I'm trying to solve the following exercise (see page 150) which should follow easily from the Arzelà-Ascoli theorem.
It says that a family of continuous functions $F$ taking values in a Banach space $(B,|\cdot|_{B})$, $F \subset C([0, T],B)$ such that
- $\dot{f}(t)$ exists for all $t \in(0, T)$ and $f \in F$,
- $\sup _{f \in F}|f(0)|_{B}<\infty$ and
- $\sup _{f \in F} \sup _{t \in(0, T)}|\dot{f}(t)|_{B}<\infty$
is precompact in the Banach space $C([0, T],B)$ equipped with the supremum norm $\sup_{t \in[0, T]}\|f(t)\|_{B}$, so in particular one can extract a sequence of elements in $F$ that converges in this norm.
In virtue of the usual Arzelà-Ascoli theorem, and since $[0,T]$ is compact, to check that $F$ is precompact it suffices to have that $\{f(t)\}_{f\in F}$ is precompact in $B$ for each $t\in [0,T]$ and that it is equicontinuous at each $t\in [0,T]$. I think I know how the equicontinuity follows from the fundamental theorem of calculus (for Banach spaces) and property 3. However I really don't know how 1-3 could imply that $\{f(t)\}_{f\in F}$ for each $t\in [0,T]$ is precompact in $B$.
I've tried to find a reference on the topic but I couldn't find one on this formulation in particular. If anyone could help, I would much appreciate it 🙂
Best Answer
This is false. If $f(t)$ is independent of $t$ for each $f \in F$ then 1) and 3) are trivially satisfied. So the result states that any bounded subset of $B$ is pre-compact. This is, of course, false in any infinite dimensional Banach space.