Suppose $T$ is an operator on $L^p[0,1]$ which takes $a.e.$ convergent sequences to $a.e.$ convergent sequences, i.e., if $f_n\rightarrow f$ $a.e.$, then $Tf_n\rightarrow Tf$ $a.e.$. I'm trying to show that $T$ is continuous in $L^p$ norm. I don't think that any one of $a.e.$ or $L^p$ convergence is implied by the other. I tried using the closed graph theorem, but to no avail. How can this be proved?
How do I show that an operator on $L^p[0,1]$ taking $a.e.$ convergent sequence to an $a.e.$ convergent sequence is always continuous?
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$\begingroup$
functional-analysis
operator-theory
lp-spaces
1 Answers
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$L^p$ convergence implies convergence in measure. Convergence in measure implies a.e. convergence of a subsequence. Using this you can apply the Closed Graph Theorem, because if $(f_n,Tf_n)\to(f,g)$, then there is a subsequence $f_{n_k}$ such that $f_{n_k}\to f$ a.e., hence by your hypothesis you have $Tf_{n_k}\to Tf$ a.e.. You also have a subsequence of $Tf_{n_k}$ converging a.e. to $g$, so $Tf=g$.
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0Clever application of Closed Graph Theorem! – 2017-02-09
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0There's one point not entirely clear to me-why are the two subsequences converging to $Tf$ and $g$ the same? I mean, it may be that $Tf_{n_k}\rightarrow Tf$ and $Tf_{n_j}\rightarrow g$ for a different subsequence. How do we conclude $Tf=g$ in that case? Maybe I'm being very silly here, but could you please clarify the point? – 2017-02-09
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0@adrija: Notice we're not just taking any subsequence of $(Tf_n)$ converging to $g$. Once we already have $Tf_{n_k}\to Tf$, we still also have $Tf_{n_k}\to g$ in $L^p$ because $Tf_n\to g$ in $L^p$, so there is a subsequence *of that subsequence* converging pointwise a.e. to $g$. Because it is a subsequence of the same subsequence converging pointwise a.e. to $Tf$, it also converges pointwise a.e. to $Tf$. You are not being silly, that is the important point to get straight. – 2017-02-09