Calulcating the norm of the exponential of a bounded operator

Question

Let $X$ be a Banach space. If $A \in L(X)$ we know that we can define $e^{At}$ via the exponential formula: $$e^{At}=\sum_{k=0}\frac{(tA)^k}{k!}$$ Now take $A=-I$ where $I$ is the identity. In the book by Engel Nagel p.74 they use the fact that $$|e^{-It}|_{L(x)}\leq e^{-t}$$ Of course by using the triangular inequality in the series one obtains $|e^{-It}|\leq e^{t |I|}=e^t$ but this is much weaker of course. How is the other inequality proved?

Answer 1

$e^{(-I)t}= \sum\limits_{k=0}^{\infty} \frac {(-t)^{k}} {k!} I=e^{-t} I$. So $\|e^{-It}\|=e^{-t}$ (It is an equality).