Non differentiability of Pettis

primitives

Udayan B. Darji

University of Louisville

Joint work with B. Bongiorno, L. Di Piazza

May 17, 2012

Preliminaries

1

Throughout X denotes a Banach space unless

otherwise indicated. ‖ ‖ denotes the norm of

this Banach space.

We consider functions from [0,1] into X. We

use the standard Lebesgue measure λ on [0,1].

Let f : [0,1] → X. Function f is strongly

measurable if there exists a sequence of simple

functions {fn} converging a.e. to f . Moreover,

if {fn} satisfies the additional property that

limn→∞∫

[0,1]‖f − fn‖dλ = 0,

then we say that f is Bochner integrable. In

such a case, we let

(B)∫

[0,1]f = lim

n→∞

∫[0,1]

fn.

Recall that (B)∫f is independent of the choice

of {fn}.2

As usual, we let X∗ denote the dual of X.

We say that f : [0,1] → X is weakly measur-

able, if x∗f is λ-measurable for all x∗ ∈ X∗.

We call a weakly measurable function f : [0,1]→X Pettis integrable if for each measurable set

A ⊆ [0,1] there is xA ∈ X such that

x∗(xA) =∫Ax∗fdλ,

for all x∗ ∈ X∗.

In the case that f is Pettis integrable, we use

(P )∫

[0,1]f

to denote the Pettis integral of f .

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Basics Properties of Bochner

Integral

and

Pettis Integrals

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We next recall some basic well-known facts and

differences between the Bochner integral and

Pettis integral.

Fact 1 For any X,

f is Bochner integrable ⇒ f is Pettis integrable.

Fact 2 For X finite dimensional (i.e., X = Rn)

we have the following:

f is Lebesgue

mf is Bochner integrable

mf is Pettis integrable

Moreover, values of∫f under all of the above

integrals coincide.

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Fact 3 There are weakly measurable functions

which are not strong measurable.

Fact 4 The following are equivalent:

• f is strongly mesurable.

• f is weakly measurable and the range of f

is essentially separable, i.e., there exists a set

E of measure zero such that f([0,1] \ E) is

separable.

Corollary 5 If X is a separable Banach space,

then strong measurability and weak measura-

bility coincide.

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Fact 6 Let f : [0,1]→ X be strongly measur-

able. Then,

f is Bochner integrable

m∫‖f‖ <∞

Fact 7 Let f : [0,1] → X be Bochner and F

be its primitive, i.e., F is the X-valued measure

on [0,1] defined by

(B)∫Af = F (A).

Then, F is countably additive, λ-continuous

and is of finite variation.

Moreover, for almost every t ∈ [0,1], we have

that

limh→0

F (t+ h)− F (t)

h= f(t).

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Fact 8 Let f : [0,1] → X be Pettis and F be

its primitive, i.e., F is the X-valued measure

on [0,1] defined by

(P )∫Af = F (A).

Then, F is countably additive, λ-continuous

and is of σ-finite variation.

Theorem 9 (Dilworth-Girardi, 1995)

Let X be any infinite dimensional Banach space.

Then, there exists Pettis integrable f : [0,1]→X such that F , the primitive of f , is nowhere

differentiable! More precisely, for every t ∈[0,1] we have that

limh→0

∥∥∥∥∥F (t+ h)− F (t)

h

∥∥∥∥∥ =∞.

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Motivation

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It is generally considered that the class P of

Pettis integrable functions is much larger than

the class B of Bochner integrable functions

whenever X is infinite dimensional.

Indeed, if X is not separable then, the class of

strongly measurable functions is much smaller

than the class of weakly measurable functions

as the range of strongly measurable functions

is essentially separable.

However, even when X is separable and infinite

dimensional class P is considered much smaller

than the class B.

How does one make this statement precise?

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A natural notion of smallness in a complete

metric space is that of first category.

The space B is a Banach space, however, the

class P enjoys no natural complete metric. Hence,

it is difficult to make a meaningful use of cat-

egory in this setting.

Moreover, given any Bochner primitive F ∈C([0,1], X), arbitrarily close to it, in the norm

of C([0,1], X), there are Pettis primitives.

However, both of B and P are vector spaces.

Our investigation was motivated by the follow-

ing question:

How large a set A ⊂ P must be so that

< B ∪ A >= P?

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A large subset of P \ B

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The concept of unconditional convergence is

central to the investigation of Pettis integral.

Let {xn} be a sequence in X. We say that

{xn} is unconditionally convergent if every re-

arrangement of {xn} converges.

If X is finite dimensional, then {xn} is uncon-

ditionally convergent iff {xn} is absolutely con-

vergent.

Theorem 10 (Dvoretzky-Rogers) If X is infi-

nite dimensional, then there is a sequence {xn}in X which converges unconditionally but does

not converge absolutely.

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Fact 11 Let {En} be a partition of [0,1] and

let {xn} be a sequence in X.

•∑∞n=1 xnχEn is Bochner integrable iff∑∞n=1 ‖xn‖λ(En).

•∑∞n=1 xnχEn is Pettis integrable iff

∑∞n=1 xnλ(En)

is unconditionally convergent.

In particular, if∑∞n=1 xnλ(En) converges un-

conditionally but not absolutely, then∑∞n=1 xnχEn

is Pettis integrable but not Bochner integrable.

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Theorem 12 (BDDi, Main theorem) There ex-

ists a subset K ⊂ P of the cardinality contin-

uum such that the following holds:

• Each f ∈ K is nowhere differentiable.

• If f1, . . . , fn ∈ K and λ1, . . . , λn ∈ R are not

all zero, then

n∑i=1

λifi /∈ B.

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Theorem 13 (BDDi, First Approximation) There

exists a subset K ⊂ P of the cardinality contin-

uum such that the following holds:

• If f1, . . . , fn ∈ K and λ1, . . . , λn ∈ R are not

all zero, then

n∑i=1

λifi /∈ B.

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Theorem 14 (BDDi, Second Approximation)

There exists a subset K ⊂ P of the cardinality

continuum such that the following holds:

• For each f ∈ K and subinterval I of [0,1]

we have that f|I is not Bochner.

• If f1, . . . , fn ∈ K and λ1, . . . , λn ∈ R are not

all zero, then

n∑i=1

λifi /∈ B.

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Proof of the First Approximation. Recall that

there is a set a A of cardinality continuum such

that each A ∈ A is an infinite subset of N and

if A,B are distinct elements of A, then A ∩ Bis finite. (Such a collection is called an almost

disjoint collection.)

Let {xn} be an unconditionally convergent se-

quence X which does not converge absolutely.

Let I1, I2, . . . be a parition of [0,1) into nonover-

lapping intervals. For each A ∈ A, we define

a function fA ∈ P \ B in the following fashion.

List A as a1, a2, . . . in an increasing order.

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Then,

fA ≡∞∑i=1

xi1

|Iai|χIai

.

Let K = {fA : A ∈ A}. Clearly, fA ∈ P \ B.

Let fA1, . . . fAn be distinct elements of K and

λ1, . . . λn ∈ R be non zeros. As the collection

A is almost disjoint, there is n ∈ A1 such that

for (n,∞) ∩ A1 ∩ Ai = ∅, for 2 ≤ i ≤ n. Then

the function∑ni=1 λifi is equal to λ1f1 on Im

for all m ∈ A1, m > n. Hence, we have that∑ni=1 λifi is not in B. End of proof of the First

Approximation.

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Proof of the Second approximation. We pro-

ceed with some notations and then lemmas.

Let n1, n2, . . . be a sequence of integers greater

than 1. Then, (n1, n2, . . .)-tree, denoted by

T (n1, n2, . . .), is defined as

T (n1, n2, . . .) = {∅}∞⋃i=1

Πij=1{0, . . . , nj − 1}.

If σ ∈ T (n1, n2, . . .), then |σ| denotes the length

of σ and if n ≤ |σ|, then σ|n denotes the restric-

tion of σ to the first n terms. If σ|n = τ , then

we say that σ is an extension of τ or τ is a

restriction of σ. If σ ∈ T (n1, n2, . . .) and i ∈ N,

then σi is a extension of σ of length n+1 whose

n+ 1 term is i. If σ ∈ T (n1, n2, . . .), then

[σ] = {τ ∈ T (n1, n2, . . .) : τ is an extension of σ.}

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Associated with (n1, n2, . . .), we also define an

interval function. More specifically, with each

σ ∈ T (n1, n2, . . .), we associate Iσ an subinter-

val of [0,1] such that the following holds.

1. For each k, {Iσ : σ ∈ T (n1, n2, . . .), |σ| =

k} is partition of [0,1] into n1 · nk non-

overlapping closed intervals.

2. For each σ ∈ T (n1, n2, . . .) with |σ| = k, we

have that {Iσl : 0 ≤ l ≤ nk+1−1} is a parti-

tion of Iσ into non-overlapping intervals of

equal length.

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Lemma 15 There exists a functionT : T (n1, n2, . . .) → X such that the followingcondition holds.

For all σ ∈ T (n1, n2, . . .), and any bijectionα : N → [σ], the sequence {xn} defined byxn = T (α(n)) converges unconditionally butnot absolutely.

Moreover, we can make the function T 1-1, ifnecessary.

Lemma 16 Let {Aσ}σ∈T (n1,n2,...)be a pairwise

disjoint system of positive measure subsets of[0,1] such that Aσ ⊆ Iσ and let T be a functionof the Lemma 15. Then,

f =∑

σ∈T (n1,n2,...)

T (σ)1

λ(Aσ)χAσ

is Pettis but Bochner on no subinterval of [0,1].

The rest of the proof of the Second Approxi-mation Theorem on the board.

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Now, the proof of the Main Theorem follows

from techniques of the above theorems and

the following important theorem of Dvoret-

zky’s Theorem:

Theorem 17 (Dvoretzky’s Theorem)

Let ε > 0 and K > 0. Then, there is N(ε,K) >

0 such that for any normed linear space X with

dimension greater than N(ε,K), there is an iso-

morphism T : EK → X such that 1 ≤ ‖T‖ <1 + ε.

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