Let $F = [a,b]^X$ be a space of real-valued, bounded functions on some set $X$ equipped with the topology of pointwise convergence, $a,b \in \mathbb{R}$. Let $(f_i)_{i \in I}$ be a collection of functions in $F$, and let $f \in F$ as well.
Suppose $f$ has the property that $$\inf_i f_i(x) \leq f(x) \leq \sup_i f_i(x)$$ for all $x \in X$.
Claim. $f$ is in the closed convex hull of $(f_i)_{i \in I}$, which we denote by $\overline{co}(f_i)$.
The claim seems very intuitive, but I'm unsure about the following argument. Suppose for contradiction that $f \notin \overline{co}(f_i)$. Then $f$ can be separated from $\overline{co}(f_i)$ by a continuous linear functional $\psi: F \to \mathbb{R}$. That means that there exists $r \in \mathbb{R}$ such that $\psi(f) < r$ and $\psi(g) \geq r$ for all $g \in \overline{co}(f_i)$. Now the claim would follow if the function $\sup_if_i \in \overline{co}(f_i)$. To show that rigorously, I suppose one must construct a net of convex combinations of $(f_i)_{i \in I}$ that converges pointwise to $\sup_if_i$, and this is where I'm stuck.
Can you please help me complete the argument? Or perhaps this line of reasoning is overkill and I'm missing something simpler. In that case, what's the simpler way?
This is far from being true. Let $f_0(x)=a$ and $f_1(x)=b$ for all $x\in X$. The convex hull of $\{f_0,f_1\}$ is the set of all constant functions on $X$ with values in $[a,b]$, which is clearly closed with respect to pointwise convergence. But there are many functions $f$ that satisfy $a\leq f(x)\leq b$ for all $x\in X$ and are not constant.
Note that in this case one has $\sup\{f_0,f_1\}\in \overline{co}(f_1,f_2)$, so your claim does not follow from this assertion.
Finally, a separating continuous linear functional in the previous example can be constructed as follows: Let $x,y\in X$ such that $f(x)< f(y)$ and let $\psi\colon [a,b]^X\longrightarrow \mathbb{R},\,g\mapsto g(x)-g(y)$. Then $\psi(g)=0$ for all $g\in\overline{co}(f_0,f_1)$ and $\psi(f)<0$.