ERRATA CORRECTION TO DENTABILITY, TREES, AND DUNFORD-PETTIS OPERATORS ON L_{1}

maria girardi zhibao hu

Pacific J. Math. 157 (1993) 389–394

Abstract. A Banach space has the complete continuity property if all its bounded subsets are midpoint Bocce dentable. We show that a lemma used in the original proposed proof of this result if false; however, we give a proof to show that the result is indeed true.

1. introduction

Throughout this paper, 𝔛 denotes an arbitrary Banach space, 𝔛^{∗} the dual space of 𝔛, B(𝔛) the closed unit ball of 𝔛, and S(𝔛) the unit sphere of 𝔛. The triple (Ω, Σ, μ) refers to the Lebesgue measure space on [0, 1], Σ^{+} to the sets in Σ with positive measure, and L_{1} to L_{1}(Ω, Σ, μ). The σ-field generated by a partition π of [0, 1] is σ(π). The conditional expectation of f ∈ L_{1} given a σ-field B is E(f|B).

A Banach space 𝔛 has the complete continuity property (CCP) if each bounded linear operator from L_{1} into 𝔛 is Dunford-Pettis (i.e. carries weakly convergent sequences onto norm convergent sequences). Since a representable operator is Dunford-Pettis, the CCP is a weakening of the Radon-Nikodým property (RNP). Recall that a Banach space has the RNP if and only if all its bounded subsets are dentable. A subset D of 𝔛 is dentable if for each ϵ > 0 there is x in D such that x ∉ \overline{co} (\{y ∈ D : ∥ x − y ∥≥ ϵ\}) . Midpoint Bocce dentability is a weakening of dentability. The subset D is midpoint Bocce dentable if for each ϵ > 0 there is a finite subset F of D such that for each x^{∗} in B(𝔛^{∗}) there is x in F satisfying:

if x = \frac{1}{2}z_{1} + \frac{1}{2}z_{2} with z_{i} ∈ D then | x^{∗}(x − z_{1}) | ≡ | x^{∗}(x − z_{2}) | < ϵ.

The following theorem is presented in [G1].

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 47B38, 46B20, 46B22, 28B99.

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theorem 1. 𝔛 has the CCP if each bounded subset of 𝔛 is midpoint Bocce dentable.

Our purpose in writing this note is to show that Lemma 2.9 in [G1] (which was used in [G1] to prove Theorem 1) is false and to provide a proof of the theorem. Lemma 2.9 asserts that if A is in Σ^{+} and f in L_{∞}(μ) is not constant a.e. on A, then there is an increasing sequence {π_{n}} of positive finite measurable partitions of A such that σ(∪π_{n}) = Σ ∩ A and for each n

μ ( ⋃ \{E : E ∈ π_{n} and \frac{∫_{E} f dμ}{μ(E)} ≥ \frac{∫_{A} f dμ}{μ(A)} \}) = \frac{μ(A)}{2} .

Example 2 shows that Lemma 2.9 is false.

EXAMPLE 2. Let f = 3χ_{[0, \frac{1}{4} )} − χ_{[ \frac{1}{4} ,1]}. Then ∫_{Ω} f dμ = 0. Suppose that \{π_{n}\} is an increasing sequence of positive finite measurable partitions of [0, 1] such that for each n

μ ( ⋃ \{E : E ∈ π_{n} and \frac{∫_{E} f dμ}{μ(E)} ≥ 0 \}) = \frac{1}{2} .

Then σ(∪π_{n}) ≠ Σ.

proof. Consider the martingale {f_{n}} given by

f_{n}(·) = E(f|σ(π_{n})) = ∑_{E∈π_{n}} ^{\frac{∫_{E} f dμ}{μ(E)}} χ_{E} (·) .

For each n ∈ ℕ put

P_{n} = ⋃ \{E : E ∈ π_{n} and ∫_{E} f dμ ≥ 0 \} and Q_{n} = P_{n} ∩ (\frac{1}{4}, 1] .

Since μ(P_{n}) = \frac{1}{2}, we have that μ(Q_{n}) ≥ \frac{1}{4}. Thus

∫_{Ω} | f_{n} − f | dμ ≥ ∫_{Q_{n}} | f_{n} − f | dμ ≥ ∫_{Q_{n}}(f_{n} − −1) dμ
≥ ∫_{Q_{n}} 1 dμ = μ(Q_{n}) ≥ \frac{1}{4} .

We know that such a martingale E(f|σ(π_{n})) converges in L_{1} norm to E(f|σ(∪π_{n})). But E(f|Σ) = f. Thus σ(∪π_{n}) ≠ Σ. □

The error in the proof of Lemma 2.9 occurred in assuming that if A is in Σ^{+} and {π_{n}} is an increasing sequence of positive measurable partitions of A such that for

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each n and each E in π_{n} the μ(E) ≤ ϵ_{n} with lim_{n} ϵ_{n} = 0, then σ(∪π_{n}) = Σ∩A. This seemingly sound assertion is not true as shown by the following counterexample. EXAMPLE 3. For n ∈ ℕ and 1 ≤ i ≤ 2^{n}, define

E_{i}^{n} = [\frac{i − 1}{2^{n+1}}, \frac{i}{2^{n+1}} ) ∪ [\frac{1}{2} + \frac{i − 1}{2^{n+1}}, \frac{1}{2} + \frac{i}{2^{n+1}} )

and

π_{n} = {E_{i}^{n} : 1 ≤ i ≤ 2^{n}} .

Clearly {π_{n}} is an increasing sequence of positive measurable partitions of [0, 1] such that for each n and each E in π_{n} the μ(E) = 2^{−n}. Let f = χ_{[0, \frac{1}{2} )}. An easy computation shows that E(f|σ(π_{n})) = \frac{1}{2}χ_{[0,1]}. We know that such a martingale E(f|σ(π_{n})) converges in L_{1} norm to E(f|σ(∪π_{n})). But E(f|Σ) = f. Thus σ(∪π_{n}) ≠ Σ. □

2. proof of theorem

Our proof of Theorem 1 uses the following observations. For f in L_{1} and A in Σ, the average value and the Bocce oscillation of f on A respectively are

m_{A}(f) ≡ \frac{∫_{A} f dμ}{μ(A)}

and

Bocce-osc f|_{A} ≡ \frac{∫_{A} |f − m_{A}(f)| dμ}{μ(A)}

observing the convention that 0/0 is 0. LEMMA 4. Fix A in Σ and f in L_{1}. There is a subset E of A with 2 μ(E) = μ(A) and

\frac{1}{2} Bocce-osc f|_{A} ≤ | m_{E}(f) − m_{A}(f) | .

Furthermore, for each subset E of A with 2 μ(E) = μ(A),

| m_{E}(f) − m_{A}(f) | ≤ Bocce-osc f|_{A} .

proof. Without loss of generality, A = Ω and ∫_{Ω} f dμ = 0 and ∫_{Ω} | f | dμ = 1. With this normalization, Bocce-osc f|_{A} = 1 and | m_{E}(f) − m_{A}(f) | = | m_{E}(f) | . Let P = [f ≥ 0] and N = [f < 0].

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The first claim now reads that \frac{1}{2} ≤ 2 | ∫_{E} f dμ | for some subset E of measure one half. Wlog μ(P) ≥ \frac{1}{2}. Partition P into 2 sets, P_{1} and P_{2}, of equal measure such that ∫_{P_{2}} f dμ ≤ ∫_{P_{1}} f dμ. Note that

1 = ∫_{Ω} | f | dμ = ∫_{P} f dμ + ∫_{N} −f dμ
= 2 ∫_{P} f dμ = 2 [∫_{P_{1}} f dμ + ∫_{P_{2}} f dμ] ≤ 4 ∫_{P_{1}} f dμ .

Since μ(P_{1}) ≤ \frac{1}{2} ≤ μ(P), we can find a set E such that P_{1} ⊂ E ⊂ P and μ(E) = \frac{1}{2}. For such a set E

\frac{1}{4} ≤ ∫_{P_{1}} f dμ ≤ ∫_{E} f dμ ,

as needed.

Normalized, the second claim reads that for each subset E of measure \frac{1}{2}

2 |∫_{E} f dμ| ≤ 1 .

Fix a subset E of measure \frac{1}{2}. Wlog ∫_{E∩N} −f dμ ≤ ∫_{E∩P} f dμ . So

|∫_{E} f dμ| = |∫_{E∩P} f dμ + ∫_{E∩N} f dμ| ≤ |∫_{E∩P} f dμ| ≤ ∫_{P} | f | dμ = \frac{1}{2} ,

as needed. □

A subset K of L_{1} satisfies the Bocce criterion if for each ϵ > 0 and B in Σ^{+} there is a finite collection F of subsets of B each with positive measure such that for each f in K there is an A in F satisfying

(∗)  Bocce-osc f|_{A} < ϵ .

Lemma 4 provides an equivalent formulation of the Bocce criterion; namely we can replace condition (∗) by the condition (∗∗) if the subset E of A has half the measure of A, then | m_{E}(f) − m_{A}(f) | < ϵ .

We now attack the proof of Theorem 1. Our proof follows mainly the proof in [G1].

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proof of theorem 1. Let all bounded subsets of 𝔛 be midpoint Bocce dentable. Fix a bounded linear operator T from L_{1} into 𝔛. It suffices to show that the subset T^{∗}(B(𝔛^{∗})) of L_{1} satisfies the Bocce criterion (this is a necessary and sufficient condition for T to be Dunford-Pettis [G2]). To this end, fix ϵ > 0 and B in Σ^{+}.

Consider the vector measure F from Σ into 𝔛 given by F (E) = T (χ_{E}). For x^{∗} ∈ 𝔛^{∗}

m_{E}(T^{∗}x^{∗}) = \frac{x^{∗}F (E)}{μ(E)}

since ∫_{E}(T^{∗}x^{∗}) dμ = x^{∗}T (χ_{E}) = x^{∗}F (E) .

Since the subset \{\frac{F (E)}{μ(E)} : E ⊂ B and E ∈ Σ^{+}\} of 𝔛 is bounded, it is midpoint Bocce dentable. Accordingly, there is a finite collection F of subsets of B each in Σ^{+} such that for each x^{∗} ∈ B(𝔛^{∗}) there is a set A in F such that if

\frac{F (A)}{μ(A)} = \frac{1}{2} \frac{F (E_{1})}{μ(E_{1})} + \frac{1}{2} \frac{F (E_{2})}{μ(E_{2})}

for some subsets E_{i} of B with E_{i} ∈ Σ^{+}, then

|\frac{x^{∗}F (E_{1})}{μ(E_{1})} − \frac{x^{∗}F (A)}{μ(A)}| = |\frac{x^{∗}F (E_{2})}{μ(E_{2})} − \frac{x^{∗}F (A)}{μ(A)}| < ϵ .

Fix x^{∗} ∈ B(𝔛^{∗}) and find the associated A in F.

At this point we turn to our new formulation of the Bocce criterion (whereas [G1] used the old formulation and Lemma 2.9).

This A ∈ F satisfies the condition (∗∗). For consider a subset E of A with μ(E) = \frac{1}{2}μ(A). Since

\frac{F (A)}{μ(A)} = \frac{1}{2} \frac{F (E)}{μ(E)} + \frac{1}{2} \frac{F (A \ E)}{μ(A \ E)}

we have that

| m_{E}(T^{∗}x^{∗}) − m_{A}(T^{∗}x^{∗}) | ≡ |\frac{x^{∗}F (E)}{μ(E)} − \frac{x^{∗}F (A)}{μ(A)}| < ϵ .

Thus T^{∗}(B(𝔛^{∗})) satisfies the Bocce criterion, as needed. □

3. closing comments

A relatively weakly compact subset of L_{1} is relatively norm compact if and only if it satisfies the Bocce criterion [G2]. Thus our new formulation of the Bocce

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criterion provides another (perhaps at times more useful) method for testing for norm compactness in L_{1} .

Fix A in Σ^{+} and f in L_{1}. Put

M_{A}(f) = sup {| m_{E}(f) − m_{A}(f) | : E ⊂ A and 2 μ(E) = μ(A)} .

This supremum is obtained. For just normalize so that A = Ω and ∫_{Ω} f dμ = 0 and ∫_{Ω} | f | dμ = 1. As Ralph Howard pointed out, next find disjoint subsets E_{1} and E_{2} of measure \frac{1}{2} and a ∈ ℝ such that

E_{1} ⊂ [f ≤ a]  and  E_{2} ⊂ [f ≥ a] .

Then M_{A}(f) will be the larger of | m_{E_{1}}(f) | and | m_{E_{2}}(f) | .

Basically, our lemma 4 says that

\frac{1}{2} Bocce-osc f|_{A} ≤ M_{A}(f) ≤ Bocce-osc f|_{A} .

These bounds are the best possible.

For the second inequality, consider the function defined on A ≡ [0, 1] by

f = χ_{[0, \frac{1}{2} )} − χ_{[ \frac{1}{2} ,1]} .

Straightforward calculations show that m_{[0, \frac{1}{2} ]}(f) = 1 and that Bocce-osc f|_{A} = 1. Thus

M_{A}(f) = Bocce-osc f|_{A} .

As for the first inequality, consider the family of functions defined on A ≡ [0, 1] by

f_{δ} = \frac{δ − 1}{δ} χ_{[0,δ)} + χ_{[δ,1]}

for 0 < δ < \frac{1}{2} . Straightforward calculations show that

M_{A}(f_{δ}) = \frac{1}{2 (1 − δ)} Bocce-osc f_{δ}|_{A} .

Actually M_{A}(f) = \frac{1}{2} Bocce-osc f|_{A} if and only if f is the zero function on A.

References

[B]. R.D. Bourgin, Geometric aspects of convex sets with the Radon-Nikodým property, Lecture

Notes in Mathematics, vol. 933, Springer-Verlag, Berlin and New York, 1983. [DU]. J. Diestel and J. J. Uhl, Jr., Vector Measures, Math. Surveys, no. 15, Amer. Math. Soc.,

Providence, R.I., 1977. [G1]. Maria Girardi, Dentability, Trees, and Dunford-Pettis operators on L_{1}, Pacific J. Math.

148 (1991), 59–79. [G2]. Maria Girardi, Compactness in L_{1}, Dunford-Pettis Operators, Geometry of Banach Spac-

es, Proc. Amer. Math. Soc. 111 (March 1991), 767–777.

girardi: Dept. of Math., University of South Carolina, Columbia, SC 29208 hu: Dept. of Math. and Statistics, Miami University, Oxford, OH 45056