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Suppose $H$ is a Hilbert space, and we have a sequence $f_k \rightharpoonup f$ in $H^*$ (the sequence in $H^*$ converges weakly in $H^*$). This implies that $f_k \rightharpoonup^* f$ in $H^*$ (weak-star convergence).

I have a question. Is it possible to commute limit and suprema like this:

$$\lim_{k \to \infty}\sup_{w \in H, \\\lVert w \rVert = 1} \langle f_k, w \rangle_{H^*, H} = \sup_{w \in H\\ \lVert w \rVert = 1} \lim_{k \to \infty} \langle f_k, w \rangle_{H^*, H}$$

Usually I would immediately say no in general, but we do have a weak-star convergence, and I read that this is much nicer than the usual one, so I wonder if it holds?

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    For Hilbert spaces (and reflexive spaces in general) the weak topology on the dual and the weak* topology are the same. Also when you are taking the sup, you mean the sup on the unit ball, right? – 2017-01-17
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    @s.harp Yes that's right, on the unit ball. – 2017-01-17
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    In that case if you have a sequence of orthonormal elements $f_k$ you have that $f_k\rightharpoonup0$ and $\|f_k\|=1$. $\sup_{w\in B_1(0)}\langle f_k,w\rangle =\|f_k\|=1$, but $\lim_k \langle f_k,w\rangle = 0$ from definition for all $w$. So the statement is untrue in infinite dimensional Hilbert spaces. But it is true in finite-dimensional spaces. – 2017-01-17

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