Let $A, B, C$ be compact convex sets in $\Bbb R^n$ such that $A + C = B + C$. The purpose of this problem is to prove that $A = B$. Define the support function $$\sigma _A (x) := \max\{\langle x, u\rangle : u ∈ A\}.$$ (a) Show that $\sigma _A$ is a convex function defined for all $x ∈ R ^n$ .
Proof
Observe: $\sigma_A((1-t)x+ty)=\max\{\langle(1-t)x+ty,u\rangle:u \in A\}=\max\{\langle(1-t)x,u\rangle+\langle ty,u\rangle :u \in A\}\leq \max\{\langle(1-t)x,u\rangle :u \in A\}+\max\{\langle ty,u \rangle :u \in A\}=(1-t)\sigma_A(x)+t\sigma_A(y)$
(b) Show that $$\sigma_{A+B}=\sigma_A +\sigma_B$$
I need help with part (b).
Let $C=\{<x,u>:u \in A\}$ , $D=\{<x,u>:u \in B\}$ and fix $x \in R^{n}$. We have $z \leq \sigma_{A}(x) \forall z \in C$ and $y \leq \sigma_{B}(x) \forall z \in D$. Hence we have $z+y \leq \sigma_{A}(x)+\sigma_{B}(x) \forall z \in C, y\in D$. Hence we have$ \sigma_{A+B}=max_{z \in C, y \in D} z+y \leq \sigma_A(x)+\sigma_B(x)$