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

Riesz-Thorin interpolation and applications

Updated May 13, 2020

Plan

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Outline: Riesz-Thorin interpolation Applications to Fourier and convolution Maximal function and Hilbert transform Weak Lp-spaces

General idea

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Suppose p ă p1, q, q1 given, T continuous linear as T: Lp Ñ Lq and T: Lp1 Ñ Lq1 Then T defined on Lp X Lp1 and, in fact, on Lp ` Lp1. Recall @p2 P rp, p1s: Lp2 Ď Lp ` Lp1 Q: Is T continuous on Lp2? Image space?

Riesz-Thorin interpolation theorem

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Theorem (Riesz-Thorin interpolation theorem) For p0, p1, q0, q1 P r1, 8s, and pX, F, µq and pY, G, νq (with ν σ-finite if q0 “ q1 “ 8), let T: pLp0 ` Lp1qpX, F, µq Ñ pLq0 ` Lq1qpY, F, νq be linear such that, for some M0, M1 P p0, 8q, @f P Lp0 : }Tf}q0 ď M0}f}p0 and @f P Lp1 : }Tf}q1 ď M1}f}p1 Then for each θ P r0, 1s and with pθ, qθ P r1, 8s defined by 1 pθ :“ 1 ´ θ p0 ` θ p1 ^ 1 qθ :“ 1 ´ θ q0 ` θ q1 we have @f P Lpθ : }Tf}qθ ď M1´θ Mθ

1 }f}pθ .

In particular, T is continuous as a map LpθpX, F, µq Ñ LqθpY, G, νq. The Lp-spaces and Lp-norms above are for C-valued functions.

Hausdorff-Young inequality

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Corollary (Hausdorff-Young inequality) Let p P r1, 2s and let q P r1, 8s obey p´1 ` q´1 “ 1. For the measure space pRd, LpRdq, λq, the Fourier transform obeys @f P L1 X Lp : }p f}q ď }f}p In particular, f ÞÑ p f extends continuously to a linear operator Lp Ñ Lq with operator norm at most one. Proof: Fourier maps L1 Ñ L8 and L2 Ñ L2 with norm one. Now 1 p “ 1 ´ θ 1 ` θ 2 ^ 1 q “ 1 ´ θ 8 ` θ 2 for θ :“ 2{q. Riesz-Thorin gives the claim. Note: Operator norm “ p1{pq´1{q (Beckner 1975). All maximizers Gaussian (Lieb 1990).

Strong Young inequality for convolution

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Corollary (Stronger Young’s inequality) Let p, q, r P r1, 8s be such that 1 p ` 1 r “ 1 ` 1 q. Then @f P L1 X Lp @g P L1 X Lr : }f ‹ g}q ď }f}p }g}r. In particular, for each g P Lp, the map Tgf :“ f ‹ g on L1 X Lr extends to a continuous linear operator Lr Ñ Lq with operator norm ď 1.

Proof of Young inequality

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Pick p P r1, 8q and g P Lp. Denote Tg :“ f ‹ g. Standard Young inequality: Tg maps L1 Ñ Lp with norm ď 1. H¨

• lder: For q with p´1 ` q´1 “ 1,

}Tgf}8 ď }f}q}g}p so Tg maps Lq Ñ L8 with norm ď 1. For p0 :“ 1, p1 :“ q, q0 :“ p and q1 :“ 8 and θ “ . . . , we get 1 p ` 1 pθ “ 1 ` 1 qθ Riesz-Thorin: Tg continuous as Lpθ Ñ Lqθ with norm ď 1. Denoting r :“ pθ and q :“ qθ, this is above claim.

Hadamard’s three lines theorem

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Theorem (Hadamard’s three lines theorem) Let F be continuous on the vertical strip tz P C: 0 ď Rez ď 1u and analytic on the interior thereof. Assume that Dc ą 0 @z P C: Rez P r0, 1s ñ |Fpzq| ď ec|z| For all θ P r0, 1s set Mθ :“ sup

zPC Rez“θ

ˇ ˇFpzq ˇ ˇ Then @θ P r0, 1s: Mθ ď M1´θ Mθ

1

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Denote gǫpzq :“ FpzqMz´1 Mz

1 exp

• ǫzpz ´ 1q

( . For z :“ x ` iy we have Rezpz ´ 1q “ xpx ´ 1q ´ y2 ď 2 ´ |z|2 so gǫpx ` iyq Ñ 0 superexponentially as |y| Ñ 8. So, |gǫ| ď 1 on the boundary of r0, 1s ˆ r´r, rs for r large. Maximum Modulus Principle: @z P C: Rez P r0, 1s ñ |gǫpzq| ď 1 This translates into ˇ ˇFpzq ˇ ˇ ď M1´Rez MRez

1

exp

• ǫpImzq2u.

Taking ǫ Ó 0 we get the claim.

Other versions and extensions

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Hadamard’s three circle theorem: version on annuli (precompose with z ÞÑ ez) Lindel¨

• f theorem: Allow for growth

ˇ ˇFpzq ˇ ˇ ď c exptepπ´ǫq|z|u. This is best possible in light of Fpzq :“ exptieiπzu General theory: Phragm´ en-Lindel¨

• f principle. Alternative

probabilistic approach via exit problem for Brownian motion.

Proof of Riesz-Thorin, key lemma

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Let SX :“ simple functions on pX, F, µq with µpsupppfqq ă 8. Same for SY on pY, G, νq. Note that SX Ď Lp @p P r1, 8s. Lemma (Key interpolation lemma) Let θ P r0, 1s. Then @f P SX @g P SY : ˇ ˇ ˇ ż pTfqgdν ˇ ˇ ˇ ď M1´θ Mθ

1}f}pθ}g}˜ qθ

where ˜ qθ is H¨

• lder dual to qθ,

1 ˜ qθ ` 1 qθ “ 1.

Proof of Lemma

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Assumption of theorem and H¨

• lder: True for θ “ 0, 1.

Let θ P p0, 1q and for f P SX define fz :“ |f|p1´zq pθ

p0 `z pθ p1 f

|f| Similarly, gz :“ |g|p1´zq

˜ qθ ˜ q0 `z ˜ qθ ˜ q1 g

|g|. Linearity of T: Tfz finite sum of functions with z-entire

• coefficients. So

Fpzq :“ ż pTfzqgzdν well defined and entire. Note Fpθq “ ş pTfqgdν.

Proof of Lemma continued ...

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Now estimate |F| on Rez “ 0: ˇ ˇFpitq ˇ ˇ ď M0}fit}p0}git}˜

qθ

As }fit}p0 “ }|f|pθ{p0}p0 “ }f}pθ{p0

pθ

and }git}˜

q0 “ }g}˜ qθ{˜ q0 ˜ qθ

, this gives @t P R: ˇ ˇFpitq ˇ ˇ ď M0}f}pθ{p0

pθ

}g}˜

qθ{˜ q0 ˜ qθ

Similarly, @t P R: ˇ ˇFp1 ` itq ˇ ˇ ď M1}f}pθ{p1

pθ

}g}˜

qθ{˜ q1 ˜ qθ

|F| grows at most exponentially with |z|. Hadamard’s theorem: ˇ ˇ ˇ ż pTfqgdν ˇ ˇ ˇ “ ˇ ˇFpθq ˇ ˇ ď ´ M0}f}pθ{p0

pθ

}g}˜

qθ{˜ q0 ˜ qθ

¯1´θ´ M1}f}pθ{p1

pθ

}g}˜

qθ{˜ q1 ˜ qθ

¯θ , This gives the statement.

Proof of Riesz-Thorin theorem

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Avoid “boundary cases” by assuming first pθ ă 8 and qθ ă 8 Then q1 :“ mintq0, q1u ă 8 and Tf P Lq1 so νp|Tf| ą ǫq ă 8 for all ǫ ą 0. Monotone Convergence extends ˇ ˇ ˇ ż pTfqgdν ˇ ˇ ˇ ď M1´θ Mθ

1}f}pθ}g}˜ qθ

to all g P Lr

• q0. Now take

g :“ |Tf|qθ´1 Tf |Tf|1t|Tf|ďau and then let a Ñ 8 to turn this into @f P SX : }Tf}qθ ď M1´θ Mθ

1 }f}pθ.

Every f P Lpθ can be written as f0 ` f1 where f0 P Lpθ X Lp0 and f1 P Lpθ X Lp1 on which T is defined. Approximating f0 and f1 by functions in SX in pθ-norm then extends above to all f P Lpθ.

Proof of Riesz-Thorin, boundary cases

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When p0 “ p1 (which is necessary for pθ “ 8) we get }Tf}qθ ď }Tf}1´θ

q0 }Tf}θ q1

by interpolation of Lp-norms. For qθ “ 8 we have ˜ qθ “ 1 and so (5) applies to all g P L1. Then (14) has to be substituted by g :“ 1t|Tf|ąau Tf |Tf|1A for A P G with νpAq ă 8. (Uses σ-finiteness of ν.) If a ě 0 is such that νpA X t|Tf| ą auq ą 0, then Lemma gives a ď M1´θ Mθ

1}f}pθ

which translates into }Tf}8 ď M1´θ Mθ

1}f}pθ

This is what we claim.

Riesz-Thorin’s theorem, convexity formulation

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Corollary Given measure spaces pX, F, µq and pY, G, νq, let T: SX Ñ L0pY, G, νq be a linear operator defined on the space SX Ď L0pX, F, µq of (C-valued) simple functions with finite-measure

• support. Then

DT :“ !` 1

p, 1 q

˘ P p0, 1s: }T}LpÑLq ă 8 ) is convex and p1

p, 1 qq ÞÑ log }T}LpÑLq is a convex function on DT. The

interval p0, 1s can be changed in r0, 1s provided ν is σ-finite. Proof: p 1

p0 , 1 q0 q, p 1 p1 , 1 q1 q P DT implies p 1 pθ , 1 qθ q P DT.

Remarks

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Stein’s interpolation theorem: Assume z ÞÑ Tz analytic subject to a growth restriction as Imz Ñ ˘8. Then Fpzq :“ ż pTzfzqgzdν still obeys Hadamard/Lindel¨

• f theorem.

Definition (Strong type pp, qq) Given p, q P r1, 8s, and operator T: Lp Ñ Lq (even just densely defined) is said to be strong type-pp, qq if }T}LpÑLq ă 8. Riesz-Thorin interpolation: strong type-pp0, q0q and pp1, q1q ñ strong type-ppθ, qθq

Other examples: maximal function

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Hardy-Littlewood maximal function f ‹pxq :“ sup

rą0

1 µpBpx, rqq ż

Bpx,rq

|f|dµ, For µ general Radon on R or Lebesgue on Rd we proved @p P p1, 8q: }f ‹}p ď c p p ´ 1}f}p by integrating a suitable maximal inequality. This did not extend to Radon measures on Rd due to reliance on Besicovich

• covering. There we still have

}f ‹}8 ď }f}8 but only the weak maximal inequality @t ą 0: µ ` |f ‹| ą t ˘ ď c t}f}1 In addition, f ÞÑ f ‹ is not linear!

Other examples: Hilbert transform

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Defined informally as Hfpxq :“ 1 π ż 1 x ´ y fpyq dy. Integral divergent for non-zero f P L1. Instead, take ǫ Ó 0 limit of Hǫfpxq :“ 1 π ż

ǫă|x´y|ă1{ǫ

1 x ´ y fpyq dy which converges for all f P Lp (1 ď p ď 8).

Hilbert transform: definition

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This works at least for some functions: Lemma For all f P L1 such that f P C1, @x P R: lim

ǫÓ0 Hǫfpxq “ 1

π ż

p0,8q

fpx ´ tq ´ fpx ` tq t dt Proof: For f P L1 ż

ǫă|x´y|ă1{ǫ

1 x ´ y fpyq dy “ ż

pǫ,1{ǫq

fpx ´ tq ´ fpx ` tq t dt. The integrand bounded near t “ 0 when f P C1.

Hilbert transform on L2

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Proposition There exists a linear isometry H: L2 Ñ L2 such that @f P L1 X L2 : Hǫf Ý Ñ

ǫÓ0 Hf in L2

Moreover, writing T for the Fourier transform on L2 and sgn for the sign-function on R, TpHfq “ p´iqsgnp¨qTf In particular, H is surjective and H´1 “ ´H.

Proof of Proposition

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Hǫ continuous L1 Ñ L1 so Fourier transform meaningful: y Hǫfpkq “ ´ ´2i π ż

ǫătă1{ǫ

sin 2πkt t dt ¯ p fpkq Since lim

MÑ8

ż M sinpatq t dt “ π 2 sgnpaq the prefactor converges to p´iqsgnpkq as M Ñ 8 and ǫ Ó 0. Integrals bounded, so convergence in L2 as well. Extends to all L2 by density of L1 X L2 in L2. Iterating we get H2 “ id so RanpHq “ L2 and H´1 “ ´H. Integral expression applies for f P C1 X L2 X L1 (dense in L2).

Extension to other Lp?

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Q: Does convergence Hǫf Ñ Hf hold in Lp for other p? A: Interpolation needs (ideally) strong type p1, 1q and p8, 8q. Problem: Hf R L1 for f P L1 in general! Lemma For f P C1 with compact support, Hf obeys lim

xÑ˘8 x Hfpxq “ 1

π ż fdλ In particular, if ş fdλ ‰ 0, then Hf R L1. Proof: homework So not strong type p1, 1q. For p “ 8 taking a very “flat” bump function shows }Hf}8 and }f}8 incomparable. So not strong type p8, 8q.

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Similarly to Hardy-Littlewood max function, we have: Proposition (Weak L1-estimate) There exists c P p0, 8q such that for all f P L1, @t ą 0: sup

ǫą0

λ ` |Hǫf| ą t ˘ ď c t}f}1. A slick proof using complex analysis (Stietljes transform etc). We will use Calder´

• n-Zygmund theory.

Riesz-Thorin will have to be replaced by Marcinkiewicz which needs weak-Lp-spaces.

Weak Lp-space

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pX, F, µq measure space, f : X Ñ R measurable. Then t ÞÑ µp|f| ą tq is the distribution function of f. Encodes p-norm via }f}p

p “

ż 8 p tp´1µp|f| ą tq dt Definition (Weak Lp-space) For f P L0 and p P p0, 8q set rfsp :“ sup

tą0

tµ ` |f| ą t ˘1{p and define the weak-Lp space on pX, F, µq by Lp,w :“

• f P L0 : rfsp ă 8

( . For p “ 8 we set rfs8 :“ }f}8 and so L8,w “ L8.

Basic properties of rfsp and Lp,w

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Lemma For all p P p0, 8s we have (1) @c P R @f P L0 : rcfsp “ |c|rfsp, (2) @f, g P L0 : rf ` gsp ď 2maxt1,1{puprfsp ` rgspq, and so Lp,w is a vector space. Moreover, (3) @f P L0 : rfsp ď }f}p (4) f “ g µ-a.e. is equivalent to rf ´ gsp “ 0. In particular, Lp Ď Lp,w for all p P p0, 8s. Proof: (1) trivial, (3) Chebyshev inequality, (4) inspection. For (2) use µp|f ` g| ą tq ď µp|f| ą t{2q ` µp|g| ą t{2q and pa ` bqr ď 2maxt1,ru´1par ` brq to get tµp|f ` g| ą tq1{p ď 2maxt1,1{pu” t 2µp|f| ą t{2q1{p ` t 2µp|g| ą t{2q1{pı ď 2maxt1,1{puprfsp ` rgspq.

Relations to Lp

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Lemma Let p P p0, 8s and f P Lp,w. If µpf ‰ 0q ă 8 then f P Lp1 for all p1 P p0, pq. In particular, for measure spaces pX, F, µq, µpXq ă 8 ñ @p1 ă p: Lp,w Ď Lp1. On the other hand, if f P Lp,w X L8, then f P Lp1 for all p1 ą p.

Topology on Lp,w

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Coarsest topology containing the set

• f P Lp,w : rf ´ f0sp ă r

( for all r P p0, 8q and all f0 P Lp,w. First countable, sequential convergence sufficient. So fn Ñ f in Lp,w if rfn ´ fsp Ñ 0. Same about being Cauchy. Lemma For each p P p0, 8s, the space Lp,w is complete.

Proof of Lemma

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By Cauchy property, choose tnkukě1 so that supm,nąnk µp|fn ´ fm| ą 2´kq ă 2´k. Then fn Ý Ñ

nÑ8 f :“ fn1 `

ÿ

kě1

pfni`1 ´ fniq µ-a.e. Fatou’s lemma implies tµ ` |fn ´f| ą t ˘1{p ď t lim sup

mÑ8

µ ` |fn ´fm| ą t ˘1{p ď lim sup

mÑ8

rfn ´fmsp and so rfn ´ fsp Ñ 0.

An equivalent norm for p ą 1

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f ÞÑ rfsp NOT a norm. For p ą 1 can be fixed: Lemma For all p P p1, 8s and non-zero f P Lp,w, }f}p,w :“ sup

APF µpAqPp0,8q

1 µpAq1´1{p ż

A

|f|dµ defines a norm on Lp,w satisfying @f P Lp,w : rfsp ď }f}p,w ď p p ´ 1rfsp Proof: homework. Fails for p ď 1: Lemma Suppose that pX, F, µq is such that µ is non-zero and non-atomic. Then the topology on L1,w cannot be normed.

Proof of Lemma

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Non-atomic: Find tAkukě1 with k ÞÑ Ak decreasing and lim

kÑ8 µpAkq “ 0

Set fn :“

n

ÿ

k“1

hk where hk :“ 1 µpAkq1AkAk`1 For t ą 0 find minimal k with t ď µpAkq´1. Then tfn ą tu Ď Ak X supppfq so tµpf ą tq ď 1 µpAkq µ ` Ak X supppfq ˘ ď 1 and thus rfns1 ď 1.

Proof of Lemma continued ...

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On the other hand, rhks1 “ µpAkq´1µpAk Akq Choose tkjujě1 such that k1 “ 1 and µpAkj Akj`1q ě 1

2µpAkjq for

all j ě 1. Then ÿ

kě1

rhks1 ě ÿ

kě1

1 µpAkq µpAk Ak´1q ě ÿ

jě1

1 µpAkjq µpAkj Akj`1q ě ÿ

kě1

1 2 “ 8 so the ratio of řn

k“1rhns1 and rřn k“1 hks1 can be made arbitrarily

• large. This rules out that c1}f} ď rfs1 ď c2}f} for norm } ¨ }.

Other strange facts

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On non-atomic spaces, for p ă 1: pLp,wq‹ “ t0u Nontrivial functionals do exist in pL1,wq‹ but they vanish on all simple functions! Such singular functionals exist (by Hahn-Banach theorem) also for p ą 1. (Non-singular ones exist as well.) Reason: fpxq :“ x´1{p not approximated by simple functions Lorentz spaces Lp,q embed weak-Lp-space for q “ 8. Idea: Instead of L8 norm of t ÞÑ tµp|f| ą tq1{p, take q-norm under 1

t 1p0,8qptqt instead.