Convergence of Operator Values implies Convergence

Question

Consider a bounded operator $T:C(X)\to C(X)$ in the sense that $\Vert Tf \Vert_{\infty}\leq C \Vert f \Vert_{\infty}$. Consider $z\in \rho(T)$ (i.e. $(T-z)^{-1}$ exists) and a bounded sequence $(f_n)_{n\in \mathbb{N}}$ such that $(T-z)f_n\to g \in C(X)$. Is there a short way to show $f_n\to (T-z)^{-1}g$? I think this is used in the proof I want to use, but I have problems finding a good argument.

Answer 1

Yes, there is. The operator $(T-z)^{-1}$ is closed because $T-z$ is, and therefore it is bounded by the closed graph theorem. Hence $$ f_n=(T-z)^{-1}(T-z)f_n\to (T-z)^{-1}g. $$