I want to prove that $\sigma(A)$, the spectrum of the linear operator $A \in B(C(K))$ which is defined by $Af = g∙f$, is equal to $\text{Im}(g)$, for $g\in C(K)$. We may assume that $K$ is a compact topological (also metric) space and that $C(K)$ is the Banach space consisting of all continuous complex functions on $K$, with the supremum norm.
First of all, since the spectrum of $A$ is the following: $$\sigma(A)=\{\lambda \in\mathbb{C}:\nexists(\lambda I-A)^{-1}\},$$ that is, the set of complex numbers such that $\lambda I-A$ is NOT INVERTIBLE, which means it is not bijective (since we talk about operators)?? I have thought of analysing for which $\lambda$ it is not injective (non-trivial kernel, which means $\lambda$ is eigenvalue of $A$), and same for surjective, but I don't know really how to do it.
Another way would be to analyse separately the point, continuous and residual spectrum of $A$ and take the union of them, but I don't know how to distinguish them, since I am a beginner in functional analysis.
I really look forward to a reply. Thank you very much in advance!!
For $h\in C(K)$, define an operator $T_h\in B(C(K))$ by $T_h(f)=h\cdot f$. We claim that $T_h$ is invertible if and only if $f$ does not attain the value $0$. The "if" part is easy: if $f$ doesn't attain $0$, then $f$ is invertible and ${T_f}^{-1}=T_{f^{-1}}$. Now suppose that $\{x\in K: f(x)=0\}$ is non-empty. For any $\epsilon>0$ it is disjoint from the closed subset $\{x\in K: |f(x)|\geq \epsilon\}$ so by Urysohn's lemma we can find $f_{\epsilon}\in C(K)$ that is identically $1$ on the former, but vanishes on the latter. Then $\|f_\epsilon\|=1$, but $\|T_h(f_\epsilon)\|\leq \epsilon$. Letting $\epsilon \rightarrow 0$ this shows that $T_h$ isn't bounded from below, hence cannot have a bounded inverse.
Applying this to your setting, we see that $\lambda I-A=T_{\lambda \cdot 1-g}$ is uninvertible if and only if $\lambda \cdot 1-g$ attains the value $0$, which happens if and only if $\lambda \in \text{Im}(g)$.