Showing that $X= \left\{ u \in C\bigl([0, \infty);E\bigr) \mid \sup_{t \geq 0} e^{-kt} \|u(t)\| < \infty \right\}$ is Banach. I am working on my degree thesis studying the Hille Yosida Theorem and as a motive to it, I want to present the Cauchy-Lipschitz-Picard Theorem. 
Haim Brezis highlights the path for a very interesting proof. A part of it, relies on showing that the space 
$$X= \left\lbrace u \in C\bigl([0, \infty);E\bigr) \mid \sup_{t \geq 0} e^{-kt} \|u(t)\| < \infty \right\rbrace$$
equipped with the norm $\|\cdot\|_X = \sup_{t \geq 0} e^{-kt} \|\cdot\|$ is a Banach space. Note that $E$ is a Banach space too.
I've made an attempt but I am not sure if it's properly stated.
Let $u_n \in X$ be a Cauchy sequence. Then, $\forall \varepsilon >0$ there exists $n_0 \in \mathbb N$ such that :
$$\|u_n-u_m\|_X < \varepsilon, \quad \forall n,m \geq n_0$$
$$\Leftrightarrow$$
$$\sup_{t \geq 0}e^{-kt}\|u_n(t)-u_m(t)\| < \varepsilon$$
Essentialy, I would like to show that $\|u_n(t)-u_m(t)\|< \varepsilon$ and then by letting $m\to \infty$ I get what I need. But, that exponential is causing issues since it's decreasing $t$ grows. 
I thought about fixing a $t_0 \in [0,\infty)$ which would then mean 
$$\sup_{t\geq 0} e^{-kt_0} \|u_n(t_0)-u_m(t_0)\| < \varepsilon$$
$$\implies$$
$$\|u_n(t_0)-u_m(t_0)\| < \varepsilon'$$
where $\varepsilon' = \varepsilon e^{kt_0}$. That yields that $\{u_n(t_0)\}$ is a Cauchy sequence in $X$ for any $t_0$ fixed.
Letting $m \to \infty$ now, leads us to $\|u_n(t_0) - u(t_0)\| < \varepsilon'$.
Now, since $u \in C\left( [0,\infty);E\right)$ then of course $u$ is continuous at every point $t \in [0,\infty)$ then it should be $\|u(t) - u(t_0)\| < \varepsilon''$ for every such $t_0$ fixed with $\varepsilon''$ depending on $t_0$, thus generalising the result.
I am afraid that my elaboration, especially at the end, isn't totally correct. Since this is a part of my Thesis, I am trying to find a perfect way around it. I hope some of my intuition is of use or even makes sense. I would highly appreciate any help or tips.
 A: Here are some remarks.


*

*You have show that for all fixed $t_0$, the sequence $(u_n(t_0))_{n\geq 1}$ is Cauchy. Maybe here a line could be added, arguing that since $E$ is a Banach space, the sequence $(u_n(t_0))_{n\geq 1}$ converges in $E$ to some $u(t_0)$.

*When you let $m$ going to infinity, you get a large inequality. But in any case you do not need that.

*For the moment, $u$ is only defined and you did not prove that this map is continuous. To this aim, you can show that $(u_n)_{n\geq 1}$ converges uniformly on all bounded interval.

*It remains to show $u_n\to u$ in $X$. Fix a positive $\varepsilon$; there is $n_0$ such that for all $n,m\geq n_0$ and $t\geq 0$,  $\lVert u_m(t)-u_n(t)\rVert\leq e^{kt}\varepsilon$. For a fixed $t$, let $m$ going to infinity to get $\lVert u(t)-u_n(t)\rVert\leq e^{kt}\varepsilon$ for all $n\geq n_0$.

A: It is not obvious that $u$ is continuous at every point $t$. What you showed is that for each point $t$, $u(t)= \lim u_n(t)$ is well-defined. You did not show that $u$ defined as such is continuous, nor that $u_n \to u$ w.r.t $\Vert \cdot \Vert_X$.
For the sake of completeness, let's repeat part of the proof in other words. What we must prove is that a sequence $u_n$ which is $\Vert \cdot \Vert_X$-Cauchy is $\Vert \cdot \Vert_X$-convergent. From now on, "Cauchy", "convergent" etc are all with respect to $\Vert \cdot \Vert_X$ unless mentioned otherwise. The norm $\Vert \cdot \Vert$ will be used as the norm on $E$. 


*

*First, let's prove that $u(t_0) = \lim u_n(t_0)$ is well-defined for all $t_0$.
Fix $t_0$. Since $u_n$ is Cauchy, there exists $N$ such that if $n,m> N$,
$$\sup e^{-kt}\Vert u_n(t)-u_m(t) \Vert < \epsilon e^{-kt_0}. $$
Therefore,
$$e^{-k t_0}\Vert u_n(t_0)-u_m(t_0) \Vert < \epsilon e^{-kt_0}, $$
and hence $\Vert u_n(t_0) - u_m(t_0) \Vert < \epsilon$ for $n,m >M$. Since $E$ is Banach, the sequence $u_i(t_0)$ is convergent and thus $u$ is well-defined.


*

*Now, let's prove that $u_n \to u$.
We have that
$$\Vert u-u_n\Vert_X \leq 2e^{ -k t_n}\Vert u(t_n)-u_n(t_n)\Vert $$
for some sequence $t_n$, due to the definition of $\Vert \cdot \Vert_X$. Now, given $\epsilon>0$, select $N$ such that $\Vert u_n -u_m\Vert_X <\epsilon/4$ for $n,m> N$. Picking $m_n$ such that $\Vert u(t_n)-u_{m_n}(t_n)\Vert <\epsilon e^{ kt_n}/4$ and $m_n>N$, we then have that
\begin{align*}
\Vert u(t_n)-u_n(t_n)\Vert& \leq \Vert u(t_n)-u_{m_n}(t_n) \Vert+\Vert u_{m_n}(t_n)-u_n(t_n)\Vert \\
&< \epsilon e^{kt_n}/4+e^{kt_n}e^{-kt_n}\Vert u_{m_n}(t_n)-u_n(t_n)\Vert \\
&< \epsilon e^{kt_n}/4+e^{kt_n}\epsilon/4.
\end{align*}
Going back to our original inequality, this implies that
$$\Vert u-u_n\Vert_X \leq 2e^{ -k t_n}\Vert u(t_n)-u_n(t_n)\Vert < \epsilon$$
for $n >N$, as we wanted to show.


*Finally, let's show that $u$ is continuous.
This is similar to showing that the uniform limit of a sequence of continuous functions is continuous. In fact, you can just mimic that proof (with minor caveats), or proceed to show that $u_n \to u$  uniformly on any $[0,T]$. Indeed, we have that
\begin{align*}
\Vert u - u_n\Vert_{\infty, [0,T]}&= \sup_{t \in[0,T]} \Vert u(t) - u_n(t)\Vert \\
& = e^{kT} e^{-kT}\sup_{t \in[0,T]} \Vert u(t) - u_n(t)\Vert \\
& = e^{kT} \sup_{t \in[0,T]} e^{-kT} \Vert u(t) - u_n(t)\Vert  \\
& \leq e^{kT} \sup_{t \in[0,T]} e^{-kt} \Vert u(t) - u_n(t)\Vert \leq e^{ kT}\Vert u -u_n\Vert_X.
\end{align*}
