Bounded sequence in dual with any weak*convergent subsequence

Let $$(E,\|\cdot\|_E)$$ be a separable normed vector space over a field $$\mathbb{K}$$ and $$E'$$ its dual.

Theorem: Each bounded sequence in $$E'$$ has a weak-* convergent subsequence.

Question: With $$(E,\|\cdot\|_E)$$ not being separable, provide an example of a bounded sequence in $$E'$$ without any weak-* convergent subsequence.

I thought about an example on the space of continuous functions over a non compact set $$(C((0,1)),\|\cdot\|_\infty)$$, where $$\|\cdot\|_\infty$$ is the supremum's norm.

But any other would be highly appreciated.

Thanks.

• Do you mean without any weak* convergent subsequence? – K.Power Feb 22 at 20:40
• Exactly! But to clarify... $\{ \varphi_n\}\rvert_{n \in \mathbb{N}} \subset E'$ is a bounded sequence of linear functionals in the dual space $E'$. with $(E,\| \cdot \|_E)$ not separable. – mathematifier Feb 22 at 20:46

Consider the sequence $$\{f_n\}\subset(\ell^\infty)^*$$ where each $$f_n$$ is the standard coordinate functional. I.e $$f_n(e_j)=1$$ if $$n=j$$ and $$0$$ otherwise. Clearly $$\|f_n\|=1$$ for all $$n\in \mathbb N$$, so it is a bounded sequence. Now we know that for $$\{f_n\}$$ to have a weak* convergent subsequence we would require some subsequence $$\{f_{n_k}\}$$ such that $$f_{n_k}(x)\to f(x)$$ for some $$f\in (\ell^\infty)^*$$ for all $$x\in \ell^\infty$$. However, for any subsequence $$\{f_{n_k}\}$$ we can construct the element $$x_{n_k}\in \ell^\infty$$ by $$f_n(x_{n_k})=0$$ if $$n\neq n_k$$ for some $$k\in \mathbb N$$, and $$f_{n_k}(x_{n_k})=(-1)^k$$. Clearly $$(f_{n_k}(x_{n_k}))\subset \mathbb R$$ is not convergent, so $$\{f_{n_k}\}$$ cannot be convergent. As this is true for any subsequence we conclude that $$\{f_n\}$$ has no convergent subsequence.
• Dear sir. Thanks for this answer. I have two questions. First: at the first line, did you mean "the sequence $\{f_n\} \subset (\ell^\infty)'$"? Second: How do you come up with $f_{n_k}(x_{n_k}) = (-1)^k$? Thanks. – mathematifier Feb 22 at 21:24
• The first is a typo, which I've corrected. To answer the second point I am just using the coordinate functionals to define $x_{n_k}$. In words $x_{n_k}$ has entries $(-1)^k$ at the $n_k$-th coordinate, and $0$ at all other coordinates. – K.Power Feb 22 at 21:27
• Oke! I see now, so it comes directly from the right choice of the element $x \in \ell ^\infty$. Right? – mathematifier Feb 22 at 21:32
• Exactly. If for each subsequence $\{f_{n_k}\}$ we can find an $x\in \ell^\infty$ such that $f(x_{n_k})$ does not converge then $\{f_{n_k}\}$ cannot be weak* convergent. The $x_{n_k}$ I chose is such an $x$. – K.Power Feb 22 at 21:35