k α fk p is a Hilbertian norm associated to the inner product k α f + α gk p i p k α fk L p (Ω) + k α gk + X

Transcription

1 436 BRUCE K. DRIVER 23. Sobolev Spaces Defnton For p [1, ], k N and Ω an open subset of R d, let W k,p loc (Ω) :={f Lp (Ω) : α f L p loc (Ω) (weakly) for all α k}, W k,p (Ω) :={f L p (Ω) : α f L p (Ω) (weakly) for all α k}, (23.1) kfk W k,p (Ω) := X k α fk p L p (Ω) 1/p f p< and (23.2) kfk W k,p (Ω) = X k α fk L (Ω) f p =. In the specal case of p =2, we wrte W k,2 loc (Ω) =:Hk loc (Ω) and W k,2 (Ω) =:H k (Ω) n whch case k k W k,2 (Ω) = k k H k (Ω) s a Hlbertan norm assocated to the nner product (23.3) (f,g) H k (Ω) = X α f α gdm. Theorem The functon, k k W k,p (Ω), s a norm whch makes W k,p (Ω) nto a Banach space. Proof. Let f,g W k,p (Ω), then the trangle nequalty for the p norms on L p (Ω) and l p ({α : α k}) mples kf + gk W k,p (Ω) = X 1/p k α f + α gk p L p (Ω) X Ω 1/p h p k α fk L p (Ω) + k α gk L p (Ω) X k α fk p L p (Ω) 1/p = kfk W k,p (Ω) + kgk W k,p (Ω). + X k α gk p L p (Ω) Ths shows k k W k,p (Ω) defned n Eq. (23.1) s a norm. We now show completeness. If {f n } n=1 W k,p (Ω) s a Cauchy sequence, then { α f n } n=1 s a Cauchy sequence n L p (Ω) for all α k. By the completeness of L p (Ω), there exsts g α L p (Ω) such that g α = L p lm n α f n for all α k. Therefore, for all φ Cc (Ω), hf, α φ = lm hf n, α φ =( 1) α lm n n h α f n,φ =( 1) α lm hg α,φ. n Ths shows α f exsts weakly and g α = α f a.e. Ths shows f W k,p (Ω) and that f n f W k,p (Ω) as n. 1/p

2 ANALYSIS TOOLS WITH APPLICATIONS 437 Example Let u(x) := x α for x R d and α R. Then u(x) p dx = σ S d 1 R 1 B(,R) r αp rd 1 dr = σ S d 1 R r d αp 1 dr = σ ( S d 1 R d αp d αp f d αp > (23.4) otherwse and hence u L p loc R d ff α<d/p.now u(x) = α x α 1 ˆx where ˆx := x/ x. Hence f u(x) s to exst n L p loc R d t s gven by α x α 1 ˆx whch s n L p loc R d ff α +1 < d/p,.e. f α < d/p 1 = d p p. Let us not check that u W 1,p loc R d provded α<d/p 1. To do ths suppose φ Cc (R d ) and >, then hu, φ = lm u(x) φ(x)dx x > ( ) =lm u(x)φ(x)dx + u(x)φ(x) x x > x = dσ(x). Snce u(x)φ(x) x x = dσ(x) kφk σ S d 1 d 1 α as and u(x) = α x α 1 ˆx e s locally ntegrable we conclude that hu, φ = u(x)φ(x)dx R d showng that the weak dervatve u exsts and s gven by the usual pontwse dervatve Mollfcatons. Proposton 23.4 (Mollfcaton). Let Ω be an open subset of R d,k N := N {},p [1, ) and u W k,p loc (Ω). Then there exsts u n Cc (Ω) such that u n u n W k,p loc (Ω). Proof. Apply Proposton wth polynomals, p α (ξ) =ξ α, for α k. Proposton Cc (R d ) s dense n W k,p (R d ) for all 1 p<. Proof. The proof s smlar to the proof of Proposton 23.4 usng Exercse 19.2 n place of Proposton Proposton Let Ω be an open subset of R d,k N := N {} and p 1, then (1) for any α wth α k, α : W k,p (Ω) W k α,p (Ω) s a contracton. (2) For any open subset V Ω, the restrcton map u u V s bounded from W k,p (Ω) W k,p (V ). (3) For any f C k (Ω) and u W k,p k,p loc (Ω), the fu Wloc (Ω) and for α k, (23.5) α (fu)= X µ α β f α β u β β α where α β := α! β!(α β)!.

3 438 BRUCE K. DRIVER (4) For any f BC k (Ω) and u W k,p loc k,p (Ω), the fu Wloc (Ω) and for α k Eq. (23.5) stll holds. Moreover, the lnear map u W k,p (Ω) fu W k,p (Ω) s a bounded operator. Proof. 1. Let φ C c (Ω) and u W k,p (Ω), then for β wth β k α, h α u, β φ =( 1) α hu, α β φ =( 1) α hu, α+β φ =( 1) β h α+β u, φ from whch t follows that β ( α u) exsts weakly and β ( α u)= α+β u. Ths shows that α u W k α,p (Ω) and t should be clear that k α uk W k α,p (Ω) kuk W k,p (Ω). Item 2. s trval Gvenu W k,p loc (Ω), by Proposton 23.4 there exsts u n Cc (Ω) such that u n u n W k,p loc (Ω). From the results n Appendx A.1, fu n Cc k (Ω) W k,p (Ω) and (23.6) α (fu n )= X β α µ α β f α β u n β holds. Gven V o Ω such that V s compactly contaned n Ω, we may use the above equaton to fnd the estmate k α (fu n )k L p (V ) X µ α β β f L (V ) α β u L n p (V ) β α C α (f,v ) X α β u L n p (V ) C α(f,v ) ku n k W k,p (V ) β α wheren the last equalty we have used Exercse 23.1 below. Summng ths equaton on α k shows (23.7) kfu n k W k,p (V ) C(f,V ) ku nk W k,p (V ) for all n where C(f,V ) := P C α(f,v ). By replacng u n by u n u m n the above nequalty t follows that {fu n } n=1 s convergent n W k,p (V ) and snce V was arbtrary fu n fu n W k,p loc (Ω). Moreover, we may pass to the lmt n Eq. (23.6) and n Eq. (23.7) to see that Eq. (23.5) holds and that kfuk W k,p (V ) C(f,V ) kuk W k,p (V ) C(f,V ) kuk W k,p (Ω) Moreover f f BC (Ω) then constant C(f,V ) maybechosentobendependent of V and therefore, f u W k,p (Ω) then fu W k,p (Ω). Alternatve drect proof of 4. We wll prove ths by nducton on α. If α = e then, usng Lemma 19.9, hfu, φ = hu, f φ = hu, [fφ] f φ = h u, fφ + hu, f φ = hf u + f u, φ showng (fu) exsts weakly and s equal to (fu)=f u + f u L p (Ω). Supposng the result has been proved for all α such that α m wth m [1,k). Let γ = α + e wth α = m, then by what we have just proved each summand n Eq. (23.5) satsfes β f α β u exsts weakly and β f α β u = β+e f α β u + β f α β+e u L p (Ω).

4 ANALYSIS TOOLS WITH APPLICATIONS 439 Therefore γ (fu)= α (fu) exsts weakly n L p (Ω) and γ (fu)= X µ α β+e f α β u + β f α β+e u = X β β α β γ For the last equalty see the combnatorcs n Appendx A.1. µ γ β f γ β u. β Theorem Let Ω be an open subset of R d,k N := N {} and p [1, ). Then C (Ω) W k,p (Ω) s dense n W k,p (Ω). Proof. Let Ω n := {x Ω :dst(x, Ω) > 1/n} B (,n), then Ω n {x Ω :dst(x, Ω) 1/n} B (,n) Ω n+1, Ω n s compact for every n and Ω n Ω as n. Let V = Ω 3,V j := Ω j+3 \ Ω j for j 1, K := Ω 2 and K j := Ω j+2 \ Ω j+1 for j 1 as n fgure 41. Then K V n Ω 5 Ω 4 Ω 3 Ω 2 1 Ω Fgure 41. Decomposng Ω nto compact peces. The compact sets K,K 1 and K 2 are the shaded annular regons whle V,V 1 and V 2 are the ndcated open annular regons. for all n and K n = Ω. Choose φ n Cc (V n, [, 1]) such that φ n =1on K n and set ψ = φ and j 1 Y ψ j =(1 ψ 1 ψ j 1 ) φ j = φ j (1 φ k ) k=1 for j 1. Then ψ j Cc (V n, [, 1]), nx ny 1 ψ k = (1 φ k ) as n k= k=1 so that P k= ψ k =1on Ω wth the sum beng locally fnte. Let > be gven. By Proposton 23.6, u n := ψ n u W k,p (Ω) wth supp(u n V n. By Proposton 23.4, we may fnd v n C c (V n ) such that

5 44 BRUCE K. DRIVER ku n v n k W k,p (Ω) /2n+1 for all n. Let v := P n=1 v n,thenv C (Ω) because the sum s locally fnte. Snce X ku n v n k W k,p (Ω) X /2 n+1 = <, n= the sum P n= (u n v n ) converges n W k,p (Ω). The sum, P n= (u n v n ), also converges pontwse to u v and hence u v = P n= (u n v n ) s n W k,p (Ω). Therefore v W k,p (Ω) C (Ω) and X ku vk ku n v n k W k,p (Ω). n= n= Notaton Gven a closed subset F R d, let C (F ) denote those u C (F ) that extend to a C functononanopenneghborhoodoff. Remark It s easy to prove that u C (F ) ff there exsts U C R d such that u = U F. Indeed, suppose Ω s an open neghborhood of F, f C (Ω) and u = f F C (F ). Usng a partton of unty argument (makng use of the open sets V constructed n the proof of Theorem 23.7), one may show there exsts φ C (Ω, [, 1]) such that Ω and φ =1on a neghborhood of F. Then U := φf s the desred functon. Theorem 23.1 (Densty of W k,p (Ω) C Ω n W k,p (Ω)). Let Ω R d be a manfold wth C boundary, then for k N and p [1, ), W k,p (Ω o ) C Ω s dense n W k,p Ω. Ths may alternatvely be stated by assumng Ω R d s an open set such that Ω o = Ω and Ω s a manfold wth C boundary, then W k,p (Ω) C Ω s dense n W k,p (Ω). Before gong nto the proof, let us pont out that some restrcton on the boundary of Ω s needed for asserton n Theorem 23.1 to be vald. For example, suppose Ω := x R 2 :1< x < 2 ª and Ω := Ω \{(1, 2) {}} and θ : Ω (, 2π) s defned so that x 1 = x cos θ(x) and x 2 = x sn θ(x), see Fgure 42. Then θ BC (Ω) W k, (Ω) for all k N yet θ can not be Fgure 42. The regon Ω along wth a vertcal n Ω.

6 ANALYSIS TOOLS WITH APPLICATIONS 441 approxmated by functons from C Ω BC (Ω ) n W 1,p (Ω). Indeed, f ths were possble, t would follows that θ W 1,p (Ω ). However, θ s not contnuous (and hence not absolutely contnuous) on the lnes {x 1 = ρ} Ω for all ρ (1, 2) and so by Theorem 19.3, θ/ W 1,p (Ω ). The followng s a warm-up to the proof of Theorem Proposton (Warm-up). Let f : R d 1 R be a contnuous functon and Ω := x R d : x d >f(x 1,...,x d 1 ) ª and C ( Ω) denote those u C Ω whch are restrctons of C functonsdefnedonanopenneghborhoodof Ω. Then for p [1, ), C ( Ω) W k,p (Ω) s dense n W k,p (Ω). Proof. By Theorem 23.7, t suffces to show than any u C (Ω) W k,p (Ω) may be approxmated by elements of C Ω W k,p (Ω). For s> let u s (x) := u(x + se d ). Then t s easly seen that α u s =( α u) s for all α and hence u s W k,p (Ω se d ) C (Ω se d ) C Ω W k,p (Ω). These observatons along wth the strong contnuty of translatons n L p (see Proposton 11.13), mples lm s ku u s k W k,p (Ω) = ProofofTheorem23.1. Proof. By Theorem 23.7, t suffces to show than any u C (Ω) W k,p (Ω) may be approxmated by elements of C Ω W k,p (Ω). To understand the man deas of the proof, suppose that Ω s the trangular regon n Fgure 43 and suppose that we have used a partton of unty relatve to the cover shown so that u = u 1 + u 2 + u 3 wth supp(u ) B. Now concentratng on Fgure 43. Splttng and movng a functon n C (Ω) so that the result s n C Ω. u 1 whose support s depcted as the grey shaded area n Fgure 43. We now smply translate u 1 n the drecton v shown n Fgure 43. That s for any small s>, let w s (x) :=u 1 (x + sv), then v s lves on the translated grey area as seen n Fgure 43. The functon w s extended to be zero off ts doman of defnton s an element of C Ω moreover t s easly seen, usng the same methods as n the proof of Proposton 23.11, that w s u 1 n W k,p (Ω). The formal proof follows along these same lnes. To do ths choose an at most countable locally fnte cover {V } = of Ω such that V Ω and for each 1,

7 442 BRUCE K. DRIVER after makng an affne change of coordnates, V =(, ) d for some > and V Ω = {(y, z) V : >z>f (y)} where f :(, ) d 1 (, ), seefgure44below. Let{η } = be a partton of Ω Fgure 44. The shaded area depcts the support of u = uη. unty subordnated to {V } and let u := uη C (V Ω). Gven δ>, we choose s so small that w (x) :=u (x+se d ) (extendedtobezerooff ts doman of defnton) maybevewedasanelementofc ( Ω) andsuchthatku w k W k,p (Ω) <δ/2. For =we set w := u = uη. Then, snce {V } s a locally fnte cover of Ω, t follows that w := P = w C Ω and further we have X ku w k W k,p (Ω) X δ/2 = δ. Ths shows = u w = X (u w ) W k,p (Ω) = and ku wk W k,p (Ω) <δ.hence w C Ω W k,p (Ω) s a δ approxmaton of u and snce δ> arbtrary the proof s complete Dfference quotents. Recall from Notaton that for h 6= h u(x) := u(x + he ) u(x). h Remark (Adjonts of Fnte Dfferences). For u L p and g L q, h u(x + he ) u(x) u(x) g(x) dx = g(x) dx = u(x) g(x he ) g(x) R d R h d R h d = u(x) h g(x) dx. R d We summarze ths dentty by ( h) = h. Theorem Suppose k N, Ω s an open subset of R d and V s an open precompact subset of Ω. (1) If 1 p<, u W k,p (Ω) and u W k,p (Ω), then dx (23.8) k h uk W k,p (V ) k uk W k,p (Ω) for all < h < 1 2 dst(v,ωc ).

8 ANALYSIS TOOLS WITH APPLICATIONS 443 (2) Suppose that 1 <p,u W k,p (Ω) and assume there exsts a constant C(V ) < such that k h uk W k,p (V ) C(V ) for all < h < 1 2 dst(v,ωc ). Then u W k,p (V ) and k uk W k,p (V ) C(V ). Moreover f C := sup V Ω C(V ) < then n fact u W k,p (Ω) and there s a constant c< such that ³ k uk W k,p (Ω) c C + kuk Lp (Ω). Proof. 1. Let α k, then k α h uk Lp (V ) = k h α uk Lp (V ) k α uk Lp (Ω) wheren we have used Theorem for the last nequalty. Eq. (23.8) now easly follows. 2. If k huk W k,p (V ) C(V ) then for all α k, k h α uk L p (V ) = k α h uk L p (V ) C(V ). So by Theorem 19.22, α u L p (V ) and k α uk L p (V ) C(V ). From ths we conclude that k β uk L p (V ) C(V ) for all < β k +1 and hence kuk W k+1,p (V ) c C(V )+kuk L p (V ) for some constant c. Notaton Gven a mult-ndex α and h 6=, let dy h α := h α. The followng theorem s a generalzaton of Theorem Theorem Suppose k N, Ω s an open subset of R d, V s an open precompact subset of Ω and u W k,p (Ω). (1) If 1 p< and α k, then k h αuk W k α (V ) kuk W k,p (Ω) for h small. (2) If 1 <p and k h αuk W k,p (V ) C for all α j and h near, then u W k+j,p (V ) and k α uk W k,p (V ) C for all α j. Proof. Snce h α = Q α h, tem 1. follows from Item 1. of Theorem and nducton on α. For Item 2., suppose frst that k =so that u L p (Ω) and k h αuk L p (V ) C for α j. Then by Proposton 19.16, there exsts {h l } l=1 R \{} and v Lp (V ) such that h l and lm l h h α u, φ = hv, φ for all φ Cc (V ). Usng Remark 23.12, hv,φ = lm h h α l u, φ =( 1) α lm hu, h α l φ =( 1) α hu, α φ whch shows α u = v L p (V ). Moreover, snce weak convergence decreases norms, k α uk Lp (V ) = kvk L p (V ) C. For the general case f k N, u W k,p (Ω) such that k h αuk W k,p (V ) C, then (for p (1, ), the case p = s smlar and left to the reader) X k h α β uk p L p (V ) = X k β h α uk p L p (V ) = k α h uk p W k,p (V ) Cp. β k β k

9 444 BRUCE K. DRIVER As above ths mples α β u L p (V ) for all α j and β k and that k α uk p W k,p (V ) = X k α β uk p L p (V ) Cp. β k Sobolev Spaces on Compact Manfolds. Theorem (Change of Varables). Suppose that U and V are open subsets of R d,t C k (U, V ) be a C k dffeomorphsm such that k α T k BC(U) < for all 1 α k and := nf U det T >. Then the map T : W k,p (V ) W k,p (U) defned by u W k,p (V ) T u := u T W k,p (U) s well defned and s bounded. Proof. For u W k,p (V ) C (V ), repeated use of the chan and product rule mples, (u T ) =(u T ) T (u T ) =(u T ) T +(u T ) T =(u T ) T T +(u T ) T ³ (u T ) (3) = u (3) T T T T +(u T )(T T ) (23.9). +(u T ) T T +(u T ) T (3) l tmes ³ z } { Xl 1 ³ (u T ) (l) = u (l) T T T + u (j) T p j ³T (l+1 j),t,...,t. Ths equaton and the boundedness assumptons on T (j) for 1 j k mples there s a fnte constant K such that (u T ) (l) lx K u (j) T for all 1 l k. By Hölder s nequalty for sums we conclude there s a constant K p such that X X α (u T ) p K p α u p T and therefore ku T k p W k,p (U) K p X U α u p (T (x)) dx. Makng the change of varables, y = T (x) and usng dy = det T (x) dx dx, we fnd ku T k p W k,p (U) K X p α u p (T (x)) dx (23.1) K p X U V α u p (y) dy = K p kukp W k,p (V ).

10 ANALYSIS TOOLS WITH APPLICATIONS 445 Ths shows that T : W k,p (V ) C (V ) W k,p (U) C (U) s a bounded operator. For general u W k,p (V ), we may choose u n W k,p (V ) C (V ) such that u n u n W k,p (V ). Snce T s bounded, t follows that T u n s Cauchy n W k,p (U) and hence convergent. Fnally, usng the change of varables theorem agan we know, kt u T u n k p L p (V ) 1 ku u n k p L p (U) as n and therefore T u = lm n T u n and by contnuty Eq. (23.1) stll holds for u W k,p (V ). Let M be a compact C k manfolds wthout boundary,.e. M s a compact Hausdorff space wth a collecton of charts x n an atlas A such that x : D(x) o M R(x) o R d s a homeomorphsm such that x y 1 C k (y (D(x) D(y))),x(D(x) D(y))) for all x, y A. Defnton Let {x } N A such that M = N D(x ) and let {φ } N be a partton of unty subordnate do the cover {D(x )} N. We now defne u W k,p (M) f u : M C s a functon such that N (23.11) kuk W k,p (M) := X (φ u) x 1 <. W k,p (R(x )) Snce k k W k,p (R(x )) s a norm for all, t easly verfed that k k W k,p (M) on W k,p (M). s a norm Proposton If f C k (M) and u W k,p (M) then fu W k,p (M) and (23.12) kfuk W k,p (M) C kuk W k,p (M) where C s a fnte constant not dependng on u. Recall that f : M R s sad to be C j wth j k f f x 1 C j (R(x), R) for all x A. Proof. Snce f x 1 has bounded dervatves on supp(φ x 1 ), t follows from Proposton 23.6 that there s a constant C < such that (φ fu) x 1 = W k,p (R(x )) f x 1 (φ u) x 1 C W k,p (R(x )) (φ u) x 1 and summng ths equaton on shows Eq. (23.12) holds wth C := max C. Theorem If {y j } K A such that M = K D(y j) and {ψ j } K s a partton of unty subordnate to the cover {D(y j )} K, then the norm K (23.13) u W k,p (M) := X (ψj u) y 1 j W k,p (R(y j)) s equvalent to the norm n Eq. (23.11). That s to say the space W k,p (M) along wth ts topology s well defned ndependent of the choce of charts and parttons of unty used n defnng the norm on W k,p (M). W k,p (R(x ))

11 446 BRUCE K. DRIVER Proof. Snce W k,p (M) s a norm, NX u W k,p (M) = NX φ u φ u W k,p (M) W k,p (M) KX NX = (ψ j φ u) yj 1 (23.14) KX NX (ψ j φ u) y 1 j W k,p (R(y j)) W k,p (R(y j)) and snce x yj 1 and y j x 1 are C k dffeomorphsm and the sets y j (supp(φ ) supp(ψ j )) and x (supp(φ ) supp(ψ j )) are compact, an applcaton of Theorem and Proposton 23.6 shows there are fnte constants C j such that (ψj φ u) yj 1 C W k,p (R(y j )) j (ψj φ u) x 1 W k,p (R(x )) C j φ u x 1 whch combned wth Eq. (23.14) mples K u W k,p (M) X NX C φ j u x 1 W k,p (R(x )) C kuk W k,p (M) where C := max P K C j <. Analogously, one shows there s a constant K< such that kuk W k,p (M) K u W k,p (M). Lemma Suppose x A(M) and U o M such that U Ū D(x), then there s a constant C< such that (23.15) u x 1 W k,p (x(u)) C kuk W k,p (M) for all u W k,p (M). Conversely a functon u : M C wth supp(u) U s n W k,p (M) ff u x 1 W < and n any case there s a fnte constant such that k,p (x(u)) (23.16) kuk W k,p (M) C u x 1 W k,p (x(u)). W k,p (R(x )) Proof. Choose charts y 1 := x, y 2,...,y K A such that {D (y )} K s an open cover of M and choose a partton of unty {ψ j } K subordnate to the cover {D(y j )} K such that ψ 1 =1on a neghborhood of Ū. To construct such a partton of unty choose U j o M such that U j Ūj D(y j ), Ū U 1 and K U j = M and for each j let η j Cc k (D(y j ), [, 1]) such that η j =1on a neghborhood of Ū j. Then defne ψ j := η j (1 η ) (1 η j 1 ) where by conventon η. Then {ψ j } K s the desred partton, ndeed by nducton one shows andnpartcular lx 1 ψ j =(1 η 1 ) (1 η l ) KX 1 ψ j =(1 η 1 ) (1 η K )=.

12 ANALYSIS TOOLS WITH APPLICATIONS 447 Usng Theorem 23.19, t follows that u x 1 W k,p (x(u)) = (ψ 1 u) x 1 W k,p (x(u)) (ψ1 u) x 1 K W k,p (R(y 1 )) X = u W k,p (M) C kuk W k,p (M) (ψj u) y 1 j W k,p (R(y j)) whch proves Eq. (23.15). Usng Theorems and there are constants C j for j =, 1, 2...,N such that KX kuk W k,p (M) C KX (ψj u) yj 1 = C W k,p (R(y (ψj u) y 1 j)) 1 y 1 yj 1 C K X C j (ψj u) x 1 W k,p (R(y 1 )) = C W k,p (R(y j)) KX C ψj j x 1 u x 1 W k,p (R(y 1 )). Ths nequalty along wth K applcatons of Proposton 23.6 proves Eq. (23.16). Theorem The space (W k,p (M), k k W k,p (M)) s a Banach space. Proof. Let {x } N A and {φ } N be as n Defnton and choose U o M such that supp(φ ) U Ū D(x ). If {u n } n=1 W k,p (M) s a Cauchy sequence, then by Lemma 23.2, u n x 1 ª n=1 W k,p (x (U )) s a Cauchy sequence for all. Snce W k,p (x (U )) s complete, there exsts v W k,p (x (U )) such that u n x 1 ṽ n W k,p (x (U )). For each let v := φ (ṽ x ) and notce by Lemma 23.2 that kv k W k,p (M) C v x 1 W k,p (x (U )) = C kṽ k W k,p (x < (U )) so that u := P N v W k,p (M). Snce supp(v φ u n ) U, t follows that N ku u n k W k,p (M) = X NX v = C C φ u nw k,p (M) NX N kv φ u n k W k,p (M) C X [φ (ṽ x u n )] x 1 NX NX φ x 1 ṽ u n x 1 W k,p (x (U )) C ṽ u n x 1 W k,p (x (U )) as n wheren the last nequalty we have used Proposton 23.6 agan. W k,p (x (U )) Trace Theorems. For many more general results on ths subject matter, see E. Sten [7, Chapter VI].

13 448 BRUCE K. DRIVER ³H d Lemma Suppose k 1, H d := x R d : x d > ª o R d,u Cc k and D sthesmallestconstantsothatsupp(u) R d 1 [,D]. Then there s a constant C = C(p, k, D, d) such that (23.17) kuk W k 1,p ( H d ) C(p, D, k, d) kuk W k,p (H d ). Proof. Wrte x H d as x =(y, z) R d 1 [, ), then by the fundamental theorem of calculus we have for any α N d 1 wth α k 1 that z (23.18) y α u(y, ) = α y u(y, z) y α u t(y, t)dt. Therefore, for p [1, ) α y u(y, ) p 2 p/q α y u(y, z) p + p p 1 where q := R d 1 [,D] mples or equvalently that 2 p/q α y u(y, z) p + 2 p 1 " y α u(y, z) p + z z D p y α u t (y, t)dt q/p α y u t (y, t) p dt z # y α u t (y, t) p dt z p 1 s the conjugate exponent to p. Integratng ths nequalty over D k α uk p L p ( H d ) 2p 1 k α uk p L p (H d ) + α+e d u p L p (H d ) k α uk p L p ( H d ) 2p 1 D 1 k α uk p L p (H d ) +2p 1 Dp 1 p from whch mples Eq. (23.17). Smlarly, f p =, then from Eq. (23.18) we fnd D p p α+e d u p L p (H d ) and agan the result follows. k α uk L ( H d ) = k α uk L (H d ) + D α+e d u L (H d ) Theorem (Trace Theorem). Suppose k 1 and Ω o R d such that Ω s acompactmanfoldwthc k boundary. Then there exsts a unque lnear map T : W k,p (Ω) W k 1,p ( Ω) such that Tu = u Ω for all u C k Ω. Proof. Choose a coverng {V } N = of Ω such that V Ω and for each 1, there s C k dffeomorphsm x : V R(x ) o R d such that x ( Ω V )=R(x ) bd(h d ) and x (Ω V )=R(x ) H d as n Fgure 45. Further choose φ C c (V, [, 1]) such that P N = φ =1on a

14 ANALYSIS TOOLS WITH APPLICATIONS 449 (Ω ) Fgure 45. Coverng Ω (the shaded regon) as descrbed n the text. neghborhood of Ω and set y := x Ω V for 1. Gven u C Ω k, we compute N ku Ωk W k 1,p ( Ω) = X (φ u) Ω y 1 = NX NX max C (φ u) x 1 W k 1,p (R(x ) bd(h d )) bd(h d ) W k 1,p (R(x ) bd(h d )) C (φ u) x 1 W k,p (R(x )) NX (φ u) x 1 W k,p (R(x ) H d ) + (φ u) x 1 C kuk W k,p (Ω) where C =max{1,c 1,...,C N }. The result now follows by the B.L.T. Theorem 4.1 and the fact that C k Ω s dense nsde W k,p (Ω). Notaton In the sequel wll often abuse notaton and smply wrte u Ω for the functon Tu W k 1,p ( Ω). Proposton (Integraton by parts). Suppose Ω o R d such that Ω s a compact manfold wth C 1 boundary, p [1, ] and q = p p 1 s the conjugate exponent. Then for u W k,p (Ω) and v W k,q (Ω), (23.19) u vdm = u vdm + u Ω v Ωn dσ Ω where n : Ω R d s unt outward pontng norm to Ω. Ω Ω W k,p (R(x ))

15 45 BRUCE K. DRIVER Proof. Equaton holds for u, v C Ω 2 and therefore for (u, v) W k,p (Ω) W k,q (Ω) snce both sdes of the equalty are contnuous n (u, v) W k,p (Ω) W k,q (Ω) as the reader should verfy. Defnton Let W k,p (Ω) :=Cc (Ω) W k,p (Ω) be the closure of C c (Ω) nsde W k,p (Ω). Remark Notce that f T : W k,p (Ω) W k 1,p Ω ³ s the trace operator n Theorem 23.23, then T W k,p (Ω) = {} W k 1,p Ω snce Tu = u Ω =for all u Cc (Ω). Corollary Suppose Ω o R d such that Ω s a compact manfold wth C 1 boundary, p [1, ] and T : W 1,p (Ω) L p ( Ω) s the trace operator of Theorem Then W 1,p (Ω) =Nul(T ). Proof. It has already been observed n Remark that W 1,p (Ω) Nul(T ). Suppose u Nul(T ) and supp(u) s compactly contaned n Ω. The mollfcaton u (x) defned n Proposton 23.4 wll be n Cc (Ω) for > suffcently small and by Proposton 23.4, u u n W 1,p (Ω). Thus u W 1,p (Ω). We wll now gve two proofs for Nul(T ) W 1,p (Ω). Proof 1. For u Nul(T ) W 1,p (Ω) defne ½ u(x) for x Ω ũ(x) = for x/ Ω. Then clearly ũ L p R d and moreover by Proposton 23.25, for v Cc (R d ), ũ vdm = u vdm = u vdm R d Ω Ω from whch t follows that ũ exsts weakly n L p R d and ũ =1 Ω u a.e.. Thus ũ W 1,p R d wth kũk W 1,p (R d ) = kuk W 1,p (Ω) and supp(ũ) Ω. Choose V Cc 1 R d, R d such that V (x) n(x) > for all x Ω and defne ũ (x) =T ũ(x) :=ũ e V (x). Notce that supp(ũ ) e Ω Ω for all suffcently small. By the change of varables Theorem 23.16, we know that ũ W 1,p (Ω) and snce supp(ũ ) s a compact subset of Ω, t follows from the frst paragraph that ũ W 1,p (Ω). To so fnsh ths proof, t only remans to show ũ u n W 1,p (Ω) as. Lookng at the proof of Theorem 23.16, the reader may show there are constants δ> and C< such that (23.2) kt vk W 1,p (R d ) C kvk W 1,p (R d ) for all v W 1,p R d. By drect computaton along wth the domnated convergence t may be shown that (23.21) T v v n W 1,p R d for all v Cc (R d ). As s now standard, Eqs. (23.2) and (23.21) along wth the densty of Cc (R d ) n W 1,p R d allows us to conclude T v v n W 1,p R d for all v W 1,p R d whch completes the proof that ũ u n W 1,p (Ω) as. Proof 2. As n the frst proof t suffces to show that any u W 1,p (Ω) may be approxmated by v W 1,p (Ω) wth Ω. As above extend u to Ω c

16 ANALYSIS TOOLS WITH APPLICATIONS 451 by so that ũ W 1,p R d. Usng the notaton n the proof of 23.23, t suffces to show u := φ ũ W 1,p R d may be approxmated by u W 1,p (Ω) wth supp(u Ω. Usng the change of varables Theorem 23.16, the problem may be on B = R(x ). But n ths case we need only defne w (y) :=w (y e d) for > suffcently small. Then supp(w ) Hd B andaswehavealreadyseenw w n W 1,p H d. Thus u := w x W 1,p (Ω), u u as wth supp(u Ω. reduced to workng wth w = u x Extenson Theorems. Lemma Let R>, B:= B(,R) R d,b ± := {x B : ±x d > } and Γ := {x B : x d =}. Suppose that u C k (B \ Γ) C(B) and for each α k, α u extends to a contnuous functon v α on B. Then u C k (B) and α u = v α for all α k. Proof. For x Γ and <d,then by contnuty, the fundamental theorem of calculus and the domnated convergence theorem, u(x + e ) u(x) = lm y x y B\Γ = lm y x y B\Γ and smlarly, for = d, u(x + e d ) u(x) = lm y x y B sgn( ) \Γ = lm y x y B sgn( ) \Γ [u(y + e ) u(y)] = v e (y + se )ds = [u(y + e d ) u(y)] = v ed (y + se d )ds = lm y x y B\Γ v e (x + se )ds lm y x y B sgn( ) \Γ u(y + se )ds v ed (x + se d )ds. d u(y + se d )ds These two equatons show, for each, u(x) exts and u(x) =v e (x). Hence we have shown u C 1 (B). Suppose t has been proven for some l 1 that α u(x) exsts and s gven by v α (x) for all α l<k.then applyng the results of the prevous paragraph to α u(x) wth α = l shows that α u(x) exts and s gven by v α+e (x) for all and x B and from ths we conclude that α u(x) exsts and s gven by v α (x) for all α l +1. So by nducton we conclude α u(x) exsts and s gven by v α (x) for all α k,.e. u C k (B). Lemma Gven any k +1 dstnct ponts, {c } k =, n R\{}, the (k +1) (k +1) matrx C wth entres C j := (c ) j s nvertble. Proof. Let a R k+1 and defne p(x) := P k j= a jx j. If a Nul(C), then kx = (c ) j a j = p (c ) for =, 1,...,k. j= Snce deg (p) k and the above equaton says that p has k +1 dstnct roots, we conclude that a Nul(C) mples p whch mples a =. Therefore Nul(C) = {} and C s nvertble.

17 452 BRUCE K. DRIVER Lemma Let B, B ± and Γ be as n Lemma and {c } k =, be k +1 dstnct ponts n (, 1] for example c = ( +1) wll work. Also let a R k+1 betheunquesoluton(seelemma23.3toc tr a = 1 where 1 denotes the vector of all ones n R k+1,.e. a satsfes kx (23.22) 1= (c ) j a for j =, 1, 2...,k. j= For u C k (H d ) C c (H d ) wth supp(u) B and x =(y,z) R d defne ½ u(y, z) f z (23.23) ũ(x) =ũ(y,z) = P k = a u(y, c z) f z. Then ũ Cc k (R d ) wth supp(ũ) B and moreover there exsts a constant M ndependent of u such that (23.24) kũk W k,p (B) M kuk W k,p (B + ). Proof. By Eq. (23.22) wth j =, kx a u(y,c ) = u(y, ) = kx a = u(y, ). Ths shows that ũ n Eq. (23.23) s well defned and that ũ C H d. Let K := {(y, z) :(y, z) supp(u)}. Snce c (, 1], f x =(y, z) / K and z < then (y, c z) / supp(u) and therefore ũ(x) =and therefore supp(ũ) s compactly contaned nsde of B. Smlarly f α N d wth α k, Eq. (23.22) wth j = α d mples ½ ( α u)(y, z) f z v α (x) := P k = a c α d ( α u)(y, c z) f z. s well defned and v α C R d. Dfferentatng Eq. (23.23) shows α ũ(x) =v α (x) for x B \ Γ and therefore we may conclude from Lemma that u Cc k (B) C k R d and α u = v α for all α k. We now verfy Eq. (23.24) as follows. For α k, p k α ũk p L p (B ) R = kx 1 z< a c α d ( α u)(y, c z) dydz d = 1 z> R d = = kx C 1 z< ( α u)(y, c z) p dydz R d = kx 1 = C c ( α u)(y, z) p dydz = C Ã kx = 1 c! k α uk p L p (B + ) ³ Pk where C := = a c α d q p/q. Summng ths equaton on α k shows there exsts a constant M such that kũk W k,p (B ) M kuk W k,p (B + ) and hence Eq. (23.24) holds wth M = M +1.

18 ANALYSIS TOOLS WITH APPLICATIONS 453 Theorem (Extenson Theorem). Suppose k 1 and Ω o R d such that Ω s a compact manfold wth C k boundary. Gven U o R d such that Ω U, there exsts a bounded lnear (extenson) operator E : W k,p (Ω) W k,p R d such that (1) Eu = u a.e. n Ω and (2) supp(eu) U. Proof. As n the proof of Theorem 23.23, choose a coverng {V } N = of Ω such that V Ω, N = V U and for each 1, there s C k dffeomorphsm x : V R(x ) o R d such that x ( Ω V )=R(x ) bd(h d ) and x (Ω V )=R(x ) H d = B + where B + s as n Lemma 23.31, refer to Fgure 45. Further choose φ Cc (V, [, 1]) such that P N = φ =1on a neghborhood of Ω and set y := x Ω V for 1. Gven u C Ω k and 1, the functon v := (φ u) x 1 may be vewed as a functon n C k (H d ) C c (H d ) wth supp(u) B. Let ṽ Cc k (B) be defned as n Eq. (23.23) above and defne ũ := φ u + P N ṽ x C c R d wth supp(u) U. Notce that ũ = u on Ω and makng use of Lemma 23.2 we learn N kũk W k,p (R d ) kφ uk W k,p (R d ) + X kṽ x k W k,p (R d ) N kφ uk W k,p (Ω) + X kṽ k W k,p (R(x )) N C (φ ) kuk W k,p (Ω) + X kv k W k,p (B + ) N = C (φ ) kuk W k,p (Ω) + X (φ u) x 1 N C (φ ) kuk W k,p (Ω) + X C kuk W k,p (Ω). W k,p (B + ) Ths shows the map u C k ( Ω) Eu := ũ C k c (U) s bounded as map from W k,p (Ω) to W k,p (U). As usual, we now extend E usng the B.L.T. Theorem 4.1 to a bounded lnear map from W k,p (Ω) to W k,p (U). So for general u W k,p (Ω), Eu = W k,p (U) lm n ũ n where u n C k ( Ω) and u = W k,p (Ω) lm n u n. By passng to a subsequence f necessary, we may assume that ũ n converges a.e. to Eu from whch t follows that Eu = u a.e. on Ω and supp(eu) U Exercses. Exercse Show the norm n Eq. (23.1) s equvalent to the norm f W k,p (Ω) := X k α fk L p (Ω).

19 454 BRUCE K. DRIVER Soluton. 23.1Ths s a consequence of the fact that all norms on l p ({α : α k}) are equvalent. To be more explct, let a α = k α fk Lp (Ω), then X a α X 1/p a α p X 1/q 1 q whle X 1/p a α p px p X a β β k 1/p [# {α : α k}] X 1/p a β. β k