Contemporary Mathematics Volume 00, 0000

Bochner vs. Pettis norm: examples and results

S.J. DILWORTH AND MARIA GIRARDI

Contemp. Math. 144 (1993) 69–80 Banach Spaces, Bor-Luh Lin and William B. Johnson, editors

Abstract. Our basic example shows that for an arbitrary infinite-dimen-sional Banach space 𝔛, the Bochner norm and the Pettis norm on L_{1}(𝔛) are not equivalent. Refinements of this example are then used to investigate various modes of sequential convergence in L_{1}(𝔛).

1. INTRODUCTION

Over the years, the Pettis integral along with the Pettis norm have grabbed the interest of many. In this note, we wish to clarify the differences between the Bochner and the Pettis norms. We begin our investigation by using Dvoretzky’s Theorem to construct, for an arbitrary infinite-dimensional Banach space, a sequence of Bochner integrable functions whose Bochner norms tend to infinity but whose Pettis norms tend to zero. By refining this example (again working with an arbitrary infinite-dimensional Banach space), we produce a Pettis integrable function that is not Bochner integrable and we show that the space of Pettis integrable functions is not complete. Thus our basic example provides a unified constructive way of seeing several known facts. The third section expresses these results from a vector measure viewpoint. In the last section, with the aid of these examples, we give a fairly thorough survey of the implications going between various modes of convergence for sequences of L_{1}(𝔛) functions.

\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society 0000-0000/00 $1.00 + $.25 per page

1991 Mathematics Subject Classification. 46E40, 28A20, 46G10.

This paper is in final form and no version of it will be submitted for publication elsewhere.

2. EXAMPLES: L_{1}(𝔛) vs. P_{1}(𝔛)

Let 𝔛 be a Banach space with dual 𝔛^{∗} and let (Ω, Σ, μ) be the usual Lebesgue measure space on [0, 1]. Let L_{1}(𝔛) = L_{1}(Ω, Σ, μ; 𝔛) be the space of 𝔛-valued Bochner integrable functions on Ω with the usual Bochner norm || · ||_{Bochner} .

A quick review of the Pettis integral. Consider a weakly measurable function f : Ω → 𝔛 such that x^{∗}f ∈ L_{1}(ℝ) for all x^{∗} ∈ 𝔛^{∗}. The Closed Graph Theorem gives that for each E ∈ Σ there is an element x_{E}^{∗∗} of 𝔛^{∗∗} satisfying x_{E}^{∗∗}(x^{∗}) = ∫_{E} x^{∗}fdμ for each x^{∗} ∈ 𝔛^{∗}. If x_{E}^{∗∗} is actually in 𝔛 for each E ∈ Σ, then we say that f is Pettis integrable with the Pettis integral of f over E being the element x_{E} ∈ 𝔛 satisfying

x^{∗}(x_{E}) = ∫_{E} x^{∗}fdμ

for each x^{∗} ∈ 𝔛^{∗}. In this note we shall restrict our attention to Pettis integrable functions that are strongly measurable. The space P_{1}(𝔛) of (equivalence classes of) all strongly measurable Pettis integrable functions forms a normed linear space under the norm

∥ f ∥_{Pettis} = sup_{x^{∗}∈B(𝔛^{∗})} ∫_{Ω} | x^{∗}(f) | dμ .

Clearly if f ∈ L_{1}(𝔛) then f ∈ P_{1}(𝔛) with

||f||_{Pettis} ≤ ||f||_{Bochner} .

If 𝔛 is finite-dimensional, then the Bochner norm and the Pettis norm are equivalent and so L_{1}(𝔛) = P_{1}(𝔛). However, in any infinite-dimensional Banach space 𝔛, the next example shows that the two norms are not equivalent on L_{1}(𝔛). Specifically, it constructs an essentially bounded sequence of Bochner-norm one functions whose Pettis norms tend to 0.

EXAMPLE 1.

Let 𝔛 be an infinite-dimensional Banach space. We now construct a sequence {f_{n}} of L_{1}(𝔛) functions which tend to zero in the Pettis norm but not in the Bochner norm.

Let {I_{k}^{n} : n = 0, 1, …; k = 1, …, 2^{n}} be the dyadic intervals on [0, 1], i.e.

I_{k}^{n} = [\frac{k−1}{2^{n}}, \frac{k}{2^{n}} ) .

Fix n ∈ ℕ. We define f_{n} with the help of Dvoretzky’s Theorem [D]. Find a 2^{n}- dimensional subspace E_{n} of 𝔛 such that the Banach-Mazur distance between E_{n} and ℓ_{2}^{2^{n}} is at most 2. So there is an operator T_{n}: ℓ_{2}^{2^{n}} → E_{n} such that the norm of T_{n} is at most 2 and the norm of the inverse of T_{n} is 1. Let {e_{k}^{n}: k = 1, … 2^{n}} be the image under T_{n} of the standard unit vectors {u_{k}^{n}: k = 1, … 2^{n}} of ℓ_{2}^{2^{n}}. Define f_{n}: Ω → 𝔛 by

f_{n}(ω) = ∑_{k=1}^{2^{n}} e_{k}^{n} 1_{I_{k}^{n}}(ω) ,

The sequence {f_{n}} has the desired properties.

Since

||f_{n}||_{Bochner} = ∑_{k=1}^{2^{n}} ∫_{I_{k}^{n}} ||e_{k}^{n}||_{𝔛} dμ = 2^{−n} ∑_{k=1}^{2^{n}} ||e_{k}^{n}||_{𝔛}

and 1 ≤ ||e_{k}^{n}||_{𝔛} ≤ 2,

1 ≤ ||f_{n}||_{Bochner} ≤ 2 .

As for the Pettis norm, fix x^{∗} ∈ B(X^{∗}). Note that the restriction y_{n}^{∗} of x^{∗} to E_{n} is an element in E_{n}^{∗} of norm at most 1. Also, the adjoint T_{n}^{∗} of T_{n} has norm at most 2. Thus

∑_{k=1}^{2^{n}} |x^{∗}(e_{k}^{n})| = ∑_{k=1}^{2^{n}} |y_{n}^{∗}(T_{n}u_{k}^{n})|

= ∑_{k=1}^{2^{n}} |(T_{n}^{∗}y_{n}^{∗})(u_{k}^{n})| ≤ ||T_{n}^{∗}y_{n}^{∗}||_{(ℓ_{2}^{2^{n}})^{∗}} || ∑_{k=1}^{2^{n}} u_{k}^{n}||_{ℓ_{2}^{2^{n}}} ≤ 2 \sqrt{2^{n}} ,

and so

∫_{Ω} |x^{∗}(f_{n})| dμ = ∑_{k=1}^{2^{n}} ∫_{I_{k}^{n}} |x^{∗}(e_{k}^{n})| dμ = 2^{−n} ∑_{k=1}^{2^{n}} |x^{∗}(e_{k}^{n})| ≤ 2 2^{− \frac{n}{2}} ,

giving that

||f_{n}||_{Pettis} ≤ 2 2^{− \frac{n}{2}} .

Thus the Pettis norm of f_{n} tends to zero. □

This difference between the Bochner and Pettis norms is further emphasized by the following minor variation of Example 1.

EXAMPLE 2.

Fix 0 < β < \frac{1}{2} and put α_{n} ≡ 2^{nβ}. Define f̃_{n}: Ω → 𝔛 by f̃_{n} = α_{n}f_{n} where f_{n} is as in Example 1. The Bochner norm of f̃_{n} tends to infinity at the rate of α_{n} but the Pettis norm of f̃_{n} tends to zero. □

A variation on this example gives a (strongly measurable) Pettis integrable function that is not Bochner integrable – working in any infinite-dimensional Banach space. The usual way of constructing such a function uses the Dvoretsky-Rogers Theorem. Recall [DR] that in any infinite-dimensional Banach space 𝔛, there is an unconditionally convergent series ∑_{n} x_{n} that is not absolutely convergent. The function f: Ω → 𝔛 given by

f(·) = ∑_{n=1}^{∞} \frac{x_{n}}{μ(E_{n})} 1_{E_{n}}(·),

(where {E_{n}} is a partition of Ω into sets of strictly positive measure) is Pettis integrable but not Bochner integrable. Our example is in the same spirit.

EXAMPLE 3.

Let 𝔛 be an infinite-dimensional Banach space. Consider the above construction. Note that the collection {I_{2}^{n}}_{n=1}^{∞} partitions [0, 1]. Let g_{n}: Ω → 𝔛 be a normalized f_{n} (from Example 1) supported on I_{2}^{n}, i.e.

g_{n}(ω) = 2^{n} f_{n}(2^{n}ω − 1) 1_{I_{2}^{n}}(ω) .

Define f: Ω → 𝔛 by

f(ω) = ∑_{n=1}^{∞} g_{n}(ω) 1_{I_{2}^{n}}(ω) .

Clearly f is strongly measurable.

But f is not Bochner integrable since for any N ∈ ℕ

∫_{Ω} ||f|| dμ ≥ ∑_{n=1}^{N} ∫_{I_{2}^{n}} ||g_{n}||_{𝔛} dμ ≥ ∑_{n=1}^{N} 2^{−n} 2^{n} = N .

However f is Pettis integrable. First note that x^{∗}f ∈ L_{1}(ℝ) for all x^{∗} ∈ 𝔛^{∗} since for a fixed x^{∗} ∈ B(X^{∗}) we have (using computations from the first example)

∫_{I_{2}^{n}} |x^{∗}f| dμ = ∫_{I_{2}^{n}} |x^{∗}g_{n}| dμ = ∫_{Ω} |x^{∗}f_{n}| dμ ≤ 2 2^{− \frac{n}{2}} ,

and so

∫_{Ω} |x^{∗}f| dμ = ∑_{n=1}^{∞} ∫_{I_{2}^{n}} |x^{∗}f| dμ ≤ 2 ∑_{n=1}^{∞} 2^{− \frac{n}{2}} = 2(\sqrt{2} + 1) .

Next fix E ∈ Σ. We know there is an element x_{E}^{∗∗} of 𝔛^{∗∗} satisfying that x_{E}^{∗∗}(x^{∗}) = ∫_{E} x^{∗}fdμ for each x^{∗} ∈ 𝔛^{∗}. To see that x_{E}^{∗∗} is actually in 𝔛, consider the sequence {s_{N}} of L_{1}(𝔛) functions on Ω where

s_{N}(ω) = ∑_{n=1}^{N} g_{n}(ω) 1_{I_{2}^{n}}(ω) .

Let x_{E}^{N} be the Bochner integral of s_{N} over E. Viewed as a sequence in 𝔛^{∗∗}, {x_{E}^{N}}_{N} converges in norm to x_{E}^{∗∗} since for any x^{∗} ∈ B(X^{∗})

|(x_{E}^{N} − x_{E}^{∗∗})(x^{∗})| = | ∫_{E} x^{∗}s_{N} dμ − ∫_{E} x^{∗}f dμ| ≤ ∫_{Ω} |x^{∗}(s_{N} − f)| dμ

= ∑_{n=N+1}^{∞} ∫_{I_{2}^{n}} |x^{∗}f| dμ ≤ 2 ∑_{n=N+1}^{∞} 2^{− \frac{n}{2}} .

Thus x_{E}^{∗∗} is actually in 𝔛, as needed. □

Janicka and Kalton [JK] showed that if 𝔛 is infinite-dimensional then P_{1}(𝔛) is never complete under the Pettis norm. The first example provides a constructive way to see this.

EXAMPLE 4.

Taking some care in our construction in Example 1, we may arrange for {E_{n}} to form a finite dimensional decomposition (FDD) of some subspace of 𝔛. To see this, first choose a basic sequence {x_{n}} in 𝔛. Take a blocking {F_{n}} of the basis so that each subspace F_{n} is of large enough dimension to find (using the finite-dimensional version of Dvoretzky’s Theorem) a 2^{n}-dimensional subspace E_{n} of F_{n} such that the Banach-Mazur distance between E_{n} and ℓ_{2}^{2^{n}} is at most 2. Then {E_{n}} forms a FDD.

Keeping with the notation from Example 1, define h_{n}: Ω → 𝔛 by

h_{n} = ∑_{k=1}^{n} f_{k} .

Clearly each h_{n} is in P_{1}(𝔛). Also, {h_{n}} is Cauchy in the Pettis norm since

||h_{n} − h_{n+j}||_{Pettis} ≤ ∑_{k=n}^{∞} ||f_{k}||_{Pettis} ≤ 2 ∑_{k=n}^{∞} 2^{− \frac{k}{2}} .

To show that h_{n} cannot converge to an element in P_{1}(𝔛), we need the following two lemmas whose proofs are given shortly.

lemma 1. If the sequence {f_{n}} of functions in P_{1}(𝔛) converges to f ∈ P_{1}(𝔛) and each f_{n} is essentially valued in some subspace 𝒴 of 𝔛, then f is also essentially valued in 𝒴. □

lemma 2. Let {f_{n}} be a sequence of P_{1}(𝔛) functions that converges in Pettis norm to f ∈ P_{1}(𝔛). Furthermore, suppose that each f_{n} is essentially valued in some subspace 𝒴 of 𝔛 and that T : 𝒴 → 𝔛 is a bounded linear operator. Then T f_{n} converges in the Pettis norm to T f. □

Suppose that h_{n} converges in the Pettis norm to h ∈ P_{1}(𝔛). Let 𝒴 ≡ ∑ ⊕E_{n} and let P_{n}: 𝒴 → 𝔛 be the natural projection from 𝒴 onto E_{n}. Note that each h_{n}, thus also h, are essentially valued in 𝒴. Applying the second lemma to the operator P_{n} and noting that P_{n}h_{m} ≡ f_{n} for m ≥ n, we have that P_{n}h ≡ f_{n}. Thus for each n ∈ ℕ

1 ≤ ||P_{n}h(ω)||_{𝔛} ≤ 2

for almost all ω. This contradicts the fact that for almost all ω

lim_{N→∞} ||h(ω) − ∑_{n=1}^{N} P_{n}h(ω)||_{𝔛} = 0 .

Thus h_{n} cannot converge to an element in P_{1}(𝔛). □

Now for the proofs of the lemmas.

proof of lemma 1. Let {f_{n}} be a sequence of P_{1}(𝔛) functions that converges in the Pettis norm to f ∈ P_{1}(𝔛). Furthermore, suppose that each f_{n} is essentially valued in some subspace 𝒴 of 𝔛 and that f is essentially valued in some subspace 𝒵 with 𝒴 ⊂ 𝒵. Since f and each f_{n} are strongly measurable, we may assume that 𝒵 is separable.

Since 𝒵 is separable, there is a sequence {z_{m}^{∗}} of functionals from 𝒵^{∗} such that an element z from 𝒵 is in 𝒴 if and only if z_{m}^{∗}(z) = 0 for each m.

Since {f_{n}} converges in the Pettis norm to f, for each z_{m}^{∗} the sequence {z_{m}^{∗}f_{n}}_{n} converges to z_{m}^{∗}f in L_{1}(ℝ) -norm. But for each m, note that z_{m}^{∗}f_{n} is zero almost everywhere and so z_{m}^{∗}f is also zero almost everywhere. Thus f is also essentially valued in 𝒴. □

proof of lemma 2. Let {f_{n}} be a sequence of P_{1}(𝔛) functions which converges in Pettis norm to f ∈ P_{1}(𝔛). Furthermore, suppose that each f_{n} is essentially valued in some subspace 𝒴 of 𝔛. Lemma 1 gives that f is also essentially valued in 𝒴. Let T : 𝒴 → 𝔛 be a bounded linear operator. Note that for a fixed x^{∗} ∈ 𝔛^{∗} we may extend T^{∗}x^{∗} to an element in 𝔛^{∗} without increasing its norm.

Note that T f (and likewise each T f_{n}) is actually in P_{1}(𝔛). To see this, first observe that for a fixed x^{∗} ∈ 𝔛^{∗} (viewing T^{∗}x^{∗} as an element in 𝔛^{∗} of the same norm)

∫_{Ω} |x^{∗}(T f)| dμ = ∫_{Ω} |(T^{∗}x^{∗})(f)| dμ ≤ ||T^{∗}x^{∗}|| ||f||_{Pettis} .

and so x^{∗}(T f) ∈ L_{1}. Next, let x_{E} be the Pettis integral of f over E ∈ Σ. Note that x_{E} ∈ 𝒴 and for each x^{∗} ∈ 𝔛^{∗}

x^{∗}(T x_{E}) = (T^{∗}x^{∗})(x_{E}) = ∫_{E}(T^{∗}x^{∗})(f) dμ = ∫_{E} x^{∗}(T f) dμ ,

and so T (x_{E}) is the Pettis integral of T f over E ∈ Σ. Thus T f is indeed in P_{1}(𝔛).

Since for x^{∗} ∈ B(X^{∗})

∫_{Ω} |x^{∗}(T f_{n} − T f)| dμ = ∫_{Ω} |(T^{∗}x^{∗})(f_{n} − f)| dμ ≤ ||T || ||f_{n} − f||_{Pettis} ,

the sequence T f_{n} converges in the Pettis norm to T f. □

3. VECTOR MEASURES

Let’s look at our examples from a vector measure viewpoint.

A Pettis integrable function f gives rise to a vector measure G_{f}: Σ → 𝔛 where G_{f}(E) is the Pettis integral of f over E. The Pettis norm of f is the semivariation of G_{f}. If f ∈ L_{1}(𝔛), then the Bochner norm of f is the variation of G_{f}.

Example 2 shows that for any infinite-dimensional Banach space 𝔛, there is a sequence of 𝔛-valued measures whose variations tend to infinity but whose semivariations tend to zero. On the other hand, Example 3 gives a 𝔛-valued measure of bounded semivariation but infinite variation.

For f ∈ P(𝔛), the corresponding measure G_{f} is μ-continuous and has a relatively compact range. In fact [DU], the completion of P_{1}(𝔛) is (isometrically) the space K(𝔛) of all μ-continuous measures G : Σ → 𝔛 whose range is relatively compact, equipped with the semivariation norm. Thus Example 4 gives a measure in K(𝔛) which is not representable by a (strongly measurable) Pettis integrable function.

4. CONVERGENCE

We now restrict our attention to sequences of L_{1}(𝔛) functions which converge (in the Bochner norm, in the Pettis norm, or weakly in L_{1}(𝔛)) to functions actually in L_{1}(𝔛). We have the following obvious relations:

(1) Bochner-norm convergence implies weak and Pettis-norm convergence;

(2) weak convergence implies neither Bochner-norm nor Pettis-norm convergence (e.g. consider the Rademacher functions in L_{1}(ℝ) );

(3) if 𝔛 is finite dimensional then the Pettis norm and the Bochner norm are

equivalent. Our examples help to clarify the other implications.

If 𝔛 is infinite-dimensional, then there is a sequence of functions which converges in Pettis norm but whose Bochner norms tend to infinity (Example 2). Thus this sequence does not converge in the Bochner norm or weakly.

What if we require the sequence to be bounded (sup_{n} ||f_{n}||_{L_{1}(𝔛)} < ∞), or uniformly integrable, or even essentially bounded (sup_{n} ||f_{n}||_{L_{∞}(𝔛)} < ∞)? Again, if 𝔛 is infinite-dimensional then Example 1 gives a sequence of essentially bounded functions which converges in Pettis norm but not in the Bochner norm.

In certain situations we can pass from Pettis-norm convergence to weak convergence. If f_{n} → f in the Pettis norm, then clearly ∫_{Ω} g(f_{n}) dμ → ∫_{Ω} g(f) dμ for each simple function g ∈ L_{∞}(𝔛^{∗}). As a first step towards weak convergence, when can we at least conclude that f_{n} → f in the σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-topology? A word of caution, the simple functions need not be dense in L_{∞}(𝔛^{∗}). For example, the L_{∞}(ℓ_{2}) function g(·) = ∑_{n=1}^{∞} e_{n}1_{I_{2}^{n}}, where {e_{n}} are the standard unit vectors of ℓ_{2}, is at least \frac{\sqrt{2}}{2} from any simple function in L_{∞}(ℓ_{2}). Our next example illustrates the fact that for any infinite-dimensional Banach space there is a bounded sequence in L_{1}(𝔛) that converges in the Pettis norm but not in the σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-topology.

EXAMPLE 5.

Consider the sequence {g_{n}} from Example 3 where

g_{n}(ω) = 2^{n} f_{n}(2^{n}ω − 1) 1_{I_{2}^{n}}(ω) .

Since ||f_{n}||_{L_{1}(𝔛)} ≤ 2, the Bochner norm of each g_{n} is at most 2.

To examine the Pettis norm, fix x^{∗} ∈ B(X^{∗}). Computations in Example 1 show that

∫_{Ω} |x^{∗}g_{n}| dμ = ∫_{I_{2}^{n}} |x^{∗}g_{n}| dμ = ∫_{Ω} |x^{∗}f_{n}| dμ ≤ 2 2^{− \frac{n}{2}}

and so

||g_{n}||_{Pettis} ≤ 2 2^{− \frac{n}{2}} .

Thus g_{n} → 0 in the Pettis norm.

As for the σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-topology, find y_{k}^{n} ∈ B(𝔛^{∗}) such that y_{k}^{n}(e_{k}^{n}) = ||e_{k}^{n}||_{𝔛}. Note that {I_{2}^{n}: n = 1, 2, … } partitions [0, 1] and that {I_{j}^{2n}: j = 2^{n} + 1, …, 2^{n+1}} partitions the interval I_{2}^{n}. Define g: Ω → 𝔛^{∗} by

g(ω) = ∑_{n=1}^{∞} ∑_{k=1}^{2^{n}} y_{k}^{n}1_{I_{2^{n}+k}^{2n}}(ω) .

Clearly g ∈ L_{∞}(𝔛^{∗}). But

∫_{Ω} g(g_{n}) dμ = ∫_{I_{k}^{n}} g(g_{n}) dμ = ∑_{k=1}^{2^{n}} ∫_{I_{2^{n}+k}^{2n}} g(g_{n}) dμ

= ∑_{k=1}^{2^{n}} ∫_{I_{2^{n}+k}^{2n}} y_{k}^{n}(2^{n} e_{k}^{n}) dμ ≥ ∑_{k=1}^{2^{n}} 2^{n} 2^{−2n} 1 = 1 .

Thus {g_{n}} does not converge in the σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-topology. □

However, a uniformly integrable Pettis-norm convergent sequence does converge in the σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-topology. To see why, consider a uniformly integrable sequence {f_{n}} such that ∫_{Ω} g(f_{n}) dμ → 0 for each simple function g ∈ L_{∞}(𝔛^{∗}). For an arbitrary g ∈ L_{∞}(𝔛^{∗}), note that there is a sequence of simple functions which converges to g in measure. So we can approximate g (in L_{∞}(𝔛^{∗}) norm) by a simple function g̃ on a subset E ∈ Σ of full enough measure that ∫_{E^{c}} ||f_{n}|| dμ is small for each n and the measure of E^{c} is also small. Note that

|∫_{Ω} g(f_{n}) dμ| ≤ ∫_{E^{c}} ||g||_{𝔛^{∗}}||f_{n}||_{𝔛} dμ + ∫_{E} ||g−g̃||_{𝔛^{∗}}||f_{n}||_{𝔛} dμ + |∫_{E} g̃(f_{n}) dμ| .

The first two integrals on the left-hand side are small for each n. The last integral converges to zero as n → ∞. Thus ∫_{Ω} g(f_{n}) dμ → 0, as needed.

A Pettis-norm convergent sequence need not be uniformly integrable (Example 2). However, if a sequence converges in the σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-topology then it is necessarily uniformly integrable. To see why this is so, recall the usual proof [DU p. 104] that a weakly convergent sequence is uniformly integrable. If a subset K of L_{1}(𝔛) is not uniformly integrable, then we can find (with the help of Rosenthal’s Lemma) a sequence {f_{n}} in K and a disjoint sequence {E_{n}} in Σ such that

∫_{E_{n}} ||f_{n}|| dμ > 2δ  and  ∫_{∪_{j≠n}E_{j}} ||f_{n}|| dμ < δ

for some δ > 0. For each n we can find l_{n} ∈ L_{∞}(𝔛^{∗}) ⊂ [L_{1}(𝔛)]^{∗} of norm one and supported on E_{n} such that ∫_{E_{n}} l_{n}(f_{n}) > 2δ. Define l: Ω → 𝔛^{∗} by

l(·) = ∑_{n=1}^{∞} l_{n}(·) 1_{E_{n}}(·) .

Clearly, l is a norm one element of L_{∞}(𝔛^{∗}) ⊂ [L_{1}(𝔛)]^{∗}. But ∫_{Ω} l(f_{n}) ↛ 0 since

∫_{Ω} l(f_{n}) dμ = ∫_{E_{n}} l_{n}(f_{n}) dμ + ∫_{F_{n}} l(f_{n}) dμ ,

where F_{n} = ∪_{m≠n}E_{m}, and so

|∫_{Ω} l(f_{n}) dμ| ≥ |∫_{E_{n}} l_{n}(f_{n}) dμ| − ∫_{F_{n}} ||l||_{𝔛^{∗}} ||(f_{n})||_{𝔛} dμ
≥ 2δ − ∫_{F_{n}} ||f_{n}|| dμ
≥ 2δ − δ = δ .

So if g_{n} → g in the σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-topology then the set {g_{n} − g}, and thus also the set {g_{n}}, are uniformly integrable.

To pass from σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-convergence to weak convergence, recall [DU] that the dual of L_{1}(𝔛) is (isometrically) L_{∞}(𝔛^{∗}) if and only if 𝔛^{∗} has the Radon-Nikodým property (RNP). Thus we have:

Fact. If 𝔛^{∗} has the RNP, then a uniformly integrable Pettis-norm convergent sequence of L_{1}(𝔛) functions also converges weakly.

As noted above, the uniform integrability condition is necessary. Our next proposition shows that it is also necessary that 𝔛^{∗} have the RNP. The proof, which uses Stegall’s Factorization Theorem, may be pulled out of a result of Ghoussoub and P. Saab [GS] characterizing weakly compact sets in L_{1}(𝔛).

Proposition. If 𝔛^{∗} fails the RNP, then there is an essentially bounded sequence of L_{1}(𝔛) functions that is Pettis-norm convergent but not weakly convergent.

Proof.

Since 𝔛^{∗} fails the RNP, there is a separable subspace 𝒴 of 𝔛 such that 𝒴^{∗} is not separable. We shall construct an essentially bounded sequence {g_{n}} of L_{1}(𝒴) functions such that g_{n} → 0 in the Pettis-norm on L_{1}(𝒴) but g_{n} ↛ 0 weakly in L_{1}(𝒴). This is sufficient.

Let Δ = {−1, 1}^{ℕ} be the Cantor group with Haar measure ν. Let {Δ_{k}^{n}: k = 1, …, 2^{n}} be the standard n-th partition of Δ. Thus Δ_{1}^{0} = Δ and Δ_{k}^{n} = Δ_{2k−1}^{n+1} ∪ Δ_{2k}^{n+1} and ν(Δ_{k}^{n}) = 2^{−n}. We will work with our underlying measure space being the Cantor group endowed with its Haar measure instead of our usual Lebesgue measure space on [0, 1].

Consider the set C(Δ) of real-valued continuous function on Δ as a subspace of L_{∞}(ℝ). Let {1_{Δ}} ∪ {h_{k}^{n}: n = 0, 1, 2, … and k = 1, …, 2^{n}} be the usual Haar basis of C(Δ), where h_{k}^{n}: Δ → ℝ is given by

h_{k}^{n} = 1_{Δ_{2k−1}^{n+1}} − 1_{Δ_{2k}^{n+1}} .

Let {e_{k}^{n}: n = 0, 1, 2, … and k = 1, …, 2^{n}} be an enumeration (lexicographically) of the usual ℓ_{1} basis and H: ℓ_{1} → L_{∞} be the Haar operator that takes e_{k}^{n} to h_{k}^{n}.

By Stegall’s Factorization Theorem [S, Theorem 4], H factors through 𝒴, i.e. there are bounded linear operators R: ℓ_{1} → 𝒴 and S: 𝒴 → L_{∞} such that H = SR. Let R̃ be the natural extension of R to a bounded linear operator from L_{1}(ℓ_{1}) to L_{1}(𝒴).

Consider the sequence {f_{m}} of L_{1}(ℓ_{1}) functions given by

f_{m}(·) = \frac{1}{m} ∑_{n=1}^{m} ∑_{k=1}^{2^{n}} h_{k}^{n}(·)e_{k}^{n} .

Let g_{m} = R̃(f_{m}). The sequence {g_{m}} of L_{1}(𝒴) functions does the job.

Since sup_{m} ||f_{m}(ω)||_{ℓ_{1}} = 1 for ν-a.e. ω, we have that {g_{m}} is essentially bounded.

To examine the Pettis norm of f_{m}, consider the continuous convex function ψ: ℓ_{∞} → ℝ where the image of x^{∗} = (α_{k}^{n}) ∈ ℓ_{∞} ≃ ℓ_{1}^{∗} (again, lexicographically ordered) is

ψ(x^{∗}) = ∫_{Δ} |x^{∗}(f_{m})| dν = \frac{1}{m} ∫_{Δ} |∑_{n=1}^{m} ∑_{k=1}^{2^{n}} h_{k}^{n}(·)x^{∗}(e_{k}^{n})| dν .

Since the image of f_{m} is supported on {e_{k}^{n}: n = 1, …, 2^{m} and k = 1, …, 2^{n}}, we consider the natural restriction map r_{m}: ℓ_{∞} → ℓ_{∞}^{2(2^{2^{m}} −1)} ⊂ ℓ_{∞} given by

r_{m}((α_{k}^{n}: n = 0, 1, 2, … and k = 1, …, 2^{n})) =
(α_{k}^{n}: n = 1, …, 2^{m} and k = 1, …, 2^{n}) .

Let K be the unit ball of r_{m}(ℓ_{∞}). The maximum of ψ over K is attained at some extreme point x_{0}^{∗} of K. Being an extreme point, x_{0}^{∗} = (ϵ_{k}^{n}) satisfies |ϵ_{k}^{n}| = 1 for each admissible n and k. So

||f_{m}||_{Pettis} = sup_{x^{∗}∈B(ℓ_{∞})} ψ(x^{∗}) = sup_{x^{∗}∈K} ψ(x^{∗})

= ψ(x_{0}^{∗}) = \frac{1}{m} ∫_{Δ} |∑_{n=1}^{m} ∑_{k=1}^{2^{n}} ϵ_{k}^{n} h_{k}^{n}(·)| dν .

Since ∑_{k=1}^{2^{n}} ϵ_{k}^{n} h_{k}^{n} behaves like the n^{th}-Rademacher function, Khintchine’s in-

equality gives that ||f_{m}||_{Pettis} decays like m^{− \frac{1}{2}} .

Now for the Pettis norm of g_{m}, note that if y^{∗} ∈ 𝒴^{∗} then

∫_{Δ} |y^{∗}g_{m}| dν = ∫_{Δ} |y^{∗}(Rf_{m})| dν
= ∫_{Δ} |(R^{∗} y^{∗})(f_{m})| dν ≤ ||R^{∗}|| ||y^{∗}|| ||f_{m}||_{Pettis} .

Thus g_{m} → 0 in the Pettis norm on L_{1}(𝒴).

To see that g_{m} ↛ 0 weakly in L_{1}(𝒴), it suffices to show that H̃f_{m} ↛ 0 weakly in L_{1}(L_{∞}) where H̃ is the natural extension of H to an operator from L_{1}(ℓ_{1}) to L_{1}(L_{∞}). But since

H̃f_{m}(·) = \frac{1}{m} ∑_{n=1}^{m} ∑_{k=1}^{2^{n}} h_{k}^{n}(·)h_{k}^{n} ,

we may view H̃f_{m} as a function from Δ into C(Δ) and thus we only need to show that H̃f_{m} ↛ 0 weakly in L_{1}(C(Δ)). To see this, just consider the linear functional l ∈ [L_{1}(C(Δ))]^{∗} given by

l(f) = ∫_{Δ}[f(ω)](ω) dν where f ∈ L_{1}(C(Δ))

and note that

l(H̃f_{m}) = \frac{1}{m} ∫_{Δ} ∑_{n=1}^{m} ∑_{k=1}^{2^{n}} h_{k}^{n}(ω)h_{k}^{n}(ω) dν = 1 .

Thus the sequence {g_{m}} does all it is supposed to do. □

We close with one last example illustrating the above proposition in the case that 𝔛 is ℓ_{1}.

EXAMPLE 6. Once again working on the Lebesgue measure space on [0, 1], consider the sequence {g_{n}} of L_{1}(ℓ_{1}) functions where

g_{n}(·) = \frac{1}{n} ∑_{k=1}^{n} r_{k}(·)e_{k}

where {e_{k}} are the standard unit vectors of ℓ_{1} and {r_{k}} are the Rademacher functions.

The sequence is uniformly integrable, indeed it is even essentially bounded by 1. As for the Pettis norm of a g_{n}, fix x^{∗} ∈ (ℓ_{1})^{∗}, say x^{∗} = (α_{j}) ∈ ℓ_{∞}. Since

∫_{Ω} |x^{∗}g_{n}| dμ = \frac{1}{n} ∫_{Ω} | ∑_{k=1}^{n} α_{k}r_{k}| dμ ,

Khintchine’s inequality shows us that the Pettis norm of g_{n} behaves like \frac{1}{\sqrt{n}}. Thus g_{n} → 0 in the Pettis norm and so also in the σ(L_{1}(𝔛), L_{∞}(𝔛^{∗}))-topology.

However, {g_{n}} does not converge weakly. For consider the vector measure G: Σ → ℓ_{∞} given by

G(E) = (μ (E ∩ [r_{j} = 1])) .

Clearly G is a μ-continuous vector measure of bounded variation. Thus G gives rise to a functional l ∈ [L_{1}(ℓ_{1})]^{∗} given by

l(f) = ∫_{Ω} f dG for f ∈ L_{1}(ℓ_{1}) ,

where we view G as taking values in ℓ_{1}^{∗}. Since

l(g_{n}) = \frac{1}{n} ∑_{k=1}^{n} l(e_{k}r_{k}) = \frac{1}{n} ∑_{k=1}^{n} l(e_{k}1_{[r_{k}=1]} − e_{k}1_{[r_{k}=−1]}) = \frac{1}{n} ∑_{k=1}^{n} \frac{1}{2} −0 = \frac{1}{2} ,

we see that {g_{n}} does not converge weakly.

Note that the above linear functional l: L_{1}(ℓ_{1}) → ℝ is continuous for the Bochner norm but not for the Pettis norm. □

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University of South Carolina, Mathematics Department, Columbia, SC 29208