A subset of the norm dual of a normed space is weak* compact if and only if it is weak* closed and norm bounded.

This is stated without a proof in:

Infinite Dimensional Analysis A Hitchhiker's Guide

Authors: Aliprantis, Charalambos D., Border, Kim Page 235 6.21.

I know that this is true for Banach spaces and one direction from weak* closed and norm bounded to weak* compact is also true. But I´m not sure if a subset of of the norm dual which is weak* compact is also norm bounded.

If not does exist a counterexample?

  • $\begingroup$ @ David C.Ullrich can u elaborate this? I tried it myself like: weak* compact sets are weak* bounded hence pointwise bounded but for employing UBP we need that X is a Banach space. Am I missing something? $\endgroup$
    – H.K.
    Feb 27, 2020 at 15:13
  • $\begingroup$ @WoolierThanThou: The uniform boundedness principle isn't valid when $X$ is not complete. $\endgroup$ Feb 27, 2020 at 15:56
  • 1
    $\begingroup$ @WoolierThanThou: But your sequence $x_n$ lives in $X^*$, not in $X$. $\endgroup$ Feb 27, 2020 at 15:58

1 Answer 1


Meanwhile I think I found a counterexample

Let $X= c_{00}$ the space of finitly supportet sequences endowed with the sup norm and define for $m\in \mathbb{N}$

$$ \delta_m: c_{00}\to \mathbb{R}; \delta_m((a_n)_n) = ma_m. $$ Then $(\delta_m)_m\subset X'$ converges weakly* to $0$. Hence $\{\delta_m:m\in \mathbb{N}\}\cup \{0\}$ is weak* compact since X' endowed with the weak* topology is a hausdorff TVS. But clearly $\{\delta_m:m\in \mathbb{N}\}$ is not norm bounded since for $m\in \mathbb{N}$ $$ ||\delta_m||_{op}=m. $$


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