2
$\begingroup$

Let $K$ be a convex cone in ${\mathbb R}^n$. A continuous function $f:K\rightarrow\mathbb R$ satisfies a Hlawka inequality if $$f(0)+f(x+y)+f(y+z)+f(z+x)\le f(x)+f(y)+f(z)+f(x+y+z),\qquad\forall x,y,z\in K.$$ The original Hlawka inequality is for the Euclidian norm $f(x)=\|x\|$. Some non-Euclidian norms don't satisfy the Hlawka inequality. Mind that the inequality is trivial if one of the vectors vanishes.

It is rather natural to assume that $f$ is positively homogeneous of degree one~; only in this degree does the inequality turn into an inequality when $x,y,z$ are colinear. Mind that now $f(0)=0$.

When $f$ is positively homogeneous of degree one and is twice differentiable in the interior of $K$, a necessary condition for the Hlawka inequality is that $f$ be convex. Incidentally, I don't how to prove this necessity without the differentiability assumption.

Besides norms, popular functions that are convex and positively homogeneous of degree one, are given as $f=-p^{1/d}$ where $p$ is a hyperbolic polynomial, homogeneous of degree $d$, and $K$ is its future cone. This result is due to Garding. Therefore we may wander whether such functions satisfy the Hlawka inequality. Unfortunately, this turns out to be false in general: take the case of $p(X)=\det X$ over ${\bf Sym}_d^+(\mathbb R)$, for which $K$ is the cone of positive semi-definite matrices. If $n\ge2$, decompose $I_d=P+Q+R$ into non-trivial, mutually orthogonal projectors. Then $f( P)=\cdots=f(Q+R)=0$, while $f(P+Q+R)=-1$.

The case $d=2$ in the example above remains open. More or less it amounts to the following question:

Let $q$ be a non-degenerate quadratic form of signature $(1,n-1)$ (it is a hyperbolic polynomial). Let $K$ be the future cone of $q$. Does $-\sqrt q$ satisfy the Hlawka inequality over $K$ ? In other words, is is true that $$\sqrt{q(x+y)}+\sqrt{q(y+z)}+\sqrt{q(z+x)}\ge\sqrt{q(x)}+\sqrt{q(y)}+\sqrt{q(z)}+\sqrt{q(x+y+z)}\quad ?$$

Edit. Eventually, I got the answer, which is positive. See this file.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.