Is zero the liminf of the norm of a weakly convergence sequence to zero in HIlbert spaces?

Question

Let $X$ Hilbert and ${x_n} \subset X$ such that $x_n \rightharpoonup x \in X$ (where this is the weak convergence in Hilbert spaces). Now of course the following doesn't happen in general $$\lim_n |x_n-x|= 0$$

But if I take the liminf $$\liminf_{n} |x_n-x|=0$$ is this true?

Answer 1

No, consider $X=L^2(0,2\pi)$ with $x_n=e^{inx}\rightharpoonup0$ (this is Riemann-Lebesgue for Fourier series) but $\|x_n\|_{L^2}$ is constant in $n$.

Answer 2

Another (well, in some sense it is actually the same) nice counterexample is given using shifts. If $S\colon \mathcal{l}^2(\mathbb{Z})\to\mathcal{l}^2(\mathbb{Z})$ denotes the right-shift, i.e., $S$ maps a sequence $(x_n)_n\in\mathcal{l}^2(\mathbb{Z})$ to $(x_{n-1})_n$, then $\lim_nS^n= 0$ weakly, even though $S$ is a unitary operator (which means that the whole sequence consists of unitary operators). In particular one sees that the unitary operators are not closed w.r.t. the weak operator topology.

I like this version better, because to me it is more tangible and intuitively clear why the operators converge to $0$ w.r.t. the weak topology.