# If X ist not Banach are weak* compact sets always norm bounded

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?

• @ 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?
– H.K.
Feb 27, 2020 at 15:13
• @WoolierThanThou: The uniform boundedness principle isn't valid when $X$ is not complete. Feb 27, 2020 at 15:56
• @WoolierThanThou: But your sequence $x_n$ lives in $X^*$, not in $X$. Feb 27, 2020 at 15:58

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 $$m\in \mathbb{N}$$ $$||\delta_m||_{op}=m.$$