Intersection of a closed convex set

[Funky1981]

Let X be a real Banach Space, C be a closed convex subset of X.

Define Lc = {f: f - a ∈ X* for some real number a and f(x) ≥ 0 for all x ∈ C} (X* is the dual space of X)

Using a version of the Hahn - Banach Theorem to show that

C = ∩ {x ∈ X: f(x) ≥ 0} with the index f ∈ Lc under the intersection Could someone help me to solve this problem, i can't see how Hahn - Banach can imply the above statement ( I used the separation version to obtain g(x)<a<g(y) for some linear functional g)

 

[Hawkeye18]

You will need the following corollary of the Hahn--Banach theorem (with sublinear functional): if $U$ is a convex open set, and $a\notin U$ then there exists a non-zero linear functional $f\in X^*$ such that $f(x)\le f(a)$ for all $x\in U$. To prove the corollary you first notice that because of translation invariance, you can assume without loss of generality that $0\in U$.

Then you define a sublinear functional $q$ to be the Minkowski functional of $U$, and define $f_0$ to be the linear functional on the one-dimensional space spanned by $a$ such that $f_0(a)=1$. Extending $f_0$ by Hahn--Banach you get a functional $f$ such that $f(x)\le q(x)$ for all $x\in X$. Then for all $x\in U$ $$ f(x) \le q(x) \le 1 = q(a),$$ so we get the desired inequality.
We still need to show that $f\in X^*$ (i.e. that $f$ is bounded), but that is easy: $U$ is open, so an open ball $B_\varepsilon \subset U$, so $q(x), q(-x)\le \frac1\varepsilon \|x\|$ ($B_\varepsilon$ is an open ball of radius $\varepsilon$ centered at $0$), and therefore $|f(x)|\le \frac1\varepsilon \|x\|$.

To get the statement you need, take $a \notin C$. Then for some $\varepsilon>0$ the whole open ball $x+B_\varepsilon$ does not intersect $C$, or, equivalently, the (open) set $U:= C+B_\varepsilon$ does not contain $a$. Applying the above corollary yo get that there exists $f\in X^*$ such that
$$ f(x)\le f(a)$$ for all $x\in U$, or, equivalently, $$f(x) \le f(a) + f(y) \qquad \forall x\in C, \ \forall y\in B_\varepsilon.$$ Noticing that for any non-zero $f\in X^*$ the point $0$ is an interior point of $f(B_\varepsilon)$ (in fact $f(B_\varepsilon)$ is an open set), we conclude that there exists real $b$ such that $$f(x) \le b <f(a) \qquad \forall x\in C.$$ The last statement means that $C$ contains the intersection of the half-spaces. The opposite inclusion is trivial.