Is there a direct way to reduct the general case of Brunn - Minkowski inequality $|A+B|^{1/n}\geqslant |A|^{1/n}+|B|^{1/n}$ for non-empty compact sets $A, B\subset \mathbb{R}^n$ (here $|\cdot|$ denotes Lebesgue measure in $\mathbb{R}^n$) to the case of convex compact sets?
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asked Aug 29, 2022 at 11:23
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$\begingroup$ Why do you need it? the nonconvex case is has a short proof. $\endgroup$– Anton PetruninCommented Aug 29, 2022 at 12:05
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3$\begingroup$ @AntonPetrunin the inequalities I am thinking about are somehow similar to BM and we have proofs in the convex case only $\endgroup$– Fedor PetrovCommented Aug 29, 2022 at 12:35
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$\begingroup$ It might be possible to do this by providing a version of the linearized (i.e., mixed volume) inequality. $\endgroup$– Deane YangCommented Aug 29, 2022 at 15:34
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2$\begingroup$ @DeaneYang but what are mixed volumes in the non-convex case? The measure $|A+tB|$ is not a polynomial in $t$ $\endgroup$– Fedor PetrovCommented Aug 29, 2022 at 16:23
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$\begingroup$ Here, by the mixed volume of $A$ and $B$, I mean $$V(A,B) = \frac{1}{n}\lim_{t\rightarrow 0+} \frac{|A + tB|-|A|}{t}.$$ The mixed volume inequality states that $$V(A,B) \ge |A|^{n-1}|B|.$$ If $A$ and $B$ are convex, then this is equivalent to the Brunn-Minkowski inequality. The mixed volume is also equal to the integral of the support function of $B$ with respect to the surface area measure of $A$, $$V(A,B) = \int_{S^{n-1}} h_B(u)\,dS_A(u). $$ But on further thought, I doubt this can be used to do what you want. $\endgroup$– Deane YangCommented Aug 29, 2022 at 23:13
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In a recent preprint of Conlon and Lim (link), they proved a discrete Brunn-Minkowski-type result (Lemma 2.1) by using “compressions” to reduce to the case where $A,B$ are well-behaved connected subsets.
By using more aggressive “compressions” and induction, I think you should be able to reduce Brunn-Minkowski to the case where $A,B$ are both simplexes. Here’s a sketch:
Basically, given a subset $U$ and codimension-1 hyperplane $H$ with normal vector $v$, we define the $H$-compression of $U$, $U’ = C_H(U)$ , by making $U’ \cap (H+tv)$ a symmetric simplex based at $tv$ with $(d-1)$-dimensional volume equal to that of $U \cap (H+tv)$.
If you use $d$-hyperplanes $H_1,\dots,H_d$ whose normal vectors form an orthonormal basis, then repeatedly applying $C_{H_1},\dots, C_{H_d}$ to a set $U$ should converge towards a simplex with the same volume as $U$, which we denote $C(U)$.
We claim that $\mu(C(A)+C(B)) \le \mu(A+B)$. It suffices to check that $\mu(C_H(A)+C_H(B)) \le \mu(A+B)$ for any hyperplane $H$, this should follow from the assumption that Brunn-Minkowski holds on $(d-1)$-dimensions which implies $C_H(A)+C_H(B) \subset C_H(A+B)$. Thus, we’ve reduced to the case where $A,B$ are simplexes and thus convex.
answered Aug 29, 2022 at 19:40