Let $(E, |\cdot|)$ be a Banach space, $k>0$ and $$ X := \{u \in C([0, \infty); E) : \sup_{t \ge 0} e^{-kt} |u(t)| < \infty\}. $$
In the proof of Theorem 7.3 from Brezis' Funtional Analysis, the author said that
$X$ is a Banach space for the norm $$ \|u\| := \sup_{t \ge 0} e^{-kt} |u(t)|. $$
Could you confirm if my attempt on the completeness of $(X, \| \cdot \|)$ is fine?
Let $(u_n)$ be a Cauchy sequence in $X$. For a fixed $t \ge 0$, we have $(u_n (t))_n$ is a Cauchy sequence and thus converges in $E$. Let $v(t) := \lim_n u_n (t) \in E$. Let's prove that $\|u_n - v\| \to 0$. Fix $\varepsilon >0$. There is $N >0$ such that $\|u_n - u_m\| <\varepsilon$ for all $n,m>N$. Then for $n>N$, $$ \begin{align*} \|u_n-v\| &= \sup_{t \ge 0} e^{-kt} |u_n(t) -v(t)| \\ &= \sup_{t \ge 0} \lim_m e^{-kt} |u_n(t) - u_m(t)| \\ &\le \sup_{t \ge 0} \limsup_m \|u_n-u_m\| \\ &= \limsup_m \|u_n-u_m\| \\ &\le \varepsilon. \end{align*} $$
It remains to verify that $v$ is continuous. Fix $t_0 > 0$. We will prove that $v$ is continuous at $t_0$. Fix $\varepsilon >0$. There is $N \in \mathbb N$ such that $\|u_N - v\| <\varepsilon$. Since $u_N$ is continuous at $t_0$, there is $0<\delta<1$ such that $|u_N (t)-u_N(t_0)| < \varepsilon$ for all $t \ge 0$ with $|t-t_0| <\delta$. By triangle inequality, we have for $t \ge 0$ with $|t-t_0| <\delta$: $$ \begin{align*} |v(t)-v(t_0)| & \le |v(t)-u_N(t)|+|u_N(t)-u_N(t_0)|+|u_N(t_0) - v(t_0)| \\ &\le \|v-u_N\|e^{kt} + \varepsilon+ \|v-u_N\|e^{kt_0} \\ &\le \|v-u_N\|e^{k(t_0+\delta)} + \varepsilon+ \|v-u_N\|e^{kt_0} \\ &\le (e^{k(t_0+1)}+e^{kt_0}+1)\varepsilon. \end{align*} $$
The claim then follows.