Duality and weak convergence

Updated April 29, 2020

Plan 2

Outline:Continuous linear functionals over Lp

Duality of Lp with Lq with p´1 ` q´1 “ 1Weak topology and convergenceUniform boundedness principleReflexivity of Lp for 1 ă p ă 8

Topological spaces looked at via continuous functions 3

Natural idea: if X topological space, study CpXq

If X = vector space, then consider linear functions

Definition (Continuous linear functionals)

A functional φ : Lp Ñ R is said to be(1) linear if it respects the linear structure of Lp, i.e.,

@f , g P Lp @a, b P R : φpaf ` bgq “ aφpf q ` bφpgq.

(2) continuous if φ´1pOq is open in Lp for every open O Ď R.Notation: pLpq‹ :“ set of continuous linear functionals on Lp

Key example 4

Lemma

Consider a measure space pX,F , µq and let p, q P r1,8s be Holderconjugate indices, i.e., p´1 ` q´1 “ 1. For any g P Lq set

φgpf q :“ż

fg dµ

Then φg P pLpq‹; i.e., φg is a continuous linear functional on Lp.

Proof: Holder givesˇ

ˇφgpf qˇ

ˇ ď }f }p }g}q

so integral well defined for all f P Lp. Linearity clear. Forcontinuity, note that φg images BXp f , rq into BRpφgpf q, r}g}qq.

Continuity and boundedness 5

Definition (Bounded linear functional)

A linear functional φ : Lp Ñ R is bounded if

Dc P r0,8q @f P Lp :ˇ

ˇφpf qˇ

ˇ ď c}f }p.

We then observe:

Lemma

For any linear functional φ : Lp Ñ R:

φ is continuous ô φ is bounded

Proof: ð proved above. Forñ note that φ´1pp´1, 1qq contains 0and thus BXp0, rq for some r ą 0. Then }f }p ă r implies|φpf q| ď 1 and, by homogeneity, |φpf q| ď r´1}f }p.

pLpq‹ as a complete normed space 6

Proposition

pLpq‹ is a linear vector space with addition and scalar multiplicationdefined by pφ` ψqpf q :“ φpf q ` ψpf q and paφqpf q :“ aφpf q.Moreover, denoting

}φ} :“ supfPLprt0u

|φpf q|}f }p

defines a norm on pLpq‹. The space pLpq‹ is complete in this norm.

Proof of Proposition 7

Linearity and properties of norm checked readily, so main taskis to show completeness.

Suppose tφnuně1 Ď pLpq‹ Cauchy in norm } ¨ }. Linearity implies

@f P Lp :ˇ

ˇφnpf q ´ φmpf qˇ

ˇ “ˇ

ˇpφn ´ φmqpf qˇ

ˇ ď }φn ´ φm} }f }p

so, for each f P Lp, the sequence tφnpf quně1 is Cauchy in R. Set

φpf q :“ limnÑ8

φnpf q

Then φ linear and obeys

|φpf q| ď c}f }p

with c :“ limnÑ8 }φn}. So φ bounded and so continuous. Itremains to show }φn ´ φ} Ñ 0 as n Ñ8 . . .

Proof of Proposition continued . . . 8

. . . For this note that, for all n ě 1 and all f P Lp,ˇ

ˇφpf q ´ φnpf qˇ

ˇ “ limmÑ8

ˇ

ˇφmpf q ´ φnpf qˇ

ˇ ď limmÑ8

}φm ´ φn} }f }p

So}φ´ φn} ď lim

mÑ8}φm ´ φn}

Taking n Ñ8, the RHS tends to zero by the assumed Cauchyproperty of tφnuně1. Hence φn Ñ φ in norm } ¨ }

Note: noting specific to Lp above, works for all normed spaces!

Riesz representation theorem 9

Recall:φgpf q :“

ż

fg dµ

Theorem

For p P p1,8q, the map g ÞÑ φg is a linear bijection Lq Ñ pLpq‹ andis an isometry,

@g P Lq : }φg} “ }g}q.

If µ is σ-finite, the same holds also for p “ 1.

Proof for 1 ă p ă 8 10

Lemma (Differentiability of Lp-norms)

For each p P p1,8q and each f , h P Lp,

ddt}f ` th}pp

ˇ

ˇ

ˇ

t“0“ p

ż

|f |p´2fh dµ

Proof: Fptq :“ |f ` th|p convex and C1,

F1ptq “ p|f ` th|p´2rf ` thsh

Convexity implies t ÞÑ F1ptq non-decreasing andF1p0q ď 1

t rFptq ´ Fp0qs ď F1ptq for any t ě 0. Integrating over µgives

pż

|f |p´2fh dµ ď}f ` th}pp ´ }f }

pp

tď p

ż

|f ` th|p´2rf ` thsh dµ

Dominated Convergence permits taking t Ó 0.

Proof for 1 ă p ă 8 11

Let φ P pLpq‹ and set

K :“

h P Lp : φphq “ 0(

Pick any f P Lp r K (WLOG φ ‰ 0). Then any f 1 P Lp is a linearcombination of f and an element from K:

f 1 “φpf 1qφpf q

f ` h where h P K

Note that if f 1 “ af ` h then φpf 1q “ aφpf q so φ determined by itsvalue at f . Key problem: choose f in an optimal way.

Proof for 1 ă p ă 8 continued . . . 12

Pick any f0 P Lp r K with φpf0q ą 0. As K closed and convex,there is h0 P K such that

}f0 ´ h0}p “ infhPK}f0 ´ h}p.

Define f :“ f0 ´ h0 and note that φpf q “ φpf0q ą 0 and

@h P K :ż

|f |p´2fh dµ “ 0.

Claim:φpf q “ }φ} }f }p.

Indeed, if ă holds then Df 1 P Lp with φpf 1q “ φpf q and}f 1}p ă }f }p. But then f 1 “ f0 ´ h for some h P K and so}f0 ´ h}p ă }f0 ´ h0}p, contradiction!

Proof for 1 ă p ă 8 completed 13

Define

gpxq :“}φ}

}f }p´1p

f pxq|f pxq|p´2

Then g P Lq and φgpf q “ }φ} }f }p. Identity above gives φgphq “ 0for h P K, so φg “ φ.For injectivity, note that, since qpp´ 1q “ p,

}g}q “}φ}

}f }p´1p

´

ż

|f |qpp´1qdµ¯1{q

“ }φ}

and so g ÞÑ φg is a bijective isometry.

Proof for p “ 1 14

Assume µ finite and let φ P pL1q‹. Then L2 Ď L1 by}f }1 ď µpXq1{2}f }2 and so

@f P L2 :ˇ

ˇφpf qˇ

ˇ ď }φ}µpXq1{2}f }2.

By Theorem for p “ 2 we get

@f P L2 : φpf q “ż

fg dµ.

Taking f :“ 1A with A :“ t|g| ą }φ} ` δuwe get }g}8 ď }φ} sog P L8.

If f P L1 then fn :“ f 1t|f |ďnu Ñ f in L1 and so φpfnq Ñ φpf q. Asfn P L2 and

ş

fngdµ Ñş

fgdµ we get φpf q “ş

fgdµ for all f P L1.

Holder: |φpf q| ď }g}8}f }1 and so }φ} ď }g}8.

Alternative line of proof 15

Alternative argument via Radon-Nikodym Theorem.

Suppose µ finite, p “ 1. Let φ P pL1q‹. Then

νpAq :“ φp1Aq, A P F

a finite measure. As |νpAq| ď }φ} µpAq, By Radon-Nikodym,three is g : X Ñ R s.t.

φp1Aq “

ż

g1Adµ

Elementary estimates: }g}8 ď }φ}.

Linearity+L1-convergence, extends from 1A to f P L1.

Extends to σ-additive for p “ 1.Extension needed for p ą 1 w/o finiteness on µ

Counterexamples for L1 16

LemmaSuppose DA P F with no non-empty measurable subsets of finitemeasure. Then φ1A “ 0 yet 1A P L8 is non-zero. Thus g ÞÑ φg as amap L8 Ñ pL1q‹ is not injective and, in particular, not isometric.

Proof: If L1 “ t0u then pL1q‹ “ t0u so assume Df P L1. Thenµp|f | ą εq ă 8 for each ε ą 0 and so

@f P L1 : µpAX tf ‰ 0uq “ 0

This means φ1A “ 0 yet 1A ‰ 0 because µpAq ą 0.

Counterexamples for L1 17

Lemma

Let X be an uncountable set, F “ 2X and µ the counting measure.Let F0 be the σ-algebra of countable and co-countable sets and µ0 thecounting measure on F0. Then

L1pµq “ L1pµ0q ^ L1pµq‹ “ L1pµ0q‹ “ L8pµq ‰ L8pµ0q

So g ÞÑ φg is not surjective as the map L8pµ0q Ñ L1pµ0q‹.

Proof: If f P L1pµq then tf ‰ 0u countable so f P L1pµ0q. HenceL1pµ0q “ L1pµq.

L8pµq contains all bounded functions, L8pµ0q only those thatare constant outside a countable set. So L8pµq ‰ L8pµ0q.

Situation in L8 18

g ÞÑ φg as a map L1 Ñ pL8q‹ is an (injective) isometry

Not surjective whenever X partitions into infinitely many setsof positive measure. Need Hahn-Banach Theorem, to bediscussed later.

Weak convergence 19

Idea: A subspace of CpXq induces a “new” topology on X

Definition (Weak topology)

Let V be a normed space over R and let V‹ be the space ofcontinuous linear functionals on V . The coarsest topologycontaining

φ´1pOq : O Ď R ^ φ P V‹(

is called the weak topology on V .

Generally not first countable and so not metrizable. Nets areneeded to describe convergence! Still:

Definition (Weakly convergent sequences)

txnuně1 Ď V is weakly convergent to x P V (denoted xnwÑ x) if

@φ P V‹ : φpxq “ limnÑ8

φpxnq

Example 20

Define fn : r0, 1s Ñ R by

fnpxq :“

#

1, if t2nxu is even,0, else.

Then fn P Lp for all 1 ď p ă 8 and so @g P Lq:ż

r0,1sg fn dλ ÝÑ

nÑ8

12

ż

r0,1sg dλ

So we think

@p P r1,8q : fnwÝÑnÑ8

12

1r0,1s in Lp

Note: tfnuně1 not convergent in Lp for any p. Weak convergencefails for p “ 8 as Dφ P pL8q‹ with φpf2nq “ 1 and φpf2n`1q “ 0.

Separation of points 21

Q: Why is the limit unique?A: Because weak-topology is Hausdorff!

Lemma (Continuous linear functionals on Lp separate)

Let p P r1,8q. Then

@f P Lp r t0u Dφ P pLpq‹ : φpf q ‰ 0

If µ is semifinite, then same true for p “ 8.

Proof: For 1 ď p ă 8, f P Lp r t0u let g :“ |f |p´2f . Then g P Lq

and φgpf q “ }f }pp ą 0.

For p “ 8, if f P L8r t0u then, by semifinitness, Dε ą 0 andDA Ď t|f | ą εuwith µpAq P p0,8q. Now take g :“ signpf q1A.

Note: True for all Banach spaces by Hahn-Banach theorem!

Weak boundedness 22

Q: Are weakly convergent sequences bounded?

DefinitionA set A Ď Lp is weakly bounded if

φ P pLpq‹ : supfPA

ˇ

ˇφpf qˇ

ˇ ă 8.

BysupfPA

ˇ

ˇφpf qˇ

ˇ ď }φ} supfPA

}f }p

if A Ď Lp is norm-bounded, then it is weakly bounded.

For converse we need . . .

Uniform boundedness principle 23

TheoremLet p P r1,8q. Then for all non-empty A Ď Lp,

´

@φ P pLpq‹ : supfPA

ˇ

ˇφpf qˇ

ˇ ă 8

¯

ñ supfPA

}f }p ă 8

In particular, every weakly bounded subset of Lp is bounded in Lp.The same holds for p “ 8 whenever µ is semifinite.

Note: True for all Banach spaces. Many different proofs(Baire-Category Theorem, or direct arguments)

Proof of Uniform boundedness principle 24

Will use the “sliding hump” argument. Let p P r1,8q andsuppose supfPA }f }p “ 8. Then Dtfnuně1 Ď A s.t.

limnÑ8

3´n}fn}p “ 8

Setgn :“

1

}fn}p´1p|fn|p´2fn

Then φgn P pLpq‹ with }φgn} “ }gn}q “ 1. Next definetσnuně1 P t´1,`1uN recursively by σ1 :“ `1 and

@n ě 2 : σn φgnpfnq´

n´1ÿ

k“1

3´kσkφkpfnq¯

ě 0

Setφ :“

ÿ

ně1

3´nσn φgn

and estimate . . .

Proof of Uniform boundedness principle continued . . . 25

ˇ

ˇφpfnqˇ

ˇ ě

ˇ

ˇ

ˇ

ˇ

3´nσn φgnpfnq `n´1ÿ

k“1

3´kσkφgkpfnqˇ

ˇ

ˇ

ˇ

´

ˇ

ˇ

ˇ

ˇ

ÿ

kąn

3´kσkφgkpfnqˇ

ˇ

ˇ

ˇ

ě 3´nˇˇφgnpfnq

ˇ

ˇ´ÿ

kąn

3´k}φgk}}fn}p

“ 3´n´

ˇ

ˇφgnpfnqˇ

ˇ´12}fn}p

¯

From φgnpfnq “ }fn}p we get |φpfnq| ě 12 3´n}fn}p Ñ8 !!!

For p “ 8 use semifiniteness of µ to find

An Ď

|fn| ą 23}fn}8

(

with µpAnq P p0,8q. Then set gn :“ 1µpAnq

1An P L1 and note that

}φgn} “ 1 and φgnpfnq ě23}fn}8. Now continue as before.

Principle of condensation of singularities 26

Corollary

Assume p P r1,8q or p “ 8 with the underlying measure semifiniteand let tφi,j : i, j ě 1u Ď pLpq‹ be such that

@i ě 1 Dfi P Lp : supjě1

|φi,jpfiq| “ 8

ThenDf P Lp @i ě 1 : sup

jě1|φi,jpf q| “ 8

Proof: homework

Another application 27

Corollary

Assume 1 ă p ă 8 and let tφnuně1 P pLpq‹ be such that

@f P Lp : φpf q :“ limnÑ8

φnpf q exists in R

Then φ P pLpq‹.

Proof: NTS continuity.

Key fact: For 1 ă p ă 8, double dual pLpq‹‹ isometric to Lp andevaluation map φ ÞÑ φpf qmember of pLpq‹ by |φpf q| ď }φ} }f }p

Assumption gives: tφnuně1 weakly bounded in pLpq‹

Uniform boundedness principe: c :“ supně1 }φn} ă 8 and so|φpf q| ď c}f }p which gives φ P pLpq‹.

Reflexive space 28

We used that Lp, for 1 ă p ă 8, adheres to:

DefinitionA normed vector space V is reflexive if the evaluation mapx ÞÑ x‹‹ defined by

@φ P V‹ : x‹‹pφq :“ φpxq

images V onto the double dual V‹‹.

Note x‹‹ always injective; in fact, isometry by |x‹‹pφq| ď }x} }φ}

L1 and L8 are NOT reflexive in general!