# Are inf and sup continuous functionals in general?

Let $$X$$ be any topological space and $$\bar{\mathbb{R}} = [-\infty, \infty]$$ with the standard topology. Is it true, in general, that the functionals $$\inf: C(X,\bar{\mathbb{R}}) \to \bar{\mathbb{R}}, f \mapsto \inf f$$ $$\sup: C(X,\bar{\mathbb{R}}) \to \bar{\mathbb{R}}, f \mapsto \sup f$$ are continuous, if $$C(X,\bar{\mathbb{R}})$$ is equipped with the compact-open topology?

If $$X$$ is compact, it should be true, if the following argument is correct (wlog. proof only for $$\inf$$ and $$X \neq \emptyset$$): Let $$a \in \mathbb{R}$$ arbitrary. Then $$\inf^{-1}([-\infty,a)) = \bigcup_{x\in X} \langle x,[-\infty,a) \rangle$$ is open. On the other hand $$\inf^{-1}((a,\infty]) = \bigcup_{\varepsilon > 0}\langle X, (a+\varepsilon, \infty] \rangle$$, which is open, because $$X$$ is compact. $$[-\infty,a)$$ and $$(a,\infty]$$ build a subbase of $$[-\infty,\infty]$$. Therefore $$\inf$$ is continuous.

What happens for arbitrary spaces $$X$$, or locally compact, etc.?

• Hmm, in fact I think if $X$ is Tychonoff and not compact, then $\sup$ is never continuous: for any basic open set $\bigcap_{i=1}^n \langle K_i, U_i \rangle$ contaning the zero function, $\bigcup K_i$ is compact and therefore not the whole space. So choose $x \notin \bigcup K_i$; then there is a function which is 0 on $\bigcup K_i$ but 1 at $x$, so its sup is at least 1, but the function is still in $\bigcap_{i=1}^n \langle K_i, U_i \rangle$. (Not sure on the details here, though...) Commented Mar 16, 2023 at 18:23

I will present a counterexample and show that in the case $$X = \mathbb{R}$$, neither $$\sup$$ nor $$\inf$$ is continuous. To see this, let us consider the sequence of functions $$f_n(x) := \arctan(x/n)$$. I claim that in the compact-open topology, $$f_n$$ converges to the constant zero function. To see this, take some subbasic open neighborhood $$V(K, U)$$ of the zero function with $$K$$ nonempty. Then since the zero function is in the neighborhood, we must have $$0\in U$$; therefore, for some $$\varepsilon > 0$$, $$(-\varepsilon, \varepsilon) \subseteq U$$. Also, $$K$$ must be bounded, so take some $$R \in \mathbb{R}^+$$ such that $$K \subseteq [-R, R]$$. Now, we know that $$\arctan(R/n) \to 0$$ as $$n\to \infty$$, so for $$n$$ sufficiently large, we have $$\arctan(R/n) < \varepsilon$$. Therefore, for $$n$$ sufficiently large, for every $$x\in K$$, we have $$|f_n(x)| \le \arctan(R/n) < \varepsilon$$, implying $$f_n(x) \in U$$. This shows that for $$n$$ sufficiently large, $$f_n \in V(K, U)$$. Since this is true for every subbasic open neighborhood of the zero function, we have shown that $$f_n \to 0$$ in the compact-open topology.
On the other hand, $$\sup(f_n) = \frac{\pi}{2}$$ for each $$n$$, while $$\sup(0) = 0$$, showing that $$\sup$$ is not continuous. Similarly, $$\inf(f_n) = -\frac{\pi}{2}$$ for each $$n$$, while $$\inf(0) = 0$$, which implies that $$\inf$$ is not continuous.