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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 $0$

Please help . Thanks in advance

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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.

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

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