# Weak convergence, together with convergence of norms, implies strong convergence in a Hilbert space.

Let $(x_n)$ be a weakly convergent sequence in a Hilbert space $H$. If $\| x_n \| \to \| x \|$, show that $x_n$ converges strongly to $x$.

### Context

This problem comes from a question in my exam paper; the original problem was incorrect.

• So every sequence of unit vectors converges to every unit vector? – Jonas Meyer Oct 22 '13 at 19:51
• Clearly false. Easy to build a counterexample: let $v\in H$, $v\ne 0$, then take $X_n=(-1)^n v$. – TZakrevskiy Oct 22 '13 at 19:51
• Thank you TZakrevskiy, thats correct ! That was question in my exam paper.. and was incorrect. – Ricardo Gomes Oct 31 '13 at 19:59
• I edited the question into what it should have been, so that it can serve as a reference for this fact. – user147263 Jan 5 '15 at 3:17
• Does strong convergence implies convergence in norm? – manhattan Mar 14 '17 at 6:19

The result you want to show should be: if $x_n$ converges to $x$ weakly and $\lVert x_n\rVert\to \lVert x\rVert$, then there is convergence in norm. To see that, expand $\lVert x_n-x\rVert^2$ and use the fact that $\langle x_n,x\rangle\to \lVert x\rVert^2$.