# How to find the domain of the support function

First the definition of the support function of set $S$: Let $S$ be a nonempty convex set. The support function $h$ of $S$ is the real-valued function defined by

$$h(x)=\sup_{s\in S}\langle s,x\rangle$$

for all $x$ for which the supremum is finite.

Let $S$ be the region in $E^2$ defined by

$$S =\{(x,y): 0 \leq y \leq x-1\}.$$

Since $S$ is unbounded, the domain $D$ of the support function $h$ of $S$ is a proper subset of $E^2$, and we find

$$D=\{ (x,y):x \leq 0 \text{ and }y \leq -x\}.$$

I couldn't understand how he could get the set $D$ in this form?

Since $S$ is homogeneous, it suffices to consider unit vectors $(u,v)$. A geometric way to $h(u,v)$ is to project $S$ onto the directed line with direction vector $(u,v)$ and find how far in the positive direction this projection stretches.
If there exists $(x,y)\in S$ that forms acute angle with $(u,v)$, then the projection stretches infinitely far (considering the shape of $S$), hence $(u,v)\notin D$. Conversely, if $(x,y)\cdot (u,v)\le 0$ for all $(x,y)\in S$, then $h(u,v)=0$.
Thus, if we expand the angular region $S$ by $90$ degrees in both directions, the result is precisely the complement of $D$.