Does weak convergence imply $L^2$-norm convergence?

Question

Let $\{f_n\}_{n \geq 1}$ be a sequence of functions in $L^2[0,1]$ converging to $f$ weakly. Does it imply that $f_n \to f$ in $L^2$-norm?

This question appeared in an exam paper on functional analysis which I failed to solve properly. Any help in this regard will be warmly appreciated.

Thanks for your time.

Answer 1

No. As $X:=L^2[0,1]$ is a separable Hilbert space there exists an orthonormal basis $\{e_n\}_{n=1}^\infty$ (this is going to be some trignometric sequence depending on whether your functions are real or complex valued). By Riesz Representation we know for any $f\in X^*$ there exists a fixed $z\in X$ such that $f(x)=\langle x,z\rangle$ for all $x\in X$. As $z=\sum_{n=1}^\infty \langle e_n,z\rangle e_n$ we must have that $\langle e_n,z\rangle\to 0$. Thus for all $f\in X^*$ we have $f(e_n)\to 0$. However, $\|e_n\|=1$ for every $n$.

Answer 2

Any orthonormal sequence converges weakly to zero but clearly is not norm convergent.