# How close is Mazur's Separation Theorem and the 2nd geometric form of HBT?

The Mazur's separation theorem, also known as "Geometric Hahn-Banach theorem", is:

Mazur's Theorem. Let $K$ be a convex set having a nonempty interior in a real normed linear vector space $X$. Suppose $V$ is a linear variety in $X$ containing no interior points of $K$. Then there is a closed hyperplane in $X$ containing $V$ but containing no interior points of $K$; i.e., there is an element $x^\ast \in X^\ast$ and a constant $c$ such that ${\displaystyle \langle x^{*},v\rangle =c}$ for all $v \in V$ and ${\displaystyle \langle x^{*},k\rangle <c}$ for all $k \in K^\circ$.

And there is, by Haïn Brézis' "Functional Analysis, Sobolev Spaces and Partial Differential Equations", this result:

Hahn–Banach Theorem (second geometric form). Let $X$ be a normed linear space. Let $A \subset X$ and $B \subset X$ be two nonempty convex subsets such that $A \cap B = \varnothing$. Assume that $A$ is closed and $B$ is compact. Then there exists a closed hyperplane that strictly separates $A$ and $B$.

When a hyperplane separates $A$ and $B$ then there is a linear functional $\phi\in X^\ast$ and a real constant $c$ such that $\langle \phi, a\rangle > c$ for all $a\in A$ and $\langle \phi, b\rangle < c$ for all $b\in B$.

Since they are both considered geometrics versions of Hahn-Banach Theorem and are very similar, is there a easy way to proof their equivalence (if they are equivalent) by, for instance, identifying the sets in the hypothesis and the respective linear functional without relying on heavy results in Functional Analysis? It seems to me that at least the latter implies the first, but I don't know how to do it.

• Mazur's Theorem is stated somewhat strangely. If $K$ is convex, then its interior is convex. Why work with $K$ when the statement is essentially about its interior? Also, if you have access to a copy of Brezis, the first geometric form of the Hahn-Banach theorem stated there is very similar (but not equivalent to) the statement of Mazur's Theorem here. – fourierwho Jul 11 '18 at 22:03