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If $\phi \in L^\infty(\mu)$, define $M_\phi:L^2(\mu)\rightarrow L^2(\mu)$ by $M_\phi f=\phi f$.

Also, I know that $M_\phi$ is bounded linear operation and $\Vert M_\phi \Vert = \Vert \phi \Vert$.

I proved that $\ker M_\phi=0$ iff $\mu(\{x:\phi(x)=0\})=0$.

I want to find that necessary and sufficient conditions on $\phi$ that $\operatorname{range} M_\phi$ be closed.

How to find that??...

hardmath
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sfsanjfafk
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  • "ker$M_\phi$=0" is not proper MathJax usage. I changed it to $\ker M_\phi=0$. $\qquad$ – Michael Hardy Mar 26 '16 at 16:28
  • $M_\phi$ is a selfadjoint operator. The range of a selfadjoint operator is closed if and only if zero is an isolated spectral point (or a regular point). The spectrum of $M_\phi$ is the essential range of $\phi$. So...? – Friedrich Philipp Mar 26 '16 at 16:31
  • But,.. $M_\phi$ is not selfadjoint operator. $(M_{\phi})^*=(M_{\bar\phi})$ – sfsanjfafk Mar 26 '16 at 16:41

1 Answers1

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If $E = \{ x : \phi(x) = 0 \}$, then a necessary and sufficient condition for the range to be closed is that there exist $\epsilon > 0$ such that $|\phi(x)| \ge \epsilon$ a.e. $[d\mu]$ on the complement of $E$.

Disintegrating By Parts
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    Reasoning (may be interesting for some people): The range of $M_\phi$ is the same as the range of $T := M_\phi |{({\rm{ker}} M\phi)^\perp}$. Since $T$ is injective, the range of $T$ is closed if and only if $T$ is (when considered as a map onto its range) an isomorphism of Hilbert spaces (here, we use the inverse mapping theorem), i.e. if there is $\epsilon > 0$ satisfying $| M_\phi x | \geq \epsilon |x|$ for all $x \perp {\rm{ker} M_\phi}$. It is not hard to see that this is equivalent to the stated condition. – PhoemueX Mar 26 '16 at 19:36