Multiplication operator

Example

Let $(\Omega, \mu)$ be a $\sigma$-finite measure space and let $\varphi \colon \Omega \to \C$ be an essentially bounded measurable function. The multiplication operator $M_\varphi$ on the Lebesgue space $L^2(\Omega, \mu)$ is $$(M_\varphi f)(x) = \varphi(x) f(x).$$

Check of the main properties:

1. $M_\varphi$ is bounded with $\norm{M_\varphi} = \norm{\varphi}_\infty$, since $\int \abs{\varphi f}^2 \, d\mu \le \norm{\varphi}_\infty^2 \int \abs{f}^2 \, d\mu$, and $\sigma$-finiteness gives sets of finite positive measure on which $\abs{\varphi}$ is close to $\norm{\varphi}_\infty$.
2. Its adjoint is $M_{\bar\varphi}$, because $\int \varphi f \bar g \, d\mu = \int f \, \overline{\bar\varphi g} \, d\mu$.
3. It is a normal operator, since $M_\varphi M_{\bar\varphi} = M_{\abs{\varphi}^2} = M_{\bar\varphi} M_\varphi$.
4. Its spectrum is the essential range of $\varphi$.

A concrete case: on $L^2[0, 1]$ with Lebesgue measure, take $\varphi(x) = x$. Then $M_\varphi$ is self-adjoint and positive, its spectrum is $[0, 1]$, and it has no eigenvalues: $x f(x) = \lambda f(x)$ almost everywhere forces $f = 0$ almost everywhere.