Show that support function of any set in $\mathbb{R}^n$ is lower semi-continuous function.

Question

Let $A \subseteq \mathbb{R}^n$.

The support function of set $A$ is defined as the following

$$
S_A(x)=\sup_{y \in A} x^Ty
$$
where $x \in \mathbb{R}^n$.

To show it is lower semi-continuous we have two choices.

1- Epi graph of a lower semi-continuous is closed.

2- Show that for any $x_c\in \mathbb{R}^n$ and $\forall \epsilon>0$ and $\forall x \in \mathbb{R}^n$ such that $\|x-x_c\| < \delta$ $\rightarrow$

$$
f(x_c)-\epsilon \leq f(x)
$$

I have tried both:

First method:

To show $epi(S_A)$ is closed we need to take a convergent sequence $(x_n,t_n)$ in $epi(S_A)$ whose limit point is $(x_l,t_l)$ and show that $(x_l,t_l)$ is indeed in the $epi(S_A)$. So let $(x_n,t_n)$ be a convergent sequence in $epi(S_A)$, then

$$
\sup_{y \in A} x_n^Ty \leq t_n
$$

If we take the limit we have

$$
\lim \sup_{y \in A} x_n^Ty \leq \lim t_n =t_l
$$

How I can show that $\sup_{y \in A} \leq t_l$ using the above statement?

Also, how I can proof the lower semi-continuity using the second method?

Answer 1

The relation $\sup_y x_n^Ty \le t_n$ implies $$ x_n^T y \le t_n \quad \forall y\in A, $$ here we can pass to the limit $$ x_l^Ty\le t_l \quad \forall y\in A, $$ now we can take supremum on the left-hand side again: $$ \sup_{y\in A} x_l^Ty \le t_l. $$