Arzelà-Ascoli corollary

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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

  1. $\dot{f}(t)$ exists for all $t \in(0, T)$ and $f \in F$,
  2. $\sup _{f \in F}|f(0)|_{B}<\infty$ and
  3. $\sup _{f \in F} \sup _{t \in(0, T)}|\dot{f}(t)|_{B}<\infty$

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 :-)

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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.

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To complement geetha290krm's answer, here is a hint for the correct case $B = \mathbb{R}$: try showing that, for differentiable functions, $\|\dot{f}\|_\infty \leq K \Rightarrow$ $f$ $K$-Lipschitz (it's actually equivalent but you only need that implication).
From there and with 2. and 3. you should be able to arrive at the conclusion that the sets $\{f(t)\mid f \in F\}$ are bounded thus precompact (but that's only true in finite dimension!), and therefore you'll be able to apply Arzelà-Ascoli once you prove the equicontinuity.