# Can A Weakly Convergent Sequence Be Unbounded in Norm

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.

• @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? – caffeinemachine Nov 28 '15 at 20:06
• You can test your sequence against $x=(1,1/2,1/3,\dots,1/n,\dots).$ – user940 Nov 28 '15 at 20:09

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.