Show that $C \subseteq D$ only if $\sigma_C(x) \leq \sigma_D(x)$ $\forall x \in \mathbb{R}^n$

Question

Let $C,D \in \mathbb{R}^n$ be two nonempty set, where $D$ is closed and convex. Show that if

$$\sigma_C(x) \leq \sigma_D(x)$$

then $C \subseteq D$.

This simply means that the support function of a smaller set is less than the bigger one provided that the bigger set is closed and convex.

Note:

The support function of a set $A \in \mathbb{R}^n$ is defined as the following

$$ \sigma_A(x)=\sup_{y \in A} x^Ty $$ where $x \in \mathbb{R}^n$.

Answer 1

Assume $C \not\subset D$. Therefore there exist a point $p$ in $C$ which is not in $D$.

Since $D$ is closed, $p$ is not in the closure of $D$. Also, since $D$ is convex, and $p$ is not on the closure of $D$, $p$ and $D$ are strongly separated. Hence, there exist $0\neq x \in \mathbb{R}^n$ for which

$$ \langle x , y \rangle < \langle x , p \rangle \,\,\,\,\, \forall y \in D $$

Since the left hand side holds for $\forall y \in D$, it holds for the following

$$ \sup_{y \in D}\langle x , y \rangle < \langle x , p \rangle $$

Since $p$ is in $C$, the above right hand side is always less than

$$ \sup_{y \in D}\langle x , y \rangle < \langle x , p \rangle \leq \sup_{p \in C}\langle x , p \rangle $$

Hence,

$$\sigma_D(x) < \sigma_C(x)$$

which contradicts with the assumption. Therefore, the claim.

Answer 2

Notice that $\sigma_A(x)=\sup_{y\in A}\sigma_{\{y\}}(x)$, so the case $C=\{c\}$ is sufficient; in other words, the goal is to prove that, if $y\notin D$, then there are some $x\in\Bbb R^n$ and some $d\in D$ such that $x^Td> x^Ty$. In fact, by Hahn-Banach there are a functional $\phi:\Bbb R^n\to\Bbb R$ and a constant $\alpha$ such that $\phi(y)< \alpha$ and $\phi(d)>\alpha$ for all $d\in D$. There is exactly one $x_\phi$ such that $\phi(v)=x_\phi^Tv$ for all $v\in\Bbb R^d$. That $x_\phi$ works.