# How to prove that the convex hull of an epigraph is closed?

$$\newcommand{\epi}{\operatorname{epi}}$$

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 $$f for some $$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.

$$D$$ 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 $$t \in [0,1]$$.