Prove that the support function of an arbitrary set $A\subset\mathbb R^n$ is convex 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
$\sigma_A(x):= \sup_{z \in A} \langle x,z \rangle$
Prove that $\sigma_A$ is convex.
Can I confirm that the inner product $\langle x, z \rangle$ is just a line orthogonal to $x$
I am confused by the meaning of "support function" and also what $\sigma_A(x)$ is.
If it's the supremum of the inner product of $x$ and $z$ (which tells us what exactly?) isn't it just going to be a single point/value? 
How can I prove this, and where has my interpretation failed me?
 A: First of all,
\begin{align*}
\sigma_A(x)=\sup_A\langle x, z\rangle=\sup\{\langle x,z\rangle\ |\ z\in A\}.
\end{align*}
To prove that $\sigma_A$ is convex, let $x,y\in\mathbb R^n$ and $t\in[0,1]$. Then
\begin{align*}
\sigma_A(tx+(1-t)y)&=\sup_{z\in A}\langle tx+(1-t)y,z\rangle \\ &=
\sup_{z\in A}\left(t\langle x,z\rangle+(1-t)\langle y,z\rangle\right) \\
&\leq
t\sup_{z\in A}\langle x,z\rangle+(1-t)\sup_{z\in A}\langle y,z\rangle \\ &=t\sigma_A(x)+(1-t)\sigma_A(y),
\end{align*}
which shows that $\sigma_A$ is convex.
A: When $f(x)=\sup_{a\in A}\ a\cdot x$, then for $x,\ y$ s.t. they are
close, there are two affine hyperplane $P_x,\ P_y$ s.t. $$(1)\ P_x\perp x\ {\rm and}\  P_y\perp y$$ $(2)$ $A,\ x$ (resp $y$) are in
different sides wrt $P_x$ (resp $ P_y$) and $(3)$ $P_x\bigcap A,\
P_y\bigcap A$ are non-empty.
Note that $f(x)= p\cdot x$ for any $p\in P_x$. When $z$ is a mid
point in $[xy]$, then
\begin{align*} f(z) &\leq \sup_{p\in P_x^+ \bigcup P_y^+}\ z\cdot p
\\ &=\frac{1}{2}\sup_{p\in P_x^+ \bigcup
P_y^+}\ x\cdot p + y\cdot p\\
&= \frac{1}{2} \sup_{p\in P_x^+\bigcup P_y^+}\ x\cdot p +
\frac{1}{2}\sup_{p\in P_x^+ \bigcup P_y^+}\ y\cdot p\\&=
\frac{1}{2}\{ f(x)+f(y)\}
\end{align*}
where $P_x^+$ is a closed half plane intersecting $A$ whose boundary
is $P_x\bigcap P_y$.
