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‎let ‎ ‎$ ‎\varphi ‎\in ‎C[0 ,‎ ‎1]‎ $‎. ‎

define: ‎

$ M‎_{‎\varphi‎} :‎ ‎L‎^{2}‎‎ [0 ,‎ ‎1]‎‎‎ ‎\longrightarrow‎ ‎L‎^{2}‎‎ [0 ,‎ ‎1]‎‎‎ $ ,‎ ‎$ ‎M‎_{‎\varphi‎} ‎‎(f ) = ‎\varphi f ‎‎$ ‎.‎ ‎

It ‎seams ‎that $ M‎_{‎\varphi} ‎‎$‎ ‎is a‎ ‎linear ‎bounded ‎operator.‎ ‎

Is‎ ‎$ ‎M‎_{‎\varphi} ‎\geq ‎0‎$ ‎if ‎only ‎if ‎ $ ‎\varphi‎ ‎\geq ‎0‎‎$‎?‎

what ‎is ‎the ‎square ‎root ‎of ‎‎$‎M‎_{‎\varphi}‎$‎?‎

I‎ ‎think ‎that‎ ‎is ‎normal, ‎is ‎it ‎right?‎

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    What is the adjoint of $M‎_{‎\varphi} $? What is its spectrum? These are both easy to answer, and will help answer your questions. – 2017-01-17

1 Answers 1

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Yes. The map $\varphi \mapsto M_{\varphi}$ from $C[0,1]$ to $\mathcal{B}(L^2[0,1])$ is an homomorphism of $C^{*}$-algebras whose image lies in the collection of normal operators. It preserves positive elements and assuming $\varphi \geq 0$, (a) square root of $M_{\varphi}$ is the operator $M_{\sqrt{\varphi}}$.