E):=sup E IA (En)d
E E'931,
1n=1
where the supremum is taken over all partitions {En} C 9971 of E, is a measure.
Definition B.76. Let (X, 9912) be a measurable space, let p : fit - [0, oo] be a measure, and let A :9931- [-oo, oo] be a signed measure. (i) A is said to be absolutely continuous with respect to p, and we write
A«p,if A(E)=0whenever EE991Iand a(E)=0. (ii) A and p are said to be mutually singular, and we write A 1 it, if there exist two disjoint sets X, , Xa E 9932 such that X = X. U Xa and for every E E 9932 we have
p(E)=p(EnX,,), A(E)=A(EnXA). Note that if A << p, then A+ « p and A- « p. Theorem B.77 (Lebesgue's decomposition theorem). Let (X, 9931) be a mea-
surable space, let A : M - [-oo, oo] be a signed measure, and let p :9912 [0, oo] be a a-finite (positive) measure. Then
A=Aac+A, with Aac << p, and there exists a measurable function u : X -> [-oo, oo] such that
dp A. (E) = JE u for all E E 9931. Moreover, if A is a -finite, then A, 1 it and the decomposition
is unique; that is, if A = Aac + As,
for some signed measures aac, A,, with aac << p and a, 1 p, then Aac = Aac
and
A, = A,.
It is implicit in the statement of the theorem that fE u dp is well-defined for every E E 9931, which implies that u+ or u is Lebesgue integrable.
B. Measures
526
We call ) and A,, respectively, the absolutely continuous part and the singular part of )A with respect to p, and often we write dam u= du
B.7. LP Spaces Let (X, WI, µ) be a measure space. Given two measurable functions n, v : X - [-oo, oo], we say that u is equivalent to v and we write
u - v if u(x) = v (x) for p a.e. x E X.
(B.15)
Note that N is an equivalence relation in the class of measurable functions. With an abuse of notation, from now on we identify a measurable function u : X -> [-oo, oo] with its equivalence class [u].
Definition B.78. Let (X, 9)1, p) be a measure space and let 1 < p < oo. Then Lp (X, fit, µ)
{u: X - [-oo, ool : u is measurable, IIuIIl,(x,yr,,.) < o }
where
,
1/p
IItIILP(X,an,p) :=
If p = oo, then
Ux Iulp
dp)
L°D (X, T1,
{u : X -> [-oo, ool : u is measurable, IIiIIL-(x,srru) < 001 where IIUIIL-(x,tnt,..) is the essential supremum esssup Jul of the function Jul,
that is, esssup Jul = inf {a E R ; Iu (x)l < a for p-a.e. x E X}.
For simplicity, and when there is no possibility of confusion, we denote the spaces L" (X, WI, µ) simply by U' (X, µ) or U' (X) and the norms IIUIILI(X,M,) by IIUIIL9(x), IIUIILP, or
Ilullp.
We denote by Lp (X, fit, lc; R) (or more simply by 17 (X; R)) the space of all functions u : X - R' whose components are in LP (X, WI, µ). We will endow L" (X, fit, µ;1P') with the norm IIuIILP(X,r,p;ttm)
IIuiIILp(X,
,w)
i=1
For 1 < p < oo sometimes it will be more convenient to use the equivalent norm
1/p
IIuhIL'(x,tm,u;Rm) .-
(Ix Jul" dµ)
B.7. LP Spaces
527
Given 1 < p < oo, the Holder conjugate exponent of p is the extended real number p' E [1, oo], defined by
p1 if l < p < oo,
p` .=
oo
ifp = 1,
1
ifp=oo.
Note that, with an abuse of notation, we have
p+--=1. Theorem B.79 (Holder's inequality). Let (X, fii, p) be a measure space, let 1 < p < oo, and let p' be its Holder conjugate exponent. If u, v : X -' [-oo, oo] are measurable functions, then (B.16)
IIUVIILI <- IILIILP IIVIILPI
In particular, if u E LP (X) and v E LP' (X), then uv E Ll (X).
The proof of Holder's inequality for 1 < p < oo is based on Young's inequality
ab < pap + -bpi,
(B.17)
which holds for all a, b > 0 and whose proof is left as an exercise.
Exercise B.80. Let (X, fit, µ) be a measure space. (i) Let 1 < p1 i ... , pn, p < oo, with »I + 1 , . . . , n . Prove that
p,
and ui E LP' (X ),
n
n
<_ [J IIuiIILPl
1=1 i=1
+ pn =
LP
i=1
(ii) Let u : X --+ R be a measurable function. Prove that (IIuIILr)1-e
IIUIILq -< (IIuII LP)B
where 1
max IIUIILP =
fx Iuvi dµ : v E LP( X), IIvIILP# < 11
sup I fX Iuvl dµ : v E L' (X) , IIVIILI < 1
I
if p < 00,
ifp = oo.
B. Measures
528
Proof. If u = 0, then both sides of (B.18) are zero, and so there is nothing to prove. Hence, without loss of generality, we may suppose that 0 < IIullL' < oo. By Holder's inequality, for every v E I?' (X) with IlvllLp < 1, Ix Iuvl dµ s 1141' IIUIIP <_ IIUIILP
,
and so
sup f Ix Iuvl dla : v E V" (X), IIvIIL,,, 5 1} <_ IIullL.
-
To prove the reverse inequality, assume first that 1 < p < oo and define lu(x)IP-1 if u(x) 0 0,
(x) :_ fo
if U (X) = 0,
so that luwl = Jul'. (IIuiiL,)_p/p'
If rp > 1, one has Iwlp' = lulp and if v :=
w, then IIvllLd = 1
Iuvl du = IlullL,. On the other hand, if p = 1, then w E L°° (X),
and Ix 11W I I
L- = 1, and Ix l uw l dp = I IuII L' This proves the first equality in
(B.18).
Finally, when p = oo, let 0 < M < IIuII L- By the finite subset property we can find a set F C X with 0 < p (F) < oo such that Iu (x)I > M for all x E F. Define µ 0
v (x) .=
F
1
ifXEF,
ifx0F.
Then IIVIIL1 = 1 and
luvl dp
mess (F) IFL
do
M,
and so sup { J Iuvldµ:vEL'(X),IIVIILI _
l
X
1)>M.
It now suffices to let M / IIuII L«> .
Corollary B.82. Let (X, )t, y) be a measure space, let 1 <_ p <_ oo, and let p' be the Holder conjugate exponent of p. Assume that µ has the finite subset property. Then for every measurable function u : X - R, (B.19)
IIUIILP = sup I fX Iuvl dp : v E L''' (X), IIvIILpI <_ 1}
Proof. Step 1: Assume that 1 < p < oo. We claim that if (B.20)
C := sup
l Ix
Iuvl dµ : v E rte' (X), IIVIILP' < 1 } < 00, JJJ
529
B.7. LP Spaces
then the set {x E X : u (x) # 0} has a -finite measure. Let x E X : lu (x)I > n } .
Xn
(B.21)
JJJ
Since
{xEX: u(x):A 0}= UXn, 00 n=1
oo for all n E N. Assume by contradiction it remains to show that IL that p (XI) = oc for some t E N and let
M:=sup {p(E): ECX1,EEfit,0
kyoo
Let co
Eoo:=
UEk. k=1
Then p (EE) = M. Note that M = oo. Indeed, if p (E.) < oo, then p (X11 E,o) = oo, and so there exists a set G C XI \ E0,, G E fit, such that 0 < p (G) < oo. But then EOOUG would be admissible in the definition of M and this would give a contradiction. Thus, p(Eo,,) = oo. Define w := uXE.. By (B.20) we have that
0o>C _ supIwvl dµ: vEL'(X), IIVILP, <1JJJ
I fx
On the other hand, taking 1
vh
XE",
(µ
(Ek))1/p'
we have that vk E & (X) with IIvkIIL,' = 1, and so
fx
C
>
1
Iwvkl dp = 1
( (Ek))1
1
(1A (El.))
1
J Jul du k
= n (p (Ek))n -, 00
as k - oo. This contradiction shows that the set {x E X : u (x) # 0} has u-finite measure.
Step 2: We prove (B.19) in the case 1 < p < 00. In view of the previous proposition, it remains to show that if tc 0 LP (X), then the right-hand side
B. Measures
530
of (B.19) is infinite. Thus, assume by contradiction that (B.20) holds. Let
{ u (x) if Iu (x)I < n and x E Xn,
vn (x)
0
otherwise,
where the Xn are the sets defined in (B.21). Then
Ix IunI" dA < n"p (Xn) < 00, and so by the previous proposition we may find vn E LP' (X) such that IIVnIILP' << 1 and
Iunvnl dµ -> IIunIILP(X) -1. Since
JX IuvnI dit >
fx IunvnI dµ ? IIunIILP(X) - 1
and IIUnIILP(X) - IIUIILP(X) = oo as n -'
oo, we obtain that C = 00. This contradiction completes the proof in the case 1 < p < oo.
Step 3: If p = oo and u V LOO (X), let un (x)
n,
u (x) otherwise.
0
Since un E LOO (X), we can now continue as in the previous step to show
that
Ix
Iuvnl dx > I Unvn dx >- IIUnIILQQ(X) -1- 00
asn -oo. As a corollary of Proposition B.81 we have the following result.
Corollary B.83 (Minkowski's inequality for integrals). Let (X, 9R,µ) and (Y, IN, v) be two measure spaces. Assume that p and v are complete and a-finite. Let u : X x Y - [0, oo] be an (9R x 9't)-measurable function and let I < p < oo. Then
III Iu (x, )I X
<-
dµ (x)IILP(Y,ryt,v)
JX
IIU (X") IILP(Y,'AL) dµ (x)
Proof. To simplify the notation, we write LP (Y) for LP (Y, 91, v). Define v (y)
Ix IU (x, y) I dµ (x) ,
y E Y.
B.7. LP Spaces
531
By Tonelli's theorem, Proposition B.81, and Remark B.11 we have that
II f lu (x, -)I dµ (x)ILP(Y) = IIVIILP(r) X
= sup
{ fy Iv (y) w (y) I dv (y) : w E JY (Y) , II W IILP' =1 }
= sup {
fy
flu (x, y) w (y) I dp (x) dv (y) : w E 1/ (Y) X
,
IIw IIL' =1
= sup { fx f I u (x, y) w (y) I dv (y) dta (x) : w E &(Y), IIW IILP' = 1 } Y l
5 fx sup UY Iu (.T, y) w (y) I dv (y)} : w E L" (Y) ,
l
=f
IIu (x, -) IILP dp. (x)
II W IILP,
dp (-T)
.
0 We now turn to the relation between different LSD spaces.
Theorem B.84. Let (X, fit, la) be a measure apace. Suppose that 1 < p < q < oo. Then (i) LP (X) is not contained in LQ (X) if and only if X contains measurable sets of arbitrarily small positive measure2,
(ii) Lq (X) is not contained in LP (X) if and only if X contains measurable sets of arbitrarily large finite measure3. Corollary B.85. Let (X, MZ, µ) be a measure space. Suppose that 1 < p < q < oo. If It (X) < oo, then Lq (X) (Z LP (X).
Theorem B.86 (Minkowski's inequality). Let (X, 91t, p) be a measure space, let 1 S p :5 oo, and let u, v : X -> [-oo, oa] be measurable functions. Then, IIu + V16 <_ IIuIIrp + IIVIILP .
In particular, if u, v E LP (X), then u + v E LP (X). By identifying functions with their equivalence classes [u], it follows from Minkowski's inequality that II'IILn is a norm on LP (X).
Theorem B.87. Let (X, TI, p) be a measure space. Then 1/ (X) is a Banach space for 1 < p < oo. 2By this we mean that for every e > 0 there is a measurable set of positive measure less than C.
3By this we mean that for every M > 0 there is a measurable set of finite positive measure greater than M.
B. Measures
532
Next we study some density results for LP (X) spaces.
Theorem B.88. Let (X, 99)1, µ) be a measure space. Then the family of all simple functions in LP (X) is dense in LP (X) for 1 < p::5 oo.
The next result gives conditions on X and p. that ensure the density of continuous functions in V (X). Theorem B.89. Let (X, 9)1, p) be a measure space, with X a normal space and p a Borel measure such that
p (E) = sup {p (C) : C is closed, C C E} = inf {p (A) : A is open, A D E} for every set E E M with finite measure. Then LP (X) fl C, (X) is dense in
LP (X) forl
space, p : fii -' [0, oo] a Borel measure, and 1 < p < oo. A measurable function u : X -> [-oo, oo] is said to belong to L10C (X) if u E LA (K) for every compact set K C X. A sequence {un} C Lo,, (X) is said to converge to u in L a (X) if u,, - u in LP (K) for every compact set K C X. Theorem B.91 (R.iesz's representation theorem in LP). Let (X,9)1, p) be a measure space, let 1 < p < oo, and let p' be its Holder conjugate exponent. Then every bounded linear functional L : LP (X) R is represented by a unique v E L'1 (X) in the sense that (B.22)
L (u) = f uv dp for every u E Lp (X) . x
Moreover, the norm of L coincides with IIVIIL,y. Conversely, every functional of the form (B.22), where v E V" (X), is a bounded linear functional on Lp (X). Thus, the dual of LP (X) may be identified with &(X). In particular, LP (X) is reflexive.
Definition B.92. Let (X, fit, p) be a measure space and let {EQ}QE J be a family of measurable sets of X. A measurable set E,, is called the essential supremum of the family {Ea}a.E J if the following hold.
(i) E. D Ea (up to a set of p-measure zero) for every a E J. (ii) If E E 811 is such that E D E. (up to a set of p-measure zero) for every a E J, then E,,,,, D E. (up to a set of p-measure zero). Definition B.93. Let (X, M1, p) be a measure space. The measure p is said to be localizable if every family of measurable sets admits an essential union. Proposition B.94. Let (X, s9)1, p) be a measure space. If p is u-finite, then it is localizable.
B.7. LA Spaces
533
Theorem B.95 (Riesz's representation theorem in L1). Let (X, fit, µ) be a measure space. Then the dual of L' (X) may be identified with L°O (X) if and only if the measure it is localizable and has the finite subset property. In particular, if p is a-finite, then the dual of L1 (X) may be identified with LOD (X).
The next result shows that L1 (X) is not reflexive, since in general its bidual is larger than L' (X). Let (X, WI, µ) be a measure space. The dual of LO0 (X) may be identified with the space ba (X, 9)1, µ) of all bounded finitely additive signed measures
absolutely continuous with respect to µ, that is, all maps A : 9l? - R such that (i) A is a finitely additive signed measure, (ii) A is bounded, that is, its total variation norm l
IAI (X) := Sup E IA (En)I 1n=1
where the supremum is taken over all finite partitions {En}1 C fit of X, I E N, is finite, (iii) A (E) = 0 whenever E E 9R and µ (E) = 0. Given a measure space (X, 9)t,µ), A E ba (X, fii, µ), and u E L°O (X), by Theorem B.36 we may find a sequence {sk} C L' (X) of simple functions that converges uniformly to u. For every E E fit, we define fE sk dA as in (B.2) with A and sk in place of It and s, respectively. Then
I fE (sk - Sm) dal
<_ 118k
- Sm II L-(X) I AI
(E)
for all k, m E N, and so there exists the limit (B.23)
lim ( sk dA. E 1 u dA :=k-+oo
It may be verified that the integral fE u dA does not depend on the particular approximating sequence {sk}.
Theorem B.96 (Riesz's representation theorem in L°D). Let (X, 9971, µ) be a measure space. Then every bounded linear functional L : L°O (X) - ]R is represented by a unique A E ba (X, 9)i,µ) in the sense that (B.24)
L (u) = J u dA for every u E L°° (X) x
.
Moreover, the norm of L coincides with IIAII Conversely, every functional of the form (B.24), where A E ba (X, M, p), is a bounded linear functional on L°° (X).
B. Measures
534
B.S. Modes of Convergence In this section we study different modes of convergence and their relation to one another.
Definition B.97. Let (X, 931, p) be a measure space and let un, u : X -> R be measurable functions. (i)
is said to converge to u pointwise fa-a.e. if there exists a set E E fit such that p (E) = 0 and
lim u (x) = u (x)
n-4oo
for all xEX\E. (ii)
is said to converge to u almost uniformly if for every e > 0 converges there exists a set E E 931 such that p (E) < e and to u uniformly in X \ E, that is, lim sup Iu n oo TEX\E
(x) - u(x)I = 0.
(iii) {un) is said to converge to u in measure if for every e > 0,
nlim µ({xEX:Iun(x)-u(x)I>e})=0. The next theorem relates the types of convergence introduced in Definition B.97 to convergence in LP (X).
Theorem B.98. Let (X, 931, p) be a measure space and let u,,, u.: X -* JR be measurable functions.
(i) If {un} converges to u almost uniformly, then it converges to u in measure and pointwise pi-a.e. (ii) If {un} converges to u in measure, then them exists a subsequence {un,, } such that {u,,k } converges to u almost uniformly (and hence pointwise p-a.e.). (iii) If {un} converges to u in LP (X), 1 < p < oo, then it converges to u in measure and there exist a subsequence {un,, } and an integrable function v such that fu. ,j converges to u almost uniformly (and
hence pointwise p-a.e.) and lung (x)Ip < v (x) for p-a.e. x E X
and/or ailk EN. Theorem B.99 (Egoroff). Let (X, MI, p) be a measure space with p finite and let u, : X -- R be measurable functions converging pointwise p-a.e. converges almost uniformly (and hence in measure). Then
In order to characterize convergence in LP, we need to introduce the notion of p-equi-integrability.
B.8. Modes of Convergence
535
Definition B.100. Let (X, 9A, p) be a measure space and let 1 < p < oo. A family F of measurable functions u : X -). [-oo, ooj is said to be p-equiintegrable if for every e > 0 there exists d > 0 such that
for all u E F and for every measurable set E C X with is (E) < S. When p = 1, we refer to 1-equi-iniegrability simply as equi-integmbility.
Theorem B.101 (Vitali's convergence theorem). Let (X, fit, p) be a measure space, let 1 < p < oo, and let un, u E LP (X). Then {u, j converges to u in LP (X) if and only if the following conditions hold: (i) {un} converges to u in measure. (ii) {u} is p-equi-integrable. (iii) For every e > 0 there exists E C X with E E D1 such that µ (E) < 0o and
IunIpdp<e
fx\E for all n.
Remark B.102. Note that condition (iii) is automatically satisfied when X has finite measure.
In view of Vitali's theorem it becomes important to understand equiintegrability.
Theorem B.103. Let (X, VA, µ) be a measure space and let.F be a family of integrable functions u : X - [-oo, oo]. Consider the following conditions:
(i) F is equi-integrable. (ii)
lim sup f
(B.25)
Jul dA = 0.
t'OO uE.7 .!{xEX: jug>t}
(iii) (De la Vail& Poussin) There exists an increasing function y : [0, oo) -> [0, ooI with lim y (t) = 00
(B.26)
t-4oo
such that (B.27)
sup
f
ue7 JX
t
y (I ul) dji < oo.
B. Measures
536
Then (ii) and (iii) are equivalent and either one implies (i). If in addition we assume that sup J dp < oo, uE.F IX X
then (i) implies (ii) (and so all three conditions are equivalent in this case).
B.9. Radon Measures Definition B.104. An outer measure µ* : P (X) -> [0, oo] is said to be regular if for every set E C X there exists a µ*-measurable set F C X such that E C F and µ* (E) = µ* (F). An important property of regular outer measures is the fact that Proposition B.9(i) continues to hold.
Proposition B.105. Let µ* : P (X)
[0, oo] be a regular outer measure. If {En} is an increasing sequence of subsets of X, then 00
* U En = lim µ* (En) n-oo n=1
Definition B.106. Let X be a topological space and let µ* : P (X) -+ [0, oo] be an outer measure.
(i) A set E C X is said to be inner regular if µ* (E) = sup {µ* (K) : K C E, K is compact}, and it is outer regular if µ* (E) = inf {µ* (A) : A D E, A is open}.
(ii) A set E C X is said to be regular if it is both inner and outer regular.
Definition B.107. Let X be a topological space and let µ* : P (X) --> [0, oo] be an outer measure.
(i) µ* is said to be a Borel outer measure if every Borel set is µ*measurable. (ii) µ* is said to be a Borel regular outer measure if µ* is a Borel outer
measure and for every set E C X there exists a Borel set F C X such that E C F and µ* (E) = µ* (F). Definition B.108. Let X be a topological space and let µ* : P (X) - [0, oo] be an outer measure. Then µ* is said to be a Radon outer measure if is a Borel outer measure, (ii) µ* (K) < oo for every compact set K C X, (i)
537
B.9. Radon pleasures
(iii) every open set A C X is inner regular, (iv) every set E C X is outer regular.
Remark B.109. Note that a Radon outer measure is always Borel regular.
We investigate the relation between Radon outer measures and Borel regular measures.
Proposition B.110. Let X be a locally compact Hausdorff space such that every open set is a-compact. Let p* : P (X) -> [0, ool be a Borel outer measure such that p* (K) < oo for every compact set K C X. Then every Borel set is inner regular and outer regular. If, in addition, p* is a Borel regular outer measure, then it is a Radon outer measure. We now introduce analogous regularity properties for measures.
Definition B.111. Let (X, 931, p) be a measure space. If X is a topological space, then the following hold.
(i) p is a Borel regular measure if it is a Borel measure and if for every set E E 9)1 there exists a Borel set F such that E C F and p (E) = it (F). (ii) A Borel measure p : fit [0, oo] is a Radon measure if (a) p (K) < oo for every compact set K C X, (b) every open set A C X is inner regular, (c) every set E E 9)1 is outer regular.
Hausdorff measures if, s > 0, represent an important class of regular Borel measures that are not Radon measures. Proposition B.112. Let X be a locally compact Hausdorff space such that every open set is a-compact. Let it : B (X) -> [0, oo] be a measure finite on compact sets. Then p is a Radon measure and every Borel set E is inner regular.
Definition B.113. Let (X, 9R) be a measurable space and let A : 9)1 [-oo, oc] be a signed measure. If X is a topological space, then A is a signed [0, oo] is a Radon measure. Radon measure if IAI : 932
If X is a topological space, then Mb (X; R) is the space of all signed finite Radon measures A : B (X) - R endowed with the total variation norm. It can be verified that Mb (X; R) is a Banach space with the norm II'IIMb(X;R) := IAI (X)
B. Measures
538
Similarly, if (X, fit) is a measurable space and X is a topological space, then A : fit - R1 is a vectorial Radon measure if each component Ai : fit -, R is a signed Radon measure. The space Mb (X; Rm) of all vectorial Radon measures A : B (X) - R n is a Banach space with the norm IIAIIM4(x;Rm) = IAI (X)
Theorem B.114 (Riesz's representation theorem in CO). Let X be a locally compact Hausdorff space. Then for every bounded linear functional L Co (X; Rm) --+ R there exists a unique A E Mb (X; R'n) such that
L (u) = J u dA for every u E CO (X; R).
(B.28)
Ix
Moreover, the norm of L coincides with IAI (X). Conversely, every functional of the form (B.28), where A E Mb (X; R), is a bounded linear functional on CO (X; Rfb).
Theorem B.115 (Riesz's representation theorem in Ce). Let X be a locally
compact Hausdorf space and let L : Cc (X) - ]R be a linear functional. Then
(i) if L is positive, that is, L (v) > 0 for all v E CC (X), then there exists a unique (positive) Radon measure µ : B (X) - [0, oo] such that
L (u) =
u dµ for every u E Cc (X) , fX (ii) if L is locally bounded, that is, for every compact set K C X there exists a constant CK > 0 such that IL (u)1:5 CK IlvIlcc(x)
for all v E CC (X) with supp v C K, then there exist two (positive) Radon measures Al, µ2 B (X) - [0, oo] such that ::
L (u) = J u dµ1 - Ix u dµ2 for every u E CC (X) x Note that since both µl and µ2 could have infinite measure, their difference is not defined in general, although on every compact set it is a well-defined finite signed measure.
B.10. Covering Theorems in R" Theorem B.116 (Besicovitch's covering theorem). There exists a constant f, depending only on the dimension N of RN, such that for every collection 7 of (nondegenerate) closed balls with (B.29)
sup { diam B : B E F} < 00
B.10. Covering Theorems in Riv
539
there exist F1,. .. , .gyp C F such that each .7;,, n family of disjoint balls in F and
is a countable
I
EC UU n=1 BEF'J
where E is the set of centers of balls in F.
Definition B.117. Given a set E C RN, a family.F of nonempty subsets of RN is said to be (i) a cover for E if
ECUF, FEF
(ii) a fine cover for E if for every x E E there exists a subfamily .Fy C F of sets containing x such that
inf {diem F : F E .Fx} = 0.
(B.30)
The following result is an important consequence of the Besicovitch covering theorem.
Theorem B.118 (Vitali-Besicovitch's covering theorem). Let E C RN and let .F be a family of closed balls such that each point of E is the center of arbitrarily small balls, that is,
inf {r : B (x, r) E F} = 0 for every x E E. Let A* : P (RN) - (0, ooj be a Radon outer measure. Then there exists a countable family {B (xn, r,,)} C F of disjoint closed balls such that
µ* E 1 U B (xn, rn) = 0. n
Theorem B.119 (Besicovitch's derivation theorem). Let µ, v : 8 (RN) [0, oo] be two Radon measures. Then there exists a Borel set M C RN, with tc (M) = 0, such that for every x e RN \ M, (B.31)
v (B (x, r)}
dv
°C (x) = lim
IA(r(B -(T, r))
and ve 11m
r-r0+
(.(z,r)) (B(x,r))
- 0,
E Ilt
B. Measures
540
where v = vac + U3,
(B.32)
Vac << µ,
v8 1 A-
Remark B.120. In the previous theorems one can use closed cubes instead of balls.
Theorem B.121. Let p : B (RN) --> [0, oo] be a Radon measure, let u : RN -+ [-oo, oo] be a locally integrable function. Then there exists a Borel set M C RN, with µ (M) = 0, such that RN \ M C {x E R" : u (x) E R} and for every x E RN \ M, lim
r+
1
(
,
r))\
J
u(y)dpc(y)=u(x). (z,r)
By enlarging the "bad" set M, we can strengthen the conclusion of the previous theorem.
Corollary B.122. Let it : B (RN) - [0, oo] be a Radon measure and let u : RN - [-oo, oo] be a locally integrable function. Then there exists a Borel set M C RN, with tt (M) = 0, such that
RN\MC {xERN: u(x) ER} and for every x E RN \ M, (B.33)
lim 1 r0+ J Iu (y) - u (x) I dµ (y) = 0. µ (ii (x, r)B(x,r)
A point x E RN for which (B.33) holds is called a Lebesgue point of u.
Corollary B.123. Let p : B (RN) - [0, oo] be a Radon measure, let 1 < p < oo, and let u E L' (RN). Then there exists a Bored set M C RN, with p (M) = 0, such that RN \ M C {x E RN : u (x) E R}, and for every x E RN \ M, (B.34)
rlim+
1
p (B (x, r))
f
B(x,r)
j u (y)
- u (x) I P dp (y) = 0.
A point x E RN for which (B.34) holds is called a p-Lebesgue point of u. By applying Theorem B.121 to XE, we obtain the following result.
B.10. Covering Theorems in RN
541
Corollary B.124. Let it : B (RN) - [0, oo] be a Radon measure and let E C RN be a Borel set. Then there exists a Bored set M C RN, with p (M) = 0, such that for every x E RN \ M,
lim p (B (x, r) f E) = XE (x) p (B (x, r))
r-0+
A point x E E for which the previous limit is one is called a point of density one for E. More generally, for any t E [0, 1] a point x E RN such that
(x, r) n E) _ t
lim p (B r-.0+ p (B (x, r))
is called a point of density t for E.
Appendix C
The Lebesgue and Hausdorff Measures The Analysis of Value must be used carefully to avoid the following two types of errors. Type I: You incorrectly believe your research is not Dull. Type H. No conclusions can be made. Good luck graduating. -Jorge Cham, www.phdcomics.com
C.1. The Lebesgue Measure In the Euclidean space RN consider the family of elementary sets
9:= {Q (x, r) : x E RN, 0 < r < oo} U {0} and define p (Q (x, r)) := rN and p (0) := 0, where
Q (r, r) := x +
rrN
(-ii)
For each set E C RN define
0 ,C (E) := inf
(rn.)N : {Q (xn, rm)} C G, E C 11t=1
UQ n=1
By Proposition B.3, Lo is an outer measure, called the N-dimensional Lebesgue outer measure. Using Remark B.4, it can be shown that
(C.i)
Go (Q (x, r)) = p (Q (x, r)) = rN
and that Go is translation-invariant, i.e., Co (x + E) = Lo (E) for all x E RN and all E C RN. The class of all Co -measurable subsets of RN is called the a-algebra of Lebesgue measurable sets, and by Caratheodory's theorem, Lp restricted to this a-algebra is a complete measure, called the N-dimensional Lebesgue measure and denoted by LN. Since LN (RN) > 543
C. The Lebesgue and Hausdorff Measures
544
oo, we conclude that GN is not a finite measure. However, it is o--finite, since
Go (Q (x, r)) = rN, by sending r
00
RN
= U Q (0, n) n=1
and GN (Q (0, n)) _ nN < oo.
Exercise C.1. Prove that for every b > 0 and for each set E C RN, 00
00
Go (E) = inf E (rn.)N : {Q (x,,, rn)} C 9, rn <,6, ECU Q (xn, rn)
.
n=1
1n=1
Conclude that Go is a metric outer measure.
It follows from the previous exercise and Proposition B.19 that every Borel subset of RN is Lebesgue measurable. It may be proved that there are Lebesgue measurable sets that are not Borel sets (see Exercise 1.45). Hence, GN : B (RN) -> 10, oo] is not a complete measure. Using the axiom of choice, it is also possible to construct sets that are not Lebesgue measurable. Exercise C.2 (A non-Lebesgue measurable set). On the real line consider
the equivalence relation x - y if z - y E Q. By the axiom of choice we may construct a set E C (0,1) that contains exactly one element from each equivalence class. Let
F:=
U
(r + E) C (-1, 2).
re(-1,1)f1Q
(1) Prove that F D (0, 1). (ii) Prove that if r, q E Q, with r # q, then (r + E) fl (q + E) _ 0. (iii) Prove that E is not Lebesgue measurable. We observe that the Lebesgue outer measure is a Radon outer measure. Indeed, outer regularity of arbitrary sets follows from (C.1) and the definition of 'C' N' while inner regularity of open sets is an immediate consequence of the fact that each open set can be written as a countable union of closed cubes with pairwise disjoint interiors.
Proposition C.3. Go is a Radon outer measure. Moreover, every Lebesgue measurable set E is the union of a Borel set and a set of Lebesgue measure zero.
Using the notation introduced in Section B.4, we have
Proposition C.4. Let N = m+ k, where N, n, m E N. Then (,Cl x G"')"` _ GN. 0
C.2. The Brunn-Minkowski Inequality and Its Applications
545
If E C RN is a Lebesgue measurable set, fit is the a-algebra of all Lebesgue measurable subsets of E, Y is a nonempty set, and 91 is an algebra on Y, then a measurable function u : E --+ Y is called Lebesgue measurable.
Proposition C.5. Let E C RN be a Lebesgue measurable set and let u : E -* R be a Lebesgue measurable function. Then there exists a Borel func-
tion v : E - R such that v (x) = u (x) for iN-a.e. x E E.
If E C RN is a Lebesgue measurable set and u : E -' J-oo, ooJ is a Lebesgue measurable function, then, whenever it is well-defined, we denote fE u simply by u dx. JE
If N = 1 and I is an interval of endpoints a and b, we also write fa u dx for fr u dz.
C.2. The Brunn-Minkowski Inequality and Its Applications We leave the following preliminary result as an exercise.
Exercise C.6. Let E, F C RN be two compact sets. (i) Construct a decreasing sequence of sets {En} and {Fn} such that 00
00
E= n E,,, En, F= n1'., n=1
n=1
and all E. and F,, consist of finite unions of rectangles with sides parallel to the axes. (ii) Prove that
LN(E" +Fn) LN(E+F) as
Theorem C.7 (Brunn-Minkowski's inequality). Let E, F C RN be two Lebesgue measurable sets such that
E+F:={x+y: xEE,yEF} is also Lebesgue measurable. Then
(f-'v (E))"' + (LN(F))"' < (.CN(E+F))N
.
C. The Lebesgue and Hausdorff Measures
546
Proof. Step 1: Assume that E and F are rectangles whose sides are parallel to the axes and let x; and yi, i = 1, ... , N, be their respective side lengths. Then N
GN (E)
N
N
= fl xi, GN (F) = II yi, £ N (E + F) = fi (xi + yi) i=1
i=1
i-1
By the arithmetic-geometric mean inequality
(n) N
N . 1 xi + yi +
i_1 xi + tfi
+
1N
N
NN
xi + 3Ji
Ti + Vi
lv
which gives the Brunn-Minkowski inequality for rectangles.
Step 2: We now suppose that E and F are finite unions of rectangles as in Step 1. The proof is by induction on the sum of the numbers of rectangles in E and in F. By interchanging E with F, if necessary, we may assume that E has at least two rectangles. By translating E if necessary, we may also assume that a coordinate hyperplane, say {XN = 01, separates
two rectangles in E. Let E+ and E- denote the union of the rectangles formed by intersecting the rectangles in E with the half spaces {XN > 01 and {XN < 01, respectively. Translate F so that
GN (Et) _ GN (Ft) GN (E)
GN (F)
where F+ and F- are defined analogously to E+ and E-. Then E+ + F+ C
{xN > 0} and E- + F- C {xN < 0}, and the numbers of rectangles in E+ U F+ and in E- U F- are both smaller than the number of rectangles in E U F. By induction and Step 1 we obtain GN (E +.F) > GN (E+ + F+) + GN (E- + F-) >
((GN
(E+))
N
(CN (F+)) x +
+ ((CN (E-)) x +
(V"" (F-))
N
)
N
= GN (E+)
1+
(F)) 1 N
= GN (E)
1+ (
GN (E))A
1+
=
N N
GN (E)
(GN (E)) 'W
(F)) A
+
(GN (E)) N
((N(E))+(N(F))*)N
C.2. The Brunn-Minkowski Inequality and Its Applications
547
Step 3: Assume next that E and F are compact sets. Let {E} and {Fn} be as in the previous exercise. Then for all n E N (,CN (E,)) *
(GN
*
(En + Fn)) + (,CN (F'n)) < It now suffices to let n - oo. Finally, if E and F are Lebesgue measurable sets, with E + F measurable, fix two compact sets K1 C E, K2 C F. Then K1 + K2 C E + F, and so 77
(,CN (,CN (KI + K2)) ' < (GN (E + F)) a N' (Ki)) W + (K2)) < Using the inner regularity of the Lebesgue measure, we obtain the desired (,CN
result.
Remark C.8. (i) Note that the hypothesis that E+F is measurable cannot be omitted. Indeed, Sierpinsky [154] constructed an example of two Lebesgue measurable sets whose sum is not measurable. However, if E and F are Borel sets, then E + F is an analytic set, and so it is measurable. We refer to [65] for more information on this topics.
(ii) Fix 0 E (0,1). By replacing E with OE and F with (1 - 0) F and using the N-homogeneity of the Lebesgue measure, we obtain that
0(GN(E))77 +(1-0) (GN(F))* -L
-L
= (GN(0E))N + (GN((1-0)F))N
(G^'(0E+(1-0)F))p. Thus, the function f (t) := (GN (tE + (1 - t) F)) is concave in [0,1]. In particular, if C C RN+1 is a convex set and E and F are the intersections of C with the hyperplanes xl = 1 and xl = 0 (we write x = (x1i x2, ... , xN+1) E RN+1), then tE + (1 - t) F is 717
the intersection of the convex hull of E and F with the hyperplane xl = t and is therefore contained in the intersection of C with the hyperplane xl = t. Hence, the function giving the nth root of the volumes of parallel hyperplane sections of an (n + 1)-dimensional convex set is concave (see [711). We used this fact in the proof of Poincare's inequality in convex domains (see Theorem 12.30).
Exercise C.S. Let
C={xER3:
VX2
ll JJJJ
Prove that C is convex but the function 9 (xl)
fR2 XE (x1, x2, x3) dx1dx2
is not concave (although its square root is by the previous remark).
C. The Lebesgue and Hausdorff Measures
548
Using the Minkowski inequality, we can prove the isodiametric inequality.
Theorem C.10 (Isodiametric inequality). Let E C RN be a Lebesgue measurable set. Then
(diarnE)N
LN (E)
<
Proof. It is enough to prove the previous inequality for bounded sets, since otherwise the right-hand side is infinite. If A > 0, we have that LN (AE) = ANON (E) and (diam (AE))N = AN diam E, so without loss of generality we may assume that diam E = 1. Let
F:={-x:xEE} By the previous remark the function f (t) := (LN (tE + (1 - t) F)) N is concave in [0,1], and so (LN
(LN
(E)) N +
(F))'v < ( LN (2E + 2 F)) N
.
\\
2 2 But LN (F) = LN (E), and so the previous inequality becomes
(E+F).
LN(E)=GN(F) <,CN
If x, y E 4E + 4F, then x = x and y
where x', x", y', y" E E.
Hence
2Ix-yI= Ix'-x"-(y' - y")I <
Ix'-x"I+IV'-y"I 5 1+1,
since diam E < 1. This shows that diam (2 E + F) < 1. Z Since E + 4F = { x, y E E} is symmetric with respect to the origin, it follows that 12E + 4F C B (0, Hence,
-
D.
LN (E) < LN
2E + 2F < LN B t
= 2N'
0 Remark C.11. Let r > 0. Since LN (B (x, r)) = oN rN for every x E RN, the isodiametric inequality shows that the greatest volume among all sets with given diameter 2r is given by the volume of the ball with radius r, that is,
max {Lo (E). E C RN, diam E = 2r} = LN (B (x, r))
C.2. The Brunn-Minkowski Inequality and Its Applications
549
Definition C.12. Given a set E C RN and an integer 0:5 n < N, we define the n-dimensional upper Minkowski content of E, £N RN : dist(x,E) < e}) ({x E M*n (E) := lim sup aN-neN-n e"0+ and the n-dimensional lower Minkowski content of E, cN ({x E RN : dist (x, E) < e})
M (E) := tun inf e-+0+
aN-neN-n
When the n-dimensional upper and lower Minkowski contents of E coincide, their common value is called the n-dimensional Minkowski content of E and
is denoted Mn (E). Next we prove the isoperimetric inequality (see also Theorem 13.40)
Theorem C.13 (Isoperiinetric inequality). Let E C RN be a Lebesgue measurable set with £N (E) < oo. Then N-1
<M;-1(8E)
LYN
AN
Proof. Since
{x E RN : dist(x,E) < s} _
+ B(0,e),
by the Brunn-Minkowski inequality
£N ({x E RN : dist (x, E) < e}) > { (1.N (E-))7 + (.,) IL N e)
N-1 1
> GN (E) + N (C N ("J) N aNe, where we have used the inequality (a + b)N > aN + NaN-lb for a, b > 0. Hence,
(C.2)
CN ({x E RN : 0 < dist (x, E) < e}) = GN ({x E RN : dist (x, E) < e}) - LN (E)
>N(CN( ))'a' C. On the other hand,
E° ) {x E RN : dist (x, RN \ E) > e } + B (0, e) , and so, again by the Brunn-Minkowski inequality and reasoning as before, LN (E0) LN ({x E RN : dist (x, RN \ E) e})
+N(fN({xERN: dirt(x,RN\E)>e})) N1c
e.
C. The Lebesgue and Hausdor(f Measures
550
It follows that
GN({xERN: 0
(C.3)
- RCN ({x E RN : dist (X, RN 1 E) > e})
>N(GN({xERN. dist(x,RN\E) >E})) "' aNS. Since the sets
{x E RN : 0 < dist (x, E) < E} and
{x ERN : 0 < dist (x, R N \ E) <e} are disjoint subsets of {x r= RN : dist (x, OE) < E}, combining the inequalities (C.2) and (C.3), we obtain
GN ({x E RN : dist (x, 8E) < F})
N (GN (E))
aN
x-i 1 +N(.C"'({xERN: dist(x,RN\E) >E})) N aN. Letting e -> 0+ and using the facts that al = 2 and
GN({xERN: dist(x,RN\E) >E})1GN(E°), we deduce
2M; -' (8E) = lim inf
£N ({x E RN : dist (x, 8E) < e})
s-.0+
> NaN [(EN (y))
E N1
+ (CN
(E°)), NIJ
-
To conclude, we now observe that if MN-' (8E) < oo, then, necessarily, GN (8E) = 0. O
Exercise C.14. Let Q' := (0,1)N-1, let f E C' (Q' , and let
E:= {(x'XN) EQ'xR: f (x') <xN< f (x)+1}. Prove that MN-' (8E) < oo and give an explicit formula.
C.3. Convolutions Given two measurable functions u : RN - R and v : RN -p R, the convolution of u and v is the function u * v defined by (C.4)
(u * v) (x) :_ IRN u (x - y) v (y) dy
for all x E RN for which the integral exists.
C.3. Convolutions
551
Theorem C.15 (Young's inequality). Let n E LP (R"), 1 < p < oo, and v E L' (R'v). Then (u * v) (x) exacts for RCN-a.e. x c RN and IIuc VIIL,(RN) <_ IIUIILD(RN) IIVIIL1(RN) .
Proof. Consider two Borel functions uo and vo such that ua (x) = u (x) and vo (x) = v (x) for CN-a.e. X E RN. Since the integral in (C.4) is unchanged if we replace u and v with uo and vo, respectively, in what follows, without loss of generality we may assume that u and v are Borel functions. Let w : RN x RN -+ R be the function defined by
w(x,y):=u(x-y), (x,y)ERNXRN. Then w is a Borel function, since it is the composition of the Borel function u with the continuous function g : RN x RN -, RN given by g (x, y) := x - y. In turn, the function (x, y) E 1$N x RN +--> u (x - y) v (y)
is Borel measurable. We are now in a position to apply Minkowski's inequality for integrals (see Corollary B.83) and Tonelli's theorem to conclude
that IIU * VIILP(RN)
= IIJRN lu (
f =f :s
N
RN
- y) v (y)I dyIL(RN)
Ilu ( - y) v (y)llf,(RN) dy Iv (y)I IIu( - y)IILP(RN) dy
= IIUIILP(RN)
I
N
Iv (y)I dy,
where in the last equality we have used the fact that the Lebesgue measure is translation invariant. Hence, u * v belongs to LP (RN), and so it is finite GN-a.e. in RN. D The following is the generalized form of the previous inequality.
Theorem C.16 (Young's inequality, general form). Let 1 < p < q < oo and let u E L" (RN) and v E L4 (RN). Then (u * v) (x) exists for LN-a.e. x E RN and IIn * VIILP(RN) <_ IIUIILP(RN) IIVIILQ(RN),
where
1+1=1+1. p
q
r
C. The Lebesgue and Hausdorff Measures
552
Proof. If r = oo, then q is the Holder conjugate exponent of p and the result follows from Holder's inequality and the translation invariance of the Lebesgue measure, while if p = 1, then r = q and the result follows from the previous theorem (with u and v interchanged). In the remaining cases, write lu (x - y) v (?a) I = (lu (x - y) IP Iv (y) I,) Iu (x - ir) I r Define
P1 := r,
P2 :=
pr
r-p
,
P3
Iv (y) I
qr
r-q
Then - + in + = 1, and so by the extended Holder inequality (see Exercise B.80) and the translation invariance of the Lebesgue measure, I(u * v) (x)I < f lu (x - y) v (y)I dy RN
f
dy) N
Iu (x - y)I'° Iv(y)1'
Taking the norm in L' (IRN) on both sides, we get
* vIILr(RN) <_ (fRIu (x - y) JRN
-I
r IlullLp(RN) IlvIl o(RN)
Iv (y)14
dydv)
.
IIuIIL!(RN) IIVIILQ(RN)
Applying the previous theorem (with p = 1), we get that the right-hand side of the previous inequality is less than or equal to I.
(IIIU1P11VPt-) III'IgIILI(RN)) r
IIUIILA'(RN) IIhIILq RN) = II1IILP(RN)
II4IILq(RN)
0
This concludes the proof.
C.4. Mollifiers Given a nonnegative bounded function cpr E L' (RN) with (C.5)
supp cp C B (0,1), I
RN
(x) dx = 1,
for every e > 0 we define
(X),
X E P.N.
The functions W. are called mollifiers. Note that supp We C B (0, e). Hence, given an open set fl C Rv and a function u E Liar (Il), we may define (C.6)
De (x) = (2s * VC) (x) = fa we (x - Y) U (Y) dy
C.4. Mollifiers
553
for x E flet where the open set fl, is given by SfE :_ {x E fl : dist (x, Ofl) > e} .
(C.7)
The function uE : Sle - R is called a mollification of u. Note that if x E Sl, then ue (x) is well-defined for all 0 < e < dist (x, 090).
Thus, it makes sense to talk about lime.0+ uE (x). We will use this fact without further mention. Remark C.17. Note that if n = RN, then S1E = RN; thus uE is defined in the entire space RN. Remark C.18. In the applications we will consider two important types of mollifiers:
(i) cp is the (renormalized) characteristic function of the unit ball, that is, 1
w (x) := aN XB(o,1) (x) , x E RN, (ii) cp is the function of class C°° defined by (C.8)
c exp (xI_1) if IxI < 1,
P (x)
if IxI > 1, where we choose c > 0 so that (C.5) is satisfied. In this case, the functions cpe are called standard mollifiers. 0
The following theorem is the first main result of this subsection.
Theorem C.19. Let S2 C RN be an open set, let W E Li (RN) be a nonnegative bounded function satisfying (C.5), and let u E L (S).
I
(i) If u E C (S2), then ue - u as e -- 0+ uniformly on compact subsets of Q.
(ii) For every Lebesgue point x E 12 (and so for LN-a.e. x E fl), as
(iii) If 1 < p < oo, then (C.9)
IIUEIILP(n,) s IIUIILP(n)
for every e > 0 and IItIILP(n) as a - 0+
(C.10) IIUEIILP(ne)
(iv) If u E L" (Sl), 1 < p < oo, then
Iue - uI' dx'
ell
0.
(foe
In particular, for any open set cZ' C Sl with dist (1k', 80) > 0, ue u in L" (fl').
C. The Lebesgue and Hausdorff Measures
554
Proof. (i) Let K C RN be a compact set. For any fixed
0 < rj < dist (K, M), let
K,i :_ {x E RN : dist (x, K) < ,i} so that K1 C 92. Note that fore > 0 sufficiently small we have that K,r C Q . Since K., is compact and u is uniformly continuous on K,1, for every p > 0 there exists 5 = 5 (,, K, p) > 0 such that P
1u(X) - u(y)I <_
aN (i + for all x, y E K,7, with I x - yI < S. Let 0 < e < min {S, ii}. Then for all x E K,
f(x_v)u(v) dy -
lug (x) - u (x)I = (C.12)
sv
(x
1 B(x.a,
E
u(x)I
y) [u (y)
- u (x)] dl!
Iu (y) - u (x)I dy,
<- Ik II00;1 f
B{x,E}
where we have used (C.5) and the fact that supp cpe C B (0, e). It follows by (C.11) that
Ius(x)-u(x)I SP for all x E K, and so IIUE - uhIC(K) < P
(ii) Let x E f be a Lebesgue point of u; that is, emO
£N JS(x,E) Iu (y) - u (x) I dy = 0.
Then from (C.12) it follows that uE (x) -+ u (x) as e 0+. (iii) To prove (C.9), it is enough to assume that u E I' (1k). If 1 <- p < oo, then by Holder's inequality and (C.5) for all x E f2E, lug
(C.13)
Ifn (APE (x
(x) I <-
-
y)) -Pr
(co (x - y)) ; u (y) dyI
(jex_) dy) (Jex_vIu(vh9 dr) 1 dy 1 (f jog (x - O) Iu(y)Ip qq'
P
9I
7
CA Mollif ers
555
and so by Tonelli's theorem and (C.5) once more
nn=flu
f lue (x) Ip dx < f Jcoe(T-v)Iu(Y)rdydx (y)Ip
f
We (x - y) dxdy
= jIlL(Y)IPd;.
On the other hand, if p = oo, then for every x E Q, Iu.(x)I
again by (C.5), and so (C.9) holds for all 1 < p:5 oo. In particular, lien sup 11 U16(0.) <_ IIuIILP(n)
-
To prove the opposite inequality, assume first that 1 < p < oo. Fix an open set SI' CC S2. Then f2' C Q. for all e > 0 sufficiently small. By part (ii), uE (x) -, u (x) as e -i 0+ for V'-a.e. X E St, and so by Fatou's lemma fw lu (x) I" dx =
Jn Letting 0' / Q, we obtain
limo Iue (x) I" dx < limonf
lu (x) I" dx < liminf 5-40+
f
foe
f
n,
IuE (x) I' dx.
Iuf (x) V' dx,
which implies that IIUIILP(n) <- liiminf IIuEIILP(nr)
and so (C.10) is satisfied.
If p = oo, then again by part (ii), ue (x) - u (x) as a -- 0+ for CN-a.e. x E fl. Hence, lu(x)I = Jim lue (x)I <- limiinf 11U116-A) for CN_a.e. X E Sl. It follows that IIuIILX(Q) <_ limonf IIuEIIL-(0.)
Hence, (C.10) holds in this case also. (iv) Fix p > 0 and find a function v E CC (ti) such that IIU - VIILP(n) < -p.
C. The Lebesgue and Hausdorff Measures
556
Since K := supp v is compact, it follows from part (i) that for every 0 < q < dist (K, M), the mollification vE of v converges to v uniformly in the compact set
K,, := {xERN: dist(x,K)Ci?}.
Since ve=v=0in0t,\K,,for0<e<,,wehave that f Ive - vip dx = f Ivr - vlp dx < (live - vII C(Kn))
I K,,I <- p,
,,
provided e > 0 is sufficiently small. By Minkowaki's inequality Ilue - UIILD(n.) <_ Ilue - ueiILn(S2,) + Ilue - vIILn(n.) + Ilv - UIILp(n.) < 2 IIu - VIILp(n) + Ilve - VIILp(sie) < 3p,
where we have used the previous inequality and (C.9) for the function u -
0
V.
More can be said about the regularity of u£ if we restrict our attention to standard mollifiers.
Theorem C.20. Let St C RN be an open set, let y E L' (RN) be defined as in (C.8), and let u E LiC (St). Then ue E COO (Re) for all 0 < E < 1, and for every multi-index a,
(u*!)(x)=f(z_v)u(v)dJ
x=
(C.14)
for all x r= Ste.
Proof. Fix x E Qe and 0 < i < dist (x, act) - e. Let ei, i = 1, ... , N, be an element of the canonical basis of RN and for every h E R, with 0 < Ihi < n, consider ue (x + hei) - ue (x) h
- fa axiWe (x - y) u (y) dy
_ fn=
(x
-
}} u (y) dy
dt - axe (x - ?!) u (y) 1
J'J (x+,,) `( axi
(x -
y + te) -
axi
(x -
dv
u (y)
where we have used Fhbini's theorem and the fact that supp We C B (0, e). Since We E C'° (RN), its partial derivatives are uniformly continuous. Hence, for every p > 0 there exists 6 = S (ii, x, p, e) > 0 such that aV,
axi (z)
BtPe
- 8xi (W)
1 + IIUIILI(B(x,e+n)) p
C.4. Mollifiers
557
for all z, y E B (x, c + i), with Iz - wi < 6. Then for 0 < IhI <_ min {ri, 5} we have
uE (x + h) - of (x) h
- in axi (x - y) u (y) dyl
<- P,
which shows that
8xi
(x) = f ax; (x - y) u (y) dy.
Note that the only properties that we have used on the function cpc are that cpE E C,,' (RN) with supp W f C B (0, e). Hence, the same proof carries over . Thus, by induction we may prove that for if we replace V. with Oa :_ every multi-index a there holds
,
(x) =
2b *
(x)
8 aE
axaE (x - y) u (y) dy
This completes the proof.
An important application of the theory of mollifiers is the existence of smooth partitions of unity.
Theorem C.21 (Smooth partition of unity). Let fl C RN be an open set and let {Ua}aEA be an open cover of Q. Then there exists a sequence C°° (Sl) of nonnegative functions such that
C
(i) each ikn has support in some Ua, o°
(ii) E On (x) = 1 for all x E St, n=1
(iii) for every compact set K C 11 there exists an integer e E N and an open set U, with K C U C 1, such that e
'On(x)=1 n=1
for alixE U. Proof. Let S be a countable dense set in ft, e.g., S := {x E QN n fl}, and consider the countable family .F of closed balls (C.15)
.F':={B(x,r): rEQ,O
Since F is countable, we may write F = {B (xn,rn)}. Since {Ua}aEA is an open cover of 1, by the density of S and of the rational numbers we have
C. The Lebesgue and Hausdorff Measures
558
that 00
fZ=
(C.16)
UB(xn,
n=1
l
21
For each n E N consider On :=
d * XB(xn,4rn)f
where the cpra are standard mollifiers (with a := 4 ). By Theorem C.20, On E C°° (RN). Moreover, if x E B (xn, ), then
2
On (x) = f
RN
W rm (X - Y) XB(yn,grn) (V) dY
4
J
(x - ?1) XB(yn,4rn) (y) dy
(x - y) dy= 1,
a
where we have used (C.5) and the fact that if x E B (xn,
B (x,
a), then
4) C B (xn' 4rn)
Since 0 < XB(xn,4rn) < 1, a similar calculation shows that 0 < On < 1. On
the other hand, if x B (xn, rn), then On (x) =
W
JRN
4 (x - y) XB(yn,4rn) (y) dy
= fB(z,!L)
(x -?I) XB(xn,4rn) (y) dy = 0,
where we have used the fact that if x
B (x' 4)
B (xn, rn), then
n B (Xni 4rn)
0.
In particular, on E C° ° (RN) and supp On C B (xn, rn). Note that in view of the definition of F, supp On C Ua n f for some a E A. Define 11 := 01 and (C.17)
'+Gn :_ (1 - (01) ... (1- On-0 On
for n > 2, n E N. Since 0 < Ok < 1 and supp ¢k C B (xk, rk) for all k E N, we have that 0 < 'fin < 1 and supp?Pn C B (xn, rn). This gives (i). To prove (ii), we prove by induction that (C.18)
Ijl+...+?Pn=1-(1-01)...(1-On)
C.4. Mollifiers
559
for all n E N. The relation (C.18) is true for n = 1, since 01 that (C.18) holds for n. Then by (C.17),
Assume
01 +...+On+'On+1 = 1 - (1 -01)...(1 -On)+tn+1 01)...(1
= 1 - (1 -
= 1 - (1 - 01)...(1
- On) + (1 -
01)...(1
- n+1)
Hence, (C.18) holds for all n E N. Since cbk = 1 in B (xk, 2) for all k E N, it follows from (C.18) that n
forallxE UB xk, L2
(C.19)
k=1
.
Thus, in view of (C.16), property (ii) holds. Finally, if K C S2 is compact, again by (C.16), we may find e E N so large that t
UB(xk, Lk-) DK, k=1
and so (iii) follows by (C.19).
Exercise C.22 (Cut-off function). Let SZ C RN be an open set and let K C St be a compact set. Prove that there exists a function cp E Cc°O (SZ)
such that 0
Theorem C.23. Let 11 C RN be an open set. Then the space C,,00 (fl) is dense in LP (SZ) for 1 < p < oo.
Proof. Let u E LP (fl) and extend it to be zero outside 11. Define
Kn:_
IxI
Then Kn is compact, Kn C Kn+l, and 00
UKn=ci. n=1
Define vn := uXKn and un := cp * vn, where the cp are standard mollifiers n
n
(with
.1). Since supp
n),
it follows that
supp un C { x E RN : dist (x, Kn) <
! } C St.
C. The Lebesgue and Hausdorff Measures
560
Hence, by the previous theorem, u" E CC° (f?). Moreover, by Minkowski's inequality IIu,a - UIILp(
)=
Ilun - UIILP(ll N)
< I Vn * V- -
Vn
* t1I
LP(RN)
< IIVf - uIILP(RN) + Ik'n *
+
-
* u - uIILP(RN) 11(p1.-
IILP(RN)
= 114uXK.. - UIILP(RN) + n * u -UII
IILP(RN)
}
where we have used (C.9). It now follows from the Lebesgue dominated convergence theorem that II UXK - UIILP(RN) -' 0
as n -} oo, while
IV2*U-UIILP(RN) by Theorem C.19. This completes the proof.
C.5. Differentiable Functions on Arbitrary Sets Let 51 C RN be an open set and let m E No. While the definitions of the spaces Cm (SI) and C00 (C') are standard, in the literature (see, e.g., (7], [24], 153], [54], [182]) there are different definitions of the spaces Cm A and C°° (Th, and unfortunately these definitions do not coincide. Here, we follow the point of view of Whitney [176].
Definition C.24. Let E C RN be an arbitrary set and let u : E -t R. Given m E No, we say that u is of class C" in E and we write u E C"' (E) if for every multi-index a, with 0 < lal < m, there exists a function ua : E -> R satisfying the following property:
For every multi-index a, with 0 <_ lal < m, for every x0 E E, and for every e > 0, there exists 6 > 0 such that (C.20)
IRa (x; y) 1 <- a Ix
-
yI'"-1C'1
for all x, y E B (xo, 5) fl E, where the function Ra : E x E -i R is defined by (C.21)
Ra (x; y)
uO (J)
ua (x) -
QI
(x
y)0
$ multi-index, 1RI<m-IaI
for all x,yEEandwhere tiou. We say that u is of class C°° in E and we write u E C°O (E) if u is of class Cm in E for every m E N0.
C.5. Differentiable Functions on Arbitrary Sets
561
Several considerations are in order. If m = 0. then Ro (x; y) = u (x) u (y) for all x, y E E. Therefore, the previous definition reduces to the definition of the continuity of u. More generally, if m E No, then all the functions %, given in the previous definition are continuous (why?). Hence,
if u E Cm (E), then u E C' (E) for all I = 0,... , m (why?). If xo E E is an isolated point, then (C.20) is automatically satisfied, no matter how the functions ua are defined at xo. If xo is an interior point of E and a is a multi-index with Ian < m, let ei be an element of the canonical basis of RN. Note that ei can be regarded as a multi-index. Taking x = xo + hei, where h E R is so small that x E E, and y = xo in (C.21) gives ua (xo + hei) = ua (x0) +
h + O (h)
,
where we have used (C.20). Letting h -, 0, it follows that 0 (xo) exists and equals u4+e, (xo). Thus, by an induction argument, we get that a (xo) _ uF (xo) for every multi-index /3 with 0 < 1,31 < m, and so (C.20), with a = 0 and y = xo, gives !3
U(X)
=
0 M1
K, 10KI .
a-U
(xo)(x-xo)'+Ro(x;xo),
where Ro (x; xo) = o (Ix - xoI') as x -' xo, which is Taylor's formula of order m at xo. Since all the functions ug are continuous, this shows that u is of class C'n in E° in the classical sense. On the other hand, if u is of class C"' in the classical sense in some open set U that contains E, then by Taylor's formula (applied to u and to all its partial derivatives of order less than or equal to m), we get that u is of class Ctm in E in the sense of Definition C.24. In particular, for open sets E the previous definition coincides with the classical one. We now restrict our attention to closed sets. By what we just remarked,
if E C RN is closed and if there exists an open set U that contains E such
that u e Cm (U), then u E C' (E). Whitney's theorem [176], which we present here without proof, proves that the opposite is also true, namely, that a function of class Cm in a closed set can be extended to a function of class Cm in RN.
Theorem C.25 (Whitney). Let C C RN be a closed set, let u E Cm (C), m E No U {oo}, and let ua : C -p R be functions given as in Definition C.24
for every multi-index a with 0 <- Ial <- m. Then there exists a function v E Cm' (RN) (in the classical sense) such that v = it in C, 8 = ua, in C for every mufti-index or, with 0 < 1a1 < m, and v is analytic in RN \ C.
C. The Lebesgue and Hausdorff Measures
562
Since the functions uQ in Definition C.24 are not uniquely defined (except in the interior E°), the previous theorem gives a different extension of u for every family of functions {uQ}a. The long-standing problem of determining whether a function u : C -t R
defined on a compact set C C RN can be extended to a function of class Cm (RN) has recently been solved by C. Fefferman (see [59]).
In view of the previous theorem, if f2 C RN is an open set and m E NoU{oo }, then one could define the space C"` (SZ) as the space of all functions
u c- C'" (i2) that can be extended to Cm (RN). Theorem C.25 is based on a decomposition of the open set RN \ C into a particular family of cubes. We present the proof here, since it turned out to be quite useful in extension theorems (see Exercises 12.5 and 12.6; see also the paper of Jones [92] and the monograph of Stein [160]).
Theorem C.26. Given an open set 1 C RN, there exists a countable family rn) In of closed cubes such that _{ !Q(xn,
!
(1) f1 = Un Q (xn, rn),
(ii) Q (xn, rn) n Q (xm, rm) _ O for all n# n, (iii) V iYr <_ dist (Q (xn, rn), 00) <_ 4y 1vr a,l (iv) if Q (xn, rn) and Q (a'm,
touch, then 4rn < rm < 4rn,
(v) for every fixed cube Q (x., rn) in F there are at most (12)N cubes in F that touch Q (xn, rn), (vi) for even fixed 0 < e < . and for every x E 11 there exist at most (12)N cubes Q (xn, (1 + e) rn) that contain x.
Proof. Let go := {Q(x , 1) : x E zN}. The family co leads to a family {JCk}kEz of collections of cubes with the property that each cube in the family G,k. gives rise to 2N cubes in the family j9k+1 by bisecting the sides.
The cubes in the family 9k have side length 1 and, in turn, diameter -2;For k E Z define Slk :=
x E 11 : 21 < dist (x, 8f2) <
Then f2 = U Stk.
hez
1Note that V 1rr.. = diamQ(xn,r.,).
LN 2
.
C.S. Differentiable Functions on Arbitrary Sets
563
Consider the family
Fo:=U{QEck:Qnnk00}. kEZ
Since each family Gk is a partition of RN, we have that
nCUQ.
(C.22)
QE.Fo
Next we claim that for every Q E .F'o,
diamQ < dist (Q, 81) < 4 diam
(C.23)
Indeed, if Q E Fo, then E Gk for some kE Z, so that diamQ 2 Moreover, since Q n ilk # 0, there exists x E Q n Stk. Thus, also by the definition of ilk,
dist (Q, Oil) < dist (x, On) < 4 2k
On the other hand, dist (Q, On) > dist (x, Oil) - diamQ >
2k-i
- 2k - 2k
,
which proves (C.23).
Note that the first inequality in (C.23) shows, in particular, that every Q E Fo is contained in il, so that also by (C.22),
SZ= U iUEfo
Thus, properties (i) and (iii) are satisfied. To obtain (ii), we construct an appropriate subfamily of F0. We begin by observing that if Q1 E ck, and Q2 E 9k2 intersect, with k1 < k2i then, necessarily, Ql D Q2.
Start from any cube Q E Jco and consider the maximal cube Q' (with respect to inclusion) in Fo that contains Q. Such a cube exists, since, by (C.23), the diameter of any cube in F'o containing Q cannot exceed 4 diam Q. Note that by the previous observation there is only one such maximal cube
Q'. Let F be the subfamily of maximal cubes in .F'o. Then (i)-(iii) hold. To prove (iv), assume that Q1, Q2 E F touch. Then diam Q2 < diet (Q2, Oil) < dist (Q1, ail) + diam Q1 < 5 diam Ql, where we have used (C.23). On the other hand, since diam Q2 = 2k diam Q1 for some integer k E Z, then, necessarily, diam Q2 < 4 diam Q1. By reversing the roles of Q1 and Q2, we obtain (iv).
Next we show that (v) holds. Fix a cube Q E F and let kE Z be such that Q E ck. In the family Qk there are only 3N - 1 cubes that touch Q.
C. The Lebesgue and Hausdorff Measures
564
Each cube in 9k can contain at most 4N cubes of F with diameter greater than or equal to diam Q. Hence (v) follows from (iv). .1
Finally, we prove (vi). For every x E Il let Q E F be such that x E intersects Q E F, then Q (x,,, (1 + e) We claim that if Q (x,,, only if Q (x,,, r,,) touches Q. Indeed, consider the union of all the cubes in F that touch Q. By (iv), the diameter of each of these cubes is greater than or equal to 16 diam Q. Since 0 < e < a, the union of these cubes contains Q (x,,, (1a+ e) r,.). By the maximality of the family F, it follows must be one of these cubes and the claim is proved. Property that Q (x,,, (vi) now follows from (v).
The family .F is called a Whitney decomposition of Q.
C.6. Maximal Functions In this section we introduce the notion of maximal function and study its properties. Throughout this section we consider LP spaces in the case that the underlying measure is the Lebesgue measure.
Definition C.27. Let u e Li (RN). The (Hardy-Litttewood) maximal C
function of u is defined by M (u) (x) :=
sup JB(xr) Iu (y) I dy IB (x, r) I >p
for all xERN .
Exercise C.28. Prove that the sets {x E RN : M (u) (x) > t},
{x E RN : MR (u) (x) > t}
are open (and so Lebesgue measurable).
Theorem C.29. Let u E LP (RN), 1 < p < oo. Then (i) M (u) (x) < oo for LN-a.e. x E RN, (ii) if p = 1, then for any t > 0, (C.24)
I {x E RN : M (u) (x) > t}j <
3N IN 1 u(x)I dx, t
(iii) if 1 < p:5 oo, then M (u) E LP (RN) and IIM (U)IILP <_ C (N,p) IIUIILa .
Proof. We begin by proving (ii). Let
At := {xERN: M(u)(x)>t}
C.6. Maximal Functions
565
By the definition of M (u), for every x E At we can find a ball B (x, rx), with
rx > 0, such that (C.25)
1
IB (x, rx)I
J B(=,r=)
Iu (y)I dy > t.
Let K C At be a compact set. Since {B (x, rx)}ZEAt is an open cover for K, we may find a finite subcover. By Lemma 13.43 there exists a disjoint finite subfamily {B (xi, ri)}z 1, such that n
KC UB(xi,3r;). i=1
Hence, by (C.25), n
I KI < 3N E I B (x., r )I <
3t
n
2 fB(s.,r:) Jul dy < a-1
1=1
3
Jul dy. t JN
Using the inner regularity of the Lebesgue measure, together with Exercise C.28, we obtain that
{xERN: M(u)(x)>t}l <
3N
t
rNJul dy.
Note that this implies, in particular, that M (u) (x) < oo for LN-a.e. x E RN. Thus (i) is proved for p = 1. In order to prove (iii), it suffices to consider 1 < p < oo, since the case p = oo is immediate from the definition of M (u). For t > 0 define
ut (x) := {
u (x)
if Iu (x)I >,
0
otherwise.
We claim that ut E L' (RN). Indeed,
J
utdy = N
f
{ SERN:Iu(x)I>Z}
\p-1 <
\ 2t J/
I{xERN: Iu(x)I>a }
IuIp dy < oo.
Moreover, since Jul < jutI + 2, we have that M (u) < M (ut) +
{xERN': M(u)(x)>t}C(xERN: M(ut)(x)>
,
and so
C. The Lebesgue and Hausdorff Measures
566
Part (ii) applied to ut E L1 (RN) now yields (C.26)
<3Nt2JANIutldy r
I{xER'V: M(u)(x)>t}I
3N2
t IIXEDv: lu(X)I> s } lul dy. Hence, using Theorem B.61 and Tonelli's theorem, we obtain
dx =
IRN (M (u) x
f°°I{xER
(M (u) x
> sds
pfW
=
o-1 I {x E RN : M (u) (x) > t} I dt
Iu(J)I dydt 2Iu(y)I
= p3N2 f Iu (F/)I f =
p3N2
p-1
tp-2 dtdy
o
11N
IRN
Iu(x)IP dx.
This proves (iii) and in turn (i) for p > 1.
0
Exercise C.30. Let f,g E L1 (RN). Assume that
Ig(x)I: h(x) for GN-a.e. x E RN, where h E L' (Rg) has the form h(x) = hl (IxI) for some nonnegative, decreasing function hl. Prove that fRN f (x) g (x - 81) dy <- IIhIILI(mN) M (f) (x)
for GN-a.e. x E RN. Hint: Consider first the case in which hl = X1o,,.j for some r > 0. Using the previous exercise and Theorem C.29, we can prove the following result.
Proposition C.31. Let 0 < a < N, 1 < p < QN. Then for every u E LP (RN),
f where
q
u
N
Ix
I
dy d.<- C (a,p, N) IIUIILP(RN) , YIN-
pN
' N - ap.
C.6. Maximal Functions
567
Proof. Define
if0
hl (t) :=
otherwise.
0
Then by the previous exercise, for every R > 0 and for LN-a.e. X E RN we have that Jlu (Y)I N-«
dy < CRa M (u) (x),
B(ZIP) Ix - YI
and so U (y) !RN Ix - YIN-«
dy <
f
Iu(y)I JB(x,R) Ix _YIN -« dy lu (y)I .' f + J \B(x,R) Ix -
YIN-Q
dy.
(fR"
\B(x,R) Ix - YI
(N- «)
dY
= CRa M (u) (x) + C IIUIILP(RJV) R«- P,
where we have used Holder's inequality. Taking
R
IIuIILP(RA)
.__
M(u)(x)+E
"
where e > 0, we get U (y)
RN Ix -
YIN-«
dV
< C IIuIIL'(jtN) (M (u) (x) + E)1
N
for GN-a.e. x E RN. Letting E -+ 0+ and taking the norm in L (RN) on both sides yields i
q
(fax
u (y) d uRN Ix y1N-«
-
1
4
dx
C IIHII
(RN
JAN
(M (u) (x))'" dx
Q
Exercise C.32. What happens if p = 1 or p = «N?
0
C. The Lebesgue and Hausdorff Measures
568
C.T. Anisotropic LP Spaces For p = (p1, ... , pN), 1 < pi < oo, i = 1, ... , N, we denote by LP (RN) the space of all measurable functions u : RN -+ R such that IIuIILp < oo, where
1uIILp I
PX
II
JR
(JIlt
(jut lu (x)l' dxi) Pi dx2
...
dXN
If pi = oo for some i, then we can still define the space LP (RN) by replacing the integral (fa I-Ip' dxi) of with esssupxfER
in the definition of
IIuIILp
Remark C.33. Note that the order in which the Lpf-norms are taken is important. If E C RN is a Lebesgue measurable set, we define LP (E) as the space of all measurable functions u : E - R such that XEU E LP (RN), where we define u (x) to be zero outside E. For simplicity of notation in what follows, we will focus on the space LP (RN). Given p = ( p i , . . . , PN)11 < pi < oo, i = I,-, N, the Holder conjugate exponent of p is the vector p', defined by where p'i is the Holder conjugate exponent of pi, i = 1,.. . , N.
Theorem C.34 (Holder's inequality). Let p = (pl, ... , pN), 1 < pi < 00, i = 1'... , N, and let p' be its Holder conjugate exponent. If u, v : RN -: R are measurable functions, then (0.27)
IIUVIILI
< IIuIILp IIuIILp,
In particular, if u E LP (RN) and v G= LP( RN), then uv E L' (RN).
Proof. By Tonelli's theorem IIuVIILI =
fe...
f fR
Iu(x)v(x)I dxldx2... dXN.
The result now follows by successive applications of Holder's inequality with 0 respect to each variable xi separately, starting from i = 1.
Exercise C.35. Let p = (pl,... , pN), 1 < pi < oo, i = 1,...,N, let u : RN " R be a measurable function, and let {E } be an increasing sequence of bounded Lebesgue measurable sets, with
C.7. Anisotropic LP Spaces
569
(i) Prove that "M IIXE,uIILP f2-400
= IIuIILP
Prove that if u E LP (RN) and 1 < pi < oo, i = 1, ... , N, then urn IIXEnu - UIILP = 0.
(iii) Does part (ii) hold if some of the exponents pi are infinite? (iv) Define u : RN R by u>z (x)
:_
a (x) if Iu (x)I < n and x E E,,, otherwise. 0
Prove that
"M IIunhILP = IIuIILP
Corollary C.36. Let p = (P1,. .. , PN), 1 < pi < oo, i = 1, ... , N, and let p' be its Holder conjugate exponent. ff u : RN -> R is a measurable function, then (C.28)
IIIILP = sup I
!RN
IuvI dx : v E &(R"), IIt'IILp < 1}
Proof. If u = 0, then both sides of (C.28) are zero, and so there is nothing to prove. Hence, without loss of generality, we may suppose that IIuIILP > 0. By Holder's inequality, fR
V
Iuvl dx :5 IIVIILP, IIUIILP :5 IIuIILP
for all v E LI' (RN) with I I V I I LP' < 1, and so
M := sup {LN IuvI d : v E L" (RN) , IIILP, !5
I
IIuIILP
To prove the reverse inequality, it suffices to consider the case that M < 00. Let 0 < m < N be the number of indexes pi that are infinite. Assume by contradiction that IIuIILP > M. Then there exists s > 0 so small that IIuIILP > M (1 + e)rn .
FornENdefine u,, : R-"' --,R by
u (x)
u(x) if Iu(x)I
otherwise.
By the previous exercise we may find an integer n E N such that (C.29)
IIUnIILP > M (1 + 07z .
C. The Lebesgue and Hausdorff Measures
570
Set R = (-n, n)N. For every 1 < i < N write R'-1 xRx x = (&/j, xi, zi') E Pi = (PI)...,Pi-1),
RN-',
(pN-ii.. ,pN),
P4' _
and
R=R;x(-n,n) xki'CR[ -'
xRxRN-i
Note that p = (p;, pi, The cases i = 1 and i = N are simpler. If 1 < pi < oO, define vi : R x RN-i _, R as 11U. (', xi, z')IILP:(Ri-1)
vi (xi, Zi") :=
m -1
,
RN-i
(xi, 4) E R X
,
11
if I1 u
# 0, and vi := 0 otherwise. For i = 1 we would
z,') II L(;,i"") (R`)
take instead
Iu'n(x1,zi)I
vi (xl, z)
P1-1 ,
II
(xl, zl) E R X
RN-1,
IILP1(R))
If pi = oo, let Ei,E :=
1
(xi, 41) E R X
PtN-i
II''nz,')II
(1+e) IMu
and consider the section {xi E R : (xi, z,') E Ei,,t }
(E:,e)z Define vi : R x RN-i
l[2 as 1
vi (xi, z,') :=
G1
XE,,, (xi, zz')
otherwise.
0
One should replace I U"' definition of the set E1,E. Finally, set
if L1 ((Ei,E),1) , 0,
with I un (x 1, zi ) I for i = 1 in the
xi, xs') II Lp:
N
V:_Evi. i=1
We leave it as an exercise to check that v is zero outside a compact set, v r= Li' (RN) n L00 (RN), IIVIILp, = 1, and
fgN IId
(1+C) "+'IIfhILP>M,
C.7. Anisotropic LP Spaces
571
where in the last inequality we have used (C.29). On the other hand,
M>
j
IuvIdx> Jilt"
and so we have reached the desired contradiction.
0
Theorem C.3T (Minkowski's inequality). Let p = (PI, ... ,PN), 1 < pi < 00, i = 1, ... , N, and let u, v : RN - R be measurable functions. Then, IIU + uIILP C II'IILP + IIvIILP
In particular, if u, v E LP (W'), then u + v E LP (RN). Proof. The result now follows by successive applications of Minkowski's inequality with respect to each variable x;, starting from i = 1. 0 By identifying functions with their equivalence classes [u], it follows from Minkowski's inequality that is a norm on LP (RN).
Theorem C.38. Let p = (pi, ... , pN), 1 C p= < oo, i = 1, ... , N. Then LP (RN) is a Banach space.
Proof. Let {u,,} c LP (RN) be a Cauchy sequence; that is, (C.30)
hm m,n->00 IIU
-4
IILp
0.
By Corollary C.36 for every 1, in, n E N, IIUm - U,IILP =
sup<1
JI IIVIILP1t 1
IIXB(0,1) IILP'
I(Um - un) VI dx
J
I(um - un)I dx. (O,1)
Thus, for every fixed I E N, letting in, n -p oo in the previous inequality and using (C.30) shows that {XB(o,1)u.,b} is a Cauchy sequence in L1 (RN). Thus, there exists a function u(i) E L1 (RN) such that XB(o,t)un u(t) in L1 (RN).
By the uniqueness of limits, we have that 0+1) (x) = u(i) (x) for GN-a.e. x r= B (0,1). Hence, we may define a measurable function u as follows. For every x E RN let l be so large that x E B (0, l) and set u (x) := u(t) (x). Using a diagonal argument, we may find a subsequence {u,,,' } of {un} (x) - u (x) for LN-a.e. x E RN. such that Fix a> 0 and find n E N such that Ilum - U-IILp < 6
C. The Lebesgue and Hausdorff Measures
572
for all m, n E N with m, n > n. Let v E LP' (RN), with IIVIILP, < 1, and let m, k E N be such that m, uk > rt. Then by Corollary C.36, NI(U,, - unk)vl dx <_ 11U. -'LRkIILP
E-
Letting k -+ oo and using Fatou's lemma gives
f
N
I(um - u)vI dx
for all m E N such that m > r and all v E LP( R
,
with II V II LP,
< 1.
Hence, by Corollary C.36 once more, we have that ZIILP <_ S
for all m E N such that m > fi. Since V.,,, E LP (RN), it follows by Minkowski's inequality that it e LP (RN). The previous inequality implies D that -u.,,, -> it in LP (RN). This concludes the proof.
C.8. Hausdorff Measures In this section we introduce the Hausdorff measure in RN. Loosely speaking, the Hausdorff measure is a measure that is adapted to measure sets of lower
dimensions in RN , say, a curve in the plane or a surface in R3. It is also used to measure fractals. For 0 <,9 < oo define
r(2+1 where r (t) is the Euler Gamma function
r (t) := fo O0 e-xxt-ldx,
0< t< oo.
Note that r(n)= (ii-1)!forall nEN. We remark that when N E N, then aN is the Lebesgue outer measure of the unit ball in RN, so that £N (B (x, r)) = aNrN for every ball B (x, r) C RN
For 0 < 6:5 oo consider the family of elementary sets
co:= {FCRN:diamF
P8(F):=as
/diamF\8 2 J
C.S. Hausdorff Measures
573
For each set E C RN we define °O
(C.31) 7{a (E) := inf
ad
'
diam E 2
fn=1
8
ao
: E C U E., dials E. < b
,
n=1
where, when s = 0, we sum only over those En # 0.
Exercise C.39. Prove that in the definition (C.31) it is possible to restrict the class of admissible sets in the covers {En.} to closed and convex sets (open and convex, respectively) and that the condition diam En < 8 can be replaced by diam En < S, without changing the value of Ws' (E). By Proposition B.3, 1-la is an outer measure. We next define the Hausdorff outer measure. Since for each set E C R-,V the function b'--t fa (E) is decreasing, there exists
N' (E) := ali m ft- (E)
(C.32)
(E) . = sup 60 xa
7{o is called the s-dimensional Hausdorff outer measure of E.
It follows from (C.31) and (C.32) that for all E C RN, x E RN, and t > 0, we have
H (x + E) = xo (E), 7fo (tE) = en.- (E) Another useful property of 1 as an exercise.
is the following result whose proof is left
Proposition C.40. Let E C RN and let $ : E - Rm be a Lipschitz function. Prove that for all e > 0,2 7'10 (W (E)) < (Lip
(E)
.
Let's prove that 7-lo is actually an outer measure.
Proposition C.41. Let 0:5 s < oo. Then fo is an outer measure. Proof. We prove only countable subadditivity. Let {En} C IIBN. Since fa is an outer measure, we have that 00
We
00
xa(En) <EN-a(E)
U En n=1
00
n=1
n=1
where in the last inequality we have used (C.32). Letting 6 -> 0+ and using (C.32) once more gives the desired inequality. O 2Hence, on the right-hand side we have the Hausdorff outer measure in 111m and on the left-hand side we have the Hausdorff outer measure in 110. We use the same notation.
C. The Lebesgue and Hausdorff Measures
574
By Caratheodory's theorem, f; restricted to the a-algebra of all l0 measurable subsets of RN is a complete measure denoted 7{9 and called the s-dimensional Hausdorff measure. Using Proposition B.19 we have
Proposition C.42. For 0 < a < oo the outer measure W8 is a metric outer measure, so that every Borel subset of RN is W,' -measurable.
Proof. Let E, F C RN, with dist (E, F) > 0, and fix 0 < S < dist (E, F). Consider any sequence {En} C RN such that E U F C U0° 1 E7z and diamEn < S for all n E N. Discard all the sets E. that do not intersect E U F. Since diam E < S < dist (E, F), each remaining set E. intersects either E or F (but not both). Thus
E diamEn' = n=11
2
J
a diamEn
E as (diamEn)s+ E. nET9
2
2
E,
xa°(E)+'Ca(F). Taking the infimum over all admissible sequences {En} yields ?V' 6 (E U F) > 7( 6(E) + 1-C6 (F)
.
To conclude, it suffices to let S - 0+ and to use (C.32).
O
Remark C.43. We remark that unlike W,",, 7-f6" is not a metric outer measure for 0 < a < N. As an example, consider the outer measure 7H6, given by 00
1 : E C U En, diam En < S
(E) := inf E"'00
E C RN.
n=1
If we take E1 = {x} and E2 = {y}, where 0 < Ix - yi < b, then to cover E1UE2 it is enough to consider one set, so Na (El U E2) = 1, while HO, (El) _ 1, NO (E2) = 1.
6(ElUE2) = 1 <2 = Ra(El)+(E2). M6 We now study the dependence of %o on the parameter a.
Theorem C.44. Let 0 <_ s < oo. Then (i) 1Co is the counting measure, (ii) x0N = IV,
(iii) -H,', 0 ifs > N. Lemma C.45. The measure WN is absolutely continuous with respect to r-N.
C.B. Hausdorff Measures
575
Proof. Fix b > 0. Given a cube Q (x, r), we have that aN
dram Q (x,Nr)aN
(V--r)N =: CNrN
2
2
and so for each set E C RN, by (C.31) and Exercise C.1, 00
00
7j6 (E) := inf E 'IN ( diem En) N : E C U En, diam En < b fn=1
J
l\
E CI(
< inf 1 00
n=1
diam QZXn, rnT:
00
E C U Q(xn, rn), r < 6
n=1
n=1
= CNG0 (E). Thus, if Go (E) = 0, then
(E) = 0 for all b > 0. Hence, fl (E) = 0.
This implies that 71N is absolutely continuous with respect to GN.
0
We turn to the proof of the theorem.
Proof. (i) Since ao = 1, we have 00
7{0(E) := inf E 1: E C U En, diamEn < b
.
n=1
EnOO
If y := card E is finite and S is sufficiently small, then we need at least
y sets En to cover E. Thus 71a (E) = card E. On the other hand, if card E is infinite, let En be a subset of E with exactly n elements. Then
(ii) Fix 5 > 0 and consider any sequence {En} C RN such that E C UO_1 En and diam En < 8 for all n E N. Since diam En = diam En, without
loss of generality, we may assume that the sets En are closed and thus Lebesgue measurable.
Using the monotonicity and subadditivity of Go together with the isodiametric inequality, we have 00
G0 (E) < L. U En < n=1
00
00
aN
GN (En) < n=1
n=1
( diam2 En ) N J \
Taking the infimum over all admissible sequences {En} gives (C.33)
Go (E) < 7{0 (E).
C. The Lebesgue and Hausdorff Measures
576
To prove the other inequality, fix e > 0 and 6 > 0 and by Exercise C.1 find a sequence of cubes {Q (x., rn)} with diameter less than S such that 00
EGN(Q(xn,rn))
For every n E N let F. be the family of all closed balls contained in Q (x,a, rn). By the Vitali-Belsicovitch covering theorem there exists a countable family {B (zr, r{'L)) } C .Fn of disjoint closed balls such that GN
rin)
Q (xn, r) \ U B
= 0.
By the previous lemma,
'Hd
Q(xn,rn),UB(x(n)rr(n)1
J
i
xN Q (xn, rn) , U B (x(n), r(.))
=
_ 0,
i
and so, since Na is an outer measure,
xN (E) <
00
00
W6
(Q (xn, rn)) = E Wg U B n=1
nn[=1 00
RN (B xin) L L rin)} n=1 i 5
00
n=1
cN (B (x)rr)))
(xr)) 1n)
[[i-
[L L aNri n=1 i 00
00
L N (Q (xn, rn))
,-o (E) + e.
n=1
i
Letting 6 -> 0+ and then e -* 0+ gives the desired result. IN into MN cubes of side length (iii) Subdivide the unit cube Q . Let 6 :_ . Then m and diameter
(Q) < M a8 -1 n=1
2m
= aA
2
1
7no-N
--+0
as m - oo, and so %o (Q) = 0 in view of (C.32). Since RN can be written
as a countable union of unit cubes, we have that to (RN) = 0, and so 1
- 0.
Remark C.46. In view of Theorem C.44, it follows that the Hausdorff outer measures fo are of interest only when 0 < a < N.
C.S. Hausdorff Measures
577
Exercise C.47. Give a direct, simple proof of the fact that f00 = Ga. Hint=1. f o r a l l
0.
Proposition C.48. Let E C RN and let 0 < s < t < oo. (i) If H' (E) = 0 for some 0 < 6:5 oo, then fl (E) = 0. (ii) If 1{0 (E) < oo, then V,, (E) = 0. (iii) If No (E) > 0, then ?{o (E) = oo.
Proof. (i) For s = 0 there is nothing to prove, so assume a > 0. Since fa (E) = 0, for every e > 0 we may find a sequence {En} C RN such that
ECU, 1E,,,diamE.<5forallnEN,and 00
a8
(diamEn)s
<
c=1 1/a
It follows that diamE, < 2 l a I =: S! and so 11 (E) < e. Since St -, 0 as a -4 0+, it follows from (C.32) that If" (E) = 0. (ii) Assume that f. (E) < oo, fix S > 0, and find a sequence {E,,} C R" such that E C U°O_1 En, diamE < S for all n E N, and as
(diamEfl)3 <x(E)+ 1 <1 (E)+1.
Then
x (E) <
at
(diamE,,lt < at (o\ t -6t.a8 (diamEn 2
n_1
<
(S\ I
t
ag
2
n_1
2
8
(No (E) + 1) .
Letting S -> 0, it follows from (C.32) that 11. (E) = 0.
0
As a consequence of the previous proposition we have the following result.
Proposition C.49. Let 0 < s < N. Then the Hausdorff measure N" is not u-finite.
Proof. Assume by contradiction that for some 0 < a < N there exists a sequence
of ?{o-measurable sets such that 00
R = UEn. n=1
C. The Lebesgue and Hausdoiff Measures
578
and if (En) < oo for all n E N. By Proposition C.48(ii) and the fact that 8 < N, we have that
% (En)=0. Hence, by Theorem C.44 (see (C.33)), Go since 00 (RN) = C0 U E. 00 = C0
0, which is a contradiction, 00
< >2'Co (En) = 0. n=i
m_1
In view of Proposition C.48 the following definition makes sense.
Definition C.50. The Hausdorff dimension of a set E C RN is defined by
dimn(E):=inf{0<_s
Exercise C.51. Let f : RN-1 -p R be a Lipschitz function. Prove that the graph of f,
Grf
{(x', f (x')) : x' E RN-'} C RN has Hausdorfi dimension N - 1 and that Vv-1(Gr f)
= surface area of Grf = J
N-
1 V+ IVf (x')12 dx'.
Prove that for every Borel set B' C RN-1,
xN-i ({(x',f (x')) : x' E B'}) = 15'
1+Ivf (x') dx'.
Throughout this text we use surface integrals over the boundary O) of sufficiently regular open sets 12 C RN. We begin with the special case
Q:=
(C.34)
{(X1
)
xN) E RN-' X R : XN > f (2)} ,
where f : RN-1 --+ R is a Lipschitz function. In view of the previous exercise, for every measurable function u : On -+ [0, oo] we have that the surface integral of u over 00 is given by (C.35)
J
N-1
u
=J
N-1
1
u (x', f (x'))
1+ IV&f (x')12 &l.
If the boundary of the open set SZ C RN is now locally Lipschitz, for every Xo E 8) there exist a neighborhood Ax0 of xo, local coordinates
C.S. Hausdorff Measures
579
y = (y', IN) E ][8N-1 x R, with y = 0 at x = x0, a Lipschitz function fxo : RN-1 - R, and r > 0 such that t) n A0 (y', yN) E Sl n Ax0 : y' E QN-1 (0, r) a yN > fxo (t') } .
Since
M C U Ax, xE&-k
we may construct a smooth partition of unity {1ji} subordinated to the family {Ax}XEBSt. For every measurable function u : &I -+ R we have that
for each n the function u, has support contained in some A.., and so by (C.35) we have that fan uzfin QN-1(0,r)
a%N-1
'u (Y', f.- (y')) *T+ (l/, f'. (g'))
Since En z1' = 1 and
(C.36)
is locally finite, we have that U
foS!
1 + I V fso (y') I2 dy'
r
dHN-1 =
,b
dhN-1
0St
Remark C.52. Without using the Hausdorff measure VN-1, one could define the surface integral as in (C.35) when fl has the form (C.34) and as in (C.36) when SZ is locally Lipschitz. In this case one should prove that the previous definition of surface integral is independent of the particular partition of unity. This approach is taken up in the book of Fleming [621.
Exercise C.53. Let M C RN be a k-dimensional manifold of class C1,
1:k
(i) Let rp be a local chart; that is, W : A -> M is a function of class C' for some open set A C Rk such that V p has maximum rank k where . is the inner product in RN. in A. Define gz? Prove that cp (A) has Hausdorff dimension k and that
1k (cp (A)) = surface area of the manifold
Appendix D
Notes Deciphering Academnese, I. "To the best of the author's knowledge" = "We were too lazy to do a real literature search. " -Jorge Chain, www.phdcomics.coni
Chapters 1, 2, and 3 draw upon the book of van Rooij and Schikhof (171].1
Chapter 1: The first proof of Lebesgue's differentiation theorem is due to Faure [561 (see also [21) for another proof). Lemma 1.24 (see 156)) is a slight modification of Riesz's rising sun lemma (see [141] and [171]). For more information and extensions on the Weierstrass function (see Theorem 1.12) we refer to the paper of Hardy [82) and the recent paper of Pinkus [138]. Exercises 1.18 and 1.19 and Proposition 1.20 are based on a paper of Krantz [101]. See also the paper of Stein and Zygmund 11611 and the monograph (160] for more information on the Zygmund space A1(R). We refer to the paper of Dovgoshey, Martio, Ryazanov, and Vuorinen (49] and the references contained therein for an extensive treatment of the Cantor function. For Exercise 1.44 see [11]. For Exercise 1.46 see [94]. Theorem 1.47 is due to Tak.4cs [164] (see also [146] and [89]), while for Exercise 1.51 see the paper of Freilich [67]. Chapter 2: Exercise 2.15 is due to Heuer [87]. For Exercise 2.22 see [17]. Theorem 2.26 is due to Katznelson and Stromberg (961. Exercise 2.29 is based on a paper of Gehring [72]. Theorem 2.31 and Exercise 2.33 are due to Josephy [93]. The proof of the Helly selection theorem (Theorem 2.35) follows [132]. For Exercises 2.41 and 2.42 see [5]. Corollary 2.43 is proved in [183].
'This is a really nice book that is not as well known as It should be. We highly recommend it.
581
D. Notes
582
Theorem 2.47 is due to Banach (14], while Theorem 2.46 and Lemma 2.48 are due to Federer [57]. Exercise 2.50 is based on a paper of Wiener [179].
Chapter 3: For Exercise 3.3 see [132]. The proofs of Theorem 3.12 and Lemmas 3.13 and 3.16 are adapted from papers of Goffman [75], Van Vleck [172], and Varberg [173]. For Exercises 3.20 and 3.21 see [132]. Exercise
3.28 is based on a paper of Lindner [109]. Exercise 3.34 is due to Botsko [22].
The author would like to thank J. Ma1y for useful conversations on the proof of Corollary 3.41. Exercise 3.43 is based on a paper of Gehring [72]. Theorem 3.44 was first proved by Krzyzewski [102] and then later independently by Serrin and Varberg (153] (see also [76] for some extensions). An alternative proof of Lemma 3.45 in the Sobolev setting due to A. Ancona may be found in [20]. Theorem 3.54 and all its corollaries were
proved by Serrin and Varberg [153]. For Exercise 3.53 see the paper of Marcus and Mizel [117].
For Theorem 3.65 see the monograph [11]. Theorem 3.68 is adapted from a paper of Josephy 1931. The proof of Corollary 3.74 uses some ideas of 1147). We refer also to the
papers of Morse [127] and of Kober [99] for more information on singular functions.
Chapter 4: The first two sections of Chapter 4 draw upon the review paper of Cesari [30]. Theorem 4.5 is due to Hilbert [90]. Our proof is due to Moore [125].The first example of this type was given by Peano in 1890 [136]. Theorem 4.15 is due to Almgren and Lieb [8]. The first part of the proof of Theorem 4.18 follows [77]. For Theorem 4.37 see [11].
Section 4.3 draws upon the books of Ambrosio and Tilli [11] and of Falconer [55]. Theorem 4.42 is due to Federer [57]. Theorem 4.54 is due to Marcus and Mizel [117] (see also [9], [106], and [126]).
Our proof of the Jordan curve theorem (Theorem 4.56) is due to Tverberg [169] (see also the paper of Maehara [111] for a different proof based on the Brouwer fixed point theorem).
Chapter 5: This chapter draws upon the book of Saks [145]. Theorem 5.33 is due to Cater [29]. Proposition 5.39 and Corollary 5.40 were proved by Hewitt [88] to which we refer for extensions to the case of unbounded intervals. For Theorem 5.42 see the paper of Winter [180].
D. Notes
583
Chapter 6: Theorem 6.15 and Exercise 6.17 follow [78]. For Theorem 6.18 see [108]. In the case in which %P is even, it was first proved independently
by Chiti [31] and by Crandall and Tartar [41]. We refer to the paper of Hajaiej (79] for extensions to the case in which is real-valued. Exercise 6.19 is based on [31]. Theorem 6.23 is due to Ryff [142], who studied the case in which E _ [0, 1]. The present extension and parts of the proof follow the papers [36] and [60]. Corollary 6.26 and Theorem 6.28 in this general form are also due to Ryff [142] (see the work of Duff [50] for the case in which u is assumed to be absolutely continuous). Exercise 6.29 is based on a paper of Cianchi [35]. We refer to the papers of Cianchi [35] and Dahlberg [43] for more information on the regularity of (u`)'. Corollary 6.30 is due to Novak [134]. For more information on the topics of this chapter and for an extensive bibliography we refer to the monographs of Kawohl [97] and Kesavan [98]; see also the book of Lieb and Loss [108].
Chapter 7 For more information on functions in BV (11) we refer to the monographs of Ambrosio, Fusco, and Pallara [10], Evans and Gariepy [54], and Ziemer [182]. The proof of Lemma 7.3 is adapted from [24]. For Exercise 7.5 see the lecture notes of Dal Maso [44]. For more information on Sobolev functions of one variable we refer to the monographs of Brezis [24] and Burenkov [28]. Exercise 7.16 is based on a paper of Helmberg [85]. Exercise 7.22 is due to Lu and Wheeden [110]. Proposition 7.23 was proved by Chua and Wheeden [33].
Chapter 8: Proposition 8.2 and Theorems 8.4 and 8.10 are due to Varberg [174]. Lemma 8.7 was proved by Flett [63]. Lemma 8.5 may be found in a paper of Schwartz (150]. In recent years there has been a renewed interest in absolutely continuous functions of several variables and the Lusin (N) property. In [113], Maly proposed an alternative definition. We refer to the papers of Csornyei [42] and of Hencl and Maljr [86] for more information and references on this topics. Theorem 8.17 is due to Lax [104] to which we refer for some historical background and references on the Brouwer fixed point theorem [26]. Theorem 8.21 is due to Varberg [175] (see also Rudin [143]). We refer to the paper of Hajlasz [80] for significant improvements.
Chapter 9: The material from this chapter is based on [52], [144], and [181]. For Exercise 9.4 see the lecture notes of Acquistapace [3].
584
D. Notes
Chapter 10: For more information on Sobolev functions we refer to the monographs [7J, [281, [541, [120], and [1821.
Exercise 10.11 is based on a paper of Simader 1155]. Theorem 10.15 is due to Meyers and Serrin [1221. Theorem 10.24 and Exercise 10.26 follow a paper of Fraenkel 166] to which we refer for more information about different definitions of regular domains (see also the monograph of Delfour and Zoldsio [47]). Exercise 10.39 is due to Kolsrud [100]. For the proof of Theorem 10.33 see Hilden 1911.
The first step of the proof of Theorem 10.55 is due to Stein (see the paper of Brezis [25] ). Theorem 10.55 is quite important for two different reasons. This characterization is often used to prove higher regularity of solutions of partial differential equations (see, e.g., [241, [531, and 1731}. It is also important because it allows us to give a definition of Sobolev spaces that does not involve derivatives. Such characterizations have been studied extensively in recent years, since they can be used to define Sobolev spaces
on metric spaces. We refer to the books of Ambrosio and Tilli [111 and Hajlasz and Koskela 1811 and to the recent survey paper of Heinonen [84] for more details on this subject. Chapter 11: Step 4 of the proof of the Sobolev-Gagliardo-Nirenberg embedding theorem follows the paper of Maggi and Villani [112]. For a proof of the results mentioned in Remark 11.5 we refer to the papers of Aubin [13] and Talenti [165]. For sharp forms of the Sobolev-Gagliardo-Nirenberg inequality we refer to the recent work of Cianchi, Fusco, Maggi, and Pratelli [38] and to the references contained therein. Exercises 11.7 (i) and (ii) and 11.16 are based on 166]. For Exercise 11.7(iii) see [121.
The proof of the Rellich-Kondrachov theorem is adapted from a paper of Serrin [152]. For Exercise 11.15 see (1521. Exercise 11.19 and the first part of Exercise 11.20 are based on a paper of Strauss [163], while the second part of Exercise 11.20 is based on a paper of Ebihara and Schonbek [51]. The author would like to thank Bill Hrusa for suggesting the reference [163]. Theorem 11.29 and Exercise 11.26 are due to Adachi and Tanaka [4]. The simplified proof of Theorem 11.29 is due to Francesco Maggi. To the author's knowledge the first result of this kind in bounded domains is due to Trudinger [168] (see also the work of Moser [130] and the recent paper of Li and Ruf [107] for some recent results). Corollary 11.36 and Exercise 11.40 follow a paper of Serrin [1511. Corollary 11.41 follows a paper of Marcus and Mizel [1161. For Exercise 11.45 see [166].
D. Notes
585
In [1151 Malt and Martio have constructed a continuous transformation
T E W 1,N (RN; RN) that does not satisfy the (N) property and whose Jacobian is zero Vv-a.e. in RN. Theorem 11.49 is due to Rademacher [140]. The proof presented here is follows the lecture notes of Lang [103). Theorem 11.50 is due to Stepanoff [162]. The proof presented here is due to Malt [114]. Chapter 12: A necessary and sufficient condition on the regularity of Oil for the existence of an extension operator is still missing except in the twodimensional case (see the paper of Jones [92]). For Exercises 12.1 and 12.2 see the book of Maz'ja [120]. Exercises 12.5, 12.6, 12.7, 12.12, and 12.14 and the proof of Theorem 12.15 follow the book of Stein 1160). Definition 12.10 is also from the book of Stein [160], but we added the condition that {l,a} is locally finite, since we were not able to show that conditions (i)-(iii) imply that is locally finite and we used this property in the proof of Theorem 12.15.
We refer to the book of Ziemer [182] for an extensive treatment of Poincare's inequality. Exercises 12.26 and 12.40, Proposition 12.29, and Theorem 12.36 are due to Smith and Stegenga [156]. The author would like to thank Bill Hrusa for suggesting Exercise 12.27. Theorem 12.30 follows a paper of Chua and Wheeden [34]. The idea of the proof comes from the paper of Bebendorf [16], who corrected a mistake in the original paper of Payne and Weinberger [135] in the case p = 2. See also the paper of Acosta and Duran [2] for the case p = 1. Chapter 13: As we already mentioned at the beginning of Chapter 13, what is covered here is just the tip of the iceberg. We refer the interested reader to the monographs [10], [54], 1581, 174], and [182) for more information on functions of bounded variation. For Exercises 13.5 and 13.8 see the book of Giusti [74]. Section 13.3 is based on a paper of Serrin [151]. Theorem 13.25 is due to Fleming and Rishel [61]. Theorem 13.42 is due to Sard [148], and the proof follows the book of Milnor [124].
Chapter 14: This chapter draws upon the books of Besov, Vin, and Nikol'skiT [18], [19].
Theorem 14.10 is due to Maz'ja and Shaposhnikova [121]. The present proof is based on that of Karadzhov, Milman, and Xiao [95]. Theorem 14.16 was first proved by Bourgain, Brezis, and Mironescu [23] and by Ponce [139]
(see also the paper of Brezis [25)). The present proof is based on that of Karadzhov, Milman, and Xiao [95].
586
D. Notes
Theorem 14.17 and Lemma 14.23 follow [19]. The proofs of Theorems 14.22, 14.29, and 14.32 are due to Solonnikov [158]. For Proposition 14.40 see [6].
Chapter 15: Theorems 15.6, 15.20, and 15.21 were first proved for bounded Lipschitz domains by Gagliardo in [69]. The proof of Theorem 15.20 is due to Solonnikov [157]. The proof of Theorem 15.21 is adapted from a paper of Uspenskii [170]. Exercise 15.27 is based on [66].
Chapter 16: Exercise 16.8 is based on [78]. Theorems 16.17 and 16.19 were proved independently by Hilden [91] and Talenti [165]. We follow here
the approach of Hilden, although the proofs are significantly simpler due to the results of Chiti [31] and Crandall and Tartar [41] on the continuity of decreasing rearrangement in L" and to recent work of Martin, Milman, and Pustylnik [119], to whom Proposition 16.21 and Step 1 of the proof of Theorem 16.19 are due. Theorem 16.18 has been significantly extended by Cianchi and Fusco [37]. For more information on the topics of this chapter and for an extensive bibliography we refer to the monographs of Kawohl [97] and Kesavan [98]; see also the book of Lieb and Loss [108].
Appendix A: This chapter draws upon [65]. Appendix B: This chapter draws upon [65], to which we refer for the proofs of all the results that cannot be found in classical texts (such as [48], [54], [64], [143]). Appendix C: The proof of the Brunn-Minkowski inequality (see Theorem C.7) follows a survey paper of Gardner [71], to which we refer for more information and an extensive bibliography. Exercise C.9 is based on a paper of Bebendorf [16]. Theorem C.26 follows the book of Stein [160]. Exercise C.30 is due to L. Tartar. The proof of Proposition C.31 follows a paper of Hedberg [83] but is presented here in a simplified form due L. Tartar. Section C.7 is based on the book of Besov, ll'in, and Nikol'skil [18].
Appendix E
Notation and List of Symbols Deciphering Academese, 11: "...remains an open question" _ '4we have no clue either."
-Jorge Cham, www.phdoomics.com
Since the number of letters and symbols (and the author's imagination) are limited, sometimes, and when there is no possibility of confusion, we use the same letter or symbol for different objects (not in the same theorem, we hope). For example, the letter C is used for constants, but also for closed sets, while the letter a is used for multi-indices, but also as a real number. A subscript on an equation number refers to that expression in the display. For example, given (E.1)
Expression 1,
Expression 2,
(E.1)2 refers to Expression 2 in (E.1).
Constants C, c: arbitrary constants that can change from line to line and that can be computed in terms of known quantities. Sets
X, Y: sets or spaces; card X: the cardinality of a set X; P (X): the family of all subsets of a set X ; F, G: family of sets or of functions. C: inclusion between two sets, with equality possible. X LY = (X \ Y) U (Y \ X): symmetric difference between the sets X and Y.
587
E. Notation and List of Symbols
588
Functional Analysis A, U: open sets; K: a compact set; B: a Borel set; C: a closed or convex set.
E, F, G: usually denote sets; 8E: boundary of E; E°: the interior of E; E: the closure of E. XE: characteristic function of the set E. T: topology; 11.11: norm.
dist: distance; diam: diameter. duality pairing. weak star convergence. C (X): space of all continuous functions; C,, (X): space of all continuous functions whose support is compact; Co (X): the closure of CC (X) in the sup norm. weak convergence;
Measure Theory 9N, 91: algebras or a-algebras; Wi ® 01: product o-algebra of 9R and
01 (not to be confused with 0 x 91); B (X): Borel a-algebra. p, v: (positive) finitely additive measures or (positive) measures;
v 1 it means that p, v are mutually singular measures; v << µ means that the measure v is absolutely continuous with respect to the measure p; dam: the Radon-Nikodym derivative of v with respect to A; p*: outer measure. A: a finitely additive signed measure or a signed measure; A+, A-, IAI: the upper, lower, and total variation measures of A. u, v, and w usually denote functions or variables. f, g, gyp, Eli, 0 usually denote functions; supp f : support of the function f ; Lip f : Lipschitz constant of the function f ; osc (f ; E): os-
cillation of f on E; f * g: convolution of the functions f and g. Mb (X; R): space of all signed finite Radon measures. ba (X, M, µ): space of all bounded finitely additive signed measures absolutely continuous with respect to A. LP (X, M, /4), LP (X, p), LP (X) are various notations for LP spaces; p': Holder conjugate exponent of p; IHILP(x), IIil[p' or 11.11p are various notations for the norm in LP.
Functions of One Variable N: the set of positive integers; No := N U {0}; Z: the integers; Q: the rational numbers; R: the real line; R:= [-oo, oo]: the extended
real line; if x E R, then x+ := max {x, 0}, x- := max {-x, 0}, fix) := x+ + x-; LxJ: the integer part of x.
E. Notation and List of Symbols
589
I: an interval; ®: the Cantor set. Go: the 1-dimensional Lebesgue outer measure; 41: the 1-dimensional Lebesgue measure. A1(R): the Zygmund space; BPV (I): the space of all functions of
bounded pointwise variation on I; AC (I): the space of all absolutely continuous functions defined on I. Given a function u, u+ (x) and u- (x) are the right and left limits of u at a point x, uj is the jump function of u, uc is the Cantor part of u, uAC is the absolutely continuous part of u, Var u is the pointwise variation of u, V (or is the indefinite pointwise variation
of u, PVar u and NVar u are the positive and negative pointwise variations of u, Vary u is the p-variation of u, essVar u is the essen-
tial variation of u, Nu E) is the Banach indicatrix (or counting function) of u over a set E, and D u (x) , D_u (x) , D+u (x), and D+u are the Dini's derivatives of u at x. µu: the Lebesgue-Stieltjes measure generated by an increasing function u; ) : the Lebesgue-Stieltjes signed measure generated by a function u with bounded pointwise variation. go,,.: the distribution function of a function u; u*: the decreasing rearrangement of u. L (-y): the length of a curve -y.
Functions of Several Variables RN: the N-dimensional Euclidean space, N > 1, for x E RN, with x = (x1, ... , xN)91 IxI :=
(xl )Z +
... + (x]1/)2.
Given x = (xl, ... , XN) E RN, for every i = 1, ... , N we denote by x; the (N - 1)-dimensional vector obtained by removing the ith component from x and with an abuse of notation we write (E.2)
x = (xa, xi) E RN-1 x R. When i = N, we will also use the simpler notation
(E.3)
x = (XI, XN) E
RN-1
xR
1When there is no possibility of confusion, we will also use xl,x2, etc, to denote different points of R"'. Thus, depending on the context, z; is either a point of RN or the ith coordinate of the point z E RN.
E. Notation and List of Symbols
590
and
lt+ _ {x = (x', XN) E
1RN-1
x R : XN > 01,
RN-i x R
: XN < 0} . = {x = (x', xN) E 6ij: the Kronecker delta; that is, b_j := 1 if i = j and 5iN := 0 RN
otherwise.
ei, i = 1, . . . , N,: the unit vectors of the standard (or canonical) orthonormal basis of RN. det: determinant of a matrix or a linear mapping. 1: an open set of RN (not necessarily bounded); v: the outward unit normal to BfZ; BN (x, r) (or simply B (xo, r)): open ball in RN of center xo and radius r; QN (xo, r) := x0 +
N
(- i,2 /\
(or simply Q (xo, r)); SN-1: unit sphere in RN. L : the N-dimensional Lebesgue outer measure; GN: the Ndimensional Lebesgue measure;
aN:=G^'(B(0,1)). Given a Lebesgue measurable set E C RN, I EI = GN (E). fo: the s-dimensional Hausdorff outer measure; H°: the s-dimensional Hausdorff measure.
Given E C RN and' : E - RM, dW (xo) is the differential of W at xo, Zcr- (xo) is the partial derivative of at xo with respect to (xo) is the directional derivative of ' at xo in the direction xi, v E SN-1, DAY (xo) is the gradient of it at xo, and J% (xo) is the Jacobian of %F at xo (for N = M). Cm (a): the space of all functions that are continuous together with their partial derivatives up to order m E 1'4o, C°°
00
(il) := fl Cm (n); M=0
Cr (Sl) and C°° (f2): the subspaces of C"' (fl) and C°° (fl), respectively, consisting of all functions with compact support. For a multi-index a = (al, ... , aN) E (No)N, c?"
8xa
_
8IaI
axi 1
... 8xN
Ial := al + ... + aN.
DK (f2): the set of all functions in CC° (fl) with support in the compact set K; D (fI): the space C°° (ft) endowed with a particular topology r; D' (C): the topological dual of D (C).
E. Notation and List of Symbols
591
In Chapter 9, T: a distribution (an element of D' (S2)); 55a: the delta Dirac with mass at xo; L" T : the derivative of the distribution T with respect to the multi-index a; suppT: the support of T; T * 0: the convolution of T with the function 0. W1"p (S2): a Sobolev space; the norm in W1,P (fl); for u E W1,P (fl), TXT : the weak (or distributional) partial derivative of u with respect to xi; Vu: the weak (or distributional) gradient of U.
Hl (Il): the Hilbert space W1.2 (Sl); Wkp (i): a higher-order Sobolev space; L1'P (S2): a Sobolev space; (W1,P (S2))': the topological dual of W 1,P (12); Wo'P (S2): the closure of C°° (11) in W' ,P (0); W-1,P' (fl): the topological dual of Wo'P (S2). cp.: a standard mollifier; uE := We * u: the mollification of u; fl,: the set {x E S2 : dist (x, 8S2) > e}. PA: a particular family of continuous piecewise affine functions. p* := N P: Sobolev critical exponent. Ll (E): an Orlicz space; CO," (ii): the space of all bounded Holder continuous functions with exponent a, E : W 1,P (S2) -- W 1,P (RN) : an extension operator.
dreg: the regularized distance. Given a function u and a set E, UE is the integral average of u over
a set E. BV (12): the space of functions of bounded variation. Given u E BV (11), Diu is the weak (or distributional) partial derivative of u with respect to xi, Du is the weak (or distributional) gradient of u, IDul is the total variation measure of the measure Du. V (u, f2): the variation of a function u in Q. P (E, fl): the perimeter of a set E in fl. O'u (x) := u (x + he-j) - u (x). Bs,P,e (RN): a Besov space; W8'P (RN): a fractional Sobolev space. Tr (u): the trace of a Sobolev function. uO: the spherically symmetric rearrangement (or Schwarz symmetric rearrangement) of a function u. M (u): maximal function of u.
Bibliography In the end, we will remember not the words of our enemies, but the silence of our friends.
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[8] F.J. Almgren and E.H. Lieb, Symmetric decreasing rearrangement is sometimes continuous, J. Amer. Math. Soc. 2 (1989), no. 4, 683-773. [9] L. Ambrosio and G. Dal Maso, A general chain rule for distributional derivatives, Proc. Amer. Math. Soc. 108 (1990), no. 3, 691-702. [10] L. Ambrosio, N. Fusco, and D. Pallara, Functions of bounded variation and free discontinuity problems, Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York, 2000.
[11] L. Ambrosio and P. Tilli, Topics on analysis in metric spaces, Oxford Lecture Series in Mathematics and its Applications, 25, Oxford University Press, Oxford, 2004.
[12] C.J. Amick, Decomposition theorems for solenoidal vector fields, J. London Math. Soc. (2) 15 (1977), no. 2, 288-296. [13] T. Aubin, Problimes isopesrimetriquee et espaces de Sobolev, J. Differential Geometry 11 (1976), no. 4, 573-598. 593
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Index absolute continuity of u", 208 of a function, 73, 241 of a measure, 522 of a signed measure, 525 absorbing set, 497 accumulation point, 494 algebra, 508 arclength of a curve, 128, 132 area formula, 100 atom, 510 background coordinates, 232 balanced set, 256, 497 Banach indicatrix, 66 Banach space, 501 Banach-Alaoglu's theorem, 504 base for a topology, 495 Besicovitch's covering theorem, 538 Besicovitch's derivation theorem, 539 bidual space, 499 Borel function, 511 boundary Lipschitz, 354 locally Lipschitz, 354 of class C, 287 uniformly Lipschitz, 354 Brouwer's theorem, 242 Brunn-Minkowski's inequality, 545
Cantor diagonal argument, 60 Cantor function, 31 Cantor part of a function, 108 Cantor set, 30 Carathdodory's theorem, 510 Cauchy sequence, 494, 498
Cauchy's inequality, 232 chain rule, 94, 145 change of variables, 98, 183, 346 for multiple integrals, 248 characteristic function, 514 closed curve, 116 simple, 116 closed set, 494 closure of a set, 494 coarea formula, 397 cofactor, 243 compact embedding, 320 compact set, 495 complete space, 494, 498 connected component, 14 exterior, 146 interior, 146 connected set, 14 continuous function, 495 continuum, 137 convergence almost everywhere, 534 almost uniform, 534 in measure, 534 in the sense of distributions, 264 strong, 494 weak, 503 weak star, 504 convergent sequence, 494 convolution, 275, 550 of a distribution, 275 counting function, 66 cover, 539 curve, 116 continuous, 116 parameter change, 115 603
Index
604
parametric representation, 115 cut-off function, 496, 559
finite width, 359 first axiom of countability, 495 FYdchet curve, 131
De Is Vallee Poussin's theorem, 173, 535 decreasing function, 3 decreasing rearrangement, 190, 478 delta Dirac, 264 dense set, 494 derivative, 8
of a distribution, 266 differentiability, 8 differentiable transformation, 233 differential, 233 Dini's derivatives, 20 directional derivative, 233 disconnected set, 14 diet-once, 493
distribution, 264 order infinite, 264 distribution function, 187, 477 distributional derivative, 215, 222, 267 distributional partial derivative, 279, 377 doubling property, 22 dual space, 499 dual spaces D' (ft), 284 Mb (X;R), 537 of W - (ft), 299
(l), 303 duality pairing, 499 Eberlein-Smulian's theorem, 505 edge of a polygonal curve, 146 Egoroff's theorem, 534 embedding, 502 compact, 503 equi-integrability, 535 equi-integrable function, 76 equivalent curves Frdchet, 131 Lebesgue, 115 equivalent function, 526 equivalent norms, 502 essential supremum, 526, 532
essential variation, 219 Euclidean inner product, 231 Euclidean norm, 232 extension domain for BV (fl), 402 for W '-P (fd), 320 extension operator, 320
Fo set, 29 Fatou's lemma, 516 fine cover, 539 finite cone, 355
Fubini's theorem, 35, 521 function of bounded pointwise variation, 39 in the sense of Cesari, 389 function of bounded variation, 377 function spaces ACv ((a, b]), 94
AC(1), 73 Agac (1), 74 AC (1; Rd), 74
B',P,e (R-), 415 B'-p-s (On), 474 BV P (I; Rd), 40 BV P (1), 39 BVP1OC (1), 40
BV ((2), 215, 377
BVI, (fl), 220 Co.- (f2), 335 C(X;Y), 495 Co(X),501 C, (X), 501 C°O (ft), 255
C'° (il), 255 Cm (E), 561 C'" (f2), 256 C," (ft), 255 CC (X), 496 D (cl), 259
DK (ft), 255 L1,P (fl), 282
LP (RN), 568 L°O (X), 526 LP (X), 526
LP,, 632 L'V (E), 331
PA, 292 Al (1), 11 W - (fl), 222 ylrs,P (RN), 44$ Wt,P (0), 279
W," (ci), 282 function vanishing at infinity, 187, 312, 477 functional locally bounded, 538 positive, 538 fundamental theorem of calculus, 85 Gd set, 29
Gagliardo's theorem, 453 Gamma function, 572 gauge, 498 geodesic curve, 133 gradient, 233
Index
7{k-rectifiable set, 143 Hahn-Banach's theorem, analytic form, 500 Hahn-Banach's theorem, first geometric form, 500 Hahn-Banach's theorem, second geometric form, 501 Hamel basis, 12 Hardy-Littlewood's inequality, 196, 482 Hausdorff dimension, 578 Hausdorff measure, 574 Hausdorff outer measure, 573 Hausdorff space, 494 Helly's selection theorem, 59 Hilbert space, 506 Hilbert's theorem, 118 Holder's conjugate exponent, 527, 568 Holder continuous function, 335 Holder's inequality, 527, 588 immersion, 502 increasing function, 3 indefinite pointwise variation, 44 infinite sum, 100 inner product, 506 inner regular set, 536 integrals depending on a parameter, 519 integration by parts, 89, 181 interior of a set, 494 interval, 3 inverse of a monotone function, 6 isodiametric inequality, 548 isoperimetric inequality, 405, 549
Jacobian, 233 Jensen's inequality, 518 Jordan's curve theorem, 146 Jordan's decomposition theorem, 524 Josephy's theorem, 55 jump function, 5
Kakutani'a theorem, 505 Katznelson-Stromberg's theorem, 50 Laplacian, 267 Lax's theorem, 243 Lebesgue integrable function, 517 Lebeegue integral of a nonnegative function, 514 of a simple function, 514 of a real-valued function, 516 Lebesgue measurable function, 545 Lebeegue measurable eat, 543 Lebesgue measure, 543 Lebeegue outer measure, 543 Lebesgue point, 540 Lebesgue's decomposition theorem, 523, 525
605
Lebesgue's dominated convergence theorem, 518 Lebesgue's monotone convergence theorem, 515
Lebesgue's theorem, 13 Lebesgue-Stieltjes measure, 157 Lebesgue-Stieltjes outer measure, 157 Leibnitz formula, 264 length function, 125 length of a curve, 118 o-finite, 118 Lipschitz continuous function, 342 local absolute continuity of a function, 74 local base for a topology, 495 local coordinates, 232 locally bounded pointwise variation, 40 locally compact space, 496 locally convex space, 498 locally finite, 496 locally integrable function, 517 locally rectifiable curve, 118 lower variation of a measure, 524 Lusin (N) property, 77, 208, 234, 340
It'-measurable set, 508 maximal function, 564 measurable function, 511, 513 measurable space, 509 measure, 509 v-finite, 509 absolutely continuous part, 526 Bore], 509 Borel regular, 537 complete, 509 counting, 516 finite, 509 finitely additive, 509 localizable, 532 nonatomic, 510 product, 520 Radon, 537 semifinite, 510 signed Radon, 537
singular part, 526 with the finite subset property, 510 measure space, 509 measure-preserving function, 202 measures mutually singular, 523, 525 metric, 493 metric space, 493 metrizable space, 497 Meyers-Serrin's theorem, 283 Minkowski content lower, 549 upper, 549 Minkowski functional, 498
Index
606
Minkowski's inequality, 531, 571 for integrals, 530 mollification, 553 mollifier, 552 standard, 553 monotone function, 3 Morrey's theorem, 335, 437 Muckenhoupt's theorem, 373 multi-index, 255
point of density one, 541 of density t, 541 pointwise variation, 39 polygonal curve, 146 positive pointwise variation, 45 precompact set, 496 principal value integral, 268 purely ?lk-unrectifiable set, 143
multiplicity of a point, 116
N-simplex, 291 negative pointwise variation, 45 neighborhood, 494 norm, 501 normable space, 501 normal space, 495 normed space, 501
open ball, 232, 493 open cube, 232 open set, 494 operator bounded,500 compact, 502 linear, 499 order of a distribution, 264 orthonormal basis, 232 outer measure, 507 Borel, 536 Borel regular, 536 metric, 511 product, 520 Radon, 536 regular, 536 outer regular set, 536 p-equi-integrability, 535 p-Lebeague point, 540 p-variation, 54 parallelogram law, 506 parameter of a curve, 115 partial derivative, 233 partition of an interval, 39 partition of unity, 496 locally finite, 497 smooth, 557 subordinated to a cover, 497 pat hwise connected set, 137 Peano's theorem, 116 perimeter of a set, 379 Poincar4's inequality, 225, 361, 405 for continuous domains, 363 for convex sets, 364 for rectangles, 363 for star-shaped sets, 3T0 in W.'D, 359
Rademacher's theorem, 343 radial function of a star-shaped domain, 370
Radon measure, 155 Radon-Nikodym's derivative, 523 Radon-Nikodym's theorem, 523 range of a curve, 116 rectifiable curve, 118 reflexive space, 505 regular set, 536 regularized distance, 358 relatively compact set, 496 Rellich-Kondrachov'a theorem, 320, 402 for continuous domains, 326 Riemann integration, 87 Riesz's representation theorem in Cc, 538 in Co, 538
in Lt, 533 in L°0,533 in V', 532 in W , in W in W
, 304
in Wo'OO, 305
Riesz's rising sun lemma, 14 rigid motion, 232 o-algebra, 508 Borel, 509 product, 512, 520 a-compact set, 496 a-locally finite, 496 ealtus function, 5 Sard's theorem, 408 Schwarz symmetric rearrangement, 479 second axiom of countability, 495 section, 521 segment property, 286 seminorm, 498 separable space, 494 sequentially weakly compact set, 505 Serrin's theorem, 389 set of finite perimeter, 379 sherically symmetric rearrangement, 479 signed Lebesgue-Stielties measure, 162 signed measure, 524
Index
bounded, 524 finitely additive, 523 simple arc, 116 simple function, 513 simple point of a curve, 116 singular function, 107, 212 Sobolev critical exponent, 312 Sobolev function, 222 Sobolev-Gagliardo-Nirenberg's embedding theorem, 312 spherical coordinates, 253 spherically symmetric rearrangement of a set, 479 star-shaped set, 370 Stepanoff's theorem, 344 strictly decreasing function, 3 strictly increasing function, 3 subharmonic function, 267 superposition, 104 support of a distribution, 271 surface integral, 578
tangent line, 119 tangent vector, 119 testing function, 259 Tonelli's theorem, 91, 125, 521 topological space, 494 topological vector space, 497 topologically bounded set, 498 topology, 494 total variation measure, 378 total variation norm, 533 total variation of a measure, 524 trace of a function, 452 upper variation of a measure, 524 Urysohn's theorem, 495 vanishing at infinity, 312 Varberg's theorem, 240 variation, 378 vectorial measure, 525 Radon, 538 vertex of a polygonal curve, 146 vertex of a symplex, 291 Vitali's convergence theorem, 535 Vitali's covering theorem, 20, 408 Vitali-Besicovitch's covering theorem, 539 weak derivative, 215, 222, 267 weak partial derivative, 279, 377 weak star topology, 503 weak topology, 503 Weierstrass's theorem, 9 weighted Poincar6's inequality, 226 Whitney's decomposition, 564 Whitney's theorem, 561
607
Young's inequality, 527, 551 Young's inequality, general form, 551
Sobolev spaces are a fundamental tool in the modern study of partial differential equations. In this book, Leoni takes a novel
approach to the theory by looking at Sobolev spaces as the natural development of monotone, absolutely continuous, and BV functions of one variable. In this way, the majority of the text wit hout t eh prerequisite o fa course in nctrona
can b a rea d analysis.
117.
e
I
The first part of this text is devoted to studying functions of one
variable. Several of the topics treated occur in courses on real analysis or measure theory. Here, the perspective emphasizes their applications to Sobolev functions, giving a very different flavor to the treatment. This elementary start to the book
makes it suitable for advanced undergraduates or beginning graduate students. Moreover, the one-variable part of the book helps to develop a solid background that facilitates the reading and understanding of Sobolev functions of several variables.
The second part of the book is more classical, although it also contains some recent
results. Besides the standard results on Sobolev functions, this part of the book includes chapters on BV functions, symmetric rearrangement, and Besov spaces. The book contains over 200 exercises.
ISBN 978-0-8218-4768-8
I
For additional information and updates on this book, visit
www.ams.org/bookpages/gsm- I OS 9
78082111847688 GSM/ 105
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