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Suppose that $(X, \|.\|)$ is an infinite dimensional Banach space. I would like to ask whether we could construct a sequence $\{x_n^*\}_{n\in \mathbb{N}}\subset X^*$ (dual space of $X$) such that:

  • $\|x_n^*\|_{X^*}=1$;

  • $\{x_n\}$ is weakly convergent to $0$.

Example. Let $H=\ell_2$ be the real Hilbert space. Then $H^*=\ell_2$ and we can choose $\{x^n\}_{n\in\mathbb{N}}\subset H^*$ as $$ x^1=(1,0, \ldots, 0) $$ $$ x^2=(0,1, \ldots, 0) $$ $$ \vdots $$ $$ x^n=(0,\ldots,0,1,0,\ldots, 0) $$ $$ \vdots $$ Then $x^n\overset{w^*}{\rightarrow} 0$ and $\|x^n\|_{H^*}=1$.

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What you can see easily in general is that the weak$^*$ closure of $S_{X^*} = \{x^* \in X^* \mid \|x^*\| = 1\}$ contains $0$, as weak$^*$-open sets are so "large" that they will allways contain a line. But I doubt that there will allways be a sequence ... –  martini Oct 22 '12 at 11:23
@martini: How can we continue? –  blindman Oct 22 '12 at 11:29
You know that your claim is true? Is there allways a sequence? As I wrote, I doubt that sequences are sufficient ... –  martini Oct 22 '12 at 11:33
The question seems poorly expressed to me; are there typos in the phrase "$\{ x_n\}$ is weakly convergent to $0$"? What is $\{ x_n\}$ in relation to $\{ x_n^\ast\}$, and why do you change from weak convergence to weak$^\ast$ convergence? Anyway, have you heard of the Josefson-Nissenzweig theorem? It might be of interest, depending on what you mean by your question. –  Philip Brooker Oct 22 '12 at 13:52
@blindman: okay, but earlier you are talking about a Banach space $X$... Anyway, for the answer to your question see Chapter XII Joe Diestel's book Sequences and Series in Banach Spaces; the entire (short) chapter is devoted to the proof of the Josefson-Nissenzweig theorem. I hope this helps :-) . –  Philip Brooker Oct 23 '12 at 4:37

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