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Let $H$ be a complex Hilbert spaces and let $(x_n)$ be a sequence in $H$ which converges to $0$ weakly.

Question. Can $(x_n)$ be unbounded in the norm? In other words, is it possible that the sequence $(\| x_n\|)$ be unbounded?

In particular, I am wondering if $(e_n)_{n=1}^\infty$ is an orthonormal basis of $H$ then whether or not the sequence $e_1, 2e_2, 3e_3, \ldots$ converges to $0$ weakly.

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    $\begingroup$ @ChrisJanjigian So here is what I have after your hint. Let $phi_n:H*\to \mathbb C$ be the image of $x_n$ under the map $H\to H^*$. Then since $(x_n)$ converges weakly, we have $(\phi_n(f))_{n=1}^\infty$ is a bounded sequence for all $f\in H^*$. Therefore, by Banach-Steinhaus, $\| \phi_n\|$ is uniformly bounded. Since $\|x_n\|=\|\phi_n\|$, we are done. Is this okay? $\endgroup$ – caffeinemachine Nov 28 '15 at 20:06
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    $\begingroup$ You can test your sequence against $x=(1,1/2,1/3,\dots,1/n,\dots).$ $\endgroup$ – user940 Nov 28 '15 at 20:09
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No because in a Banach space weak topology has the same bounded sets as the topology induced by the norm. This relies on the Banach–Steinhaus theorem. Certainly weakly convergent sequences are bounded in the weak topology.

Let $X$ be a Banach space. Suppose that $A\subset X$ is bounded in the weak topology. Then $A$ regarded as a subset of $X^{**}$ is nothing but a pointwise-bounded set of continuous linear functionals on $X^*$. By the Banach–Steinhaus theorem it must be uniformly bounded, hence $A$ is norm-bounded as $X$ is embedded into $X^{**}$ isometrically.

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