TITLE: Show that $C \subseteq D$ only if $\sigma_C(x) \leq \sigma_D(x)$ $\forall x \in \mathbb{R}^n$
QUESTION [1 upvotes]: 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 $x \in \mathbb{R}^n$.

REPLY [1 votes]: 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.
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