AnIntroduccToSobolevSpaces[1]

AN INTRODUCTION TO SOBOLEV SPACES

Steve Taylor, Montana State University.

Preface: These notes were written to supplement the graduate level PDE course atMontana State University. Sobolev Spaces have become an indispensable tool in the theoryof partial differential equations and all graduate-level courses on PDE's ought to devotesome time to the study of the more important properties of these spaces. The object of thesenotes is to give a self-contained and brief treatment of the important properties of Sobolevspaces. The main aim is to give clear proofs of all of the main results without writing anentire book on the subject! Why did I write these notes? Much of the existing literature onthe subject seems to fall into two categories, either long treatises on the subject with themost general assumptions possible (and thus unsuitable for part of a PDE course), or verysketchy discussions confined to a chapter of a PDE text.

CONTENTS:

1. The Spaces W j , p () and W0j , p ().......................................... 12. Extension Theorems........................................................... 103. Sobolev Inequalities and Imbedding Theorems...................... 134. Compactness Theorems...................................................... 205. Interpolation Results......................................................... 256. The Spaces H k() and H0k()............................................. 277. Trace Theorems................................................................ 29Appendix: Some Spaces of Continuous Functions....................... 34References................................................................................ 35

In these notes, is a domain (i.e. an open, connected set) in Rn .

1. The Spaces W j , p () and W0j , p ()

Definitions: Suppose 1 p < . Then(i) Llocp () = {u: u Lp (K) for every compact subset K of }(ii) u is locally integrable in if u Lloc1 () .(iii) Let u and v be locally integrable functions defined in . We say that v is

the th weak derivative of u if for every C0()

1

uD dx = (1)| | v dx ,

and we say that Du = v in the weak sense.(iv) Let u and v be in Llocp () . We say that v is the th strong derivative of u if

for each compact subset K of there exists a sequence { j} in C| | (K) such that j u in Lp (K) and D j v in Lp (K).

THEOREM 1 If Du = v and Dv = w in the weak sense then D+ u = w in the weaksense.

PROOF Let C0 () and = D . Then

uD + dx = (1)| | v dx = (1)| | vD dx = (1)| |+| | w dx .

Definition (mollifiers): Let C0 (Rn ) be such that(i) Supp B1(0) , (recall that "supp" denotes the support of a function, and

Br (p) denotes an open ball of radius r and center p).(ii) (x) dx =1 ,(iii) (x) 0 .If > 0 then we set (provided that the integral exists)

Ju(x ) =1 n

( x y

)u(y) dy .

Ju is called a mollifier of u. Note that if u is locally integrable in and if K is acompact subset of then Ju C

(K) provided that < dist(K,). Suppose now thatu Lloc

p () . Clearly

Ju(x ) = (y)u(x y) dyB1(0) ,so for p > 1 we have (if 1 / p +1 / q = 1)

| Ju(x)| {(y)}1/ q{(y)}1/ p |u(x y)| dyB1 (0)

2

( ({(y)}1/ q )q dx)1/ q ( ({ (y)}1/ p| u(x y)|)p dy)1/ pB1(0)B1 (0) .

Hence | Ju(x)|p (y)|u(x y)|p dyB1 (0) , and this trivially holds if p = 1 too. Integratingthis, we see that

| Ju(x)|pK dx (y) |u(x y)|p dx dyKB1(0) (y) |u(x)|p dx dy

K0B1 (0)= |u(x)|p dx

K0 ,where K0 is a compact subset of , K Interior(K0 ) and < dist(K,K0 ) . i.e. we have

|| Ju||L p (K) ||u||Lp (K0 ) . (1)

LEMMA 2 If u Llocp () and K is a compact subset of then || Ju u||Lp (K ) 0 as 0 .

PROOF Let K0 be a compact subset of where K Interior(K0 ) and let < dist(K,K0 ) . Let > 0 and let w C(K0 ) be such that || u w||Lp (K0 ) < . Thenapplying (1) to u w , we obtain

|| Ju Jw||Lp (K ) < . (2)

But J w(x) w(x) = (y){w(x y) w(x)} dyB1(0 ) , and this goes to zero uniformly on Kas 0. Hence, if is sufficiently small, we have

|| Jw w||Lp (K ) < . (3)

Hence, by (2) and (3)

|| Ju u||Lp (K ) ||w u||Lp ( K) +|| Ju Jw||Lp (K ) +|| Jw w||Lp ( K) < 3 . (4)

Since is arbitrary, || Ju u||Lp (K ) 0 as 0.

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The proof of the following theorem contains some other important approximatingproperties of mollifiers.

THEOREM 3 Suppose that u and v are in Llocp () . Then Du = v in the weak sense if and

only if Du = v in the strong Lp sense.

PROOF Suppose that Du = v in the strong Lp sense. Let C0() and letK = supp . Let > 0 and take C| |(K) so that || u||Lp ( K) < and|| D v||Lp (K ) < . Then

| uD dxK (1)| | v dxK || D dxK (1)| | D dxK |

+ | (u )D dx|K +| (v D ) dxK |

||u ||Lp ( K) || D ||Lq (K ) +||v D ||Lp ( K) || ||Lq (K ) (||D ||Lq (K ) +|| ||Lq (K ) ),

where q is the conjugate exponent of p (if p = 1 then q = and if p > 1 then1 / p +1 / q = 1). But is arbitrary, so the LHS must be zero. So Du = v in the weaksense.

Conversely, suppose that Du = v in the weak sense and let K be a compact subsetof . Then Ju C

(K) if < dist(K,) and we have for all x in K

D Ju(x) = n Dx(x y

)u(y) dy

= n (1)| | Dy(

x y

)u(y) dy

= n (x y

)v(y) dy

= Jv(x).

But by Lemma 2, || Ju u||Lp (K ) 0 and || D Ju v||Lp ( K) =|| Jv v||Lp ( K) 0 as 0.Thus Du = v in the strong sense.

Definitions (i) |u| j, p = ( | D u(x)| p dx| | j )1/ p .(ii) C j, p () = {u C j(): | u| j, p < }.(iii) H j , p () = completion of C j, p () with respect to the norm | | j, p .

4

H j , p () is called a Sobolev space. We will encounter other such spaces as well.Recall that the completion of a normed linear space is a larger space in which all Cauchysequences converge (i.e. it is a Banach space). It is constructed by first defining a space ofequivalence classes of Cauchy sequences. Two Cauchy sequences {xm}, {ym} are said tobe in the same equivalence class if lim

m || xm ym ||= 0 . A member x of the old space is

identified with the equivalence class of the sequence {x, x,x, . . .} of the new space and inthis sense the new space contains the old space. Further, the old space is dense in itscompletion. Moreover, if a normed linear space X is dense in a Banach space Y, then Y isthe completion of X.

Recall that for 1 p < , Lp () is the completion of C0() with respect to theusual "p norm". This knowledge allows us to see what members of H j , p () "look like".Members of Lp () are equivalence classes of measurable functions with finite p norms,two functions being in the same equivalence class if they differ only on a set of measurezero.

Suppose that {um} is a Cauchy sequence in C j, p (). Then for || j , {D um} is aCauchy sequence in Lp (). Hence, there are members u of Lp () such that Dum uin Lp (). Hence, according to our definition of strong derivatives, u0 is in Lp () and uis the strong derivative of u0 . Hence we see that

H j , p () = {u Lp (): u has strong Lp derivatives of order j in Lp () and there exists asequence {um} in C j, p () such that Dum Du in Lp ()}.

Definition W j , p () = {u Lp (): the weak derivatives of order j of u are inLp ()}

Note that by Theorem 3, an equivalent definition of W j , p () is obtained by writing "strongderivatives" instead of "weak derivatives". Because of this, we see easily thatH j , p () W j , p() . In fact, H j , p () = W j, p (). This is not obvious because formembers of H j , p () we can find sequences of nice functions such that Dum Du inthe topology of Lp (), while according to our definition of strong derivatives, such limitsexist only in the topology of Lloc

p () for members of W j , p () . Before proving thatH j , p () = W j, p (), we need the concept of a partition of unity.

5

LEMMA 4 Let E Rn and let G be a collection of open sets U such that

E {U: U G} . Then there exists a family F of non-negative functions f C0(Rn ) suchthat 0 f (x) 1 and

(i) for each

f F , there exists

U G such that supp f U ,(ii) if K E is compact then supp f K is non-empty for only finitely many

f F ,(iii)

f (x)f F =1 for each x E (because of (ii), this sum is finite),(iv) if

G ={1, 2 , . . .} where each i is bounded and i E then the familyF of such functions can be constructed so that

F = {f 1, f 2, . .} andsupp f j j .

The family of functions F is called a partition of unity subordinate to the cover G.

PROOF Suppose first that E is compact, so there exists a positive integer N such thatE i =1

N Ui , where each

Ui G . Pick compact sets Ei Ui such that E i =1N Ei . Letgi = J i Ei , where i is chosen to be so small that supp gi Ui . Then gi C0

(Ui ) andgi > 0 on a neighborhood of Ei . Let g = gii =1

N , and let S = supp g i =1N Ui . If < dist(E,S) then k = J S is zero on E and h = g + k C(Rn ). Further, h > 0 on Rnand h = g on E. Thus

F = {f i : f i = gi / h} does the job.If E is open, let

Ei = E B i (0) {x: dist(x,E) 1i}.

Thus Ei is compact and E = i =1N Ei . Let

Gi be the collection of all open sets of the formU [Interior(Ei +1) Ei 2 ], where

U G and E0 = E1 = . The members of

Gi providean open cover for the compact set Ei Interior(Ei1 ), so they possess a partition of unity

F i with finitely many elements. We let

s(x) = g(x)gF ii =1

and observe that only finitely many terms are represented and that s > 0 on E. Now we letF be the collection of all functions of the form

6

f (x) =g(x)s(x) , x E

0, x E

This F does the job.If E is not open, note that any partition of unity for U is a partition of unity for

E.For the proof of (iv), let H be the partition of unity obtained above and let f i = sum

of functions h in H such that supp h i , but supp h j , j < i . Note that each h isrepresented in one and only one of these sums and that the sums are finite since each i isa compact subset of E. Thus the functions f i provide the required partition of unity.

THEOREM 5 (Meyers and Serrin, 1964) H j , p () = W j, p ().

PROOF We already know that H j , p () W j , p() . The opposite inclusion followsif we can show that for every u W j ,p and for every > 0 we can find w C j, p such thatfor || j , || D w Du||L p () < .

For m 1 let

m = {x : || x||< m, dist(x,) > 1m

}

and let 0 = 1 = . Let { m} be the partition of unity of part (iv), Theorem 4,subordinate to the cover {m+ 2 m}. Each u m is j times weakly differentiable and hassupport in m +2 m . As in the "conversely" part of the proof of Theorem 3, we can pick

m > 0 so small that wm = J m (um ) has support in m +3 m 1 and |wm um | j, p < 2m .Let w = m =1

wm . This is a C

function because on each set m +2 m we havew = wm 2 + wm 1 + wm + wm +1 + wm +2 . Further,

|| D w Du||L p () =||m =1 D (wm u m)||L p () m =1

|| D (wm u m)||Lp () m =1

/ 2m = .

7

Remarks(i) The proof shows that in fact C() C j, p () is dense in W j , p () .(ii) Clearly members of C() C j, p () are not necessarily continuous on oreven bounded near . It would be very useful to have the knowledge thatC( ) C j , p () or C j( ) C j, p () is also dense in W j , p () . But the followingexample shows that this cannot always be expected.

Problem 1 Let = {(x, y) : 1 < x2 + y2 < 2, y 0 if x > 0}, i.e. an annulus minus thepositive x-axis. Let w(x, y) = , the angular polar coordinate of (x,y). Clearly w is inW1,1() because it is a bounded continuously differentiable function. Show that we cannotfind a C1( ) such that |u |1,1 < 2pi . (Note that is the whole annulus).

The reason for the failure of the domain in Problem 1 is that the domain is on eachside of part of its boundary. The following definition expresses the idea of a domain lyingon only one side of its boundary.

Definition A domain has the segment property if for each x there exists anopen ball U centered at x and a vector y such that if z U then z + ty for0 < t < 1.

We will not need the following theorem, so we don't prove it. For a proof, see Adam'sbook. However, see Lemma 9 for the simpler version of the result that we will need.

THEOREM 6 If has the segment property then the set of restrictions to of functionsin C0

(Rn ) is dense in Wm, p () .

THEOREM 7 Change of Variables and the Chain Rule. Let V, be domainsin Rn and let T: V be invertible. Suppose that T and T 1 have continuous, boundedderivatives of order j . Then if u W j ,p () we have

v = u o T W j , p (V) and thederivatives of v are given by the chain rule.

PROOF Let y denote coordinates in and let x denote coordinates in V( y = T(x) ). If f Lp() then

foT Lp(V ) because

| foT|p dxV = | f |p J dy const. | f |p dy (5)

8

(Here J is the Jacobian of T 1 ).If u W j ,p (), let {um} be a sequence in C j, p () converging to u in W j , p () and

set

vm = umoT . By the chain rule, if || j

Dxvm = (Dyum )oT R ,

where the R , are bounded terms involving T and its derivatives. But for || jDy

u Lp ()

(Dyu)oT Lp (V) (Dyu)oTR , Lp (V) since the R , are bounded.Further,

|| Dx vm (Dyu)oTR , ||Lp (V ) =|| (Dyum Dyu)oTR , ||Lp (V )

||(Dyum Dyu)oTR , ||Lp (V )

const. ||(Dyum Dyu)oT||Lp (V ) const. || Dyum Dyu||Lp ()

by (5). So ( = 0 case),

vm v = uoT in Lp (V ) and

Dxvm (Dyu)oT R , in

Lp (V ). This shows that v W j, p (V) and

Dxv = (Dyu)oT R , .

Definition W0j , p () =completion of C0() with respect to the norm | | j, p .

Remarks (i) Clearly W0j , p () W j , p() because C0() C j , p().(ii) Saying that f W0j, p () is a generalized way of saying that f and itsderivatives of order j 1 vanish on . e.g. W01, p () W2, p () is auseful space for studying solutions of the Dirichlet problem for secondorder elliptic PDE's.(iii) C0j() W0j, p () because if f C0j(), we know that if issufficiently small then J f C0 () and J f f in | | j, p norm.

Problem 2 Show that W j , p (Rn ) = W0j, p (Rn ) . Hint: Why is it enough to show that C j, p (Rn ) W0j , p (Rn ) ?

Problem 3 Show that if is a domain in Rn , f W0j, p () and if f is extended to bezero outside then the new function is in W j , p (Rn ).Problem 4 Show that if y C1[0,1] and y(0) = y(1) = 0 then y W01, p (0,1) . Use thisfact to show that for any f Lp(0,1) there is a unique y W01, p (0,1) W2, p(0,1) such thaty"y = f . Hint: Solve the problem first with f C0(0,1) and then take limits.

9

2. Extension Theorems

Most of the important Sobolev inequalities and imbedding theorems that we willderive in the next section are most easily derived for the space W0

j , p () which (seeProblem 3) can be viewed as being a subspace of W j , p (Rn ). Direct derivations of theseresults for the spaces W j , p () are tedious and difficult because of the boundary behaviorof the functions (Adams uses the direct derivation approach in his book). In this section weinvestigate the existence of extension operators that allow us to extend functions inW j , p () to be functions in W j , p (Rn ). This will allow us to easily deduce the Sobolevimbedding theorems for the spaces W j , p () from the corresponding results for W j , p (Rn ).

LEMMA 8 Let u Rn and f Lp(Rn ) . Set f (x) = f (x +u) . Then lim 0 f = f inLp (Rn ) .

PROOF Given > 0, let C0(Rn ) be such that || f ||L p < . Since uniformly on a sufficiently large ball containing the supports of all (say, for 1), wecan pick so small that || ||Lp < . Then

|| f f ||Lp || f ||L p +|| ||Lp +|| f ||L p < 3 .

LEMMA 9 Let R+n

= {x Rn : xn > 0} . C(R +n ) C j, p (R+n ) is dense in W j , p (R+n ).

PROOF Suppose f is in W j , p (R+n ) let > 0 and pick C(R+n ) C j, p (R+n ) so that|| D D f ||Lp ( R+n ) < for all || j . We take the vector of Lemma 8 to beu = (0,0,0, . . ,1) and define functions Lp (Rn ) as

(x ) = D (x) , xn > 00 , xn 0

Observe that for each > 0, C(R +n ) C j, p (R+n ) . By Lemma 8, we can pick > 0so that, for all || j , || ||Lp ( Rn ) < . But this implies that || D D ||Lp (R+n ) < .Hence

|| D D f ||Lp ( R+n ) || D D ||Lp (R+n ) +||D D f ||Lp ( R+n ) < 2 .

10

LEMMA 10 There exists a linear mapping E0: Wj, p (R+n ) W j, p (Rn) such that E0 f = f

in R+n and |E0 f | j , pRn C| f | j, pR+n , where C depends on only n and p.

PROOF If f C(R +n ), define

E0 f (x) =f (x) , xn 0

ck f (x1, x2, . . , xn 1, kxn )k =1j +1 , xn < 0

where the constants ck are chosen so that E0 f (x) C j (Rn ) , i.e.

(k)m ckk =1j+1 =1, m = 0,1,2, . . , j .

It is easy to check that there is a constant C depending on only n and p such that

|| D E0 f ||Lp (Rn ) C|| D f ||Lp ( R+n ) . (6)

If now f W j, p (R+n ) , take a sequence f m C(R +n ) C j, p (R+n ) converging to f inW j , p (R+n ) (we can do this by Lemma 9). Then f m is a Cauchy sequence and (6) impliesthat E0 f m is a Cauchy sequence in W j , p (Rn ). We denote the limit by E0 f . Since|| D E0 f m ||Lp (Rn ) C|| D f m ||Lp (R+n ) , taking limits shows that f satisfies (6).

Definition A domain is of class Cm if can be covered by bounded open sets jsuch that there are mappings j : j B , where B is the unit ball centered at the originand

(i) j( j ) = B R+n(ii) j( j ) = B R+n(iii) j Cm( j) and j1 Cm(B ).

(Because of (iii), all derivatives of order m of j and its inverse are bounded).

THEOREM 11 If is a bounded domain of class Cm then there exists a bounded linearextension operator E:Wm , p() Wm , p (Rn ) .

11

PROOF Since is compact (boundaries are always closed), we might as wellassume that the number of sets j covering is a finite number N . Let U = j =1N j andlet d = dist(,U ). Setting 0 = {x : dist(x,) > d / 2} , we see that0,1,2 , . . , N cover . These sets also cover , which is compact, so by the firstpart of the proof of Lemma 4, there exists a finite partition of unity 0,1,2 , . . ,N for and supp j j . Recall that the support of a function is the closure of the set on whichthat function is non-zero. Hence, supp j is even bounded away from j .

Let f Wm, p () . Then f j Wm, p ( j ), so by our chain rule theorem(Theorem 7)

w j = ( f j ) o j1 Wm, p (R+n B). Clearly supp w j is bounded away fromB , so we can extend w j to be a member of Wm, p (R+n ) by letting it be zero in R+n B . Wecan further extend w j to all of R

n by use of the extension operator E0 of Lemma 10. Let

w j = E0w . If

3. Sobolev Inequalities and Imbedding Theorems

THEOREM 12 If Rn satisfies the cone condition (with height h and opening ) and ifp > 1, mp > n then Wm, p () CB() and there is a constant C depending on only , h,n and p such that for all u Wm , p() , sup|u| C|u|m , p .

Note: does not have to be bounded as Friedman suggests in his Theorem 9.1!

PROOF Initially, suppose that u is in C m, p (). Let g C(R) be such that g( t) =1if t 1 / 2 and g( t) = 0 if t 1. Let x and let (r, ) denote polar coordinates centeredat x. Here, = (1, 2 , . . , n1 ) denotes the angular coordinates and we can describe thecone with vertex x in polar coordinates as Vx ={(r,) : 0 r h, A}. Clearly, we have

u(x) = r {g(r / h)0h u(r,)} dr

=

(1)m(m 1)! r

m 1 mrm {g(r / h)0

h u(r, )} dr ,after m-1 integrations by parts. Next, we integrate with respect to the angular measure dS ,noting that the left-hand-side becomes a constant times u(x).

u(x) = c r m 1 m

r m {g(r / h)0h u(r, )}drdSA

= c rm n m

rm {g(r / h)0h u(r, )} r n1drdSA

= c rm n m

r m {g(r / h)u(r, )}dVV x .Applying Hlder's inequality to this, we obtain

|u(x)| const.||r m n ||Lq (V x ) || m

rm {g(r / h)u(r,)}||Lp (V x ) const. || rm n||Lq (V x ) |u|m, p .

But r m n is in Lq(Vx) if n 1+ (m n)q > 1, which is the case because q = pp1 andmp > n . Thus, we obtain sup|u| C|u|m , p . To extend this result to arbitrary u Wm , p(),take a sequence {uk} of functions in C m, p () converging to u in the | |m, p norm. Thensup|u j uk | C|u j uk |m , p , showing that the sequence is a Cauchy sequence in CB().

13

Thus u is in CB() and taking the limit of sup|u j | C|u j |m , p shows that u satisfies the sameinequality.

Problem 5. Modify the proof to show that the theorem also applies to the case p = 1,m = n .

Problem 6. Show that a similar theorem holds for W0m, p () and the cone condition is

not required. Note that here we can even conclude that W0m, p () {u CB ( ) : u = 0 on

}.

COROLLARY 13 If Rn satisfies the cone condition (with height h and opening ) and if p > 1, (m k)p > n then Wm, p () CBk() and there is a constant C dependingon only , h, n, k and p such that for all u Wm , p() sup| |k | D

u| C|u|m , p .

PROOF Apply the previous theorem to the derivatives Du for || k .

Problem 7 What can you conclude if p = 1 and m k = n ? See Problem 5.Problem 8 What is the corresponding theorem for W0

m, p () ? See Problem 6.

THEOREM 14 If Rn is any domain and p > n then W01, p () C0, ( ) , where =1 np and there exists a constant C depending on only p and n such that for allu W0

1, p()

|u(x) u(y)||| x y|| C || Diu||Lp ()i=1

n .

PROOF Let u C0 () . We might as well assume that u C0 (Rn ) . Let

d =|| x y|| , Sx = Bd (x), Sy = Bd (y) and S = Sx Sy. Then

|u(x) u(y)| vol(S) = |u(x) u(y)| dzS

|u(x) u(z)|+|u(z) u(y)| dzS

|u(x) u(z)| dzSx + |u(z) u(y)| dzSy

But if (r, ) are the polar coordinates of z in a coordinate system centered at x, we get|u(x) u(z)| 0r | u | d , which implies

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|u(x) u(z)| dzSx | u | d rn 1drdS0

r0d | u | d r

n1drdS0d0d

=

dn

n| u | d0

d dS=

dn

n1 n| u |

n 1d0

d dS=

dn

n1 n| u | dzSx

dn

n||1n ||Lq (Sx ) ||

u ||Lp (Sx )

where q = pp 1

. A simple calculation shows that

||1 n||Lq (Sx ) = const. d1

n

p

and it is easy to see that

|| u ||Lp (Sx ) const. || Diu||Lp ()i =1n .

Also, vol(S) = const. dn and the integral over Sy can be estimated in a similar fashion.Putting this together yields

|u(x) u(y)| Cd1n

p || Diu||Lp ()i=1

n

which is precisely the inequality that we wanted. Further, we know from Theorem 12applied to Rn that sup|u| C|u|1, p . Combining this with the previous inequality shows thatfor u C0

() we have || u||C 0, ( ) C|u|1, p . Thus, if we now let u W01, p() and take asequence {um} of functions in C0() converging to u in | |1,p norm, it follows that {um}converges in C0, ( ) . Thus u C0, ( ) , and taking limits shows that u satisfies theinequality in the statement of the theorem.

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THEOREM 15 If Rn is any domain and p < n then W01, p () Lr () , wherer =

npn p

and there exists a constant C depending on only p and n such that for allu W0

1, p()

|| u||Lr () C || Diu||L p ()i =1

n .

Remark The proof relies on a simple generalization of Hlder's inequality which canbe proved by induction by using Hlder's inequality. The inequality states

|u1u2u3 . . um | dx ||u1||L p1 ||u2||Lp2 . . ||um ||Lpm (7)where 1

p1+

1p2

+1p3

+ . . . +1pm

= 1.

PROOF of Theorem 15. It suffices to prove the result for u C01(Rn ). First we prove

the result for the case p = 1. For each i we have

|u(x)| | Diu| dxi

xi | Diu| dxi

.Multiplying these n inequalities together and taking the n 1 th root gives

|u(x)|n

n 1 ( | Diu| dxi

) 1n1i =1

n (8)

Observe that | Diu| dxi

does not depend on xi , but it does depend on all n 1 of theremaining variables. We integrate each side of (8) with respect to x1 and use thegeneralized Hlder inequality with pi = m = n 1 to obtain

|u(x)|n

n1 dx1

( | D1u| dx1

) 1n 1 ( | Diu| dxi

) 1n 1i= 2

n dx1

( | D1u| dx1

) 1n1 ( | Diu| dxi

dx1

) 1n1i = 2

n .

16

The RHS is still a product of n 1 functions of x2 , so we integrate each side with respectto x2 , again applying (7) with pi = m = n 1. Continuing in this manner, we finally obtain

|u(x)|n

n1 dxRn ( | Diu| dxRn

i =1

n )1

n1

i.e. || u||L

n

n1 ( | Diu| dxRn

i =1

n )1n (geometric mean)

1n

|Diu| dxRni=1

n (arithmetic mean)

Here we have used the fact that an arithmetic mean is no less than a geometric mean of thesame numbers. This proves the result for the case p = 1.

For p > 1, let = (n 1)pn p

= 1 + n(p 1)n p

. Since > 1 and u C01(Rn ), it follows

that |u| C01(Rn) . Clearly

Di |u| = (n 1) pn p

|u|n( p1)

n p (Diu) .

We apply the p = 1 case to |u| and obtain

( |u|np

n p dxRn )

n 1n

1n

(n 1)pn p

|u|n( p1)

n p | Diu| dxRni =1

n

(n 1)pn(n p) ( (|u|

n( p1)n p )

pp1 dx

Rn )p1p || Diu||Lp

i =1

n=

(n 1)pn(n p) ( |u|

npn p dx

Rn )p1

p || Diu||L pi =1

n

Hence

( |u|np

n p dxRn )

n pnp

(n 1) pn(n p) || Diu||Lpi =1

n ,

which is the desired result. As usual, to obtain the same result for a function u W01, p(),

we just take a sequence of functions in C01 converging to u.

17

Remark According to the theorem, W01, p () Lr (), where r is given above. But

obviously W01, p () Lp (), so by the following interpolation lemma, W01, p () Lq ()

for all q satisfying p q r . If is bounded then clearly this holds for all q satisfying1 q r .

LEMMA 16 If s q r and Ls () Lr () , then Lq () and

|| ||Lq || ||Ls || ||Lr1 ,

where = s(r q)q(r s) .

PROOF Apply Hlder's inequality to the integral of | |q , using the facts that| |(1 )q L

r

(1 )q and | | q L

s

q.

Problem 9 Modify the proof of Theorem 15 to show that if p = n >1 then

W01, p () Lr () for every r p . Hint: First prove this for the case r > n

2

n 1 by setting

= r n 1n

and applying the p = 1 result to |u| . The r nn 1

norm that shows up on the

RHS after applying Hlder's inequality (as we did in our proof above) can be estimated interms of the n = p norm and the r norm by use of Lemma 16. Finally, obtain the result forall r p by applying Lemma 16 to the result that you have just proved.

COROLLARY 17 For every domain in Rn there exists a constant C depending ononly n and p such that

a) if kp < n then W0k, p () Lnp

n kp () and for each u W0k , p ()

|| u||L

npnkp

C|u|k , p

b) if kp > n then W0k, p () Cm , ( ) , where m is the integer satisfying0 < k m n

p

PROOF a) If || k 1 and u W0k , p then Du W01, p , which is contained in Lnp

n p

by Theorem 15. Thus W0k, p

W0k 1,

npn p

. Iterating this process once more, we find

W0k, p

W0k 2,

npn2 p

. Continuing the iterations culminates in the desired result.

b) Since (k m 1)p < n , case (a) implies that W0k m 1, p Lnp

n (k m1)p= L

n

(1 ). Thus if

u W0k , p

and || m +1, then Du W0k m 1, p . Hence W0k, p W0m +1, n(1 )

. But this shows

that if u W0k , p

and || m then Du W01, n(1 )

, which is contained in C0, ( ) byTheorem 14. Thus W0

k, p () Cm , ( ) .

Remarks A few "particular cases" have been left out because they require separateproofs (see Problem 10 below). They are:

i) If kp = n and p > 1 then W0k, p () Lq () for all q satisfying p q < .ii) If kp = n and p = 1 (so that k = n) then W0k, p () CB ( ) .

iii) If kp > n , p > 1 and np

is an integer then W0k, p () W0

k n

p,q

() for all q satisfying

p q < .

iv) If kp > n and p = 1 (so np

is obviously an integer) then W0k, p () CBk n ( ).

All of the particular cases listed above have the appropriate norm inequalities associatedwith them.

Problem 10 Use the results of Problems 5, 6, 7, 8 and 9 to prove the particular caseslisted above.

COROLLARY 18 If is a bounded C1 domain in Rn (or any other domain such thatthere exists a bounded extension operator E : W1, p () W1, p (Rn) ) then the statementsconcerning the spaces W0

k, p () in Corollary 17 and in the remark following the corollaryalso apply to the spaces W k, p () . However, the constant C may also depend on .

19

PROOF The cases for k = 1 dealt with in Theorems 14 and 15 are easily seen tohave their counterparts here because of the extension operator. Inspection of the proof ofCorollary 17 shows how the results for k >1 may be derived from the results for k = 1without any additional assumptions on the domain.

Remark One can show that extension operators exist for Lipschitz domains and evendomains satisfying certain cone conditions (see the remarks following the proof ofTheorem 11). This Sobolev imbedding theorem is thus valid for such domains.

Definition Let A and B be Banach spaces. If A B , we say that A is continuouslyimbedded in B (in symbols, this is written A\o(\s\do3( ), )B)if there is a constant Csuch that || x||B C|| x||A for all x A .

The theorems in this section provide examples of imbeddings and are called

Sobolev Imbedding Theorems. e.g. W01, p () \o(\s\do3( ), ) L

npn p

for p < n . It is easy to see that A\o(\s\do3( ), )B is equivalent to the identity mapping fromA into B being continuous (i.e. bounded).

4. Compactness Theorems

Definition Suppose that A\o(\s\do3( ), )B. We say that A is compactly imbeddedin B if every sequence bounded in A has a subsequence that converges in B.

e.g. If K is compact then any bounded sequence in C1(K) is a set of equicontinuousfunctions so, by the Arzela-Ascoli Theorem, it has a subsequence that converges in C(K) .i.e. C1(K) is compactly imbedded in C(K) .

Recall that if A and B are Banach spaces and if M : A B is a bounded linearmapping then M is said to be compact if for every bounded sequence {xm} in A thesequence {Mxm} has a subsequence that converges. Thus, saying that A is compactlyimbedded in B is equivalent to saying that the identity mapping from A into B is compact. Itis easy to see that if M : A B and P : B C are bounded linear mappings and A, B andC are Banach spaces then PM is compact if one of the mappings A or B is compact.Consequently, we obtain the very useful result that if A\o(\s\do3( ), )B and B\o(\s\do3(), )C then the imbedding A\o(\s\do3( ), )C is compact if one of the other twoimbeddings is compact.

20

LEMMA 19 Suppose that is a bounded domain. Ifa) 0 < 1 then Cm, ( ) is compactly imbedded in Cm( ).b) 0 < < 1 then Cm, ( ) is compactly imbedded in Cm, ( ) .

PROOF It suffices to prove the results for m = 0 because, once this is done, we canapply this case to the derivatives of the functions and deduce the result for general m.. Let{ f j} be a sequence in C0, ( ) such that || f j ||C0, M . But this implies| f j(x) f j (y)| M || x y|| , showing that the sequence is a bounded, equicontinuous setof functions. By the Arzela-Ascoli Theorem, there exists a subsequence { f jk } thatconverges in C( ). Thus C0, ( ) is compactly imbedded in C( ).

We show below that the same subsequence also converges in C0, ( ). Supposethat C0, ( ). Then

[ ]0, = sup| (x) (y)|

||x y|| = sup| (x) (y)|

|| x y||

| (x) (y)|1

21

([ ]0, )

(max | | )1

We apply this to f jk f j r , noting that [ f j k f jr ]0, [ f jk ]0, + [ f jr ]0, 2M , and obtain

[ f j k f jr ]0, 2M

(max | f jk f jr | )1

,

showing that the subsequence is a Cauchy sequence in C0, ( ) (because it converges inC( )). Thus the subsequence converges in C0, ( ).

COROLLARY 20 If is bounded, kp > n and 0 < k m np

COROLLARY 21 If is a bounded C1 domain (or any other domain for which thereis a bounded extension operator E : W1, p () W1, p (Rn) ), kp > n and 0 < k m n

p

(ii) Every sequence in E has a convergent subsequence.(iii) E is totally bounded.

The Theorem gives us two other very useful characterizations of compact mappings andimbeddings.

THEOREM 23 If is bounded and p < n, then W01, p () is compactly imbedded inLq() for all q < np

n p.

PROOF Consider first the case q = 1. Let A be a bounded set in W01, p () . We may

consider the members of A as members of W1, p (Rn ) with supports contained in . LetAh = {Jhu : u A}. Note that we have

| Jhu(x)| h n ( x zh )|u(z)| dz h n (max)|| u||L1 ()and | Di Jhu(x)| hn 1 | Di( x zh )||u(z)| dz h n1(max| Di|)||u||L1 () .Since is bounded, || u||L1 const.||u||L p . The inequalities above show that Ah is abounded equicontinuous set of functions in C( ). By the Arzela-Ascoli Theorem, everysequence in Ah has a subsequence that converges in C( ). Obviously, such subsequencesalso converge in L1(), so we see that Ah is totally bounded in L1().

If u A then

u(x) Jhu(x) = (z)(u(x) u(x hz)) dz|z|1

= (z) r u(x rz

|| z|| ) dr0h| |z|| dz

|z|1 .

Thus |u(x) Jhu(x)| (z) | Diu(x r z|| z|| )|i =1n dr0

h ||z || dz|z |1 .

Integrating this with respect to x, we find

23

|u(x) Jhu(x)| dx (z) | Diu(x r z||z||)|Rni =1n dx dr0

h ||z || dz|z |1

= (z) | Diu(x)|i =1

n dx dr0h| |z|| dz

|z|1

= (z)h||z|| | Diu(x)|i=1

n dx dz|z|1

h |Diu(x)|i=1

n dx hB, (9)

where B is a constant depending on our bound of members of A in W01, p () (again we use

the fact that the L1 norm is weaker than the Lp norm on a bounded domain).Let > 0. Since Ah is totally bounded in L

1(), we can cover Ah by a finitenumber of balls Bi of radius / 2. Let h =

2B. By (9), if Jhu Bi , then u is contained in

a ball of radius centered at the center of Bi . Thus, A is covered by a finite number ofballs of radius . i.e. A is totally bounded in L1(). Thus W01, p () is compactly imbeddedin L1().

Suppose W01, p (). Then Lnp

n p by Theorem 15 and we get from Lemma 16

(with s = 1 and r = npn p

) that

|| ||Lq || ||L1 || ||L

npn p

1 C|| ||L1 (|| || Di ||L pi =1

n )1

Now let {um} be a bounded sequence in W01, p () and assume |um |1, p M . Since W01, p ()is compactly imbedded in L1(), we can extract a subsequence {um j} that converges inL1(). Applying the inequality above to um j umk , noting that |um j umk |1, p 2M , weobtain

|| um j umk ||Lq const.||um j umk ||L1 ,

showing that the subsequence is a Cauchy sequence in Lq() . Hence the subsequenceconverges in Lq() and W01, p () is compactly imbedded in Lq() .

24

COROLLARY 24 If kp < n and is bounded then W0k, p () is compactly imbeddedin Lq() for all q < np

n kp.

PROOF W0k, p () is continuously imbedded in W0

1,np

n (k 1) p (), which is compactlyimbedded in Lq() if q < np

n kp, by Theorem 23.

COROLLARY 25 The same compactness results hold for W k, p () if is a bounded,C1 domain (or any other type of bounded domain for which there is an extension operatorE : W1, p () W1, p (Rn) .

PROOF See the proof of Corollary 21.

Remark The case kp = n is missing from the previous results. But since W k, p iscontinuously imbedded in W k, r for all r < p (provided that the domain is bounded), itfollows from Corollary 25 that W k, p is compactly imbedded in Lq() for all q < . Thesame applies to W0

k, p ().

5. Interpolation Results

The following results are very useful in PDE theory. We make use of Theorem 26in our proof of Grding's Inequality in our study of elliptic problems.

THEOREM 26 Let u W0k , p () . Then for any > 0 and any 0

First suppose that u C02 (R) and consider an interval (a,b) of length b a = . If

y (a,a + / 3) and z (b / 3,b), then by the Mean Value Theorem there is ap (a,b) such that

| u ( p)|=| u(z ) u(y)z y

| 3

(|u(z)|+|u(y)|)

Consequently, for every x (a,b), we obtain

| u (x)|=| u ( p) + u (t) dtp

x | 3 (|u(z)|+|u(y)|) + | u (t)| dtab .

Integrating with respect to y and z over the intervals (a, a + / 3) and (b / 3,b)respectively, we obtain

| u (x)| | u (t)| dta

b + 18 2 |u(t)| dtab ,

so by Hlder's inequality and the inequality (A + B)p 2p1(A p + B p ),

| u (x)|p 2p 1({ | u (t)| dta

b }p + (18)p 2p { | u(t)| dtab }p )

2p1({ | u (t )|p dta

b }{ 1 dtab }p1 + (18)p

2 p{ |u(t)|p dt

a

b }{ 1 dtab }p1 )= 2p1 ( p1 | u (t)|p dt

a

b + (18)p p+1 |u(t)|p dtab ).

Integrating this with respect to x over the interval (a,b) gives

| u (x)| p dxa

b = 2 p1( p | u (t)|p dtab + (18)p

p|u(t)|p dt

a

b ).We now subdivide R into intervals of length and obtain by adding all of theseinequalities that

| u (x)|p dx

2p 1( p | u (t)| p dt

+ (18)p p |u(t)|p dt ) (11)

26

Suppose now that u C0 (Rn ) . Then we can apply (11) to u regarded as a function of xi

and integrate with respect to the remaining variables to obtain

| uxi |p dx

Rn 2p1( p | 2u

x i2|p dx

Rn + (18)p

p|u|p dx

Rn )Taking the pth root of this and using (Ap + Bp )1/ p A + B, we obtain (10). (Actually, wedon't quite obtain (10). We actually obtain the inequality (10) for 2 instead of . Butsince is an arbitrary positive constant, (10) holds). Finally, (as usual) to obtain the resultfor u W0

(), we take a sequence of functions in C0 converging to u.

COROLLARY 27 The interpolation inequality stated in Theorem 26 also applies tomembers of W k, p () , provided that is a bounded C2 domain (or any other domain forwhich there is a bounded extension operator E : W2, p () W2,p (Rn ) . Here the constant Cmay also depend on p and .

PROOF Because of the extension operator, an inequality of the form (10) holds forfunctions in W2, p (). The full result follows by induction from this case.

6. The Spaces H k() and H0k().

Definitions H0k() = W0k, 2() and H k() = W k, 2(). These spaces are Hilbert spaces

with inner product

(u,v)k = Du(x)Dv (x) dx|| k .

If = Rn , we get a very useful representation of such functions in terms of theFourier-Plancherel Transform. Recall that for functions f L2 (Rn ) , we define the Fourier-Plancherel transform of f as

f () = limR

(2pi ) n/ 2 eix . f (x) dx||x | | R .

27

The limit exists in the topology of L2(Rn ), || f ||L2 =|| f ||L2 , and f can be recovered by usingthe inversion formula

f (x) = limR

(2pi ) n/ 2 e ix . f ( ) d| | || R .

Again, the limit here exists in the topology of L2(Rn ). The reason for the limits in theseformulas is that the integrands are not necessarily in L1(Rn ). Clearly, the Fourier-Plancherel transform is an isometric (i.e. norms are equal) isomorphism (i.e. boundedlinear mapping with a bounded inverse) from L2(Rn ) onto L2(Rn ).

Integration by parts shows that for f C0(Rn ), the transform of fx j

is i j f ( ).From this, we see by induction that the transform of D f is (i ) f (). If now we letf H k(Rn ) and take a sequence of C0 functions converging to f, we find that (i ) f ()is in L2(Rn ) for all || k and the transform of D f is (i ) f (). Thus, we see that iff H k(Rn ) then f H k(Rn ), where

H k(Rn ) ={g L2 (Rn ) : (1+|| ||) k g( ) L2(Rn )}.

It is easy to see that C0

is dense in H k and if g is in C0

then g is the transform of aninfinitely differentiable, rapidly decaying function f (a function in the Schwarz class, to beprecise). Taking limits, we see that if g is in H k then g is the transform of a function fbelonging to H k . Further, if we define an inner product on H k as

(u,v) H k = 2u( )v () d| |k ,

we find that || f ||k =|| f || H k . Thus, the Fourier-Plancherel transform is an isometricisomorphism from H k onto H k . Questions about functions in H k are thus transformedinto equivalent (and often simpler) questions about functions in H k .

Problem 12 Consider the initial value problem for the wave equation

d2udt 2

= u , u(0) = f H k , dudt

(0) = g H k 1 .

28

(We think of u as being a function of t taking values in H k ). Construct a candidate u for asolution using Fourier transforms.a) Show that u is a continuous H k -valued function of t.b) Show that u is a continuously differentiable H k 1-valued function of t.c) Show that du

dt is a continuously differentiable H k 2 -valued function of t and that

d2udt 2

= u in H k 2 .

d) How large does k have to be in order for u to be a classical (i.e. C2 ) solution.Hints: Clearly it suffices to answer the equivalent questions about u . Use the DominatedConvergence Theorem to help you answer a), b), c). For d), use the Sobolev ImbeddingTheorem.

7. Trace Theorems.

In PDE Theory, one often needs to know how functions behave on boundaries ofdomains. If f is a function defined on a domain , we call the restriction of f to thetrace of f. If all we know about f is that it is in some Lp space, then the trace of f is notwell-defined because has measure zero. However, if kp > n and is a bounded C1

domain in Rn , then we know by Corollary 18 that functions in W k, p () are continuous on and thus they have well-defined traces that are bounded functions. In this section, weconcern ourselves with the important case kp < n.

In the following results, a vector x in Rn is denoted by x = ( x , xn ) , where x belongs to Rn1.

LEMMA 27 If u W1,1 (Rn ) , then for every R , the function v( x ) = u( x , ) is inL1(Rn1) , and

|| v||L1( Rn1 ) ||u||L1 (Rn ) +|| Dnu||L1 (Rn ) .

Remark. One needs to be careful when talking about traces of equivalence classes offunctions. The trace certainly exists for u C0

(Rn ) . For u W1,1 (Rn ) , we know that wecan find a sequence of functions in C0

(Rn ) that converges to u. The norm inequalityasserted in the lemma shows that the sequence of traces of these functions converges inL1(Rn1) . It is in this sense that the trace of u exists in L1(Rn1) .

29

PROOF It suffices to prove the result for the case = 0 and u C0 (Rn ) . By theMean Value Theorem for integrals

|u( x , xn )| d x dxnRn1

0

1

= |u( x ,)| d x Rn1

for some [0,1]. But

|u( x ,0)|=|u( x , ) Dnu( x ,t) dt0

||u( x ,)|+ | Dnu( x ,t)| dt

0

1

.

Integrating this over Rn1 gives

|| v||L1( Rn1 ) |u( x ,)| d x Rn1 + | Dnu( x ,t)| dt

0

1

Rn1 d x

= |u( x , )| d x Rn1

0

1

dt + | Dnu( x ,t)| dt0

1

Rn1 d x .

This completes the proof of the lemma.

LEMMA 28 If u W1, p(Rn ) where p < n, then for every R , the functionv( x ) = u( x , ) is in Lr (Rn 1) , where

r =(n 1)pn p

= 1+ n(p 1)n p

and there is a constant C depending on only n and p such that

|| v||Lr (Rn1) C |u|1, pRn

.

PROOF We can assume that p > 1 because the p = 1 case is dealt with in theprevious lemma. We first show that if u W1, p(Rn ) then w =|u|r W1,1 (Rn ) and

|| w||L1 const. || Du||Lpr 1 ||u||Lp , || Diw||L1 const. || Du||Lpr . (12)

30

It suffices to prove this result for the case u C0 (Rn ) . Let q = p / (p 1) . Then

(r 1)q = np / (n p) , so by the Sobolev Imbedding Theorem (Th. 15),

|| |u|r 1 ||Lqq const. ||Du||L pnp / (n p)

and combining this with Hlder's Inequality, we get the first of (12):

|| w||L1 = |u|r dx = |u|r1 |u| dx ||u||Lp || |u|r 1 || Lq const.|| Du||Lpr 1 ||u||Lp .Since Diw = r|u|r1 Diu , we obtain the second of (12):

|| Diw||L1 = r|| |u|r 1 ||Lq || Diu||Lp const. || Du||L pr .

We now apply Lemma 27 to w and immediately obtain the inequality

|| v||Lr (Rn1) const.(|| Du||Lpr1 ||u||L p +|| Du||Lpr )1/ r const.(|| Du||Lp11/ r ||u||Lp1/ r +||Du||L p )

const.(||u||Lp +||Du||Lp ).

LEMMA 29 If u Wk , p (Rn ) where kp < n , then for every R , the functionv( x ) = u( x , ) is in Lr (Rn 1) , where

r =(n 1)pn kp

and there is a constant C depending on only n, k and p such that

|| v||Lr (Rn1) C |u|k , pRn

.

PROOF By Sobolev's Imbedding Theorem (Th. 15) applied to the first orderderivatives of u, we have u W1,np / (n(k 1)p )(Rn) . Now apply Lemma 28.

31

Reminder: Parametrized Surface Integrals.

If X(u) = (x1(u1,u2 ), x2 (u1,u2 ), x3 (u1,u2 )) is a parametrization for a smooth surfaceS in R3 , it is well-known from elementary calculus that one may integrate functions definedon S using the formula

f (x) dSS = f o X(u) K(u) du1du2

,

where is the domain of X and

K(u) =|| Xu1 Xu2

|| = (((x3, x2 )(u1,u2 ) )2 + ((x1,x3 )(u1,u2 )

)2 + ((x1, x2 )(u1,u2 ))2 )1/ 2 .

Differential Geometry yields a generalization of this formula. Suppose now that is a domain in Rn1 and that X: Rn is a parametrization for a smooth hypersurface S .Then surface integrals over S may be calculated using

f (x) dSS = f o X(u) K(u) du

,

where

K(u) = ( ( (x1,x2,K, x k,K, xn )(u1,u2 ,K,un ))2

k =1

n )1/ 2 .

Here the x k notation means that the xk term does not appear.

THEOREM 30 Suppose that is bounded and is of class C k . If u Wk , p () wherekp < n , then the restriction v of u to is in Lr () , where

r =(n 1)pn kp

and there is a constant C depending on only n, k and p and such that

|| v||Lr () C |u|k , p .

32

Remark. See the remark following the statement of Lemma 27 for clarification of thephrase "restriction v of u to ". The same remark applies because we know that anextension operator E: W k , p() Wk , p (Rn ) exists, and thus the restriction to offunctions in C0

(Rn ) is dense in W k, p ().

PROOF Let E: W k , p() Wk , p (Rn ) be the extension operator of Theorem 11.Since any u Wk , p () is associated with an element Eu Wk , p (Rn ), we might as well juststudy the properties of the trace on of C0(Rn ) functions.

Let j and j be as in the definition of Ck domains. Since is compact, we

might as well assume that there is a finite number of the j , 1 j N , covering . Let j 1 j N , be a partition of unity for subordinate to this cover. If u C0 (Rn ) , then

( ju)o j1 C0k (B) and we can extend

( ju)o j1 to be in C0k (Rn ) by defining thefunction to be zero outside B. By Lemma 29, the trace w j of

( ju)o j1 on the hyperplaneP: yn = 0 satisfies

|| w j ||Lr (P ) C|( ju)o j1|k , pB Cj |u|k , pRn

,

where C depends on only n, p, and k and C j is independent of u..

X j (y) = j1(y1,K,yn1,0) is a parametrization for the hypersurface S j = () j andwe may estimate the trace

v j = wj o j of ju on this hypersurface using thisparametrization (see the "reminder" preceding the statement of the theorem).

|v j(x)|r dSS j = |w j (y)|r K j(y) dy

PB Rj | wj |r dy

P B ,

where Rj = max(K j ). Comparing this to the preceding inequality, we see that there is aconstant M j independent of u such that

|| v j ||Lr (S j ) M j |u|k, pRn

.

The function v satisfies a similar inequality because v = v j . Finally, it is clear that theresult holds for arbitrary u Wk , p () (see the remark following the statement of thetheorem).

33

Problem 13 Modify the proof of Lemma 27 to show that if u W1, p(Rn ), then for every R , the function v( x ) = u( x , ) is in Lp (Rn1) , and there exists a constant Kdepending only on n and p such that|| v||Lp (Rn1 ) K|u|1, pR

n

.

Problem 14 Deduce from the previous problem and Lemma 28 that the function v ofLemma 28 belongs to Lq(Rn1 ) for all q satisfying p q r .

Appendix: Some Spaces of Continuous Functions.

Here, we define the spaces of continuous functions that appear in these notes. Caution:Notation and definitions of such function spaces vary from text to text. Recall that westated that is a domain in Rn . The connectedness of is not needed in the followingdefinitions, so we need only assume that is an open subset of Rn .

1. C() is the set of functions continuous in .2. C( ) is the set of functions continuous in .3. C k () is the set of functions which have derivatives of order k that are

continuous in .4. C k ( ) is the set of functions in C( ) which have derivatives in of order k

that can be extended to be members of C( ).5. C() is the set of functions in C k () for all k.6. C( ) is the set of functions in C k ( ) for all k.7. C0() is the set of functions in C() that have supports that are compact subsets

of (recall that the support of a function is the closure of the set on which thefunction fails to vanish). Since is open, such functions necessarily vanish in aneighborhood of the boundary of .

8. C0k () is the set of functions in C k () that have supports that are compact subsets

of .9. C0

() is the set of functions in C0k () for all k.10. CB() is the set of bounded functions in C(). This is a Banach space when

equipped with the "sup norm".11. CB( ) is the set of bounded functions in C( ). This is a Banach space when

equipped with the "sup norm". If is bounded, this space coincides with C( ).

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12. CBk () is the set of functions in CB() with derivatives of order k belonging to

CB(). This is a Banach space if we define the norm of a member f of this space assup|| k, x | D

f (x)|.13. CB

k ( ) is the set of functions in both CBk () and C k ( ). This is a Banach space,equipped with the same norm as in (12). If is bounded, this space coincides withC k ( ).

14. C k , ( ) , where 0 < 1, is the set of functions in CBk ( ) that have derivatives oforder k that are uniformly Hlder continuous with exponent . C k , ( ) is aBanach space with norm

|| f ||C k, = sup|| k, x | D f (x)|+[ f ]k, ,

where [ f ]k , = supx , y, x y, | |= k

| D f (x) D f (y)||| x y|| .

References

The results stated in these notes appear in most texts on Sobolev spaces, includingthose listed below. However, there are many different proofs of the results. For thisreason, the key lemmas and theorems that appear in these notes are listed below with areference to the source which has a proof that most resembles the proof in these notes.

Th. 1 [Fr], Th. 6.1L. 2 [Fr], Th. 6.2Th. 3 [Fr], Th. 6.2L. 4 [Zi], L. 2.3.1Th. 5 [Fr], Th. 6.3; [Ad], Th. 3.16.Th. 6 [Ad], Th. 3.18.Th. 7 [Ad], Th. 3.35. (see [Zi], Th. 2.2.2 for Lipschitz changes of variables).Th. 11 [Ad], Th. 4.26; [G.T.], Th. 7.25.Th. 12 [Fr], Th. 9.1Th. 14 [Fr], Th. 9.2Th. 15 [G.T.], Th. 7.10L. 19 [Ad], Th. 1.31

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Th. 22 [Ru], Appendix A4Th. 23 [G.T.], Th. 7.22Th. 26 [G.T.], Th. 7.27; [Ad], Th. 4.13

There is obviously no room here for a complete bibliography. For a more complete list ofreferences, the reader should refer to the bibliographies of the texts listed below.

Ad. R. A. Adams, Sobolev Spaces, Academic Press, 1975.

Fr. A. Friedman, Partial Differential Equations, Krieger, 1983.

G.T. D. Gilbarg and N. S. Trudinger, Elliptic Partial Differential Equations of SecondOrder, Springer-Verlag, 1983.

Ma. V. G. Maz'ja, Sobolev Spaces, Springer-Verlag, 1985.

Ru. W. Rudin, Functional Analysis, MacGraw-Hill, 1973.

Zi. W. P. Ziemer, Weakly Differentiable Functions, Springer-Verlag, 1989.

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