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.


  • $\begingroup$ Do you mean without any weak* convergent subsequence? $\endgroup$ – K.Power Feb 22 '19 at 20:40
  • $\begingroup$ 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. $\endgroup$ – rarc Feb 22 '19 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.

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  • $\begingroup$ 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. $\endgroup$ – rarc Feb 22 '19 at 21:24
  • $\begingroup$ 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. $\endgroup$ – K.Power Feb 22 '19 at 21:27
  • $\begingroup$ Oke! I see now, so it comes directly from the right choice of the element $x \in \ell ^\infty$. Right? $\endgroup$ – rarc Feb 22 '19 at 21:32
  • 2
    $\begingroup$ 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 $x$. $\endgroup$ – K.Power Feb 22 '19 at 21:35
  • $\begingroup$ Amazing answer. Thanks. +up $\endgroup$ – rarc Feb 22 '19 at 21:37

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