Let $\emptyset \neq D$ a open set in $\mathbb{C}$ and $J: D \to B(E)$ a continuos function such that $J(\lambda) - J(\mu) = (\mu - \lambda)J(\lambda) J(\mu)$ where $E$ is Banach space.
We must show that:
If exists $ \lambda_{0} \in D$ such that $J(\lambda_{0})$ is invertible operator then $J$ is a resolvent function for some $T \in B(E)$.
My attempt:
Let $ \lambda_{0} \in D$ such that $J(\lambda_{0})$ is invertible operator. If $\lambda_{0} \neq 0$ we can write $J(\lambda_{0}) = (I - \lambda_{0}A)^{-1}$.
Case $\lambda_{0} = 0$. We use that the set of linear, continous and invertibles operators is open in $B(E)$, so by continuity of $J$, we have that there exist a neighborhood $V$ of $0$ such that for all $\lambda \in V$ we have that $J(\lambda)$ is invertible too. So we change $\lambda_{0}$ for another $\tilde{\lambda} \in V$, and we again have that $J(\tilde{\lambda}) = (I - \tilde{\lambda}A)^{-1}$.
So we always can assume that $J(\lambda_{0}) = (I - \lambda_{0}A)^{-1}$.
I showed that $J$ is holomorphic in $D$. But I stuck here. Someone have some hint?
Thank you!
For simplicity, suppose that $0\in D$ and $T = -J(0)^{-1}$. We prove that for any $z\in D$, $J(z) = (zI - T)^{-1}$.
Note that by definition of $J$, $J(z)J(w) = J(w)J(z)$ for $w\in D$, $z\in D$. Also, $J(z) = -zJ(z)J(0) + J(0)$. Hence \begin{align*} (zI - T)J(z) = zJ(z) + zTJ(z)J(0) - TJ(0)= I. \end{align*} Similarly, $J(z)(zI - T) = I$. That is, $J(z) = (zI - T)^{-1}$.
You should know how to modify it when $J(z_0)$ is invertible, and $z_0 \neq 0$.