Subset of Hilbert space contains zero Let $H$ be a Hilbert space with ONB $\{b_1, b_2, \cdots \}$. Let $x_n = \frac{1}{n}\sum_{i=1}^{n^2} b_i$.
Let $A=\overline{\operatorname{co}(\{x_1, x_2, x_3, \cdots \})}$, the closure of the convex hull of the $x$'s.

I want to prove that $0\in A$.

So I need to find a sequence $y_n\in \operatorname{co}(\{x_1, x_2,\cdots \})$ such that $\| y_n \|\to 0$. I've tried all that I could think of, e.g. $y_n = 1/n x_n + (1-1/n) x_{n-1}$, but in each case I found $\|y_n\| \to 1$.
Any hint would be MUCH appreciated.

Hint: Say $f,g$ are in some $L^2$ space, $||f||_2=||g||_2=1$, and $f$ and $g$ have disjoint support. Then $||(f+g)/2||_2=1/\sqrt 2$. Similarly for $(f_1+\dots+f_n)/n$.
Now if $n$ is much much larger than $m$ then your $x_n$ and $x_m$ "almost" have disjoint support. Show that $$\lim_{n\to\infty}\langle x_n,x_m\rangle=0.$$
