Let $A$ be a bounded operator on a Banach space $X$, and let $\sigma$ be an isolated part of $\sigma(A)$, that is, both $\sigma$ and $\tau:=\sigma(A)\backslash\sigma$ are closed. $P_{\sigma}(A)$ is the Riesz projection of $A$ corresponding to $\sigma$, \begin{equation} P_{\sigma}(A) = \int_{\Gamma} (\lambda - A)^{-1} d\lambda, \end{equation} where $\Gamma$ is a Cauchy contour around $\sigma$ on $\mathbb{C}$.
Then if $N$ is an $A$-invariant subspace of $X$, such that $\sigma(A\rvert_{N}) \subset \sigma$, then how can I show that $N \subset Im P_{\sigma}(A)$?
I have made some attempts using Cauchy contours around $\tau$ and using properties of the Riesz projections, but always seem to hit a wall because of the restriction to $N$, since the resolvents $R_{\lambda}(A)$ and $R_{\lambda}(A\rvert_{N})$ are different operators.
I figured it out.
Let $\tilde{A} := A\rvert_{N}$. Write $P_{\sigma}(A) = f(A)$ for $f \in\mathcal{F}(A)$ an analytic function on a neighbourhood of $\sigma(A)$ such that $f(z)=1$ on a neighbourhood of $\sigma$ and $f(z)=0$ on a neighbourhood of $\tau$. Moreover, we can write
\begin{equation} f(\lambda) = \sum_{j=1}^{\infty} \alpha_{j} \lambda^{j}, \end{equation} and then \begin{equation} f(A) = \sum_{j=1}^{\infty} \alpha_{j} A^{j}. \end{equation} Now take $x\in N$, \begin{equation} P_{\sigma}(A)x = f(A)x = \sum_{j=1}^{\infty} \alpha_{j}A^{j}x = \sum_{j=1}^{\infty} \alpha_{j} \tilde{A}^{j} x = f(\tilde{A})x = x, \end{equation} since $f \in\mathcal{F}(\tilde{A})$ and $f(\tilde{A})=I\rvert_{N}$, and this proves the result.