It is known that minimum or maximum of a function does not always exist but the supremum/infimum usually tends to exist.
Example 1: For example, if we consider $X$ as the set or rational numbers with $a_n\in X$ and consider the sequence
$$a_{n+1}=\frac{a_n}{2}+\frac{1}{a_n}$$
this sequence tends to $\sqrt{2}$ which is not rational but a real number. This means that rational numbers are not compact and supremum or infimum is possible not to be achieved in the rational numbers. However such supremum or infimum can be achieved in the real numbers. For example
$$\{a\mbox{ rational}:a^2<2\}$$
and the supremum is $\sqrt{2}$.
Example 2: Similarly on real numbers $X=\mathbb{R}$
$$\arg\sup 1-e^{-x}$$
doesnt exist but if one extends the real number set to extended real number set then $$\arg\sup 1-e^{-x}=\infty$$.
Question: Let us consider the closed ball
$$ X=\{g:D(g,f)\leq \epsilon\} $$ of density functions, i.e. $f$ and $g$ are probability density functions, e.g. on $\mathbb{R}$ and $\epsilon$ is a small positive number. Here $D$ is some distance for example KL-divergece.
This closed ball is defined on an infinite dimensional space and closed balls on infinite dimensional spaces are not compact. This suggests that supremum and infimum do not necessarily exist in the space.
Here I am trying the make an analogy with the two examples above and try to find the supremum ouside of $X$, namely on some $Y\supset X$. At this point I am confused because the ball is not open, it is closed and something outside this closed ball will not satisfy the condition $D(g,f)\leq \epsilon$.
Regarding Noah's answer: of course there is a functional of density functions and supremum is searched over this functional. For example: $$T=\int_{\mathbb{R}}\delta g \mathrm{d}\mu\quad\mbox{for some}\,\,\delta$$
Is there any common notation kind of $Y=$extended closed ball such that all supremums/infimums exist in this space? Here I am talking about the argument, namely a density function which provides the supremem like in the second example, where it was provided by $\infty$.
Would the extension with the set $Y=\{g:D(g,f)\leq \epsilon^{'}\}$ with $\epsilon^{'}>\epsilon$ guarantee the existence of supremums/infimums defined on $X$?
Is there any case where $Y$ needs to extend to include non-density functions for the existence of supremums/infimums?
Let's take a step back, to more generality: you're asking what we can conclude about the existence of suprema/infima (and maxima/minima) of a functional defined on some subset of a metric space.
First of all, if the functional in question isn't continuous, then there's basically nothing that can be said.
Assuming we restrict attention to continuous functionals $H: A\subseteq X\rightarrow \mathbb{R}$, then we have the following easy results:
If $A$ is compact, then $\max(H(A))$ exists. (Ditto for $\inf$.)
If $A$ is contained in some compact set $B\subseteq A$, then $\sup(H(A))$ exists. (Ditto for $\sup$.)
These assumptions are necessary. For instance, consider $X=\mathbb{R}, H=id, A=(0, 1)$. Then $\sup(H(A))$ and $\inf(H(A))$ exist, but $\max(H(A))$ and $\min(H(A))$ do not. Meanwhile, if we take $X$ to be Baire space with its usual metric, then given any nonempty open subset $A$ of $X$ there is a continuous surjection from $A$ to $\mathbb{R}$.
Meanwhile, if our target space is $\mathbb{R}\cup\{+\infty,-\infty\}$, then suprema/infima always exist, regardless of what $X, A, H$ are.
Re: your most recent comment: there, we were extending the codomain. If we make the codomain compact - that is, pass from values-in-$\mathbb{R}$ to values-in-the-extended-reals - then every functional has a supremum, continuous or no, from a compact set or no.
But you're asking about expanding the domain a little. This in general isn't possible: if I have a subset $A\subseteq X$ of a complete metric space, which is not contained in any compact subset of $X$, then I may very well have a continuous surjection from $A$ to $\mathbb{R}$. Such a surejection obviously has no supremum (unless I expand the codomain).
This domain/codomain issue, by the way, was why I misunderstood your question at first.