For the pointwise supremum of affine functions that are global underestimators of a convex function $f$,
$$h(x) = \text{sup}\{g(x)\ |\ g\ \text{affine}, \ g(z)\le f(z)\}$$
It can be shown that $h(x) = f(x)$ for $x \in \text{int dom }f$
Furthermore, if $f$ is closed, then $h(x)=f(x)$ for all $x \in \text{dom }f$
My question is why is it not the case that $h(x) = f(x)$ for all $x \in \text{dom }f$ when $f$ is open?
What I am thinking is that the pointwise supremum somehow applies closure to $f$ so that:
$$\text{epi }h = \text{cl epi }f$$
I believe this would imply that $h$ is equal to $f$ except at the point in boundary of the domain. I can't seem to prove this though. Thank you for reading and any help!
Let $f : \mathbb{R}_+ \to \mathbb{R}$ be defined by: $$f(x) = \begin{cases}1 & \text{if } x = 0 \\ 0 & \text{if } x > 0.\end{cases}$$ Note that $f$ is open and convex.
Let's look at $h(0)$. What affine functions $g(x)=ax+b$ satisfy $g(z) \leq f(z)$ for all $z \geq 0$? You need $a\leq 0$ as otherwise $g(z)>f(z)$ for $z>-b/a$. You also need $b \leq 0$ as otherwise $g(z)>f(z)$ for sufficiently small $z$. Therefore, $g(0) \leq 0$, and $h(0)=0$.