So, I am trying to solve the following statement: Suppose that $\sigma(a) \subseteq \{\lambda \in \mathbb{C}: \text{Re} \lambda < 0\}$. Show that there exist $M, \omega \in \mathbb{R}$ with $\omega > 0$ such that \begin{equation*} ||\exp(ta)|| \leq Me^{-\omega t} \text{ for every } t \in [0, \infty). \end{equation*}
I already kinda see where $\omega$ comes from, namely for any $\lambda \in \sigma(a)$ we have $|e^{t\lambda}| = e^{\text{Re}(t\lambda)} = e^{t\text{Re}(\lambda)}$. We know that $e^{t\lambda} \in \sigma(\exp(ta))$ by the Spectral Mapping Theorem. Since $\sigma(a) \subseteq \{\lambda \in \mathbb{C}: \text{Re} \lambda < 0\}$, and since $\sigma(a)$ is compact (thus closed), there exists $\omega \in \mathbb{R}_{>0}$ such that $\text{Re } \lambda \leq -\omega$ for any $\lambda \in \sigma(a)$. This shows that $|e^{t\lambda}| = e^{t\text{Re}(\lambda)} \leq e^{-\omega t}$. I assume we need to use some properties of the Banach algebra like $||ab|| \leq ||a||\thinspace||b||$ but I do not see how to find a constant $M$ that it time independent. Could anyone help me?
Let $r$ and $\sigma$ denote the spectral radius and the spectrum, respectively. We use the following result:
If $f:(0,\infty) \to \mathbb{R}$ is continuous and subadditive, then $$ \lim_{t \to \infty} \frac{f(t)}{t} ~~{\rm exists ~(and}= \inf\{\frac{f(t)}{t}: t>0\}). $$
For each $a$ we have $r(\exp(a))>0$, and $t \mapsto \log\|\exp(t a)\|$ is continuous and subadditive. By Gelfand's formula we have $$ \log(r(\exp(a)))= \lim_{n \to \infty} \frac{1}{n} \log\|\exp(na)\|. $$ Hence $$ \log(r(\exp(a)))= \lim_{t \to \infty} \frac{1}{t} \log\|\exp(ta)\|. $$ The spectral mapping theorem yields $\sigma(\exp(a))=\exp(\sigma(a))$, thus $$ r(\exp(a))=\max\{\exp(\Re \lambda): \lambda \in \sigma(a)\}=\exp(\max \{\Re \lambda: \lambda \in \sigma(a)\}). $$ We obtain $$ \max \{\Re \lambda: \lambda \in \sigma(a)\}= \lim_{t \to \infty} \log\frac{1}{t} \|\exp(ta)\|. $$ If $\max \{\Re \lambda: \lambda \in \sigma(a)\} < 0$ we can choose $\omega>0$ with $\max \{\Re \lambda: \lambda \in \sigma(a)\} < -\omega < 0$. For some $t_0>0$ we get $$ \forall t \ge t_0: ~ \frac{1}{t} \log\|\exp(ta)\| \le -\omega. $$ Hence $$ \forall t \ge t_0: ~ \|\exp(ta)\| \le \exp(-\omega t). $$ For $t \in [0,t_0]$ we have $\|\exp(ta)\| \le M \exp(-\omega t_0)$ for some $M\ge 1$ (since $[0,t_0]$ is compact). Now, $$ \forall t \ge 0: ~ \|\exp(ta)\| \le M\exp(-\omega t). $$