Post on 26-Jul-2021
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.