Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function and define the multiplication operator $M_{f}:\text{dom}(M_{f})\subset L^{2}(\mathbb{R}) \rightarrow L^{2}(\mathbb{R})$ as $M_{f}g = fg$ for all $g \in \text{dom}(M_{f}) := \left\{g \in L^{2}(\mathbb{R}): fg \in L^{2}(\mathbb{R})\right\}$. Show that if $f$ is bounded and strictly monotonic, then $\text{sup}_{x\in \mathbb{R}}f(x) \in \sigma_{c}(M_{f})$. Here $\sigma_{c}(M_{f})$ denotes the continuous spectrum of $M_{f}$.
Here is what I´ve done: I know that $\sigma(M_{f}) = \overline{\left\{f(x) : x \in \mathbb{R}\right\}}$. Since $M_{f}$ is self-adjoint, I also know that $\sigma_{r}(M_{f}) = \emptyset$ and $\sigma_{p}(M_{f}) = \left\{\lambda \in \mathbb{R} : \mu\left(\left\{x \in \mathbb{R} : f(x) = \lambda\right\}\right)>0\right\}$. So I end up with $\sigma_{c}(M_{f})= \sigma(M_{f})\setminus\sigma_{p}(M_{f})$. How do I show that $\text{sup}f \in \sigma_{c}(\mathbb{R})$, any ideas?
I'm assuming you are using Lebesgue measure. Since $f$ is strictly monotonic, $\sigma_p(M_f)=\varnothing$. Thus $$\sigma_c(M_f)=\sigma(M_f)=\overline{f(\mathbb R)}.$$
As $f$ is real valued, continuous, and strictly monotonic, $$\sup\{f(x):\ x\in\mathbb R\}=\lim_{x\to\infty}f(x)\in\overline{f(\mathbb R)}=\sigma(M_f)=\sigma_c(M_f).$$