Sobolev Spaces on Domains (Teubner-Texte zur Mathematik; 137)

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TEUBN ER -TEXTE zur Mathematik Victor I. Burenkov

Sobolev Spaces on Domains

B. G. Teubner Stuttgart Leipzig

Band 137

TEUBNER-TEXTE zur Mathematlk Band 137

V. I. Burenkov

Sobolev Spaces on Domains

TEUBNER-TEXTE zur Mathematik Herausgegeben von Prof. Dr. Jochen Broning, Berlin Prof. Dr. Herbert Gajewski, Berlin Prof. Dr. Herbert Kurke, Berlin Prof. Dr. Hans Triebel, Jena

Die Reihe soil sin Forum fOr BeitrAge zu aktuellen Problemstellungen der Mathematik sein. Besonderes Anliegen let die Verbffentlichung von Darstellungen unterschiedlicher methodischer AnsAtze, die das Wechselsplel zwischen Theorie and Anwendungen sowle zwischen Lehre and Forschung reflektleren. Thematlache Schwerpunkte sind Analysis, Geometrie and Algebra. In den Texten sollen sich sowohl Lebendigkeit and Originalitat von SpezieNorlesungen and Seminaren ale such Diskussionsergebnlsse aus Arbeitsgruppen widerspiegein. TEUBNER-TEXTE erscheinen In deutscher odor englischer Sprache.

Sobolev Spaces on Domains By Prof. Dr. Victor I. Burenkov University of Wales, Cardiff

M

B. G.Teubner Stuttgart

Leipzig 1998

Preface The book is based on the lecture course "Fnction spaces", which the author gave for more than 10 years in the People's Friendship University of Russia (Moscow). The idea to write this book was proposed by Professors H. Triebel and H.A. SchmeiBer in May-June 1993, when the author gave a short lecture course for post-graduate students in the Friedrich-Schiller University Jena. The initial plan to write a short book for post-graduate students was transformed to wider aims after the work on the book had started. Finally, the book is intended both for graduate and post-graduate students and for researchers, who are interested in applying the theory of Sobolev spaces. Moreover, the methods used in the book allow us to include, in a natural way, some recent results, which have been published only in journals. Nowadays there exist numerous variants and generalizations of Sobolev spaces and it is clear that this variety is inevitable since different problems in real analysis and partial differential equations give rise to different spaces of Sobolev type. However, it is more or less clear that an attempt to develop a theory, which includes all these spaces, would not be effective. On the other hand, the basic ideas of the investigation of such spaces have very much in common.

For all these reasons we restrict ourselves to the study of Sobolev spaces themselves. However, we aim to discuss the main ideas in detail, and in such a way that, we hope, it will be clear how to apply them to other types of Sobolev spaces.

We shall discuss the following main topics: approximation by smooth functions, integral representations, embedding and compactness theorems, the problem of traces and extension theorems. The basic tools of investigation will be mollifiers with a variable step and integral representations. Mollifiers with variable step are used both for approximation by smooth functions and for extension of functions (from open sets in R" in Chapter 6 and from manifolds of lower dimensions in Chapter 5). All approximation and extension operators constructed in these chapters are the best possible in

PREFACE

6

the sense that the derivatives of higher orders of approximating and extending functions have the minimal possible growth on approaching the boundary. Sobolev's integral representation is discussed in detail in Chaper 3. It is used in the proofs of the embedding theorems (Chapter 4) and some essential estimates in Chapter 6. An alternative proof of the embedding theorems, without application of Sobolev's integral representation, is also given.

The direct trace theorems (Chapter 5) are proved on the basis of some elementary identities for the differences of higher orders and the definition of Nikol'skii-Besov spaces in terms of differences only. The author pays particular attention to all possible "limiting" cases, including the cases p = oc in approximation theorems, p = 1 in embedding theorems and p = 1, oc in extension theorems. There are no references to the literature in the main text (Chapters 1-6): all relevant references are to be found in Chapter 7, which consists of brief notes and comments on the results presented in the earlier chapters.

The proofs of all statements in the book consist of two parts: the idea of the proof and the proof itself. In some simple or less important cases the proofs are omitted. On the other hand, the proofs of the main results are given in full detail and sometimes alternative proofs are also given or at least discussed. The one-dimensional case is often discussed separately to provide a better understanding of the origin of multi-dimensional statements. Also sharper results for this case are presented. It is expected that the reader has a sound basic knowledge of functional analysis, the theory of Lebesgue integration and the main properties of the spaces L9(SI). It is desirable, in particular, that he/she is accustomed to applying Holder's and Minkowski's inequalities for sums and integrals. The book is otherwise self-contained: all necessary references are given in the text or footnotes. Each chapter has its own numeration of theorems, corollaries, lemmas, etc. If you are reading, say, Chapter 4 and Theorem 2 is mentioned, then Theorem 2 of Chapter 4 is meant. If we refer to a theorem in another chapter, we give the number of that chapter, say, Theorem 2 of Chapter 3. For more than 30 years the author participated in the famous seminar "The theory of differentiable functions of several variables and applications" in the Steklov Institute of Mathematics (Moscow) headed at different times by Professors S.L. Sobolev, V.I. Kondrashov, S.M. Nikol'skiT, L.D. Kudryavtsev and O.V. Besov. He was much influenced by ideas discussed during its work and,

in particular, by his personal talks with Professors S.M. Nikol'skiI and S.L. Sobolev.

It is a pleasure for the author to express his deepest gratitude to the partic-

PREFACE

7

ipants of that seminar, to his friends and co-authors, with whom he discussed the general plan and different parts of the book. I am grateful to my colleagues in the University of Wales Cardiff: Professor W.D. Evans, with whom I have had many discussions, and Mr. D.J. Harris, who has thoroughly read the manuscript of the book. I would also like to mention Dr. A.V. Kulakov who has actively helped in typing the book in 'IBC. Finally, I express my deepest love, respect and gratitude to my wife Dr. T.V. Tararykova who not only typed in TIC a considerable part of the book but also encouraged me in all possible ways.

Moscow/Cardiff, November 1997

V.I. Burenkov

Contents Notation and basic inequalities 1

11

Preliminaries

15

1.1

15

1.2 1.3

..... .... ... ...... Weak derivatives .. .. .. ....... .... . .. .. .... Sobolev spaces (basic properties) ..... ... .. . ..... Mollifiers ... . ... .. .. ..

.

2 Approximation by infinitely differentiable functions Approximation by C000-functions on R"

2.2 2.3 2.4 2.5 2.6

The best possible approximation with preservation of boundary

39

.

.

values

.

.

... . .. ..

.

.

.

.

.

.

.

... ... .. .. .. ... .

3 Sobolev's integral representation 3.2 3.3 3.4 3.5

28

. .... .. .. .... 39 Nonlinear mollifiers with variable step .. . ..... . ...... 42 Approximation by C°°-functions on open sets .. . ....... 47 Approximation with preservation of boundary values .. .... 56 Linear mollifiers with variable step ...... . .. .. ... 60

2.1

3.1

18

..... .. ..... Star-shaped sets and sets satisfying the cone condition The one-dimensional case ... .. .....

73

81 81

..... 92 Multidimensional Taylor's formula ... .. . ......... 100 Sobolev's integral representation .... ...... ... ..... 104 Corollaries .. .. ......... .. . ...... . .. .. ... .

.

111

4 Embedding theorems

119

4.1

Embeddings and inequalities .................... 119

4.2

The one-dimensional case ....

4.3

Open sets with quasi-resolvable, quasi-continuous, smooth and

4.4 4.5

Estimates for intermediate derivatives . ..... .. ....... 160

. ..... ...... ......

127

Lipschitz boundaries .......... ..... . .... .. .. 148 Hardy-Littlewood-Sobolev inequality for integral of potential type177

CONTENTS

10

4.6 4.7

Embeddings into the space of continuous functions ....... 181 Embeddings into the space Lq

. .............. . ... 186

5 Trace theorems

197

5.1 5.2

Notion of the trace of a function ... .. ... .. ..... ... 197 Existence of the traces on subspaces ... ........ .... . 201

5.3 5.4 5.5

Description of the traces on subspaces

Nikol'skil-Besov spaces .......... .. .... .. .... .

. ...... .. ...

.

Traces on smooth surfaces ... ....... .... .. .... .

6 Extension theorems 6.1 6.2 6.3 6.4

.

202 214 238

247

The one-dimensional case ........... ...... . .. .. 247 Pasting local extensions .......... . .. ... . .. .. 264 ... 269 Extensions for sufficiently smooth boundaries .. .. .

.

Extensions for Lipschitz boundaries .. ..

.

.

.

.

.

.. ..

. .

.

.. 271

7 Comments

289

Bibliography

296

Index

311

Notation and basic inequalities We shall use the following standard notation for sets: the set of all natural numbers, N 1o-- the set of all nonnegative integers, Z the set of all integers, R - the set of all real numbers. C - the set of all complex numbers. x No - the set of multi-indices (rs is the natural number Rlo = Nlo x n

which will be used exclusively to denote the dimension), R" = llt

lR,

n

B(x, r) - the open ball of radius r > 0 centered at the point x E R", '11 (S2 C R") - the complement of SZ in R", S2 (52 C Rn) - the closure of 12, S2 (S2 C R") - the interior of 0, 126 (9 c R", 6 > 0) - the 6-neighborhood of 0 (S26 = U:en B(x, 6)),

126={xE9 :list(x,812)>6} (S2CR",d>0) (for each 0CR" 16={xES2:dist(x,8S2)>6}). For a E 1%, a 96 0, we shall write:

the (ordinary) derivative of the function f of order a

D° f and DW f

\ B +,,

q)

the weak derivative of the function f of order a W

(see section 1.2).

For an arbitrary nonempty set fl C R" we shall denote by: C(S2) - the space of functions continuous on 0, C6(SZ) - the Banach space of functions f continuous and bounded on 11 with the norm III 11C(n) = sup If (x) I,

NOTATION

12

C(S2) - the Banach space of functions uniformly continuous and bounded on 12 with the same norm. For a measurable nonempty set 11 C R" we shall denote by: LD(S2) (1 < p < oo) - the Banach space 1 of functions f measurable 2 on

) such that the norm Of IIL,(n) _ (J I J I9dx)

1

< 00,

n

L,°(S2) - the Banach space of functions f measurable on S2 such that the norm If IIL-(n) = ess sup If (x)I = sEn

inf

- sup If W1 < 00

w:measw=0 yEn\w

(in the case in which meas 1 > 0 3 ; if meas S2 = 0, then we set IIf IIL (n) = 0).

For an open nonempty set f2 C R" we shall denote by: L, (QZ) (1 < p < oo) - the set of functions defined on SZ such that for each compact K C SZ f E Ln(K),5 C1 (fl) (l E N) - the space of functions f defined on 0 such that Va E 1V$ where Ial = al + . + a" _ l and b'x E 11 the derivatives (D° f)(x) exist and D* f E C(O), Cb(9) (l E N) - the Banach space of functions f E Cb(fl) such that Va E N where j al = I and Vx E SZ the derivatives (D° f)(x) exist and DI f E Cb(1l), with the norm IIf IIo(n) = If 11c(n) +

E II D°f IIc(n), IaI=t

i As usual when saying a "Banach space" we ignore here the fact that the condition IIfIIL,(n) = 0 is equivalent to the condition f - 0 on Cl (i.e., f is equivalent to 0 on fl e=: mess {x E Cl : f (x) # 0) = 0) and not to the condition f = 0 on Cl. To be strict we ought to call it a "semi-Banach space" (and it will be necessary to keep this fact in mind in Section 4.1) or consider classes of equivalent functions instead of functions. The same applies to the and W,(fl) below. spaces 2 "Measurable" means "measurable with respect to Lebesgue measure." All the integrals thoughout the book are Lebesgue integrals. s We need to do so because otherwise if meal f i = 0, then by the convention sup 0 = -oo we have ess sup If (x) I = -oo.

:en

If Cl C R" is an open set, then for f e C(fl) IIfIIc(n) = IIfIIL.,(n)-

a fk - f in La (fl) as k - oo means that for each compact K C fl fl, -i f in L,(K).

13

NOTATION C'(Sl) (1 E N) - the Banach space of functions f E C(Q) such that Va E 1`l0` where Jai = I and dx E Sl the derivatives (D° f)(x) exist and D* f E C(S1), with the same norm. 00

C°°(Sl) = n o(Q) -- the space of infinitely continuously differentiable 1=0

functions on Sl, Co (Sl) - the space of functions in C' (0) compactly supported in Q,

tit'n(Q) (1 E N, I < p < oo) - Sobolev space, which is the Banach space of functions f E L,(Sl) such that Va E t where jai = I the weak derivatives DW f exist on Q and Dc ,,,f E L,(1), with the norm JIDWfllr.,(n)

Ill lw;(n) = 101=1

(see Section 1.3),

wl,(Q) (I E N.1 < p < cc) the semi-normed Sobolev space, which is the semi-Banach space of functions f E L;a(St) such that Va E N where Jai = 1 the weak derivatives Dw f exist on Q and DW f E L,(Sl), with the semi-norm IllIlw,(n) =

E IIDwfIILP(n) Ia1=1

(see Section 1.3),

W, (f2) (1 E N, I < p < oo) - the Banach space of functions f E Lp(11) such that Va E No' where I al < l the weak derivatives Dw f exist on Sl and D. ('f E L,(Sl), with the norm IllIIfpP(n) = E IIDwf lILo(n) I°I 0 and Vx E R" we set w5(x) = a!,w(s).

Definition 1 Let S2 C R" be a measurable set and 6 > 0. For a function f defined on 11 and such that f E L1(S2 fl B) for each ball B, the operator A6 A6.1z (a mollifier with step (or radius) 6) is defined by the equality !: `dx E R" (A6f)(.r) = (wd * fo)(a-) =

II

(

y).f (y)dy =

f

fo(x - 6>) w(z)dz.

B(0,1) (1.2)

We recall that for each function f under consideration A6f E CO°(R"),

VnEl D°A6f = 6-IQI (D°y) i *.fo

(1.3)

supp A6f C (supp f )J.

(1.4)

on R" and 1 Here and in the sequel fe denotes the extension of f by zero outside Cl: fo(x) = f(x)

for r E fl and fo(x) = 0 for x E `fl.

CHAPTER 1. PRELIMINARIES

16

We note also that on S26

r (A6f)(x) = (w6* f)(x) =

f(x - bz)w(z)dz,

J

B(o,1)

and We E No"

D°A6f = a-1°1(Dow)6 * f. If 5l C lR is an open set and f E Ll°`(S1), then A6f E C°°(S2a) and

A6f -a f a.e.2 on S2 as d -a 0+ (if f E C(Q). then the convergence holds everywhere on 5l). For

I a,10 then Dwf on 526.

D''(A6f) =

(1.21)

Idea of the proof. Use Lemma 4. Proof. Using the properties of mollifiers (Section 1.1), we can write

D"(A6f) = D"-°(D°(A6f)) = D"-°(A6(DWf)) =

810I-I1l(D'r-Qw)6

* Dwf

on S26 (we note that Dw f E L'" (n))-

Example 6 If 1 C R" is an open set, 52 96 R", then (1.20) does not hold on 2, because, for f = 1 on 52, A6(D° f) = 0 on 52 and Dw(A6 f)

0 on 52 \ Q6.

In Definition 2 the weak derivative is defined directly (not by induction as ie ordinary derivative). Therefore the question arises as to whether a weak erivative DWf, where Q < a, i3 # a, exists, when a weak derivative D' Of xists. In general the answer is negative as the following example shows.

Example 7 Set d(xl, x2) E R2 A XI, x2) = sgn x1 + sgn x2. Then derivatives (e )w and (e )°, do not exist (see Example 2, while (e)w = 0 on R2. Idea of the proof. Direct calculation starting with Definition 2. Nevertheless, in some important cases we can infer the existence of derivatives of lower order.

Lemma 5 Let 52 C R" be an open set, I E N, I > 2, f E Ll°`(52) and suppose

that for some j = l,n a weak derivative (4)w exists on 52. Then Vm E N satisfying m < 1 a weak derivative (gym )w also exists on 52. 10 Here and in the sequel y > a means that ' 3 > a3 for j =fin. We note also that j = 1,n means j E {1,...,n}.

1.2. WEAK DERIVATIVES

27

Idea of the proof. Apply the inequality

Iaf

c' (IIfllL,(Q)

IIL,(Q)

+ la f

L,(Q)

where f E C'(Q), Q is any open cube with faces parallel to the coordinate planes, which is such that Q C Sl and cl > 0 is independent of f. (See footnote 3 in Section 3.1.) Proof. For sufficiently large k E N the functions fk = A, f E C' (Q). By (1.5)

and Lemma 4 fk - f in L1(Q) and &-fk 11

axe

-

ma II

axe

L,(Q)

< C,

Consequently,

lim

k,a-,oo

=A(

a in LI (Q). Moreover,

)

(04 - fall L,(Q) + I

j, -' Ox '

8xJ L, (Q)

Ifk

BxJ

_ O'f' 8xt

II

L,(Q) /

= 0.

11

Because of the completeness of L1(Q) there exists a function gQ E L1(Q) such

that

-* gQ in L1(Q) as k -+ oo. Since fk -4 f in L1(Q) as well, by

Definition 3 it follows that gQ is a weak derivative of order I with respect to x; on Q. We note that if Q1 and Q2 are any intersecting admissible cubes then gQ, _ gQ, almost everywhere on Q, n Q2 , since both gQ, and gQ, are weak derivatives

of f on Q1nQz. Consequently, there exists a function g E Lla(Il) such that g = gq almost everywhere on each admissible cube Q and g is a weak derivative

of f on Q. Hence, by Section 1.2 g is a weak derivative of f of order I with respect to x3 on SI.

Lemma 6 Let n > 2, SI C R" be an open set, I E N, I > 2, f E L;°°(O) and suppose that Va E N satisfying lal = I a weak derivative DO Of exists on S1. Then V O E N satisfying 0 < 1,31 co cae°IIf(x)IIZ(R") 0 does not depend on x. Moreover, if n > 2, then for some t E R",

where I{I=1,e>0andb'xEK-{xER":x00, I11-eI<E} jDa(jxjµjlogjxjjv)j ? c9

jxja-1Q1jlogjxjI"-O,

where cg > 0 does not depend on x. Finally, use that for some clo, c,, > 0 I/2

f g(axl)dx = cio f 9(p)p"-idp, B(0,1/2)

f

1/2

g(IxD)dx = cii

B(0,1/2)nK

0

f g(p)p"-idp. 0

Example 9 Let 1 < p < oo.

Under the suppositions of Example 8 jxjµ(logjxj)" E Wo(eB(0, 2)) if, and only if, p < -n/p, v E R or p = -n/p, v < -1/p. On the other hand, (xjµ(logjxj)" E wp(CB(0, 2)) and if, and only if, in

thecase p0No,e:p, x;) and ( f.L (x) ( < M. Integrating by parts (which is possible because dx0> E c0) the function f (x0), ) is locally absolutely

continuous on 520)(x0))) show that the ordinary derivative a (existing thus almost everywhere on 0) is a weak derivative (E )w on a If 12 is convex, then to obtain the converse result use Lemma 4 and (1.7) to prove that Vx, y E S2 and 0 < 6 < dist ([x, y], 8f2) the following inequalities for the mollifier A6 with a nonnegative kernel are satisfied is

I(Asf)(x) - (Aof)(y)I -< II V A6fIIC(Iz,vuIx - yI 15 When writing II v 911c(c) we mean that I I V 911c(c) = I I I v 91 IIC(c) =

J-1

(II v 911L.,(o) is understood in a similar way).

L99 1 2 ) 1 1 2 11

f

(c)

1.3. SOBOLEV SPACES (BASIC PROPERTIES) = II A6 vw f Ilc([x, ]) Ix - y1:5 II

35

V. f II L-([x,y)6) I x - yl

Y15 MIX - yl

< II V. f IIL (n)IX - y1 = II v f

(note also that for f E C(Sl) fl wl(Q) the gradient of exists a.e. on Q and v f = vw f on Sl ). Now it is enough to pass, applying (1.5), to the limit as

6->0+. Corollary 3 If SZ C R" is a convex open set, then g E w.(Sl) if, and only if, it is equivalent to a function f satisfying (1.29) with some M > 0. (Given a function g, the function f is defined uniquely.) Moreover, denote by M' the minimal possible value of M in (1.29). Then M' and, hence, V9 M' < II9IIw- (n) < n M*.

Idea of the proof. The first statement is just a reformulation of Lemma 9 for the case of convex open sets. The second one follows from the definitions of II9IIwi (n) and vwg.

Lemma 10 (Minkowski's inequality for Sobolev spaces) Let S1 C W" be an open set and A C Wn a measurable set, l E N, 1 < p < oc. Moreover, suppose that f is a function measurable on Sl x A and that f y) E WP(I) for almost every y E A. Then 11f f(x,y)dyll wo (n) S A

f

pn)dy

(1.31)

Ilf(x,y)llw,(

A

(the norm Ilf (x, y) II wo(n) is calculated with respect to x).

Idea of the proof. Use Lemma 3 and Minkowski's inequality for Lp(Sl). Proof. Let the right-hand side of (1.31) be finite, then by Holder's inequality for each compact K C 11

f ( f if (x, y)ldx)dy < oo A

K

and 1(1 IDwf(x, y)Idx)dy < oo A

K

Va E N where Ial = I. Hence by Fubini's theorem the function f, being measurable on K x A, belongs to Ll(K x A). Now the inequality (1.31) follows from Lemma 3 and Minkowski's inequality for Lp(SQ): 11f f(x, y)dy11 Wp(n) = 11f f (x, y)dy11Lp(n)

+ E 11Dw f f (x, y)dyll Lv(n)

CHAPTER 1. PRELIMINARIES

36

f IIf(x,y)IIL,(n)dy+E f II(Dwf)(x,y)IIL,(n)dy= f IIf(x,y)Ilwp(n)dy. I°I=1 A

A

A

Lemma 11 (Multiplication by Co -functions) Let S2 C R" be an open set, I E N, I 0 such that V f E W,(0) (1.32)

114 16"(f)) - II f a

-

'

IIL00(-1,1)= II sgnx

IIL.(-1,1) > 2

However, by Lemmas 1- 2 it follows that Co (R") is dense in W0O (R") in a weaker sense, namely,'df E W;° (R") functions W. E C' (W), s E N, exist such

that

ova - f in woo i(R"), II'PsIIW;,(R") -' IIfIIw (R") as8-100. 2 Thus *y(R") = W, (R"), where T pl(R") is the closure of Co (R") in WW (R").

CHAPTER 2. APPROXIMATION BY Coo-FUNCTIONS

42

2.2

Nonlinear mollifiers with variable step

We start by presenting four variants of smooth partitions of unity, which will be constructed by mollifying discontinuous ones.

Lemma 3 Let K C R" be a compact set, s E N, Slk (: R", k = 1, ..., s, be open sets and 9

K C U Stk.

(2.4)

k=1

Then functions 1Jik E Co (Slk), k = 1, ..., s, exist such that 0 < ok < 1 and a

t tpk = 1

K.

on

(2.5)

k=1

Idea of the proof. Without loss of generality we may assume that the 1k are bounded. There exists b > 0 such that K C G =_ 6 (S1k)a. Set Gk = (1k)6 \ k=1 k-1

a

U (Slm)6 and consider the discontinuous partition of unity: E co,, = Xc on k=1

m=1

R". Mollifying it establishes the equality t A¢Xck = AIXc on R", which ] 2 k=1

implies (2.5), where 'pk = Al Xck. (Here A6 is a mollifier with a nonnegative kernel.) 0 Lemma 4 Let SZ C R" be an open set and S1k C R", k E N, be bounded open sets such that 00

C S1k+1,

k E N,

U Slk = 0.

(2.6)

k=1

Then functions 1Pk E Co (Sl), k E,/'N, exist such that Gk C 3upp Wk C Gk-1 U Gk U Gk+l

,

(2.7)

where Gk=Slk\Ilk- I (fork=0 we set ilk= 0),0 0 such that Vx E R" and Vk E Z ID°+'k(x)I .5

k

(2.12)

CHAPTER 2. APPROXIMATION BY C°°-FUNCTIONS

44

Idea of the proof. The same as in Lemma 4. Now the Ilk are defined via the k

Gk: Qk = U

m=-oo

G," and ok = 2-k-3. Estimate (2.12) follows from the equality

10i

D°V,k = 0k (D°W ek) * XGk on Q \ (Ilk_1)ek and the analogous equality on (Dk)ek-i . 0

Remark 3 Sometimes it is more convenient to suppose that the functions &k in Lemmas 4 and 5 are defined on R" and supp'k C Il. (We shall use the same notation ikk E Co (Il) in this case also). Then equality (2.8) can be written in the following form: 00 E Ok = Xn (the same refers to equality (2.11)). k=1

Remark 4 There may exist an integer ko = k0(Il) > 1 such that Gk = 0 for k < ko (in this case we assume that '1k - 0) and (2.11) takes the form 00

00

E V)k=EOk=1 on k=-oo

Q.

(2.13)

k=ko

For Il = R" we shall apply the following analogue of Lemma 5.

Lemma 6 For nonpositive k E Z let

Go={xEW':Ixj 0) such that the properties (2.7) and Vc E No (2.12) are satisfied, 0 < tPk < 1 and 00

0

k=-oo

k=-oo

E4 =E Vk = 1

R".

(2.14)

(supp Qek C (Gk-1 U Gk U Gk+1)ek

(2.15)

on

Idea of the proof. The same as in Lemma 5. 0

Remark 5 Note that in Lemmas 4 - 6

Moreover, in the case of Lemma 4 for any arbitrarily small ryk > 0, k E N, one can construct functions 'kk, k E N, satisfying the requirements of Lemma 4

such that suppikk C (Gk)7k,

Ok = 1 on (Gk),.

(2.16)

2.2. NONLINEAR MOLLIFIERS WITH VARIABLE STEP

45

To do this it is enough to replace pk defined by (2.9) by pk = min{ dist (Gk, 8(Gk_l U Gk U Gk+1)),'Yk}. a

In the case of Lemmas 5 and 6 for any fixed y > 0 one can construct functions yk, satisfying the requirements of those lemmas, such that suPPiPk C (Gk)ry2-k,

lPk = 1 on (Gk),2-k.

(2.17)

Remark 6 From (2.15) it follows, in particular, that the multiplicity of the covering {supp iik} in Lemmas 4-6 is equal to 2, i.e., Vx E Q there are at most 2 sets supp tbk containing x and there exists x E Il such that there are exactly 2 sets supp t/ik containing x. (From (2.7) it follows only that the multiplicity of this covering does not exceed 3.) Of course 2 is the minimal possible value (if supp ,bk J Gk and the multiplicity of covering is equal to 1, then iik = XGk). Moreover, from (2.15) it follows that for 6 E (0,'-g] the multiplicity of the covering {(supp

Vlk)62_k

} is also equal to 2.

In Chapter 6 we shall need a variant of Lemma 5 for 0 = {x E R" : x,, > W(x1,...,x"-1)}, where W is a function of class Lip 1 on R"-1, - that variant will be formulated there.

Let Q C R" be an open set and Qk C S2, k E N, be bounded open sets, possessing the properties (2.6), Gk = ilk \ S1k_1. Suppose that pk is defined by (2.9) and {k}kEN is the partition of unity in Lemma 4 defined by (2.10).

Definition 1 Let b = {bk}kEN, where

0 f in WW(Sl), m = 0,1, ...,1- 1,

IIcvkHlw, (a) -+ IIf IIwd,(nl

ask-4 oo. Later, in Section 2.6, we shall see that bf E W,(Sl) (or &(Q)) functions gyp, E COO (11) fl WD(S1), gyp, E C°°(11) fl ?71 (11) respectively, exist, which depend

linearly on f, do not depend on p and are such that for 1 < p < oo

G, -Y f in Wy(1l),

(2.27)

in C'(c) respectively. Moreover, the functions W. may possess additional useful properties.

We shall deduce the statement of this theorem, in the case in which 1 < p < oc, from a much more general result, which holds for a wide class of semi-normed linear spaces Z(Sl) of functions defined on SZ with semi-norms II' 11z(n) such that Co (Sl) C Z(fl) C Li°°(Sl). Let Zo(fl) denote the subspace of Z(Sl) that consists of all functions f E Z(fl), which are compactly supported in

SZ. Moreover, let Z(SZ) denote the space of all functions f E Lr(12), which are such that VW E C01(fl) we have cof E Z(Sl). From these definitions it follows, in particular, that Co (Sl) C Zo(fl) C (L()o(SI) and

C°°(Sl) C Z(°`(Sl) C L10°(Sl).

"Under additional assumptions on fl (see Theorem 6 of Chapter 4), IIfIIw, (n) 5 M II/IIw;(n), m = 1,..,1- 1, where M is independent of f, and this statement follows from the density of COO (0) nW.(fl) in W,(fl). However, for arbitrary opens sets it is not so (see Examples 8 - 9 of Chapter 4), and this statement needs a separate proof. We also note that

it is possible that f ¢ W.'(11). In that case also ,k V Wa (fl) but f - wk E W, '(fl) and

f -Ws -0in WD (Sl) as k-4oo.

2.3. APPROXIMATION BY Coo-FUNCTIONS ON OPEN SETS

49

Remark 9 For any I E No we have (C1)'a(c2) = (C,)'°°(c2) = C1(cl). Moreover, for 1 < p:5 oo the following equivalent definition of the space (Wp)'-(Q) can be given: (WP)" (c2) = (f E Li°`(c1): for each open set G compactly embedded into 1l f E Wp(G)}.

Theorem 2 Let Q C R" be an open set and suppose that the semi-Banach space Z(n) satisfies the following conditions:

1) Co (cl) C Z(c2) C Li°L(1), 2) (Minkowski's inequality) if A C 1R function measurable on c2 x A, then

II f f (x, y)dyl z(n) 5 A

f

is a measurable set and f is a

IIf (x, y)Ilz(n)dy,

A

3) if p E C0 00(Q) and f E Z(cl), then W f E Z(S2),

4) all functions f E Zo(S2) are continuous with respect to translation, i.e.,

u oIIfo(x +

h) - f(x)IIz(n) = 0.

(2.28)

Then C°°(Q) is dense in Z"(cu) (and, hence, C°°(Q) f1 Z(Q) is dense in Z(11)), i.e., V f E Z11(c2) functions cp, E C°°(c2) fl Z'-(c2), 8 E N, exist such that

cp, - f in Z(c2)

(2.29)

ass -4 oo.

Idea of the proof. Apply Minkowski's inequality to the right-hand side of the equality (Baf)(x) - f(x) =00E

f

(fk(x - bkz) - fk(x)) w(z) dz,

(2.30)

k-1B(o,1)

where fk = I k f and the mollifier BI is constructed with the help of a nonnegative kernel of mollification, and prove the inequality It Baf - f Itz(n)5 F, w(bk, fk)Z(n) k=1

(2.31)


50

Here

w(b, f)z(n) = sup Ilfo(x + h) - f (x)llz(n) Ihj 0

such that Vh E R" satisfying IhI < 'y the function fo( + h) - f E Z(SZ). Let us suppose, in addition, that y < dist (suppf, 6Q), then suppfo( + h) C (suppf )Ihi C Sl and fo( + h) E Zo(SZ). For, first of all fo( + h) E Z'-(Q). Consider, furthermore, a function of "cap-shaped" type 'n E C0 00(Q) such that n = 1 on (see Section 1.1), then by definition of Z'°°(1) rlfo(' + h) E Zo(SZ).

Let A(h) = Il fo(x+h) -f (x)IIz(n) for h E B(0, ry). Condition 4) means that the function A is continuous at the point 0. Moreover, A E C(B(0, ry)). Indeed,

let u E B(0, ry). Then, by the continuity of the semi-norm, in order to prove

that A(h) -+ A(u) as h i u it is enough to prove that fo(x + h) - f (x) fo(x + u) - f (x) or fo(x + h) - fo(x + u) = go(x + h - u) - g(x) -r 0 ash -

u

where g(x) = fo(x + u). And this is valid because g E Zo(SZ). 2. Let us consider the mollifiers B1, which are constructed with the help

"

of any nonnegative kernel. Since 00 E )k = 1 on SZ and f wdx = 1 we have k=1

NnEi1

B(0,1)

00

(Baf)(x) - f (x) = E,((A6k (Okf )) - Wk(x)f (x)) k=1

00

_ F I

(fk(x - 6kz) - fk(x))w(z)dz =

k=1 B(0,1)

Fk(x),

k=1

where fk = 'kf and Fk(x) = f (fk(x - akz) - fk(x)) w(z)dz. B(0,1)

By 3) and (2.15) we have that fk, Ft E Zo(S1) and supp fk, supp Fk C Gk_1 U Gk U Gk+1. Applying Minkowski's inequality for infinite sums (which holds besause of the completeness of the space Z(Il)) we have 00

IIFkllz(n)

II Baf - f Ilz(n) 0 there exist akl) > 0, k E N, such that Wk E (0, akl)) we have w(bk, 1kfo)w;(R") < e 2-k, m = 0, ...,1 - 1, and hence IIBaf - f 11w,,_.(n)

< s, m = 0, ...,1- 1.

Furthermore, for Va E No satisfying j al = I by (2.25), Lemma 4 of Chapter 1 and Leibnitz' formula we have 00

E (*)EAj,,(D--0,0kD.#f), 0 w(bk,

D°-dV

k Dwfo)LW(R')

k=1

Since by Lemma 11 of Chapter 1 D`011k Dw f0 E W.'-"l (R") as in the proof of Lemma 1 we establish that w(bk, D°-p k L'w0 f0)Loo(R°) < M2bkIID°-1 Vk DwfolI

W:181(R") ,

where M2 is independent of f and k. Consequently, there exist ak2) E (0, ak1)), k E N, such that dbk E (0, ak2)) we have w(bk, D°-",Pk DwfO)Loo(R") < c 2-k-"(1 + E 1)-1 101=1

and, hence,

II E 00 A6.(D°-O*k Dwf)IIL,o(n) < e 2-"(1 + E 1)-1. k=1

I°I=1

If Q = a, then since 10k, k E N, and the kernel of mollification is nonnegative we have 00

00

IIEIA6k(PkD,°of)IIIL-(n)

II

k=1

k=1

00

00

S

0, there exist s E N and 5ks E (0,ak4)) such that IlBd,f - fIIw,1(n) <e and

IIfUIW,(n) -s < IIBa.fllw.,(n) and the statement of Theorem 1 in the case p = oo follows.

Corollary 1 Let 11 C R" be an open set, l E No

Then C°°(S2) is dense in

(WP)'0°(&1) where 1 < p < oo and in C'(S2).

Idea of the proof. Apply Theorem 2 to Z'«(S2) _ (WW)'°°(SZ) and Z(SZ) _ C'(Sl).

Remark 11 If p = oo, then C°°(SI) fl W/,(SZ) is not dense in W;°(1l) (see Remark 2).

Remark 12 The crucial condition in Theorem 2 is condition 4). It can be proved that under some additional unrestrictive assumptions on Z(Sl) the density of C°°(Il) in Z'°°(n) (or the density of C°°(a)flZ(SZ) in Z(Sl)) is equivalent to condition 4).

Remark 13 Theorem 2 is applicable to a very wide class of spaces Z(fl), which are studied in the theory of function spaces. We give only one example. Consider positive functions ao, as E C(fl) (a E 1%, lal = 1) and the weighted Sobolev space Wp.{°o)(S2) characterized by the finiteness of the norm IlaofIIL,(n)+

lla.DwfIIL,(n)

By Theorem 2 it follows that C°°(fl) fl WP(°Ql(S2) is dense in this space for 1 < p < oo without any additional assumptions on weights ao and a.. Such generality is possible due to the fact that the continuity with respect to translation needs to be proved only for functions in this weighted Sobolev space, which are compactly supported in 0.

CHAPTER 2. APPROXIMATION BY Coo-FUNCTIONS

56

Now we give one more example of an application of Theorem 2, in which the spaces Z'°`(SI) (and not only Z(SZ)) are used.

Example 1 Let SI C R" be an open set, then dµ E C(SI) and de > 0 there exist µE E C°°(Q) such that Vx E SI we have µ(x) < p,(x) < µ(x) + e.

To prove this it is enough to set µe = Bg(,u + Z) with 3 = b(z, µ + a) in the proof of Theorem 2 for Z(S)) = C(SI) (hence, Zb°`(SI) = C(SI)) and apply inequality (2.34).

2.4

Approximation with boundary values

preservation

of

In Theorem I it is proved that for each open set SI C R" and Vf E W'(SI) (1 < p < oo) functions W. E C°°(i) n WW(SI), s E N, exist such that (2.27) holds. In this section we show that it is possible to choose the approximating functions cp, in such a way that, in addition, they and their derivatives of order a E No satisfying jal < 1 have in some sense the same "boundary values" as the approximated function f and its corresponding weak derivatives. The problem of existence and description of boundary values will be discussed in Chapter 5. Here we note only that for a general open set SI C R" it may happen that the boundary values do not exist and even for "good" 0 boundary values of weak derivatives of order a satisfying lad = 1, in general, do not exist. For this reason in this section we speak about coincidence of boundary values without studying the problem of their existence - we treat the coincidence as the same behaviour, in some sense, of the functions f and cp, (and their derivatives) when approaching the boundary of SI. Theorem 3 Let SZ C R" be an open set, I E N, 1 0

(2.35)

ass -* oo. For p = oo this assertion is valid V f E C'(SI). Corollary 2 Let SI C R" be an open set, fI 34 R" and I E N. Then V f E U1 (fl) functions cp, E COD(A) n s E N, exist, which depend linearly on f and

2.4. PRESERVATION OF BOUNDARY VALUES

57

are such that, besides (2.27) where p = oo, s

D*v.Ion = D°f Ian,

(al < 1.

Idea of the proof. Choose any positive p E C(12) such that

(2.36)

lim

y-rx,yEf

µ(y) = oo

forall xE812. Proof. One may set, for example, p(x) = dist (x, 812)-1. For a continuous function II - 11C(n) = II - IIL,(n), therefore, from (2.35) it follows that for some

M>OVsENandVyE0

ID°w9(y) - D°f (y)I S M(.u(y))-1.

Passing to the limit as y -+ x E all, y E S2 we arrive at (2.36).

Corollary 3 Let 12 C R" be an open set, 1 E N, 812 E C1 and 1 < p < oc. Then V f E W(1l) functions cp, E C°°(Q) n WW(Q), s E N, exist such that, besides (2.27), D°wal an = Dwf Ian,

j al < 1 - 1.

(2.37)

Idea of the proof. Take again µ(x) = dist (x, 812)-'. By Chapter 5 it is enough to consider the case, in which 12 = R+ = {x E R" : x" > 01 and µ(x) = x;1. In this case the statement follows by Lemma 13 of Chapter 5.

Remark 14 The function p in Theorem 3 can have arbitrarily fast growth when approaching M. Let, for instance, u(x) = g(Q(x)), where e(x) = dist (x, 80) and g E C((0, oo)) is any positive, nonincreasing function. Then

forl 0 such that df E Z0(SZ) Ikof Ilz(n) 0 depending only on µ77x (and, thus, independent of bk), such that IIIFkIIZ(n) < CkIIFkIIz(n) < CkW6,t(fk)Z(ft)

and (2.40) follows (without loss of generality we can assume that ck > 1).

Choosing Ve > 0 positive numbers 5k in such a way that in this case w6,t(fk)z(n) < e2_kckl (instead of (2.33)), we establish, besides (2.34), the inequality II (B3f - f)µllz(n) < c.

Remark 15 From the above proof it follows that b'µ1i...,µm E C' (Q) (m E N) and Vf E Zb0C(Sl) functions cp, E C°°(Q) f1 Zf0C(c), s E N, exist such that

II(f - e)WIIz(f:) -> 0,

i = 1,...,in.

as s -+ oo. (Theorem 4 corresponds to m = 2, it, = 1, µ2 = µ.)

Idea of the prof of Theorem 3. Apply Theorem 4 and Remark 15 to the space Z(Q) = WW(c) and to a set of the weight functions (D'µ,)I,I- 6i 6

'IIA6xj -xjIIL,(G)IIxjII;P(G6)

=If zjw(z) dz 16lli+m I R

II A6f - f IIL,(G) II f II W'(G6)

IUII.'Imoo

mess G mess

G6' PL

=

zjw(x) dz I

R

Thus, if, in addition to (1.1), w(z) < 0 if zj < 0 and w(z) > 0 if zj > 0, then cl is the best possible constant in inequality (2.43).


62

Example 3 Let n = 1, G = St = R, p = 2 and f E W2 (R). Then by the properties of the Fourier transform

IIA6f - fIIL,(R) = (2ir)26II6-'((Fw)(6f) - (Fw)(0))(Ff)(f)IIL,(R). We have

6-1((Fw)(bf) - (Fw)(0)) -* (Fw)'(0)t

ash -0+ and sup 6>0

6-'I (Fw) (6f) - (Fw)(0)I

m ax I (Fw)'(z)I

Therefore, by the dominated convergence theorem

II6-'((Fw)(6f) - (Fw)(0))(Ff)(f)IIL2(R)

- I(Fw)'(0)I IIe(Ff)(f)IIL,(R) = I Jzw(z)dz I IIf'IIL,(R). R

Hence, if f zw(z) dz 0 0 and f E W2 (1R) is not equivalent to zero, then for R

some c3 > 0 (independent of 6) and II A6f -f IIL,(R) >- c36 for sufficiently small J.

Let us make now a stronger assumption: f E WP(f) where I > 1. In this case, however, in general we cannot get an estimate better than IIA6f - f II L,(c) = O(5)

(which is the same as for I = 1), if for some j E {1, ..., n} f z,w(z) dz # 0, R

as Examples 2 - 3 show. Thus, in order to obtain improvement of the rate of convergence of A6f to f for the functions f E Wp(f2) where 1 > 1, some moments of the kernel of mollification need to be equal to zero.

Lemma9 LetnCW bean open set, 1 0 and G C fl be a measurable set such that G6 C S2. Assume that the kernel of the mollifier As satisfies, besides (1.1), condition (2.45). Then V f E w,(S)) and Va E N IID°(Asf) - DafUL..,(o) 2, we define the operators b.0 (E6), where a E I ` and jal =1, with the domain (Wi)'O°(12): b.-(E6) =

((aI)w'

...

(E6).

Lemma 16 Let 0 C R" be an open set, 1 E ICI, 0 < b < s and f E (Wi)i0c(Sl) Then Va E N' satisfying dal < I 00

(D°(E6)) f = E (D°,0k)A62-Ik1f k=-oo

Idea of the proof. Induction. For Ian = 1 by (2.55)

l\ax,) w (Ea))f = (((

(A62-ikif+V)kaxj(A62-ikif))

00

k=-oo 00

00

-kA62-iki (e ii) k=-oo

on S2, because by (2.7) and (1.19) ikk

371

_ W

k=-oo

821

A62-ikif

(Ab2_j,,jf) = tPkAd2-iki (e )W on 11. E3

Remark 21 For the mollifiers A6 we have (Dw(A6)) f =_ 0 but, only on 126 (see Section 1.5), while for the mollifiers E6 in general (Dw(E6)) f $ 0 even on 526, but on the whole of S2 the quantity (Dw*(E6))f is in some sense small (because 00

E D°9fik = 0 on S2) and, as we shall see below, it tends to 0 in Lp(S2) fast k=-oo

enough under appropriate assumptions on f .

Remark 22 On the base of Lemma 16 we define for Va E N satisfying a 96 0 the operator E6°) with the domain L"(1) directly by the equality

Ea°if = E (D°*k)A62_Iki f.

(2.65)

k=-00

Lemma 17 Let SZ C W' be an open set and 0 1, then V f E w,(1l) II (Do(Eaf

))p101-'IIL,(n) 0 and m > 1, a function f E WP(12) exists such that, whatever are the functions W. E C°°(12), s E N, satisfying 1) and 2), for some Va E I' satisfying Ial = m IIL,(n) = cc.

(2.90)

Idea of the proof. As in the proof of Corollary 3, by Lemma 13 of Chapter 5, propety 2) follows from relation (2.86). 0 The most direct application of Theorem 7, for the case in which p = oo, is a construction of the so-called regularized distance. We note that for an open set S2 C R", SZ # R1, the ordinary distance p(x) = dist (x, 812), x E 12, satisfies Lipschitz condition with constant equal to 1:

I p(x) - p(y)I 5 Ix - yI,

x, y E S2.

(2.91)

(This is a consequence of the triangle inequality.) Hence, by Lemma 8 of Chapter 1 (2.92) p E wolo(12), Ivpl < 1 a.e. on 12. The simplest examples show (for instance, p(x) = 1- Ixi) for 12 = (-1, 1) C R) that in general the function p does not possess any higher degree of smoothness than follows from (2.91) and (2.92).


78

Theorem 10 Let Sl C R" be an open set, 11 36 R. Then Vl E (0,1) a function ea E C°°(11) (a regularized distance) exists, which is such that (1 - 6)e(x) < ea(x) < e(x),

x E 0,

(2.93)

I Pa(x) - e6(y) 15 Ix - yI,

x, y E Sl,

(2.94)

on fl

I oea(x)I < 1

(2.95)

and We E N satisfying IaI > 2 and Vx E Q I (D°ea)(x)I 5

cool-lalg(x)1-IaI

,

(2.96)

where cc, depends only on a.

Idea of the proof. In order to construct the regularized distance it is natural to regularize, i.e., to mollify, the ordinary distance. Of course, one needs to apply mollifiers with variable step. Set ea = aE"g and choose appropriate a, b > 0. Here EM is a mollifier defined by (2.50) where l = 1 and the kernel of mollification w is nonnegative.

Proof. First let Da = E6e. Since p E w;o(Q), from (2.81) and footnote 4 on the page 12 it follows that sup Ioa(x) - e(x)I e(x)-1 < c156 'En

or bxESl (1 - c156)0(x) 5 Do(x) 5 (1 + ctsb)e(x),

where c15 > 0 depends only on n. Moreover, from (2.82) it follows that Va E N satisfying a 96 0 sup

IDaoa(x)le(x)IaI-1
0 is independent of f. These inequalities were used in the proof of Lemmas 5-6 of Chapter 1.

CHAPTER 3. SOBOLEV'S INTEGRAL REPRESENTATION

88

Remark 4 We note a simple particular case of the integral representation (3.17): if w is absolutely continuous on [a, b], w(a) = w(b) = 0 and fwdx = 1, a

then for each f such that f' is absolutely continuous on [a, b], for all x E [a, b] b

b

(3.24)

f'(x) = - f w'(y)f (y) dy + f A(x, y)f"(y) dy. a

a

It follows that

f

6

If'(x)I 0 and h > 0 if Vx E 0 there exists 7 a cone Kx C 0 with the point x as vertex congruent to the cone K. Moreover, an open set Sl C R" satisfies the cone condition if for some r > 0 and h > 0 it satisfies the cone condition with the parameters r and h. Example 1 The one-dimensional case is trivial. Each domain SZ = (a, b) C R is star-shaped with respect to a ball (__ interval). An open set ft = 6 (ak, bk), k=1

where s E N or s = oo and (ak, bk) fl (am, bm) = 0 for k # m, satisfies the cone condition if, and only if, inf (bk - ak) > 0.

Example 2 A star (with arbitrary number of end-points) in R2 is star-shaped with respect to its center and with respect to sufficiently small balls (= circles) centered at its center. It also satisfies the cone condition.

Example 3 A convex domain 11 C R" is star-shaped with respect to each point y E fl and each ball B C C. A domain fZ is convex if, and only if, it is star-shaped with respect to each point y E 0. Example 4 The domain ci C R2 inside the curve described by the equation lxi If + lxs l' = 1 where 0 < y < 1 (the astroid for y = 2/3) is star-shaped with respect to the origin, but it is not star-shaped with respect to any ball B c Cl. It does not also satisfy the cone condition.

Example 5 The union of domains, which are star-shaped with respect to a given ball, is star-shaped with respect to that ball. The union (even of a finite number) of domains star-shaped with respect to different balls in general is not star-shaped with respect to a ball. In contrast to it the union of a finite number of open sets satisfying the cone condition satisfies the cone condition. Moreover, the union of an arbitrary number of open sets satisfying the cone condition with the same parameters r and h satisfies the cone condition. 7"Vx E !2" can be replaced by "Vx E Il" or by "Vx E Oft" and this does not affect the definition.


94

Example 6 The domain 0 = {z E Rn : IxI7 < x < 1,

Jxj < 1}, where for ry > 1 is star-shaped with respect to a ball and satisfies

Y = (xl, ...,

the cone condition. For 0 < y < 1 it is not star-shaped with respect to a ball. Furthermore, it cannot be represented as a union of a finite number of domains, which are star-shaped with respect to a ball, and does not satisfy the cone condition.

Example 7 The domain 0 = {x E Rn : -1 < xn < 171-1, 171 < 1} satisfies the cone condition for each 'y > 0. It is not star-shaped with respect to a ball, but can be represented as a union of a finite number of domains, which are star-shaped with respect to a ball.

Example 8 The domain SZ = ((XI, x2) E R2 : either - 2 < x1 < 1 and - 2 < x2 < 2, or 1 < x1 < 2 and - 2 < x2 < 1} is star-shaped with respect to the ball B(0,1). For 0 < b < %F2 - 1 the domain fl D B(0,1), but it is not star-shaped with respect to the ball B(0,1). (It is star-shaped with respect to some smaller ball.)

Lemma 1 An open set S2 C R" satisfies the cone condition if, and only if, there exist s E N, cones Kk, k = 1, ._a, with the origin as vertex, which are mutually congruent and open sets SZk, k = 1, ..., s, such that ,

U

k=1

2)b'xEIk the cone' x+KkCQ. Idea of the proof. Sufficiency is clear. To prove necessity choose a finite number

of congruent cones Kk, k = 1, ..., s, with the origin as a vertex, whose openings are sufficiently small and which cover a neighbourhood of the origin, and consider the sets of all x E 1 for which x + Kt C Q. 0 Proof. Necessity. Let fl satisfy the cone condition with the parameters r, h > 0. We consider the cone K(rl, h1) defined by (3.34), where h1 < h and r1 < r is such that the opening of the cone K(rl, h1) is half that of the cone K(r, h). Furthermore, we choose the cones Kk, k = 1, ..., s, with the origin as a vertex, which are congruent to K(rl, h1) and are such that B(0, h1) C 6 Kk. Hence, k=1

Vx E Sl the cone Ks of the cone condition contains x + Kk for some k. Denote by Gk the set of all x E St, for which K. contains x + Kk. Finally, there exists b= > 0 such that Vy E B(x, 6) we have y + Kk C fl. Consequently, the open 8 Here the sign + denotes a vector sum. The cone z + Kk is a translation of the cone Kk and its vertex is x.

3.2. STAR-SHAPED SETS AND THE CONE CONDITION

95

sets S2k = IJ B(x, a=), k = 1, ..., s, satisfy conditions 1) and 2). zEG4

Let a domain SZ C R" be star-shaped with respect to the point x0. For t; E S"'', where S"-' is the unit sphere in R", set w(t;) = sup{p > 0 : xo+pt; E S2}. Then

52={XER":x=xo+pt where DES"-'. 0 - w(f) 2w(()sin2 sin B-7

'0(07C-0-2) ein(9-ry

- w(f)

-ry

sin

d(f,+])Con

Vt;, 7 E 51-1 such that -y < Q I

'(q) I = R2 '

(a-7)

d(f, n).

Hence, given e > 0, there exists (e) > 0 such that V , r! E S"-1: -y < b(e) we have Iw(f) - W(17)1:5 (R2 cot,0+e)d(t;,7)

= (RZ

Rz

- r2 +e)d(la,r7).

Now let t; and' be arbitrary points in S"-1, f 96 ri. We choose on the circle, centered at x0 and passing through f and 17, the points to = f 1 -< ... -< f,,,-j -< t;,,, = 17 such that all the angles between the vectors O6_1 and 01, i = 1, m, are less than 6(e). Then m

Iw(f) - w('7)I 0 such that bk -+ 0+ as k -4 oo and

L

(A6k(a )w)(x+th)dt- L_)x+th)dt

almost everywhere on Sty. Moreover, by (1.5) (A,, f)(x + h) -+ f (x + h) and (A6k f)(x) -> f (x) almost everywhere on Sty. Thus, passing to the limit as k - oo, we obtain the desired equality for almost every x E Sty and, since y > 0 was arbitrary, for almost every x E St . 11

We note that one can prove similarly that if f E L10C(1) and `da E N satisfying jai = I there exist weak derivatives D° Of on St, then for almost every x E StjhI i

h°

(Dw°f)(x) h° + I

f (x + h) _ I°I 0, k E N, such that 'Yk -+ 0 as

k ->oo,wehave

S.,,, (x) -1 Sa(x) a.e. on ft[,q.

(3.49)

Now passing to the limit as k -> oo in equality for Al f with 'Yk replacing ry, by (3.46), (3.47) and (3.49) we obtain that (3.45) holds almost everywhere on S2[k[. Since U 11[o[ = f2 we establish that (3.45) is valid almost everywhere on

Q.0

6>0

20 In the last inequality in the expression 116 - nj* the sign - denotes vector subtraction

of sets inR",i.e.,A-B=(zER":z=x-y where xEA,VEB). Clearly, flj - n6' C B(o, 8-1) - B(o,6-') =

B(o,2a-1).

3.5. COROLLARIES

3.5

111

Corollaries

Corollary 9 In Theorems 4, 5 let conditions (3.37), (3.44) respectively, be replaced by

fdx=1.

supp w C B,

w E Co (S2),

(3.50)

Rn

Then Vf E CI(Q) for every x E 11 and Vf E (Wi)'-(fl) for almost every x E S2

f (x) = f (E B

(_ aDy a

Ial 0 and h > 0 and K. be the cone of that condition. Suppose that Vx E fl W. E Co (R"),

suppw: C IT.,

Jiz(v)dY = 1,

(3.73)

Rn

I E N, 0 E No and lal < 1. Then V f E C1(Q) for every x E 1 and V f E (W11)(f2) for almost every x E 12

[(x-y)°w=(y)])f(y)dy

(D"f)(x) =J K.

io 0 depends only on n and 1. Remark 20 Compared with (3.54) inequality (3.76) is valid for a wider class of open sets satisfying the cone condition. On the other hand, (3.54) is a sharper version of (3.76) (for in the right-hand side of (3.54) f If I dx replaces f if I dx) B a for a narrower class of domains star-shaped with respect to the ball B.

Idea of the proof. Let K = K(r, h) if h > r and K = K(h, h) if h < r, and let B(xo, r2) be the ball inscribed into the cone K. Here r2 = rh( r + h +r)-1 >-


118

r(l+f)-' if h > r and r2 = h(1+f)-' if h < r. Hence B(xo, rl(l+f)-') C K. Moreover, let w be a fixed function defined by (3.50). Suppose that in (3.74) wx is defined by: Wx(y) = (i+ 22)-1(y-£y)), where £_ = A1(xo) and

A. is a linear transformation such that Kx = A.(K). Following the proof of estimates (3.56) and (3.57), establish that b'x E Sl and Vy E K.

(-1 )'IQH Ifll D,+R[(x - y)awx(y)]I < Ml (h

r1)'-ih-101r1

(3.78)

Ial 0 such that Vf E Z, II9IIz, 0 such that Vf E Z1

inf IIf - hllz, 5 csllf Ilzi.

heo,

' The sign + denotes the vector sum of sets. 2In this case, in general, the embedding operator is not unique. However, one may easily

verify that for different embedding operators, say I, and 12, Vf E Z, we have IIIiflIz, _ 1112flIs,-

CHAPTER 4. EMBEDDING THEOREMS

122

Lemma 2 Let Z1 and Z2 be semi-Banach spaces such that Z1 C Z2. Suppose that for any fk E Z1, k E N, g1 E Z1 and g2 E Z2 lim fk = 91 in Z1i lim fk = g2 in Z2 k-+oo

k-+oo

91 - 92 E 01.

(4.10)

Then (4.9) is satisfied.

Idea of the proof. Apply the Banach closed graph theorem to the factor spaces Zl = Z1/81 and Z2 = Z_2/81 and 3 the embedding operator I : Z1 -> Z2. Proof. We recall that Z1 is a Banach space and b'/ E Z1 If 112, = 11f 11z"

(4.11)

where f is an arbitrary element in f. (If fl, f2 E f, then 11f, 11 = 11f211.) As for Z2 it is, in general, a semi-Banach space and

IIJ II2, =h0i inf If - hllz2

(4.12)

where f E j. (The right-hand side does not depend on the choice of f E f.) From Z1 C Z2 it follows that Z1 C Z2 and by (4.10) the corresponding embedding operator I is closed. For, let 1k E ZZ1, k E N, 91 E Z1, 92 E Z2 and

k

Ik=91

in Z1, k Ifk=kl Ik=g2 in Z2.

Suppose that fk E fk, 91 E 91 and g2 E 92 Then fk - 91 E It - 91, fk - 92 E fk - 92 and

limllft -91IIz, = 0,

lim (hnf I1fk-92-hIlz,) = 0.

Therefore, Vk E N there exists hk E 81 such that 1lim Ilfk-92-hkllz, = 0. Thus fk - hk -+ 92 in Z2 ask -+ oo. Moreover, since Ilft - hk - 91IIz, = lift - 9111z we also have that fk - hk --1 91 in Z, as k -3' oo. By (4.10) 91 - g2 E 81 and, hence, g1 = 92 Now by the Banach closed graph theorem the operator I is bounded: for some c4 > 0 we have b'/ E Zl 111112,=IIIJII2,:5 04II!II2,.

Consequently, by (4.10) and (4.12) the desired inequality (4.9) follows. 3 The spaces Z'1 and Z2 consist of nonintersecting classes f C Z1, f C Z2 respectively, such that fl, f2 E 1 . = fl - f2 E 01, fl - fs E B2 respectively.

4.1. EMBEDDINGS AND INEQUALITIES

123

Corollary 1 If, in addition to the assumptions of Lemma 2, (4.8) is satisfied, then Z, C Z2 is equivalent to Z, C:; Z2. Idea of the proof. Apply Remark 4.

Corollary 2 In addition to the assumptions of Lemma 2, let 01 C 02.

(4.13)

Then there exists c5 > 0 such that V f E Z1 IIfIIz2 0 such that V f E Z CO IIf112 0 such that 'If E Z1(f2) IIfIIzs(n) < cs II/IIz,(n)

(4.23) (4.24)

and, hence, (4.23) is equivalent to ZA(0) C4 Z2(11).

(4.25)

4.1. EMBEDDINGS AND INEQUALITIES

125

Idea of the proof. Apply Corollary 2. Proof. If fk E Z1(12), k E N, gl E ZI(f ), 92 E Z2(S2), and

urn fk = gi in Zi(I), k fk = 92 in Z2(Q),

(4.26)

then by (4.22)

lim fk = gl in L"(9),

k-+oo

lim fk = 92 in L1°`(SI) kloo

(4.27)

and gi - g2 E 9(e) = 9z,(n). Hence, (4.24) follows from (4.13) From (4.21) it follows that Z2(S2) + Bz,(n) = Z2(St). Moreover, for each g E Z2(11), for which g- f E Oz,(n) we have II9IIz2(n) = IIf1Iz2(n) Consequently,

(4.23) coincides with (4.6), (4.24) coincides with (4.7) and, hence, (4.23) is equivalent to (4.25).

Corollary 4 Let 9 C R" be an open set and Z(i2) a semi-normed vector space equipped with two semi-norms and complete with respect to both of them.

Moreover, suppose that conditions (4.19) and (4.21) are satisfied. Then these semi-norms are equivalent. Idea of the proof. Apply Lemma 3 to the semi-normed vector spaces Z1(il) and Z2(0), which are the same set Z(1l), equipped with the given semi-norms. Finally, we collect together all the statements about equivalence of inequalities, embeddings and continuous embeddings in the case of Sobolev spaces.

Theorem l Let l E N, m E No, m 0 independent of f. 2. If b - a = oo, then inequality (4.33) is equivalent to ii;(m)1LP(a,b) (b - a)"-'". This is equivalent to Iifw')IILp(a,b) 0. It follows that for each function f whose derivative f' is locally absolutely continuous If'(x)I 0 and applying Definition 4 of Chapter 1. In the sequel we shall need a more general inequality: for 1 < p:5 oc 1(1-1)

11+1 IIfwIIL,(-oo,oo)

(4.52)

To prove it we apply Holder's inequality to the integral representation (3.24) and get If'(x)I 0 there exists C(c) such that

5 C(c)

IllIILp(a,b) + e

for each f E W, (a, b). Consequently, Vk, s E N II(fk)y,'") - (fs);p"`)IILp(a,b) 0 we choose a in such a way that 2eK < 6. Since fk is a Cauchy sequence in Lp(a, b), there exists N E N such that C(e) Ilfk - fsIILp(a,b) < s if k,s > N. Hence, bk,s > N we have II(fk)W'") - (fs)w'")IILp(a,b) < 6, i.e., the sequence (fk)(W'") is Cauchy in Lp(a, b). Because of the completeness of Lp(a, b) there exists g E Lp(a, b) such that (fk)(W") -+ gas k -+ oo in Lp(a, b). Since the weak differentiation operator is closed (see Section 1.2), g is a weak derivative

of order m off on (a, b). Consequently, fk -4 f in Wpm (a, b) as k -+ oo. 2. By (4.41) and (4.43) it follows that dk, s E N II(fk)y,'") - (fs)w'' llL; (a,b) 0, and, in particular, f (x) = (x - a) e, then there is equality in inequality (4.65) and, consequently, in (4.59). Let f > 0 and f < 0 on (a, oo) in the case of inequality (4.60). Then by Remark 13 there is equality in the first inequality (4.66). Furthermore, if n_1)p' (- f')P = M2(f on (a, oo), M2 > 0, then there is equality in the second ii Let f and g be measurable on (a, b) and 1 < p < oo. The equality IIf9IIL,(a,b) = IIfIIL,(a,b)II9IIL,,(a,b)

holds if, and only if, Alf I° = B191° almost everywhere on (a, b) for some nonnegative A and B.


142

inequality (4.66). All solutions f E Lp(a, oo) of this equation have the from f (x) = e-"(x-a) for some a > 0. A more sophisticated argument of similar type explains the choice of testfunctions f (x) = 1 + E(y - a)" in the case of inequality (4.57). Corollary 10 (inequalities with a small parameter multiplying the norm of a derivative) Let -oo < a < b < oo,1 < p < oo. 1) If -oo < a < b < coo, then Ve E (0, Eo), where Eo = ((p' + 1) If IIL (a,b) 5 W+ 1) °E

2) If -oo < a < b = oo, then

IIfIIL°(n,b)

(b - a)) ', (4.67)

+EIIfwIIL°(a,b)

VE E (0, oo)

(4.68)

If (a, b)

_

(-oo, oo), then Ve E (0, oo) (p')-'(2E)-

(4.69)

IIfIIL°(-aa,aa)+6IIfwIIL°(-aa,aa)

The constants in inequalities (4.68) and (4.69) are sharp.

Idea of the proof. Apply the proofs of inequalities (4.40) and (4.42). Verify exp(-`-dr that there is equality in (4.68) for f (x) = -6) and in (4.69) for

f (x) = exp (-

Qa

r II) . See also Remarks 15 -16 and 18 below. C3

Corollary 11 (multiplicative inequalities) Let -oo < a < b:5 oo, 1

a, and F(x) _ f (2a - x), if x < a. Then 00 P

IIfIILo (0,00) = IIFIIL-(-00,00) 0 and in inequality (4.84) IIfIIW,(a,b) can be replaced by Ilfw')IIL,(a,b)

Idea of the proof. To prove (4.81) apply Corollary 12 and Holder's inequality if

b - a < oo and the inequality IIfIIL,(a,b) s IIfIIL,(a,b)IIf11L1 (a,b),

whereifb-a=ooandp 0, then from (4.83) or (4.84) it follows that fk, -i f in WW (a, b).

2. Ifb-a W(2) - MI! - 91.

4.3. CLASSES OF OPEN SETS

151

(For, if p(y') < gyp(.t) - Ml - 91, then (9, sp(y)) E 0.) Similarly, Kz n W C` Q implies that cp(y) < ap(t) + MI± - 91 and (4.89) follows.

We note also that the tangent of the angle at the common vertex of both cones K= and K= is equal to may.

Example 5 Let 52 = {(x1,x2) E R2 : x2 < cp(xl)}, where tp(x1) _ Ix, 17 if x1 < 0, cp(x1) = xi if x1 > 0 and 'y > 0. Then the function cp satisfies a Lipschitz condition on R if, and only if, y < 1, while 0 has a Lipschitz boundary in the sense of the above definition for each y > 0.

Example 6 Let y > 0. Both the domain 521 = { x E R" : 1xI" < x,, < 1, Ixl < 1 } and the domain 522 = { x E R" : -1 < x,, < 1x1 < 1 } have a Lipschitz boundary if, and only if, y > 1. (Compare with Examples 6 and 7 of Chapter 3.)

Lemma 4 If an open set 52 C R" has a Lipschitz boundary with the parameters d, D, x and M, then both 52 and `S2 satisfy the cone condition with the parameters r, h depending only on d, Al and n. Idea of the proof. Let z E Vj n 852 and x = ad(z). Consider the cones K= and K= defined in Example 4, where 1P is replaced by tp,. Proof. By Example 4 we have K= n A, (V;) c.\;(V; n 52),

K= n a; (V,) c A, (V; n' 52).

By 1) bid - aid > 2d and (4.90) implies that there exist r, h > 0 depending only on d, M and n such that ) (V, n c) and \j(Vj n` S2) satisfy the cone condition with the parameters r and h. (The cone condition is satisfied for the largest cone with vertex the origin, which is contained in the intersection of the cone K(d, defined by (3.34) and the infinite rectangular block x1, ..., x"_1 > 0). SinceM) aj is a composition of rotations, reflections and translations, the sets Vj n 52, Vj n` S2 and, hence, the sets 52 and `S2 also satisfy the cone condition with the parameters r and h.

Example 7 Let 52 = Q1 U Q2, where Qt and Q2 are open cubes such that the intersection iU1 n Z72 consists of just one point. Then both i and `S2 satisfy the cone condition, but the boundary of S2 is not Lipschitz. (It is not even resolvable.)

Lemma 5 A bounded domain i C R" star-shaped with respect to the ball B C Cl has a Lipschitz boundary with the parameters depending only on diam B, diam 52 and n.


152

Idea of the proof. Apply the proof of Lemma 2 of Chapter 3 and Example 1. 0 Proof. Let SI be star-shaped with respect to the ball B(xo, r) and z E BSI. We consider the conic body V. = U (y, z) and the supplementary infinite cone yEB(xo,r)

V= _

U

(y, z) , where (y, z)" _ {z + Q(z - y) : 0< @< oo} is an open ray

yE B(xo,r)

that goes from the point z in the direction of the vector yd. Then V2 CSI and by the proof of Lemma 2 of Chapter 3 V2 C` N. Without loss of generality we assume that the vector x is parallel to the axis Ox, hence, z = (xo, zn), where to = (xo,1, ...) xo,n-1) and z,+ > xo,,,, and consider the parallelepiped U2 = {y E R" : xo,n < y,, < 2zn - xo,n, y E U,}, where U= = {y E Rn-1 : jy; - xo,;j < Z, i = 1, ..., n - 1}. Then Vp E U, the ray that goes from the point (y, xo,n) in the direction of the vector x intersects

the boundary 8) at a single15 point, which we denote by y = (y, p(p)). In particular, p(I) = zn. Since the tangent of the angle at the common vertex of Vs and V2 is greater where R2 = max Ixo - yl, it follows (see Example 1) that than or equal to VEOn

y E U=. We note that if 9 E U,, then the conic body V, contains the cone Ky with the point y as a vertex, whose axis is parallel to Ox,, and which is congruent to the cone defined by (3.34) with the parameters xo,n. Moreover, V. contains the supplementary infinite cone Ky. The tangent of the angle at the common vertex of these cones is equal to r 2 iP OT x0.n

2(p(s)-xo,n+ ? Ir-flU

4R,

x-

U.4

Consequently (see Example 1), 1'P (.t) - V(9)

Moreover, since V, C a and Vs C` SI, we have x0,, + z < V(2) < 2zn - xo,n z 2 E Us . We note also that B(z, E) C Us C B(z, (Rl + (n - 1)2(x)2){).

(4.94)

Finally, we consider a minimal covering of R" by open balls of radius (Its multiplicity is less than or equal to 2".) Denote by B1, ..., B, a collection of those of them, which covers the 6-neigbourhood of the boundary BSI. Each of e.

'5 Suppose that q E 8fl, rr # y and Q = P. If nn > yn, then y E V,r C fl. If nn < yn, then y E Vs C° 3f. In both cases we arrive at a contadiction since y E M.

153


these balls is contained in a ball of the radius z centered at a point of 852. Since Vz E 8Sl we have U= D B(z, a), we can choose Us ..., Us,, where zk E 8i2 in such a way that UEk D Bk. Consequently, the parallelepipeds UZ, , ..., U, cover the B-neigbourhoods of 852. From (4.94) it follows that the multiplicity of this (2)2)

covering does not exceed x = 2' (1 + r 15 of Chapter 3.)

+ (n - 1)2

in )

.

(See footnote

Thus, 0 has a Lipschitz boundary with the parameters d = e, D =

diam il, M = 4 f

< 4D and x.

Lemma 6 1. A bounded open set it C R" satisfies the cone condition if, and only if, there exist s E N and elementary bounded domains 12k, k = 1, ..., s, with

Lipschitz boundaries with the same parameters such that S2 = U Stk. k=1

2. An unbounded open set 52 C R" satisfies the cone condition if, and only if, there exist elementary bounded domains ilk, k E N, with Lipschitz boundaries

with the same parameters such that 00

1) iz = U Qk, k=1

and 2) the multiplicity of the covering x({ilk}k 1) is finite.

Idea of the proof. To prove the necessity combine Lemma 4 of Chapter 3 and Lemma 5. Note that if the boundaries of the elementary domains ilk, k = !-,s are Lipschitz with the parameters dk, Dk and Mk then they are Lipschitz with the parameters d = inf dk, D = sup Dk, M = sup Mk as well if d > 0, D < 00 k=1s

k=1,-s

k=Ti

and M < oo. To prove the sufficiency apply Lemma 4 and Example 5 of Chapter

3.0 Remark 19 If in Lemma 6 ilk are elementary bounded domains with Lipschitz boundaries with the same parameters d, D, M, then fl satisfies the cone condition with the parameters r, h depending only on d and M. Remark 20 If we introduce the notion of an open set with a quasi-Lipschitz boundary in the same manner as in the case of a quasi-continuous boundary, then by Lemma 6 this notion coincides with the notion of an open set satisfying the cone condition. If we define an open set satisfying the quasi-cone condition,

then this notion again coincides with the notion of an open set satisfying the cone condition.


154

Lemma 4 of Chapter 3 and Lemma 6 allow us to reduce the proofs of embedding theorems for open sets satisfying the cone condition to the case of bounded domains star-shaped with respect to a ball or to the case of elementary bounded domains having Lipschitz boundaries. To do this we need the following lemmas about addition of inequalities for the norms of functions.

Lemma 7 Let mo E N, 1 < p1, ...pm, q < oo and let f? = 6 Slk, where Slk C k=1

max pm ands E N or s = co

R" are measurable sets, s E N for q
0, m = 1, ..., mo, for each k mo

II9IIL,(nk) 1 and

oM

k=1 <s

xD

IIfIIL,,,,(n) 0 such that inequality (4.102) holds. Idea of the proof. Apply Lemma 4 and, if SZ is unbounded, Remark 7 of Chapter

3 and Corollary 13. 0


159

Proof. Let 1 satisfy the cone condition with the parameters r, h. By Lemma 4

and Remark 7 of Chapter 3, St = U 1k, where s E N for bounded 1, s = 00 k=1

for unbounded Sl, and S1k are bounded domains star-shaped with respect to the balls Bk C Wk C Stk. Moreover, 0 < M1 < diam Bk < diam Slk < M2 < oo and x({Slk}k=1) < M3 < oo, where MI, M2 and M3 depend only on n, r and h. If Sl is bounded, then IIDwfIILq(nk) 0 such that inequality (4.102) holds. Idea of the proof. Apply Lemma 6, Remark 19 and the proof of Lemma 9.


160

Proof. Let Q satisfy the cone condition with the parameters r, h. By Lemma a

6 and Remark 19, St = U IZk, where s E N for bounded Q and s = oo k=1

for unbounded St. Here Qt are bounded elementary domains with Lipschitz boundaries with the same parameters d, D, M depending only on n, r and h. Moreover, x({SZk}k=1) < Mg, where M3 also depends only on n, r and h. If SZ is bounded , then as in the proof of Lemma 10 we have inequalities (4.103) and

(4.104). Let fl be unbounded. Suppose that n, 1, m, p, q are fixed (p < q). Then in (4.101) c26(G) = c28(d, D, M). Hence, Vk E N IIDWf IILq(nk) < c26(d, D, M) If II W;(nk)

and, by Corollary 13,

I

IID.'f IILq(nk) 5 M c28(d, D, M) IIf Ilwgn) = cze(r, h)11f11W;(n)

4.4

Estimates for intermediate derivatives

Theorem 6 Let I E N, 8 E N satisfy 1(31 < l and let 1:5 p:5 oo. 1. If 11 C R" is an open set having a quasi-resolved boundary, then Vf E WW (n)

IIDwfIIL,(n) 5 cas Ilfllwl(n),

(4.105)

where c28 > 0 is independent of f. 2. If SZ C R" is a bounded domain having a quasi-resolved boundary and the ball B C Sl, then df E wP(S2) IIDmfIIL,(n) 0 is independent of f. 3. If 11 C R" is a bounded open set having a quasi-continuous boundary, then be > 0 there exists cgo(e) > 0 such that df E WW(S1) (n) +e llfllw;(n)

(4.107)

Idea of the proof. Apply successively the one-dimensional Theorem 2 to prove

(4.105) and (4.106) for an elementary bounded domain Q with a resolved boundary. In the general case apply Lemma 9 and the proof of Lemma 7. Deduce inequality (4.107) from Theorem 8 and Lemma 13 below.

4.4. ESTIMATES FOR INTERMEDIATE DERIVATIVES

161

Proof. 1. Suppose that S2 is a bounded elementary domain (4.87) with the parameters d, D. By inequality (4.35) it follows that VQ E N satisfying IQI < 1 and V2 E W

MI \ ll(De, f)( x,

'

s

a

) + II \ 8xn w

\Dw.f

(

x'

) II

J'

0 < 6 < 2 and MI depends only on 1, p, b and D. (We recall that +p(2) - an < D.) where Q = (.8 1,

By the theorem on the measurability of integrals depending on a parameter1 ' both sides of this inequality are functions measurable on W. Therefore, taking Lp-norms with respect to ± over W and applying Minkowski's inequality for sums and integrals, we have IID ,f IIL,(n) an and set or = (x0,1, ..., xo,n_1 i an), 6 = min{ , f). Then Q(xo, 6) C B(xo, r) and the parallelepiped G = {x E Rn : 1 xj - xoi 1 < 6, an - 6 < xn < xo,n + b} C S2. Applying inequality (4.108) in the case, in which ,0 = 0, p = 1 and SZ, Q(o, 6) are replaced by G, Q(xo, 6) respectively, we get IllIIL,(c) -< M4 (11f11L,(Qt=o,o)) + Ilf ll.,(c))

5 M5(IIIIIL,(B)+IIf114(c)),

where M4 and M5 are independent of f. Hence, from (4.108) it follows that (4.105) holds for each Q E No' satisfying 1131 < 1.

4. From step 3 and the proofs of Lemmas 7 and 9 it follows that for each bounded domain 1 having a resolved boundary

(t s

IIDwflIL,(n) 5 Ms

11f11 L,(8;) + I1f111,(n)),

(4.109)

j=1

where s E N and Bj are arbitary balls in n fl Vj. (Vj ands are as in the definition of SZ having a resolved bondary.)

Let the ball B c fl. We choose m E N and the ball B0 in such a way that B0 C B fl Cl fl V,n. By step 3 and Holder's inequality it follows that II!IIL,(Bm) b", j E N and00F, mess (1 fl V;) < x meal 0. 1=1

The last example shows that for bounded open sets having a quasi-resolved boundary inequality (4.107) does not necessarily hold. 00 Example 11 Let 1 < p:5 oo, w = U (2-(2a+1), 2-2s) and fl = { (xl, x2) E R2 :

s=0

xiEwif05X2 0 such that 119'I1Lp(u) 0. By Lemma 3 and Remark 7 of Chapter 3 0 = U S1k, where each k

S1k is a domain star-shaped with respect to a ball of radius rl whose diameter does not exceed 2h, and the multiplicity of the covering does not exceed 6"(1 + p)n. By (4.112) and Holder's inequality it follows that for all k

(h) n-1

((h )'-1+" IIDwfIILp(flk) eo and ea < E < co. Let c32 = Then IIDwf lILp(n) s X32 (e0)-)

(E _< X32

\ E*/' Q

E-'

IllIILp(fl) +EO IllIlw;(fl), If IILp(n) + E 11f 11."(n),

and (4.113) again follows. Finally, inequality (4.114) follows from (4.113) in the same way as inequality (4.41) follows from (4.40). 19 We apply the formula

f 9(IxI) dx = on f 9(e)Qn-1 d9, B(O,r)

0

where on is the surface area of the unit sphere in R.

(4.116)


168

Corollary 16 Let 1 E N, Q E N satisfy 0 < IQI < 1 and let 1 0andbf EW,(B,) II Dwf II L,(B,) 0 is independent off and r. Idea of the proof. Apply (4.112) where B = SZ = B?.

R.emark 22 The statement about equivalence of this inequality, the relevant inequality with a parameter and the multiplicative inequality, analogous to the one-dimensional Corollary 7, also holds.

Remark 23 Let I E N, m E No, m < 1. By Section 4.1 inequality (4.105) for all ,Q E No satisfying 1,31 = m is equivalent to the embedding WW (1l) Ci W742)

.

(4.118)

Next we pass to the problem of compactness of this embedding and start by recalling the well-known criterion of the precompactness of a set in Ln(f2), where SZ C R" is a measurable set and 1 < p < oo. We shall write fo for the

extension by 0 of the function f to R": fo(x) = f (x) if x E 0 and fo(x) = 0 if x 0 f2. The set S is precompact if, and only if, i) S is bounded in L,(f)), ii) S is equicontinuous with respect to translation in Ly(ft), i.e., lim sup I)fo(x + h) - f (x) IIL,(n) = 0 h-0 fES

and iii) in the case of unbounded fl, in addition, l iM )ES

IIf II L,(n\Br) = 0

Lemma 12 Let I E N0, 1 < p < 00. Moreover, let fl C R" be an open set and S C W,1(0). Suppose that 1) S is bounded in Wa(ft), 2) all suupp IIf IIw,,(n\n,) = 0,

3) aim sup IIf (x + h) - f (xIIw:(nihi) = 0 fES

and 4) in the case of unbounded Cl, in addition, lim sups IIf II wa(n\B.) = 0. r-+00 fe

Then the set S is precompact in WP(C).

4.4. ESTIMATES FOR INTERMEDIATE DERIVATIVES

169

Idea of the proof. Apply the inequality Ilfo(x + h) - f (x)IIL,(sz) 0 and S = (f e Wp(12) : IIf Ilw;(n) < Ml}. By inequality (4.79) for almost all x E W

and 0 < y < d 7°II1(x, < M,-y1

IIa\).(I,

)IILv(m(s)-d,w(s))

where M2 is independent of f and y. By the theorem on the measurability of integrals depending on a parameter (see footnote 17) both sides of this inequality are functions measurable on W.


170

Therefore, taking LP -norms with respect to x over W and applying Minkowski's inequality, we have V f E S 16

IIfIIL,(:z\(n-rya.))

M27" IIlIIWW(n) llfk - f3IIL,(n) > 2P. Hence, again every subsequence of {fk}kEN is divergent. If measf1k < oo, one may just set fk = (messfIk) on Stk. If mess Qk = oo, let fk(x) = n( 4) on cck, where r) E Co (R" ),'7(x) = 1 if lxl < 1, and rk > 1 are chosen in such a way that mess (f k f1 B(0, rk)) > 1 . A more sophisticated example shows that embedding (4.118) for bounded domains having a guasi-resolved boundary can also be non-compact.

Example 14 Let 1 < p < oo and 0 be the domain in Example 11. Then the azk+l embedding WP (Q) C- Lp(S2) is not compact. For, let fk(x1, x2) = 2 x2 if 2-(2k+1) < x1 < 2-2k and fk(x1, x2) = 0 for all other (x1, x2) E fI . Then the sequence { fk}kEN is bounded in Wp (1): IIfkIIW;(n) = 1 + (p + 1)-P. However, it does not contain a subsequence convergent in Lp(S2) since Ilfk - fmllt,(n) _

2P (p+1)-P ifkAm.


172

Lemma 13 Let l E N, m E No, m < 1, 1 < p, q < oo and let Sl C R" be an open set. 1. If the embedding (4.120) W,(0) C.; WQ (I) is compact , then de > 0 there exists c34(e) > 0 such that Vf E Wp(sl) Ill Ilwa (n) 0 (4.121) holds and the embedding W,()) C; Lp(Q) is compact, then embedding (4.120) is also compact.

Idea of the proof. 1. Suppose that inequality (4.121) does not hold for all e > 0, i.e., there exist co > 0 and functions fk E Wp(sl), k E N, such that Ilfkllw;(n) =1 and IIfkIIW; (n) > kIIfkIIL,(n)+e0llfkIIw(n)

Obtain a contradiction by proving that lim

(4.122)

= 0 and, consequently,

liminfllfkllw, (n) 2! co.

2. Given a bounded set in Wp(1), it follows that it contains a sequence { fk}kEN convergent in L,,()). Applying inequality (4.121) to fit - f, , prove that iim Ilk - fs II W; (n) = 0. 13 k

Proof. 1. Since IIfkIIw,.(n) = 1, by (4.120) it follows that IIfkIIw, (n) - co lim inf IIfkllVl (n) = CO. k-+00

k-.o0

P

Since embedding (4.120) is compact, there exists a subsequence fk, converging to a function f in W, ,'n(11). The function f is equivalent to 0 since fk, -r 0 in Lp(O). 20 This contradicts the inequality II!lIW; (n) = lim IIfkIIW (n) >- co

2. Let M2 > 0 and S = { f E Wp(fl) : Ilf IIw,(n) 0, take c =

Since fk is Cauchy there exists N E N such

Thus, Vk,s > N

that Vk,s > N we have Ilk - faII L,(n) < 2 (c34 (.L))

Ilk - fall w, (n) < b, i.e., the sequence fk is Cauchy in Wq (SZ). By the completeness of WQ (SZ) there exists a function f E W, '(Q) such that fk -> f in W."' (SZ) as k -a oo.

Corollary 17 Let1EN,mENo,m 0 depends only on n and 1.

Idea of the proof. Apply the Leibnitz formula, Holder's inequality and Theorem 6. E3

Proof. If fk E c- (n) fl Wpk (Q), k = 1, 2, then starting from (C') D"f1 D°-0f2,

D°(f1 f2) = 0 0 is independent of f. Idea of the proof. Direct application of the proof of Lemma 15. Lemma 16 Let I E N, 1 < p < oo and let SZ C 1R" be an open set with a quasiresolved boundary. Moreover, let g = (91,...,g") ft -+ R", gk E C'(52), k = 1, ..., n. Suppose that Va E Z satisfying 1 < IaI 0. Furthermore, let g(fl) be also an open set with a quasi-resolved boundary. Then V f E Wp(1) C37IIfIIww(9(n)) 0 are independent off and p.

Remark 24 By the assumptions of the lemma on g it follows that there exists the unique inverse transform g('1) = (9(1-1), ..., 9n(-')) : g(f2) - Sl such that

gk-1) E C'(g(cI)), k = 1,...,n. Moreover, Va E K satisfying 1 < IaI 0. YE9(n)

Idea of the proof Apply the formula for derivatives of f (g), keeping in mind that for weak derivatives, under the assumptions of Lemma 16 on g, it has the same form as for ordinary derivatives, i.e.,

Dw(f(9)) = E (Dmf)(9)

D"g...D71o19,

(4.127)

01

where ryk E NJ and cp,.,......,., are some nonnegative integers. Apply also Theorem 6 of Chapter 4. Proof. Let a E 1Vo and IaI = 1. By (4.127), Minkowski's inequality and Theorem 6 it follows that

IID.*(f(9))IIL,(n) < M1 E II (Dwf)(9)IIL,(n) = (y = 9W) 0:50,101>1


176

M,

II(Dff)(y)IDx(g-1(y))I

IILp(9(n))

d1

< M2 > IIDWfIILp(9(n)) :5 M3 II f Ilwa(9(n)) 95x,181>_1

where Ml, M2i M3 > 0 are independent off and p. Hence, the second inequality (4.126) is proved in a similar way. 0

Remark 25 From the above proof it follows that p C38

C39 (inf I Dx (x) I)

IIDagIIc(n),

1maa 1, 1 < p < oo and let 0 C R" be an open set satisfying the cone condition. Suppose that f E Lp(S2), the weak derivatives (e ) w, j = 1, ..., n, exist on I and are in Lp(SZ). Then '13 E N satisfying (aI = l the weak derivatives L), Of also exist on 0 and II Dwf IIL,(n) 0 depends only on n, p and q.

Remark 28 By applying inequality (4.134) to f (ex), where f E Lp(R" ), f * 0, is fixed and 0 < e < oo, one may verify that if A (4.134) does not hold for any choice of c41.

n(-L + 1), then inequality

We give a sketch of the proof of Theorem 10 based on the properties of

maximal functions. 22

Lemma 17 Let n E N, p < n. Then for all functions f measurable on Rn

VxER" andVr>0

f

Ix - yl-"If (y)I dy

C43 r"-µ(M f)(x),

(4.136)

B(z,r)

where c43 > 0 depends only on n and p. For f E LiOO(R") the maximal function Mf is defined by

(M j) (x) =

x E R". r >o mess B(x, r) JB(z,r) If d11,

For almost all x E R" If (x)I < (Mf)(z) < oo. Moreover, M f is measurable on R" and for 1 0 such that V f E La (R" ) IIMf IIL,(R,) < C42 Ilf IIL,(R-).

(If p = 1, this inequality does not hold.)

(4.135)

179

4.5. HARDY-LITTLEWOOD-SOBOLEV INEQUALITY

Idea of the proof. Split the ball B(x, r) into a union of spherical layers S(x, r2-1) = B(x, r2-k) \ B(x, r2-k-1), k E N0, and estimate f If I dy S(x,r2-k)

via the maximal function MI.

Lemma 18 Let n E N, 1 0 depends only on n, p and µ. IxI-µ * f into an integral over Idea of the proof. Split the integrals defining B(x, r) and an integral over `B(x, r). Applying inequality (4.136) to the first

integral and using Holder's inequality to estimate the second one via IIf IIL,(R^),

establish that IxI-µ * f I < Ml r"-µ(Mf)(x) + M2 r-(µ-P)IIf IIL,(a ),

where M1, M2 depend only on n, p and µ. Finally, minimize with respect to r. Idea of the proof of Theorem 10. Apply inequalities (4.137) and (4.135). Proof. Since (Mf)(x) < oo for almost all x E R", the convolution IxI-A * f exists, by (4.137), almost everywhere on R" and, by Remark 27, is measurable on R". Since Mf is also measurable on R", taking Lq-norms in inequality (4.137) and taking into account (4.135), we get

-a II IxI

* fIIL9(Rn) s C44IIf'IL,(Rn) IIMfIIL,(Rn) e. Then f E LL(R"), but IxI

IxI_ p (1n IxI)-1 if

For this reason we consider the case in which 1 < p < oo for functions in Lp(R") with compact supports.

Theorem 11 Let 1 < p < oo, f E Lp(R"), f x 0. If Q < -L, then for each compact SI C R"

fQIXl* f I l/ I

Ill IIL

,(R"d--

< oo.

(4.141)

Idea of the proof (in the case j3 < Vft ). Suppose that if IIL,(R*) = 1, the case in which IIf IIL,(R^) # 1 being similar. Following the proof of Lemma 18, establish the inequality

I IxI-A * f 15 M1 rp (Mf)(x) + (Q" I lnri)' +M2i where 0 < r < 1 and M1, M2 depend only on n, p. Take r = and apply inequality (4.47). 0

(X))P)- In

25 Let E, F C II" be measurable sets, k be a function measurable on E x F and (K f)(y) _ f k(x, y) f (y) dy. Then for 1 < q < 00 E

IIKIIL,(F)_,L,(E) = II IIk(z,y) IIL,,.(E) IIL,,. (F)

(4.139)

IIKIIL,(F)NL,(E) = IIIIk(x,Y)IIL,,,,(E) IIL..,,(F)

(4.140)

and for 1:5 p:5 00

4.6. EMBEDDINGS INTO THE SPACE C

181

Corollary 19 If 0 < µ < p', then b'Q > 0

fexp (t

lxr_a

* fI")

dx < oo.

A

Idea of the proof. Apply the elementary inequality aµ _< 6a > 0, 6 > 0, which follows from (4.47).

+ BaP', where

Remark 31 If 33 < e1 , there is a simpler and more straightforward way of proving inequality (4.141), based on expanding the exponent and application of Young's inequality for convolutions (4.116).

Example 17 Let 1 2. Then f E LP(W') and M, I In IxIl < jxI -d * f < M21 In IxII", 0 < jxj

i,

where Ml, M2 > 0 are independent of x. Idea of the proof. To obtain the lower estimate it is convenient to estimate jxj- * f from below via the integral over B(0,1) \ B(0, Y). To get the upper estimate one needs to split the integral defining 1x1-A * f into integrals over B(0, \ B(0, 21x1), B(0, 21x1) \ B(x, ll) and B(x, and to estimate them 2) z) separately.

Remark 32 This example shows that the exponent p' in inequality (4.141) is sharp. Indeed, if µ > p', then for < ry 0 and for each compact fl C R".

A more sophisticated example can be constructed showing that for 6 > n Theorem 11 does not hold.

4.6

Embeddings into the space of continuous functions

Theorem 12 Let I E N, 1 p for 1 P, then applying (4.116), we have Vx E K II Ix - YI` "IIL,,.,(K) < II IZI'

"IIL,,(B(O,D))

D

a" _ an f e(!-n)p'+n-1 dLoJ i = ( Alp) 0

Df- n

4.6. EMBEDDINGS INTO THE SPACE C

183

If p = 1 and 1 > n, then II Ix - yI` "IIL-.r(K) 1,1 < 1 < a and f E Co (R"). We define the - iq,_,). exponents q1, ..., qq-1 by 1 =9jn(1 9- 1), 1 = n(1 0 - 9, ), ..., 1 = n(1 D

i

Applying (4.155) successively, we get n

of

IIfIIL,.(R")_<M1 m,=11

...

n

n

< M1

II 8xm,IL,I(R")
t}. Clearly f' (x) = g(Ix1). The following properties of f' are essential: IIf'IIL,(R") = IIf IIL,(R") , 15 P:5 00

(4.156)

Ilowf'IIL,(R") 0 is independent off and E. Furthermore, 3(i-m-n(n-y))

IIDwf IIL,(n) 0 is independent of f. If 0 = R" or, more generally, Q is an arbitrary infinite cone defined by

Q={xEW:x=pv,0<e 0, IIDwfIIL,(n) 5 IIDwTfIIL,(R^) <M1y-'IITIIo

m + n (D - a) also follows for 1 < p,q < oo and f1 = R" by comparison of the differential dimensions of spaces WW(R") and Wa (R"). See footnote 14 of Chapter 1. With slight modifications a similar argument works for open sets f1 34 R".


196

If SZ is unbounded, consider, as in step 1, the disjoint balls B(xk, p) and

set ft(x) = (!). As in Example 1, fk does not contain a subsequence convergent in Wy (S2).

Next let us consider in more detail the case I = p. By Theorem 14, for an open set 0 satisfying the cone condition, it follows that Wp (S2) C L,(S2) for a each p < q < oo. However, Wp (0) ¢ L,,. This statement may be improved in the following way.

Theorem 15 Let 1 0 depending on n, p and the parameters r, h > 0 of the cone condition such that V f E Wp (S2), f x 0,

J

exp (c541

IIf II

fa

JP)

dx
g° in Ll°`(R") as k -+ oo and, hence, for the function h, defined by h(u, v) =gn(u), (u, v) E R", we have fk(.,v) -* v) in L'°C(Rm) for all v E B1. f v) On the other hand, v E B1. This follows since by the Fatou and Fubini theorems and condition 1)

J

(liminfJ Ifk(u,v)- f(u,v)Idu)dv BN

B1

< lim inf f k-+oo

(f I fk(u, v) - f (u, v) I du) dv

B1

= lim

k- oo

J

B,v

Ifk(u,v)-f(u,v)Idudv=0.

BN x B1

Thus f v) is equivalent to v) on Rm for almost all v E B1. Consequently, by Fubuni's theorem, 2 f is equivalent to h on Rm x B1. Furthermore, by the continuity of a semi-norm, on letting s -+ oo in (5.2), where f is replaced by fk - f we get Ilfk(', v) -

v)II L,(K) 5 c1(K)Ilfk - f IIZ(R").

2Fbr, let e" = {(u,v) E Rm x 61 : f(u,v) # h(u,v)} and e, (v) = {u E Rm : f(u,v) # h(u,v)}. Then meas"e" = f (measme,"(v))dv = 0. B,

CHAPTER 5. TRACE THEOREMS

200

Therefore

Ilh(', v) - 9IIL,(K) = Ilh(', v)

- h(., O)IIL,(K) n-m P

l > n - m for p=1,

(5.4)

i.e., if, and only if, Wy(R"-m) C3 C(R"-m).

Idea of the proof. If (5.4) is satisfied, write the inequality corresponding to embedding (5.5) for functions f (u, ) with fixed u, and take Lp norms with respect to u. Next use Theorem 1. If (5.4) is not satisfied, starting from Example 8 of Chapter 1, construct counter-examples, considering the functions fe(u, v) = Ivlp 771(u) 7k(v) if I < n PM and g.1(u, v) = I in lvI I" nl (u) *i2(v) if

I = "pm, 1 < p < oo. Here 77, E Co (R"),, E Co (R"-m) are "cap-shaped" functions such that nl = 1 on B1, rh = 1 on B1, where B1, B1 are the unit balls in Rm, R"-m respectively. Proof. Sufficiency. Let (5.4) be satisfied. First suppose that 1 0, 1 < p, 9 < oo. Then the norms 6 II IIB,,,(R") corresponding to different a E N satisfying a> I are equivalent. Idea of the proof. Denote temporarily semi-norms (5.9) and (5.10) corresponding to a by II II(°). It is enough to prove that II II(°) and II II(°+') are equivalent on L (R") where a > 1. Since Iloh+'fIILp(R") 2IIAAfIIL,(R"), it follows that II

< 211

ll(°)

To prove the inverse inequality start with the case

0 < I < 1, a = 1 and apply the following identity for differences ohf = 2 1&21.f - 2 Ohf,

(5.10)

which is equivalent to the obvious identity 6 x - 1 = 2(x2 - 1) - 2(x - 1)2 for polynomials. To complete the proof deduce a similar identity involving A"f, Ashf and ph+l f. Proof. 1. Suppose that 0 < l < 1 and Ill 11(2) < oo. By (5.11) we have IlohfllL,(R") c

Since 11Ahf IIL,(R^) 0. From (5.12) it follows,

after substituting 2h = 77, that ',(e) < 2'-'V5(2e) + 2-' 11f 1j(2),

and a similar argument leads to the same inequality (5.13). then 2. If (x - 1)° = 2-0(x2 - 1)° + (x - 1)0-2-0 (X2 - 1)° = 2°(x2 - 1)-0 + P°-1(x)(x - 1)°}',

where

P°-1(x) = -2-°(x - 1)-1((x + 1)° - 2°) = -2-0 s=1

Hence,

An f = 2'A &f + P°-,(Eh)Or+l f

3

) (x - 1)8-1.

205

5.3. NIKOL'SKII-BESOV SPACES and, since IIEhIIL,(R^)-*LP(R^) = 1,

II' hf IIL,(R") 0, a E N, a > l,1 < p, 9 < oo. The norm 00

IIfIIBl(R^) = IllIILp(R")+ (f

dt1 1

),

(5.22)

1

0

is an equivalent norm on the space B,,e(R").

Idea of the proof. Since, clearly, Ilohf1lL9(R^) < wo(jhl, f)p, the estimate 11f11 Bp B(R*) < M1II f II B B(R^), where M1 is independent of f , follows directly

by taking spherical coordinates. To obtain an inverse estimate apply inequalities (5.19) and (5.14). Proof. In fact, by (5.19)

0

r f (wa(61f)p)8 l

`0 `

) ) < M2IItn_1-,vn

n

f

InI-"Ilonf IILp(R^) d'jII

L. (50)'

Inl<e

where M2 is independent of f. Since I > 0, the assumptions of Corollary 2 are satisfied and by (5.14) IIf il(l) B' .O R") 0,1 < p, 0 < oo. Then Bp a(R") is a Banach space. 9 Idea of the proof. Obviously B,,e(R") is a normed vector space. To prove the completeness, starting from the Cauchy sequence {fk}kEN in BB,e(R"), deduce, using the completeness of Lp(R") and 10 Lpg(RIn), that there exist functions f E L,(R") and g E Lp,e(R2") such that fk - f in Lp(R") andlhl-I-*fk(x) -+ g(x, h) in Lp,B(R2n ). Choosing an appropriate subsequence {fk. }aEN, prove that

g(x, h) = Ihl-1-I f (x) for almost all x, h E R" and thus fk - f in B, e(R"). 9 See footnote 1 on page 12. 10 Lp,,(R2") is the space of all functions g measurable on Re", which are such that II9IIL,,.(Ra..) = II

00.

5.3. NIKOL'SKII-BESOV SPACES

211

Lemma 8 Let I > 0. The norm =11(1+IKI2`)2(Ff)(OIIL2(R-)

(5.25)

is an equivalent norm on the space B2,2(Rn).

Idea of the proof. Apply Parseval's equality (1.26) and the equality

1)°(Ff)(C) = (tie' 2)° (sin h2

(F(Def))(f) = for f E L2(R"). Proof. Since a is equivalent to

+

\

/ 0, the norm IIIIIez.x(R")

IhI-21IIohf

IIL2(R.)

R"

_

(f R^

f

where

IhI-21-"sin' 2t dh.

R"

If n = 1, then after substituting h = ', we have 00

AI(S) = MIIfI21, M1 = f ItI-2-'sin2o 2 dt < oo, -00

since l > 0 and a > 1. If n > 1, we first substitute h = Afq, where A( is a rotation in R" such that h = If Ini, and afterwards q _'I. Hence An(t) =

J R^

I77I-21 nsin2o

dq = M"ICI21, Mn =

J

ItI-'-"sin2o 21 dt.

R^

If t i , = It11Tk, k = 2,..., n, then ItI = It1I

1 + ITI2, where ITI = (E Tk) k=2

Hence, applying (4.116), we have ao

Mn=M1 f R^-1

1+In2-21 nd-r=M1an-1f 0

1+p2-2-2 de 0,. 1 0, 0 < e < 1, 1 < p, 9, 91, 92 < oo, then

Bpei(R") C P'81

Brpe(R")

C BPI-

Hence the parameter 9, which is also a parameter describing smoothness, is a weaker parameter compared with the main smoothness parameter 1.

Remark 6 If I E N, 1 min {p, 2}, 92 < max {p, 2}, the corresponding embeddings do not hold. Remark 7 The following norms are equivalent to II f I I BD,,(R" ) IIfIIBk)0(R") = IIfIIL,,(R")+IIfII,a

(R")+

k=5,6,7,8,

5.3. NIKOL'SKII-BESOV SPACES where

2(f

II!II ,,(R.) _

213

(IIz

DtifIILa(RI))8 dh I

1a1=m

1

I

I

I

lhlGN

f

o,(n)

ti-

0

J

t

H

-v'(f 1_te)(POW IILa(R^) ! 0dtt

Ilfll() bpl.#(RI) -

ti-m

J=1

0

H

n

R - L(f (

Ilflld'

=1

woj t+

8x

wP

hp(30) B it I It

0

Here m E 1%, m < l < a+m, 0 < H < oo , ee is the unit vector in the direction of the axis Oxi and wQa gyp) is the modulus of continuity of the function cp of order a in the direction of the axis Ox,. If 0 = oo, then, as in Definition 2, the integrals must be replaced by the appropriate suprema. There also exist other eqivalent ways of defining the space B,,e(R ): with the help of Fourier transforms (not only for p = 0 = 2 as in Lemma 8), with the help of the best approximations by entire functions of exponential type, by means of the theory of interpolation, etc.

Remark 8 It can be proved that Wp(Ye)

1 Boo,P (r), l > p, 1 < p < oo,

and

°n(1-1) 0 is independent of f. Idea of the proof. Take spherical coordinates and apply Corollary 6 and Lemma

2.0 Proof. After setting h = pt;, where p =IhI and p = i and applying (5.28) and (5.14) we get t+

I = II IhI-1(°hf)(x)IIL9,h(1t") - I IIp

Il E S", substituting

(A' f)(x)IILp.a(0'-)IILp.E(s"-')

!Q

< (l 11)! II IIp

1+' rr'

(

).(X

drll Lp.e(O,oo)II

0 11-A P)!11111.-1+1+P

r / r'-1I(01 )w(x+ST)I drIILp,.(0,.)IILp.E(S"-1) 0

(l - a)(l - 1)! II IIr 81;1) Since

< nt-1, where a E 1% and Ia! = 1, we have, for almost all x E R", \81;1/w(z)I - I j1=1

j)=1

'9xj+)(z)I

11 If n = I = 1,p > 1, then cls = p' and (5.29) is equivalent to (5.13), where a = 0 and f is replaced by f;,.


216

E a',I(D°f)w(x)I < n-' E I(D°f)w(Z)I. i°I=1

I°I=1

Consequently, by Minkowski's inequality, n1_1l1_av

I

(l - p)(l 1

1

11

1).

101=1

T nD`(

D° f)w( x + T )

3

_ (1-p)(1 D

Lp..(O.oo) Lp.1(S°-1) 1

1)1

I II(D°f)w(x+y)IILp.V(o:') _

(l

11

ID

np)(l

1)!IIfIIw,(').

Theorem 3 Let 1, m, n E N, m oo. By a standard limiting procedure (see, for example, the proof of Theorem 1 t_ n-m of Chapter 4), it follows, since the space Bp ' (RI) is complete, that there g in Bp n '(RI). m By exists a function g E By ' , (R"`) such that Remark 2 the function g is a trace of the function f on R'". Moreover, IltrfIlBD

(RT)

(5.33)

< M IIfIIWW(R")

Since inequality (5.6) is already proved, it is enough to show that the inequality IIf(',0)IIbp

< M2IIfIIWo(R") n-m(Rm)

holds V f E CO0 (R") n W, (R" ), where M2 is independent of f .

2. Let l = 1, m = 1, n = 2, 1 < p < oo and f E C' (R) n Wp(R2). By (5.31) and (5.26) we get II(Au,hf)(u,O)IILp..(R) < II(Av,hf)(u+h,0)IIL,,,,(R) +II(Au,hf)(u, h)IIL,,.(R) + II(ov,hf)(u, 0)IIL,,U(R) < 2 II(Av,hf)(u, 0)IIL,,r(R) + IhI

. II

BL (u, h)IILP.r(R)

Hence, applying Fubuni's theorem and inequality (5.29), we get II Ihl-'II(ou,hf)(u,0) IIL,..(R) IIL,,h(R)

< 2 11II IhI-'(Av,hf)(u,0) IIL,.,,(R) IIL,.,,(R) + II2U-

2p'

2p'

11112(U, v)II L,.,(R)IIL,.%(R) + IIOUIIL,(RI)
1, we apply the identity (Au,hf)(u, 0) _

)' (1) (Ov lhlnf)(u +) h, 0) a=o

- (-1)1

(%X ,hf)(u, Alhlrl),

(5.36)

A=I

where r) E S"-", which also follows from (5.35) if we replace x by E4,h and y by EvJhl,,. Taking spherical coordinates in Rm and using equality (4.116), we get l1f(',0)'1bv

Rm)

M7(IIII

fO'IhInf)(u,0)IILp.h(Rm)IIL,,.(Rm)

IhI-t+

13 This follows from the obvious identity for polynomials

(-1)'(Z - 1)'(v -1)' + (x - 1)'(1 - (-1)'(v - 1)') a=a

\ a)Za(v /

if x is replaced by Ed.h and v by Eh.

(5.35)

A=1

5.4. DESCRIPTION OF THE TRACES ON SUBSPACES

+E

219

(D(7,olf)(u, Alhl77)IIL,,,,(Rm)IIL,,u(Rm))

II II

171=1 A=1

=a.M7(lllle `+°

+::ll lien a-' 171=1 A=1

Here M7 is independent of f. Taking Lp norms with respect to q E and applying inequality (5.29), we get _i

Ilf(,0)Ilbv

"°m(Rm)

Sn-m-1

1

G On°mQn,M7(ll II Ihl-`(oV,h)(U, O)IIL,,h(w m)IIL,,u(R=)

IIII(D(7,o)f)(u,Av)IIL,,(R^-m)IIL,,.(R=))

+ I71=1 A=1

< M8(II 1

Il(D(0,0)f)(u,v)IIL,,,,(R--m)IIL,,,.(RV)

I0I=1

+E

IID(7io)(u, )tv)IIL,(R")) 1. Suppose that the functions A, v E

have

compact supports and satisfy the equality 1a z E Rn.

(5.37)

f (Ohf)(x)A(h) dh = f (Oh f)(x)v(h) dh.

(5.38)

A(z)

- (-1)1_k(k) knv\/, k=1

Then bf E Li°C(Rn) for almost all x E Rn

Rn

R-

14 We note that from (5.37) it follows that f z'A(z) dz = 0, s = 1, ...,1- 1. R"


220

Idea of the proof. Notice that from (5.38) it follows that

f A(h) dh = (-1)1+1 Rn

J v(h) dh,

(5.39)

Rn

expand the difference Oh f in a sum and use appropriate change of variables for each term of that sum. Proof. By (5.37) and (5.38)

f(f)(x)(h) dh Rn

_ E(-1)1-k (k) J f (x + kh)v(h) dh + (-1)h f (x) J v(h) dh k=1

Rn

R.

1 f f(x+z)v(k)dz- f(x)J A(z)dz k=1

=

f

Rn

an

(f (x + z) - f (x))A(z) dz = f ('&hf)(x)A(h) dh. Rn

Rn

Let w E Co (R") and let w6 where 3 > 0 be defined by wa(x) = -nw(17). We the operator defined by A6,W f = w6 * f for f E L a(R" ). (If, denote by

in addition, suppw C B(0,1) and f wdx = 1, then A6,_ = A6 is a standard Rn

mollifier, considered in Chapters 1 and 2).

Lemma 12 Let I E N,1 0 is independent off and 5. Idea of the proof. Notice that

(A6,%f)(x)-f(x) = f (f(x-z8)-f(x))A(z)dz = f (Di:6f)(x)v(z)dz (5.41) R's

15 If I = 1, then A = v and (5.40) coincides with (1.8).

R-


221

since by (5.39) f adz = 1, and apply Minkowski's inequality for integrals and R"

(5.21). 0

Proof. Since the functions J1(-a) and v(-a) also satisfy (5.37), equality (5.38) still holds if we replace µ(h) and v(h) by a(-a) and v(-6). After substituting

h = -zb we obtain (5.41). Let r > 0 be such that suppv C B(0, r). By Minkowski's inequality for integrals and (5.21) IIA` zbf IIL,(R")jv(z)I dz < SUP IIAhf

II A6,Af - f IIL9(R')

Ihl 0 are independent off and b.

Idea of the proof. Inequality (5.42) is a direct corollary of (5.40) because in this case f (A * µ) dx = f A dx f ,u dx = 1. If f µ dx = 0, starting from the R'

R"

Rn

equality

R"

(A6,a.µf)(x) = f ( f (f (x - zb - Cb) - f (x))A(z) dz) /i(C) df, R.

R.

argue as in the proof of Lemma 12. O Idea of the proof of the inverse trace theorem. Define the "strips" Gk by Gk = {v E

R"-m

:

2-k-1 < Ivi < 2-k}, k E Z.

Consider an appropriate partition of unity (see Lemma 5 of Chapter 2), i.e., functions tpk E Co (RI), k E Z, satisfying the following conditions: 0 < k < 1, 00

E '+Gk(v) = 1, v # 0,

kc-0o

Gk C SUPP?bk C {v E 1r_- : 8 2-k-1 < Ivl

2-k}


222

C Gk-1 U Gk U Gk+1

(5.44)

and

I(D7Y'k)(v)I :5C14 2kI"I,

k E Z, v E

Ir-m,

7E

where c14 > 0 is independent of v and k.

L

(5.45)

n

Keeping in mind Definition 2 of Chapter 2, for g E Bp (T9) (u, v) =

N-'",

°m

(R'") set (5.46)

16k(v)(A2-k,W9)(u),

k=1

where (5.47)

w=A*A

and the function A is defined by equality (5.37), in which n is replaced by m and v E C0 (Wn) is a fixed function satisfying 16 f vdu = (-1)1+1 Prove that g is a trace of Tg on RI by applying Definition 1 and property (1.8). To estimate IIT9IIL,(Rn) apply inequality (1.7). Estimate IID°T9IIL,(R^), . To do this where a = (fl, y), Q E IV, -y E Alo-m and IaI = 1, via 1191j(',)--. B, a (R")

differentiate (5.46) term by term, apply inequalities (2.58), (5.42) and (5.43) and the estimate IIDyV)kIIL,(R--m) < C152k(IYI-1

(5.48)

),

where c15 > 0 is independent of k, which follows directly from (5.45). Proof. 1. By the properties of the functions 1Iik it follows that the sum in (5.42) is in fact finite. Moreover, s+1

(Tg)(u,v) _

*k(v)(A2-'%,Wg)(u) on Ii"" x G,

(5.49)

k=s-1

and (Tg)(u, v) = 0 if IvI > 18. Hence Tg E C°°(R" \ R"`) and Va = (Q, ry) where ,B E N , 7 E

N`o-m

00

(D°(T9))(u, v) = F (D7 k)(v)D,6((Az-4w9)(u)) k=1

By (5.39) and the properties of convolutions it follows that f w du = 1. If 1 = 1, then A = v. In this case one may consider an arbitrary w E Co (R'") satisfying f w du = 1. R-


223

00

_ E(D"*k)(v)2k101(A2-k,A.DsA9)(u)

(5.50)

k=1

since, by the properties of mollifiers and convolutions, D0(A2-k,A.A9) = 2k101A2-k,D8(A.A)9 = 2k101A2-k,A.DDA9

2. Let IvI < 16. By (5.44) tik(v) = 0 if k < 0. Hence Ek(v) = 1.

(5.51)

k=1

Let s = s(v) be such that 2-'-' < IvI < 2-'. Then by (5.51), (5.44), (5.42) and Minkowski's inequality 8+1

(T9)(', v) - 90 IIL,(Rm) = II

0A:(v)(A2-k,A.A9 k=s-1

9)IILp(Rm)

s+1

s+1

_

0k(v)IIA2-k,A.A9 - 9IILp(Rm) < MI E wt(2-k, 9)p k=s-1

k=s-i 8+1

< M2

L.r

2k((-"pm)wl(2-k, 9)p

k=s--1

s+1

< M3 I vI t

n pm E

2k(1- p )wl (2-k,

9)p,

k=s-1

where M1, M2, M3 are independent of g and v. Since the function g E BB

° (WTh ), by Lemma 6 it follows that the quantity

e D! )w` (2-k, 9)p -+ 0 as k - oo if 1 < p < oo and is bounded if p = coo. Hence

II(T9)(', v) - 9(')IIL,(Rm) =

o(Ivl'-'),

1 < p < oo

(5.52)

and

II(T9)(', v) -

O(Ivl')

(5.53)

as v -+ 0 (hence s -+ oo). In particular, by Definition 1, if follows that g is a

trace of Tg on R.


224

3. By (1.7) 00 Etkk(v)IIA2-k,W9IIL,(Rm)

2. Therefore (D(°'1)(T9))(u,v) = (D"t l)(v)(A2-lA,.%9)(u), IvI > & Consequently, by (2.58), (5.42), (5.48) IID

(p")

(T9)IIL,(Rmxb

)

0.


229

Lemma 13 Let 1, m, n E N, m < n,1 ' n'" if p > 1 and I > n - m if p = 1. Suppose that A is a nonnegative function measurable on R"-"`, which is such that 1IAIILp(&) = oo for each e > 0. Moreover, let f E I40c(R"), for -f E N_' satisfying I ryl = I the weak derivatives f exist on R" and IIAf IILp(Rn) + F, II D;p°'7)f IILp(R-) < 001-f1=1

Then f IRm = 0.

Idea of the proof. Using the embedding Theorem 12 and the proof of Corollary 20 of Chapter 4, establish that there exists a function G, which is equivalent to f on R" and is such that the function IIG(., v)IILp(Rm) is uniformly continuous

on RI-1. Proof. Let us consider the case p < oo, the case p = oo being similar. By Theorem 6 of Chapter 4 f E LL(R"), hence, f E W,(R") and for almost all u E R' we have f (u, ) E W,(R"'m). By Theorem 12 of Chapter 4 there exists a function E C(R"-'") such that Vv E R"-' Ig4(v)I 0. This is t' 0 2II' < IIafIIL,(R^) < oo. Hence A = 0, i.e., impossible because IIakIL,(a.) = oo and v)IIL,(R-) = 0 and by Definition 1 f IRm = 0. 0 lim

The next theorem deals with the case p = 1, 1 = n - m, which was not considered in Theorem 3.

Theorem 5 Let m, n E N, m < n. Then tr, Wl -'"(R'") = L1(Rm).

(5.69)

Idea of the proof. The direct trace trace theorem follows from Theorem 2 and, in particular, from inequality (5.6). To prove the inverse trace (- extension) theorem it is enough to construct an extension operator T : L1(Rm) -> Wi (R"`+' )

and iterate it to obtain an extension operator T : L1(Rm) - Wl -'(R"). However, it is more advantageous to give a direct construction for arbitrary n > m. Start from an arbitrary sequence {6k}kEZ of posivite numbers 6k satisfying M 6k+1 oo). Thus by Definition 1 g is a trace of Tg on Rm.


232

3. Since (Tg)(u, v) = 0 if IvI > pi + a and pi +a < P0, we have 00

IIT9IIL1(R^) = ll F,0 kilA66,,W911L,(Rm)llL,(R^-m) k=1 00

< M6 II E'Pkll

I19liL,(Rm) < M7µoII9IIL,(Rm) 0, 1 < p:5 oo and b > 0. Then for each open set SZ C R" and V f E B' (Q) satisfying supp f C 06 (5.101)

IIfoIIB;(R") < C23 IIf IIB,(n)

where c23 > 0 is independent off and Q.

Idea of the proof. Note that for IhI < 2 p

IlohfOllL,(R") < I IOhpJ IIL,(fl0.

1) + IIOh JOIILy(`ltelhl) -

IIL9(S2nlhl)

0

Proof. From the definition of the spaces B,(c) we have IIfoIIb',(R")
1, respec-

0

tively (5.72) for I = p = 1. Namely, the sum E in (5.46), (5.72) must be k=1

00

replaced by E , where ko is such that k=ko

supp Toh C (supp h)a x B(0, d). 0

(5.105)

Proof. 1. By Corollary 18 of Chapter 4 f tljj E WW(Vj n 11) and by Lemma 16 of Chapter 4 (f1/ij)(A 1)) E WI (A,(Vj n ()). Since supp (ft/ij)(A(-1)) C Aj(Vj n n), the extension 21 by 0 to R" of the function (fiPj)(A(71)) is such that E W,(R"). Hence by Theorem 2 there exists a trace hj of this function on R"-1 and therefore on Wj* = Aj(Vj n 80). This means that

gj = hj(Aj) is a trace of ft/ij on V n BSI. So by Definition 3 g = E gj is a j=1

trace of f on 852. Moreover, if I > 1, then by Theorem 3 1101 ,-1

BP P(Vjn0n)

-1(Re-1) 1, then as in Remark 15 one may state that there exists a bounded linear extension operator 1-1

T:flBp' °(852)-,Wp())nC°°(52), a=0

satisfying the inequalities

II

`

18k(T{9a}) II 8Uk

C c24 1-1 Lo(ft)

119a11B,,

k > 1,

(5.109)

3=0

and

8't(T {k'}) IIPfc-1(

8s,

- 9k) II

1-' Lp(f1)

S 025E 119311 1-.-1 By °(on) 8=0

0 < k < 1,

(5.110)

where e(x) = dist (x, 8f2) and c24, c25 > 0 are independent of ga.

In (5.109) the exponent k - l cannot be replaced by k - l - e for any e > 0. If p = 1, then a similar statement holds. (We recall that in this case the extension operator T is nonlinear.)

Remark 21 The problem of the traces on smooth m-dimensional manifolds where m < n - 1 may be treated similarly, though technically this is more n-1

complicated. Suppose that 52 C R" is an open set such that SZ = U Fm, m=0

where Fm are m-dimensional manifolds in the class C' and rm n r1, = 0 if m & p. (Some of I'm may be absent.) Let, for example, 1 n - pl, then for each f E W, (1) the trace off on rm exists.


246

Moreover, the traces of the weak derivatives Daf also exist if dal < l - npm. For this reason the total trace and the total trace space are defined by

(

n-pl<m 0, there exists b > 0 such that for any finite system of disjoint

intervals (a;l), f3;1)) C [a, b] and (a;2), (2)) C [b, c] satisfying the inequalities E(p;') a;')) < d, j = 1, 2, the inequalities E If (a;')) - f (Q;f))I < 2, j = 1, 2,

-

i This means that (T f)(z) = f (z), if z e i2.

CHAPTER 6. EXTENSION THEOREMS

248

hold. Now let (ai, Qi) C [a, b] be a finite system of disjoint intervals satisfying ai) < J. If one of them contains b, denote it by (a*, Then

If(ai) - f(Qi)I 5 +I f (b)

If(ai) - f(Qi)I + If(a") - f(b)I

- f (,6*) l +

If (ai) - f (Qi) I < e.

(If there is no such interval (a', Q'), then the summands If (a') - f (0*) 1 and If (b) - f (,Q')I must be omitted.)

Lemma 2 Let I E N,1 < p < oo, -oo < a < b < oo, f E 14, (a, b) and g E WP(b, c). Then the pasted function

h=

on (a, b), l9 on (b, c).

belongs to WP (a, c) if, and only if,

f (3)(b-) = g.(') (b+), s = 0, 1, ..., l - 1, where f,(,,')(b-) and gw')(b+) are boundary values of of Chapter 1). If (6.2) is satisfied, then

(6.2)

and gw") (see Remark 6

IIhIIWW(a,c) 5 IIf II W,(a,b) + 11911Wp(b,c)

(6.3)

Idea of the proof. Starting from Definition 4 and Remark 6 of Chapter 1, apply Lemma 1. Proof. Let f, and gi be the functions, equivalent to f and g, whose derivatives exist and are absolutely continuous on [a, b], [b, c] respectively.

Then f;') (b) =

and g( )(b) = g.(') (b+), s = 0,1, ...,1 - 1. If (6.2) is

satisfied, then the function on [a, b], on [b, c]

is such that h('-1) exists and is absolutely continuous on [a, b]. Consequently, the weak derivative h;;) exists on (a, b) and 1 If 92) (1)

hw

on (a, b), on (b, c).

6.1. THE ONE-DIMENSIONAL CASE

249

Hence, inequality (6.3) follows. If (6.2) is not satisfied, then for any function h2 defined on [a, b], coinciding with fl on [a, b) and with gl on (b, c], the ordinary derivative h(21-1)(b) does not 4-1) does not exist on (a, c) and h is not in exist. Hence, the weak derivative Wy1)(a,c).

Lemma 3 Let I E N, 1 < p < oc. Then there exists a linear extension operator T : W, (oo, 0) -+ WW(-oo, coo), such that (6.4)

IITIIw;(-.,u)-,wa(-...) < 81.

Idea of the proof. If l = 1, it is enough to consider the reflection operator, i.e., to set (6.5) (T1f)(x) = f (-x), X>0. If l > 2, define (T2)(x) for x > 0 as a linear combination of reflection and

dilations: (T2f)(x)

_

(6.6)

ak(TIf)(Qkx) = E akf(-Qkx), k=1

k=1

where Qk > 0 and ak are chosen in such a way that (T2f )(W8) (0+) = As) (04

3 = 0,1, ---,l - 1.

Verify that IIT2jIwl(oo,o)-.wW(_c ,ao) < oo and choose Qk

(6.7)

k = 1, ...,1, in

order to prove (6.1). Proof. Equalities (6.7) are equivalent to

Eak(-Qk)' = 1, s = 0, 1, ..., l - 1.

(6.8)

k=1

Consequently, by Cramer's rule and the formula for Van-der-Monde's determinant, 11 (Qi - Qj) I 1 1, 1 E N,

Qp,) = 1.

(6.16)

Lemma 4 Let I E Nl,1 < p < oo. Then i

/

1 1+ Qv t11 p< if IITIIwy(-oo,0).,wW(-00,00) < 1+ Qp,a.

(6.17)

(If p = 00, then (1 + QP 1) o must be replaced by Qo,1.)

Idea of the proof. Apply the inequality 1

(Ilf Ilwo(_oo,o) + IITf Illpvo(o,o.)) p 0, the extension operator Te setting TE f = g£ for x E (0, oo), where gE E Wp(0, 00) is any function, which is such that ffk)(0-), k = 0,..., l - 1, and II9EIIw;(0,oo) 5

Gp 1(f (0-), ..., f( .1-1)(0-)) + e 11f Ilwo(-oo,o)

(6.20)


253

Proof. 1. The second inequality (6.18) is trivial since IIh11L°(-00,00) _ { (IIh11L°(-o,o) +

Ilh(')IIL°(-oO,0))p

p

+(Ilhllr,°(o,.) +

(Ilhllwp(-,,,,,o)

+

2. It follows from (6.18) and (6.19) that for each ao, ..., a(_) E R such that Iaol + ... + Ia(-) I > 0

IITIIwp(-.,o)-.w;(-.,.) =

> (1+

IIT!

sup

t

IIJ

IlWp(-00,00)

IIWp(-.,O)

(IITfllwp(o,oo)\\p

sup JEWp(-06.0):

> 1+ G+p,(ao,...,a(_,)) p

IIf IIwp(-oo,o )

J 1

sup

°

°(-00,0)

/EWp(-W-0)'

1lfllwl

(G-( ao, ... , ai_t Y) and we arrive at the first inequality (6.17). 3. Given e > 0 by (6.18) and (6.20) we have I19eIIWp(0,oo)

sup

IITehI - 20 for each

extension operator T. On the other hand it is clear that for the extension i operator T, defined by (6.5) IITi IIw;(-oo,o)-+wo(-oo,oo) = 2' .

Remark 2 Note also that if the norm in the space W,(a, b) is defined by b

v

IIfDwp(a,b) = (f

(If(x)IP+Ifw')(x)I")dx)

a

(see Remark 8 of Section 1.3), then

inf

(1 + (QPI) )P)o,

where QI; P is defined by (6.14) - (6.15) with II . II') replacing II - II. This follows from the proof of Lemma 4 and the equality IITillw;(-oa,ao) =

((IITf Ilyvo(,,,o))P + (IITillwo(o.00))P)'.

Lemma 5 Let I E N, 1 < p:5 oo and f E Wp(0, oo). Then IlfIlw (o,00)

-

Il

k=0

f

(6.21)

(0+) k.

xkilL,(o,a,).

Idea of the proof. Apply Taylor's formula and Holder's inequality. Proof. Let f E Wp(0, co). Then for almost every x E (0, oo) !-1

f (x) = E ka0

(k) fW

xk

(k±)

+ (1 11)! f(x -

u)'-1

du,

0

k = 0, 1, ... ,1 - 1, are the boundary values of the weak where the derivatives f). (See formula (3.10) and comments on it in Section 3.1). Hence, by the triangle inequality for each a > 0

II

k=0

< IIf IIL f (k)(°+) k! xk ll Lp(o,a) -

+

( 1 1 1)!

ill (x-u)'-'f.() (u) dull L,(O,a)' 0


255

By Holder's inequality x

.

(lx-1)P

f (x - u)1-' f (()(u) du IILP(O,a)

IIJW''ll/.P(0,2)IILP(O.a)

< II (I - 1)p' + 1 /

0

II

k=0

k!

IILP(0,a)

IIhIIWW(-oo,0)


256

= II(x+a)'- x1jjL,(o,.) > II(x+a)'IIL,(o,a) - IIX'IIL,(o,a) II (x + a)' II w;(-a,o)

_

(21p+1-1)a-1 ,

1 + (!p+ 1)p

>

Ilx' I I wa(o,a)

2'-1 1

,

2'-1 _z 1 0>2

>

1

2

1

3

1a(1-° ,+(P+1) °)

s P.

Hence by (6.17) inequality (6.23) follows.

Remark 3 Note also that there exists a constant cl > 1 such that 2 IITIIw;(-ao,o)_*w,,(-o,.) >- c1, l > 2,

1 < p < oo,

(6.24)

for every extension operator T. For cl = 7 this follows from the inequality

1 1, w is any nonnegative infinitely differentiable kernel of mollification (see Section 1.1) and y is a sufficiently small positive number. Apply Young's inequality (4.138) and the equality 2),

II (w

*

Ilw

+, IIL,o(R).

(6.27)

Proof. Leta=L(l+4)(1+y)-1. By Section 1.1nECo (R),0 0 satisfying e7 < 2, then (1 + y)k < ek(l + e obtain (6.25). Finally we note that (6.27) follows from

(1 + Y)kE-k

< 2 lk and so

b+ 12

W

e

#

x

e

x - 'dy

W

a-i z-b-1

-(

r

J

E

2

+,(z)dz =W, (x-b-2)-low,+

E)l 2

--a-3

since the terms of the right-hand side have disjoint supports.

Corollary 5 In the one-dimensional case `dl E N there exists a nonnegative infinitely differentiable kernel of mollification p satisfying (1.1) such that I1(k)(x)I < (41)k, x E R, k = 0, ...,1.

(6.28)

Idea of the proof. Define q by (6.26), where a = b = 0 and E = 1, and apply the equality If *9IIL,(R) =IIfIIL,(R) - II9IIL,(R) for non-negative f,g E L1(R).

Lemma 8 There exists c2 > 0 such that for all 1, m E N, m < 1,1 1 and q > p. Choose intervals (ak, bk), k = 1, ... , s, in such a way that bk - at = 1, (a, b) = U (ak, bk) and the multiplicity of the covering k=1

{(ak, bk)}k=1 is equal to 2. By (6.29) IIfwm)IIL,(a.,b.) set

_

!

1)1

11TfIIw°(-oo,oo) (6.40)

IIfIIW;(a,b)

(

1

ll

(6.41)

IIhllw;(-oo,a) >- 1. 0

(6.42)

Q1-1,P b-a), apply inequality (4.50) and the relation inf

he Wj (-oo.a): h(a-)=I

First proof. It follows from (6.40), (6.41) and (6.39) that IITII

- (I

II9IIw;(-oo,oo)

IIQt 1;PIILo(,1)


262

2 41-1(1 - 1)! (b -

a)-1+-,

II9IIw;(-oo,oo)

where g = T f . By inequality (4.50) 2 II9IIL,II9w`'IILp(Too,oo) < 2 II9IIWW(-oo,oo)

II9w`-1)IILP(-oo,oo)

Consequently (1-1)

II9w

ir

(1)

2 +1

IIL,(-oo,oo) + II9w IILp(-oo,oo) 5

(-oo,oo) = II9w

II W,

II9II W;(-oo,oo)

and

IITII

-

4 ' -7' ( 1+ r

21)1 II9w)-1)IIW, (-oo,oo)

Since fw1-1) = 1 and g E Wp(-oo, oo), by Lemma 2, gw-1)(a-) = 1. Hence by (6.42) II9w(-))Ilwp(-oo,oo)

t

AEwvnf

IIhlIww(-oo,a)

1

h(.-)=l

Thus by Stirling's formula

IITII?

41_I (1- 1)! (b

- a)-1+

21r(l - 1) 4 1-1 t

e

4(7r+2) l >

2,

(e)

it+2

+2

J-1(1- I

(e

(4)111(b-a)-'+p >

4(7r+2)f e

l1(

(l-1) 1-1 (b-a)- '+

ba )-1+

0.12(4)111(b-a)-1+ 1

e

and we obtain (6.38) with 0.12 replacing Finally we note that (6.42), by Holder's inequality, follows from (3.8): 0+1

a+1

1 = Ih(a-)I 5 f IhI dy + f IhwI dy 5 IIhhIwp(-oo,oo) 13 o

a

Now for 1, n E N and 1 0 and of E Wp(R")

IILg(R") < A If II W (R') II Dmf

(6.43)


263

It follows from Chapter 4 that p < q < oo and 1,61 < l - n(I - 1) or

q=ooandl6I -

C' (Q, p, g,1.0)

sun

(6.45)

(9.Q)Ebf+.n.v C'(R", p, q, 1, l3)

Idea of the proof. Prove (6.44) by applying an arbitrary extension operator T and inequality (6.43) where A = C' (R", p, q,1, (t). Proof. For all (q, /3) E Mt,n,a IIDWfIIL,(n) 5 IID1(Tf)IIL,(R-) s C'(R, p, q, 1,13) IITfIIw/(Rn) < C'(R", p, q,1,13) IITIIw;(n)-,wo(Rn) Ilf II WW(n)-

Hence, C'(S1,p,q,1,/3) 0 such that

( c8 (1

l1

+ (b -

a)1-1

l) < if

c1s (1 + (b

ll

l

- a)1-1) .

(6.47)

The estimate from below is proved in the same manner as for the space W.(a, b). When proving estimates from above, the operator T2 defined by (6.6) must be replaced by T2 defined by (T2 f) (0) = f (O-) and (T2 f) (x) _ 1+1

1+1

k=1

k=1

E ak f (-,9kx), x > 0 , where ,Bk > 0 and E ak(-Qk)' = 1, s = 0,1, ... ,1.

In that case (T2 f)(')(0+) = f (')(0-), s = 0,1, ... ,1, which ensures that T2 f E C1(-oo, oo) for each f E G 1(-0o, 0). Moreover, (IT2IIUi(_00 o)iZ!l(_. .l < 161. The rest of the proof is the same as for the space WW (a, b).

6.2

Pasting local extensions

We pass to the multidimensional case and start by reducing the problem of extensions to the problem of local extensions.

6.2. PASTING LOCAL EXTENSIONS

265

Lemma 13 Let I E N, 1 0. If s = oo, suppose, in addition, that the multiplicity of the covering x - x({Uj};=1) is finite. Suppose that for all j = !-,s there exist bounded extension operators

T; : Wp(Sa n U,) -> WP(L(6.48) where Wp(S2 n U,) _ (f E Wp(S2 n Uj) : supp f c S2 n U,}. Ifs = oo, suppose also that sup IIT, 11 < oo. Then there exists a bounded extension operator jEN

T : Wp(S2) -* W,(R" ).

(6.49)

'

Moreover, (6.50)

1ITh 5 cIo sup IITifl,

j=

where c10 > 0 depends only on n, 1, 6 and x.

If all the T, are linear, then T is also linear.

Idea of the proof. Assuming, without loss of generality, that (U,)5 n 11 0 0 construct functions Oj E C°°(R"), j = !-,s such that the collection { ?}'=1 is a partition of unity corresponding to the covering {Uj)J=1, i.e., the following s

properties hold: 0 < '0j < 1, supp zlij C Uj, t O2 = 1 on S2 and Va E No j=1

satisfying jal < 1,

M1, where M1 depends only on n, l and 6.

For f E Wp(SZ) set

Tf = > ej Tj(f+,j)

on W.

(6.51)

j=1

(Assume that OjTj(fOj) = 0 on c(U,)). 0 Proof. 1. Let rlj E C°°(R") be "cap-shaped" functions satisfying 0 < ri, < 1, nj = 1 on (U,)s, rj = 0 on °((Uj)g) and ID°r)j(x)h 5 M26-1°1, a E No, a

where M2 depends only on n and a. (See Section 1.1.) Then 1 < E rl? < x j=1


266

a

a

on U (U;)

Further, let

E C6 (R"),

1 on 12, 77 = 0 on `(U (Uj) s ). One

1=1

7=1

can construct functions tkj by setting iOi _ ?7t t) & 77j2) - on U (U,)1 assuming

that O, = 0 on c(U(U;)1). :=1

2. The operator T defined by (6.51) is an extension operator. For, let x E Q. If X E supp V)j for some j, then 1,b (x)(Tj(f 1G1))(x) = OJ2 (x) f (x). If x 0 supp le,, e

then -0,(x)(T,(f-G;))(x) = 0 = 1,,(x)f(x). So (Tf)(x) = E'+GJ2 (x)f(x) _ j=1

f(x). 3. Let a E No and Ia1=1. If s E N, then

Do(Tf) _ ED.(?PjTi(f ik1)) on It".

(6.52)

i=1

If s = oo, then (6.52) still holds, because on `(U (U;)¢) both sides of (6.52)

are equal to 0 and dx E U (U;)1 the number of sets (Uj) j intersecting the i=1

ball B(x, 1) is finite. Otherwise there exists a countable set of Uj,, s E N, satisfying (UJ,)t n B(x, z) 36 0. Hence x E U,i,, and we arrive to a contradiction since x({U1},q0=1) < oo. Consequently, there exists S. E N such that supp (1G,T,(ftli,)) n B(x, z) 54 0 for j > sy. So OiTT(fOj) on B(x,1).

Tf 7=1

Hence, 00

Dm(Tf) _ E Dw(1G1Ti(f ,Oi)) _ E D"('O;T;(f O,)) i=1

on B(x,

2).

i=1

Therefore by the appropriate properties of weak derivatives (see Section 1.2) (6.52) with s = oo follows.

4. Let aENi, and a=0orjal=1. In (6.51),for all xEW1,and in (6.52), for almost all x E R", the number of nonzero summands does not exceed x. Hence, by Holder's inequality for finite sums,

ID."(Tf)IP < W-' t IDw(0tT, (f1, ))IP j=1

6.2. PASTING LOCAL EXTENSIONS

267

almost everywhere on R" and

f IDD(Tf)I°dx < W- > f ID°('jTj(fbj)) IPdx.

R.

i=1 Rn

Therefore, taking into account Remark 8 of Chapter 1, we have

IITfIIW;(Rn) < M3 t II P Ti(fuj)

flwp(R')

j=1

where M3 depends only on n, l and x. Since supp u'j C U, applying Corollary 18 of Chapter 4, we have IIVGjTj(ftj) IIwp(si) suppT,fcUj.

(6.53)

In this case the operator T may be constructed in a simpler way with the help

J=1, i.e., E of = 1 on ). We assume that

of a standard partition of unity

3=1

T,(f p1)(x)=0ifxEUj and set

Tf

T,(f,,i) on R. j=1

(6.54)


268

The operator T is an extension operator. For, let x E 1. If x E Uj, then (T.i(f Vi))(x) = b1(x) f(x), and if x V Uj, then (T,(f'j))(x) = 0 =O9(x) f(x). Thus (T f)(x) = E Oj(x) f (x) = f (x). Note also that for f E Wy(0), because i=1

of (6.53), we have T,(fiP) E Wp(Rn) and II Ti(f'Pi)

11 2' (f0,) Ilwp(u,).

Further we consider a bounded elementary domain H C R" with a C1- or Lipschitz boundary with the parameters 0 < d < D < oo, 0 < M < oo, which by Section 4.3 means that

H={xEitt":an<xn 0 depends only on n, I and M.


270

Idea of the proof. Let H- _ {x E R"; -00 < x, < tp(x), x E W}, (Tof)(x) = f (x) for x E H, (Tof)(x) = 0 for x E H-\H. Moreover, let (Af)(x) = f(a(x)), where (a(x))k = xk, k = 1, ... , n - 1, (a(x))" = xn + W(r) and (T2 f) (x) = 1

i akf (x, -Qkxn) for I E W, xn > 0, where flk > 0 and ak are defined by k=1 (6.8). Set

T = A-' T2 A To

(6.60)

and apply Lemma 16 and Remark 25 of Chapter 4. Proof. If f E W,(H), then Tof E Wp(H-) and IITofIIww(H-) = IIfUIww(H) Hence,

IIToIIIVP(H)4wp(H-) = 1. Since A(H-) = Q- = {x E Rn : x E W, xn
b1, we have I'(xl,x2,...,xn-1)- '(y1,y2,...,yn-1)I = IW(xl,x2,...,xn-1)-V(bi,y2,...,yn-1)I :5 I W(xl, x2, ..., xn_1)-p(b1, x2, ..., xn-1)I+IW(b1, x2, ..., xn-1)-co(bt, y2, ..., tin-1) I a

< M(y1 - x1) + M((x2 - y2)2 + ... + (xn-1 - yn_1)2)' < MIx - 91. Repeating this procedure with respect to the variables x2i ..., xn we obtain a function, which coincides with cp on W and satisfies a Lipschitz condition on


272

with the same constant M as the function cp. We denote it also by W and consider the domain Il defined by (6.62) and the operatorT satisfying (6.64). 2. For f E WW(H) let To f be the extension of f by zero to Q. Since suppf \ H (1 V, we have To f E W, (Q) and IIToIIw;(n) = II!Ilwp(H) Hence IIToIIww;(H)-.wp(n) = 1. Next we observe that TTo : 4' (H) -* Wp(R") and IITT011rVP(H)-,WP(V) :5 IITIIWo(n)- w (R^) < C13.

Thus Lemma 14 is applicable and the statement of Lemma 17 follows. 0 Our next aim is to construct a bounded linear extension operator (6.64) for 9 defined by (6.62), (6.63).

LetG=R"\S2={xEl(t":xn>cp(x)}. We set Gk = {x E G: 2-k-' < en(x) < 2-k},

k E Z,

where N(x) = xn - V(x)

is the distance from x E G to 8G = 8SZ in the direction of the axis Oxn. First we need an appropriate partition of unity. 3 Lemma 18 There exists a sequence of nonnegative functions 10k satisfying the following conditions: 00 1 1)

k=-oo

forxEG,

(6.66)

for x V G,

(6.67)

2) G = U suPP''k k=-oo

and the multiplicity of the covering {supptpk}kEZ is equal to 2,

3) Gk csupp'Ok c Gk_, U Gk u Gk+,, 4)

ID°lPk(x)I 5

C14(a)2k1a1,

k E Z,

x E R' , k E Z, a E K,

(6.68) (6.69)

where C14 (a) > 0 depends only on a. 3 In Lemmas 18- 25 below fl is always a domain defined by (6.62) - (6.63) and G = Ir \1I.

6.4. EXTENSIONS FOR LIPSCHITZ BOUNDARIES

273

Idea of the proof. Apply the proof of Lemma 5 of Chapter 2.

With the help of the partition of unity constructed in Lemma 18 we define an extension operator in the following way: f (x)

E

for x E S2,

00

(T f)(x) _

k=-oo

'Pk(x) fk(x)

for x E G,

(6.70)

where w(z)dz

f(x - 2-k2,x, -

fk(x) =

J Q1^

= A-12kn f w(2k(t - y), A-12k(xn - yn))f (y) dy.

(6.71)

it.

Here a

A=16(M+1)

(6.72)

and w E Co (R") is a kernel of mollification satisfying

suppwC{xEB(0,1):xn>2}

(6.73)

and

f B(0,1)

w(z) dz = 1;

f

w(z)z° dz = 0, a E Non, 0 < jal < 1.

(6.74)

B(0,1)

Now let us show that the operator T is well defined. First, we assume that Ok(x)fk(x) = 0 for x V supplik even if fk(x) is not defined. On the other hand, if x E suppok, fk(x) is defined. This is a consequence of the following inequality, which holds for x E supp ik and z E supp w since by (6.68) Bn(x) < 2-k+1 and by (6.73) IzI < 1,zn >- 1

:

xn - A2-kzn -1V(x - 2-kz) = xn -'P(x) + 1P(s - 2-kz) - A2-kzn < 2-k+1 +M2 -k121- A2-kzn2-k+1 + M2-k I zj -A2 -kZ" < 2-k 2+M-A) 2 O, h > 0 consider an open cylinder centered at the point x of radius r and height h C (x, r, h) _ (y E lr : y E B(2, r), I xn - y, ,l < Al. 2

Because of (6.73) the value fk(x) is determined by the values f (y) for y belonging to the cylinder Cyk = /r((a,x" - 4A2-k),2-k, 'A2-k),

which is centered at the point (2, xn- A2-k) translated with respect to x in the a in (6.71) w(2k(2 - y), A-12k(xn - yn)) direction of the set Q. This follows since

can be nonzero only if 2k1 - yI < 1 and .1 < A-'2k(xn - yn) < 1. For this reason (T f)(x), x E G, can be looked at as an inhomogeneous mollification of the function f, for which both the step and the translation are variable. Thus the extension operator T is closely related to the mollifiers with variable steps considered in Chapter 2. Note also that on G the operator T is an integral operator:

(Tf)(x) =

J

K(x, y)f(y) dy, x E G,

n

with kernel

K(x, y) =A-' 00E k=-ao

1Pk(x)2knw(2k(2 A-'2k(xn

(6.76)


xn -

Lemma 19 Let f E L, °°(Q), x E G and x'

275 4A[!n(x)). Then the

value (T f)(x) is determined by the values f (y) for y belonging to the cylinder Cx = C(x*, 4Bn(x), 4APn(x))) C CZ C Q.

(6.77)

Idea of the proof. Apply (6.76) and Remark 11. Proof. Let x E G. Choose the unique m E N such that x E G,n. Then zkk(x) = 0

if k V {m - 1, m, m + 1} and the value (T f)(x) is determined by the values fk(x) where k = m - 1, m, m + 1. By Remark 11 those values are determined m+I

by the values f (y) for y E U C=,k. Hence J .,t - JI < 2-' k=m-I A2-m-2

gAPn(x)
0 depends only on n, l and M. Moreover, Va E l there exists a function 6 g0, independent of k, such that IID*fk-gullLp(Gk) 0 follows from the first equality and inequality (6.81) with a = 0. Next let P. (C x)=

f

c

(_1)Ia1+171

By (3.52), as in steps 3-4, it follows from the second equality for D°fk that

(D°fk)(x) = ga(x) + L, A-a"2kIOIRry (x), 171=1-la1

where A-a"2klal

ga(x) =

f

P.(u(z),x)(D°w)(z)dz

B(0,1)

and R(°k (x) is obtained from R7,k(x) by replacing Dw f, I by D°+7 f, I - IaI respectively. By (6.74) f zO(D°w)(z) dz = 0 if IaI > I, or IaI < l and ,0 O a. B(0,1)

If IaI < l and Q = a, then

f z°(D°w)(z) dz

Hence g° = 0 for

B(0,1)

IaI>Iand ga(x) =

A-Q"2k1*1Ds (Pa(u(z),x))I:.O

= Pa(x,x)

(6.89)

for IaI < 1. With this choice of ga inequality (6.82) with IaI > 0 follows as in step 5. O


280

Remark 13 In the above proof the functions g° defined for a = 0 by (6.86) and for 0 < Jai < l - 1 by (6.89) are the first summands in the integral representations (3.51), (3.52) respectively, where B, w are replaced by Bz, w= respec-

tively. Since Vx E Gk we have B(x'(k), M12'k) C B= = B(x'(k), 4 C B(x' (k), M2 2-k), where x' (k) _ (x, x" - 2 A 2-k) and Ml, M2 > 0 depend only on n. These inclusions explain why one may expect estimate (6.82) to hold with appropriate g°,k. The choice of the ball B(x',4pf,(x)), independent of k and "compatible" with B(x'(k), M2 2-1*), allows us to construct a function g°, for which inequality (6.82) holds and which is independent of k.

Remark 14 In the proof of Lemma 22 (Section 4) we have applied property

(6.73) for Jal < l - 1. The fact that it holds also for Jai = l allows us to prove the following local variant of (6.82) for p = oo : dx E Gk and Va E N satisfying lal 0 depend only on n, 1, M and a. "If any neighbourhood of x contains infinitely many interval components of es, then yl # x. Otherwise, for a point yi, which is sufficiently dose to x, we have yj = x,, and the argument becomes much simpler.


283

Idea of the proof. 1. To prove (6.94) first observe that, as in the proof of Lemma 13 of Chapter 2,

IIT!IIL,(n) 5 2(

(6.97)

IIfkIIL,(ak))

k=-oo

then apply inequality (6.81) and the fact that the multiplicity of the covering {Stk}kEZ is finite.

2. To prove (6.95) apply (6.92) and (6.93). Estimate I. as in step 1. To estimate I,# where Q 34 a apply inequalities (6.69) and (6.82). In the case of inequality (6.96) use also the inequality x - cp(a) < M, 2-k on Gk where M1 is independent of.k. 0 Proof. 1. Since the sum (6.70) for each x E G contains at most two nonzero terms by Holder's inequality

x IITf 11%(G) 5 2P-' f (E I kfklP) dx. C

f=f=

k=-x

Furthermore,

x

G k.-x

x

m+l

k=-xGm k=m-1

m+1

m=-xk=m-lGm

x

k+1

f=j

k=-xm=k-1GT

x

k=-xC

k

and inequality (6.97) follows since 0 < 1'k < 1. Consequently, by (6.81)

IITf IIL,(G) 5 2 c1s (E If IILn(nk))

2 clam Ilf IIL,(n),

k=-x where xn is the multiplicity of the covering {Stk}kEz. which, by Remark 12, does not exceed log2(8b).

2. Suppose that a E No satisfies Ial = 1. Then we consider equality (6.92). As in step 1 IIIaaIIL,(G) UK (x), where UK = dist (x, 8K) and K C G is the infinite cone defined by yn > op(z) + MI± - gl, y E R". The desired inequality follows since B(x, (1 + M)-I (x,, - V(:t))) C K, which is clear because

by E B(z, (1+M)-'(xn-w(2))) we have yn-W(s)-Ml2-gl = yn-xn+xn-iv(y)-Ml2-gl > -(1 + M)Iz - yl + (xn -,'(2))) > 0.


286

In the case of inequality (6.101) the argument is similar. One should only take into consideration (6.102) and (6.96) and note that the appropriate weighted analogue of Lemma 17 is also valid. Finally, let S2 = R°- and suppose that for some e > 0 and for some extension operator (6.100) we have IIx;,°1-1-`D°(T f)IIL,(R .) < oo for all f E Wp(R"-) and for all a E l satisfying Ial = m > I + e. First suppose that l > 1. Let

q E Bp ° (R"-') \ BB ` ' (R`). By Theorem 3 of Chapter 5 there exists a function f E 4i'p(R")such that f g. By Lemma 2 T f *_ = f g. I

R

R

R

(R+), by the trace theorem (5.68), g E ° (R"-1) and Since T f E W' n.=n we have arrived at a contradiction. If I = p = 1, the argument is similar: one should consider g E L1 (R"`') \ Bi (R"-') and apply Theorem 5 of Chapter 5 instead of Theorem 3 of that chapter. Remark 15 The extension operator constructed in the proof of Theorem 3 satisfies (6.100) and (6.101). So, by the last statement of that theorem, it is the best possible extension operator in the sense that the derivatives of higher BB1+E

orders of T f on `S2 have the minimal possible growth on approaching 011.

Remark 16 The extension operator constructed in the proof of Theorem 3 is such that for all m E 1 satisfying m < 1 we have T : WD (11) -* WD (R"). Remark 17 Now we describe an alternative way of proving of the first statement of Theorem 3. Let 0 be defined by (6.62) and (6.63). It is possible to get an extension operator (6.100) by "improving" the extension operator (6.61) . To do this we replace x" - p(2), which in general is only a Lipschitz function, by the infinitely differentiable function A(x) = 2(1 + M)ej (x), where p# is the regularized distance constructed in Theorem 10 of Section 2.6. By (6.102) we have

x" - W(2) < A(x) < 2 (1 + M)(x" - cp(z))

(6.103)

and

a E K.

(6.104)

(Tf)(x) = Eakf(x,x" - (1 +Qk)o(x)),

(6.105)

ID°0(x)I S c, (x" So we set 1+1

k=1 1+1

1+1

k=1

k=1

where ,8k > 0 and E ak(-008 = 1, s = 0, ...,1. (Hence E ak(1 + ,Qk)' _ 0, s = 1, ...,1.) By using formula (4.127), expanding for f E Cc0(i) the derivative Dsf(x,x" - (1 + 6k)A(x)) by Taylor's formula with respect to the point


287

(I, cy(I)) E OcI and applying (6.103), one can prove that Lemmas 22 and inequalities (6.94) and (6.95) are valid for this extension operator as well. The rest is the same as in the proof of Theorem 3. The extension operator (6.105) cannot be "the best possible" because, in general, T f V C°°(`1Z). On the other hand in (6.105) it is possible to replace 00

1+1

the sum E by the sum E and 10 choose 13k > 0 and ak in such a way that k=1

k=1 00

00

E IakI II3kl' < oo and E ak(-13k)' = 1 for all s E N0. This gives an operator k=1

k=1

(independent of 1) such that (6.100) is satisfied for all 1 E N.

Remark 18 The extension operator (6.29) -- (6.30) in contrast to the extension operator described in Remark 17, is also applicable to the spaces defined in Remark 26 of Chapter 4, i.e., for I E N, 1 < p < oo and an open set S2 C R" with a Lipschitz boundary T : Wy.....(l) -4

W,.....f(IItn).

(6.106)

To prove this for n defined by (6.62) - (6.63), following the same scheme, one needs to prove an analogue of (6.95) for wP 1(Sl). This can be established with the help of an integral representation, which involve only unmixed derivatives e , j = 1, ..., n. W

Remark 19 The supposition "0 has a Lipschitz boundary" in Theorem 3 is sharp in the following sense: for each 0 < ry < 1 there exists an open set SZ with a boundary of the class 11 Lip-1, which is such that the extension operator (6.100) does not exist, as the following example shows.

Example1 Letn>1,IEN, 11andq=ooifl>D,p>1orl>n,p=1. Consider the function f6(x) = xn where 5 E R \ N0. Then f6 E W,(C1) if, and only if, 6 > l - v + np 1(1 - y because 1

1

II.f6IIWo(l.,) < 00 b

r/ J

I

(6-l)p+n-1

jtj<sn xn6-1)P dt) dxn = vn_1

xn

dxn < 00.

0

0

(This is also true for l = 0, i.e., for Lp(54).) Let I < p, the cases l = a and l > v

being similar. If-v+"p1(1-ry) =l-a+np1(1-,-y)

Sobolev Spaces on Domains (Teubner-Texte zur Mathematik; 137)

Sobolev spaces on domains

Sobolev Spaces on Riemannian Manifolds

Sobolev spaces

Sobolev spaces

Sobolev Spaces on Riemannian Manifolds

Topology & Sobolev Spaces

Spectral synthesis in sobolev spaces

Sobolev spaces in mathematics I: Sobolev type inequalities

137

Arbeitsbuch Mathematik zur Physik

Arbeitsbuch Mathematik zur Physik

Mathematik zwischen Objekten[zur Philosophie der Mathematik]

Sobolev Spaces in Mathematics I: Sobolev Type Inequalities: BD I

L^(inf)-bounds for elliptic equations on Orlicz-Sobolev spaces

Nonlinear analysis on manifolds: Sobolev spaces and inequalities

Lectures on Elliptic and Parabolic Equations in Sobolev Spaces

Lectures on elliptic and parabolic equations in Sobolev spaces

Sobolev Spaces on Metric Measure Spaces: An Approach Based on Upper Gradients

Sobolev Spaces (Springer Series in Soviet Mathematics)

An introduction to Sobolev spaces and interpolation spaces

An Introduction to Sobolev Spaces and Interpolation Spaces

Sobolev Spaces. Pure and applied Mathematics

An Introduction to Sobolev Spaces and Interpolation Spaces

An Introduction to Sobolev Spaces and Interpolation Spaces

An Introduction to Sobolev Spaces and Interpolation Spaces

Sobolev Spaces, Their Generalizations and Elliptic Problems in Smooth and Lipschitz Domains

Nonlinear Potential Theory and Weighted Sobolev Spaces

A first course in Sobolev spaces

137 секунд

Inequalities Based on Sobolev Representations

Sobolev Spaces on Domains (Teubner-Texte zur Mathematik; 137)