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If $\psi:[a,b]\to\mathbb C$ is a (measurable) function such that $\psi h$ is in $L^2[a,b]$ for all $h\in L^2[a,b]$, then must $\psi$ be essentially bounded?

The converse direction is clear: If $\psi$ is essentially bounded, then $\|\psi h\|_2\leq \|\psi\|_\infty\|h\|_2$ for all $h\in L^2$, and thus the multiplication operator $M_\psi:h\mapsto \psi h$ is defined and bounded. The question here is whether, without prior assumptions on $\psi$, simply having $M_\psi(L^2)\subseteq L^2$ implies that $\psi$ is bounded.

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What is the question here? –  Thomas E. Feb 3 '13 at 9:26
    
prove that Essential supremum of the function ψ is finite. –  Alexander Osorio Feb 3 '13 at 9:31
    
Why do you think it is true? –  Michael Greinecker Feb 3 '13 at 9:40

1 Answer 1

If $\psi h$ is in $L^2$ for all $h\in L^2$, define $M_\psi:L^2\to L^2$ by $M_\psi h=\psi h$. Then $M_\psi$ is a linear operator on $L^2$, and the closed graph theorem can be used to show that $M_\psi$ is bounded. Suppose that $\{(h_n,\psi h_n)\}$ is a sequence in the graph of $M_\psi$ converging to $(h,k)\in L^2\times L^2$. We want to see that $k=\psi h$. This can be seen from the fact that convergence in $L^2$ norm implies pointwise a.e. convergence of a subsequence. (There's a subsequence of $\{h_n\}$ converging pointwise a.e. to $h$, and a subsequence of $\{\psi h_n\}$ converging pointwise a.e. to $k$, and these can be coordinated.)

So $M_\psi$ is bounded. Suppose that $|\psi(x)|\geq c$ on a set $E$ of positive measure. Then $\|M_\psi \chi_{E}\|_2\geq C \|\chi_{E}\|_2$, which implies that $c\leq M_\psi$. Thus $\|\psi\|_\infty\leq \|M_\psi\|$.

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+1. (I deleted my answer as I made a confusion). –  Davide Giraudo Jan 1 at 11:40

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