An isomorphic version of the Busemann–Petty problem for arbitrary measures
An isomorphic version of the Busemann–Petty problem for arbitrary measures
复制标题
任意测量的 Busemann-Petty 问题的同构版本
DOI：
10.1007/s10711-014-0016-x
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发表时间：
2014
影响因子：
0.5
通讯作者：
A. Zvavitch
中科院分区：
文献类型：
--
作者：
A. Koldobsky;A. Zvavitch
The Busemann–Petty problem for an arbitrary measure $$\mu $$μ with non-negative even continuous density in $${\mathbb R}^n$$Rn asks whether origin-symmetric convex bodies in $${\mathbb R}^n$$Rn with smaller $$(n-1)$$(n-1)-dimensional measure $$\mu $$μ of all central hyperplane sections necessarily have smaller measure $$\mu .$$μ. It was shown in Zvavitch (Math Ann 331:867–887, 2005) that the answer to this problem is affirmative for $$n\le 4$$n≤4 and negative for $$n\ge 5$$n≥5. In this paper we prove an isomorphic version of this result. Namely, if $$K,M$$K,M are origin-symmetric convex bodies in $${\mathbb R}^n$$Rn such that $$\mu (K\cap \xi ^\bot )\le \mu (M\cap \xi ^\bot )$$μ(K∩ξ⊥)≤μ(M∩ξ⊥) for every $$\xi \in {\mathbb S}^{n-1},$$ξ∈Sn-1, then $$\mu (K)\le \sqrt{n}\ \mu (M).$$μ(K)≤nμ(M). Here $$\xi ^\bot $$ξ⊥ is the central hyperplane perpendicular to $$\xi .$$ξ. We also study the above question with additional assumptions on the body $$K$$K and present the complex version of the problem. In the special case where the measure $$\mu $$μ is convex we show that $$\sqrt{n}$$n can be replaced by $$cL_n,$$cLn, where $$L_n$$Ln is the maximal isotropic constant. Note that, by a recent result of Klartag, $$L_n \le O(n^{1/4})$$Ln≤O(n1/4). Finally we prove a slicing inequality $$\begin{aligned} \mu (K)\le C n^{1/4} \max _{\xi \in {\mathbb S}^{n-1}} \mu (K \cap \xi ^\perp )\ \mathrm{vol}_{n}(K)^{\frac{1}{n}} \end{aligned}$$μ(K)≤Cn1/4maxξ∈Sn-1μ(K∩ξ⊥)voln(K)1nfor any convex even measure $$\mu $$μ and any symmetric convex body $$K$$K in $${\mathbb R}^n,$$Rn, where $$C$$C is an absolute constant. This inequality was recently proved in Koldobsky (Adv Math 254:33–40, 2014) for arbitrary measures with continuous density, but with $$\sqrt{n}$$n in place of $$n^{1/4}.$$n1/4.
The Busemann–Petty problem for an arbitrary measure $$\mu $$μ with non-negative even continuous density in $${\mathbb R}^n$$Rn asks whether origin-symmetric convex bodies in $${\mathbb R}^n$$Rn with smaller $$(n-1)$$(n-1)-dimensional measure $$\mu $$μ of all central hyperplane sections necessarily have smaller measure $$\mu .$$μ. It was shown in Zvavitch (Math Ann 331:867–887, 2005) that the answer to this problem is affirmative for $$n\le 4$$n≤4 and negative for $$n\ge 5$$n≥5. In this paper we prove an isomorphic version of this result. Namely, if $$K,M$$K,M are origin-symmetric convex bodies in $${\mathbb R}^n$$Rn such that $$\mu (K\cap \xi ^\bot )\le \mu (M\cap \xi ^\bot )$$μ(K∩ξ⊥)≤μ(M∩ξ⊥) for every $$\xi \in {\mathbb S}^{n-1},$$ξ∈Sn-1, then $$\mu (K)\le \sqrt{n}\ \mu (M).$$μ(K)≤nμ(M). Here $$\xi ^\bot $$ξ⊥ is the central hyperplane perpendicular to $$\xi .$$ξ. We also study the above question with additional assumptions on the body $$K$$K and present the complex version of the problem. In the special case where the measure $$\mu $$μ is convex we show that $$\sqrt{n}$$n can be replaced by $$cL_n,$$cLn, where $$L_n$$Ln is the maximal isotropic constant. Note that, by a recent result of Klartag, $$L_n \le O(n^{1/4})$$Ln≤O(n1/4). Finally we prove a slicing inequality $$\begin{aligned} \mu (K)\le C n^{1/4} \max _{\xi \in {\mathbb S}^{n-1}} \mu (K \cap \xi ^\perp )\ \mathrm{vol}_{n}(K)^{\frac{1}{n}} \end{aligned}$$μ(K)≤Cn1/4maxξ∈Sn-1μ(K∩ξ⊥)voln(K)1nfor any convex even measure $$\mu $$μ and any symmetric convex body $$K$$K in $${\mathbb R}^n,$$Rn, where $$C$$C is an absolute constant. This inequality was recently proved in Koldobsky (Adv Math 254:33–40, 2014) for arbitrary measures with continuous density, but with $$\sqrt{n}$$n in place of $$n^{1/4}.$$n1/4.