Is the support function always unique for a convex set?

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 $C=D$.
• In regards to etiquette, isn't it incorrect to assume what we need to prove? I refer to the line where you write "since $\sigma_D = \sigma_C$" and go on from there. – diabloescobar Jan 21 '15 at 16:08
• You have to prove the equivalence $\sigma_C=\sigma_D$ $\Leftrightarrow$ $C=D$. The part "$\Leftarrow$" is trivial, and the other part is shown in my answer, i.e. there I have proven $C=D$ under the assumption $\sigma_C=\sigma_D$. Istn't that what you asked for? Sry when I misunderstood something... – sranthrop Jan 21 '15 at 17:20
Every half-plane can be written as $H(r,x):=\{z:\ r\geq\langle x,z\rangle\}$, for some $r$ and some $x$. Moving $r$ parallel translates the boundary of the half-space. If $H(r,x)$ contains the set $A$ then $$A\subset H(\sigma_A(x),x)\subset H(r,x).$$
Therefore if $C$ is convex then $C$ is equal to the intersection of all $H(\sigma_C(x),x)$. From this it follows that $\sigma_C$ determines $C$.