On J. Borwein’s Concept of Sequentially Reflexive Banach Spaces
먼저, ℓ1 subspace가 X에 존재하지 않는다면 X는 Borwein 순서 반사성(sequential reflexivity)인 것으로 보인다.
반대의 경우를 고려해 보면, 만약 Y가 ℓ1과 동형이며 Banach 공간 X의 부분공간이라면, X는 Borwein 순서 반사성이 아니라는 결론을 얻는다.
결과적으로, 본 논문에서는 Borwein의 순차적 반사성을 만족하는 Banach 공간이 ℓ1 subspace를 포함하지 않는다는 것을 증명한다.
ℓ1 subspace가 존재하면 Borwein의 순차적 반사성은 유지되지 않는다.
On J. Borwein’s Concept of Sequentially Reflexive Banach Spaces
arXiv:math/9201233v1 [math.FA] 9 Oct 1991On J. Borwein’s Concept of Sequentially Reflexive Banach Spacesby P. ØrnoA Banach space X is reflexive if the Mackey topology τ(X∗, X) on X∗agrees with thenorm topology on X∗. Borwein [B] calls a Banach space X sequentially reflexive providedthat every τ(X∗, X) convergent sequence in X∗is norm convergent.
The main result in[B] is that X is sequentially reflexive if every separable subspace of X has separable dual,and Borwein asks for a characterization of sequentially reflexive spaces. Here we answerthat question by provingTheorem.
A Banach space X is sequentially reflexive if and only if ℓ1 is not isomorphicto a subspace of X.Proof: Assume first that ℓ1 is not isomorphic to a subspace of X and let {x∗n}∞n=1be a weak∗-null sequence in X∗for which the sequence {< x∗n, xn >}∞n=1 converges tozero for every weakly null sequence {xn}∞n=1 in X; by the easy Lemma 2.1 in [B] it isenough to check that such a sequence {x∗n}∞n=1 must converge in norm to zero. If not,by passing to a subsequence we can select a sequence {xn}∞n=1 in the unit ball of X with{< x∗n, xn >}∞n=1 bounded away from zero.By passing to a further subsequence, wecan assume by Rosenthal’s theorem [R], [D, chpt.
XI] on Banach spaces which do notcontain isomorphs of ℓ1 that {xn}∞n=1 is weakly Cauchy. Since {x∗n}∞n=1 converges weak∗to zero, by passing to further subsequences and replacing {xn}∞n=1 with a subsequenceof differencesx2n−x2n−12, we can assume moreover that {xn}∞n=1 is weakly null.Thiscontradiction completes the proof of the first direction.To go the other way, suppose that Y is a subspace of X which is isomorphic to ℓ1and let {en}∞n=1 be the image of the unit vector basis under some isomorphism from ℓ1onto Y .
Define a bounded linear operator from Y into L∞[0, 1] by mapping en to the n-thRademacher function rn. By the injective property of L∞[0, 1], this operator extends to abounded linear operator T from X into L∞[0, 1].
Let r∗n be the n-th Rademacher function inL1[0, 1] considered as a subspace of L∞[0, 1]∗. Thus the sequence {r∗n}∞n=1, being equivalentto an orthonormal sequence in a Hilbert space, converges weakly to zero.
Since L∞[0, 1] hasthe Dunford-Pettis property (cf. [D, p. 113]), {r∗n}∞n=1 converges in the Mackey topology tozero, a fortiori {T ∗r∗n}∞n=1 converges τ(X∗, X) to zero.
But < T ∗r∗n, en >=< r∗n, rn >= 1,so {T ∗r∗n}∞n=1 does not converge to zero in norm.References[B] J. Borwein, Asplund spaces are “sequentially reflexive”, (preprint). [D] J. Diestel, Sequences and series in Banach spaces, Springer-Verlag Graduate Texts inMathematics 92, (1984).
[R] H. P. Rosenthal, A characterization of Banach spaces containing ℓ1, Proc.Nat.Acad. Sci.
71, (1974), 2411–2413.
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