17. Weak and Strong Derivatives and Sobolev Spaces for This Section, Let

330 BRUCE K. DRIVER†

17. Weak and Strong Derivatives and Sobolev Spaces For this section, let Ω be an open subset of Rd, p,q,r [1, ],Lp(Ω)= p p p ∈ ∞ L (Ω, Ω,m) and Lloc(Ω)=Lloc(Ω, Ω,m), where m is Lebesgue measure on Rd B B d p Bp and Ω is the Borel σ —algebraonΩ. If Ω = R , we will simply write L and Lloc Bp d p d for L (R ) and Lloc(R ) respectively. Also let

f,g := fgdm h i ZΩ for any pair of measurable functions f,g : Ω C such that fg L1(Ω). For example, by Hölder’s inequality, if f,g is deﬁned→ for f Lp(Ω) and∈ g Lq(Ω) p h i ∈ ∈ when q = p 1 . The following simple but useful remark will be used (typically without further− comment) in the sequel.

1 1 1 Remark 17.1. Suppose r, p, q [1, ] are such that r− = p− + q− and ft f p q ∈ ∞ r → in L (Ω) and gt g in L (Ω) as t 0, then ftgt fg in L (Ω). Indeed, → → → ftgt fg = (ft f) gt + f (gt g) k − kr k − − kr ft f gt + f gt g 0 as t 0 ≤ k − kp k kq k kp k − kq → → d p p Definition 17.2 (Weak Differentiability). Let v R and f L (Ω)(f Lloc(Ω)) p ∈p ∈ ∈ then ∂vf is said to exist weakly in L (Ω)(Lloc(Ω)) if there exists a function g Lp(Ω)(g Lp (Ω)) such that ∈ ∈ loc (17.1) f,∂vφ = g, φ for all φ C∞(Ω). h i −h i ∈ c (w) The function g if it exists will be denoted by ∂v f. (By Corollary 9.27, there is at 1 (w) most one g L (Ω) such that Eq. (17.1) holds, so ∂v f is well defined.) ∈ loc 1 (w) 1 Lemma 17.3. Suppose f L (Ω) and ∂v f exists weakly in L (Ω). Then ∈ loc loc (w) (1) suppm(∂v f) suppm(f), where suppm(f) is the essential support of f relative to Lebesgue⊂ measure, see Definition 9.14. (w) (2) If f is continuously differentiable on U o Ω, then ∂v f = ∂vf a.e. on U. ⊂ Proof. (1) Since (w) ∂ f,φ = f,∂vφ =0for all φ C∞(Ω supp (f)), h v i −h i ∈ c \ m (w) and application of Corollary 9.27 shows ∂v f =0a.e. on Ω suppm(f). (w) \ (w) So by Lemma 9.15, Ω supp (f) Ω supp (∂v f), i.e. supp (∂v f) \ m ⊂ \ m m ⊂ suppm(f). 1 (2) Suppose that f U is C and let ψ Cc∞(U) which we view as a function d | ∈d in Cc∞(R ) by setting ψ 0 on R U. By Corollary 9.24, there exists ≡ \ γ Cc∞(Ω) such that 0 γ 1 and γ =1in a neighborhood of supp(ψ). ∈ ≤ ≤d 1 d Then by setting γf =0on R supp(γ) we may view γf Cc (R ) and so by standard integration by parts\ (see Lemma 9.25) and∈ the ordinary product rule, (w) ∂ f,ψ = f,∂vψ = γf,∂vψ h v i −h i −h i (17.2) = ∂v (γf) ,ψ = ∂vγ f + γ∂vf,ψ = ∂vf,ψ h i h · i h i REAL ANALYSIS LECTURE NOTES 331

wherein the last equality we have used ψ∂vγ =0and ψγ = ψ. Since Eq. (17.2) is true for all ψ Cc∞(U), an application of Corollary 9.27 with (w) ∈ (w) h = ∂v f(x) ∂vf(x) and µ = m shows ∂v f(x)=∂vf(x) for m —a.e. x U. − ∈

1 d (w) Lemma 17.4 (Product Rule). Let f Lloc(Ω),v R and φ C∞(Ω). If ∂v f 1 (w) ∈ 1 ∈ ∈ exists in Lloc(Ω), then ∂v (φf) exists in Lloc(Ω) and (w) (w) ∂ (φf)=∂vφ f + φ∂ f a.e. v · v d 1 d Moreover if φ Cc∞(R ) and F := φf L (herewedefine F on R by setting d ∈ (w) ∈ (w) 1 d F =0on R Ω ), then ∂ F = ∂vφ f + φ∂v f exists weakly in L (R ). \ · Proof. Let ψ C∞(Ω), then ∈ c (w) φf, ∂vψ = f,φ∂vψ = f,∂v (φψ) ∂vφ ψ = ∂ f,φψ + ∂vφ f,ψ −h i −h i −h − · i h v i h · i (w) = φ∂ f,ψ + ∂vφ f,ψ . h v i h · i This proves the first assertion. To prove the second assertion let γ Cc∞(Ω) such ∈ d that 0 γ 1 and γ =1on a neighborhood of supp(φ). So for ψ Cc∞(R ), using ≤ ≤ ∈ ∂vγ =0on supp(φ) and γψ C∞(Ω), we find ∈ c F, ∂vψ = γF,∂vψ = F, γ∂vψ = (φf) ,∂v (γψ) ∂vγ ψ h i h i h i h − · i (w) = (φf) ,∂v (γψ) = ∂ (φf) , (γψ) h i −h v i (w) (w) = ∂vφ f + φ∂ f,γψ = ∂vφ f + φ∂ f,ψ . −h · v i −h · v i (w) (w) This show ∂v F = ∂vφ f + φ∂v f as desired. · p Lemma 17.5. Suppose p [1, ),v Rd and f L (Ω). ∈ ∞ ∈ ∈ loc p (w) p (1) If there exists fm m∞=1 Lloc(Ω) such that ∂v fm exists in Lloc(Ω) for { } ⊂ p all m and there exists g L (Ω) such that for all φ C∞(Ω), ∈ loc ∈ c lim fm,φ = f,φ and lim ∂vfm,φ = g,φ m m →∞h i h i →∞h i h i (w) p then ∂v f exists in Lloc(Ω) and ∂vf = g. (w) p (2) If ∂v f exists in Lloc(Ω) then there exists fn Cc∞(Ω) such that fn f p ∈ (w) p→ in L (K) (i.e. limn f fn Lp(K) =0)and ∂vfn ∂v f in L (K) →∞ k − k → for all K @@ Ω. Proof. (1) Since (w) f,∂vφ = lim fm,∂vφ = lim ∂ fm,φ = g, φ m m v h i →∞h i − →∞h i h i (w) p for all φ C∞(Ω),∂v f exists and is equal to g L (Ω). ∈ c ∈ loc (2) Let K0 := and ∅ c Kn := x Ω : x n and d(x, Ω ) 2/n { ∈ | | ≤ ≥ } o (so Kn Kn+1 Kn+1 for all n and Kn Ω as n or see Lemma ⊂ ⊂ o ↑ →∞ 8.10) and choose ψn C∞(K , [0, 1]) using Corollary 9.24 so that ψn =1 ∈ c n on a neighborhood of Kn 1. Given a compact set K Ω, for all sufficiently − ⊂ 332 BRUCE K. DRIVER†

large m, ψmf = f on K and by Lemma 17.4 and item 1. of Lemma 17.3, we also have (w) (w) (w) ∂ (ψmf)=∂vψm f + ψm∂ f = ∂ f on K. v · v v This argument shows we may assume suppm(f) is a compact subset of Ω in which case we extend f to a function F on Rd by setting F = f on Ω c p d (w) (w) and F =0on Ω . This function F is in L (R ) and ∂v F = ∂v f. Indeed, d d if φ Cc∞(R ) and ψ Cc∞(R ) is chosen so that supp(ψ) Ω, 0 ψ 1 ∈ ∈ ⊂ ≤ ≤ and ψ =1in a neighborhood of suppm(f), then

F, ∂vφ = F, ψ∂vφ = f,∂v (ψφ) h i h i h i = ∂(w)f,ψφ = ψ∂(w)f,φ −h v i −h v i (w) p d which shows ∂v F exist in L (R ) and (w) (w) (w) ∂v F = ψ∂v f =1Ω∂v f. d Let χ C∞(B0(1)) with d χdm =1and set δk(x)=n χ(nx). Then ∈ c R there exists N N and a compact subset K Ω such that fn := F δn ∈ R ⊂ ∗ ∈ Cc∞(Ω) and supp(fn) K for all n N. By Proposition 9.23 and the (w) ⊂ ≥ deﬁnition of ∂v F,

∂vfn(x)=F ∂vδn(x)= F (y)∂vδn(x y)dy ∗ d − ZR (w) (w) = F, ∂v [δn(x )] = ∂ F, δn(x ) = ∂ F δn(x). −h − · i h v − · i v ∗ (w) (w) p Hence by Theorem 9.20, fn F = f and ∂vfn ∂v F = ∂v f in L (Ω) as n . → → →∞

d p Definition 17.6 (Strong Differentiability). Let v R and f L , then ∂vf is p ∈ ∈p said to exist strongly in L if the limt 0 (τ tvf f) /t exists in L , where as above → − −(s) τvf(x):=f(x v). We will denote the limit by ∂v f. − p d (s) p (w) It is easily verified that if f L ,v R and ∂v f L exists then ∂v f exists (w) (s) ∈ ∈ ∈ d and ∂v f = ∂v f. To check this assertion, let φ Cc∞(R ) andthenusingRemark 17.1, ∈ (s) 1 τ tvf f ∂v f φ = L — lim − − φ. t 0 t · → Hence

(s) τ tvf f τtvφ φ ∂v f φdm = lim − − φdm =lim f − dm d · t 0 d t t 0 d t ZR → ZR → ZR d d = 0 fτtvφdm = f 0τtvφdm = f ∂vφdm, dt| d d · dt| − d · ZR ZR ZR wherein we have used Corollary 5.43 to diﬀerentiate under the integral in the fourth (w) (s) equality. This shows ∂v f exists and is equal to ∂v f. What is somewhat more (w) (s) surprising is that the converse assertion that if ∂v f exists then so does ∂v f. The next theorem is a generalization of Theorem 10.36 from L2 to Lp. Theorem 17.7 (Weak and Strong Diﬀerentiability). Suppose p [1, ),f ∈ ∞ ∈ Lp(Rd) and v Rd 0 . Then the following are equivalent: ∈ \{ } REAL ANALYSIS LECTURE NOTES 333

p d (1) There exists g L (R ) and tn n∞=1 R 0 such that limn tn =0 and ∈ { } ⊂ \{ } →∞

f( + tnv) f( ) d lim · − · ,φ = g, φ for all φ Cc∞(R ). n t →∞h n i h i ∈ (w) p d (2) ∂v f exists and is equal to g L (R ), i.e. f,∂vφ = g, φ for all d ∈ h i −h i φ Cc∞(R ). p p ∈ p d d L L (3) There exists g L (R ) and fn Cc∞(R ) such that fn f and ∂vfn g as n . ∈ ∈ → → (s) →∞ p d (4) ∂v f exists and is is equal to g L (R ), i.e. ∈ f( + tv) f( ) Lp · − · g as t 0. t → → Moreover if p (1, ) any one of the equivalent conditions 1. — 4. above are implied by the following∈ ∞ condition.

10. There exists tn n∞=1 R 0 such that limn tn =0and { } ⊂ \{ } →∞ f( + tnv) f( ) sup · − · < . n t ∞ ° n °p ° ° ° ° Proof. 4. = 1. is simply° the assertion that° strong convergence implies weak convergence. ⇒ d 1. = 2. For φ Cc∞(R ), ⇒ ∈ f( + tnv) f( ) φ( tnv) φ( ) g, φ =lim · − · ,φ = lim f, · − − · n t n t h i →∞h n i →∞h n i φ( tnv) φ( ) = f, lim · − − · = f,∂vφ , n t h →∞ n i −h i wherein we have used the translation invariance of Lebesgue measure and the dominated convergence theorem. d 2. = 3. Let φ C ( , ) such that d φ(x)dx =1and let φm(x)= c∞ R R R d ⇒ ∈ d m φ(mx), then by Proposition 9.23, hm := φm f C∞(R ) for all m and ∗R ∈

∂vhm(x)=∂vφm f(x)= ∂vφm(x y)f(y)dy = f, ∂v [φm (x )] ∗ d − h − − · i ZR = g, φm (x ) = φm g(x). h − · i ∗ p d p d By Theorem 9.20, hm f L (R ) and ∂vhm = φm g g in L (R ) as m . → ∈ ∗ → →∞ This shows 3. holds except for the fact that hm need not have compact support. d To ﬁxthisletψ Cc∞(R , [0, 1]) such that ψ =1in a neighborhood of 0 and let ∈ ψ(x)=ψ(x) and (∂vψ) (x):=(∂vψ)(x). Then

∂v (ψhm)=∂vψhm + ψ∂vhm = (∂vψ) hm + ψ∂vhm p p so that ψhm hm in L and ∂v (ψhm) ∂vhm in L as 0. Let fm = ψ hm → → ↓ m where m is chosen to be greater than zero but small enough so that

ψ hm hm + ∂v (ψ hm) ∂vhm < 1/m. k m − kp k m → kp d p Then fm Cc∞(R ),fm f and ∂vfm g in L as m . ∈ → → →∞ 334 BRUCE K. DRIVER†

3. = 4. By the fundamental theorem of calculus ⇒ τ tvfm(x) fm(x) fm(x + tv) fm(x) − − = − t t 1 1 d 1 (17.3) = f (x + stv)ds = (∂ f )(x + stv)ds. t ds m v m Z0 Z0 Let 1 1 Gt(x):= τ stvg(x)ds = g(x + stv)ds − Z0 Z0 which is deﬁned for almost every x and is in Lp(Rd) by Minkowski’s inequality for integrals, Theorem 7.27. Therefore 1 τ tvfm(x) fm(x) − − Gt(x)= [(∂vfm)(x + stv) g(x + stv)] ds t − − Z0 and hence again by Minkowski’s inequality for integrals, 1 1 τ tvfm fm − − Gt τ stv (∂vfm) τ stvg p ds = ∂vfm g p ds. t − ≤ k − − − k k − k ° °p Z0 Z0 ° ° Letting° m in this° equation implies (τ tvf f) /t = Gt a.e. Finally one more ° ° − application→∞ of Minkowski’s inequality for integrals− implies, 1 τ tvf f − − g = Gt g p = (τ stvg g) ds t − k − k − − ° °p °Z0 °p ° ° ° ° ° ° 1 ° ° ° ° τ stvg °g p ds. ° ≤ k − − k Z0 By the dominated convergence theorem and Proposition 9.13, the latter term tends to 0 as t 0 and this proves 4. → (10. = 1. when p>1) This is a consequence of Theorem 16.27 which asserts, by ⇒ f( +tnv) f( ) w p d passing to a subsequence if necessary, that · − · g for some g L (R ). tn → ∈

Example 17.8. Thefactthat(10) does not imply the equivalent conditions 1 — 4 in Theorem 17.7 when p =1is demonstrated by the following example. Let f := 1[0,1], then f(x + t) f(x) 1 − dx = 1[ t,1 t](x) 1[0,1](x) dx =2 t t − − − ZR ¯ ¯ ZR ¯ ¯ | | ¯ ¯ for t < 1. For¯ contradiction sake,¯ suppose¯ there exists g L1(R¯,dm) such that | | ¯ ¯ ∈ f(x + tn) f(x) lim − = g(x) in L1 n t →∞ n for some sequence tn n∞=1 as above. Then for φ Cc∞(R) wewouldhaveonone hand, { } ∈ 1 f(x + tn) f(x) φ(x tn) φ(x) − φ(x)dx = − − f(x)dx φ0(x)dx =(φ(0) φ(1)), t t →− − ZR n ZR n Z0 while on the other hand,

f(x + tn) f(x) − φ(x)dx g(x)φ(x)dx. t → ZR n ZR REAL ANALYSIS LECTURE NOTES 335

These two equations imply

(17.4) g(x)φ(x)dx = φ(0) φ(1) for all φ Cc∞(R) − ∈ ZR and in particular that g(x)φ(x)dx =0for all φ Cc(R 0, 1 ). By Corollary R ∈ \{ } 9.27, g(x)=0for m —a.e. x R 0, 1 and hence g(x)=0for m —a.e. x R. But this clearly contradictsR Eq.∈ (17.4).\{ This} example also shows that the unit∈ ball in L1(R,dm) is not sequentially weakly compact. We will now give a couple of applications of Theorem 17.7. 1 w Proposition 17.9. Let Ω R be an open interval and f Lloc(Ω). Then ∂ f 1 ⊂ ˜ ∈ exists in Lloc(Ω) iﬀ f has a continuous version f which is absolutely continuous on w ˜ ˜ all compact subintervals of Ω. Moreover, ∂ f = f 0 a.e., where f 0(x) is the usual pointwise derivative.

Proof. If f is locally absolutely continuous and φ Cc∞(Ω) with supp(φ) [a, b] Ω, then by Corollary 14.32, ∈ ⊂ ⊂ b b b f 0φdm = f 0φdm = fφ0dm + fφ = fφ0dm. − |a − ZΩ Za Za ZΩ w w 1 This shows ∂ f exists and ∂ f = f 0 Lloc(Ω). w 1∈ Now suppose that ∂ f exists in Lloc(Ω), let [a, b] be a compact subinterval of Ω, ψ Cc∞(Ω) such that ψ =1on a neighborhood of [a, b] and 0 ψ 1 on Ω and deﬁ∈ne F := ψf. By Lemma 17.4, F L1 and ∂(w)F exists in L1≤and≤ is given by ∈ (w) (w) ∂ (ψf)=ψ0f + ψ∂ f. (w) From Theorem 17.7 there exists Fn Cc∞(R) such that Fn F and Fn0 ∂ F in L1 as n . Hence, using the fundamental∈ theorem of calculus,→ → →∞ x x (w) (17.5) Fn(x)= F 0 (y)dy ∂ F (y)dy =: G(x) n → Z−∞ Z−∞ 1 as n . Since Fn G pointwise and Fn F in L , it follows that F = ψf = G a.e.→∞ In particular, G→= f a.e. on [a, b] and→ we conclude that f has a continuous version on [a, b], this version, G [a,b], is absolutely continuous on [a, b] and by Eq. (17.5) and fundamental theorem| of calculus for absolutely continuous functions (Theorem 14.31), (w) (w) G0(x)=∂ F (x)=∂ f(x) for m a.e. x [a, b]. ∈ Since [a, b] Ω was an arbitrary compact subinterval and continuous versions of a measurable⊂ functions are unique if they exist, it follows from the above considera- tions that there exists a unique function f˜ C(Ω) such that f˜ = G for all such ∈ |[a,b] pairs ([a, b],G) as above. The function f˜ then satisﬁes all of the desired properties.

Because of Lemma 17.3, Theorem 17.7 and Proposition 17.9 it is now safe to (w) simply write ∂vf for the ordinary partial derivative of f, the weak derivative ∂v f (s) and the strong derivative ∂v f. Deﬁnition 17.10. Let X and Y be metric spaces. A function f : X Y is said to be Lipschitz if there exists C< such that → ∞ Y X d (f(x),f(x0)) Cd (x, x0) for all x, x0 X ≤ ∈ 336 BRUCE K. DRIVER† and said to be locally Lipschitz if for all compact subsets K X there exists ⊂ CK < such that ∞ Y X d (f(x),f(x0)) CK d (x, x0) for all x, x0 K. ≤ ∈ 1 d Proposition 17.11. Let f L (R ) such that essential support, suppm(f), is ∈ compact. Then there exists a Lipschitz function F : Rd C such that F = f a.e. (w) d → iﬀ ∂v f exists and is bounded for all v R . ∈ Proof. Suppose f = F a.e. and F is Lipschitz and let p (1, ). Since F is Lipschitz and has compact support, for all 0 < t 1, ∈ ∞ | | ≤ f(x + tv) f(x) p F (x + tv) F (x) p − dx = − dx C˜ v p , d t d t ≤ | | ZR ¯ ¯ ZR ¯ ¯ ¯ ¯ ¯ ¯ where C˜ is a¯ constant depending¯ on the size¯ of the support of¯f and on the Lipschitz ¯ ¯ ¯ ¯ (w) constant, C, for F. Therefore Theorem 17.7 may be applied to conclude ∂v f exists in Lp and moreover,

F (x + tv) F (x) (w) lim − = ∂v f(x) for m —a.e.x. t 0 t →

Since there exists tn n∞=1 R 0 such that limn tn =0and { } ⊂ \{ } →∞ (w) F (x + tnv) F (x) d ∂v f(x) = lim − C v for a.e. x R , n t ≤ | | ∈ →∞ ¯ n ¯ ¯ ¯ ¯ ¯ ¯ (w¯ ) ¯ d ¯ it follows¯ that ∂v ¯ f ¯C v for all v R .¯ ≤ | | ∈ ∞ Conversely,° let φm°be an approximate δ — function sequence as used in the ° ° d proof of Theorem° 17.7° and fm := f φm. Then fm Cc∞(R ),∂vfm = ∂vf φm, ∗ ∈ ∗ ∂vfm ∂vf < and therefore, | | ≤ k k∞ ∞ 1 d 1 fm(y) fm(x) = fm(x + t(y x))dt = (y x) fm(x + t(y x))dt | − | dt − − · ∇ − ¯Z0 ¯ ¯Z0 ¯ ¯ 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ (17.6) ¯ y x fm(x + t(y ¯ x))¯ dt C y x ¯ ≤ | − |·|∇ − | ≤ | − | Z0 where C is a constant independent of m. By passing to a subsequence of the

φm m∞=1 if necessary, we may assume that limm fm(x)=f(x) for m —a.e. { } →∞ x Rd. Letting m in Eq. (17.6) then implies ∈ →∞ f(y) f(x) C y x for all x, y / E | − | ≤ | − | ∈ where E Rd is a m — null set. It is now easily verified that if F : Rd C is defined by⊂F = f on Ec and → f(x) if x/E F (x)= ∈ limy x f(y) if x E ( y/→E ∈ ∈ defines a Lipschitz function F on Rd such that F = f a.e. Lemma 17.12. Let v Rd. ∈ 1 1 (1) If h L and ∂vh exists in L , then d ∂vh(x)dx =0. ∈ R R REAL ANALYSIS LECTURE NOTES 337

1 1 1 p q (2) If p, q, r [1, ) satisfy r− = p− + q− ,f L and g L are functions ∈ ∞ p q ∈ ∈ such that ∂vf and ∂vg exists in L and L respectively, then ∂v(fg) exists in r L and ∂v(fg)=∂vf g + f ∂vg. Moreover if r =1we have the integration by parts formula, · ·

(17.7) ∂vf,g = f,∂vg . h i −h i 1 1 d 1 d (3) If p =1,∂vf exists in L and g BC (R ) (i.e. g C (R ) with g and ∈ ∈ 1 its ﬁrst derivatives being bounded) then ∂v(gf) exists in L and ∂v(fg)= ∂vf g + f ∂vg and again Eq. (17.7) holds. · · d Proof. 1) By item 3. of Theorem 17.7 there exists hn Cc∞(R ) such that 1 ∈ hn h and ∂vhn ∂vh in L . Then → → d d ∂vhn(x)dx = 0 hn(x + tv)dx = 0 hn(x)dx =0 d dt| d dt| d ZR ZR ZR and letting n proves the ﬁrst assertion. →∞ d 2) Similarly there exists fn,gn Cc∞(R ) such that fn f and ∂vfn ∂vf in p ∈ q → → L and gn g and ∂vgn ∂vg in L as n . So by the standard product rule → → r →∞ and Remark 17.1, fngn fg L as n and → ∈ →∞ r ∂v(fngn)=∂vfn gn + fn ∂vgn ∂vf g + f ∂vg in L as n . · · → · · →∞ r It now follows from another application of Theorem 17.7 that ∂v(fg) exists in L and ∂v(fg)=∂vf g + f ∂vg. Eq. (17.7) follows from this product rule and item 1. when r =1. · · d 1 3) Let fn Cc∞(R ) such that fn f and ∂vfn ∂vf in L as n . Then ∈ 1 → → 1 →∞ as above, gfn gf in L and ∂v(gfn) ∂vg f + g∂vf in L as n . In → d → · →∞ particular if φ Cc∞(R ), then ∈

gf,∂vφ =limgfn,∂vφ = lim ∂v (gfn) ,φ n n h i →∞h i − →∞h i = lim ∂vg fn + g∂vfn,φ = ∂vg f + g∂vf,φ . n − →∞h · i −h · i

This shows ∂v(fg) exists (weakly) and ∂v(fg)=∂vf g + f ∂vg. Again Eq. (17.7) holds in this case by item 1. already proved. · ·

1 1 1 p q Lemma 17.13. Let p, q, r [1, ] satisfy p− + q− =1+r− ,f L ,g L ∈ ∞ ∈ ∈ and v Rd. ∈ r p (1) If ∂vf exists strongly in L , then ∂v(f g) exists strongly in L and ∗

∂v(f g)=(∂vf) g. ∗ ∗ q r (2) If ∂vg exists strongly in L , then ∂v(f g) exists strongly in L and ∗

∂v(f g)=f ∂vg. ∗ ∗ p d d (3) If ∂vf exists weakly in L and g Cc∞(R ), then f g C∞(R ),∂v(f g) exists strongly in Lr and ∈ ∗ ∈ ∗

∂v(f g)=f ∂vg =(∂vf) g. ∗ ∗ ∗ 338 BRUCE K. DRIVER†

Proof. Items 1 and 2. By Young’s inequality and simple computations:

τ tv(f g) f g τ tvf g f g − ∗ − ∗ (∂vf) g = − ∗ − ∗ (∂vf) g t − ∗ t − ∗ ° °r ° °r ° ° ° ° ° ° ° τ tvf f ° = − − (∂vf) g ° ° ° t − ∗ ° °· ¸ °r ° ° °τ tvf f ° − − (∂vf) g ≤ ° t − k k°q ° °p ° ° which tends to zero as t 0. The second item° is proved analogously,° or just make ° ° use of the fact that f g→= g f and apply Item 1. ∗ ∗ d Using the fact that g(x ) Cc∞(R ) and the deﬁnition of the weak derivative, − · ∈

f ∂vg(x)= f(y)(∂vg)(x y)dy = f(y)(∂vg(x )) (y)dy ∗ d − − d − · ZR ZR = ∂vf(y)g(x y)dy = ∂vf g(x). d − ∗ ZR Item 3. is a consequence of this equality and items 1. and 2.

17.1. Sobolev Spaces.

α 1 Notation 17.14. Let ∂ be deﬁnedasinNotation9.10andf Lloc(Ω). We say ∂αf exists weakly in L1 (Ω) iﬀ there exists g L1 (Ω) such that∈ loc ∈ loc α α f,∂ φ =( 1)| | g, φ for all φ C∞(Ω). h i − h i ∈ c As usual g is unique if it exists and we will denote g by ∂αf.

Deﬁnition 17.15. For p [1, ],k N and Ω an open subset of Rd, let ∈ ∞ ∈ W k,p(Ω):= f Lp(Ω):∂αf Lp(Ω) (weakly) for all α k { ∈ ∈ | | ≤ } and deﬁne α f k,p := ∂ f p . k kW (Ω) k kL (Ω) α k |X|≤ k,p Theorem 17.16. The space W (Ω) with the norm k,p is a Banach space. k·kW (Ω) k,p α Proof. Suppose that fn n∞=1 W (Ω) is a Cauchy sequence, then ∂ fn n∞=1 is a Cauchy sequence in L{ p(Ω} ) for⊂ all α k. By the completeness of Lp{(Ω), there} p p | | ≤ α exists gα L (Ω) such that gα = L — limn ∂ fn for all α k. Therefore, for ∈ →∞ | | ≤ all φ C∞(Ω), ∈ c α α α α α f,∂ φ = lim fn,∂ φ =( 1)| | lim ∂ fn,φ =( 1)| | lim gα,φ . n n n h i →∞h i − →∞h i − →∞h i α α This shows ∂ f exists weakly and gα = ∂ f a.e.

Example 17.17. Let Ω be an open subset of Rd, then H1(Ω):=W 1,2(Ω) is a Hilbert space with inner product deﬁned by

(f,g)= f gdm¯ + f gdm.¯ · ∇ · ∇ ZΩ ZΩ d k,p d Proposition 17.18. Cc∞(R ) is dense in W (R ) for all 1 p< . ≤ ∞ REAL ANALYSIS LECTURE NOTES 339

Proof. The proof of this proposition is left an exercise to the reader. How- ever, note that the assertion 2. implies 3. in Theorem 17.7 essentially proves the statement when k =1and the same smooth approximations used in the proof of Theorem 17.7 will work here as well.

17.2. Hölder Spaces.

Notation 17.19. Let Ω be an open subset of Rd,BC(Ω) and BC(Ω¯) be the bounded continuous functions on Ω and Ω¯ respectively. By identifying f BC(Ω¯) ∈ with f Ω BC(Ω), we will consider BC(Ω¯) as a subset of BC(Ω). For u BC(Ω) and 0 <β| ∈ 1 let ∈ ≤ u(x) u(y) u u := sup u(x) and [u]β := sup | − β | . k k x Ω | | x,y Ω x y ∈ x=∈y ½ | − | ¾ 6 35 If [u]β < , we say u is β — Hölder continuous with holder exponent β and we let ∞ 0,β C (Ω):= u BC(Ω):[u]β < { ∈ ∞} denote the space of Hölder continuous functions on Ω. For u C0,β(Ω) let ∈ (17.8) u 0,β := u u +[u]β. k kC (Ω) k k Remark 17.20. If u : Ω C and [u]β < for some β>1, then u is constant on → ∞ each connected component of Ω. Indeed, if x Ω and h Rd then ∈ ∈ u(x + th) u(x) β − [u]βt /t 0 as t 0 t ≤ → → ¯ ¯ ¯ ¯ which shows ∂ u(x)=0¯ for all x Ω.¯If y Ω is in the same connected component h ¯ ¯ as x, then by Exercise 15.5 there∈ exists as∈ smooth curve σ :[0, 1] Ω such that σ(0) = x and σ(1) = y. So by the fundamental theorem of calculus→ and the chain rule, 1 d 1 u(y) u(x)= u(σ(t))dt = 0 dt =0. − dt Z0 Z0 This is why we do not talk about Hölder spaces with Hölder exponents larger than 1.

1 Exercise 17.1. Suppose u C (Ω) BC(Ω) and ∂iu BC(Ω) for i =1, 2,...,d. 0,1∈ ∩ ∈ Show [u]1 < , i.e. u C (Ω). ∞ ∈ Theorem 17.21. Let Ω be an open subset of Rd. Then (1) BC(Ω¯) isaclosedsubspaceofBC(Ω). (2) Every element u C0,β(Ω) has a unique extension to a continuous function (which we will still∈ denote by u) on Ω¯. Therefore we may identify C0,β(Ω) with a subspace of BC(Ω¯). We may also write C0,β(Ω) as C0,β(Ω¯) to em- phasize this point. 0,β 0,β (3) The function u C (Ω) u 0,β [0, ) is a norm on C (Ω) ∈ → k kC (Ω) ∈ ∞ which make C0,β(Ω) into a Banach space.

35If β =1,uis is said to be Lipschitz continuous. 340 BRUCE K. DRIVER†

Proof. 1. The first item is trivial since for u BC(Ω¯), the sup-norm of u on Ω¯ agrees with the sup-norm on Ω and BC(Ω¯) is complete∈ in this norm. 2. Suppose that [u]β < and x0 ∂Ω. Let xn ∞ Ω be a sequence such ∞ ∈ { }n=1 ⊂ that x0 = limn xn. Then →∞ β u(xn) u(xm) [u]β xn xm 0 as m, n | − | ≤ | − | → →∞ showing u(xn) n∞=1 is Cauchy so that u¯(x0) := limn u(xn) exists. If yn n∞=1 { } →∞ { } ⊂ Ω is another sequence converging to x0, then β u(xn) u(yn) [u]β xn yn 0 as n , | − | ≤ | − | → →∞ showing u¯(x0) is well defined. In this way we define u¯(x) for all x ∂Ω and let u¯(x)=u(x) for x Ω. Since a similar limiting argument shows ∈ ∈ β u¯(x) u¯(y) [u]β x y for all x, y Ω¯ | − | ≤ | − | ∈ it follows that u¯ is still continuous. In the sequel we will abuse notation and simply denote u¯ by u. 3. For u, v C0,β(Ω), ∈ v(y)+u(y) v(x) u(x) [v + u]β =sup | − β − | x,y Ω x y x=∈y ½ | − | ¾ 6 v(y) v(x) + u(y) u(x) sup | − | | β − | [v]β +[u]β ≤ x,y Ω x y ≤ x=∈y ½ | − | ¾ 6 and for λ C it is easily seen that [λu]β = λ[u]β. This shows [ ]β is a semi-norm 0,β ∈ · on C (Ω) and therefore C0,β (Ω) defined in Eq. (17.8) is a norm. 0,β k·k 0,β To see that C (Ω) is complete, let un n∞=1 be a C (Ω)—Cauchy sequence. ¯ { } ¯ Since BC(Ω) is complete, there exists u BC(Ω) such that u un u 0 as n . For x, y Ω with x = y, ∈ k − k → →∞ ∈ 6 u(x) u(y) un(x) un(y) | − β | = lim | − β | lim sup[un] lim un C0,β (Ω) < , x y n x y ≤ n ≤ n k k ∞ | − | →∞ | − | →∞ →∞ and so we see that u C0,β(Ω). Similarly, ∈ u(x) un(x) (u(y) un(y)) (um un)(x) (um un)(y) | − − − | = lim | − − − | x y β m x y β | − | →∞ | − | lim sup[um un]β 0 as n , ≤ m − → →∞ →∞ showing [u un]β 0 as n and therefore limn u un C0,β (Ω) =0. − → →∞ →∞ k − k Notation 17.22. Since Ω and Ω¯ are locally compact Hausdorff spaces, we may define C0(Ω) and C0(Ω¯) as in Definition 8.29. We will also let 0,β 0,β 0,β 0,β C (Ω):=C (Ω) C0(Ω) and C (Ω¯):=C (Ω) C0(Ω¯). 0 ∩ 0 ∩ It has already been shown in Proposition 8.30 that C0(Ω) and C0(Ω¯) are closed subspaces of BC(Ω) and BC(Ω¯) respectively. The next proposition describes the relation between C0(Ω) and C0(Ω¯).

Proposition 17.23. Each u C0(Ω) has a unique extension to a continuous function on Ω¯ given by u¯ = u ∈on Ω and u¯ =0on ∂Ω and the extension u¯ is in C0(Ω¯). Conversely if u C0(Ω¯) and u ∂Ω =0, then u Ω C0(Ω). In this way we ∈ | | ∈ may identify C0(Ω) with those u C0(Ω¯) such that u ∂Ω =0. ∈ | REAL ANALYSIS LECTURE NOTES 341

Proof. Any extension u C0(Ω) to an element u¯ C(Ω¯) is necessarily unique, since Ω is dense inside Ω¯. So∈ define u¯ = u on Ω and u¯∈=0on ∂Ω. We must show u¯ is continuous on Ω¯ and u¯ C0(Ω¯). For the continuity assertion∈ it is enough to show u¯ is continuous at all points in ∂Ω. For any >0, by assumption, the set K := x Ω : u(x) is a { ∈ | | ≥ } compact subset of Ω. Since ∂Ω = Ω¯ Ω,∂Ω K = and therefore the distance, \ ∩ ∅ δ := d(K,∂Ω), between K and ∂Ω is positive. So if x ∂Ω and y Ω¯ and ∈ ∈ y x <δ,then u¯(x) u¯(y) = u(y) <which shows u¯ : Ω¯ C is continuous. | − | | − | | | → This also shows u¯ = u = K is compact in Ω and hence also in Ω¯. {| | ≥ } {| | ≥ } Since >0 was arbitrary, this shows u¯ C0(Ω¯). ∈ Conversely if u C0(Ω¯) such that u ∂Ω =0and >0, then K := x Ω¯ : u(x) ∈is a compact subset of| Ω¯ which is contained in Ω since ∈ | | ≥ ∂Ω K = . Therefore K is a compact subset of Ω showing u Ω C0(Ω¯). © ∩ ∅ ª | ∈ Definition 17.24. Let Ω be an open subset of Rd,k N 0 and β (0, 1]. Let BCk(Ω) (BCk(Ω¯)) denote the set of k — times continuously∈ ∪ di{ ff}erentiable∈ functions u on Ω such that ∂αu BC(Ω) (∂αu BC(Ω¯))36 for all α k. Similarly, let k,β ∈ k ∈ α | | ≤ BC (Ω) denote those u BC (Ω) such that [∂ u]β < for all α = k. For u BCk(Ω) let ∈ ∞ | | ∈ α u k = ∂ u u and k kC (Ω) k k α k |X|≤ α α u k,β = ∂ u u + [∂ u]β. k kC (Ω) k k α k α =k |X|≤ |X| k k,β Theorem 17.25. The spaces BC (Ω) and BC (Ω) equipped with Ck(Ω) and k k·k k,β respectively are Banach spaces and BC (Ω¯) isaclosedsubspaceof k·kC (Ω) BCk(Ω) and BCk,β(Ω) BCk(Ω¯). Also ⊂ k,β k,β k,β α C (Ω)=C (Ω¯)= u BC (Ω):∂ u C0(Ω) α k 0 0 { ∈ ∈ ∀ | | ≤ } is a closed subspace of BCk,β(Ω). k α Proof. Suppose that un n∞=1 BC (Ω) is a Cauchy sequence, then ∂ un n∞=1 is a Cauchy sequence in {BC}(Ω) for⊂ α k. Since BC(Ω) is complete, there{ exists} α | | ≤ gα BC(Ω) such that limn ∂ un gα u =0for all α k. Letting u := g0, ∈ k →∞ k α − k | | ≤ we must show u C (Ω) and ∂ u = gα for all α k. This will be done by induction on α .∈If α =0there is nothing to| prove.| ≤ Suppose that we have |l | | | α verified u C (Ω) and ∂ u = gα for all α l for some l

36To say ∂αu BC(Ω¯) means that ∂αu BC(Ω) and ∂αu extends to a continuous function on Ω¯. ∈ ∈ 342 BRUCE K. DRIVER†

It is easy to check that BCk(Ω¯) is a closed subspace of BCk(Ω) and by using Exercise 17.1 and Theorem 17.21 that that BCk,β(Ω) is a subspace of BCk(Ω¯). The k,β k,β fact that C0 (Ω) is a closed subspace of BC (Ω) is a consequence of Proposition 8.30. k,β k,β To prove BC (Ω) is complete, let un ∞ BC (Ω) be a k,β — { }n=1 ⊂ k·kC (Ω) Cauchy sequence. By the completeness of BCk(Ω) just proved, there exists u k ∈ BC (Ω) such that limn u un Ck(Ω) =0. An application of Theorem 17.21 then α →∞αk − k shows limn ∂ un ∂ u C0,β (Ω) =0for α = k and therefore limn u →∞ k − k | | →∞ k − un k,β =0. kC (Ω) 17.3. Exercises. Exercise 17.2. Let p [1, ),αbe a multi index (if α =0let ∂0 be the identity operator on Lp), ∈ ∞ α p n α p n D(∂ ):= f L (R ):∂ f exists weakly in L (R ) { ∈ } and for f D(∂α) (the domain of ∂α) let ∂αf denote the α — weak derivative of f. (See Notation∈ 17.14.) (1) Show ∂α is a densely defined operator on Lp, i.e. D(∂α) is a dense linear subspace of Lp and ∂α : D(∂α) Lp is a linear transformation. (2) Show ∂α : D(∂α) Lp is a closed→ operator, i.e. the graph, → Γ(∂α):= (f,∂αf) Lp Lp : f D(∂α) , { ∈ × ∈ } is a closed subspace of Lp Lp. (3) Show ∂α : D(∂α) Lp ×Lp is not bounded unless α =0. (The norm on D(∂α) is taken to⊂ be the→Lp —norm.) Exercise 17.3. Let p [1, ),f Lp and α be a multi index. Show ∂αf exists ∈ ∞ p ∈ n p weakly (see Notation 17.14) in L iff there exists fn Cc∞(R ) and g L such α p ∈ ∈ that fn f and ∂ fn g in L as n . Hint: One direction follows from Exercise→ 17.2. For the other→ direction, see→∞ the proof of Theorem 17.7. Exercise 17.4. Folland 8.8 on p. 246. 1 Exercise 17.5. Assume n =1and let ∂ = ∂e1 where e1 =(1) R = R. 1 ∈ (1) Let f(x)= x , show ∂f exists weakly in Lloc(R) and ∂f(x)=sgn(x) for m —a.e.x. | | 1 (2) Show ∂(∂f) does not exists weakly in Lloc(R). (3) Generalize item 1. as follows. Suppose f C(R, R) and there exists a finite ∈ 1 set Λ := t1

1 d Exercise 17.8. Suppose,f L (R ) and ∂vf exists weakly and ∂vf =0in ∈ loc L1 (Rd) for all v Rd. Then there exists λ C such that f(x)=λ for m —a.e. loc ∈ ∈ x Rd. Hint: See steps 1. and 2. in the outline given in Exercise 17.9 below. ∈ Exercise 17.9. (A generalization of Exercise 17.8.) Suppose Ω is a connected open d 1 α n subset of R and f Lloc(Ω). If ∂ f =0weakly for α Z+ with α = N +1, then f(x)=p(x) for m —a.e.∈ x where p(x) is a polynomial∈ of degree| at| most N. Here is an outline.

(1) Suppose x0 Ω and >0 such that C := Cx () Ω and let δn be a ∈ 0 ⊂ sequence of approximate δ — functions such supp(δn) B0(1/n) for all n. α α ⊂ Then for n large enough, ∂ (f δn)=(∂ f) δn on C for α = N +1. Now ∗ ∗ | | use Taylor’s theorem to conclude there exists a polynomial pn of degree at most N such that fn = pn on C. (2) Show p := limn pn exists on C andthenletn in step 1. to show there exists a polynomial→∞ p of degree at most N such→∞ that f = p a.e. on C. (3) Use Taylor’s theorem to show if p and q are two polynomials on Rd which agree on an open set then p = q. (4) Finish the proof with a connectedness argument using the results of steps 2. and 3. above. d d 1 Exercise 17.10. Suppose Ω o R and v,w R . Assume f Lloc(Ω) and that 1 ⊂ ∈ ∈ ∂v∂wf exists weakly in Lloc(Ω),show∂w∂vf also exists weakly and ∂w∂vf = ∂v∂wf. (1,1) 1 Exercise 17.11. Let d =2and f(x, y)=1x 0. Show ∂ f =0weakly in Lloc ≥ 1 despite the fact that ∂1f does not exist weakly in Lloc!