$\newcommand{\epi}{\operatorname{epi}}$
Let $F:[a,b] \to [0,\infty)$ be continuous, and let $\epi F$ be the epigraph of $F$.
How to prove that the convex hull of $\epi F$ is closed?
This shouldn't be hard but I am struggling with finding a neat argument:
I know that the the convex hull of a compact set in $\mathbb R^2$ is closed, but here $\epi F$ is unbounded. However, since $f<M$ for some $M$, we can just take $$D=\operatorname{conv}\big( \epi F \cap ([a,b] \times [0,M])\big) \cup \big([a,b] \times [M,\infty)\big).$$
$\epi F \cap ([a,b] \times [0,M])$ is closed (as the intersection of two closed sets) and bounded, so its convex hull is closed.
We "broke" $\operatorname{conv} (\epi F)$ into two parts, a bounded one and an unbounded one.
The only remaining part is to prove that $D$ is convex, and even though this is fiarly intuitive, I am having trouble filling out the details.
Best Answer
$D$ is the union of two convex sets, $A = \operatorname{conv}\bigl(\operatorname{epi} F \cap ([a,b]\times [0,M])\bigr)$ and $B = [a,b]\times [M, +\infty)$. Thus we need only check that the segment connecting a point in $A$ to a point in $B$ is contained in $D$ (and we can assume that neither point lies in the intersection).
Take $p_1 = (x_1,y_1), p_2 = (x_2, y_2) \in D$ with $y_1 < M < y_2$ and let $p(t) = (1-t)p_1 + tp_2$ for $t \in [0,1]$. Then there is $t_M \in (0,1)$ such that $(1 - t_M)y_1 + t_M y_2 = M$.
Let $p_M = p(t_M)$. Then $p_M \in \bigl(\operatorname{epi} F\bigr) \cap B \subset A \cap B$, hence we have $p(t) \in A$ for $t \in [0,t_M]$ and $p(t) \in B$ for $t \in [t_M,1]$. Thus $p(t) \in D = A\cup B$ for all $t \in [0,1]$.