17. Weak and Strong Derivatives and Sobolev Spaces For this section, let Ω be an open subset of R d, p,q,r [1, ], L p (Ω) =

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1 33 BRUCE K. DRIVER 17. Weak and Srong Derivaives and Sobolev Spaces For his secion, le Ω be an open subse of R d, p,q,r [1, ], L p (Ω) = L p (Ω, B Ω,m) and L p loc (Ω) =Lp loc (Ω, B Ω,m), where m is Lebesgue measure on B R d and B Ω is he Borel σ algebraonω. If Ω = R d, we will simply wrie L p and L p loc for L p (R d ) and L p loc (Rd ) respecively. Also le hf,gi := fgdm for any pair of measurable funcions f,g : Ω C such ha fg L 1 (Ω). For example, by Hölder s inequaliy, if hf,gi is defined for f L p (Ω) and g L q (Ω) when q = p p 1. The following simple bu useful remark will be used (ypically wihou furher commen) in he sequel. Remark Suppose r, p, q [1, ] are such ha r 1 = p 1 + q 1 and f f in L p (Ω) and g g in L q (Ω) as, hen f g fg in L r (Ω). Indeed, kf g fgk r = k(f f) g + f (g g)k r kf fk p kg k q + kfk p kg gk q as Definiion 17.2 (Weak Differeniabiliy). Le v R d and f L p (Ω) (f L p loc (Ω)) hen v f is said o exis weakly in L p (Ω) (L p loc (Ω)) if here exiss a funcion g L p (Ω) (g L p loc (Ω)) such ha (17.1) hf, v φi = hg, φi for all φ Cc (Ω). The funcion g if i exiss will be denoed by v (w) f. (By Corollary 9.27, here is a mos one g L 1 loc (Ω) such ha Eq. (17.1) holds, so (w) v f is well defined.) Lemma Suppose f L 1 loc (Ω) and (w) v f exiss weakly in L 1 loc (Ω). Then (1) supp m ( v (w) f) supp m (f), where supp m (f) is he essenial suppor of f relaive o Lebesgue measure, see Definiion (2) If f is coninuously differeniable on U o Ω, hen v (w) f = v f a.e. on U. Proof. (1) Since h v (w) f,φi = hf, v φi =for all φ Cc (Ω \ supp m (f)), and applicaion of Corollary 9.27 shows v (w) f =a.e. on Ω \ supp m (f). So by Lemma 9.15, Ω \ supp m (f) Ω \ supp m ( v (w) f), i.e. supp m ( v (w) f) supp m (f). (2) Suppose ha f U is C 1 and le ψ Cc (U) which we view as a funcion in Cc (R d ) by seing ψ on R d \ U. By Corollary 9.24, here exiss γ Cc (Ω) such ha γ 1 and γ =1in a neighborhood of supp(ψ). Then by seing γf =on R d \ supp(γ) we may view γf Cc 1 (R d ) and so by sandard inegraion by pars (see Lemma 9.25) and he ordinary produc rule, h v (w) f,ψi = hf, v ψi = hγf, v ψi (17.2) = h v (γf),ψi = h v γ f + γ v f,ψi = h v f,ψi Ω

2 REAL ANALYSIS LECTURE NOTES 331 wherein he las equaliy we have used ψ v γ =and ψγ = ψ. Since Eq. (17.2) is rue for all ψ Cc (U), an applicaion of Corollary 9.27 wih h = v (w) f(x) v f(x) and µ = m shows v (w) f(x) = v f(x) for m a.e. x U. Lemma 17.4 (Produc Rule). Le f L 1 loc (Ω), v Rd and φ C (Ω). If v (w) f exiss in L 1 loc (Ω), hen (w) v (φf) exiss in L 1 loc (Ω) and (w) v (φf) = v φ f + φ v (w) f a.e. Moreover if φ Cc (R d ) and F := φf L 1 (herewedefine F on R d by seing F =on R d \ Ω ), hen (w) F = v φ f + φ v (w) f exiss weakly in L 1 (R d ). Proof. Le ψ Cc (Ω), hen hφf, v ψi = hf,φ v ψi = hf, v (φψ) v φ ψi = h v (w) f,φψi + h v φ f,ψi = hφ v (w) f,ψi + h v φ f,ψi. This proves he firs asserion. To prove he second asserion le γ Cc (Ω) such ha γ 1 and γ =1on a neighborhood of supp(φ). So for ψ Cc (R d ), using v γ =on supp(φ) and γψ Cc (Ω), we find hf, v ψi = hγf, v ψi = hf, γ v ψi = h(φf), v (γψ) v γ ψi = h(φf), v (γψ)i = h v (w) (φf), (γψ)i This show (w) v = h v φ f + φ (w) v F = v φ f + φ (w) f as desired. v f,γψi = h v φ f + φ v (w) f,ψi. Lemma Suppose p [1, ), v R d and f L p loc (Ω). (1) If here exiss {f m } m=1 Lp loc (Ω) such ha (w) v f m exiss in L p loc (Ω) for all m and here exiss g L p loc (Ω) such ha for all φ C c (Ω), lim m hf m,φi = hf,φi and lim h vf m,φi = hg,φi m hen v (w) f exiss in L p loc (Ω) and vf = g. (2) If v (w) f exiss in L p loc (Ω) hen here exiss f n Cc (Ω) such ha f n f in L p (K) (i.e. lim n kf f n k L p (K) =)and vf n v (w) f in L p (K) for all Ω. Proof. (1) Since hf, v φi = lim hf m, v φi = lim m m h (w) v f m,φi = hg, φi for all φ Cc (Ω), v (w) f exiss and is equal o g L p loc (Ω). (2) Le K := and K n := {x Ω : x n and d(x, Ω c ) 2/n} (so K n Kn+1 o K n+1 for all n and K n Ω as n or see Lemma 8.1) and choose ψ n Cc (Kn, o [, 1]) using Corollary 9.24 so ha ψ n =1 on a neighborhood of K n 1. Given a compac se K Ω, for all sufficienly

3 332 BRUCE K. DRIVER large m, ψ m f = f on K and by Lemma 17.4 and iem 1. of Lemma 17.3, we also have v (w) (ψ m f)= v ψ m f + ψ m v (w) f = v (w) f on K. This argumen shows we may assume supp m (f) is a compac subse of Ω in which case we exend f o a funcion F on R d by seing F = f on Ω and F =on Ω c. This funcion F is in L p (R d ) and v (w) F = v (w) f. Indeed, if φ Cc (R d ) and ψ Cc (R d ) is chosen so ha supp(ψ) Ω, ψ 1 and ψ =1in a neighborhood of supp m (f), hen hf, v φi = hf, ψ v φi = hf, v (ψφ)i = h v (w) f,ψφi = hψ v (w) f,φi which shows v (w) F exis in L p (R d ) and (w) v F = ψ v (w) f =1 Ω v (w) f. Le χ Cc (B (1)) wih R χdm =1and se δ R d k (x) =n d χ(nx). Then here exiss N N and a compac subse K Ω such ha f n := F δ n Cc (Ω) and supp(f n ) K for all n N. By Proposiion 9.23 and he definiion of v (w) F, v f n (x) =F v δ n (x) = F (y) v δ n (x y)dy R d = hf, v [δ n (x )]i = h v (w) F, δ n (x )i = v (w) F δ n (x). Hence by Theorem 9.2, f n F = f and v f n v (w) F = v (w) f in L p (Ω) as n. Definiion 17.6 (Srong Differeniabiliy). Le v R d and f L p, hen v f is said o exis srongly in L p if he lim (τ v f f) / exiss in L p, where as above τ v f(x) :=f(x v). We will denoe he limi by (s) v f. I is easily verified ha if f L p,v R d and v (s) f L p exiss hen v (w) f exiss and v (w) f = v (s) f. To check his asserion, le φ Cc (R d ) andhenusingremark 17.1, v (s) f φ = L 1 lim φ. Hence τ v f f (s) τ vf f v f φdm = lim φdm =lim f τ vφ φ dm R d R d R d = d d fτ v φdm = f d R d R d τ v φdm = f v φdm, d R d wherein we have used Corollary 5.43 o differeniae under he inegral in he fourh equaliy. This shows v (w) f exiss and is equal o v (s) f. Wha is somewha more surprising is ha he converse asserion ha if v (w) f exiss hen so does v (s) f. The nex heorem is a generalizaion of Theorem 1.36 from L 2 o L p. Theorem 17.7 (Weak and Srong Differeniabiliy). Suppose p [1, ), f L p (R d ) and v R d \{}. Then he following are equivalen:

4 REAL ANALYSIS LECTURE NOTES 333 (1) There exiss g L p (R d ) and { n } n=1 R\{} such ha lim n n = and lim hf( + nv) f( ),φi = hg, φi for all φ Cc (R d ). n n (2) v (w) f exiss and is equal o g L p (R d ), i.e. hf, v φi = hg, φi for all φ Cc (R d ). (3) There exiss g L p (R d ) and f n Cc (R d L ) such ha f p L n f and v f p n g as n. f exiss and is is equal o g L p (R d ), i.e. (4) (s) v f( + v) f( ) L p g as. Moreover if p (1, ) any one of he equivalen condiions above are implied by he following condiion. 1. There exiss { n } n=1 R\{} such ha lim n n =and sup n f( + n v) f( ) n <. p Proof. 4. = 1. is simply he asserion ha srong convergence implies weak convergence. 1. = 2. For φ Cc (R d ), hg, φi = lim hf( + nv) f( ),φi = lim hf, φ( nv) φ( ) i n n = hf, lim n n φ( n v) φ( ) i = hf, v φi, n wherein we have used he ranslaion invariance of Lebesgue measure and he dominaed convergence heorem. 2. = 3. Le φ Cc (R d, R) such ha R φ(x)dx = 1 and le φ R d m (x) = m d φ(mx), hen by Proposiion 9.23, h m := φ m f C (R d ) for all m and v h m (x) = v φ m f(x) = v φ m (x y)f(y)dy = hf, v [φ m (x )]i R d = hg, φ m (x )i = φ m g(x). By Theorem 9.2, h m f L p (R d ) and v h m = φ m g g in L p (R d ) as m. This shows 3. holds excep for he fac ha h m need no have compac suppor. To fix hisleψ Cc (R d, [, 1]) such ha ψ =1in a neighborhood of and le ψ (x) =ψ( x) and ( v ψ) (x) :=( v ψ)( x). Then v (ψ h m )= v ψ h m + ψ v h m = ( v ψ) h m + ψ v h m so ha ψ h m h m in L p and v (ψ h m ) v h m in L p as. Le f m = ψ m h m where m is chosen o be greaer han zero bu small enough so ha kψ m h m h m k p + k v (ψ m h m ) v h m k p < 1/m. Then f m Cc (R d ),f m f and v f m g in L p as m. n

5 334 BRUCE K. DRIVER 3. = 4. By he fundamenal heorem of calculus τ v f m (x) f m (x) = f m(x + v) f m (x) = 1 1 d 1 (17.3) ds f m(x + sv)ds = ( v f m )(x + sv)ds. Le 1 1 G (x) := τ sv g(x)ds = g(x + sv)ds which is defined for almos every x and is in L p (R d ) by Minkowski s inequaliy for inegrals, Theorem Therefore τ v f m (x) f m (x) 1 G (x) = [( v f m )(x + sv) g(x + sv)] ds and hence again by Minkowski s inequaliy for inegrals, τ v f m f m p 1 1 G kτ sv ( v f m ) τ sv gk p ds = k v f m gk p ds. Leing m in his equaion implies (τ v f f) / = G a.e. Finally one more applicaion of Minkowski s inequaliy for inegrals implies, τ v f f 1 g = kg gk p = (τ sv g g) ds p 1 kτ sv g gk p ds. By he dominaed convergence heorem and Proposiion 9.13, he laer erm ends o as and his proves 4. (1. = 1. when p>1) This is a consequence of Theorem which assers, by passing o a subsequence if necessary, ha f( + nv) f( ) n w g for some g L p (R d ). Example Thefacha(1 ) does no imply he equivalen condiions 1 4 in Theorem 17.7 when p =1is demonsraed by he following example. Le f := 1 [,1], hen f(x + ) f(x) R dx = 1 1[,1 ] (x) 1 [,1] (x) dx =2 R for < 1. For conradicion sake, suppose here exiss g L 1 (R,dm) such ha f(x + n ) f(x) lim = g(x) in L 1 n n for some sequence { n } n=1 as above. Then for φ C c (R) wewouldhaveonone hand, f(x + n ) f(x) φ(x n ) φ(x) 1 φ(x)dx = f(x)dx φ (x)dx =(φ() φ(1)), R n R n while on he oher hand, R f(x + n ) f(x) φ(x)dx n R g(x)φ(x)dx. p

6 REAL ANALYSIS LECTURE NOTES 335 These wo equaions imply (17.4) g(x)φ(x)dx = φ() φ(1) for all φ Cc (R) R and in paricular ha R R g(x)φ(x)dx =for all φ C c(r\{, 1}). By Corollary 9.27, g(x) =for m a.e. x R\{, 1} and hence g(x) =for m a.e. x R. Bu his clearly conradics Eq. (17.4). This example also shows ha he uni ball in L 1 (R,dm) is no sequenially weakly compac. We will now give a couple of applicaions of Theorem Proposiion Le Ω R be an open inerval and f L 1 loc (Ω). Then w f exiss in L 1 loc (Ω) iff f has a coninuous version f which is absoluely coninuous on all compac subinervals of Ω. Moreover, w f = f a.e., where f (x) is he usual poinwise derivaive. Proof. If f is locally absoluely coninuous and φ Cc (Ω) wih supp(φ) [a, b] Ω, hen by Corollary 14.32, b f φdm = f φdm = Ω a b a fφ dm + fφ b a = fφ dm. Ω This shows w f exiss and w f = f L 1 loc (Ω). Now suppose ha w f exiss in L 1 loc (Ω), le [a, b] be a compac subinerval of Ω, ψ Cc (Ω) such ha ψ =1on a neighborhood of [a, b] and ψ 1 on Ω and define F := ψf. By Lemma 17.4, F L 1 and (w) F exiss in L 1 and is given by (w) (ψf) =ψ f + ψ (w) f. From Theorem 17.7 here exiss F n Cc (R) such ha F n F and Fn (w) F in L 1 as n. Hence, using he fundamenal heorem of calculus, (17.5) F n (x) = x F n(y)dy x (w) F (y)dy =: G(x) as n. Since F n G poinwise and F n F in L 1, i follows ha F = ψf = G a.e. In paricular, G = f a.e. on [a, b] and we conclude ha f has a coninuous version on [a, b], his version, G [a,b], is absoluely coninuous on [a, b] and by Eq. (17.5) and fundamenal heorem of calculus for absoluely coninuous funcions (Theorem 14.31), G (x) = (w) F (x) = (w) f(x) for m a.e. x [a, b]. Since [a, b] Ω was an arbirary compac subinerval and coninuous versions of a measurable funcions are unique if hey exis, i follows from he above consideraions ha here exiss a unique funcion f C(Ω) such ha f = G [a,b] for all such pairs ([a, b],g) as above. The funcion f hen saisfies all of he desired properies. Because of Lemma 17.3, Theorem 17.7 and Proposiion 17.9 i is now safe o simply wrie v f for he ordinary parial derivaive of f, he weak derivaive v (w) f and he srong derivaive v (s) f. Definiion Le X and Y be meric spaces. A funcion f : X Y is said o be Lipschiz if here exiss C< such ha d Y (f(x),f(x )) Cd X (x, x ) for all x, x X

7 336 BRUCE K. DRIVER and said o be locally Lipschiz if for all compac subses K X here exiss C K < such ha d Y (f(x),f(x )) C K d X (x, x ) for all x, x K. Proposiion Le f L 1 (R d ) such ha essenial suppor, supp m (f), is compac. Then here exiss a Lipschiz funcion F : R d C such ha F = f a.e. iff v (w) f exiss and is bounded for all v R d. Proof. Suppose f = F a.e. and F is Lipschiz and le p (1, ). Since F is Lipschiz and has compac suppor, for all < 1, p p f(x + v) f(x) dx = F (x + v) F (x) dx C v p, R d R d where C is a consan depending on he size of he suppor of f and on he Lipschiz consan, C, for F. Therefore Theorem 17.7 may be applied o conclude v (w) f exiss in L p and moreover, F (x + v) F (x) lim = v (w) f(x) for m a.e.x. Since here exiss { n } n=1 R\{} such ha lim n n =and v (w) f(x) = lim F (x + n v) F (x) n n C v for a.e. x Rd, i follows ha v (w) f C v for all v R d. Conversely, le φ m be an approximae δ funcion sequence as used in he proof of Theorem 17.7 and f m := f φ m. Then f m Cc (R d ), v f m = v f φ m, v f m k v fk < and herefore, 1 f m (y) f m (x) = d d f m(x + (y x))d = 1 (y x) f m (x + (y x))d (17.6) 1 y x f m (x + (y x)) d C y x where C is a consan independen of m. By passing o a subsequence of he {φ m } m=1 if necessary, we may assume ha lim m f m (x) =f(x) for m a.e. x R d. Leing m in Eq. (17.6) hen implies f(y) f(x) C y x for all x, y / E where E R d is a m null se. I is now easily verified ha if F : R d C is defined by F = f on E c and ( f(x) if x/ E F (x) = f(y) if x E lim y x y/ E defines a Lipschiz funcion F on R d such ha F = f a.e. Lemma Le v R d. (1) If h L 1 and v h exiss in L 1, hen R R d v h(x)dx =.

8 REAL ANALYSIS LECTURE NOTES 337 (2) If p, q, r [1, ) saisfy r 1 = p 1 + q 1,f L p and g L q are funcions such ha v f and v g exiss in L p and L q respecively, hen v (fg) exiss in L r and v (fg)= v f g + f v g. Moreover if r =1we have he inegraion by pars formula, (17.7) h v f,gi = hf, v gi. (3) If p =1, v f exiss in L 1 and g BC 1 (R d ) (i.e. g C 1 (R d ) wih g and is firs derivaives being bounded) hen v (gf) exiss in L 1 and v (fg)= v f g + f v g and again Eq. (17.7) holds. Proof. 1) By iem 3. of Theorem 17.7 here exiss h n Cc (R d ) such ha h n h and v h n v h in L 1. Then v h n (x)dx = d R d h n (x + v)dx = d d R d h n (x)dx = d R d and leing n proves he firs asserion. 2) Similarly here exiss f n,g n Cc (R d ) such ha f n f and v f n v f in L p and g n g and v g n v g in L q as n. So by he sandard produc rule and Remark 17.1, f n g n fg L r as n and v (f n g n )= v f n g n + f n v g n v f g + f v g in L r as n. I now follows from anoher applicaion of Theorem 17.7 ha v (fg) exiss in L r and v (fg)= v f g + f v g. Eq. (17.7) follows from his produc rule and iem 1. when r =1. 3) Le f n Cc (R d ) such ha f n f and v f n v f in L 1 as n. Then as above, gf n gf in L 1 and v (gf n ) v g f + g v f in L 1 as n. In paricular if φ Cc (R d ), hen hgf, v φi = lim hgf n, v φi = lim h v (gf n ),φi n n = lim h vg f n + g v f n,φi = h v g f + g v f,φi. n This shows v (fg) exiss (weakly) and v (fg)= v f g + f v g. Again Eq. (17.7) holds in his case by iem 1. already proved. Lemma Le p, q, r [1, ] saisfy p 1 + q 1 =1+r 1,f L p,g L q and v R d. (1) If v f exiss srongly in L r, hen v (f g) exiss srongly in L p and v (f g) =( v f) g. (2) If v g exiss srongly in L q, hen v (f g) exiss srongly in L r and v (f g) =f v g. (3) If v f exiss weakly in L p and g C c (R d ), hen f g C (R d ), v (f g) exiss srongly in L r and v (f g) =f v g =( v f) g.

9 338 BRUCE K. DRIVER Proof. Iems 1 and 2. By Young s inequaliy and simple compuaions: τ v (f g) f g ( v f) g = τ v f g f g ( v f) g r τ v = f f ( v f) g r τ v f f ( v f) kgk q p which ends o zero as. The second iem is proved analogously, or jus make use of he fac ha f g = g f and apply Iem 1. Using he fac ha g(x ) Cc (R d ) and he definiion of he weak derivaive, f v g(x) = f(y)( v g)(x y)dy = f(y)( v g(x )) (y)dy R d R d = v f(y)g(x y)dy = v f g(x). R d Iem 3. is a consequence of his equaliy and iems 1. and Sobolev Spaces. Noaion Le α be definedasinnoaion9.1andf L 1 loc (Ω). We say α f exiss weakly in L 1 loc (Ω) iff here exiss g L1 loc (Ω) such ha hf, α φi =( 1) α hg, φi for all φ Cc (Ω). As usual g is unique if i exiss and we will denoe g by α f. Definiion For p [1, ], k N and Ω an open subse of R d, le and define W k,p (Ω) :={f L p (Ω) : α f L p (Ω) (weakly) for all α k} kfk W k,p (Ω) := X α k k α fk L p (Ω). Theorem The space W k,p (Ω) wih he norm k k W k,p (Ω) is a Banach space. Proof. Suppose ha {f n } n=1 W k,p (Ω) is a Cauchy sequence, hen { α f n } n=1 is a Cauchy sequence in L p (Ω) for all α k. By he compleeness of L p (Ω), here exiss g α L p (Ω) such ha g α = L p lim n α f n for all α k. Therefore, for all φ Cc (Ω), hf, α φi = lim hf n, α φi =( 1) α lim n n h α f n,φi =( 1) α lim hg α,φi. n This shows α f exiss weakly and g α = α f a.e. Example Le Ω be an open subse of R d, hen H 1 (Ω) :=W 1,2 (Ω) is a Hilber space wih inner produc defined by (f,g) = f ḡdm + f ḡdm. Ω Proposiion Cc (R d ) is dense in W k,p (R d ) for all 1 p<. Ω r

10 REAL ANALYSIS LECTURE NOTES 339 Proof. The proof of his proposiion is lef an exercise o he reader. However, noe ha he asserion 2. implies 3. in Theorem 17.7 essenially proves he saemen when k =1and he same smooh approximaions used in he proof of Theorem 17.7 will work here as well Hölder Spaces. Noaion Le Ω be an open subse of R d,bc(ω) and BC( Ω) be he bounded coninuous funcions on Ω and Ω respecively. By idenifying f BC( Ω) wih f Ω BC(Ω), we will consider BC( Ω) as a subse of BC(Ω). For u BC(Ω) and <β 1 le kuk u := sup u(x) and [u] β := sup x Ω x,y Ω x6=y ½ u(x) u(y) x y β If [u] β <, we say u is β Hölder coninuous wih holder exponen 35 β and we le C,β (Ω) :={u BC(Ω) :[u] β < } denoe he space of Hölder coninuous funcions on Ω. For u C,β (Ω) le (17.8) kuk C,β (Ω) := kuk u +[u] β. Remark If u : Ω C and [u] β < for some β>1, hen u is consan on each conneced componen of Ω. Indeed, if x Ω and h R d hen u(x + h) u(x) [u] β β / as which shows h u(x) =for all x Ω. If y Ω is in he same conneced componen as x, hen by Exercise 15.5 here exiss as smooh curve σ :[, 1] Ω such ha σ() = x and σ(1) = y. So by he fundamenal heorem of calculus and he chain rule, 1 d 1 u(y) u(x) = d u(σ())d = d =. This is why we do no alk abou Hölder spaces wih Hölder exponens larger han 1. Exercise Suppose u C 1 (Ω) BC(Ω) and i u BC(Ω) for i =1, 2,...,d. Show [u] 1 <, i.e. u C,1 (Ω). Theorem Le Ω be an open subse of R d. Then (1) BC( Ω) isaclosedsubspaceofbc(ω). (2) Every elemen u C,β (Ω) has a unique exension o a coninuous funcion (which we will sill denoe by u) on Ω. Therefore we may idenify C,β (Ω) wih a subspace of BC( Ω). We may also wrie C,β (Ω) as C,β ( Ω) o emphasize his poin. (3) The funcion u C,β (Ω) kuk C,β (Ω) [, ) is a norm on C,β (Ω) which make C,β (Ω) ino a Banach space. ¾. 35 If β =1,uis is said o be Lipschiz coninuous.

11 34 BRUCE K. DRIVER Proof. 1. The firs iem is rivial since for u BC( Ω), he sup-norm of u on Ω agrees wih he sup-norm on Ω and BC( Ω) is complee in his norm. 2. Suppose ha [u] β < and x Ω. Le {x n } n=1 Ω be a sequence such ha x = lim n x n. Then u(x n ) u(x m ) [u] β x n x m β as m, n showing {u(x n )} n=1 is Cauchy so ha ū(x ) := lim n u(x n ) exiss. If {y n } n=1 Ω is anoher sequence converging o x, hen u(x n ) u(y n ) [u] β x n y n β as n, showing ū(x ) is well defined. In his way we define ū(x) for all x Ω and le ū(x) =u(x) for x Ω. Since a similar limiing argumen shows ū(x) ū(y) [u] β x y β for all x, y Ω i follows ha ū is sill coninuous. In he sequel we will abuse noaion and simply denoe ū by u. 3. For u, v C,β (Ω), ½ ¾ v(y)+u(y) v(x) u(x) [v + u] β = sup x,y Ω x y β x6=y ½ ¾ v(y) v(x) + u(y) u(x) sup x,y Ω x y β [v] β +[u] β x6=y and for λ C i is easily seen ha [λu] β = λ[u] β. This shows [ ] β is a semi-norm on C,β (Ω) and herefore k k C,β (Ω) defined in Eq. (17.8) is a norm. To see ha C,β (Ω) is complee, le {u n } n=1 be a C,β (Ω) Cauchy sequence. Since BC( Ω) is complee, here exiss u BC( Ω) such ha ku u n k u as n. For x, y Ω wih x 6= y, u(x) u(y) u n (x) u n (y) x y β = lim n x y β lim sup[u n ] lim ku nk C n n,β (Ω) <, and so we see ha u C,β (Ω). Similarly, u(x) u n (x) (u(y) u n (y)) (u m u n )(x) (u m u n )(y) x y β = lim m x y β lim sup[u m u n ] β as n, m showing [u u n ] β as n and herefore lim n ku u n k C,β (Ω) =. Noaion Since Ω and Ω are locally compac Hausdorff spaces, we may define C (Ω) and C ( Ω) as in Definiion We will also le C,β (Ω) :=C,β (Ω) C (Ω) and C,β ( Ω) :=C,β (Ω) C ( Ω). I has already been shown in Proposiion 8.3 ha C (Ω) and C ( Ω) are closed subspaces of BC(Ω) and BC( Ω) respecively. The nex proposiion describes he relaion beween C (Ω) and C ( Ω). Proposiion Each u C (Ω) has a unique exension o a coninuous funcion on Ω given by ū = u on Ω and ū =on Ω and he exension ū is in C ( Ω). Conversely if u C ( Ω) and u Ω =, hen u Ω C (Ω). In his way we may idenify C (Ω) wih hose u C ( Ω) such ha u Ω =.

12 REAL ANALYSIS LECTURE NOTES 341 Proof. Any exension u C (Ω) o an elemen ū C( Ω) is necessarily unique, since Ω is dense inside Ω. So define ū = u on Ω and ū =on Ω. We mus show ū is coninuous on Ω and ū C ( Ω). For he coninuiy asserion i is enough o show ū is coninuous a all poins in Ω. For any >, by assumpion, he se K := {x Ω : u(x) } is a compac subse of Ω. Since Ω = Ω \ Ω, Ω K = and herefore he disance, δ := d(k, Ω), beween K and Ω is posiive. So if x Ω and y Ω and y x <δ,hen ū(x) ū(y) = u(y) < which shows ū : Ω C is coninuous. This also shows { ū } = { u } = K is compac in Ω and hence also in Ω. Since > was arbirary, his shows ū C ( Ω). Conversely if uª C ( Ω) such ha u Ω = and >, hen K := x Ω : u(x) is a compac subse of Ω which is conained in Ω since Ω K =. Therefore K is a compac subse of Ω showing u Ω C ( Ω). Definiion Le Ω be an open subse of R d,k N {} and β (, 1]. Le BC k (Ω) (BC k ( Ω)) denoe he se of k imes coninuously differeniable funcions u on Ω such ha α u BC(Ω) ( α u BC( Ω)) 36 for all α k. Similarly, le BC k,β (Ω) denoe hose u BC k (Ω) such ha [ α u] β < for all α = k. For u BC k (Ω) le kuk C k (Ω) = X k α uk u and α k kuk C k,β (Ω) = X k α uk u + X [ α u] β. α k α =k Theorem The spaces BC k (Ω) and BC k,β (Ω) equipped wih k k C k (Ω) and k k C k,β (Ω) respecively are Banach spaces and BCk ( Ω) isaclosedsubspaceof BC k (Ω) and BC k,β (Ω) BC k ( Ω). Also C k,β (Ω) =C k,β ( Ω) ={u BC k,β (Ω) : α u C (Ω) α k} is a closed subspace of BC k,β (Ω). Proof. Suppose ha {u n } n=1 BCk (Ω) is a Cauchy sequence, hen { α u n } n=1 is a Cauchy sequence in BC(Ω) for α k. Since BC(Ω) is complee, here exiss g α BC(Ω) such ha lim n k α u n g α k u =for all α k. Leing u := g, we mus show u C k (Ω) and α u = g α for all α k. This will be done by inducion on α. If α = here is nohing o prove. Suppose ha we have verified u C l (Ω) and α u = g α for all α l for some l<k.then for x Ω, i {1, 2,...,d} and R sufficienly small, a u n (x + e i )= a u n (x)+ Leing n in his equaion gives a u(x + e i )= a u(x)+ i a u n (x + τe i )dτ. g α+ei (x + τe i )dτ from which i follows ha i α u(x) exiss for all x Ω and i α u = g α+ei. This complees he inducion argumen and also he proof ha BC k (Ω) is complee. 36 To say α u BC( Ω) means ha α u BC(Ω) and α u exends o a coninuous funcion on Ω.

13 342 BRUCE K. DRIVER I is easy o check ha BC k ( Ω) is a closed subspace of BC k (Ω) and by using Exercise 17.1 and Theorem ha ha BC k,β (Ω) is a subspace of BC k ( Ω). The fac ha C k,β (Ω) is a closed subspace of BC k,β (Ω) is a consequence of Proposiion 8.3. To prove BC k,β (Ω) is complee, le {u n } n=1 BCk,β (Ω) be a k k C k,β (Ω) Cauchy sequence. By he compleeness of BC k (Ω) jus proved, here exiss u BC k (Ω) such ha lim n ku u n k C k (Ω) =. An applicaion of Theorem hen shows lim n k α u n α uk C,β (Ω) =for α = k and herefore lim n ku u n k C k,β (Ω) = Exercises. Exercise Le p [1, ), αbe a muli index (if α =le be he ideniy operaor on L p ), D( α ):={f L p (R n ): α f exiss weakly in L p (R n )} and for f D( α ) (he domain of α ) le α f denoe he α weak derivaive of f. (See Noaion ) (1) Show α is a densely defined operaor on L p, i.e. D( α ) is a dense linear subspace of L p and α : D( α ) L p is a linear ransformaion. (2) Show α : D( α ) L p is a closed operaor, i.e. he graph, Γ( α ):={(f, α f) L p L p : f D( α )}, is a closed subspace of L p L p. (3) Show α : D( α ) L p L p is no bounded unless α =. (The norm on D( α ) is aken o be he L p norm.) Exercise Le p [1, ), f L p and α be a muli index. Show α f exiss weakly (see Noaion 17.14) in L p iff here exiss f n Cc (R n ) and g L p such ha f n f and α f n g in L p as n. Hin: One direcion follows from Exercise For he oher direcion, see he proof of Theorem Exercise Folland 8.8 on p Exercise Assume n =1and le = e1 where e 1 =(1) R 1 = R. (1) Le f(x) = x, show f exiss weakly in L 1 loc (R) and f(x) =sgn(x) for m a.e.x. (2) Show ( f) does no exiss weakly in L 1 loc (R). (3) Generalize iem 1. as follows. Suppose f C(R, R) and here exiss a finie se Λ := { 1 < 2 < < N } R such ha f C 1 (R \ Λ, R). Assuming f L 1 loc (R), show f exiss weakly and (w) f(x) = f(x) for m a.e. x. Exercise Suppose ha f L 1 loc (Ω) and v Rd and {e j } n j=1 is he sandard basis for R d. If j f := ej f exiss weakly in L 1 loc (Ω) for all j =1, 2,...,n hen vf exiss weakly in L 1 loc (Ω) and vf = P n j=1 v j j f. Exercise Show Proposiion generalizes as follows. Le Ω be an open subse of R d and f L 1 loc (Ω), hen here exiss a locally Lipschiz funcion F : Ω C such ha F = f a.e. iff v (w) f exiss and is locally bounded for all v R d. (Here we say v (w) f is locally bounded if for all compac subses K Ω, here exiss a consan M K < such ha (w) f(x) M K for m a.e.x K. v

14 REAL ANALYSIS LECTURE NOTES 343 Exercise Suppose, f L 1 loc (Rd ) and v f exiss weakly and v f =in L 1 loc (Rd ) for all v R d. Then here exiss λ C such ha f(x) =λ for m a.e. x R d. Hin: See seps 1. and 2. in he ouline given in Exercise 17.9 below. Exercise (A generalizaion of Exercise 17.8.) Suppose Ω is a conneced open subse of R d and f L 1 loc (Ω). If α f =weakly for α n + wih α = N +1, hen f(x) =p(x) for m a.e. x where p(x) is a polynomial of degree a mos N. Here is an ouline. (1) Suppose x Ω and >such ha C := C x ( ) Ω and le δ n be a sequence of approximae δ funcions such supp(δ n ) B (1/n) for all n. Then for n large enough, α (f δ n )=( α f) δ n on C for α = N +1. Now use Taylor s heorem o conclude here exiss a polynomial p n of degree a mos N such ha f n = p n on C. (2) Show p := lim n p n exiss on C andhenlen in sep 1. o show here exiss a polynomial p of degree a mos N such ha f = p a.e. on C. (3) Use Taylor s heorem o show if p and q are wo polynomials on R d which agree on an open se hen p = q. (4) Finish he proof wih a connecedness argumen using he resuls of seps 2. and 3. above. Exercise Suppose Ω o R d and v,w R d. Assume f L 1 loc (Ω) and ha v w f exiss weakly in L 1 loc (Ω), show w v f also exiss weakly and w v f = v w f. Exercise Le d =2and f(x, y) =1 x. Show (1,1) f =weakly in L 1 loc despie he fac ha 1 f does no exis weakly in L 1 loc!