[Math] Does $L^2$ strong convergence and bounded $L^\infty$ imply convergence in $L^\infty$ along subsequences

convergence-divergencefunctional-analysisfunctions

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?

Best Answer

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 $n$.

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