When thinking of some other problem, I stumbled upon the following innocently looking question that is natural enough to have been considered (and, possibly, solved) many years ago. However my attempts to search the literature for an answer resulted in next to nothing.

Let $K\subset\mathbb R^n$ be a convex cone and let $K^*=\{y:\langle x,y\rangle\ge 0\text{ for every }x\in K\}$ be its dual cone. Suppose that $K\supset K^*$ (or, if you prefer, even that $K=K^*$). What is the minimal possible ratio $\frac{|K\cap B|}{|B|}$ where $B$ is the unit ball in $\mathbb R^n$ when $n$ is large?

The answer should, probably, be of order $2^{-n}$ (positive orthant) but the best clean lower bound I can prove myself with my "homemade tools" is $(\sqrt 2+1)^{-n}$ (it can be improved a bit further to something like $2.317^{-n}$ but the argument gets somewhat messy and it is clear that this way won't lead to the optimal estimate).

Any help would be appreciated.

  • 3
    $\begingroup$ Not an answer, because it is more of a maximal ratio. If $n=\frac{N(N+1)}2$ and $K$ is the cone of positive symmetric $N\times N$ matrices, then the ratio is of order $3^{n/2}$. See Mikael's solution to my question The probability for a symmetric matrix to be positive definite, mathoverflow.net/questions/118481/… $\endgroup$ – Denis Serre Apr 24 '18 at 18:32
  • 4
    $\begingroup$ @DenisSerre The maximal ratio is known to be achieved by the circular cone (all directions making an angle of $\pi/4$ or less with some given one), which is easily seen to be about $2^{-n/2}$. For some reason that totally eludes me all papers I found (3 or 4) in which the maximum is discussed keep total silence about the other extremum. $\endgroup$ – fedja Apr 24 '18 at 23:29
  • $\begingroup$ @fedja Did you ask Artem? $\endgroup$ – Piotr Hajlasz Apr 25 '18 at 16:30
  • $\begingroup$ @PiotrHajlasz And Dima too, though not Christos yet. $\endgroup$ – fedja Apr 26 '18 at 14:15
  • $\begingroup$ You would perhaps find interesting the following question: mathoverflow.net/q/299770/121665. I don't have time to think about it. $\endgroup$ – Piotr Hajlasz May 13 '18 at 17:35

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I thought the following simple restatement of a bit narrower problem might help. Suppose that the convex cone $K$ is polyhedral: \begin{equation} K=\{x\in\R^n\colon a_i\cdot x\ge0\ \;\forall i\in\{1,\dots,N\}\}, \end{equation} where $a_1,\dots,a_N$ are nonzero vectors in $\R^n$ and $\cdot$ denotes the dot product (the case of a general convex cone $K$ can hopefully be done by approximation). Then, by Farkas' lemma (say), the dual cone $K^*$ is the conical span of the $a_i$'s. So, the condition $K\supseteq K^*$ means that $a_i\cdot a_j\ge0$ for all $i,j$, that is, all the angles between the vectors $a_i$ are $\le\pi/2$.

If now $N\le n$, then it should be comparatively easy to move the $a_i$'s so that they become pairwise orthogonal and in the process the cone $K^*$ (which is the conical span of the $a_i$'s) only increases, so that $K$ only decreases, and then so does its "spherical angle" measure $\mu(K):=\frac{|K\cap B|}{|B|}$. Thus, in the case when $N\le n$, we will have $\mu(K)\ge1/2^n$, as desired.

Then it will "only" remain to consider the case when $N>n$. Then one may try to find movements of $a_i$'s which will bring all of them into the conical hull of $n$ of the $a_i$'s without increasing $\mu(K)$; this certainly seems much more difficult to do than the above. Another thing to try here may be to embed the $a_i$'s into, or approximate them by vectors in, or otherwise map them into, $\R^N$.


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