Weak convergence in Hilbert space and $\|x\|=\liminf\|x_n\|$ implies strong convergence. The norm is (sequentially) weakly lower-semicontinuous: if 
$x_n$ converges weakly to $x$, then 
$\Vert x\Vert \leq \liminf_{n\to \infty}\Vert x_n\Vert$ ,
and this inequality is strict whenever the convergence is not strong.
This is from wikipedia and now I wonder how to prove the last statement. 
Thanks a lot.
 A: The first statement is well-known (see the link in comments).
The second statement seems to be false.
For example, let $(e_n)$ be a complete orthonormal system in $\ell^2$ (or in any separable Hilbert space), and define
$$
x_{2n} := e_n,
\qquad
x_{2n+1} := 0,
\qquad\forall n.
$$
Then $x_n \rightharpoonup 0$ and $\liminf \|x_n\| = 0$, but $(x_n)$ does not strongly converge to $0$.
What can be proved is that, if $\lim\|x_n\| = \|x\|$ and $x_n\rightharpoonup x$, then $x_n\to x$ strongly.
A: Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in a Hilbert space $\langle H, (.,.)\rangle$, which converges weakly to $x \in H$. Then we have
$$
\big|\big(x_n,\frac{x}{\|x\|}\big)\big| \leq \|x_n\|
$$
Applying the $\liminf_{n\in\mathbb{N}}$ on both sides yields
$$
\|x\| \leq \liminf_{n\in\mathbb{N}} \|x_n\|
,
$$
since $\big|\big(x_n,\frac{x}{\|x\|}\big)\big| \to \big|\big(x,\frac{x}{\|x\|}\big)\big| = \|x\|$.
Let $\|x\| = \liminf_{n\in\mathbb{N}} \|x_n\|$.
There is a subsequence of $(x_{n(i)})_{i\in\mathbb{N}}$ such that $\lim_{i\in\mathbb{N}}\|x_{n(i)}\| = \|x\|$. Clearly $(x_{n(i)})_{i\in\mathbb{N}}$ converges also weakly to $x$. Hence
\begin{align*}
\|x-x_{n(i)}\|^2 &= (x-x_{n(i)},x-x_{n(i)}) = (x,x) - 2(x_{n(i)},x) + (x_{n(i)},x_{n(i)}) \\
 &= \|x\|^2 + \|x_{n(i)}\|^2 - 2(x_{n(i)},x) \\ &\to \|x\|^2 + \|x\|^2 - 2(x,x) = 0
\end{align*}
So there is at least a subsequence $(x_{n(i)})_{i\in\mathbb{N}}$ which converges strongly to $x$. Unfortunately, we cannot get more, as Rigel showed in his answer.
