Does $L^2$ strong convergence and bounded $L^\infty$ imply convergence in $L^\infty$ along subsequences? Take a sequence of functions $f_n \in L^2([0,1])$ such that
$f_n \rightarrow f$ in $L^2([0,1])$ (convergence in $L^2$ norm) and
$|f_n|_{L^\infty},|f|_{L^\infty} <R$ (uniformly bounded in $L^\infty$ norm).
Question:
Do we get from this that there is a subsequence $f_{n_k}$ such that
$f_{n_k} \rightarrow f$ in $L^\infty$ norm?
If not:
Are there any assumptions such that this is true?
 A: Let $f_n = 1_{[0,{1 \over n}]}$. Then $f_n \to f=0$ in $L_2$ (since $\|f_n-f\|_2 = { 1 \over \sqrt{n}}$), and $\|f_n\|_\infty = 1, |f\|_\infty = 0$, so we can take $R=2$. However no subsequence can converge uniformly to zero since $\|f_n-f\|_\infty = 1$ for all $n$.
A: If you do have convergent subsequences they will all have to converge to the $L_2$ limit, because convergence in $L_\infty$ implies convergence in $L_2$. In other words, your sequence would have to converge in $L_\infty$. Boundedness in $L_\infty$ only gives you a weak$^*$ convergent subsequence. To get strong convergence in $L_\infty$ you need boundedness in a space that's compactly embedded into $L_\infty$, e.g. $W^{2,1}([0,1])$. 
A: Let $f_{n}(t)$ be $0$ on $(-\infty,0]$, climb linearly to $1$ on $[0,1/n]$, remain $1$ on $[1/n,1-1/n]$, ramp linearly back down to $0$ on $[1-1/n,1]$ and remain $0$ on $[1,\infty)$. Then $\{ f_{n}\}_{n=1}^{\infty}$ converges in $L^{2}(\mathbb{R})$ to the characteristic function $\chi_{(0,1)}$. There is no subsequence $\{ f_{n_{k}}\}_{k=1}^{\infty}$ of $\{ f_{n}\}_{n=1}^{\infty}$ which can converge in $L^{\infty}$ because the limit function would have to be continuous in order for the sequence of continuous functions $\{ f_{n_{k}}\}_{k=1}^{\infty}$ to converge in $L^{\infty}(\mathbb{R})$.
