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I know the following theorem from the lecture:

Let $X$ be a seperable Banachspace. Then $\overline{B(0)}$ is weakly* sequentially compact in $X'$.

Since it is specified that $X$ has to be separable, I want to look at an example where $\overline{B_1(0)}$ is not necessarily weakly* sequentially compact in $X'$, if we choose a Banachspace $X$ that is not separable. I found out from a book that $\overline{B_1(0)}$ is not weakly* sequentially compact in $(l^\infty)'$. $l^\infty$ is not separable ( I showed that already), but how can we now show that $\overline{B_1(0)}$ is not weakly* sequentially compact in $(l^\infty)$'?

Yasuduck
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1 Answers1

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One way to show a space is not sequentially compact is to explicitly give an example of a sequence without any convergent subsequence.

Here, we can look at the sequence $(\varphi_n)_{n\in \mathbb{N}}$, where $\varphi_n(x) = x_n$ for all $x \in \ell^{\infty}(\mathbb{N})$. Clearly $\varphi_n \in \overline{B_1(0)}$ for every $n$. To see that $(\varphi_n)$ contains no weak$^{\ast}$-convergent subsequence, consider an arbitrary subseqence $(\varphi_{n_k})_{k \in \mathbb{N}}$. Let $M = \{ n_k : k \in \mathbb{N}\}$, and define $x\in \ell^{\infty}(\mathbb{N})$ by

$$x_n = \begin{cases}\quad 0 &, n \notin M, \\ (-1)^k &, n = n_k.\end{cases}$$

Then $\varphi_{n_k}(x) = x_{n_k} = (-1)^k$, so $\lim\limits_{k\to\infty} \varphi_{n_k}(x)$ doesn't exist, whence $(\varphi_{n_k})$ is not a weak$^{\ast}$-convergent sequence. So no subsequence of $(\varphi_n)$ is weak$^{\ast}$-convergent, and it follows that $\overline{B_1(0)}$ is not weak$^{\ast}$-sequentially compact.

Daniel Fischer
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