Proving that a family of exponential operators has a uniform bound, without semigroup theory. Set $T\in B(X)$ for some arbitrary Banach space $X$. Define the family of operators $\{e^{tT}\}_{t\in\mathbb R}$ via the holomorphic functional calculus. I have been able to prove, using basic facts about the functional calculus that $\frac{d}{dt}e^{tT}=Te^{tT}$, and that the family forms a group under multiplication.
What I am now trying to prove is that if $\sigma(T)\subset\{z:Rez<0\}$, then $\sup_{t\geq 0} \|e^{tT}\|\leq M$ for some positive $M$. In the finite dimensional case, I know we can decompose into Jordan's normal form to get the result, but I do not know how to pursue that avenue for an arbitrary bounded operator. My idea is to somehow use the spectral mapping theorem. This gives us that the spectral radius $\rho (e^{tT})\to 0$ as $t\to \infty$, and that this convergence is exponential. Unfortunately, as $X$ need not be a Hilbert space and $T$ need not be normal we do not know that $\rho(e^{tT})=\|e^{tT}\|$. What I am hoping is that there is some way we can extend the estimation lemma to the holomorphic functional calculus, at least in certain cases of which this is one, which would then give the result, but I have not been able to find such a thing.
From scouring the net I know that this is an archetypal problem of continuous semi-group theory, but I encountered this while studying Rudin's Functional Analysis, specifically in relation to his chapter on Banach algebras and the holomorphic functional calculus, which makes me believe that there is a simpler solution using basic theory from these areas. Any help would be appreciated.
 A: Say $\gamma$ is a closed curve in the left complex half-plane that "encloses" $\sigma(T)$. Then, if I'm remembering this stuff correctly, $$e^{tT}=\frac1{2\pi i}\int_\gamma e^{tz}(zI-T)^{-1}\,dz.$$
But we have $$\left|e^{tz}\right|\le1\quad(t>0,z\in\gamma)$$and$$\sup_{z\in\gamma}||(zI-T)^{-1}||<\infty.$$
If that's right I bet it's the argument Rudin had in mind; he's a big fan of using CIF for the functional calculus.
Edit: I felt a little guilty posting this, taking advantage of the fact that the OP said $\sigma(T)\subset\{x+iy:x<0\}$; the argument doesn't  quite work if $\sigma(T)$ is just contained in the closed left half-plane. But aha, in that case the result is  not true! Let $$T=\begin{bmatrix}0&1\\0&0\end{bmatrix};$$since $T^2=0$ we have $$e^{tT}=\begin{bmatrix}1&t\\0&1\end{bmatrix}.$$
Edit 2: I also felt that the "right" argument should show that $||e^{tT}||\le1$. But that's not true either: Let $A_\delta=2T-\delta I$, where $T$ is as in the previous edit. Then $\lim_{\delta\to0^+}||e^{A_\delta}||=||e^{2T}||>1$.
Looks like CIT gives the whole truth.
