Find the spectrum of the multiplication operator $Mf(x) = m(x) f(x)$ on $L^2(\mathbb{R})$ Let $M$ be a multiplication operator on $L^2(\mathbb{R})$ defined by $$Mf(x) = m(x) f(x)$$ where $m(x)$ is continuous and bounded. Prove that $M$ is a bounded operator on $L^2(\mathbb{R})$ and that its spectrum is given by $$\sigma(M) = \overline{\{m(x) : x \in \mathbb{R}\}}.$$ Can $M$ have eigenvalues? 
My partial answer is below:
First, observe that for $f \in L^2(\mathbb{R})$, $$\| Mf\|^2 = \|m(x) f(x) \|^2_{L^2} = \int_\mathbb{R} (m(x))^2 (f(x))^2 \leq \int_\mathbb{R} \|m\|_\infty (f(x))^2 = \|m\|_\infty \|f\|_{L^2}^2.$$ Since $m$ is continuous, $\|m\|_\infty = R$ for some constant $R$, so $M$ is bounded. 
Define the set $X$ as $X = \overline{\{m(x) : x \in \mathbb{R}\}}$. I show $X \subset \sigma(M)$ by contrapositive. Suppose $\lambda \in \rho(M)$. Then $(M-\lambda I)$ is invertible. For $g\in L^2(\mathbb{R})$, $(M-\lambda I)g(x) = m(x)g(x) - \lambda g(x)$. Clearly, the inverse of this operator is $\frac{1}{M - \lambda I}$. As the inverse is well-defined and bounded, $m(x)g(x) - \lambda g(x) \not= 0$, which implies that $\lambda \not= m(x)$ for any $x\in\mathbb{R}$. Thus, $\lambda \not\in X$. Since the spectrum is closed, it follows that $X \subseteq \sigma(M)$. 
I'm not sure how to proceed to show that $\sigma(M) \subseteq X$. 
 A: $X$ is closed ans bounded, hence compact. If $\lambda \notin X$ the there exists $\epsilon >0$ such that $|\lambda -y| >\epsilon$ for all $y \in X$. Define $T$ by $Tf(x)=\frac 1 {m(x)-\lambda} f(x)$. Then $T$ is a bounded operator with $\|T\| \leq \frac 1 {\epsilon}$. [This is proved exactly the way you proved that $\|M\| \leq \|m\|_{\infty}$].
Now verify that $S(M-\lambda I)=(M-\lambda I)S=I$. Thus $M-\lambda I$ is invertible and $\lambda$ is not in the spectrum of $M$.   
A: Your reasoning that $X\subset \sigma(M)$ is not correct. Your proof only yields
that the spectrum of $M$ contains values $\lambda$ for which $\{x: m(x)=\lambda\}$
has positive measure: The equation $(M-\lambda I)g=f$ implies
$(m(x)-\lambda)g(x)=f(x)$ only for almost all $x$.
Here is a more expanded proof. Let $x_0$ be given and suppose $m(x_0)\in \rho(M)$. Take $\epsilon>0$. Since $m$ is continuous, there is $\delta>0$ such that $|m(x)-m(x_0)| \le \epsilon$ for all $|x-x_0|\le\delta$. Define $f(x):=\chi_{(x_0-\delta,x_0+\delta)}(x)$.
Then there is $g$ such that $(M-m(x_0)I)g=f$ or $(m(x)-m(x_0))g(x)=f(x)$ for almost all $x$. This implies $m(x)\ne m(x_0)$ for almost all $x$, 
as otherwise the equation $(M-m(x_0)I)g=f$ would have multiple solutions.
Then
$$
|g(x)| = |m(x)-m(x_0)| ^{-1} f(x)\ge \epsilon^{-1} f(x)
$$
for almost all $x$, which implies $\|g\|_{L^2} \ge \epsilon^{-1} \|f\|_{L^2}$. Now, $\epsilon>0$ was arbitrary, and $(M-m(x_0)I)^{-1}$ cannot be bounded, hence $m(x_0)\in \sigma(M)$.
