# $\{x_n\}$ sequence in Hilbert space st for every $x$ , $\langle x_n,x \rangle\to 0$ , then some subsequence of $\{x_n\}$ converge to $0$?

Let $$\{x_n\}$$ be a sequence in a Hilbert space $$H$$ such that for every $$x \in H$$ , the complex sequence $$\{\langle x_n,x\rangle\}$$ converges to $$0$$ . Then is it true that $$\{x_n\}$$ has a subsequence which converges to $$0$$ ? If not true in general , is it at least true in $$l^2 (\mathbb N)$$ ?

The only things I can show are : $$\{x_n\}$$ is bounded , hence has a weakly convergent subsequence . Moreover , there is a subsequence $$\{x_{k_n}\}$$ of $$\{x_n\}$$ such that $$\Big\{\dfrac {x_{k_1}+...+x_{k_n}}{n}\Big\}$$ converges to $$0$$

Let $x_n=e_n$ be a standard orthonormal basis for the Hilbert space. Then the sequence of inner products with each fixed vector tends to zero but the sequence does not converge.

• "in fact is eventually constant 0" Really?
– zhw.
Mar 8 '17 at 16:52
• Sorry, that was silly. Mar 8 '17 at 16:54