Consider the Banach space $C([0,1])$ of continuous functions on the closed unit interval, equipped with the supremum norm $\lVert\cdot\rVert_\infty$. Let the bounded linear operator $A: C([0,1]) \to C([[0,1])$ be defined by $$(Af)(t) = g(t)f(t), \qquad t \in [0,1],\ g \in C([0,1]).$$ How would I find the resolvent set $\rho(A)$?
My attempt at a solution:
$\rho(A)$ is the set of scalars $\lambda \in \mathbb{C}$ for which $A-\lambda I$ is invertible (where $I$ is the identity operator on $C([0,1])$). $(A-\lambda I)f(t) = Af(t) - \lambda f(t) = g(t)f(t) - \lambda f(t) = (g(t) - \lambda)(f(t))$, so to get $f(t)$ back we need to divide by $g(t)-\lambda$.
I would therefore assume that $$(A-\lambda I)^{-1}f(t) = \frac{f(t)}{g(t) - \lambda},$$ which exists whenever $g(t) \neq \lambda$ for any $t \in [0,1]$.
Beyond that I'm stuck. I think I'm close but I can't quite make the leap. Any guidance would be appreciated!
You are indeed very close. To proceed, it is simpler if you consider necessity and sufficiency separately.
Sufficiency: Assume first that $g(t)\ne\lambda$ for any $t\in[0,1]$. You already found a candidate for an inverse in this case. Show that it is really the inverse, and that it is bounded.
Necessity: Assume now that there is some $t_0\in[0,1]$ such that $g(t_0)=\lambda$. Show that $A-\lambda I$ cannot be onto by observing that all functions in the range have some property in common, and give an example of a function in $C([0,1])$ which does not have this property.