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