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Let $T\colon L^2[0,1] \to L^2[0,1]$ be a bounded linear map of Hilbert spaces such that if $f\in L^2[0,1]$ is continuous then so is $Tf$. Show that there is a constant $C$ such that $$\sup_{x\in[0,1]}|Tf(x)| \leq C \sup_{x\in[0,1]}|f(x)|$$

I have no idea how to go about. Please at least give hints.

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    $\begingroup$ What is the relation of the question to the heading? $\endgroup$ – Thomas Feb 15 '16 at 16:26
  • $\begingroup$ So the question says: if $T$ is bounded in $L^2$ norms, then it is bouded in $L^\infty$ norms? $\endgroup$ – GEdgar Feb 15 '16 at 16:27
  • $\begingroup$ @GEdgar: A little more, I think: Note that we're not using the essential supremum here. We also requires $C^0$ to be an invariant subspace of $T$. $\endgroup$ – Roland Feb 15 '16 at 16:33
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Let us consider the canonical embedding $i:C[0,1]\to L^2[0,1]$.

Your hypothesis implies that $S=i^{-1} T i:C[0,1]\to C[0,1]$ is well-defined. Moreover, since $i^{-1}$ is closed, it is not difficult to check that $S$ is a closed operator. Thus $S$ is bounded by the closed graph theorem: $$\|Tf\|_\infty \leq \|S\|\cdot \|f\|_\infty.$$

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  • $\begingroup$ Perhaps canonical embedding $C \to L^2$ $\endgroup$ – GEdgar Feb 15 '16 at 16:59
  • $\begingroup$ @GEdgar You're right. Otherwise the solution is right. $\endgroup$ – Xiamoi Wang Feb 15 '16 at 17:05

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