How do I show that given $a$ continuous on $[0,1]$, then the operator $$A_a:L^2(0,1) \to L^2(0,1)$$ with $$A_a: f \to af$$ has spectrum exactly $a([0,1])$?
I can show that $a([0,1])^c \subset \text{Res}(A_a)$, but I do not see how to show the reverse.
To show that $a([0,1])\subseteq \sigma(A_a)$, let $\lambda \in a([0,1])$ so that there exists $x\in [0,1]$ such that $\lambda=a(x)$. Then, for every $\epsilon > 0$, there exists $\delta(\epsilon) > 0$ such that $|a(x)-a(y)| < \epsilon$ whenever $x\in [0,1]$ and $|x-y| < \delta(\epsilon)$. Let $\chi_{\epsilon}$ be the characteristic function of $[0,1]\cap\chi_{(x-\epsilon,x+\epsilon)}$. Then $$ \|(A-\lambda I)\chi_{\epsilon}\|_{L^2} \le \epsilon\|\chi_{\epsilon}\|_{L^2}. $$ From this it follows that $(A-\lambda I)$ cannot have a bounded inverse. Hence $\lambda\in\sigma(A)$, which proves that $a([0,1])\subseteq \sigma(A_a)$. You have shown that $a([0,1])^c \subseteq \sigma(A_a)^c$, or $\sigma(A_a)\subseteq a([0,1])$. So $a([0,1])=\sigma(A_a)$.