Does Pointwise Convergence imply L2 Convergence in L2?

Question

Its clear that convergence in $L^2$ does not necessarily imply pointwise convergence a.e but does this work the other way too? I considered the characteristic function $\chi_{[n, n+1]}$ which I guess pointwise converges to 0 in $L^2$ but not for $x \in (0,1)$ I think but I want the example for the other way around for something that converges pointwise but not in $L^2$ if the function in $L^2$ is bounded?

Answer 1

If I am understanding your question properly, then the answer is no. Consider the sequence of indicator functions $f_n:=\sqrt{n}\mathbb 1_{(0,1/n)}$. This sequence converges $a.e$ to the $0$ function on $(0,1)$, but $$\|f_n\|_2^2=n\int_{(0,1)}\mathbb 1_{(0,1/n)}(x)dx=1.$$ This is a counterexample to both of your questions as elucidated in your comment on your question. Note that this does not contradict the dominated convergence theorem, because $(f_n)$ is not bounded above a.e. (I assumed this is what you meant by the uniform boundedness theorem). As pointed out in the comments by @nejimban, if $f_n\to f$ a.e., and $\|f_n\|_2\to \|f\|_2$, then $\|f_n-f\|_2\to 0$. See here for a proof.