Show that the support function of a bounded set is continuous. The support function of a set $A \in \mathbb{R}^n$ is defined as the following
$$
S_A(x)=\sup_{y \in A} x^Ty
$$
where $x \in \mathbb{R}^n$.
Show that the support function of a bounded set is continuous.
I tried the following:
Let $A$ be a bounded set in $\mathbb{R}^n$ and $y \in A$. So $\|y\| \leq M$ where $M>0$. (If $M=0 \rightarrow \|y\|=0 \rightarrow y=0 \rightarrow S_A(x)=0\rightarrow S_A(x)$ is continuous $\forall x$).
Let $\|x-x_c\|<\delta=\frac{\epsilon}{M}, \,\,\,\forall \epsilon>0$ be a neighborhood of $x_c$ where $x_c \in \mathbb{R}^n$.
I need to show that 
$$
|S_A(x)-S_A(x_c)|=|\sup_{y \in A} x^Ty-\sup_{y \in A} x_c^Ty|<\epsilon
$$
How can I proceed?
 A: Using Cauchy-Schwarz inequality we can bound $(x^{T}-x_c^{T})y$ 
$(x^{T}-x_c^{T})y\leq \|x^{T}-x_c^{T}\|\|y\| $
Since $\|x^{T}-x_c^{T}\| < \delta$, we have the following
So
$$(x^{T}-x_c^{T})y \leq \|x^{T}-x_c^{T}\|\|y\| < \delta \|y\|$$
So $x_c^{T}y < \delta \|y\|+x_c^{T}y$ for all $y$ in $A$.
Take sup over $y$
$$\sup_{y \in A} x_c^{T}y < \sup_{y \in A} (\delta \|y\|+x_c^{T}y) \leq \sup_{y \in A} \delta \|y\|+ \sup_{y \in A}x_c^{T}y$$
So 
$$S_A(x) < \delta M + S_A(x_c)$$
Do the same for $(x_c^{T}-x^{T})y$ to get 
$$S_A(x_c) < \delta M + S_A(x)$$
Combine them to get the following
$|S_A(x)-S_A(x_c)|<\delta M=\epsilon$.
A: Here's a solution which reduces to another problem on this site: Supremum is continuous over equicontinuous family of functions
Step 1: define for each $a\in A$ the function $f_a(x) = x \cdot a$. Each of these is continuous.
Step 2: Show that $S_A$ is the supremum of the family $\{f_a\}$
Step 3: Show that because $A$ is bounded, the family is not only continuous but equicontinuous.
Then conclude using the link above!
