27. Sobolev Inequalities 27.1

ANALYSIS TOOLS WITH APPLICATIONS 493

27. Sobolev Inequalities 27.1. Morrey’s Inequality.

d 1 d Notation 27.1. Let S − be the sphere of radius one centered at zero inside R . d 1 d For a set Γ S − ,x R , and r (0, ), let ⊂ ∈ ∈ ∞ Γx,r x + sω : ω Γ such that 0 s r . ≡ { ∈ ≤ ≤ } So Γx,r = x + Γ0,r where Γ0,r is a cone based on Γ, seeFigure49below.

Figure 49. The cone Γ0,r.

d 1 Notation 27.2. If Γ S − is a measurable set let Γ = σ(Γ) be the surface “area” of Γ. ⊂ | |

Notation 27.3. If Ω Rd is a measurable set and f : Rd C is a measurable function let ⊂ → 1 fΩ := f(x)dx := f(x)dx. − m(Ω) Ω ZΩ Z By Theorem 8.35, r d 1 (27.1) f(y)dy = f(x + y)dy = dt t − f(x + tω) dσ(ω) Γx,r Γ0,r 0 Z Z Z ZΓ and letting f =1in this equation implies d (27.2) m(Γx,r)= Γ r /d. | | d 1 Lemma 27.4. Let Γ S − be a measurable set such that Γ > 0. For u 1 ⊂ | | ∈ C (Γx,r), 1 u(y) (27.3) u(y) u(x) dy |∇ d | 1 dy. − | − | ≤ Γ x y − ΓZx,r | |ΓZx,r | − | 494 BRUCE K. DRIVER†

d 1 Proof. Write y = x + sω with ω S − , then by the fundamental theorem of calculus, ∈ s u(x + sω) u(x)= u(x + tω) ωdt − ∇ · Z0 and therefore, s u(x + sω) u(x) dσ(ω) u(x + tω) dσ(ω)dt | − | ≤ 0 Γ |∇ | ZΓ Z Z s d 1 u(x + tω) = t − dt |∇ d | 1 dσ(ω) 0 Γ x + tω x − Z Z | − | u(y) u(y) = |∇ d | 1 dy |∇ d | 1 dy, y x − ≤ x y − ΓZx,s | − | ΓZx,r | − | wherein the second equality we have used Eq. (27.1). Multiplying this inequality d 1 by s − and integrating on s [0,r] gives ∈ d r u(y) m(Γx,r) u(y) u(y) u(x) dy |∇ d | 1 dy = |∇ d | 1 dy | − | ≤ d x y − Γ x y − ΓZx,r ΓZx,r | − | | | ΓZx,r | − | which proves Eq. (27.3).

Corollary 27.5. Suppose d

0,r (0, ) and S − 1 ≤∞ ∈ B | | ∈ ∞ u C (Γx,r). Then ∈ (27.4) u(x) C( Γ ,r,d,p) u 1,p | | ≤ | | k kW (Γx,r ) where 1/p 1 1/p 1 d− p 1 − 1 d/p C( Γ ,r,d,p):= max , − r − . | | Γ 1/p r p d · | | Ã µ − ¶ ! Proof. For y Γx,r, ∈ u(x) u(y) + u(y) u(x) | | ≤ | | | − | and hence using Eq. (27.3) and Hölder’s inequality, 1 u(y) u(x) u(y) dy + |∇ d | 1 dy | | ≤ − | | Γ x y − ΓZx,r | |ΓZx,r | − | 1 1 1 (27.5) u p 1 p + u p q L (Γx,r ) L (Γx,r ) L (Γx,r ) d 1 L (Γx,r ) ≤ m(Γx,r) k k k k Γ k∇ k k x k | | | − ·| − p where q = p 1 as before. Now − r 1 q d 1 d 1 q q = dt t − t − − dσ(ω) d 1 L (Γ0,r ) k − k 0 |·| Z ZΓ ¡ ¢ r p r d 1 d 1 1 p 1 − = Γ dt t − − − = Γ dt t− p 1 | | | | − Z0 Z0 and since ¡ ¢ d 1 p d 1 − = − − p 1 p 1 − − ANALYSIS TOOLS WITH APPLICATIONS 495 we ﬁnd p 1 1/q − 1 p 1 p d p 1 p d p−1 1 p (27.6) q = − Γ r − = − Γ r − . k d 1 kL (Γ0,r) p d | | p d | | |·| − µ − ¶ µ − ¶ Combining Eqs. (27.5), Eq. (27.6) along with the identity,

1 1 1/q d 1/p (27.7) 1 q = m(Γ ) = Γ r /d − , L (Γx,r ) x,r m(Γx,r) k k m(Γx,r) | | shows ¡ ¢ 1 1/p d 1/p 1 p 1 − 1 d/p u(x) u p Γ r /d − + u Lp(Γ ) − Γ r − | | ≤ k kL (Γx,r ) | | Γ k∇ k x,r p d | | | | µ − ¶ ¡ ¢ 1/p 1 1/p 1 d− p 1 − 1 d/p = u p + u Lp(Γ ) − r − . Γ 1/p k kL (Γx,r ) r k∇ k x,r p d | | " µ − ¶ # 1/p 1 1/p 1 d− p 1 − 1 d/p max , − u 1,p r − . ≤ Γ 1/p r p d k kW (Γx,r ) · | | Ã µ − ¶ !

Corollary 27.6. For d N and p (d, ] there are constants α = αd and β = βd ∈ ∈ ∞ such that if u C1(Rd) then for all x, y Rd, ∈ ∈ p 1 −p 1/p p 1 (1 d ) (27.8) u(y) u(x) 2βα − u Lp(B(x,r) B(y,r)) x y − p | − | ≤ p d k∇ k ∩ ·| − | µ − ¶ where r := x y . | − | d 1 Proof. Let r := x y ,V := Bx(r) By(r) and Γ, Λ S − be chosen so that | − | ∩ ⊂ x + rΓ = ∂Bx(r) By(r) and y + rΛ = ∂By(r) Bx(r), i.e. ∩ ∩ 1 1 Γ = (∂Bx(r) By(r) x) and Λ = (∂By(r) Bx(r) y)= Γ. r ∩ − r ∩ − − Also let W = Γx,r Λy,r, see Figure 50 below. By a scaling, ∩ Γx,r Λy,r Γx,1 Λy,1 βd := | ∩ | = | ∩ | (0, 1) Γx,r Γx,1 ∈ | | | | is a constant only depending on d, i.e. we have Γx,r = Λy,r = β W . Integrating the inequality | | | | | | u(x) u(y) u(x) u(z) + u(z) u(y) | − | ≤ | − | | − | over z W gives ∈ u(x) u(y) u(x) u(z) dz + u(z) u(y) dz | − | ≤ − | − | − | − | ZW ZW β = u(x) u(z) dz + u(z) u(y) dz Γx,r  | − | | − |  | | WZ WZ   β u(x) u(z) dz + u(z) u(y) dz . ≤ Γx,r  | − | | − |  | | ΓZ ΛZ  x,r y,r    496 BRUCE K. DRIVER†

Λ Γ

Λ Λ Γ Γ

Figure 50. The geometry of two intersecting balls of radius r := x y . Here W = Γx,r Λy,r and V = B(x, r) B(y, r). | − | ∩ ∩

Hence by Lemma 27.4, Hölder’s inequality and translation and rotation invariance of Lebesgue measure,

β u(z) u(z) u(x) u(y) dz + dz  |∇ d | 1 |∇ d |1  | − | ≤ Γ x z − z y − | | ΓZ | − | ΛZ | − |  x,r y,r  β  1  1 u p q + u p q ≤ Γ k∇ kL (Γx,r )k x d 1 kL (Γx,r ) k∇ kL (Λy,r)k y d 1 kL (Λy,r) | | µ | − ·| − | − ·| − ¶ (27.9) 2β 1 u p q ≤ Γ k∇ kL (V )k d 1 kL (Γ0,r ) | | |·| − p where q = p 1 is the conjugate exponent to p. Combining Eqs. (27.9) and (27.6) − 1 gives Eq. (27.8) with α := Γ − . | | Theorem 27.7 (Morrey’s Inequality). If d

1 d Proof. For p< and u Cc (R ), Corollaries 27.5 and 27.6 imply ∞ ∈ u(y) u(x) u d C u 1,p d and | − | C u p d BC(R ) W (R ) 1 d L (R ) k k ≤ k k x y − p ≤ k∇ k | − | ANALYSIS TOOLS WITH APPLICATIONS 497 which implies [u]1 d C u Lp(Rd) C u W 1,p(Rd)and hence − p ≤ k∇ k ≤ k k

(27.11) u 0,1 d C u W 1,p(Rd). k kC − p (Rd) ≤ k k

1,p d 1 d Now suppose u W (R ), choose (using Exercise 21.2) un Cc (R ) such that 1,p d∈ ∈ un u in W (R ). Then by Eq. (27.11), un um 0,1 d 0 as m, n → k − kC − p (Rd) → →∞ 0,1 d d 0,1 d d and therefore there exists u∗ C − p (R ) such that un u∗ in C − p (R ). ∈ → Clearly u∗ = u a.e. and Eq. (27.10) holds. 1, d If p = and u W ∞ R , then by Proposition 21.29 there is a version u∗ of u which∞ is Lipschitz∈ continuous. Now in both cases, p< and p = , the ¡ ¢ d ∞ ∞ sequence um := u ηm = u∗ ηm C∞ R and um u∗ uniformly on compact ∗ ∗ ∈ → subsets of d. Using Eq. (27.3) with u replaced by u along with a (by now) R ¡ ¢ m standard limiting argument shows that Eq. (27.3) still holds with u replaced by u∗. The proofs of Eqs. (27.4) and (27.8) only relied on Eq. (27.3) and hence go through without change. Similarly the argument in the ﬁrst paragraph only relied on Eqs. (27.4) and (27.8) and hence Eq. (27.10) is also valid for p = . ∞ d Corollary 27.8 (Morrey’s Inequality). Suppose Ω o R such that Ω is compact C1-manifold with boundary and d

(27.12) u∗ 0,1 d C u W 1,p(Ω), k kC − p (Ω) ≤ k k where C = C(p, d, Ω).

Proof. Let U be a precompact open subset of Rd and E : W 1,p(Ω) W 1,p(Rd) be an extension operator as in Theorem 25.35. For u W 1,p (Ω) with→d

u∗ 0,1 d U ∗ 0,1 d C Eu W 1,p(Rd) C u W 1,p(Ω) . k kC − p (Ω) ≤ k kC − p (Rd) ≤ k k ≤ k k

d 1,d d 1,d d The following example shows that L∞(R ) W (R ), i.e. W (R ) contains unbounded elements. Therefore Theorem 27.76⊆ and Corollary 27.8 are not valid for p = d. It turns out that for p = d, W 1,d Rd embeds into BMO(Rd) —thespace of functions with “bounded mean oscillation.” ¡ ¢ 1 d Example 27.9. Let u(x)=ψ(x)loglog 1+ x where ψ Cc∞(R ) is chosen so | | ∈ d 1,d d that ψ(x)=1for x 1. Then u/L∞³(R ) while´ u W (R ). Let us check | | ≤ ∈ ∈ this claim. Using Theorem 8.35, one easily shows u Lp(Rd). A short computation shows, for x < 1, that ∈ | | 1 1 1 u(x)= ∇ 1 1+ 1 ∇ x log 1+ x x | | | | | | 1³ ´ 1 1 = xˆ 1+ 1 1 − x x log 1+ x µ | | ¶ | | | | ³ ´ 498 BRUCE K. DRIVER† where xˆ = x/ x and so again by Theorem 8.35, | | d d 1 1 u(x) dx 2 dx |∇ | ≥ x <1  x + x log 1+ 1  Zd Z| | | | | | x R | | d  1 ³ ´ d 1 2 d 1 σ(S − ) r − dr = . ≥ r log 1+ 1 ∞ Z0 Ã r ! 27.2. Rademacher’s Theorem. ¡ ¢ Theorem 27.10. Suppose that u W 1,p(Ω) for some d

lim u(y) u(x) pdy =0for x Ω E. r 0 − |∇ −∇ | ∈ \ ↓ B(Zx,r) Fix a point x Ω E and let v(y):=u(y) u(x) u(x) (y x) and notice that v(y)= u(y∈) \ u(x). Applying Eq. (27.13)− to−∇v then implies· − ∇ ∇ −∇ u(y) u(x) u(x) (y x) | − −∇ · − | (1 d ) C u( ) u(x) Lp(B(x,r) B(y,r)) r − p ≤ k∇ · −∇ k ∩ · 1/p p (1 d ) C u(y) u(x) dy r − p ≤ |∇ −∇ | · ÃZB(x,r) ! 1/p

d 1 1/p d/p p (1 d ) = Cσ S − r u(y) u(x) dy r − p  − |∇ −∇ |  · B(Zx,r) ¡ ¢    1/p

d 1 1/p p = Cσ S − u(y) u(x) dy x y  − |∇ −∇ |  ·| − | B(Zx,r) ¡ ¢   which shows u is diﬀerentiable at x and u(x)=w- u(x). ∇ ∇ Theorem 27.11 (Rademacher’s Theorem). Let u be locally Lipschitz continuous d on Ω o R . Then u is diﬀerentiable almost everywhere and w-∂iu = ∂iu a.e. on Ω. ⊂

(w) d Proof. By Proposition 21.29 ∂i u exists weakly and is in ∂iu L∞(R ) for i =1, 2,...,d. The result now follows from Theorem 27.10. ∈ ANALYSIS TOOLS WITH APPLICATIONS 499

27.3. Gagliardo-Nirenberg-Sobolev Inequality. In this section our goal is to prove an inequality of the form: 1 d (27.14) u Lq C u Lp( d) for u Cc (R ). k k ≤ k∇ k R ∈ For λ>0, let uλ(x)=u(λx). Then

q q q dy uλ Lq = u(λx) dx = u(y) d k k d | | d | | λ ZR ZR d/q and hence uλ Lq = λ− u Lq . Moreover, uλ(x)=λ( u)(λx) and thus k k k k ∇ ∇ d/p uλ Lp = λ ( u)λ Lp = λλ− u Lp . k∇ k k ∇ k k∇ k 1 d If (27.14) is to hold for all u Cc (R ) then we must have ∈ d/q 1 d/p λ− u Lq = uλ Lq C uλ p d = Cλ − u Lp for all λ>0 k k k k ≤ k∇ kL (R ) k∇ k which is only possible if (27.15) 1 d/p + d/q =0, i.e. 1/p =1/d +1/q. − dp Notation 27.12. For p [1,d], let p∗ := d p with the convention that p∗ = if ∈ − ∞ p = d. That is p∗ = q where q solves Eq. (27.15). d Theorem 27.13. Let p =1so 1∗ = d 1 , then − 1 d d 1 d 2 (27.16) u 1∗ = u ∂iu(x) dx d− u 1 k k k k d 1 ≤ d | | ≤ k∇ k − i=1 R Y µZ ¶ for all u W 1,1(Rd). ∈ 1 d 1,1 d Proof. Since there exists un Cc (R ) such that un u in W (R ), asimple ∈ → 1 d limiting argument shows that it suﬃces to prove Eq. (27.16) for u Cc (R ). To help the reader understand the proof, let us give the proof for d 3 ∈ﬁrst and with the 1/2 ≤ constant d− being replaced by 1. After that the general induction argument will be given. (The adventurous reader may skip directly to the paragraph containing Eq. (27.17.) (d =1,p∗ = ) By the fundamental theorem of calculus, ∞ x x u(x) = u0(y)dy u0(y) dy u0(x) dx. | | ≤ | | ≤ R | | ¯Z−∞ ¯ Z−∞ Z ¯ ¯ Therefore u L u¯0 1 , proving¯ the d =1case. k k ∞ ≤ k ¯kL ¯ (d =2,p∗ =2)Applyingthesameargumentasabovetoy1 u(y1,x2) and → y2 u(x1,y2),, → ∞ ∞ u(x1,x2) ∂1u(y1,x2) dy1 u(y1,x2) dy1 and | | ≤ | | ≤ |∇ | Z−∞ Z−∞ ∞ ∞ u(x1,x2) ∂2u(x1,y2) dy2 u(x1,y2) dy2 | | ≤ | | ≤ |∇ | Z−∞ Z−∞ and therefore

2 ∞ ∞ u(x1,x2) ∂1u(y1,x2) dy1 ∂2u(x1,y2) dy2. | | ≤ | | · | | Z−∞ Z−∞ 500 BRUCE K. DRIVER†

Integrating this equation relative to x1 and x2 gives

2 2 ∞ ∞ u L2 = u(x) dx ∂1u(x) dx ∂2u(x) dx k k R2 | | ≤ | | | | Z µZ−∞ ¶µZ−∞ ¶ 2 ∞ u(x) dx ≤ |∇ | µZ−∞ ¶ which proves the d =2case. 1 2 3 (d =3,p∗ =3/2) Let x =(y1,x2,x3),x =(x1,y2,x3), and x =(x1,x2,y3). Then as above, ∞ i u(x) ∂iu(x ) dyi for i =1, 2, 3 | | ≤ | | Z−∞ and hence 1 3 2 3 ∞ i u(x) 2 ∂iu(x ) dyi . | | ≤ | | i=1 Y µZ−∞ ¶ Integrating this equation on x1 gives, 1 1 2 3 2 3 ∞ 1 ∞ i u(x) 2 dx1 ∂1u(x ) dy1 ∂iu(x ) dyi dx1 R | | ≤ | | | | Z µZ−∞ ¶ Z i=2 µZ−∞ ¶ 1 Y 1 2 3 2 ∞ ∞ i ∂1u(x) dx1 ∂iu(x ) dx1dyi ≤ | | | | i=2 µZ−∞ ¶ Y µZ−∞ ¶ wherein the second equality we have used the Hölder’s inequality with p = q =2. Integrating this result on x2 and using Hölder’s inequality gives 1 1 1 2 2 2 3 ∞ 3 u(x) 2 dx1dx2 ∂2u(x) dx1dx2 dx2 ∂1u(x) dx1 ∂3u(x ) dx1dy3 R2 | | ≤ R2 | | R | | R2 | | Z µZ ¶ 1 Z µZ−∞ 1 ¶ µZ 1 ¶ 2 2 2 ∂2u(x) dx1dx2 ∂1u(x) dx1dx2 ∂3u(x) dx . ≤ 2 | | 2 | | 3 | | µZR ¶ µZR ¶ µZR ¶ One more integration of x3 and application of Hölder’s inequality, implies 1 3 3 2 2 3 u(x) 2 dx ∂iu(x) dx u(x) dx 3 | | ≤ 3 | | ≤ 3 |∇ | R i=1 R R Z Y µZ ¶ µZ ¶ proving the d =3case. d i For general d (p∗ = d 1 ), as above let x =(x1,...,xi 1,yi,xi+1 ...,xd). Then − − ∞ i u(x) ∂iu(x ) dyi | | ≤ | | µZ−∞ ¶ and 1 d d 1 d ∞ − d 1 i (27.17) u(x) − ∂iu(x ) dyi . | | ≤ | | i=1 Y µZ−∞ ¶ Integrating this equation relative to x1 and making use of Hölder’s inequality in the form d d

(27.18) fi fi d 1 ° ° ≤ k k − °i=2 °1 i=2 °Y ° Y ° ° ° ° ANALYSIS TOOLS WITH APPLICATIONS 501

(see Corollary 9.3) we ﬁnd

1 1 d 1 d d 1 d − − d 1 i u(x) − dx1 ∂1u(x)dx1 dx1 ∂iu(x ) dyi | | ≤ | | ZR µZR ¶ ZR i=2 µZR ¶ 1 Y 1 d 1 d d 1 − i − ∂1u(x)dx1 ∂iu(x ) dx1dyi ≤ 2 | | µZR ¶ i=2 µZR ¶ 1 Y 1 1 d 1 d 1 d d 1 − − i − = ∂1u(x)dx1 ∂2u(x) dx1dx2 ∂iu(x ) dx1dyi . 2 | | 2 | | R R i=3 R µZ ¶ µZ ¶ Y µZ ¶ Integrating this equation on x2 and using Eq. (27.18) once again implies,

1 d d 1 d − d 1 (27.19) u(x) − dx ∂iu(x) dx1dx2 ...dxd d | | ≤ d | | ZR i=1 µZR ¶ Y 1 d d d 1 d 1 u(x) dx − = u(x) dx − . ≤ d |∇ | d |∇ | i=1 R R Y µZ ¶ µZ ¶ This estimate may now be improved on by using Young’s inequality (see Exercise d 1 d d 27.1) in the form ai d i=1 ai . Indeed by Eq. (27.19) and Young’s inequality, i=1 ≤

Q d P 1 d d 1 u d ∂iu(x) dx ∂iu(x) dx k k d 1 ≤ d | | ≤ d d | | − i=1 R i=1 R Y µZ ¶ X µZ ¶ 1 d 1 = ∂iu(x) dx √d u(x) dx d d | | ≤ d d |∇ | R i=1 R Z X Z 502 BRUCE K. DRIVER† wherein the last inequality we have used Hölder’s inequality for sums,

d d 1/2 d 1/2 2 ai 1 ai = √d a . | | ≤ | | | | i=1 Ãi=1 ! Ãi=1 ! X X X The next theorem generalizes Theorem 27.13 to an inequality of the form in Eq. (27.14). Theorem 27.14. If p [1,d) then, ∈ 1/2 p(d 1) 1,p d p (27.20) u Lp∗ d− − u L for all u W (R ). k k ≤ d p k∇ k ∈ − 1 d 1,p d Proof. As usual since Cc (R ) is dense in W R it suﬃces to prove Eq. 1 d 1 d s 1 d s (27.20) for u Cc (R ). For u Cc (R ) and s>1, u Cc (R ) and u = s 1 ∈ ∈ ¡| |¢ ∈ s ∇ | | s u − sgn(u) u. Applying Eq. (27.16) with u replaced by u andthenusing Holder’s| | inequality∇ gives | | s 1 s 1 s 1 u 1 d− 2 u = sd− 2 u − u 1 k| | k ∗ ≤ k∇ | | k1 k| | ∇ kL s s 1 (27.21) u Lp u − Lq ≤ √d k∇ k ·k| | k p where q = p 1 . We will now choose s so that s1∗ =(s 1)q, i.e. − − q 1 1 s = = 1 = q 1∗ 1 1∗ d 1 − q 1 d 1 1 p − − − − p (d 1) p (d ³1) ´d 1 = − = − = p∗ − . p (d 1) d (p 1) d p d − − − − For this choice of s, s1∗ = p∗ =(s 1)q and Eq. (27.21) becomes − 1/1∗ 1/q p∗ s p∗ (27.22) u dm u Lp u dm . d | | ≤ √d k∇ k · d | | ·ZR ¸ ·ZR ¸ Since 1 1 d 1 p 1 p(d 1) d(p 1) = − − = − − − 1∗ − q d − p dp d p 1 = − = , pd p∗ Eq. (27.22) implies Eq. (27.20).

Corollary 27.15. Suppose Ω Rd is bounded open set with C1-boundary, then ⊂ for all p [1,d) and 1 q p∗ there exists C = C(Ω,p,q) such that ∈ ≤ ≤ u q C u 1,p . k kL (Ω) ≤ k kW (Ω) Proof. Let U beaprecompactopensubsetofRd such that Ω¯ U and E : ⊂ W 1,p (Ω) W 1,p Rd be an extension operator as in Theorem 25.35. Then for → u C1(Ω) W 1,p(Ω), ∈ ∩ ¡ ¢ u p C Eu p d C (Eu) p d C u 1,p , k kL ∗ (Ω) ≤ k kL ∗ (R ) ≤ k∇ kL (R ) ≤ k kW (Ω) i.e.

(27.23) u p C u 1,p k kL ∗ (Ω) ≤ k kW (Ω) ANALYSIS TOOLS WITH APPLICATIONS 503

Since C1(Ω) is dense in W 1,p(Ω), Eq. (27.23) holds for all u W 1,p(Ω). Finally for ∈ all 1 q

Theorem 27.16. Let p [1, ] and u W 1,p Rd . Then ∈ ∞ ∈ 1,p 0,1 d (1) Morrey’s Inequality. If p>d,then W¡ ¢ C − p and → u∗ 0,1 d C u W 1,p(Rd). k kC − p (Rd) ≤ k k

(2) When p = d there is an L∞ — like space called BMO (which is not deﬁned in these notes) such that W 1,p BMO. → (3) GNS Inequality. If 1 p0 is some arbitrarily small number. When s = k + α with k N0 and 0 α<1 we will write Ck,α(Ω) simply as Cs(Ω). Warning, although C∈k,1(Ω) Ck≤+1(Ω) it is not true that Ck,1(Ω)=Ck+1(Ω). ⊂

Theorem 27.18 (Sobolev Embedding Theorems). Suppose Ω = Rd or Ω Rd is ⊂ bounded open set with C1-boundary, p [1, ),k,l N with l k. ∈ ∞ ∈ ≤ k,p k l,q dp (1) If p 0 q p − d andthereisaconstantC< such that ∞ k,p u k l,q C u k,p for all u W (Ω) . k kW − (Ω) ≤ k kW (Ω) ∈ k,p k (d/p) (2) If p>d/k,then W (Ω) C − + (Ω) and there is a constant C< such that → ∞ k,p u k (d/p)+ C u W k,p(Ω) for all u W (Ω) . k kC − (Ω) ≤ k k ∈ Proof. 1. (p

1 1 1 Deﬁne pj inductively by, p1 := p and pj := p . Since = it is easily ∗ j∗ 1 pj pj 1 d − − − checked that 1 = 1 l > 0 since p

k,p k 1,p1 k 2,p2 k l,pl W (Ω) W − (Ω) W − (Ω) ... W − (Ω) . → → → This proves the ﬁrst item of the theorem. The following lemmas will be used in the proofofitem2.

Lemma 27.19. Suppose j N and p d and j>d/p(i.e. j 1 if p>dand j 2 if p = d) then ∈ ≥ ≥ ≥ j,p j (d/p) W (Ω) C − + (Ω) → and there is a constant C< such that ∞ (27.25) u j (d/p) C u j,p . k kC − + (Ω) ≤ k kW (Ω) Proof. By the usual methods, it suﬃces to show that the estimate in Eq. (27.25) holds for all u C∞ Ω¯ . For p>dand∈ α j 1, | | ≤¡ ¢− α α ∂ u 0,1 d/p C ∂ u 1,p C u j,p k kC − (Ω) ≤ k kW (Ω) ≤ k kW (Ω) and hence

u j d/p := u j 1,1 d/p C u j,p k kC − (Ω) k kC − − (Ω) ≤ k kW (Ω) which is Eq. (27.25). When p = d (so now j 2), choose q (1,d) be close to d so that j>d/qand qd ≥ ∈ q∗ = d q >d.Then − j,d j,q j 1,q∗ j 2,1 d/q∗ W (Ω) W (Ω) W − (Ω) C − − (Ω) . → → → j,d j 2,α Since d/q∗ 0 as q d, we conclude that W (Ω) C − (Ω) for any α (0, 1) which we summarizes↓ ↑ by writing → ∈ j,d j (d/d) W (Ω) C − (Ω) . →

Proof. Continuation of the proof of Theorem 27.18. Item 2., (p>d/k) . If p d, the result follows from Lemma 27.19. So nos suppose that d>p>d/k ≥ and choose the largest l such that 1 l p and let q = dp , i.e. q ≤ d pl solves q d and − ≥ 1 1 l d d = or = l q p − d q p − Then d d k,p k l,q k l (d/q) k l ( p l) k ( p ) W (Ω) W − (Ω) C − − + (Ω)=C − − − + (Ω)=C − + (Ω) → → as desired. Remark 27.20 (Rule of thumb.). Assign the “degrees of regularity” k (d/p) to − + the space W k,p and k + α to the space Ck,α. If k,p k,α X, Y W : k N0,p [1, ] C : k N0,α [0, 1] ∈ ∈ ∈ ∞ ∪ ∈ ∈ with deg (X) deg (Y ), then X Y. reg ≥© reg → ª © ª k,p k ,q d d d d Example 27.21. (1) W W − iff k k iff iff → − p ≥ − − q ≥ p − q 1 1 = . q ≥ p d ANALYSIS TOOLS WITH APPLICATIONS 505

k,p 0,α d (2) W C iﬀ k p α. ⊂ − + ≥ ³ ´ 27.5. Compactness Theorems.

Lemma 27.22. Suppose Km : X Y are compact operators and K → k − Km 0 as n then K is compact. kL(X,Y ) → →∞ Proof. Let xn ∞ X be given such that xn 1. By Cantor’s diagonal- { }n=1 ⊂ k k ≤ ization scheme we may choose x0 xn such that ym := lim Kmx0 Y exists { n} ⊂ { } n n ∈ for all m. Hence →∞

Kx0 Kx0 = K (x0 x0 ) K (x0 x0 ) k n − k k n − k ≤ k n − k K Km x0 x0 + Km (x0 x0 ) ≤ k − kk n − k k n − k K Km + Km (x0 x0 ) ≤ k − k k n − k and therefore,

lim sup Kxn0 Kx0 K Km 0 as m . l,n k − k ≤ k − k → →∞ →∞

d d Lemma 27.23. Let η Cc∞(R ),Cηf = η f, Ω R be a bounded open set ∈ ∗ ⊂ with C1-boundary, V be an open precompact subset of Rd such that Ω¯ U and ⊂ E : W 1,1(Ω) W 1,1(Rd) be an extension operator as in Theorem 25.35. Then to → 1,1 every bounded sequence u˜n n∞=1 W (Ω) there has a subsequence un0 n∞=1 such { } ⊂ d { } that CηEun0 is uniformly convergent to a function in Cc R .

Proof. Let u := Eu˜ and C := sup u 1,1 d which is ﬁnite by assumption. n n n W (R ) ¡ ¢ 1,1 d k k ¯ d So un n∞=1 W (R ) is a bounded sequence such that supp(un) U U @@ R for{ all n.} Since⊂ η is compactly supported there exists a precompact⊂ open⊂ set V such ¯ d that U V and vn := η un Cc∞(V ) Cc∞ R for all n. Since, ⊂ ∗ ∈ ⊂ vn L η L un L1 η L¡ ¢un L1 C η L and k k ∞ ≤ k k ∞ k k ≤ k k ∞ k k ≤ k k ∞ Dvn L = η Dun L η L Dun 1 C η L , k k ∞ k ∗ k ∞ ≤ k k ∞ k kL ≤ k k ∞ it follows by the Arzela-Ascoli theorem that vn n∞=1 has a uniformly convergent subsequence. { }

Lemma 27.24. Let η Cc∞(B(0, 1), [0, )) such that d ηdm =1,ηm(x)= n ∈ ∞ R m η(mx) and Kmu =(Cη Eu) Ω. Then for all p [1,d) and q [1,p∗), m | ∈ R ∈ lim Km i B(W 1,p(Ω),Lq(Ω)) =0 m →∞ k − k where i : W 1,p(Ω) Lq(Ω) is the inclusion map. → 1 Proof. For u C (U) let vm := ηm u u, then ∈ c ∗ −

vm(x) ηm u(x) u(x) = ηm(y)(u(x y) u(x))dy | | ≤ | ∗ − | d − − ¯ZR ¯ ¯ ¯ y ¯ ¯ = η(y) u(x ) ¯ u(x) dy ¯ d − m − ¯ZR ¯ ¯ h 1 i ¯ ¯1 ¯ y ¯ dy y η(y) dt u(x¯ t ) ≤ m d | | 0 ∇ − m ZR Z ¯ ¯ ¯ ¯ ¯ ¯ 506 BRUCE K. DRIVER† and so by Minikowski’s inequality for integrals, 1 1 y vm Lr dy y η(y) dt u( t ) k k ≤ m d | | 0 ∇ · − m Lr ZR Z ° ° 1 ° 1 ° (27.26) y η(y)dy u r u 1,r d . ° L °W (R ) ≤ m d | | k∇ k ≤ m k k µZR ¶ By the interpolation inequality in Corollary 9.23, Theorem 27.14 and Eq. (27.26) with r =1, λ 1 λ vm q vm 1 vm −p k kL ≤ k kL k kL ∗ 1 λ 1 λ 1/2 p(d 1) − vm 1,1 d d− − vm Lp ≤ mλ k kW (R ) d p k∇ k · − ¸ λ λ 1 λ Cm− vm 1,1 d vm −1,p d ≤ k kW (R ) k kW (R ) λ λ 1 λ Cm− vm 1,1 d vm −1,p d ≤ k kW (R ) k kW (R ) λ λ 1 λ C(p, U )m− vm 1,p d vm −1,p d ≤ | | k kW (R ) k kW (R ) λ C(p, U )m− vm 1,p d ≤ | | k kW (R ) where λ (0, 1) is determined by ∈ 1 λ 1 λ 1 1 = + − = λ 1 + . q 1 p − p p ∗ µ ∗ ¶ ∗ Now using Proposition 11.12,

vm 1,p d = ηm u u 1,p d k kW (R ) k ∗ − kW (R ) ηm u 1,p d + u 1,p d 2 u 1,p d . ≤ k ∗ kW (R ) k kW (R ) ≤ k kW (R ) Putting this all together shows λ Kmu u q Kmu Eu Lq C(p, U )m− Eu 1,p d k − kL (Ω) ≤ k − k ≤ | | k kW (R ) λ C(p, U )m− u 1,p ≤ | | k kW (Ω) from which it follows that λ Km i 1,p q Cm− 0 as m . k − kB(W (Ω),L (Ω)) ≤ → →∞

Theorem 27.25 (Rellich - Kondrachov Compactness Theorem). Suppose Ω Rd 1 ⊂ isaprecompactopensubsetwithC -boundary, p [1,d) and 1 q

1,p Proof. If un ∞ is contained in the unit ball in W (Ω) , then by Lemma { }n=1 27.23 Kmun n∞=1 has a uniformly convergent subsequence and hence is convergent q { } 1,p q in L (Ω). This shows Km : W (Ω) L (Ω) is compact for every m. By Lemma 1,p →q 27.24, Km i in the L W (Ω) ,L (Ω) — norm and so by Lemma 27.22 i : → W 1,p (Ω) Lq (Ω) is compact. → ¡ ¢ k,p k ,q Corollary 27.26. The inclusion of W (Ω) into W − (Ω) is compact provided 1 1 l d pl dp l 1 and q > p d = −dp > 0, i.e. q

k,p Proof. Case (i) Suppose =1,q [1,p∗) and un n∞=1 W (Ω) is bounded. α 1,p ∈ { } ⊂ Then ∂ un n∞=1 W (Ω) is bounded for all α k 1 and therefore there exist { } ⊂ | α| ≤ − q a subsequence uñ n∞=1 un n∞=1 such that ∂ uñ is convergent in L (Ω) for all { } ⊂ { } k 1,q α k 1. This shows that uñ is W − (Ω) — convergent and so proves this case.| | ≤ − { } 1 1 1 Case (ii) >1. Let p˜ be defined so that = − . Then p˜ p − d k,p k +1,p˜ k ,q W (Ω) W − (Ω) W − (Ω). ⊂ ⊂⊂ k,p k ,q and therefore W (Ω) W − (Ω). ⊂⊂ Example 27.27. It is necessary to assume that The inclusion of L2([0, 1]) L1([0, 1]) is continuous (in fact a contraction) but not compact. To see this, take→ 2 2 1 un n∞=1 to be the Haar basis for L .Thenun 0 weakly in both L and L so if { } → 1 un ∞ were to have a convergent subsequence the limit would have to be 0 L . { }n=1 ∈ On the other hand, since un =1, un 2 = un 1 =1and any subsequential limit would have to have norm| one| and ink particulark k k not be 0. Lemma 27.28. Let Ω be a precompact open set such that Ω¯ isamanifoldwith C1 — boundary. Then for all p [1, ),W1,p(Ω) is compactly embedded in Lp(Ω). ∈ ∞ d 1,p Moreover if p>dand 0 β<1 p , then W (Ω) is compactly embedded in 0,β 1≤,p − C (Ω). In particular, W (Ω) L∞(Ω) for all d

pwe may choose q = p to learn W (Ω) L (Ω). ⊂⊂ Case 2, p [d, ). For any p0 [1,d), we have ∈ ∞ ∈ 1,p 1,p p W (Ω) W 0 (Ω) L 0∗ (Ω). → ⊂⊂ p0d 1,p q Since p0∗ = as p0 d, we see that W (Ω) L (Ω) for all q< . d po ↑∞ ↑ ⊂⊂ ∞ Moreover by− Morrey’s inequality (Corollary 27.8) and Proposition26.13 we have 1,p 0,1 d 0,β W (Ω) C − p (Ω) C (Ω) which completes the proof. → ⊂⊂ k,p k d δ Remark 27.29. Similar proofs may be given to show W C − p − for all δ>0 provided k d > 0 and k d δ>0. ⊂⊂ − p − p − Lemma 27.30 (Poincaré Lemma). Assume 1 p , Ω is a precompact open ≤ ≤∞ subset of Rd such that Ω is a manifold with C1-boundary. Then exist C = C(Ω,ρ) such that 1,p (27.27) u uΩ p C u p for all u W (Ω), k − kL (Ω) ≤ k∇ kL (Ω) ∈ where uΩ := udm is the average of u on Ω as in Notation 27.3. −Ω Proof. ForR sake of contradiction suppose there is no C< such that Eq. 1,p ∞ (27.27) holds. Then there exists a sequence un ∞ W (Ω) such that { }n=1 ⊂ un (un)Ω p >n un p for all n. k − kL (Ω) k∇ kL (Ω) Let un (un)Ω un := − . un (un)Ω Lp(Ω) 1,p k − k Then un W (Ω), (un)Ω =0, un Lp(Ω) =1and 1= un Lp(Ω) >n un Lp(Ω) ∈ k 1k k k k∇ k for all n. Therefore un Lp(Ω) < n andinparticularsup un W 1,p(Ω) < and k∇ k n k k ∞ hence by passing to a subsequence if necessary there exists u Lp (Ω) such that ∈ 508 BRUCE K. DRIVER†

p p un u in L (Ω). Since un 0 in L (Ω), it follows that un is convergent in 1,p→ ∇ 1,p → p W (Ω) and hence u W (Ω) and u = limn un =0in L (Ω). Since k,p ∈ ∇ →∞ ∇ u =0,u W (Ω) for all k N and hence u C∞ (Ω) and u =0implies ∇ ∈ ∈ ∈ ∇ u is constant. Since uΩ =0we must have u 0 which is clearly impossible since ≡ u Lp(Ω) =limn un Lp(Ω) =1. k k →∞ k k Theorem 27.31 (Poincaré Lemma). Let Ω be a precompact open subset of Rd and p [1, ]. Then ∈ ∞ 1,p u Lp 2diam(Ω) u Lp for all u W (Ω). k k ≤ k∇ k ∈ 0 d Proof. Without loss of generality assume Ω =[ m, m] and u Cc∞(Ω). Then by the fundamental theorem of calculus, − ∈

x1 M u(x) = ∂1u(y1,x2,...,xd)dy1 ∂1u(y1x2,...,xd) dy | | | M | ≤ M | | Z− Z− and hence by Holder’s inequality, M p p 1 p u(x) (2M) − ∂1u(y,x2,...,xd) dy1. | | ≤ M | | Z− Integrating this equation over x implies,

p p 1 p p p u p (2M)(2M) − ∂1u(x) dx =(2M) ∂1u(x) dx k kL ≤ | | | | Z Z and hence

u Lp 2M ∂1u Lp 2diam(Ω) u Lp . k k ≤ k k ≤ k∇ k

27.6. Fourier Transform Method. See L2 — Sobolev spaces for another proof of the following theorem.

Theorem 27.32. Suppose s>t 0, un n∞=1 is a bounded sequence (say by 1) s d ≥ { } d in H (R ) such that K = nsupp(un) @@ R . Then there exist a subsequence ∪ t d vn n∞=1 un ∞ which is convergent in H (R ). { } ⊂ { }n=1 Proof. Since

α α iξ x α iξ x ∂ξ uˆn(ξ) = ∂ξ e− · un(x)dx = ( ix) e− · un(x)dx Rd Rd − ¯ αZ ¯ ¯Z ¯ ¯ ¯ ¯ x 2 u 2 C¯ u¯ s d C ¯ ¯ ¯ ¯ L (K) n L ¯α ¯n H (R ) α ¯ ≤ k¯ k k k ≤ ¯ k ¯ k ≤ ¯ uˆn and all of it’s derivatives are uniformly bounded. By the Arzela-Ascoli theorem and Cantor’s Diagonalization argument, there exists a subsequence vn ∞ { }n=1 ⊂ un ∞ such that vˆn and all of its derivatives converge uniformly on compact { }n=1 subsets in ξ —space. If vˆ(ξ):=limn vˆn(ξ), then by the dominated convergence theorem, →∞

2 s 2 2 s 2 2 (1+ ξ ) vˆ(ξ) dξ = lim (1+ ξ ) vˆ (ξ) dξ lim sup v s d 1. n n H (R ) ξ R | | | | n ξ R | | | | ≤ n k k ≤ Z| |≤ →∞ Z| |≤ →∞ 2 2 s Since R is arbitrary this implies vˆ L ((1 + ξ ) dξ) and v Hs(Rd) 1. Set 1 ∈ | | s d k k ≤ gn := v vn while v = − v.ˆ Then gn ∞ H (R ) andwewishtoshow − F { }n=1 ⊂ ANALYSIS TOOLS WITH APPLICATIONS 509

t d 2 t gn 0 in H (R ). Let dµt (ξ)=(1+ ξ ) dξ, then for any R< , → | | ∞

2 2 gn t = gˆ(ξ) gˆn(ξ) dµt (ξ) k kH | − | Z 2 2 = gˆ(ξ) gˆn(ξ) dµt (ξ)+ gˆ(ξ) gˆn(ξ) dµt (ξ) . ξ R | − | ξ R | − | Z| |≤ Z| |≥

The ﬁrst term goes to zero by the dominated convergence theorem, hence

2 2 lim sup gn Ht lim sup gˆ(ξ) gˆn(ξ) dµt (ξ) n k k ≤ n ξ R | − | →∞ →∞ Z| |≥ 2 s 2 (1 + ξ ) = lim sup gˆ gˆn(ξ) | 2| s t dξ n ξ R | − | (1 + ξ ) − →∞ Z| |≥ | | 1 2 lim sup 2 s t gˆ gˆn(ξ) dµs (ξ) ≤ n (1 + R ) − ξ R | − | →∞ Z| |≥ 1 2 lim sup 2 s t gn g Ht ≤ n (1 + R ) − k − k →∞ s t 1 − 4 0 as R . ≤ 1+R2 → →∞ µ ¶

27.7. Other theorems along these lines. Another theorem of this form is de- rived as follows. Let ρ>0 be ﬁxed and g Cc ((0, 1) , [0, 1]) such that g(t)=1for ∈ t 1/2 and set τ(t):=g(t/ρ). Then for x Rd and ω Γ we have | | ≤ ∈ ∈

ρ d [τ(t)u(x + tω)] dt = u(x) dt − Z0 and then by integration by parts repeatedly we learn that

ρ ρ t2 u(x)= ∂2 [τ(t)u(x + tω)] tdt = ∂2 [τ(t)u(x + tω)] d t t 2 Z0 Z0 ρ t3 = ∂3 [τ(t)u(x + tω)] d = ... − t 3! Z0 ρ tm =( 1)m ∂m [τ(t)u(x + tω)] d − t m! Z0 ρ tm 1 =( 1)m ∂m [τ(t)u(x + tω)] − dt. − t (m 1)! Z0 − 510 BRUCE K. DRIVER†

Integrating this equation on ω Γ then implies ∈ ρ m 1 m m t − Γ u(x)=( 1) dω ∂t [τ(t)u(x + tω)] dt | | − γ 0 (m 1)! mZ Z ρ − ( 1) m d m d 1 = − dω t − ∂ [τ(t)u(x + tω)] t − dt (m 1)! t − Zγ Z0 m ρ m ( 1) m d m (m k) k d 1 = − dω t − τ − (t) ∂ u (x + tω) t − dt (m 1)! k ω γ 0 k=0 − Z Z X µ ¶ h i m ρ m ¡ ¢ ( 1) m d m k m (m k) k d 1 = − dω t − ρ − g − (t) ∂ωu (x + tω) t − dt (m 1)! γ 0 k − Z Z k=0 µ ¶ h i m m X ¡ ¢ ( 1) m k m m d (m k) k = − ρ − y x − g − ( y x ) ∂ u (y) dy (m 1)! k | − | | − | y[x k=0 Γx,ρ − − X µ ¶ Z h ³ ´ i and hence m m ( 1) m k m m d (m k) k u(x)= − ρ − y x − g − ( y x ) ∂ u (y) dy Γ (m 1)! k | − | | − | y[x k=0 Γx,ρ − | | − X µ ¶ Z h ³ ´ i and hence by the Hölder’s inequality, 1/q 1/p m m p ( 1) m k m q(m d) k u(x) C(g) − ρ − y x − dy ∂ u (y) dy . Γ (m 1)! k y[x | | ≤ " Γx,ρ | − | # " Γx,ρ − # | | − k=0 µ ¶ Z Z ¯³ ´ ¯ X ¯ ¯ From the same computation as in Eq. (25.4) we ﬁnd ¯ ¯ ρ q(m d)+d q(m d) q(m d) d 1 ρ − y x − dy = σ (Γ) r − r − dr = σ (Γ) | − | q (m d)+d ZΓx,ρ Z0 − pm d ρ p −1 = σ (Γ) − (p 1). pm d − − provided that pm d>0 (i.e. m>d/p) wherein we have used − p p (m d)+d (p 1) pm d q (m d)+d = (m d)+d = − − = − . − p 1 − p 1 p 1 − − − This gives the estimate

1/q p 1 p 1 − − p pm d p q(m d) σ (Γ)(p 1) − σ (Γ)(p 1) m d/p y x − dy − ρ p = − ρ − . | − | ≤ pm d pm d "ZΓx,ρ # · − ¸ · − ¸ Thus we have obtained the estimate that p 1 −p m C(g) σ (Γ)(p 1) m d/p m k m k u(x) − ρ − ρ − ∂ u . y[x p | | ≤ Γ (m 1)! pm d k − L (Γx,p) | | − · − ¸ k=0 µ ¶ ° ° X ° ° 27.8. Exercises. ° ° d 1 Exercise 27.1. Let ai 0 and pi [1, ) for i =1, 2,...,d satisfy i=1 pi− =1, then ≥ ∈ ∞ d d P 1 pi ai a . ≤ p i i=1 i=1 i Y X ANALYSIS TOOLS WITH APPLICATIONS 511

Hint: This may be proved by induction on d making use of Lemma 2.27. Alterna- tively see Example 9.11, where this is already proved using Jensen’s inequality.

27.1. We may assume that ai > 0, in which case d d d p d d 1 i 1 pi 1 ln ai i=1 p ln ai ln a pi ai = e i=1 = e i e i = a . ≤ p p i i=1 P P i=1 i i=1 i Y X X This was already done in Example 9.11.