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Sobolev inequalities Updated June 5, 2020 Plan 2 Outline: - PowerPoint PPT Presentation

Sobolev inequalities Updated June 5, 2020 Plan 2 Outline: Gagliardo-Nierenberg-Sobolev inequality L 1 -Sobolev and L 2 -Sobolev inequalities Sobolev embedding theorems Compact embedding and Rellich-Kondrachovs theorem Boost in regularity


  1. Sobolev inequalities Updated June 5, 2020

  2. Plan 2 Outline: Gagliardo-Nierenberg-Sobolev inequality L 1 -Sobolev and L 2 -Sobolev inequalities Sobolev embedding theorems Compact embedding and Rellich-Kondrachov’s theorem Boost in regularity via Moser iteration

  3. Gagliardo-Nirenberg-Sobolev inequality 3 Morrey’s inequality: regularity for f P W 1, p p R d q with p ą d Q: What happens for p ă d ? Theorem For all d ě 2 and all p P r 1, d q there is c p d , p q P p 0, 8q such that @ f P C 8 c p R d q : } f } p ‹ ď c p d , p q} ∇ f } p where p ‹ is the Sobolev conjugate of p defined by pd p ‹ : “ d ´ p Proved independently by E. Gagliardo and L. Nirenberg for p “ 1 and by S.L. Sobolev for 1 ă p ă d

  4. First some remarks 4 p ‹ : “ pd d ´ p the only index for which this can hold. Indeed, take f t p x q : “ f p x { t q . Then d d p ´ 1 } ∇ f } p p ‹ } f } p ‹ ^ } ∇ f t } p “ t } f t } p ‹ “ t If } f } p ‹ ď c } ∇ f } p is to hold for f t for all t ą 0, then we must have 1 “ 1 p ´ 1 p ‹ d This forces p ď d . The case p “ d excluded in d ě 2 by f ǫ p x q : “ log p ǫ ` | x |q g p x q with g P C 8 c p R d q s.t. supp p g q Ď B p 0, 1 { 2 q and g p 0 q ‰ 0. Then } ∇ f ǫ } d ď c p log p 1 { ǫ qq 1 { d yet } f ǫ } 8 ě | g p 0 q| log p 1 { ǫ q so f ÞÑ } f } 8 {} ∇ f } d not bounded on C 8 c p R d q .

  5. L 1 -Sobolev inequality sufficient 5 For d “ 1 the inequality holds by @ f P C 8 } f } 8 ď } f 1 } 1 c p R q : ş p´8 , x s f 1 d λ (as supp p f q compact). Proved by writing f p x q “ The case p “ 1 fundamental for all d ě 1: Lemma Let d ě 2 and suppose there is c p d q P p 0, 8q such that @ f P C 8 c p R d q : } f } d ´ 1 ď c p d q} ∇ f } 1 d Then Sobolev inequality holds for all p P r 1, d q (and p ‹ as in (1) ) with c p d , p q : “ c p d q d ´ 1 d ´ pp

  6. Proof of Lemma 6 First extend L 1 -Sobolev inequality to f P C 1 c p R d q . c p R d q , set ˜ Pick r ě 1 and, given f P C 8 f : “ f | f | r ´ 1 . Then ∇ ˜ f “ rf | f | r ´ 2 ∇ f (and so ˜ f P C 1 c p R d q ) which shows › › @ D d ´ 1 “ } ˜ d ´ 1 ď c } ∇ ˜ › f | f | r ´ 2 ∇ f › | f | r ´ 1 , | ∇ f | } f } r f } f } 1 “ cr 1 “ cr d d r q P r 1, 8s gives H¨ older’s inequality with indices ˜ p , ˜ › › | f | r ´ 1 › @ D › | f | r ´ 1 , | ∇ f | q “ } f } r ´ 1 ď p } ∇ f } ˜ p p r ´ 1 q } ∇ f } ˜ q ˜ ˜ Now set set r : “ d ´ 1 d ´ p p and observe that this implies d d ´ 1 “ p ‹ ^ ˜ p p r ´ 1 q “ r ˜ q “ p Putting these together we get the claim.

  7. Proof of L 1 -Sobolev inequality 7 We thus focus on the proof of L 1 -Sobolev inequality. This will be derived from ź d }B i f } 1 { d @ f P C 8 c p R d q : } f } d ´ 1 ď d 1 i “ 1 from which we get L 1 -Sobolev via › |B i f | 2 ¯ 1 { 2 › › › ´ d ź d ÿ d ÿ › › ď 1 › › 1 1 }B i f } 1 { d › › ? ? › |B i f | › 1 ď “ } ∇ f } 1 › › 1 d d d 1 i “ 1 i “ 1 i “ 1 based on arithmetic-geometric/Cauchy-Schwarz inequalities. The above inequality holds for d “ 1 directly. We prove the other cases by induction. Assume it holds for dimension d ´ 1 and let us prove it for d . . .

  8. Proof of L 1 -Sobolev inequality continued ... 8 Write points of R d as p x , z q where x P R and z P R d ´ 1 . Set θ : “ d ´ 2 d ´ 1 “ θ d ´ 1 d d ´ 1 . Then d ´ 2 ` p 1 ´ θ q and so by H¨ older: ż ˇ ´ż ˇ d ´ 1 ´ż ˇ ¯ d ´ 2 ¯ ˇ ˇ ˇ 1 d d ´ 1 @ x P R d : d ´ 1 d z ď d ´ 2 d z ˇ f p x , z q ˇ ˇ f p x , z q ˇ ˇ f p x , z q ˇ d z d ´ 1 ż ˇ Now ˇ @ x P R d : ˇ f p x , z q ˇ d z ď }B 1 f } 1 Induction assumption and multivariate H¨ older in turn yield ż ´ż ˇ ż ¯ d ´ 2 ź d ź d ˇ › › 1 1 d ´ 1 d ´ 1 d x ď ˇ f p x , z q ˇ d ´ 2 d z › B i f p x , ¨q › d ´ 1 d ´ 1 d x ď }B i f } 1 1 i “ 2 i “ 2 Putting together we get ż ź d 1 d d ´ 1 d λ ď d ´ 1 | f | }B i f } 1 i “ 1 which readily gives the claim.

  9. Sobolev embedding 9 Corollary Let d ě 2 . Then č L p 1 p R d q . W 1, p p R d q Ď @ p P r 1, d q : p ď p 1 ď dp d ´ p Proof: Sobolev inequality gives W 1, p p R d q Ď L p ‹ . By definition W 1, p p R d q Ď L p . Now apply interpolation of L p norms.

  10. L 2 -Sobolev inequality 10 The case p “ 2 frequently used, but excluded in d “ 2. This is by including L p -norm on r.h.s. as in: Theorem Let d ě 2 . For each q P r 2, 2 d 2 d d ´ 2 q (where d ´ 2 is interpreted as 8 when d “ 2 ) there is ˜ c p d , q q P p 0, 8q such that ` ˘ @ f P C 8 c p R d q : } f } q ď ˜ c p d , q q } f } 2 ` } ∇ f } 2 2 d ˜ c p d , q q is bounded in d ě 3 , the bound extends to q “ d ´ 2 .

  11. L 2 -Sobolev in d ě 3 11 d ´ 2 q “ p ‹ for p “ 2. As p ‹ ą p , find θ P r 0, 1 s such that Let q P r 2, 2 d 1 p ‹ ` 1 ´ θ θ 2 . Interpolation of L p -norms: q “ } f } q ď } f } θ p ‹ } f } 1 ´ θ ď c p d , 2 q} ∇ f } θ 2 } f } 1 ´ θ 2 2 ´ ¯ ď c p d , 2 q θ } ∇ f } 2 ` p 1 ´ θ q} f } 2 where arithmetic-geometric inequality used in the last step.

  12. L 2 -Sobolev in d ě 2 12 Need some Fourier transform facts: Lemma For each f P C 8 c p R d q , ż ˇ ˇ ˇ 2 d k “ } f } 2 ˇp p 1 ` 4 π 2 | k | 2 q 2 ` } ∇ f } 2 f p k q 2 Moreover, for all f P C 8 c p R d q there is c P p 0, 8q s.t. ˇ ˇ c @ k P R d : ˇp ˇ ď f p k q p 1 ` 4 π 2 | k | 2 q d . In particular, p f P L 1 . Proof: Write x ∇ f p k q “ ´ 2 π i k p f p k q and use that Fourier transform is an isometry in L 2 . For second part, take ˜ f : “ p 1 ´ ∆ q d f to get above with c : “ } p f } 8 ď } p f } 1 .

  13. Proof of L 2 -Sobolev in d ě 2 13 Take q P p 2, 2 d d ´ 2 q and let p P r 1, 2 q be the H¨ older dual and let c p R d q . Then x ÞÑ f p´ x q is thus the Fourier transform of p f P C 8 f . Hausdorff-Young inequality gives ´ż ¯ 1 { p p 1 ` 4 π | k | 2 q p { 2 ˇ ˇ ˇ p p 1 ` 4 π | k | 2 q ´ p { 2 d k } f } q ď } p ˇp f } p “ f p k q ´ż ¯ 1 { 2 ´ ż ¯ 1 ˇ ˇ p ´ 1 2 . p ˇp ˇ 2 d k p 1 ` 4 π 2 | k | 2 q ´ 2 ´ p d k p 1 ` 4 π 2 | k | 2 q ď f p k q 2 by H¨ older’s inequality with parameters 2 { p and 2 ´ p . The 2 p integral converges when 2 ´ p ą d which is equivalent 2 d to q ă d ´ 2 . The claim follows from above lemma.

  14. L 2 -Sobolev embedding 14 Corollary For all d ě 2 , č W 1,2 p R d q Ď L q p R d q . 2 ď q ă 2 d d ´ 2

  15. Compact embedding 15 Theorem (Rellich-Kondrachov) Let d ě 2 and let O Ď R d be bounded and open. Given p P r 1, d q let q P r 1, p ‹ q , where p ‹ : “ dp d ´ p . Then every non-empty bounded set C Ď W 1, p 0 p O q is has a compact closure in L q .

  16. Criterion of compactness in L p 16 Lemma (Kolmogorov-Riesz compactness theorem) Let p P r 1, 8q and let C Ď L p p R d q be non-empty. Then C is totally bounded in L p p R d q if and only if it is bounded, sup } f } p ă 8 f P C tight r Ñ8 sup lim } f 1 B p 0, r q c } p “ 0 f P C and, denoting f h p x q : “ f p x ` h q for h P R d , obeys } f h ´ f } p “ 0. lim ǫ Ó 0 sup sup | h |ă ǫ f P C Note: Called Riesz-Tamarkin, Kolmogorov-Riesz, or Fr´ echet-Kolmogorov theorem A.N. Sudakov showed that boundedness redundant.

  17. Proof of Lemma 17 Suffices to show that C contains a finite δ -net for each δ ą 0. Let χ ǫ by our standard mollifier and set � ( C ǫ : “ p f 1 B p 0,1 { ǫ q q ‹ χ ǫ : f P C . As } χ ǫ } 1 “ 1, Minkowski’s integral inequality implies › › › › › p f 1 B p 0,1 { ǫ q q ‹ χ ǫ ´ f › › p f 1 B p 0,1 { ǫ q q ´ f › p ď p ` sup } f h ´ f } p | h |ă ǫ So suffices to find a δ -net in C ǫ . But H¨ older gives › › › ∇ p f 1 B p 0,1 { ǫ q ‹ χ ǫ q › 8 ď } f } p } ∇ χ ǫ } q and so, since }p f 1 B p 0,1 { ǫ q q ‹ χ ǫ } 8 ď } f 1 B p 0,1 { ǫ q } 1 ď c } f } p , we get C ǫ is an equicontinuous family As supp pp f 1 B p 0,1 { ǫ q q ‹ χ ǫ q Ď B p 0, ǫ ` 1 { ǫ q , the claim follows from a-Ascoli in sup-norm and then also L p -norm by compact Arzel` support. For converse, see the notes.

  18. Proof of Rellich-Kondrachov’s Theorem 18 Let C Ď W 1,2 0 p O q . Tightness trivial as O bounded. Sobolev implies 1 q ´ 1 q ´ 1 1 p ‹ } f } p ‹ ď c λ p O q p ‹ } ∇ f } p } f } q ď λ p O q so C bounded also in L q . For the last condition find θ s.t. θ P r 0, 1 s by 1 q “ 1 ´ θ ` θ p ‹ . Then p ‹ ď p 2 c q θ } f h ´ f } 1 ´ θ } f h ´ f } q ď } f h ´ f } 1 ´ θ } f h ´ f } θ } ∇ f } θ p . 1 1 As 1 ´ θ ą 0 because q ă p ‹ , it suffices to show } f h ´ f } 1 “ 0 h Ñ 0 sup lim f P C Writing ż ˇ ˇ ˇ ˇ ˇ f h p x q ´ f p x q ˇ ď | h | ˇ ∇ f p x ` th q ˇ d t we get } f h ´ f } 1 ď | h | diam p O q} ∇ f } 1 ď | h | diam p O q λ p O q 1 ´ 1 p } ∇ f } p Since sup f P C } ∇ f } p ă 8 , we are done.

  19. Boost in regularity via example 19 Let d ě 2, pick O Ď R d bounded open and consider ! ) 1 2 } ∇ f } 2 2 : f P C 8 inf c p O q ^ } f } r “ 1 We will take r ą 1 in what follows even though r “ 1 very interesting too. For r “ 2, this describes base frequency of a drum shaped as O .] Q: Is there a minimizing f (with f P L r and ∇ f P L 2 )? Q: And if so, how regular is it?

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