Spectrum of multiplication operator by the independent variable in $L^2$ If $\mu$ is a regular Borel measure on $\mathbb{C}$ with compact support $K$, define $N_\mu$ on $L^2(\mu)$ by $N_\mu f=zf$ (the multiplication by the indipendent variable).
An exercise in "Conway" asks to show that $\sigma(N_\mu)=K$.
For the inclusion $K \subseteq \sigma(N_\mu)$ I see that if $\lambda \in K$, then by definition of support if $O$ is a neighbourhood of $\lambda$, $\mu(O)>0$.
Hence $N_\mu-\lambda$ is not invertible in $L^\infty(\mu)$ because for every $f \in L^2(\mu)$ I will have $(N_\mu - \lambda) f = zf- \lambda f$ and somehow (please help me to formalize better this passage) there will be a "non negligible" zero.
Can you give me a hand please?
Thank you!
 A: Assuming that $\mu$ is a regular Borel measure implies that $\mu F < \infty$ for any compact subset $F$. Suppose $K$ is the support of $\mu$ and that $\lambda \in K$. Then, for every $r \in (0,\infty)$, the closed disk $D_{r}[\lambda]$ centered at $\lambda$ of radius $r$ is compact and, therefore, has finite $\mu$ measure. Let $f_{r}=\chi_{D_{r}[\lambda]}$ be the characteristic function of $D_{r}[\lambda]$. Then $f_{r} \in \mathcal{D}(M_{z})$ and $\|f_{r}\| \ne 0$. Furthermore,
$$
               \|(M_{z}-\lambda I)f_{r}\|\le r\|f_{r}\|.
$$
The means that $\lambda\in\sigma(M_{z})$ because $\lambda\in\rho(M_{z})$ gives the existence of a constant $m > 0$ such that $\|(M_{z}-\lambda I)f\| \ge m\|f\|$ for all $f \in \mathcal{D}(M_{z})$, which is clearly contradicted by the above. Hence,
$$
                   K \subseteq \sigma(M_{z}).
$$
On the other hand, suppose $\lambda \in \mathbb{C}\setminus K$. Then there exists $r > 0$ such that $\mu D_{r}[\lambda]=0$. That means that $1/(z-\lambda)$ is uniformly bounded by $1/r$ on $K$, which is enough to imply that $M_{1/(z-\lambda)}$ is a bounded operator on $L^{2}_{\mu}$. If $f \in L^{2}_{\mu}$, then $M_{1/(z-\lambda)}f \in \mathcal{D}(M_{z})$ and $(M_{z}-\lambda I)M_{1/(z-\lambda)}f=f$. Likewise, if $f \in \mathcal{D}(M_{z})$, then $M_{1/(z-\lambda)}(M_{z}-\lambda I)f=f$. So $\lambda\in\rho(M_{z})$, which gives
$$
       \mathbb{C}\setminus K\subseteq \rho(M_{z})=\mathbb{C}\setminus \sigma(M_{z})\\
        \implies \sigma(M_{z}) \subseteq K.
$$
Thus $\sigma(M_{z})=K$.
