Given an arbitrary set $A ⊂ \mathbb{R}^n$ , the support function associated with the set $A$
$ σ_A : \mathbb{R}^n \to \mathbb{R} ∪ \{+\infty\}$ is defined as
$\sigma_A(x):= \sup_{z \in A} \langle x,z \rangle$
Let $C, D ⊂ \mathbb{R}^n$ be nonempty, closed and convex sets.
How can I show $C = D$ if and only if $\sigma_C(x) = \sigma_D(x) \forall x ∈ \mathbb{R}^n$
If we assume $\sigma_C(x) = \sigma_D(x)$ and $ C \not\subset D$ then $\exists \space \bar z \not\subset D$ such that $\bar z \in C$. Using the projection theorem is there a way to show a contradiction perhaps?
You can also use the orthogonal projection to show $\sigma_C\equiv\sigma_D$ $\Rightarrow$ $C=D$ (the other implication is trivial...). To do this fix a point $c\in C$. Since $D$ is closed and convex, there exists the orthogonal projection of $c$ onto $D$, i.e. there is some $d\in D$ such that \begin{align*} \langle c-d,x-d\rangle\leq0\qquad\forall x\in D. \end{align*} This implies $\langle c-d,x\rangle\leq\langle c-d,d\rangle$ for all $x\in D$. Taking the supremum over $D$ on the left hand side yields \begin{align*} \sigma_D(c-d)\leq\langle c-d,d\rangle, \end{align*} and since $\sigma_D=\sigma_C$ we obtain \begin{align*} \sigma_C(c-d)\leq\langle c-d,d\rangle, \end{align*} which implies that $\langle c-d,x\rangle\leq\langle c-d,d\rangle$ for each $x\in C$. In particular, for $x=c$ we obtain \begin{align*} 0\geq\langle c-d,c\rangle-\langle c-d,d\rangle=||c-d||^2, \end{align*} and finally $c=d$. Thus, we have shown that $C\subset D$. Interchanging the róles of $c$ and $d$ will yields $D\subset C$ and hence $C=D$.