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.

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    $\begingroup$ 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 $X$.. $\endgroup$
    – kneidell
    Jun 27, 2019 at 9:51

1 Answer 1


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 $X$.

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    $\begingroup$ After projecting I get the conditions $z\geqslant 0$ and $z>e^y-1$, right? this is not an upper half-plane. $\endgroup$ Jun 27, 2019 at 9:46
  • $\begingroup$ @FedorPetrov: not really, that would be a cross-section. Projection is given by $z \geqslant \max\{0,e^x+e^y-1\}$ for some $x$. $\endgroup$ Jun 27, 2019 at 10:00
  • $\begingroup$ @MatthewDaws: Indeed, I messed up the example. I'll try to fix it soon. $\endgroup$ Jun 27, 2019 at 10:17
  • $\begingroup$ @MatthewDaws: Now it should work. Thanks! $\endgroup$ Jun 27, 2019 at 10:55
  • $\begingroup$ @FedorPetrov: Now it should work. Thanks! $\endgroup$ Jun 27, 2019 at 10:55

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