Functional Analysis – Do Weak Convergence and Convergence of Norms Imply Convergence in L^2?

Question

Let $(f_n)_n\subseteq L^2(0,1)$ s.t.
$$
f_n \rightharpoonup f, \qquad\qquad \Vert f_n\Vert_2 \to \Vert f\Vert_2
$$
where $\rightharpoonup$ means weak convergence. Is it true that $f_n \to f$ strongly, i.e. in $L^2$?

I don't know how to start and I really would like some hints on how to solve it. Actually, I do not know: is it true? I don't manage to find any counterexample…

Is this somehow related to the well-known fact that a.e. convergence + convergence of norms in $L^p$ do imply convergence in $L^p$ (Rudin, R&CA, ex. 17 pag. 73)?

Thanks.

Answer 1

It's almost surely a duplicate, but I think answering is shorter than finding the corresponding one.

Hint: we have $\lVert f-f_n\rVert_{L^2}^2=\lVert f\rVert_{L^2}^2-2\langle f,f_n\rangle+\lVert f_n\rVert_{L^2}^2$. The second term converges to $2\lVert f\rVert_{L^2}^2$.