Weak star compactness and the sphere

Question

If $X$ is a dual space, the ball $B_X$ is weak star compact. Is the sphere generally weak star compact?

I think the answer is no because in $\ell^2$, the sequence $(e_n)_n$ converges weakly to zero, and so no subsequence converges weakly to an element with norm one.

The context for the question is the following statement from Wikipedia: Let $X$ be a $C^*$ algebra and with unit. The set of states (continuous linear functionals $f$ on $X$ such that $||f||=1$ and $f(x)\geq 0$ when $x\geq 0$) is a weakly compact set.

Your argument is right; unit sphere in $\ell^{2}$ is not weakly compact. (weak=weak$^{*}$ by reflexivity. — Kavi Rama Murthy, Dec 13 '18 at 06:33

BigbearZzz · Answer 1 · 2018-12-13T17:07:48.877

3

You are correct.

For a reflexive space $X$, it is not difficult to prove that the weak closure of the unit sphere $S_{X^*}$ is the closed unit ball $B_{X^*}$ (you can read the proof here). Weak and weak* topology coincide on the dual space of a reflexive space.

edited Dec 13 '18 at 17:07

answered Dec 13 '18 at 06:53

BigbearZzz

15,530

score 2 · Accepted Answer · answered Dec 13 '18 at 07:12

You are right. The sphere is not in general weak star compact. Thus, the family of states is a special case. To see this, we can show that $f$ is a state if and only if $f$ is a positive linear functional such that $$ f(1_X) = 1 $$ where $1_X$ is the unit of $X$. That is, every state attains its norm at the unit and this condition is also sufficient. Since weak star limit $f$ of sequence of states $f_n$ is still positive and $$ f(1_X) = \lim_{n\to\infty} f_n(1_X) = 1,$$ it holds that $f$ is also a state.

Weak star compactness and the sphere

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