西南大学学报 (自然科学版)  2019, Vol. 41 Issue (10): 51-55.  DOI: 10.13718/j.cnki.xdzk.2019.10.007
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  • 平面凸体的α周长及等周不等式    [PDF全文]
    郭欢欢1, 李德宜1,2, 邹都1     
    1. 武汉科技大学 理学院, 武汉 430081;
    2. 武汉科技大学 湖北名师工作室, 武汉 430081
    摘要:Brunn-Minkowski不等式是凸几何分析的重要研究内容.目前,关于体积等几何量的Brunn-Minkowski不等式已广为人知,并在数学各个分支中扮演着重要的角色.关于凸体表面积的Brunn-Minkowski不等式作为Aleksandrov-Fenchel不等式的特殊情况也得到确证.但在Lp Brunn-Minkowski理论中,Lp表面积测度的Brunn-Minkowski不等式仍是一个重要的公开问题,不论是对0<p<1,还是p>1的情形,都没有行之有效的方法来证明相关猜测.基于Minkowski加法,利用单调有界定理和积分中值定理研究了平面凸体的α-周长,提出了两凸体关于α-周长的Brunn-Minkowski型不等式,并对两凸体分别为正n边形和单位圆盘的情形给出了证明.
    关键词凸体    α-周长    Brunn-Minkowski不等式    

    由Lutwak引入并在众多数学家的推动下,Brunn-Minkowski理论核心在近20年内发展成Lp Brunn-Minkowski理论(参见文献[1-13]).该理论中的根本性和基础性概念之一是凸体的Lp表面积测度(参见文献[14]).设p$ {\mathbb{R}}$K$ {\mathbb{R}}^2$中含原点于内部的凸体,则凸体KLp表面积测度SP(K,·)是单位球面Sn-1上的Borel测度.对任意Borel子集ωSn-1

    $ {S_p}\left( {K,\omega } \right) = \int\limits_{{g_K}^{ - 1}\left( \omega \right)} {{{\left( {{g_K}\left( x \right) \cdot x} \right)}^{1 - p}}{\rm{d}}{\mathscr{H}}\left( x \right)} $

    其中gK是定义在∂K上的广义Gauss映射.围绕Lp表面积的Minkowski问题(即Lp Minkowski问题)是Lp Brunn-Minkowski理论的基石(参见文献[15-17]). Lp表面积测度的总值Sp(KSn-1)称为KLp表面积. L1表面积是熟知的表面积,而L0表面积是体积,精确地讲,S0(K)=nV(K). Brunn-Minkowski理论的核心之一是Brunn-Minkowski不等式(参见文献[18-23]).

    KLn维欧氏空间中的凸体,则

    $ V{\left( {K + L} \right)^{\frac{1}{n}}} \ge V{\left( K \right)^{\frac{1}{n}}} + V{\left( L \right)^{\frac{1}{n}}} $ (1)

    等式成立当且仅当KL位似.就凸体的Lp表面积,张高勇提出了如下猜想:

    KL$ {\mathbb{R}}^n$中含原点于内部的凸体,0<p<1,则

    $ {S_p}{\left( {K + L} \right)^{\frac{1}{{n - p}}}} \ge {S_p}{\left( K \right)^{\frac{1}{{n - p}}}} + {S_p}{\left( L \right)^{\frac{1}{{n - p}}}} $ (2)

    n=2时,改记Lα(K)=S1-α(K),称为凸体Kα-周长.张高勇猜想的平面情形改写为如下不等式:

    $ {L_\alpha }{\left( {K + L} \right)^{\frac{1}{{1 + \alpha }}}} \ge {L_\alpha }{\left( K \right)^{\frac{1}{{1 + \alpha }}}} + {L_\alpha }{\left( L \right)^{\frac{1}{{1 + \alpha }}}} $ (3)

    其中0<α<1.本文就此问题做初步的讨论,将证明如下结果:

    定理1  若KL分别是质心在原点的正n边形域和圆盘,则不等式(3)成立.

      对平面上含原点于内部的凸体K,由α-周长的定义可知:对任意t>0,有

    $ {L_\alpha }{\left( {tK} \right)^{\frac{1}{{1 + \alpha }}}} = t{L_\alpha }{\left( K \right)^{\frac{1}{{1 + \alpha }}}} $

    由此正齐次性,不等式(3)的等价形式是

    $ {L_\alpha }{\left( {{t_1}K + {t_2}L} \right)^{\frac{1}{{1 + \alpha }}}} \ge {t_1}{L_\alpha }{\left( K \right)^{\frac{1}{{1 + \alpha }}}} + {t_2}{L_\alpha }{\left( L \right)^{\frac{1}{{1 + \alpha }}}} $

    其中t1t2>0.基于此事实,不妨假设K是单位圆盘B的外接正n边形域.并证明:

    $ {L_\alpha }{\left( {K + \varepsilon B} \right)^{\frac{1}{{1 + \alpha }}}} \ge {L_\alpha }{\left( K \right)^{\frac{1}{{1 + \alpha }}}} + \varepsilon {L_\alpha }{\left( B \right)^{\frac{1}{{1 + \alpha }}}} $ (4)

    其中ε>0.

    由于Lα为旋转不变量,不妨设K的一个顶点在x轴上,于是K的边上的单位外法向量角度为

    $ \begin{array}{*{20}{c}} {{\theta _i} = \frac{{\rm{ \mathsf{ π} }}}{n} + \frac{{2{\rm{ \mathsf{ π} }}\left( {i - 1} \right)}}{n}}&{i = 1, \cdots ,n} \end{array} $

    进而可知

    $ \begin{array}{*{20}{c}} {{h_k}\left( {{\theta _i}} \right) = 1}&{{S_K} = \sum\limits_{i = 1}^n 2 \tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right){\delta _{{\theta _i}}}} \end{array} $

    这里S(K,·)简写为SK表示表面积,δθi表示集中于θi的概率点测度.从而可以得到凸体Kα-周长为

    $ \begin{array}{l} {L_\alpha }\left( K \right) = \int_{{s^1}} {{h_K}} {\left( {{\theta _i}} \right)^\alpha }{\rm{d}}S\left( {K, \cdot } \right) = \sum\limits_{i = 1}^n {{h_K}} {\left( {{\theta _i}} \right)^a}{S_K}\left( {\left\{ {{\theta _i}} \right\}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;n \cdot {1^\alpha } \cdot 2\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right) = 2n\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right) \end{array} $ (5)

    可直接计算圆盘

    $ {L_\alpha }\left( {\varepsilon B} \right) = \int_{{s^1}} {h_{\varepsilon B}^\alpha } {\rm{d}}S\left( {\varepsilon B, \cdot } \right) = \int_0^{2{\rm{ \mathsf{ π} }}} {{\varepsilon ^\alpha }} \cdot \varepsilon {\rm{d}}\theta = 2{\rm{ \mathsf{ π} }}{\varepsilon ^{1 + \alpha }} $ (6)

    接下来,计算Kε=K+εBα-周长Lα(Kε).当$ {\rm{ - }}\frac{\pi }{n} \le \theta \le \frac{\pi }{n}$时,由支撑函数的定义

    $ \begin{array}{l} {h_{{K_\varepsilon }}}\left( \theta \right) = {h_K}\left( \theta \right) + {h_{\varepsilon B}}\left( \theta \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\left( {\frac{1}{{\cos \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}},0} \right) \cdot \left( {\cos \theta ,\sin \theta } \right) + \varepsilon = \frac{{\cos \theta }}{{\cos \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}} + \varepsilon \end{array} $

    支撑函数hKε(θ)会出现两种情况:当θ=θi时,hKε(θ)=1+ε;当θ$ \left( { - \frac{{\left( {2k - 1} \right)\pi }}{n}, \frac{{\left( {2k - 1} \right)\pi }}{n}} \right), k \in {\mathbb{Z}} $时,hKε(θ)=$ \frac{{{\rm{cos}}\theta }}{{{\rm{cos}}\left( {\frac{\pi }{n}} \right)}} + \varepsilon $.则

    $ {S_{{K_\varepsilon }}} = \sum\limits_{i = 1}^n 2 \tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right){\delta _{{\theta _i}}} + \varepsilon {{\mathscr{H}}^1} $

    其中$ {\mathscr{H}}^1$是单位圆周上的弧长测度.由于K在关于绕原点转角$ \frac{{2k\pi }}{n}\left( {k \in \mathbb{Z}} \right)$的旋转变换上是不变的,故

    $ \begin{array}{l} {L_\alpha }\left( {{K_\varepsilon }} \right) = \int_{{s^1}} {h_{{K_\varepsilon }}^\alpha } {\rm{d}}S\left( {{K_\varepsilon }, \cdot } \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;n\varepsilon \int_{\left( { - \frac{{\rm{ \mathsf{ π} }}}{n},\frac{{\rm{ \mathsf{ π} }}}{n}} \right]} {{h_{{K_\varepsilon }}}} {(\theta )^\alpha }{\rm{d}}S\left( {{K_\varepsilon }, \cdot } \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{i = 1}^n {{h_{{K_\varepsilon }}}} {\left( {{\theta _i}} \right)^\alpha }{S_{{K_\varepsilon }}}\left( {\left\{ {{\theta _i}} \right\}} \right) + {\smallint _{{S^1}\backslash \left\{ {{\theta _i}:i = 1,2, \cdots ,n} \right\}}}{h_{{K_\varepsilon }}}{(\theta )^\alpha }{\rm{d}}S\left( {{K_\varepsilon }, \cdot } \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{i = 1}^n {{h_{{K_\varepsilon }}}} {\left( {{\theta _i}} \right)^\alpha }{S_{{K_\varepsilon }}}\left( {\left\{ {{\theta _i}} \right\}} \right) + n\int_{\left( { - \frac{{\rm{ \mathsf{ π} }}}{n},\frac{{\rm{ \mathsf{ π} }}}{n}} \right)} {{h_{{K_\varepsilon }}}} {(\theta )^\alpha }{\rm{d}}S\left( {{K_\varepsilon }, \cdot } \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;n{\left( {1 + \varepsilon } \right)^\alpha } \cdot 2\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right) + n\varepsilon \int_{ - \frac{{\rm{ \mathsf{ π} }}}{n}}^{\frac{{\rm{ \mathsf{ π} }}}{n}} {{{\left( {\frac{{\cos \theta }}{{\cos \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}} + \varepsilon } \right)}^\alpha }} {\rm{d}}\theta = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;2n{\left( {1 + \varepsilon } \right)^\alpha }\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right) + 2n\varepsilon \int_0^{\frac{{\rm{ \mathsf{ π} }}}{n}} {{{\left( {\frac{{\cos \theta }}{{\cos \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}} + \varepsilon } \right)}^\alpha }} {\rm{d}}\theta \end{array} $ (7)

    已知$ \frac{{{\rm{cos}}\theta }}{{{\rm{cos}}\left( {\frac{\pi }{n}} \right)}}$≥1,结合积分中值定理得到

    $ \int_0^{\frac{{\rm{ \mathsf{ π} }}}{n}} {{{\left( {\frac{{\cos \theta }}{{\cos \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}} + \varepsilon } \right)}^\alpha }} {\rm{d}}\theta \ge \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right){(1 + \varepsilon )^\alpha } $ (8)

    由(5)-(8)式,并对结果等价处理,得

    $ \begin{array}{l} {L_\alpha }{\left( {{K_\varepsilon }} \right)^{\frac{1}{{1 + \alpha }}}} - {L_\alpha }{(K)^{\frac{1}{{1 + \alpha }}}}\\ = {\left[ {2n{{\left( {1 + \varepsilon } \right)}^\alpha }\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right) + 2n\varepsilon \int_0^{\frac{{\rm{ \mathsf{ π} }}}{n}} {{{\left( {\frac{{\cos \theta }}{{\cos \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}} + \varepsilon } \right)}^\alpha }} {\rm{d}}\theta } \right]^{\frac{1}{{1 + \alpha }}}} - {\left( {2n\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)} \right)^{\frac{1}{{1 + \alpha }}}} \ge \\ \;\;\;{\left[ {2n{{\left( {1 + \varepsilon } \right)}^\alpha }\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right) + 2{\rm{ \mathsf{ π} }}\varepsilon {{(1 + \varepsilon )}^\alpha }} \right]^{\frac{1}{{1 + \alpha }}}} - {\left( {2n\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)} \right)^{\frac{1}{{1 + \alpha }}}} = \\ \;\;\;{\left[ {2{\rm{ \mathsf{ π} }}{{\left( {1 + \varepsilon } \right)}^\alpha }\frac{{\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}}{{\frac{{\rm{ \mathsf{ π} }}}{n}}} + 2{\rm{ \mathsf{ π} }}\varepsilon {{(1 + \varepsilon )}^\alpha }} \right]^{\frac{1}{{1 + \alpha }}}} - {\left( {2{\rm{ \mathsf{ π} }}\frac{{\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}}{{\frac{{\rm{ \mathsf{ π} }}}{n}}}} \right)^{\frac{1}{{1 + \alpha }}}} \end{array} $ (9)

    为了完成证明,需证明

    $ {\left[ {2{\rm{ \mathsf{ π} }}{{\left( {1 + \varepsilon } \right)}^\alpha }\frac{{\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}}{{\frac{{\rm{ \mathsf{ π} }}}{n}}} + 2{\rm{ \mathsf{ π} }}\varepsilon {{\left( {1 + \varepsilon } \right)}^\alpha }} \right]^{\frac{1}{{1 + \alpha }}}} - {\left( {2{\rm{ \mathsf{ π} }}\frac{{\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}}{{\frac{{\rm{ \mathsf{ π} }}}{n}}}} \right)^{\frac{1}{{1 + \alpha }}}} \ge {\left( {2{\rm{ \mathsf{ π} }}} \right)^{\frac{1}{{1 + \alpha }}}} \cdot \varepsilon $ (10)

    $ u = \frac{{\tan \left( {\frac{\pi }{n}} \right)}}{{\frac{\pi }{n}}}$,取(10)式不等号左侧部分得到

    $ f\left( u \right) = {(2{\rm{ \mathsf{ π} }})^{\frac{1}{{1 + \alpha }}}}{(1 + \varepsilon )^{\frac{1}{{1 + \alpha }}}}{(u + \varepsilon )^{\frac{1}{{1 + \alpha }}}} - {(2{\rm{ \mathsf{ π} }}u)^{\frac{1}{{1 + \alpha }}}} $ (11)

    对(11)式求导,得

    $ \begin{array}{l} f'\left( u \right) = {\left( {{{\left( {2{\rm{ \mathsf{ π} }}} \right)}^{\frac{1}{{1 + \alpha }}}}{{(1 + \varepsilon )}^{\frac{1}{{1 + \alpha }}}}{{(u + \varepsilon )}^{\frac{1}{{1 + \alpha }}}} - {{(2{\rm{ \mathsf{ π} }}u)}^{\frac{1}{{1 + \alpha }}}}} \right)^\prime } = \\ \;\;\;\;\;\;\;\;\;\;\;\left( {\frac{{{{\left( {2{\rm{ \mathsf{ π} }}{u^{ - \alpha }}} \right)}^{\frac{1}{{1 + \alpha }}}}}}{{1 + \alpha }}} \right)\left( {{{\left( {\frac{{u + \varepsilon u}}{{u + \varepsilon }}} \right)}^{\frac{1}{{1 + \alpha }}}} - 1} \right) \end{array} $

    因为$ u = \frac{{\tan \left( {\frac{\pi }{n}} \right)}}{{\frac{\pi }{n}}}$≥1,ε,所以

    $ f'\left( u \right) > 0 $ (12)

    $ g\left( x \right) = \frac{{\tan \left( {\frac{1}{x}} \right)}}{{\frac{1}{x}}} $ (13)

    对(13)式求导,得

    $ g'(x) = \tan \left( {\frac{1}{x}} \right) - \frac{1}{x}{\sec ^2}\left( {\frac{1}{x}} \right) < 0 $ (14)

    由(12),(14)式可以得到复合函数f(g(x))单调递减.

    求极限

    $ \mathop {\lim }\limits_{n \to \infty } f\left( {g\left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {{{\left( {2{\rm{ \mathsf{ π} }}{{(1 + \varepsilon )}^\alpha }\frac{{\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}}{{\frac{{\rm{ \mathsf{ π} }}}{n}}} + 2{\rm{ \mathsf{ π} }}\varepsilon {{(1 + \varepsilon )}^\alpha }} \right)}^{\frac{1}{{1 + \alpha }}}} - {{\left( {2{\rm{ \mathsf{ π} }}\frac{{\tan \left( {\frac{{\rm{ \mathsf{ π} }}}{n}} \right)}}{{\frac{{\rm{ \mathsf{ π} }}}{n}}}} \right)}^{\frac{1}{{1 + \alpha }}}}} \right] = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{(2{\rm{ \mathsf{ π} }})^{\frac{1}{{1 + \alpha }}}}\varepsilon $

    证毕.

    本文给出了0<p<1时平面情形下特殊凸体的证明,该结果对Lp表面积测度的Brunn-Minkowski不等式及相关不等式的研究具有重要参考意义.

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    The α-Length of Planar Convex Bodies and Isoperimetric Inequalities
    GUO Huan-huan1, LI De-yi1,2, ZOU Du1     
    1. College of Science, Wuhan University of Science and Technology, Wuhan 430081, China;
    2. Master Studio of Hubei Province, Wuhan University of Science and Technology, Wuhan 430081, China
    Abstract: The Brunn-Minkowski inequality is an important research content of convex geometry analysis. At present, the Brunn-minkowski inequality about volume and other geometric quantities is widely known and plays an important role in various branches of mathematics. Brunn-Minkowski inequality of convex body surface area as a special case of Aleksandrov-Fenchel inequality has also been confirmed. But in Lp Brunn-Minkowski theory, the Brunn-minkowski inequality of Lp surface area measurement is still an important open problem. There is no effective method to prove the related conjecture for 0 < p < 1 and p > 1. In this paper, based on the addition of Minkowski, the monotone bounded theorem and integral mean value theorem are used to study the α-perimeter of convex body in the plane. The Brunn-Minkowski type inequality about α-perimeter is put forward and proved when two convex bodies are a regular n polygon and a unit disc, respectively.
    Key words: convex body    α-perimeter    Brunn-Minkowski inequality    
    X