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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