Riesz-Thorin interpolation and applications Updated May 13, 2020 - PowerPoint PPT Presentation

Riesz-Thorin interpolation and applications Updated May 13, 2020

Plan 2 Outline: Riesz-Thorin interpolation Applications to Fourier and convolution Maximal function and Hilbert transform Weak L p -spaces

General idea 3 Suppose p ă p 1 , q , q 1 given, T continuous linear as T : L p Ñ L q and T : L p 1 Ñ L q 1 Then T defined on L p X L p 1 and, in fact, on L p ` L p 1 . Recall L p 2 Ď L p ` L p 1 @ p 2 P r p , p 1 s : Q: Is T continuous on L p 2 ? Image space?

Riesz-Thorin interpolation theorem 4 Theorem (Riesz-Thorin interpolation theorem) For p 0 , p 1 , q 0 , q 1 P r 1, 8s , and p X , F , µ q and p Y , G , ν q (with ν σ -finite if q 0 “ q 1 “ 8 ), let T : p L p 0 ` L p 1 qp X , F , µ q Ñ p L q 0 ` L q 1 qp Y , F , ν q be linear such that, for some M 0 , M 1 P p 0, 8q , @ f P L p 0 : } Tf } q 0 ď M 0 } f } p 0 and @ f P L p 1 : } Tf } q 1 ď M 1 } f } p 1 Then for each θ P r 0, 1 s and with p θ , q θ P r 1, 8s defined by 1 : “ 1 ´ θ 1 : “ 1 ´ θ ` θ ` θ ^ p θ p 0 p 1 q θ q 0 q 1 we have @ f P L p θ : } Tf } q θ ď M 1 ´ θ M θ 1 } f } p θ . 0 In particular, T is continuous as a map L p θ p X , F , µ q Ñ L q θ p Y , G , ν q . The L p -spaces and L p -norms above are for C -valued functions.

Hausdorff-Young inequality 5 Corollary (Hausdorff-Young inequality) Let p P r 1, 2 s and let q P r 1, 8s obey p ´ 1 ` q ´ 1 “ 1 . For the measure space p R d , L p R d q , λ q , the Fourier transform obeys @ f P L 1 X L p : } p f } q ď } f } p In particular, f ÞÑ p f extends continuously to a linear operator L p Ñ L q with operator norm at most one. Proof: Fourier maps L 1 Ñ L 8 and L 2 Ñ L 2 with norm one. Now p “ 1 ´ θ q “ 1 ´ θ 1 ` θ 2 ^ 1 ` θ 8 1 2 for θ : “ 2 { q . Riesz-Thorin gives the claim. Note: Operator norm “ p 1 { p q ´ 1 { q (Beckner 1975). All maximizers Gaussian (Lieb 1990).

Strong Young inequality for convolution 6 Corollary (Stronger Young’s inequality) Let p , q , r P r 1, 8s be such that 1 p ` 1 r “ 1 ` 1 q . Then @ f P L 1 X L p @ g P L 1 X L r : } f ‹ g } q ď } f } p } g } r . In particular, for each g P L p , the map T g f : “ f ‹ g on L 1 X L r extends to a continuous linear operator L r Ñ L q with operator norm ď 1 .

Proof of Young inequality 7 Pick p P r 1, 8q and g P L p . Denote T g : “ f ‹ g . Standard Young inequality: T g maps L 1 Ñ L p with norm ď 1. older: For q with p ´ 1 ` q ´ 1 “ 1, H¨ } T g f } 8 ď } f } q } g } p so T g maps L q Ñ L 8 with norm ď 1. For p 0 : “ 1, p 1 : “ q , q 0 : “ p and q 1 : “ 8 and θ “ . . . , we get 1 p ` 1 “ 1 ` 1 p θ q θ Riesz-Thorin: T g continuous as L p θ Ñ L q θ with norm ď 1. Denoting r : “ p θ and q : “ q θ , this is above claim.

Hadamard’s three lines theorem 8 Theorem (Hadamard’s three lines theorem) Let F be continuous on the vertical strip t z P C : 0 ď Re z ď 1 u and analytic on the interior thereof. Assume that Re z P r 0, 1 s ñ | F p z q| ď e c | z | D c ą 0 @ z P C : For all θ P r 0, 1 s set ˇ ˇ ˇ F p z q ˇ M θ : “ sup z P C Re z “ θ Then M θ ď M 1 ´ θ M θ @ θ P r 0, 1 s : 1 0

9 Denote � ( g ǫ p z q : “ F p z q M z ´ 1 M z 1 exp ǫ z p z ´ 1 q . 0 For z : “ x ` i y we have Re z p z ´ 1 q “ x p x ´ 1 q ´ y 2 ď 2 ´ | z | 2 so g ǫ p x ` i y q Ñ 0 superexponentially as | y | Ñ 8 . So, | g ǫ | ď 1 on the boundary of r 0, 1 s ˆ r´ r , r s for r large. Maximum Modulus Principle: @ z P C : Re z P r 0, 1 s ñ | g ǫ p z q| ď 1 This translates into ˇ ˇ � ˇ F p z q ˇ ď M 1 ´ Re z M Re z ǫ p Im z q 2 u . exp 1 0 Taking ǫ Ó 0 we get the claim.

Other versions and extensions 10 Hadamard’s three circle theorem: version on annuli (precompose with z ÞÑ e z ) Lindel¨ of theorem: Allow for growth ˇ ˇ ˇ F p z q ˇ ď c exp t e p π ´ ǫ q| z | u . This is best possible in light of F p z q : “ exp t ie i π z u General theory: Phragm´ en-Lindel¨ of principle. Alternative probabilistic approach via exit problem for Brownian motion.

Proof of Riesz-Thorin, key lemma 11 Let S X : “ simple functions on p X , F , µ q with µ p supp p f qq ă 8 . Same for S Y on p Y , G , ν q . Note that S X Ď L p @ p P r 1, 8s . Lemma (Key interpolation lemma) Let θ P r 0, 1 s . Then ż ˇ ˇ ˇ ˇ ˇ ď M 1 ´ θ M θ @ f P S X @ g P S Y : ˇ p Tf q g d ν 1 } f } p θ } g } ˜ q θ 0 where ˜ q θ is H¨ older dual to q θ , 1 ` 1 “ 1. q θ ˜ q θ

Proof of Lemma 12 older: True for θ “ 0, 1. Assumption of theorem and H¨ Let θ P p 0, 1 q and for f P S X define f z : “ | f | p 1 ´ z q p θ p 0 ` z p θ p 1 f | f | Similarly, ˜ q θ q θ q 1 g ˜ g z : “ | g | p 1 ´ z q q 0 ` z ˜ ˜ | g | . Linearity of T : Tf z finite sum of functions with z -entire coefficients. So ż F p z q : “ p Tf z q g z d ν ş well defined and entire. Note F p θ q “ p Tf q g d ν .

Proof of Lemma continued ... 13 Now estimate | F | on Re z “ 0: ˇ ˇ ˇ F p i t q ˇ ď M 0 } f i t } p 0 } g i t } ˜ q θ As } f i t } p 0 “ }| f | p θ { p 0 } p 0 “ } f } p θ { p 0 q 0 “ } g } ˜ q θ { ˜ q 0 and } g i t } ˜ , this gives p θ q θ ˜ ˇ ˇ ˇ ď M 0 } f } p θ { p 0 } g } ˜ q θ { ˜ q 0 ˇ F p i t q @ t P R : p θ q θ ˜ Similarly, ˇ ˇ ˇ ď M 1 } f } p θ { p 1 } g } ˜ q θ { ˜ q 1 ˇ F p 1 ` i t q @ t P R : p θ ˜ q θ | F | grows at most exponentially with | z | . Hadamard’s theorem: ż ˇ ˇ ´ ¯ 1 ´ θ ´ ¯ θ ˇ ˇ ˇ ˇ M 0 } f } p θ { p 0 } g } ˜ q θ { ˜ q 0 M 1 } f } p θ { p 1 } g } ˜ q θ { ˜ q 1 ˇ F p θ q ˇ ď ˇ p Tf q g d ν ˇ “ , p θ p θ q θ ˜ ˜ q θ This gives the statement.

Proof of Riesz-Thorin theorem 14 Avoid “boundary cases” by assuming first p θ ă 8 and q θ ă 8 Then q 1 : “ min t q 0 , q 1 u ă 8 and Tf P L q 1 so ν p| Tf | ą ǫ q ă 8 for all ǫ ą 0. Monotone Convergence extends ż ˇ ˇ ˇ ˇ ˇ ď M 1 ´ θ M θ p Tf q g d ν 1 } f } p θ } g } ˜ ˇ q θ 0 to all g P L r q 0 . Now take g : “ | Tf | q θ ´ 1 Tf | Tf | 1 t| Tf |ď a u and then let a Ñ 8 to turn this into } Tf } q θ ď M 1 ´ θ M θ @ f P S X : 1 } f } p θ . 0 Every f P L p θ can be written as f 0 ` f 1 where f 0 P L p θ X L p 0 and f 1 P L p θ X L p 1 on which T is defined. Approximating f 0 and f 1 by functions in S X in p θ -norm then extends above to all f P L p θ .

Proof of Riesz-Thorin, boundary cases 15 When p 0 “ p 1 (which is necessary for p θ “ 8 ) we get } Tf } q θ ď } Tf } 1 ´ θ q 0 } Tf } θ q 1 by interpolation of L p -norms. For q θ “ 8 we have ˜ q θ “ 1 and so (5) applies to all g P L 1 . Then (14) has to be substituted by Tf g : “ 1 t| Tf |ą a u | Tf | 1 A for A P G with ν p A q ă 8 . (Uses σ -finiteness of ν .) If a ě 0 is such that ν p A X t| Tf | ą a uq ą 0, then Lemma gives a ď M 1 ´ θ M θ 1 } f } p θ 0 which translates into } Tf } 8 ď M 1 ´ θ M θ 1 } f } p θ 0 This is what we claim.

Riesz-Thorin’s theorem, convexity formulation 16 Corollary Given measure spaces p X , F , µ q and p Y , G , ν q , let T : S X Ñ L 0 p Y , G , ν q be a linear operator defined on the space S X Ď L 0 p X , F , µ q of ( C -valued) simple functions with finite-measure support. Then !` 1 ) ˘ p , 1 P p 0, 1 s : } T } L p Ñ L q ă 8 D T : “ q is convex and p 1 p , 1 q q ÞÑ log } T } L p Ñ L q is a convex function on D T . The interval p 0, 1 s can be changed in r 0, 1 s provided ν is σ -finite. Proof: p 1 p 0 , 1 q 0 q , p 1 p 1 , 1 q 1 q P D T implies p 1 p θ , 1 q θ q P D T .

Remarks 17 Stein’s interpolation theorem: Assume z ÞÑ T z analytic subject to a growth restriction as Im z Ñ ˘8 . Then ż F p z q : “ p T z f z q g z d ν still obeys Hadamard/Lindel¨ of theorem. Definition (Strong type p p , q q ) Given p , q P r 1, 8s , and operator T : L p Ñ L q (even just densely defined) is said to be strong type- p p , q q if } T } L p Ñ L q ă 8 . Riesz-Thorin interpolation: strong type- p p 0 , q 0 q and p p 1 , q 1 q ñ strong type- p p θ , q θ q

Other examples: maximal function 18 Hardy-Littlewood maximal function ż 1 f ‹ p x q : “ sup | f | d µ , µ p B p x , r qq r ą 0 B p x , r q For µ general Radon on R or Lebesgue on R d we proved p } f ‹ } p ď c @ p P p 1, 8q : p ´ 1 } f } p by integrating a suitable maximal inequality. This did not extend to Radon measures on R d due to reliance on Besicovich covering. There we still have } f ‹ } 8 ď } f } 8 but only the weak maximal inequality ` ˘ ď c | f ‹ | ą t @ t ą 0: t } f } 1 µ In addition, f ÞÑ f ‹ is not linear!

Other examples: Hilbert transform 19 Defined informally as ż Hf p x q : “ 1 1 x ´ y f p y q d y . π Integral divergent for non-zero f P L 1 . Instead, take ǫ Ó 0 limit of ż H ǫ f p x q : “ 1 1 x ´ y f p y q d y π ǫ ă| x ´ y |ă 1 { ǫ which converges for all f P L p (1 ď p ď 8 ).

Hilbert transform: definition 20 This works at least for some functions: Lemma For all f P L 1 such that f P C 1 , ż ǫ Ó 0 H ǫ f p x q “ 1 f p x ´ t q ´ f p x ` t q @ x P R : lim d t π t p 0, 8q Proof: For f P L 1 ż ż f p x ´ t q ´ f p x ` t q 1 x ´ y f p y q d y “ d t . t ǫ ă| x ´ y |ă 1 { ǫ p ǫ ,1 { ǫ q The integrand bounded near t “ 0 when f P C 1 .