Let $f \in C(\Bbb R)$ be a bounded continuous function on $\Bbb R.$ Consider the multiplication operator $M_f : L^2 (\Bbb R) \longrightarrow L^2 (\Bbb R)$ defined by $M_f (g) = fg, g \in L^2 (\Bbb R).$ What will be $\sigma (M_f),$ the spectrum of the operator $M_f\ $?
By definition, $$\sigma (M_f) = \left \{\lambda \in \Bbb C\ |\ (M_f - \lambda I) \notin GL \left (L^2(\Bbb R) \right ) \right \}.$$ Now let $\lambda \in \rho (M_f),$ the resolvent of the operator $M_f.$ That implies $(M_f - \lambda I) \in GL \left (L^2 (\Bbb R) \right ).$ When can it happen? If $\lambda \notin \text {ran}\ f$ then clearly it will be the case because then $M_{(f - \lambda)^{-1}}$ will serve as the inverse of $(M_f - \lambda I).$ What about the converse? If $(M_f - \lambda I) \in GL \left (L^2 ( \Bbb R ) \right )$ can we always say that $\lambda \notin \text {ran}\ f\ $? I think it is not necessarily the case for otherwise $\sigma (M_f) = \left (\rho (M_f) \right )^c = \text {ran}\ f,$ which is not necessarily closed. So I guess that $\sigma (M_f) = \overline {\text {ran}\ f}.$ But how to prove it rigorously?
Any help in this regard will be warmly appreciated. Thanks for your time.
Your guess is correct. The spectrum is the closure of the range $R$ of $f$.
If $\lambda \notin \overline R$ then the equation $(M_f-\lambda I)g=h$ has a unique solution $g=\frac h {f-\lambda}$ for every $h \in L^{2}(\mathbb R)$ and $\|g\|_2 \leq M\|h\|_2$ where $M=\frac 1 {\inf \{|f(x)-\lambda|: 0 \leq x \leq 1\}}$. Hence $\lambda \notin \sigma (M_f)$.
Now suppose $M_f-\lambda I$ has a bounded inverse. Then $h (\frac 1 {f-\lambda}) \in L^{2}$ for every $h \in L^{2}$. By a well known result in Functional Analysis (which has appered many times on this site) this implies that $\frac 1{f-\lambda}$ is essentially bounded. By continuity this implies that $\lambda \notin \overline R$