In the book: Functional Analysis, Sobolev Spaces and Partial Differential Equations of Haim Brezis, there is Theorem 7.3 (Cauchy,Lipschitz,Picard), page 184. In this Theorem, there is a space
$X = \{ u \in C([0,+\infty);E): \displaystyle \sup_{t\geq 0} e^{-kt}||u(t)|| < \infty \}$,
where $E$ is a Banach Space. I want to show that $X$ is a Banach Space for the norm
$||u||_X = \displaystyle \sup_{t \geq 0} e^{-kt}||u(t)||$.
I try to do it according to the definition. That is, to show that any Cauchy sequence of $X$ converges in $X$. But I can't get it done. Any hint/help would be grateful. Thank you.
Let $(u_n)_n$ be a Cauchy sequence in $X$, i.e. for all $\varepsilon>0$ there is $n_0$ s.t. for all $t\geq 0$ and $n,m\geq n_0$ $$e^{-kt}\Vert u_n(t)-u_m(t)\Vert\leq\varepsilon.$$ This implies that $(u_n(t))_n$ is a Cauchy sequence in the Banach space $E$ and hence there is a limit which we denote by $u(t)$ such that $$u_n(t)\to u(t)\quad\text{in }E.$$ We find $$e^{-kt}\Vert u_n(t)-u(t)\Vert\leq e^{-kt}\Vert u_n(t)-u_m(t)\Vert+e^{-kt}\Vert u_m(t)-u(t)\Vert\leq 2\varepsilon$$ where for all $n\geq n_0$ and $m$ chosen large enough (depending on $t$).
Thus, $u_n\to u$ in $X$.