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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.