Let $A$ be a bounded linear operator on a complex Hilbert space. Given a rational function $f$ which has no poles inside $\sigma(A)$, show that $\sigma(f(A))=\{f(\lambda)\mid\lambda\in\sigma(A)\}$.
If $\lambda\in\sigma(A)$ then take $v\neq0$ s.t. $Av=\lambda v$. Then by linearity $f(A)v=f(\lambda)v$, so $f(\lambda)$ is in the spectrum of $f(A)$. It follows $\sigma(f(A))\subset f(\sigma(A))$.
However, I am struggling with the reverse inclusion, starting with $\lambda\in f(\sigma(A))$.