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 .
3
Basics Properties of Bochner
Integral
and
Pettis Integrals
4
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
9
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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