Show that the supremum of a sequence of continuous functions is continuous

Question

Let $E$ be a $\mathbb R$-Banach space, $(f_n)_{n\in\mathbb N}\subseteq C^0([0,\infty), E)$ and $$g_n(t):=\sup_{s\in[0,\:t]}\left\|f_n(s)\right\|_E\;\;\;\text{for }t\ge0.$$ Note that $(g_n)_{n\in\mathbb N}\subseteq C^0([0,\infty), E)$. Now, let $$h(t):=\sup_{n\in\mathbb N}g_n(t)\;\;\;\text{for }t\ge0.$$

Assuming that $$\sup_{s\in[0,\:t]}\left\|f_m(s)-f_n(s)\right\|_E\xrightarrow{m,\:n\:\to\:\infty}0\tag1\;\;\;\text{for all }t\ge0,$$ are we able to show that $h$ is continuous?

I've found the claim in a book, but I don't see how we can prove it. Let $t\ge0$ and $\varepsilon>0$. By $(1)$, $$\sup_{s\in[0,\:t]}\left\|f_m(s)-f_n(s)\right\|_E<\varepsilon\;\;\;\text{for all }n\ge N\tag2$$ for some $N\in\mathbb N$ and hence $$|g_n(t)-g_N(t)|\le\sup_{s\in[0,\:t]}\left\|f_n(s)-f_N(s)\right\|_E<\varepsilon\;\;\;\text{for all }n\ge N\tag3.$$ Now, \begin{equation}\begin{split}|h(s)-h(t)|&\le\sup_{n\in\mathbb N}|g_n(s)-g_n(t)|\\&\le\sup_{n\in\mathbb N}|g_n(s)-g_N(s)|+\sup_{n\in\mathbb N}|g_n(t)-g_N(t)|+\sup_{n\in\mathbb N}|g_N(s)-g_N(t)|\end{split}\tag4\end{equation} for all $s\ge0$, but I don't see, for example, how we can estimate the first term on the right-hand side of $(4)$ (we only know that it is at most $\max(g_1(s),\ldots,g_{N-1}(s),\varepsilon)$). So, how do we need to argue?

Answer 1

Fix $T>0$. Since $(f_n|_{[0,T]})_{n \in \mathbb{N}}$ is a Cauchy sequence in $C([0,T],E)$, it follows from the the Arzela-Ascoli theorem that $(f_n|_{[0,T]})_{n \in \mathbb{N}}$ is equicontinuous.

Now fix $t \in [0,T)$. As $h$ is clearly non-decreasing, we have

$$h(t) \leq h(t+s) \qquad \text{for all $s \geq 0$}. \tag{1} $$

On the other hand,

$$\begin{align*} h(t+s) &= \max \left\{ \sup_{n \in \mathbb{N}} \sup_{r \leq t} |f_n(r)|, \sup_{n \in \mathbb{N}} \sup_{t \leq r \leq t+s} |f_n(r)| \right\} \\ &\leq \max \bigg\{ h(t), \sup_{n \in \mathbb{N}} \sup_{t \leq r \leq t+s} |f_n(r)-f_n(t)| + \underbrace{\sup_{n \in \mathbb{N}}|f_n(t)|}_{\leq h(t)} \bigg\} \\ &\leq h(t) + \sup_{n \in \mathbb{N}} \sup_{t \leq r \leq t+s} |f_n(r)-f_n(t)| \end{align*}$$

Because of the equicontinuity of $(f_n|_{[0,T]})_{n \in \mathbb{N}}$, the second term on the right-hand side gets small if we choose $s>0$ sufficiently small. Combining this estimate with $(1)$, we conclude that $h$ is right-continuous at $t$. The proof of the left-continuity goes exactly along the same lines; I leave it to you.