Does weak convergence imply pointwise convergence? [duplicate]

Question

Let $f_n,f:\mathbb R^d\to[0,\infty)$ with integral $1$ over $\mathbb R^d$. Suppose that $$ \int_{\mathbb R^d}\phi(x)\,f_n(x)\,d x\,\to\,\int_{\mathbb R^d}\phi(x)\,f(x)\,d x $$ as $n\to\infty$ for all $\phi:\mathbb R^d\to\mathbb R$ continuous and bounded.

Can I say that there exists a strictly increasing sequence $(m_n)_{n\in\mathbb N}$ such that $$f_{m_n}(x)\to f(x) \textrm{ for a.e. }x\in\mathbb R^d$$ as $n\to\infty$?

I know this would be the case for convergence in total variation, since it is equivalent to convergence of densities in $L^1$ and so there exists a subsequence which is convergent almost everywhere. Is weak convergence sufficient?

Answer 1

Weak convergence in $L^p$ does not imply pointwise $a.e.$ convergence.

Take the following counterexample: $f_n\in L^p([0,1])$ defined by $f_n = \sin(n\pi x)$. Then $f_n\rightharpoonup0$ in $L^p[0,1]$ (for $p\ge 1$ ) but it does not converge poitwise at all $x \in [0,1]$ (and you cannot extract any subsequence).