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.
No, because in a Banach space the 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.