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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'