Let $A$ be a bounded self adjoint operator and $f$ be a continuous fucntion on $\sigma(A)$. Let $\lambda \in \mathrm{Ran}(f)$. Prove that there are $\psi\in H,$ with $\|\psi\|=1$ and $\|(f(A)-\lambda)\psi\|$ arbitrarily small so that $\lambda \in \sigma(f(A))$.
This is an exercise from Reed & Simons' book chap 7, Q8.
I have managed to show the existence of $\psi$, but I am stuck at showing that leads to $\lambda \in \sigma(f(A))$. I am wondering if anyone could help me with that?
I think I might have solved it by my own. We will prove it by contradiction. Since we have shown the existence of $\psi.$ Then, we can find a sequence $\psi_n$ with $\|\psi_n\|=1$ and $\|(f(A)-\lambda) \psi_n\|\rightarrow 0.$ Assume $\lambda\in \rho(f(A)),$ then $(f(A)-\lambda)^{-1}$ is bounded. $$ 1 = \|\psi_n\| = \|(f(A)-\lambda)^{-1}(f(A)-\lambda)\psi_n\| \\ \leq \|(f(A)-\lambda)^{-1}\|\|(f(A)-\lambda)\psi_n\| \rightarrow 0. $$ We derive the contradiction.