# Intersection of a closed convex set

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 cant 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)

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.
$$f(x)\le f(a)$$ 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.