Suppose that $T$ is a bounded operator on some banach space $X$. We can define $e^T=\sum\limits_{n=0}^\infty T^n/n!$ which converges, since $T$ is bounded.
Is It true that if $\alpha\in\sigma (T)$ then $e^\alpha\in \sigma(e^T)$?
I managed to show that if $T-\alpha I$ is not injective then $e^T-e^\alpha I$ is not injective. But I was not able to show that if $T-\alpha I$ is not surjective then $e^T-e^\alpha I$ is not surjective. (Although I did manage to prove It when $ e^x$ is changed for a polynomial $P(X)$).
I also managed to solve It when we changed $T$ for an arbitrary element of a commutative banach álgebra. Since we can find a complez homomorfism such that $f(T)=\alpha$, and since complex homomorfism are continuous It follows that $f(e^T)=e^\alpha$.