# Proof of closure, convex hull and minimal cone of dual set

In my studies of convexity I have recently come across the following:

Let $V=\mathbb{R}^d ; d \geq 1$ be a Euclidean space and $S \neq \emptyset$ be a non empty set of the vector space $V$, where we define its dual as usual $S^* = \{y\in V | \forall x \in S ; \langle y,x \rangle \geq 0 \}$. Now I have already proved that $S^*$ is always a closed convex cone, now I wish to prove the following: $S^* = (\bar{S})^* = {(conv(S))}^*= {(cone(S))}^*$ where "conv" is the usual convex hull and "cone" is the minimal convex cone containing $S$ and $\bar{S}$ is the closure of $S$.

I have proved that the dual set is always a closed convex cone, is there a way to prove these equalities I have written, I seem to be stuck on this. All help is appreciated.

Assume that $S$ has nonzero vector

Note that for $x\neq 0\in S$ $H_x:=\{ y|y\cdot x\geq 0\}$ is closed half space That is $S^\ast$ is intersection of these closed half space Hence $S^\ast$ is closed convex In further closed half space is cone so that $S^\ast$ is cone

(1) $S^\ast = (\overline{S})^\ast$ : From this argument we have that $$S\subset \overline{S}\Rightarrow S^\ast \supseteq (\overline{S})^\ast$$

If $x_n\in S$ and $x_n\rightarrow x$ let $$v\in \bigcap_{n} H_{x_n}$$

so that $$v\cdot x_n\geq 0 \Rightarrow v\cdot x\geq 0$$

That is $v\in H_x$ so that $\bigcap_{n} H_{x_n} \subset H_x$

(2) $S^\ast =({\rm conv}\ S)^\ast$ : $x_i\in S$ then $v:=\sum_{i=1}^n c_ix_i\in {\rm conv}\ S$ where $c_i\geq 0,\ \sum_i c_i=1$ Assume that $$w\in \bigcap_i H_{x_i}$$ Then $$w\cdot v =\sum_i c_ix_i\cdot w \geq 0\Rightarrow w\in H_v$$

(3) $({\rm conv}\ S)^\ast =({\rm cone}\ S)^\ast$ : For $v\in {\rm cone}\ S$, we have $$w\in {\rm conv}\ S,\ tw=v,\ t>0$$

Hence $$H_v=H_w$$