Let $\varphi \in C[0, 1]$. Define $M_{\varphi}:L^{2}[0,1]\toL^2[0,1]$, $M_{\varphi}(f)= \varphi f$. It seems that $ M_{\varphi}$ is a linear bounded operator.
Questions
- Is $ M_{\varphi} \geq 0$ if only if $ \varphi \ge 0$?
- What is the square root of $M_{\varphi}$?
- I think that $T$ is normal, is it right?
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 $M_{\sqrt{\varphi}}$.