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Let $C$ be a compact convex set in $\mathbb R^n$ that intersects the strictly positive orthant $\mathbb R_+^n$. Does $C\cap \mathbb R_+^n$ have to contain a point $x$ such that some vector $v\in\mathbb R_+^n$ is normal to $C$ at $x$?

Here, a vector $v$ is normal to a convex set $C$ at $x$ iff for all $y\in C$, $\langle v,y \rangle\le \langle v,x\rangle$.

Alexander Pruss
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  • Presumably you want $x \in \partial C$? – Joseph O'Rourke Feb 16 '21 at 16:33
  • Doesn't that automatically follow, since no supporting hyperplane goes through an interior point? – Alexander Pruss Feb 16 '21 at 16:42
  • OK, good; "contain" led to me wonder if I was misinterpreting. – Joseph O'Rourke Feb 16 '21 at 17:03
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    Just take the point in $C\cap \mathbb R^n_+$ with the maximal product of coordinates. – fedja Feb 16 '21 at 21:06
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    I think fedja is right. You maximize $f(x_1,...,x_n)= x_1\cdots x_n$ on $C\cap\Bbb R^n_+$, and the normal vector at that maximizer (which exists because $f$ is a convex function on a convex set) is the gradient of $f$ at that point, which lies in the positive orthant. – M. Winter Feb 16 '21 at 22:18
  • @MWinter: Actually, $f$ isn't convex (except coordinate-wise), but it is continuous and $C$ is compact, so $f$ has a maximum on $C\cap [0,\infty)^n$. That maximum is obviously not attained at any point with a zero coordinate, so the maximum is in $C\cap (0,\infty)^n$. – Alexander Pruss Feb 17 '21 at 04:57

1 Answers1

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While fedja gave an answer in the comments, here is a different approach that yields a more general answer (the non-negative orthant is self-dual).

Proposition: Let $C\subseteq \mathbb R^n$ be a compact convex set that meets some closed convex cone $K$. Then $C\cap K$ contains a point with a non-zero normal in the dual of $K$. If $C$ meets the interior of $K$ and the dual of $K$ has non-empty interior, then the point can be taken to have a non-zero normal in the interior of the dual of $K$.

Proof: Suppose $K=\{ z : \left< x,z \right> \ge 0 \text{ for all }x\in A \}$ for a countable set $A$ (this works for any separable Banach space). Let $B$ be the basis of the largest linear space contained in $K$. Thus if $\left< x,z \right> = 0$ for all $x\in A\cup B$, then $z=0$.

Enumerate $A\cup B=\{a_1,a_2,\dots\}$. Define the sequence of non-empty compact sets $C_i$ by letting $C_0 = C\cap K$ and $C_i$ be the set of points of $C_{i-1}$ that maximize $\left< a_i, \cdot \right>$. Fix $z\in\bigcup_i C_i$.

I claim that $C\cap (z+K) = \{ z \}$. For suppose $w \in K$ and $z+w \in C\cap (z+K)$. The maximality conditions imply that $\left< a_i, w \right>\le 0$ for all $i$. Hence $w=0$.

Let $H$ be a separating closed half-space with $z$ on its boundary that contains $C$ and the closure of whose complement contains $z+K$. The outward normal of $H$ is then in the dual of $K$.

Now suppose $C$ meets the interior of $K$ at some point $z_0$ and $K^*$ has non-empty interior. Let $K_\epsilon$ be an $\epsilon$-widened closed convex cone, e.g., the cone generated by the compact set $(K\cap H)+\bar B(0;\epsilon)$ (Minkowski sum) where $H$ is a hyperplane through a point in the interior of $K^*$ and normal to the vector from the origin to that point. Then for sufficiently small $\epsilon>0$, $C\cap (z_0+K_\epsilon) \subseteq C\cap \operatorname{Int} K$. Applying the already proved part to $-z_0+C$ and $K_\epsilon$, we get a point $z$ in $C$ with a non-zero normal in $(K_\epsilon)^*$. But $\operatorname{Int} (K^*) \supseteq (K_\epsilon)^* \backslash \{ 0 \}$.

Alexander Pruss
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