# $T\colon L^2[0,1] \to L^2[0,1]$ be a bounded linear map of Hilbert spaces such that if $f\in L^2[0,1]$ is continuous then so is $Tf$. [closed]

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.

• What is the relation of the question to the heading? – Thomas Feb 15 '16 at 16:26
• So the question says: if $T$ is bounded in $L^2$ norms, then it is bouded in $L^\infty$ norms? – GEdgar Feb 15 '16 at 16:27
• @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 $T$. – Roland Feb 15 '16 at 16:33

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: $$\|Tf\|_\infty \leq \|S\|\cdot \|f\|_\infty.$$
• Perhaps canonical embedding $C \to L^2$ – GEdgar Feb 15 '16 at 16:59