On the German Wikipedia page on the Banach-Saks property, they claim that every Banach space with the Banach-Saks property is reflexive but that the converse is not true. There should be a proof due to T. Nishiura and D. Waterman, but unfortunately, I can't find a proof of this interesting statement. It would be appreciated if someone could give me an online available reference (if possible). Unfortunately, my library at the university doesn't have a copy.

Thx for your help

math

  • 2
    Studia Math. and Fund. Math. + a dozen other worthwhile series (such as Banach's collected works) can be found on the homepage of the Polish Virtual Library. It takes some time to get used to browsing these sites but it's worth it... – t.b. Dec 24 '11 at 20:49
up vote 4 down vote accepted

Here is Nishiura and Waterman's paper.

Albert Baernstein gives a counter-example for the converse in the paper On reflexivity and summability, Studia Math, Volume 42, 91-94.

  • Thx for your link! I just wanted to be sure that the reference on Wikipedia was correct. – math Dec 24 '11 at 15:13

For the record: A Banach space $X$ has the Banach-Saks property if given a bounded sequence $(x_n)$ in $X$ there is a subsequence $(y_n)$ of $(x_n)$ such that the sequence $(\sigma_n)=(n^{-1}\sum\limits_{k=1}^n y_k)$ is norm convergent.

In Diestel's Sequences and Series in Banach spaces, the following outline for the proof that a Banach space with the Banach-Saks property is reflexive is given:

1) The Banach-Saks property is an isomorphic invariant.

2) If $X$ has the Banach-Saks property, then so do all of its closed linear subspaces.

3) $\ell_1$ does not have the Banach-Saks property.

4) If $(x_n)$ is weakly Cauchy and if ${\rm norm }\,\,\lim\limits_{n\rightarrow\infty} n^{-1} \sum\limits_{i=1}^n x_n$ exists, then $(x_n)$ is weakly convergent.

5) Conclude that a space with the Banach Saks property is reflexive.

Note by Rosenthal's $\ell_1$ theorem (Every bounded sequence in the Banach space $X$ has a weakly Cauchy subsequence if and only if $X$ contains no isomorphic copy of $\ell_1$), if $X$ has the Banach-Saks property, then by 1), 2), and 3), every bounded sequence has a weakly Cauchy subsequence. From 4), then, it follows that every bounded sequence in $X$ has a weakly convergent subsequence. Thus, the unit ball of $X$ is weakly compact by the Eberlein-Smulian Theorem; and so $X$ is reflexive.

  • Thx for your outline of the proof! But I accept the answer of Davide Giraudo, just because I needed the original paper of Nishiura and Waterman or another reference for a complete proof. I hope you understand. Nevertheless I upvoted also your answer. – math Dec 24 '11 at 15:11

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