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Let $M_{\phi}$ be a multiplication operator $M_{\phi}:L^{2}\left(\mu\right)\rightarrow L^{2}\left(\mu\right)$ defined by $M_{\phi}f=\phi f$.

  1. Show that $\ker M_{\phi}=0$ if and only if $\mu\left(\left\{ x:\phi\left(x\right)=0\right\} \right)=0$.

  2. Give necessary and sufficient conditions on $\phi$ that $\mbox{ran}M_{\phi}$ be closed.

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Is this homework? If so, you should tag it as such. –  Christopher A. Wong Nov 18 '12 at 21:33
    
It is not a homework. But it is a question from Conway, "A Course in Functional Analysis". –  CrisSer Nov 18 '12 at 21:39
    
I'm too lazy to write the proof. Here are my results: $M_\phi$ have non trivial kernel iff $\mu(\phi^{-1}(\{0\}))>0$ $M_\phi$ have closed image iff $\mu(\phi^{-1}(\mathrm{Ball}_\mathbb{C}(0,r)\setminus\{0\}))=0$ for some $r>0$ –  no identity Nov 22 '12 at 19:30

1 Answer 1

  1. Let $N=\{\phi=0\}$. We have $\phi f=0$ a.e. if and only if $f=0$ a.e. on $N^c$. If the latter set has null complement, then $f=0$ in $L^2$, which means $M\phi$ is injective. Otherwise, there is a nonzero function $f$ which vanishes a.e. on $N^c$.

  2. Decompose $L^2(\mu)=L^2(\mu,N)\oplus L^2(\mu,N^c)$. Since $M_\phi$ kills the first component, its range is the same as the range of restriction to $L^2(\mu,N^c)$. This restriction in injective by part 1. Thus, it is closed iff it has a left inverse, or equivalently, if it has a positive lower bound $c$: namely $\|M_\phi f\|\ge c\|f\| $ for all $f\in L^2(\mu, N^c)$. The latter is equivalent to $|\phi|\ge c$ on $N^c$.

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