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.

  • $\begingroup$ Are you sure it's supposed to be the norm closure and not maybe the weak closure? $\endgroup$ Jan 11, 2018 at 15:27
  • $\begingroup$ Yes the norm closure, @DavidC.Ullrich $\endgroup$
    – user519975
    Jan 11, 2018 at 15:29
  • 1
    $\begingroup$ Assume $0 \notin A$. Reread Hahn-Banach. $\endgroup$ Jan 11, 2018 at 15:37
  • $\begingroup$ @DanielFischer Touche... $\endgroup$ Jan 11, 2018 at 15:55
  • $\begingroup$ Which version of Hahn Banach do you refer to @DanielFischer ? $\endgroup$
    – user519975
    Jan 11, 2018 at 16:10

1 Answer 1


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 $$\lim_{n\to\infty}\langle x_n,x_m\rangle=0.$$

  • $\begingroup$ No! The point is you need to take $n_2$ muuch larger than $n_1$, then $n_3$ mmuch much larger than $n_2$, and consider $y=(x_{n_1}+\dots x_{n_N})/N$. If the $x_{n_j}$ were orthogonal what would $||y||$ be? You can make them as close to orthogonal as you want... $\endgroup$ Jan 11, 2018 at 17:04
  • $\begingroup$ It's a norm-closed, hence weakly closed, subset of the closed unit ball, so it's weakly compact. Not norm compact, since $(x_n)$ has no norm-convergent subsequence (since $x_n$ and $x_m$ are close to orthogonal if $m$ is much larger than $n$, so that $||x_n-x_m||$ is close to $\sqrt 2$.) If this is in regard to Daniel Fischer's comments you might note that $\{0\}$ is compact. $\endgroup$ Jan 12, 2018 at 16:12

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