I am studying Rudin's Functional Analysis and come across a step in proving Theorem 13.35 (e): $Q(t)$ be a semigroup of operators that is strongly continuous (i.e. $\lim_{t\rightarrow 0}||Q(t)x-x|| = 0)$ and we want to show that $Q(t)x = \lim_{\epsilon \rightarrow 0} \exp(tA_\epsilon)x$.
$$\exp(tA_\epsilon) = e^{-t/\epsilon}\exp(\frac{t}{\epsilon}Q(\epsilon)) = e^{-t/\epsilon}\sum_n \frac{{t^n}Q(n\epsilon)}{\epsilon^n n!}$$ Rudin says we replace the norm of this sum by the sum of the norms, apply the estimate $||Q(t)|| \leq C^{\gamma t}$ to obtain $$||\exp(tA_\epsilon)|| \leq C \exp\{\frac{t}{\epsilon}(e^{\textbf{t} \epsilon \gamma} -1)\} < C \exp(te^{\gamma t})$$
I fail to see how we have the first inequality (I don't see the $t$ in the exponential) in my calculation below: $$ ||e^{-t/\epsilon}\sum_n \frac{{t^n}Q(n\epsilon)}{\epsilon^n n!}|| \leq e^{-t/\epsilon}\sum_n \frac{{t^n} ||Q(n\epsilon)||}{\epsilon^n n!} \leq e^{-t/\epsilon}\sum_n \frac{{t^n}Ce^{\gamma n \epsilon}}{\epsilon^n n!} = Ce^{-t/\epsilon}\exp{(\frac{t}{\epsilon} e^{\gamma \epsilon}}) $$ I wonder how does the $t$ come up in the exponential? Do we need another estimate to choose a specific $\gamma$? Thank you!
Actually I think this is Rudin's mistake, here is the correction. \begin{equation} \begin{aligned} ||\exp\{tA_{\epsilon}\}|| &= ||e^{-t/\epsilon}\exp\{\frac{t}{\epsilon}Q(\epsilon)\}||= ||e^{-t/\epsilon} \sum_{n=0}^{\infty}\frac{t^n Q(n\epsilon)}{\epsilon^n n!}||\\ &\leq e^{-t/\epsilon}\sum_{n=0}^{\infty}\frac{t^n ||Q(n\epsilon)||}{\epsilon^n n!}\leq e^{-t/\epsilon}\sum_{n=0}^{\infty}\frac{t^n Ce^{\gamma \epsilon n}}{\epsilon^n n!} \ \textit{(by part (1))} \\& = Ce^{-t/\epsilon}\exp(\frac{te^{\gamma \epsilon}}{\epsilon}) = C\exp(\frac{t}{\epsilon}(e^{\gamma \epsilon} -1 )) \end{aligned} \end{equation}