# Convergence in $L^p_{loc}$ implies convergence of a subsequence in $L^\infty$

Let $$\Omega \subset \mathbb{R}^n$$ be bounded or unbounded. Suppose we have a sequence $$\{f_n\} \in L^p_{loc}(\Omega)$$ such that $$f_n \rightarrow f$$ in $$L^p_{loc}(\Omega)$$ for $$f \in L^p_{loc}(\Omega)$$.

Now suppose further that $$f \in L^\infty(\Omega)$$. Since $$L^\infty(\Omega) \subset L^p_{loc}(\Omega)$$, I am wondering if it is possible to always extract a subsequence of $$\{f_n\}$$, call it $$\{f_{n_k}\}$$, such that $$f_{n_k} \rightarrow f$$ in $$L^\infty(\Omega)$$. I think this is possible but I cannot convince myself that we can always choose a convergence subsequence such that all its elements belong to $$L^\infty$$.

• @AnneBauval Sorry I should have been more specific. I meant convergence in $L^\infty$ implies pointwise convergence a.e. Feb 20 at 2:20
• @AnneBauval I have corrected my post. Feb 20 at 23:35

Not possible. Take $$f_n = \chi_{ [1/n,\ 1+1/n] }$$.
• Doesn't each $f_n$ belong to $L^\infty$? Feb 19 at 19:59
• @AnneBauval If we take the sequence itself as a subsequence don't we have $f_n \rightarrow f$ in $L^\infty$? Feb 20 at 2:19
• No: $\|f_n-f\|_\infty=1$. Feb 20 at 6:40