I'm confused by the Arzelà–Ascoli theorem since I think a set $F\subset C(X,Y)$ ($X$ compact metric space, $Y$ Banach space) is relatively compact already if $F$ is relatively compact at any point $x$ and if $F$ is equicontinuous at any $x$.
The boundedness I think is only relevant if $Y$ is finite dimensional to be able to extract a convergent subsequence. But if $F$ is already relatively compact for each $x$. One can already extract a convergent subsequence, of any sequence, for any $x$.
Also, I think the relative compactness of $F$ for any $x \in X$ can be relaxed to only a dense subset. Is that also true?
Best Answer
For a Banach space $Y$, a subset is relatively compact if, and only if, it is totally bounded. Totally bounded subsets are always bounded, and in a finite-dimensional Banach space $Y$, a subset is totally bounded if, and only if, it is bounded.
It follows from this that for a compact metric space $X$ and a finite-dimensional Banach space $Y$, a collection $F \subseteq C(X,Y)$ is pointwise relatively compact if, and only if, it is pointwise bounded. This is how the Arzelà-Ascoli theorems relate between the cases where $Y$ is finite-dimensional and where $Y$ is not assumed to be finite-dimensional.
It is also possible to relax the pointwise relative compactness assumption to only hold on a dense subset. In fact, if equicontinuity of the collection $F\subseteq C(X,Y)$ is assumed, and for a subset $A$ of $X$, it is the case that each $x \in A$ has the property that $\{f(x) : f\in F\}$ is relatively compact, then the same property holds for every point in the closure of $A$. This follows from the equivalent characterisation of totally bounded and relatively compact subsets, using the equicontinuity of $F$, and a standard estimate argument.
As a remark, all of these properties can be extended to the case where $X$ is a compact topological space.