Does weak convergence in $L^2$ implies convergence almost everywhere along subsequence?

Solution 1:

The example by David Mitra is perfectly valid, but it may be a little hard to see exactly what happens to $\sin nx$ as $n\to\infty$. Here's a slightly modified example: Rademacher functions $$r_n(x) = \operatorname{sign}\sin (2^n \pi x), \quad x\in [0,1]$$ These converge to $0$ weakly in $L^2$, but $|r_n|=1$ a.e.

To prove weak convergence, first consider $\int r_n g$ when $g$ is continuous, putting the contributions of adjacent $\pm $ intervals against each other. Then conclude by density (using the fact that $\|r_n\|_2=1$ and Hölder's inequality).

For complex-valued functions, $\exp(i nx)$ is a good counterexample.