Sufficient condition for $0$ to be in a closed convex hull in a Hilbert space. 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.  
 A: I'm not sure if this answers your question, but it might help.
The idea here is that a closed convex set is completely characterised by its
support function.
Suppose $C \subset \mathbb{H}$ is a convex set. Let the support function be given by
$\sigma_C(h) = \sup_{c \in C} \langle h, c \rangle$.
Then $0 \in \overline{C}$ iff $\sigma_C(h) \ge 0$ for all $h$. One direction is obvious, the other direction follows from Hahn Banach.
If $C= \operatorname{co} \{ a_k \}$, then $\sigma_C(h) = \sup_k \langle h, a_k \rangle$.
Hence, if $\lim_k \langle h, a_k \rangle = 0$ for all $h$, then clearly
$\sigma_C(h) \ge 0$ for all $h$, hence $0 \in \overline{C}$.
Aside: In general, for convex sets $A,B$, we have $\overline{A} \subset \overline{B}$ iff
$\sigma_A(h) \le \sigma_B(h)$ for all $h$.
A: copper.hat's answer is spectacular.  John Ma's observation was also dead on, and provides a slick answer that I would like to elaborate on.  (I suspect that at their root both answers are equivalent.)  
Since we are in a Hilbert space $\mathcal{H}$, every linear functional is of the form $\langle \cdot, x \rangle$ for some $x \in \mathcal{H}$.  The hypotheses of the problem can then be rephrased as $$ \phi(a_n) \to 0 \text{ as } n \to \infty \text{ for all } \phi \in \mathcal{H}^*. $$  That is, $a_n$ converges to $0$ in the weak topology.  Hence, by Mazur's lemma there is a sequence of convex combinations of the $a_n$ that converge strongly to $0$.  That is, $0$ is in the closed convex hull of $\left\{ a_n \right\}$.  
