I am working on the following problem:
Let $\mathcal{H}$ be a Hilbert space, let $\left\{a_n\right\}_{n=1}^\infty \subset \mathcal{H}$ be a sequence such that $||a_n|| = 1$, and consider the closed convex hull of $\left\{ a_n \right\}$, $$\mathcal{C} = \overline{\left\{ \sum_{i}^{< \infty} c_i a_i \;\colon\; c_i \geq 0 \text{ and } \sum c_i = 1 \right\}}.$$
(a) If $\mathcal{H} = \text{span}\left\{ a_n \right\}$, then $\mathcal{H}$ is finite-dimensional. (b) $$\text{If } \lim_{n \to \infty} \left<a_n, x\right> = 0 \text{ for all } x \in \mathcal{H}, \text{ then } 0 \in \mathcal{C}.$$
I already proved (a). I presume that (a) might have something to do with proving (b), but I am not sure. Notice that if $\mathcal{H}$ is finite-dimensional, then the hypotheses of (b) are absurd, for they force $||a_n|| \to 0$. I thought then maybe I could approach this by contradiction and show that if we assume $0 \notin \mathcal{C}$, then $\mathcal{H} = \text{span}\left\{a_n\right\}$ (or, at least, $\text{span}\left\{a_n\right\}$ is closed and thus a Hilbert space), and we would be done, but I haven't made any progress on this front. Another thought I had was to just try to write down explicitly a sequence in $\mathcal{C}$ that converges to $0$. The most obvious thing feels to me something like $$x_n = \frac{1}{n} \sum_{i = 1}^n a_i.$$ Then $$||x_n||^2 \leq \frac{1}{n} + \frac{2}{n^2} \sum_{i \neq j} |\left< a_i, a_j \right>|. $$ We would be done if we could show $|\left< a_i, a_j \right>| < \epsilon$ for all $i \neq j$ and $i$, $j$ sufficiently large, but my question here
uniformly convergent subsequence of bounded linear operators on a Hilbert space?
seems to preclude this method, even after restricting to subsequences to avoid pathological situations. So I'm stuck. Any help would be appreciated.
-Thanks.
EDIT: I just caught a mistake, I should have written $$||x_n||^2 \leq \frac{1}{n} + \frac{2}{n^2} \sum_{i < j} |\text{Re}\left< a_i, a_j \right>|. $$ This shouldn't fix anything though, because Martin's answer on the page linked above still gives a nice counterexample.