368 BRUCE K. DRIVER†
19. Weak and Strong Derivatives 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 . − p p Definition 19.1. A sequence un ∞ L (Ω) is said to converge to u L (Ω) { }n=1 ⊂ loc ∈ loc if limn u un Lq(K) =0for all compact subsets K Ω. →∞ k − k ⊂ The following simple but useful remark will be used (typically without further comment) in the sequel. 1 1 1 Remark 19.2. 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 → → 19.1. Basic Definitions and Properties. d p p Definition 19.3 (Weak Differentiability). Let v R and u L (Ω)(u Lloc(Ω)) p ∈p ∈ ∈ then ∂vu is said to exist weakly in L (Ω)(Lloc(Ω)) if there exists a function g Lp(Ω)(g Lp (Ω)) such that ∈ ∈ loc (19.1) u, ∂vφ = g, φ for all φ C∞(Ω). h i −h i ∈ c (w) d α The function g if it exists will be denoted by ∂v u. Similarly if α N0 and ∂ is α p p ∈ as in Notation 11.10, we say ∂ u exists weakly in L (Ω)(Lloc(Ω)) iff there exists g Lp(Ω)(Lp (Ω)) such that ∈ loc α α u, ∂ φ =( 1)| | g,φ for all φ C∞(Ω). h i − h i ∈ c α n More generally if p(ξ)= aαξ is a polynomial in ξ R , then p(∂)u exists α N ∈ weakly in Lp(Ω)(Lp (Ω))| i|ff≤there exists g Lp(Ω)(Lp (Ω)) such that loc P ∈ loc (19.2) u, p( ∂)φ = g, φ for all φ C∞(Ω) h − i h i ∈ c and we denote g by w p(∂)u. − 1 By Corollary 11.28, there is at most one g Lloc(Ω) such that Eq. (19.2) holds, so w p(∂)u is well defined. ∈ − d 1 Lemma 19.4. Let p(ξ) be a polynomial on R ,k=deg(p) N, and u Lloc(Ω) 1 ∈ ∈ such that p(∂)u exists weakly in Lloc(Ω). Then (1) suppm(w p(∂)u) suppm(u), where suppm(u) is the essential support of u relative− to Lebesgue⊂ measure, see Definition 11.14. k (2) If deg p = k and u U C (U, C) for some open set U Ω, then w p(∂)u = p (∂) u a.e. on U. | ∈ ⊂ − Proof. ANALYSIS TOOLS WITH APPLICATIONS 369
(1) Since
w p(∂)u, φ = u, p( ∂)φ =0for all φ C∞(Ω supp (u)), h − i −h − i ∈ c \ m an application of Corollary 11.28 shows w p(∂)u =0a.e. on Ω supp (u). So by Lemma 11.15, Ω supp (u)− Ω supp (w p(∂)u), i.e.\ m \ m ⊂ \ m − suppm(w p(∂)u) suppm(u). − ⊂ k (2) Suppose that u U is C and let ψ Cc∞(U). (We view ψ as a function d | d ∈ in Cc∞(R ) by setting ψ 0 on R U.) By Corollary 11.25, there exists ≡ \ γ Cc∞(Ω) such that 0 γ 1 and γ =1in a neighborhood of supp(ψ). ∈ ≤ ≤d k d Then by setting γu =0on R supp(γ) we may view γu Cc (R ) and so by standard integration by parts\ (see Lemma 11.26) and∈ the ordinary product rule, w p(∂)u, ψ = u, p( ∂)ψ = γu,p( ∂)ψ h − i h − i −h − i (19.3) = p(∂)(γu) ,ψ = p(∂)u, ψ h i h i wherein the last equality we have γ is constant on supp(ψ). Since Eq. (19.3) is true for all ψ Cc∞(U), an application of Corollary 11.28 with h =w p(∂)u p (∂) u and∈ µ = m shows w p(∂)u = p (∂) u a.e. on U. − − −
Notation 19.5. In light of Lemma 19.4 there is no danger in simply writing p (∂) u for w p(∂)u. So in the sequel we will always interpret p(∂)u in the weak or “dis- tributional”− sense. 1 Example 19.6. Suppose u(x)= x for x R, then ∂u(x)=sgn(x) in Lloc (R) 2 2 | | ∈ 1 while ∂ u(x)=2δ(x) so ∂ u(x) does not exist weakly in Lloc (R) . 1 2 Example 19.7. Suppose d =2and u(x, y)=1y>x. Then u L R , while ∈ loc ∂x1y>x = δ (y x) and ∂y1y>x = δ (y x) and so that neither ∂xu or ∂yu exists − − − ¡ ¢ weakly. On the other hand (∂x + ∂y) u =0weakly. To prove these assertions, 2 2 notice u C∞ R ∆ where ∆ = (x, x):x R . So by Lemma 19.4, for any polynomial∈ p (ξ) without\ constant term, if p (∂∈) u exists weakly then p (∂) u =0. However, ¡ ¢ © ª
u, ∂xφ = φx(x, y)dxdy = φ(y, y)dy, h − i − − Zy>x ZR
u, ∂yφ = φy(x, y)dxdy = φ(x, x)dx and h − i − Zy>x ZR u, (∂x + ∂y)φ =0 h − i from which it follows that ∂xu and ∂yu can not be zero while (∂x + ∂y)u =0. 1 On the other hand if p(ξ) and q (ξ) are two polynomials and u Lloc (Ω) is a 1 ∈ function such that p(∂)u exists weakly in Lloc (Ω) and q (∂)[p (∂) u] exists weakly 1 1 in Lloc (Ω) then (qp)(∂) u exists weakly in Lloc (Ω) . This is because u, (qp)( ∂) φ = u, p ( ∂) q( ∂)φ h − i h − − i = p (∂) u, q( ∂)φ = q(∂)p (∂) u, φ for all φ C∞ (Ω) . h − i h i ∈ c 1 2 Example 19.8. Let u(x, y)=1x>0 +1y>0 in Lloc R . Then ∂xu(x, y)=δ(x) and ∂ u(x, y)=δ(y) so ∂ u(x, y) and ∂ u(x, y) do not exist weakly in L1 2 . y x y ¡ ¢ loc R However ∂y∂xu does exists weakly and is the zero function. This shows ∂y∂xu may 1 ¡ 2¢ exists weakly despite the fact both ∂xu and ∂yu do not exists weakly in Lloc R . ¡ ¢ 370 BRUCE K. DRIVER†
1 Lemma 19.9. Suppose u Lloc (Ω) and p(ξ) is a polynomial of degree k such that 1 ∈ p (∂) u exists weakly in Lloc (Ω) then (19.4) p (∂) u, φ = u, p ( ∂) φ for all φ Ck (Ω) . h i h − i ∈ c Note: The point here is that Eq. (19.4) holds for all φ Ck (Ω) not just φ ∈ c ∈ Cc∞ (Ω) . Proof. Let φ Ck (Ω) and choose η C (B (0, 1)) such that η(x)dx =1 c c∞ Rd ∈ d ∈ and let η (x):= − η(x/ ). Then η φ C∞ (Ω) for sufficiently small and ∗ ∈ c R p ( ∂)[η φ]=η p ( ∂) φ p ( ∂) φ and η φ φ uniformly on compact sets− as ∗0. Therefore∗ by− the→ dominated− convergence∗ theorem,→ ↓ p (∂) u, φ = lim p (∂) u, η φ = lim u, p ( ∂)(η φ) = u, p ( ∂) φ . h i 0 0 ↓ h ∗ i ↓ h − ∗ i h − i
1 d 1 (w) Lemma 19.10 (Product Rule). Let u Lloc(Ω),v R and φ C (Ω). If ∂v u 1 (w) ∈ 1 ∈ ∈ exists in Lloc(Ω), then ∂v (φu) exists in Lloc(Ω) and (w) (w) ∂ (φu)=∂vφ u + φ∂ u a.e. v · v 1 1 d Moreover if φ Cc (Ω) and F := φu L (herewedefine F on R by setting F =0 d ∈ (w) ∈ (w) 1 d on R Ω ), then ∂ F = ∂vφ u + φ∂v u exists weakly in L (R ). \ · Proof. Let ψ C∞(Ω), then using Lemma 19.9, ∈ c (w) φu, ∂vψ = u, φ∂vψ = u, ∂v (φψ) ∂vφ ψ = ∂ u, φψ + ∂vφ u, ψ −h i −h i −h − · i h v i h · i (w) = φ∂ u, ψ + ∂vφ u, 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 = (φu) ,∂v (γψ) ∂vγ ψ h i h i h i h − · i (w) = (φu) ,∂v (γψ) = ∂ (φu) , (γψ) h i −h v i (w) (w) = ∂vφ u + φ∂ u, γψ = ∂vφ u + φ∂ u, ψ . −h · v i −h · v i (w) (w) This show ∂v F = ∂vφ u + φ∂v u as desired. · d q Lemma 19.11. Suppose q [1, ),p(ξ) is a polynomial in ξ R and u Lloc(Ω). ∈ q ∞ ∈ q ∈ If there exists um m∞=1 Lloc(Ω) such that p (∂) um exists in Lloc(Ω) for all m { } q ⊂ and there exists g L (Ω) such that for all φ C∞(Ω), ∈ loc ∈ c lim um,φ = u, φ and lim p (∂) um,φ = g, φ m m →∞h i h i →∞h i h i q then p (∂) u exists in Lloc(Ω) and p (∂) u = g. Proof. Since
u, p (∂) φ =limum,p(∂) φ = lim p (∂) um,φ = g, φ m m h i →∞h i − →∞h i h i q for all φ Cc∞(Ω),p(∂) u exists and is equal to g Lloc(Ω). Conversely∈ we have the following proposition. ∈ ANALYSIS TOOLS WITH APPLICATIONS 371
Proposition 19.12 (Mollification). Suppose q [1, ),p1(ξ),...,pN (ξ) is a col- d q ∈ ∞ lection of polynomials in ξ R and u Lloc(Ω) such that pl(∂)u exists weakly in q ∈ ∈ Lloc(Ω) for l =1, 2,...,N. Then there exists un Cc∞(Ω) such that un u in q q ∈ → L (Ω) and pl (∂) un pl (∂) u in L (Ω) for l =1, 2,...,N. loc → loc d Proof. Let η C (B(0, 1)) such that d ηdm =1and η (x):= η(x/ ) c∞ R − ∈ 1 be as in the proof of Lemma 19.9. For any function f Lloc (Ω) , >0 and c R ∈ x Ω := y Ω : dist(y, Ω ) > , let ∈ { ∈ }
f (x):=f η (x):=1Ωf η (x)= f(y)η (x y)dy. ∗ ∗ − ZΩ Notice that f C∞(Ω ) and Ω Ω as 0. ∈ ↑ ↓ Given a compact set K Ω let K := x Ω :dist(x, K) . Then K K as ⊂ { ∈ ≤ } ↓ 0, there exists 0 > 0 such that K0 := K 0 is a compact subset of Ω0 := Ω 0 Ω (see↓ Figure 38) and for x K, ⊂ ∈
f η (x):= f(y)η (x y)dy = f(y)η (x y)dy. ∗ − − ZΩ ZK Therefore, using Theorem 11.21,
Ω
0
Figure 38. The geomentry of K K0 Ω0 Ω. ⊂ ⊂ ⊂
f η f p = (1K f) η 1K f p (1K f) η 1K f p d 0 as 0. k ∗ − kL (K) k 0 ∗ − 0 kL (K) ≤ k 0 ∗ − 0 kL (R ) → ↓ q Hence, for all f L (Ω),f η C∞(Ω ) and ∈ loc ∗ ∈ (19.5) lim f η f Lp(K) =0. 0 k ∗ − k ↓ q q Now let p(ξ) be a polynomial on Rd,u L (Ω) such that p (∂) u L (Ω) and ∈ loc ∈ loc v := η u C∞(Ω ) as above. Then for x K and < 0, ∗ ∈ ∈
p(∂)v (x)= u(y)p(∂x)η (x y)dy = u(y)p( ∂y)η (x y)dy − − − ZΩ ZΩ = u(y)p( ∂y)η (x y)dy = u, p(∂)η (x ) − − h − · i ZΩ (19.6) = p(∂)u, η (x ) =(p(∂)u) (x). h − · i 372 BRUCE K. DRIVER†
From Eq. (19.6) we may now apply Eq. (19.5) with f = u and f = pl(∂)u for 1 l N to find ≤ ≤ N
v u p + pl(∂)v pl(∂)u p 0 as 0. k − kL (K) k − kL (K) → ↓ l=1 X For n N, let ∈ c Kn := x Ω : x n and d(x, Ω ) 1/n { ∈ | | ≤ ≥ } o (so Kn Kn+1 Kn+1 for all n and Kn Ω as n or see Lemma 10.10) ⊂ ⊂ o ↑ →∞ and choose ψn C∞(K , [0, 1]), using Corollary 11.25, so that ψn =1on a ∈ c n+1 neighborhood of Kn. Choose n 0 such that Kn+1 Ω and ↓ ⊂ n N
v u p + pl(∂)v pl(∂)u p 1/n. k n − kL (Kn) k n − kL (Kn) ≤ l=1 X Then un := ψn v C∞(Ω) and since un = v on Kn we still have · n ∈ c n N
(19.7) un u p + pl(∂)un pl(∂)u p 1/n. k − kL (Kn) k − kL (Kn) ≤ l=1 X o Since any compact set K Ω is contained in Kn for all n sufficiently large, Eq. (19.7) implies ⊂
N
lim un u p + pl(∂)un pl(∂)u p =0. n k − kL (K) k − kL (K) →∞ " l=1 # X The following proposition is another variant of Proposition 19.12 which the reader is asked to prove in Exercise 19.2 below.
Proposition 19.13. Suppose q [1, ),p1(ξ),...,pN (ξ) is a collection of poly- d q ∈q ∞d q nomials in ξ R and u L = L R such that pl(∂)u L for l =1, 2,...,N. ∈ ∈ d ∈ Then there exists un Cc∞ R such that ∈ ¡ ¢ ¡ ¢ N lim un u p + pl(∂)un pl(∂)u p =0. n k − kL k − kL →∞ " l=1 # X Notation 19.14 (Difference quotients). For v Rd and h R 0 and a function ∈ ∈ \{ } u : Ω C, let → u(x + hv) u(x) ∂hu(x):= − v h for those x Ω such that x+hv Ω. When v is one of the standard basis elements, ∈ h∈ h ei for 1 i d, we will write ∂ u(x) rather than ∂ u(x). Also let ≤ ≤ i ei hu(x):= ∂hu(x),...,∂hu(x) ∇ 1 n be the difference quotient approximation¡ to the gradient.¢ d p Definition 19.15 (Strong Differentiability). Let v R and u L , then ∂vu is p h ∈ p ∈ said to exist strongly in L if the limh 0 ∂v u exists in L . We will denote the limit (s) → by ∂v u. ANALYSIS TOOLS WITH APPLICATIONS 373
p d (s) p (w) It is easily verified that if u L ,v R and ∂v u L exists then ∂v u exists (w) (s) ∈ ∈ ∈ and ∂v u = ∂v u. The key to checking this assetion is the identity,
h u(x + hv) u(x) ∂v u, φ = − φ(x)dx h i d h ZR φ(x hv) φ(x) h (19.8) = u(x) − − dx = u, ∂ vφ . d h h − i ZR (s) h p d Hence if ∂v u =limh 0 ∂v u exists in L and φ Cc∞(R ), then → ∈ (s) h h d ∂v u, φ =lim∂v u, φ =limu, ∂ vφ = 0 u, φ ( hv) = u, ∂vφ h i h 0h i h 0h − i dh| h · − i −h i → → wherein Corollary 7.43 has been used in the last equality to bring the derivative (w) (s) past the integral. This shows ∂v u exists and is equal to ∂v u. What is somewhat (w) more surprising is that the converse assertion that if ∂v u exists then so does (s) 2 p ∂v u. Theorem 19.18 is a generalization of Theorem 12.36 from L to L . For the reader’s convenience, let us give a self-contained proof of the version of the Banach - Alaoglu’s Theorem which will be used in the proof of Theorem 19.18. (This is the same as Theorem 18.27 above.) Proposition 19.16 (Weak- Compactness: Banach - Alaoglu’s Theorem). Let X ∗ be a separable Banach space and fn X∗ be a bounded sequence, then there exist ˜ { } ⊂ asubsequence fn fn such that lim fn(x)=f(x) for all x X with f X∗. n { } ⊂ { } →∞ ∈ ∈ Proof. Let D X be a countable linearly independent subset of X such ⊂ that span(D)=X. Using Cantor’s diagonal trick, choose f˜n fn such that { } ⊆ { } λx := lim f˜n(x) exist for all x D. Define f : span(D) by the formula n R →∞ ∈ →
f( axx)= axλx x D x D X∈ X∈ where by assumption #( x D : ax =0 ) < . Then f :span(D) R is linear { ∈ 6 } ∞ → and moreover f˜n(y) f(y) for all y span(D). Now → ∈ f(y) =lim f˜n(y) lim sup f˜n y C y for all y span(D). | | n | | ≤ n k kk k ≤ k k ∈ →∞ →∞ Hence by the B.L.T. Theorem 4.1, f extends uniquely to a bounded linear functional on X. We still denote the extension of f by f X∗. Finally, if x X and y span(D) ∈ ∈ ∈
f(x) f˜n(x) f(x) f(y) + f(y) f˜n(y) + f˜n(y) f˜n(x) | − | ≤ | − | | − | | − | f x y + f˜n x y + f(y) f˜n(y) ≤ k kk − k k kk − k | − k 2C x y + f(y) f˜n(y) 2C x y as n . ≤ k − k | − | → k − k →∞ Therefore lim sup f(x) f˜n(x) 2C x y 0 as y x. n | − | ≤ k − k → → →∞ p Corollary 19.17. Let p (1, ] and q = p 1 . Then to every bounded sequence p ∈ ∞ − p un n∞=1 L (Ω) there is a subsequence u˜n n∞=1 and an element u L (Ω) such {that} ⊂ { } ∈ q lim u˜n,g = u, g for all g L (Ω) . n →∞h i h i ∈ 374 BRUCE K. DRIVER†
Proof. By Theorem 15.14, the map p q v L (Ω) v, (L (Ω))∗ ∈ → h ·i ∈ is an isometric isomorphism of Banach spaces. By Theorem 11.3, Lq(Ω) is separable for all q [1, ) and hence the result now follows from Proposition 19.16. ∈ ∞ Theorem 19.18 (Weak and Strong Differentiability). Suppose p [1, ),u ∈ ∞ ∈ Lp(Rd) and v Rd 0 . Then the following are equivalent: ∈ \{ } p d (1) There exists g L (R ) and hn n∞=1 R 0 such that limn hn =0 and ∈ { } ⊂ \{ } →∞ hn d lim ∂v u, φ = g, φ for all φ Cc∞( ). n R →∞h i h i ∈ (w) p d (2) ∂v u exists and is equal to g L (R ), i.e. u, ∂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 un Cc∞(R ) such that un u and ∂vun g as n . ∈ ∈ → → (s) →∞ p d h p (4) ∂v u exists and is is equal to g L (R ), i.e. ∂v u g in L as h 0. ∈ → → Moreover if p (1, ) any one of the equivalent conditions 1. — 4. above are implied by the following∈ ∞ condition. h ∞ n 10. There exists hn n=1 R 0 such that limn hn =0and supn ∂v u p < . { } ⊂ \{ } →∞ ∞ ° ° Proof. 4. = 1. is simply the assertion that strong convergence implies° weak° convergence. ⇒ d 1. = 2. For φ Cc∞(R ), Eq. (19.8) and the dominated convergence theorem implies⇒ ∈ hn hn g, φ = lim ∂v u, φ = lim u, ∂ φ = u, ∂vφ . n n v h i →∞h i →∞h − i −h i d 2. = 3. Let η C ( , ) such that d η(x)dx =1and let ηm(x)= c∞ R R R d ⇒ ∈ d m η(mx), then by Proposition 11.24, hm := ηm u C∞(R ) for all m and R∗ ∈
∂vhm(x)=∂vηm u(x)= ∂vηm(x y)u(y)dy = u, ∂v [ηm (x )] ∗ d − h − − · i ZR = g, ηm (x ) = ηm g(x). h − · i ∗ p d p d By Theorem 11.21, hm u L (R ) and ∂vhm = ηm 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 (ψ hm)=∂vψ hm + ψ ∂vhm = (∂vψ) hm + ψ ∂vhm p p so that ψ hm hm in L and ∂v (ψ hm) ∂vhm in L as 0. Let um = 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 um Cc∞(R ),um u and ∂vum g in L as m . 3. = ∈4. By the fundamental→ theorem→ of calculus →∞ ⇒ um(x + hv) um(x) ∂hu (x)= − v m h 1 1 d 1 (19.9) = u (x + shv)ds = (∂ u )(x + shv)ds. h ds m v m Z0 Z0 ANALYSIS TOOLS WITH APPLICATIONS 375 and therefore, 1 h ∂ um(x) ∂vum(x)= [(∂vum)(x + shv) ∂vum(x)] ds. v − − Z0 So by Minkowski’s inequality for integrals, Theorem 9.27, 1 h ∂ um(x) ∂vum (∂vum)( + shv) ∂vum ds v − p ≤ k · − kp Z0 ° ° and letting m° in this equation° then implies →∞ 1 ∂hu g g( + shv) g ds. v − p ≤ k · − kp Z0 ° ° By the dominated convergence° ° theorem and Proposition 11.13, the right member of this equation tends to zero as h 0 and this shows item 4. holds. → (10. = 1. when p>1) This is a consequence of Corollary 19.17 (or see Theorem ⇒ w 18.27 above) which asserts, by passing to a subsequence if necessary, that ∂hn u g v → for some g Lp(Rd). ∈ Example 19.19. Thefactthat(10) does not imply the equivalent conditions 1 — 4 in Theorem 19.18 when p =1is demonstrated by the following example. Let u := 1[0,1], then u(x + h) u(x) 1 − dx = 1[ h,1 h](x) 1[0,1](x) dx =2 h h − − − ZR ¯ ¯ ZR ¯ ¯ | | ¯ ¯ for h < 1. On¯ the other hand¯ the distributional¯ derivative of u is¯ ∂u(x)=δ(x) δ(x| |1) which¯ is not in L1. ¯ − − Alternatively,ifthereexistsg L1(R,dm) such that ∈ u(x + hn) u(x) lim − = g(x) in L1 n h →∞ n for some sequence hn n∞=1 as above. Then for φ Cc∞(R) we would have on one hand, { } ∈
u(x + hn) u(x) φ(x hn) φ(x) − φ(x)dx = − − u(x)dx R hn R hn Z Z 1 φ0(x)dx =(φ(0) φ(1)) as n , →− − →∞ Z0 while on the other hand,
u(x + hn) u(x) − φ(x)dx g(x)φ(x)dx. h → ZR n ZR These two equations imply
(19.10) 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 ∈ \{ } 11.28, 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.∈ (19.10).\{ This} example also shows that the unit∈ ball in L1(R,dm) is not weakly sequentially compact. Compare with Example 18.24. 376 BRUCE K. DRIVER†
Proposition 19.20 (A weak form of Weyls Lemma). If u L2(Rd) such that 2 d α 2 d ∈ f := u L (R ) then ∂ u L R for α 2. Furthermore if k N0 and 4 ∈ ∈ | | ≤ ∈ ∂βf L2 Rd for all β k, then ∂αu L2 Rd for α k +2. ∈ | | ≤ ¡ ¢∈ | | ≤ d Proof.¡By¢ Proposition 19.13, there exists u¡n ¢ Cc∞ R such that un u and 2 d ∈ → ∆un ∆u = f in L R . By integration by parts we find → ¡ ¢ 2 ¡ ¢ (un um) dm =( ∆(un um), (un um))L2 (f f,u u)=0as m, n d |∇ − | − − − →− − − →∞ ZR 2 and hence by item 3. of Theorem 19.18, ∂iu L for each i. Since ∈ 2 2 u L2 = lim un dm =( ∆un,un)L2 (f,u) as n k∇ k n d |∇ | − →− →∞ →∞ ZR we also learn that 2 u 2 = (f,u) f 2 u 2 . k∇ kL − ≤ k kL ·k kL Let us now consider d d 2 2 ∂i∂jun dm = ∂jun∂i ∂jundm d | | − d i,j=1 R i,j=1 R X Z X Z d d 2 = ∂jun∂j∆undm = ∂j un∆undm − d d j=1 R j=1 R X Z X Z 2 2 = ∆un dm = ∆un L2 . d | | k k ZR Replacing un by un um in this calculation shows − d 2 2 ∂i∂j(un um) dm = ∆(un um) L2 0 as m, n d | − | k − k → →∞ i,j=1 R X Z 2 d and therefore by Lemma 19.4 (also see Exercise 19.3), ∂i∂ju L R for all i, j and ∈ d ¡ ¢ 2 2 2 ∂i∂ju dm = ∆u L2 = f L2 . d | | k k k k i,j=1 R X Z 2 d h 2 d Let us now further assume ∂if L R . Then for h R 0 ,∂i u L (R ) ∈ ∈ \{ } ∈ and ∆∂hu = ∂h∆u = ∂hf L2( d) and hence by what we have just proved, i i i R ¡ ¢ ∂α∂hu = ∂h∂αu L2 and ∈ i i ∈ 2 2 ∂h∂αu C ∂hf + ∂hf ∂hu i L2(Rd) ≤ i L2 i L2 · i L2 α 2 |X|≤ ° ° h° ° ° ° ° ° i ° ° ° °2 ° ° ° ° C ∂if 2 + ∂if 2 ∂iu 2 ≤ k kL k kL ·k kL with the last bound being independenth of h =0. Therefore applyingi Theorem 19.18 α 2 d 6 again we learn that ∂i∂ u L (R ) for all α 2. The remainder of the proof, which is now an induction argument∈ using the| above| ≤ ideas, is left as an exercise to the reader.
Theorem 19.21. Suppose that Ω is a precompact open subset of Rd and V is an open precompact subset of Ω. ANALYSIS TOOLS WITH APPLICATIONS 377
p p h (1) If 1 p< ,u L (Ω) and ∂iu L (Ω), then ∂i u Lp(V ) ∂iu Lp(Ω) ≤ ∞ ∈1 c ∈ k k ≤ k k for all 0 < h < 2 dist(V,Ω ). (2) Suppose that| |1
hn hn vϕdm = lim ∂ uϕdm =lim u∂− ϕdm n i n i →∞ →∞ Ω VZ ΩZ Z
= u∂iϕdm= u∂iϕdm. − − ΩZ VZ p Therefore ∂iu = v L (V ) and ∂iu p v p CV . Finally if C := ∈ k kL (V ) ≤ k kL (V ) ≤ supV Ω CV < , then by the dominated convergence theorem, ⊂⊂ ∞ ∂iu Lp(Ω) = lim ∂iu Lp(V ) C. k k V Ω k k ≤ ↑
We will now give a couple of applications of Theorem 19.18.
Lemma 19.22. Let v Rd. 1 ∈ 1 (1) If h L and ∂vh exists in L , then d ∂vh(x)dx =0. R ∈ 1 1 1 p q (2) If p, q, r [1, ) satisfy r− = p− + q− ,f L and g L are functions ∈ ∞ p R 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, · ·
(19.12) ∂vf,g = f,∂vg . h i −h i 378 BRUCE K. DRIVER†
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. (19.12) holds. · · d Proof. 1) By item 3. of Theorem 19.18 there exists hn Cc∞(R ) such that 1 ∈ hn h and ∂vhn ∂vh in L . Then → → d d ∂vhn(x)dx = 0 hn(x + hv)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 19.2, 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 19.18 that ∂v(fg) exists in L and ∂v(fg)=∂vf g + f ∂vg. Eq. (19.12) 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. (19.12) holds in this case by item 1. already proved. · · 1 1 1 p q Lemma 19.23. 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. ∗ ∗ ∗ Proof. Items 1 and 2. By Young’s inequality (Theorem 11.19) and simple computations:
τ hv(f g) f g τ hvf g f g − ∗ − ∗ (∂vf) g = − ∗ − ∗ (∂vf) g h − ∗ h − ∗ ° °r ° °r ° ° ° ° ° ° ° τ hvf f ° = − − (∂vf) g ° ° ° h − ∗ ° °· ¸ °r ° ° °τ hvf f ° − − (∂vf) g ≤ ° h − k k°q ° °p ° ° which tends to zero as h 0. The second item° is proved analogously,° or just make use of the fact that f g→= g f and apply° Item 1. ° ∗ ∗ ANALYSIS TOOLS WITH APPLICATIONS 379
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.
19.2. The connection of Weak and pointwise derivatives.
Proposition 19.24. Let Ω =(α, β) R be an open interval and f L1 (Ω) such ⊂ ∈ loc that ∂(w)f =0in L1 (Ω). Then there exists c C such that f = c a.e. More loc ∈ generally, suppose F : Cc∞(Ω) C is a linear functional such that F (φ0)=0for →d all φ Cc∞(Ω), where φ0(x)= φ(x), then there exists c C such that ∈ dx ∈
(19.13) F (φ)= c, φ = cφ(x)dx for all φ C∞(Ω). h i ∈ c ZΩ Proof. Before giving a proof of the second assertion, let us show it includes the (w) first. Indeed, if F (φ):= φfdm and ∂ f =0, then F (φ0)=0for all φ C∞(Ω) Ω ∈ c and therefore there exists c C such that R ∈ φfdm = F (φ)=c φ, 1 = c φfdm. h i ZΩ ZΩ But this implies f = c a.e. So it only remains to prove the second assertion.
Let η Cc∞(Ω) such that Ω ηdm =1. Given φ Cc∞(Ω) Cc∞ (R) , let x∈ ∈ ⊂ ψ(x)= (φ(y) η(y) φ, 1 ) dy. Then ψ (x)=φ(x) η(x) φ, 1 and ψ C (Ω) R 0 c∞ as the reader−∞ should− check.h Therefore,i − h i ∈ R 0=F (ψ)=F (φ φ, η η)=F (φ) φ, 1 F (η) − h i − h i which shows Eq. (19.13) holds with c = F(η). This concludes the proof, however it will be instructive to give another proof of the first assertion. 1 (w) Alternative proof of first assertion. Suppose f Lloc(Ω) and ∂ f =0 ∈ (w) and fm := f ηm as is in the proof of Lemma 19.9. Then fm0 = ∂ f ηm =0, ∗ 1∗ so fm = cm for some constant cm C. By Theorem 11.21, fm f in Lloc(Ω) and therefore if J =[a, b] is a compact∈ subinterval of Ω, → 1 cm ck = fm fk dm 0 as m, k . | − | b a | − | → →∞ − ZJ So cm m∞=1 is a Cauchy sequence and therefore c := limm cm exists and f = { } →∞ limm fm = c a.e. →∞ Theorem 19.25. Suppose f L1 (Ω). Then there exists a complex measure µ on ∈ loc Ω such that B
(19.14) f,φ0 = µ(φ):= φdµ for all φ C∞(Ω) −h i ∈ c ZΩ iff there exists a right continuous function F of bounded variation such that F = f a.e. In this case µ = µF , i.e. µ((a, b]) = F (b) F (a) for all Proof. Suppose f = F a.e. where F is as above and let µ = µF be the associated measure on Ω. Let G(t)=F (t) F ( )=µ(( ,t]), then using Fubini’s theorem and theB fundamental theorem− of calculus,−∞ −∞ f,φ0 = F, φ0 = G, φ0 = φ0(t) 1( ,t](s)dµ(s) dt −h i −h i −h i − −∞ ZΩ ·ZΩ ¸ = φ0(t)1( ,t](s)dtdµ(s)= φ(s)dµ(s)=µ(φ). − −∞ ZΩ ZΩ ZΩ Conversely if Eq. (19.14) holds for some measure µ, let F (t):=µ(( ,t]) then working backwards from above, −∞ f,φ0 = µ(φ)= φ(s)dµ(s)= φ0(t)1( ,t](s)dtdµ(s)= φ0(t)F (t)dt. −h i − −∞ − ZΩ ZΩ ZΩ ZΩ This shows ∂(w) (f F )=0and therefore by Proposition 19.24, f = F + c a.e. for − some constant c C. Since F + c is right continuous with bounded variation, the proof is complete.∈ 1 w Proposition 19.26. 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 integration by parts, Corollary 16.32,∈ ⊂ ⊂ 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) and a Ω. Define F C (Ω) by x w ∈ ∈ F (x):= a ∂ f(y)dy. Then F is absolutely continuous on compacts and therefore by fundamental theorem of calculus for absolutely continuous functions (Theorem R w 16.31), F 0(x) exists and is equal to ∂ f(x) for a.e. x Ω. Moreover, by the first part of the argument, ∂wF exists and ∂wF = ∂wf, and∈ so by Proposition 19.24 there is a constant c such that f˜(x):=F (x)+c = f(x) for a.e. x Ω. ∈ Definition 19.27. Let X and Y be metric spaces. A function u : X Y is said to be Lipschitz if there exists C< such that → ∞ Y X d (u(x),u(x0)) Cd (x, x0) for all x, x0 X ≤ ∈ and said to be locally Lipschitz if for all compact subsets K X there exists ⊂ CK < such that ∞ Y X d (u(x),u(x0)) CK d (x, x0) for all x, x0 K. ≤ ∈ 1 Proposition 19.28. Let u Lloc(Ω). Then there exists a locally Lipschitz function ∈ 1 u˜ : Ω C such that u˜ = u a.e. iff ∂iu Lloc(Ω) exists and is locally (essentially) bounded→ for i =1, 2,...,d. ∈ ANALYSIS TOOLS WITH APPLICATIONS 381 Proof. Suppose u =˜u a.e. and u˜ is Lipschitz and let p (1, ) and V be ∈ ∞ a precompact open set such that V¯ W and let V := x Ω :dist(x, V¯ ) . c ⊂ ∈ ≤ Then for um un u = um un 2K/m 0 as m . k − k k − k∞ ≤ → →∞ Therefore, un converges uniformly to a continuous function u˜V . The next theorem is from Chapter 1. of Maz’ja [2]. Theorem 19.29. Let p 1 and Ω be an open subset of Rd,x Rd be written as d 1 ≥ ∈ x =(y, t) R − R, ∈ × d 1 Y := y R − :( y R) Ω = ∈ { }× ∩ 6 ∅ p p and u L (Ω). Then ∂tu exists© weakly in L (Ω) iff there is aª version u˜ of u such that ∈ ∂u˜(y,t) for a.e. y Y the function t u˜(y, t) is absolutely continuous, ∂tu(y, t)= ∈ → ∂t a.e., and ∂u˜ < . ∂t Lp(Ω) ∞ Proof.° For° the proof of Theorem 19.29, it suffices to consider the case where °d ° d 1 Ω =(0, 1) . Write x Ω as x =(y,t) Y (0, 1) = (0, 1) − (0, 1) and ∂tu for ∈ ∈ × × the weak derivative ∂ed u. By assumption ∂tu(y,t) dydt = ∂tu ∂tu < | | k k1 ≤ k kp ∞ ZΩ and so by Fubini’s theorem there exists a set of full measure, Y0 Y, such that ⊂ 1 ∂tu(y,t) dt < for y Y0. | | ∞ ∈ Z0 t So for y Y0, the function v(y, t):= 0 ∂tu(y, τ)dτ is well defined and absolutely ∈ ∂ continuous in t with v(y, t)=∂tu(y,t) for a.e. t (0, 1). Let ξ C∞(Y ) and ∂t R ∈ ∈ c η C∞ ((0, 1)) , then integration by parts for absolutely functions implies ∈ c 1 1 ∂ v(y, t)η˙(t)dt = v(y,t)η(t)dt for all y Y0. − ∂t ∈ Z0 Z0 Multiplying both sides of this equation by ξ(y) and integrating in y shows ∂ v(x)η˙(t)ξ(y)dydt = v(y, t)η(t)ξ(y)dydt = ∂tu(y,t)η(t)ξ(y)dydt. − ∂t − ZΩ ZΩ ZΩ Using the definition of the weak derivative, this equation may be written as u(x)η˙(t)ξ(y)dydt = ∂tu(x)η(t)ξ(y)dydt − ZΩ ZΩ and comparing the last two equations shows [v(x) u(x)] η˙(t)ξ(y)dydt =0. − ZΩ Since ξ Cc∞(Y ) is arbitrary, this implies there exists a set Y1 Y0 of full measure such that∈ ⊂ [v(y,t) u(y, t)] η˙(t)dt =0for all y Y1 − ∈ ZΩ from which we conclude, using Proposition 19.24, that u(y, t)=v(y, t)+C(y) for t Jy where md 1 (Jy)=1, here mk denotes k — dimensional Lebesgue measure. In∈ conclusion we− have shown that t (19.17) u(y, t)=˜u(y,t):= ∂tu(y, τ)dτ + C(y) for all y Y1 and t Jy. ∈ ∈ Z0 ANALYSIS TOOLS WITH APPLICATIONS 383 We can be more precise about the formula for u˜(y,t) by integrating both sides of Eq. (19.17) on t we learn 1 t 1 1 1 C(y)= dt ∂τ u(y, τ)dτ u(y, t)dt = (1 τ) ∂τ u(y,τ)dτ u(y, t)dt 0 0 − 0 0 − − 0 Z 1 Z Z Z Z = [(1 t) ∂tu(y, t) u(y,t)] dt − − Z0 and hence t 1 u˜(y, t):= ∂τ u(y, τ)dτ + [(1 τ) ∂τ u(y,τ) u(y, τ)] dτ − − Z0 Z0 which is well defined for y Y0. ∈ For the converse suppose that such a u˜ exists, then for φ C∞ (Ω) , ∈ c ∂u˜(y, t) u(y, t)∂tφ(y, t)dydt = u˜(y, t)∂tφ(y, t)dtdy = φ(y, t)dtdy − ∂t ZΩ ZΩ ZΩ wherein we have used integration by parts for absolutely continuous functions. From ∂u˜(y,t) this equation we learn the weak derivative ∂tu(y, t) exists and is given by ∂t a.e. 19.3. Exercises. Exercise 19.1. Give another proof of Lemma 19.10 base on Proposition 19.12. Exercise 19.2. Prove Proposition 19.13. Hints: 1. Use u as defined in the proof d of Proposition 19.12 to show it suffices to consider the case where u C∞ R p d α p d d ∈ ∩ L R with ∂ u L R for all α N0. 2. Then let ψ Cc∞ (B(0, 1), [0, 1]) ∈ ∈ ∈ ¡ ¢ such that ψ =1on a neighborhood of 0 and let un(x):=u(x)ψ(x/n). ¡ ¢ ¡ ¢ Exercise 19.3. 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 Definition∈ 19.3.) (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 19.4. Let p [1, ),f Lp and α be a multi index. Show ∂αf exists ∈ ∞ p ∈ n p weakly (see Definition 19.3) in L iff there exists fn Cc∞(R ) and g L such α p ∈ ∈ that fn f and ∂ fn g in L as n . Hints: See exercises 19.2 and 19.3. → → →∞ Exercise 19.5. Folland 8.8 on p. 246. 1 Exercise 19.6. 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. | | 384 BRUCE K. DRIVER† 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) Suppose x0 Ω and >0 such that C := Cx ( ) Ω and let ηn be a ∈ 0 ⊂ sequence of approximate δ — functions such supp(ηn) B0(1/n) for all n. α α ⊂ Then for n large enough, ∂ (f ηn)=(∂ f) ηn 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 19.10. 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 19.11. 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!