If $T: E \rightarrow E$ is a bounded linear operator, then $e^T$ is as well On page 39 of Differential Dynamical Systems (revised edition) by Meiss, the author gives the following Lemma.

Lemma 2.8: If $T$ is a bounded linear operator, then $e^T$ is as well.

We are assuming (I think) that $T: E \rightarrow E$, where $E$ is a normed vector space over $\mathbb{R}$. I have a couple doubts about the proof of the lemma, which goes as follows:

Proof: Choose an arbitrary $x \in E$ and consider the value of $e^T(x)$. By definition (2.25) [$e^T \equiv \sum_{k=0}^{\infty} \frac{T^k}{k!}$], this is a series whose terms are elements of $E$. The norm of this series is bounded by the sum of the norms of each term. By the definition of the operator norm, for any $x$,
$$ |T(x)| \leq \|T\| |x|, $$
$$ |T^k(x)|  = |T(T^{k-1}(x))| \leq \|T\| |T^{k-1}(x)| \leq \cdots \leq \|T\|^k |x|. $$


Consequently, each of the terms in the series $e^T(x)$ can be bounded by
$$ \left|\frac{T^k(x)}{k!} \right| \leq \frac{\|T\|^k}{k!} |x| = M_k. $$


The series of real numbers
$$ \sum_{k=0}^{\infty} M_k = \sum_{k=0}^{\infty} \frac{\|T\|^k}{k!}|x| = e^{\|T\|}|x| $$
converges for any finite value of $\|T\|$. By the Weierstrass M-test $|e^T(x)| \leq e^{\|T\|}|x|$ and the series for $e^T(x)$ converges uniformly in $x$. Moreover, $\|e^T\| \leq e^{\|T\|}$, so the exponential is a bounded operator. $\qquad \square$

My questions: 

*

*How are we allowed to use the Weierstrass M-test here? Don't we require that the norm of each term in the series is bounded uniformly in $x$, i.e. $\left|\frac{T^k(x)}{k!}\right| \leq M_k$ for all $x \in E$?  (I am going by the version of the Weierstrass M-Test that I stated in this previous post.


*Don't we also need to assume that $E$ is a Banach-space? (This was an assumption I used in the proof I linked.)
 A: You certainly need to assume $E$ is a Banach space, else you'd have to make the rather odd assertion: "$e^T$ exists for all $T$", and I think most mathematicians would only ever work with such an assertion in the wider context of a Banach space.
As for the $M$-test, you can be satisfied with showing that $e^T(x)$ converges locally uniformly. I expect this is what the author meant, as it can be strengthened to "uniform convergence over bounded subsets", which is strong enough for a lot of purposes. For example, you might fix a radius $R$ and use the $M$-test on the set $B_R(0)$, which is sufficient to get continuity and boundedness with $\|e^T\|\le e^{\|T\|}$ by bounding $|x|\le R$.
It is not true even in $E=\Bbb R$ that $e^x$ converges uniformly over the whole space. Rather we have uniform convergence over any bounded subset.
A: Yes we do need to assume that $E$ is a Banach space so that $L(E)$ is also a Banach space. The sum $e^T = \sum_{n = 0}^{\infty}\frac{1}{n!}T^n$ converges absolutely in the Banach space $L(E)$ since $\sum_{n = 0}^{\infty}\frac{1}{n!}\|T^n\| \leq \sum_{n = 0}^{\infty}\frac{1}{n!}\|T\|^n = e^{\|T\|} < \infty$. In particular, the sum converges in $L(E)$-norm.
