Eigenfunction in functional calculus

Question

Let $X$ be a complex Banach space, $A\in L(X)$ and $F$ be an analytic function in a neighborhood of $\sigma(A)$. Now I want to show that if $x\in X$ is an eigenfunction of $A$ corresponding to the eigenvalue $\lambda$, then $x$ is an eigenfunction of $F(A)$ corresponding to the eigenvalue $F(\lambda)$. My first approach is to write $$F(A)=\frac{1}{2\pi i}\int_{\partial \Omega}{\frac{F(\lambda)}{\lambda I-A}}d\lambda$$ where $\Omega$ is such that $\sigma(A)\subset \Omega$.

Answer 1

Let $F(z) = \sum_{n=0}^\infty a_n (z-\lambda_0)^n$, with $\lambda$ inside the radius of convergence (you could just take $\lambda=\lambda_0$). Then $$ \begin{split} F(A)x &= \left(\sum_{n=0}^\infty a_n (A-\lambda_0 I)^n x\right) \\ &= \sum_{n=0}^\infty a_n (A-\lambda_0 I)^n x = \sum_{n=0}^\infty a_n (\lambda-\lambda_0)^n x \\ &= \left(\sum_{n=0}^\infty a_n (\lambda-\lambda_0)^n x\right) = F(\lambda)x. \end{split} $$

Alternatively, with your approach. Notice that $(zI-A)x=(z-\lambda)x$, so that $(zI-A)^{-1}x=(z-\lambda)^{-1}x$. Hence $$ F(A)x = \frac{1}{2\pi i} \int_{\partial\Omega} \frac{F(z)}{zI-A}x \,dz = \frac{1}{2\pi i} \int_{\partial\Omega} \frac{F(z)}{z-\lambda}x \,dz = F(\lambda)x. $$

Answer 2

Properties of the functional calculus give $$ F(\mu)-F(\lambda)=G(\mu,\lambda)(\mu-\lambda) \\ F(A)-F(\lambda)I= G(A,\lambda)(A-\lambda I). $$ Therefore, if $(A-\lambda I)x=0$, then $F(A)x=F(\lambda)x$.