Riesz-Thorin interpolation and applications

Updated May 13, 2020

Plan 2

Outline:Riesz-Thorin interpolationApplications to Fourier and convolutionMaximal function and Hilbert transformWeak Lp-spaces

General idea 3

Suppose p ă p1, q, q1 given, T continuous linear as

T : Lp Ñ Lq

andT : 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 4

Theorem (Riesz-Thorin interpolation theorem)

For p0, p1, q0, q1 P r1,8s, and pX,F , µq and pY,G, νq (with ν σ-finiteif q0 “ q1 “ 8), let T : pLp0 ` Lp1qpX,F , µq Ñ pLq0 ` Lq1qpY,F , νqbe 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 by1pθ

:“1´ θ

p0`

θ

p1^

1qθ

:“1´ θ

q0`

θ

q1we have

@f P Lpθ : }Tf }qθď M1´θ

0 Mθ1 }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 5

Corollary (Hausdorff-Young inequality)

Let p P r1, 2s and let q P r1,8s obey p´1 ` q´1 “ 1. For the measurespace pRd,LpRdq, λq, the Fourier transform obeys

@f P L1 X Lp : }pf }q ď }f }p

In particular, f ÞÑ pf extends continuously to a linear operatorLp Ñ Lq with operator norm at most one.

Proof: Fourier maps L1 Ñ L8 and L2 Ñ L2 with norm one. Now

1p“

1´ θ

1`

θ

2^

1q“

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 6

Corollary (Stronger Young’s inequality)

Let p, q, r P r1,8s be such that

1p`

1r“ 1`

1q

.

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 extendsto a continuous linear operator Lr Ñ Lq with operator norm ď 1.

Proof of Young inequality 7

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.

Holder: 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

1p`

1pθ“ 1`

1qθ

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 8

Theorem (Hadamard’s three lines theorem)

Let F be continuous on the vertical strip tz P C : 0 ď Rez ď 1u andanalytic 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 setMθ :“ sup

zPCRez“θ

ˇ

ˇFpzqˇ

ˇ

Then@θ P r0, 1s : Mθ ď M1´θ

0 Mθ1

9

Denotegεpzq :“ FpzqMz´1

0 Mz1 exp

εzpz´ 1q(

.

For z :“ x` iy we have Rezpz´ 1q “ xpx´ 1q ´ y2 ď 2´ |z|2 sogεpx` iyq Ñ 0 superexponentially as |y| Ñ 8.

So, |gε| 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 |gεpzq| ď 1

This translates intoˇ

ˇFpzqˇ

ˇ ď M1´Rez0 MRez

1 exp

εpImzq2u.

Taking ε Ó 0 we get the claim.

Other versions and extensions 10

Hadamard’s three circle theorem: version on annuli(precompose with z ÞÑ ez)Lindelof theorem: Allow for growth

ˇ

ˇFpzqˇ

ˇ ď c exptepπ´εq|z|u.

This is best possible in light of

Fpzq :“ exptieiπzu

General theory: Phragmen-Lindelof principle. Alternativeprobabilistic approach via exit problem for Brownian motion.

Proof of Riesz-Thorin, key lemma 11

Let SX :“ simple functions on pX,F , µqwith µpsupppf qq 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

pTf qgdνˇ

ˇ

ˇď M1´θ

0 Mθ1}f }pθ

}g}qθ

where qθ is Holder dual to qθ ,

1qθ`

1qθ“ 1.

Proof of Lemma 12

Assumption of theorem and Holder: True for θ “ 0, 1.Let θ P p0, 1q and for f P SX define

fz :“ |f |p1´zq pθp0`z pθ

p1f|f |

Similarly,

gz :“ |g|p1´zq qθq0`z qθ

q1g|g|

.

Linearity of T: Tfz finite sum of functions with z-entirecoefficients. So

Fpzq :“ż

pTfzqgzdν

well defined and entire. Note Fpθq “ş

pTf qgdν.

Proof of Lemma continued . . . 13

Now estimate |F| on Rez “ 0:ˇ

ˇFpitqˇ

ˇ ď M0}fit}p0}git}qθ

As }fit}p0 “ }|f |pθ{p0}p0 “ }f }

pθ{p0pθ

and }git}q0 “ }g}qθ{q0qθ

, this gives

@t P R :ˇ

ˇFpitqˇ

ˇ ď M0}f }pθ{p0pθ

}g}qθ{q0qθ

Similarly,

@t P R :ˇ

ˇFp1` itqˇ

ˇ ď M1}f }pθ{p1pθ

}g}qθ{q1qθ

|F| grows at most exponentially with |z|. Hadamard’s theorem:

ˇ

ˇ

ˇ

ż

pTf qgdνˇ

ˇ

ˇ“

ˇ

ˇFpθqˇ

ˇ ď

´

M0}f }pθ{p0pθ

}g}qθ{q0qθ

¯1´θ´

M1}f }pθ{p1pθ

}g}qθ{q1qθ

¯θ,

This gives the statement.

Proof of Riesz-Thorin theorem 14

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 εq ă 8 forall ε ą 0. Monotone Convergence extends

ˇ

ˇ

ˇ

ż

pTf qgdνˇ

ˇ

ˇď M1´θ

0 Mθ1}f }pθ

}g}qθ

to all g P Lrq0 . Now take

g :“ |Tf |qθ´1 Tf|Tf |

1t|Tf |ďau

and then let a Ñ8 to turn this into

@f P SX : }Tf }qθď M1´θ

0 Mθ1 }f }pθ

.

Every f P Lpθ can be written as f0 ` f1 where f0 P Lpθ X Lp0 andf1 P Lpθ X Lp1 on which T is defined. Approximating f0 and f1 byfunctions in SX in pθ-norm then extends above to all f P Lpθ .

Proof of Riesz-Thorin, boundary cases 15

When p0 “ p1 (which is necessary for pθ “ 8) we get

}Tf }qθď }Tf }1´θ

q0}Tf }θ

q1

by interpolation of Lp-norms. For qθ “ 8we have qθ “ 1 and so(5) applies to all g P L1. Then (14) has to be substituted by

g :“ 1t|Tf |ąauTf|Tf |

1A

for A P G with νpAq ă 8. (Uses σ-finiteness of ν.) If a ě 0 issuch that νpAX t|Tf | ą auq ą 0, then Lemma gives

a ď M1´θ0 Mθ

1}f }pθ

which translates into

}Tf }8 ď M1´θ0 Mθ

1}f }pθ

This is what we claim.

Riesz-Thorin’s theorem, convexity formulation 16

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 spaceSX Ď L0pX,F , µq of (C-valued) simple functions with finite-measuresupport. Then

DT :“!

` 1p , 1

q

˘

P p0, 1s : }T}LpÑLq ă 8

)

is convex and p 1p , 1

q q ÞÑ log }T}LpÑLq is a convex function on DT. Theinterval p0, 1s can be changed in r0, 1s provided ν is σ-finite.

Proof: p 1p0

, 1q0q, p 1

p1, 1

q1q P DT implies p 1

pθ, 1

qθq P DT.

Remarks 17

Stein’s interpolation theorem: Assume z ÞÑ Tz analytic subjectto a growth restriction as Imz Ñ ˘8. Then

Fpzq :“ż

pTzfzqgzdν

still obeys Hadamard/Lindelof theorem.

Definition (Strong type pp, qq)

Given p, q P r1,8s, and operator T : Lp Ñ Lq (even just denselydefined) 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 18

Hardy-Littlewood maximal function

f ‹pxq :“ suprą0

1µpBpx, 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 cp

p´ 1}f }p

by integrating a suitable maximal inequality. This did notextend to Radon measures on Rd due to reliance on Besicovichcovering. There we still have

}f ‹}8 ď }f }8

but only the weak maximal inequality

@t ą 0 : µ`

|f ‹| ą t˘

ďct}f }1

In addition, f ÞÑ f ‹ is not linear!

Other examples: Hilbert transform 19

Defined informally as

Hf pxq :“1π

ż

1x´ y

f pyqdy.

Integral divergent for non-zero f P L1.Instead, take ε Ó 0 limit of

Hεf pxq :“1π

ż

εă|x´y|ă1{ε

1x´ y

f pyqdy

which converges for all f P Lp (1 ď p ď 8).

Hilbert transform: definition 20

This works at least for some functions:

Lemma

For all f P L1 such that f P C1,

@x P R : limεÓ0

Hεf pxq “1π

ż

p0,8q

f px´ tq ´ f px` tqt

dt

Proof: For f P L1

ż

εă|x´y|ă1{ε

1x´ y

f pyqdy “ż

pε,1{εq

f px´ tq ´ f px` tqt

dt.

The integrand bounded near t “ 0 when f P C1.

Hilbert transform on L2 21

Proposition

There exists a linear isometry H : L2 Ñ L2 such that

@f P L1 X L2 : Hεf ÝÑεÓ0

Hf in L2

Moreover, writing T for the Fourier transform on L2 and sgn for thesign-function on R,

TpHf q “ p´iqsgnp¨qTf

In particular, H is surjective and H´1 “ ´H.

Proof of Proposition 22

Hε continuous L1 Ñ L1 so Fourier transform meaningful:

yHεf pkq “´

´2iπ

ż

εătă1{ε

sin 2πktt

dt¯

pf pkq

Since

limMÑ8

ż M

0

sinpatqt

dt “π

2sgnpaq

the prefactor converges to p´iqsgnpkq as M Ñ8 and ε Ó 0.Integrals bounded, so convergence in L2 as well. Extends to allL2 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? 23

Q: Does convergence Hεf 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

limxÑ˘8

x Hf pxq “1π

ż

f dλ

In particular, ifş

f dλ ‰ 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.

24

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 : supεą0

λ`

|Hεf | ą t˘

ďct}f }1.

A slick proof using complex analysis (Stietljes transform etc).We will use Calderon-Zygmund theory.

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

Weak Lp-space 25

pX,F , µqmeasure space, f : X Ñ R measurable. Then

t ÞÑ µp|f | ą tq

is the distribution function of f . Encodes p-norm via

}f }pp “ż 8

0p tp´1µp|f | ą tqdt

Definition (Weak Lp-space)

For f P L0 and p P p0,8q set

rf sp :“ suptą0

tµ`

|f | ą t˘1{p

and define the weak-Lp space on pX,F , µq by

Lp,w :“

f P L0 : rf sp ă 8(

.

For p “ 8we set rf s8 :“ }f }8 and so L8,w “ L8.

Basic properties of rf sp and Lp,w 26

Lemma

For all p P p0,8s we have(1) @c P R@f P L0 : rcf sp “ |c|rf sp,

(2) @f , g P L0 : rf ` gsp ď 2maxt1,1{puprf sp ` rgspq,and so Lp,w is a vector space. Moreover,(3) @f P L0 : rf sp 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 andpa` 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 `

t2

µp|g| ą t{2q1{pı

ď 2maxt1,1{puprf sp ` rgspq.

Relations to Lp 27

Lemma

Let p P p0,8s and f P Lp,w. If µpf ‰ 0q ă 8 then f P Lp1 for allp1 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 28

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 inLp,w if rfn ´ f sp Ñ 0. Same about being Cauchy.

LemmaFor each p P p0,8s, the space Lp,w is complete.

Proof of Lemma 29

By Cauchy property, choose tnkukě1 so thatsupm,nąnk

µp|fn ´ fm| ą 2´kq ă 2´k. Then

fn ÝÑnÑ8

f :“ fn1 `ÿ

kě1

pfni`1 ´ fniq µ-a.e.

Fatou’s lemma implies

tµ`

|fn´ f | ą t˘1{p

ď t lim supmÑ8

µ`

|fn´ fm| ą t˘1{p

ď lim supmÑ8

rfn´ fmsp

and so rfn ´ f sp Ñ 0.

An equivalent norm for p ą 1 30

f ÞÑ rf sp NOT a norm. For p ą 1 can be fixed:

LemmaFor all p P p1,8s and non-zero f P Lp,w,

}f }p,w :“ supAPF

µpAqPp0,8q

1µpAq1´1{p

ż

A|f |dµ

defines a norm on Lp,w satisfying

@f P Lp,w : rf sp ď }f }p,w ďp

p´ 1rf sp

Proof: homework. Fails for p ď 1:

LemmaSuppose 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 31

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

limkÑ8

µpAkq “ 0

Set

fn :“nÿ

k“1

hk where hk :“1

µpAkq1AkrAk`1

For t ą 0 find minimal k with t ď µpAkq´1. Then

tfn ą tu Ď Ak X supppf q so

tµpf ą tq ď1

µpAkqµ`

Ak X supppf q˘

ď 1

and thus rfns1 ď 1.

Proof of Lemma continued . . . 32

On the other hand,

rhks1 “ µpAkq´1µpAk r Akq

Choose tkjujě1 such that k1 “ 1 and µpAkj r Akj`1q ě12 µpAkjq for

all j ě 1. Then

ÿ

kě1

rhks1 ěÿ

kě1

1µpAkq

µpAk r Ak´1q

ěÿ

jě1

1µpAkjq

µpAkj r Akj`1q ěÿ

kě1

12“ 8

so the ratio ofřn

k“1rhns1 and rřn

k“1 hks1 can be made arbitrarilylarge. This rules out that c1}f } ď rf s1 ď c2}f } for norm } ¨ }.

Other strange facts 33

On non-atomic spaces, for p ă 1: pLp,wq‹ “ t0u

Nontrivial functionals do exist in pL1,wq‹ but they vanish on allsimple functions!

Such singular functionals exist (by Hahn-Banach theorem) alsofor p ą 1. (Non-singular ones exist as well.)

Reason: f pxq :“ 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-normunder 1

t 1p0,8qptqt instead.