# On the geometric Hahn-Banach theorem

Let $$X \subset \ R ^ n$$ be a closed convex set and let $$L$$ be a straight line such that $$X \cap L = \emptyset$$. Does there exist a hyperplane containing $$L$$ that does not intersect $$X$$ ?

In the classical Hahn-Banach theorem we require $$X$$ to be open.

• I assume you mean affine hyperplane? If you take $X$ to be any convex set containing $0$ then you have no chance of finding a hyperplane which doesn't intersect $X$.. Jun 27, 2019 at 9:51

Not necessarily: consider a 3-D space, let $$L$$ be the $$x$$ axis, and let the convex set be $$X=\bigl\{(x,y,z):z\geqslant (\max\{0,y + e^x\})^2\bigr\}.$$ This is the region above the graph of a convex function; here's a 3-D plot. After projecting on the $$yz$$ plane, the convex set becomes $$X' = \{(y, z) : y < 0 , \, z \geqslant 0\} \cup \{(y, z) : y \geqslant 0, z > y^2\}.$$ There is no line in the $$yz$$ plane that contains the origin and does not intersect $$X'$$, and hence there is no hyperplane that contains the $$x$$ axis and does not intersect $$X$$.
• After projecting I get the conditions $z\geqslant 0$ and $z>e^y-1$, right? this is not an upper half-plane. Jun 27, 2019 at 9:46
• @FedorPetrov: not really, that would be a cross-section. Projection is given by $z \geqslant \max\{0,e^x+e^y-1\}$ for some $x$. Jun 27, 2019 at 10:00