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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.

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    $\begingroup$ Does this have anything to do with strong vs. weak convergence? A comment on the linked page makes me ask. $\endgroup$
    – Doug
    Jun 12, 2015 at 2:28
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    $\begingroup$ (b) is Mazur's lemma en.wikipedia.org/wiki/Mazur's_lemma $\endgroup$
    – user99914
    Jun 12, 2015 at 2:30

2 Answers 2

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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 $h$.

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  • $\begingroup$ What do you mean by the supremum of a set of complex numbers? I thought maybe it was just a typo and you meant $\sup_{c \in C} |\left\langle h, c \right\rangle |$ but then your condition for $0 \in \overline{C}$ is trivial. $\endgroup$
    – Doug
    Jun 28, 2015 at 20:19
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    $\begingroup$ @Doug: let $\sigma_C(h) = \sup_{c \in C} \operatorname{re} \langle h, c \rangle$. $\endgroup$
    – copper.hat
    Jun 28, 2015 at 21:07
  • $\begingroup$ Would you mind giving some more details why the Hahn-Banach theorem implies if $\sigma_C(h) \geq 0$ for all $h$, then $0 \in \overline{C}$? Specifically, how convexity is used here. Many thanks. $\endgroup$
    – Doug
    Jun 29, 2015 at 3:16
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    $\begingroup$ @Doug: If $0$ is not in the closure, you can find a functional $\phi$ separating the compact set $\{0\}$ from the closure of $C$. That is, there are $\alpha, \beta$ such that $\operatorname{re} \phi(x) \le \alpha < \beta \le \operatorname{re} \phi(0) = 0$ for all $x \in \overline{C}$. That is, $\sigma_C(\phi) < 0$. $\endgroup$
    – copper.hat
    Jun 29, 2015 at 4:14
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    $\begingroup$ @Doug: No, you just need to show that every point of $A$ is contained in the closure of $B$. $\endgroup$
    – copper.hat
    Jun 29, 2015 at 16:14
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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 $\left\{ a_n \right\}$.

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