Let $A$ be a normed real space and $G$ a closed convex subset of $A$.
How do I show that $G$ is the intersection of all the closed halfspaces in $A$ containing $G$?
What I know:
A halfspace is $H_{f,c}=\{a\in A: f(a)\leq c\}$ for $f\in A^*$ and $c\in\mathbb{R}$. So I want to show that $G=\bigcap_{f\in A^*,G\subset H_f}H_f$.
What we want thus, is to show that $\bigcap_{f\in A^*,G\subset H_f}H_f\subset G$, since clearly $G\subset \bigcap_{f\in A^*,G\subset H_f}H_f$.
Further we know for all $a,b\in G$ that for all $\lambda\in[0,1]$ we have $\lambda a+(1-\lambda)y\in G$. How do I use these to prove the statement?
Edit: The separation theorem states: for a real or complex normed space $X$ with $A,B\subset X$ non-empty and disjoint convex sets we have
1. $A$ is open $\implies$ $\exists f\in X^*,\gamma\in\mathbb{R}$ s.t. $$\Re f(a)<\gamma\leq\Re f(b)$$ for $a\in A,b\in B$
2. $A$ is compact and $B$ is closed $\implies$ $\exists f\in X^*,\gamma\in\mathbb{R},\delta>0$ s.t. $$\Re f(a)\leq\gamma-\delta<\gamma+\delta\leq\Re f(b)$$ for $a\in A,b\in B$.
Let $x\in A\backslash G$. Since $\{x\}$ is compact and $G$ is closed, we have (by part 2 of the separation theorem) that there exist $f\in X^*$, $\gamma\in\mathbb{R}$ and $\delta>0$ such that $$f(x)\leq\gamma-\delta<\gamma+\delta\leq f(g)$$ for $g\in G$.
Then we see that $-f(x)\geq\delta-\gamma$. So $x\notin H_{-f,-\gamma}=\{a\in A:-f(a)\leq-\gamma\}$.
Let $g\in G$. Then $-f(g)\leq\gamma-\delta$, so we have $g\in H_{-f,-\gamma}=\{a\in A:-f(a)\leq-\gamma\}$.
Since all points outside of $G$ are not element of a halfspace and all elements of $G$ are, we find that $G=\bigcap \{H_{f,\gamma}:f\in A^*,\gamma\in\mathbb{R},G\subset H_{f,\gamma}\}$.