Poincaré不等式, 传输不等式和对数Sobolev不等式是研究测度集中性的有力工具^{[1, 2]}.在这三类不等式中, 对数Sobolev不等式强于Talagrand传输不等式, Talagrand传输不等式不等式又强于Poincaré不等式, 具体例子可分别参看文献[4, 3].在文献[5]中, Qian, Ma, Zhang证明了Boltzmann测度在维数$ n\geq 3 $固定时关于参数$ h>0 $满足一致的对数Sobolev不等式, 但是并没有讨论$ n = 2 $的情形.在Ma与Zhang的文献[6]中, 作者针对$ n = 2 $的情形给出了Boltzmann测度比较精确的谱系, 而且有趣的是:当$ h\rightarrow\infty $时, 谱系以$ h $的速率趋于无穷大, 即庞加莱常数以$ 1/h $的速率趋于$ 0 $.本文将给出在$ n = 2 $时, 对数Sobolev常数$ C_{LS}(\mu_h) $关于$ h>0 $的一致非$ 0 $上下界, 这在一定程度上也能表明三个不等式之间的强弱关系.下面先介绍一下Boltzmann测度以及相关不等式.

Boltzmann测度  设$ S^{n-1} $是$ \mathbb{R}^n(n\geq2) $上的单位球面, $ \mu $是$ S^ {n-1} $上的标准Lebesgue测度, $ i.e.\ \mu = \sigma_{n-1}/s_{n-1} $, 其中$ \sigma_{n-1} $为单位球面$ S^{n-1} $上的均匀测度, $ s_{n-1}: = n\pi^{n/2}/\Gamma(1+n/2) $为单位球面面积(归一化因子).对任意的$ h>0 $和$ e_1 = (1, 0, \dots, 0)\in S^{n-1} $, 单位球面$ S^{n-1} $上的概率测度$ \mu_h $有如下表达式:

$ d\mu_{n, h}(x) = \frac{e^{h\langle e_1, x\rangle}}{c_n(h)}d\mu(x), \quad x\in S^{n-1}. $

其中, $ c_n(h) $为归一化因子.称该概率测度$ \mu_h $为外磁场下的Boltzmann测度.特别$ n = 2 $时, $ S^1 $即为环, 我们简记为$ S $, 对应测度$ \mu_{2, h} $简记为$ \mu_h $, 此即为本文所考虑的含参变量$ h $的环上的Boltzmann测度.

下面介绍几个相关不等式:设$ M $是一个完备黎曼流形, 其上的测地度量记为$ d $, $ \nabla $为$ M $上的梯度. $ \mathcal{M}_1(M) $为$ M $上的概率测度空间.

Poincaré不等式  我们称测度$ \mu\in \mathcal{M}_1(M) $满足Poincaré不等式(记$ \mu\in PI(C) $), 若对任意光滑函数$ f: M\to\mathbb{R} $, 都存在非负常数$ C $使得

$ \begin{align*} \label{P1} \text{Var}_{\mu}(f) = \int_{M} f^2 d\mu-(\int_M fd\mu)^2 \le C\int_M |\nabla f|^2 d\mu. \end{align*} $

记$ C_{\rm P}(\mu) $为最佳Poincaré常数.

$ {\rm {{L^p}}} $传输不等式  称测度$ \mu $满足$ {\rm {{L^p}}} $传输不等式, 若对任意的$ \nu = f^2\mu\in \mathcal{M}_1(M) $, 都存在非负常数$ C $使得

$ W_p^2(\nu, \mu)\leq2C{\rm Ent}_{\mu}\left( f^2\right). $

其中$ W_p(\nu, \mu) $是测度$ \nu $和$ \mu $的$ {\rm {{L^p}}} $-Wasserstein距离, 其定义如下:

$ W_p(\nu, \mu) = \inf\limits_\pi{( \iint\limits_{M^2}^{}d^p(x, y)\pi(dx, dy))}^{1/p}. $

这里$ \pi $是$ M\times M $上的概率测度, 其边缘分布为$ \mu, \nu $.记$ C_{WpH}(\mu) $为满足该不等式的最佳常数.

对数Sobolev不等式   称测度$ \mu $满足对数Sobolev不等式, 若对任意光滑函数$ f: M\to\mathbb{R} $, 都存在非负常数$ C $使得

$ \begin{align*} {\rm Ent}_{\mu}(f^2)\leq2C\int_{M}|\nabla_Mf|^2d\mu\notag \end{align*} $

成立, 其中$ {\rm Ent}_{\mu}(f^2): = \mu(f^2\log f^2)-\mu(f^2)\log(\mu(f^2)) $为函数$ f^2 $关于$ \mu $的熵.记$ C_{LS}(\mu) $为最佳对数Sobolev常数.

本文沿用[7, 8]中的降维方法:设$ \nu_h $为$ \mu_h $在映射$ x\rightarrow d(e_1, x) $下的像测度, 则$ \nu_h $为$ [0, \pi]$上的概率测度, 其概率密度为 $ \rho_{2, h} = \frac{d\nu_h}{d\theta} = \frac{1}{C_2(h)}e^{h\cos\theta} $其中, $ C_2(h) = \int_0^{\pi}e^{h\cos\theta}d\theta $为归一化因子.

在处理环上测度的对数Sobolev不等式时可参考如下的降维定理.

定理2.1 [8]   设$ \mu $为环$ S $上的均匀测度, $ M $为$ S $上的概率测度, 其定义如下

$ M(dy) = \varphi(d(y, e_1))\mu(dy), \quad y\in S. $

其中, $ \varphi $是非负可测的.设$ M $在映射$ y\rightarrow d(y, e_1) $下的像测度为$ \nu $, 则最佳对数Sobolev常数满足$ C_{LS}(\nu)\leq C_{LS}(M)\leq C_{LS}(\nu)+\frac{1}{\lambda^{DD}(\nu)} $其中, $ \lambda^{DD}(\nu) $是满足$ [0, \pi]$上Dirichlet边界条件的$ \nu $的第一特征值.

$ \lambda^{DD}(\nu): = \inf\left\lbrace \frac{\int_{0}^{\pi}\left( f'\right) ^2d\nu}{{\nu(f^2)}}:f(0) = f(\pi) = 0, f \text{不是常数} \right\rbrace $

而对于本文中所考虑的Boltzmann测度对应的$ \lambda^{DD}(\nu_h) $也有如下估计.

引理2.2^[6]  对于任意$ h>0 $, 对任意的光滑函数$ f: S\to\mathbb{R} $, $ \nu_{h} $为Boltzmann测度$ \mu_{h} $在映射$ x\rightarrow d(e_1, x) $下的像测度, 则Dirichlet边界条件下的第一特征值$ \lambda^{DD}(\nu_h) $满足

$ \dfrac{1}{\lambda^{DD}(\nu_h)}\leq\min\left\lbrace \frac{7}{h}, 1\right\rbrace. $

在上面两个结论的基础上, 本文的讨论将主要集中在$ [0, \pi]$上的一维测度$ \nu_h $上.针对一维测度, Barthe和Roberto在[9]中给出了关于对数Sobolev不等式的刻画, 现陈述如下.

定理2.3^[9]   设$ \mu_B, \nu_B $是$ \mathbb{R} $上的Borel测度, 其中$ \mu_B(\mathbb{R}) = 1 $且$ d\nu_B(x) = n(x)dx $, $ n(x) $为一绝对光滑函数.设$ m $是测度$ \mu_B $上的中位数, 且对任意光滑函数$ f:\mathbb{R}\rightarrow\mathbb{R} $, 满足

$ {\rm Ent}_{\mu_B}(f^2)\leq 2C_{LS}\int (f')^2d\nu_B. $

其中, $ C_{LS} $为最佳对数Sobolev常数.则有$ \max(b_-, b_+)\leq 2C_{LS}\leq 4\max(B_-, B_+) $, 其中

$ \begin{align*} b_+& = \sup\limits_{x>m}\mu_B([x, \infty))\log\left(1+\frac{1}{2\mu_B([x, \infty))}\right)\int_{m}^{x}\frac{1}{n}\\ B_+& = \sup\limits_{x>m}\mu_B([x, \infty))\log\left(1+\frac{e^2}{\mu_B([x, \infty))}\right)\int_{m}^{x}\frac{1}{n}\\ b_-& = \sup\limits_{x<m}\mu_B((-\infty, x])\log\left(1+\frac{1}{2\mu_B((-\infty, x])}\right)\int_{x}^{m}\frac{1}{n}\\ B_-& = \sup\limits_{x<m}\mu_B((-\infty, x])\log\left(1+\frac{e^2}{\mu_B((-\infty, x])}\right)\int_{x}^{m}\frac{1}{n}. \end{align*} $

注意到, 对于任意$ 0\leq y\leq 1/2 $, 都有$ \log \left(1+\frac{e^2}{y}\right)\leq \frac{\log(1+2e^2)}{\log2}\log\left( 1+\frac{1}{2y}\right) \leq 4\log\left(1+\frac{1}{2y}\right) $因此, $ B_+\leq 4b_+ $且$ B_-\leq 4b_- $.不难发现, 此时最佳对数Sobolev常数$ C_{LS} $满足$ \max(b_-, b_+)\leq 2C_{LS}\leq 16\max(b_-, b_+) $.本文的目的仅在于给出常数阶, 故此证明也只需去估计$ b_-, b_+ $.为此还需下面几个估计.

引理2.4   设$ C_2(h): = \int_0^{\pi}e^{h\cos\theta}d\theta $, 则

$ \begin{align*} 2&\leq C_2(h)\leq 2\pi, \qquad 0<h<1;\\ \frac{e^h}{2\sqrt h}&\leq C_2(h)\leq\frac{3e^h}{\sqrt h}, \qquad h\geq 1. \end{align*} $

证 a) 当$ 0<h<1 $时,

$ \begin{align*} &C_2(h) = \int_0^{\pi/2}e^{h\cos\theta}d\theta +\int_{\pi/2}^{\pi}e^{h\cos\theta}d\theta\leq\frac{\pi}{2}(e+1)<2\pi;\\ &C_2(h)\geq\int_0^{\pi}e^{h\cos\theta}\sin\theta d\theta = \frac{e^h-e^{-h}}{h}\geq 2. \end{align*} $

b) 当$ h\geq 1 $时, 一方面

$ C_2(h) = \int_{-1}^1\frac{e^{ht}}{\sqrt{1-t^2}}dt \geq\int_{0}^{1}\frac{e^{ht}}{\sqrt{1-t}}dt\geq\frac{\sqrt{2}e^h}{\sqrt{h}}\int_{0}^{1}e^{-u^2}du \geq\frac{\sqrt{2}e^h}{\sqrt{h}}e^{-1}\geq\frac{e^h}{2\sqrt{h}}; $

另一方面, 利用$ \int_0^xe^{t^2}dt\leq e^{x^2}, \ \int_{0}^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2} $,

$ \begin{align*} C_2(h) = &\int_{-1}^{0}\frac{e^{ht}}{\sqrt{1-t^2}}dt+\int_{0}^{1}\frac{e^{ht}}{\sqrt{1-t^2}}dt \leq\int_{-1}^{0}\frac{e^{ht}}{\sqrt{1+t}}dt+\int_{0}^{1}\frac{e^{ht}}{\sqrt{1-t}}dt\\ = &\frac{2e^{-h}}{\sqrt{h}}\int_{0}^{\sqrt{h}}e^{v^2}dv+\frac{2e^{h}}{\sqrt{h}}\int_{0}^{\sqrt{h}}e^{-v^2}dv \leq\frac{2}{\sqrt{h}}+\frac{2e^{h}}{\sqrt{h}}\frac{\sqrt{\pi}}{2} \leq\frac{3e^{h}}{\sqrt{h}}. \end{align*} $

引理2.5    $ \{h(1-\cos m_h):h>0\} $关于$ h $一致有界, 且满足

$ \begin{align*} \frac{h}{4e^2}&\leq h(1-\cos m_h)\leq h, \qquad\qquad\qquad h<1\\ \frac{1}{64}&\leq h(1-\cos m_h)\leq \log 2+2<3, \quad h\geq 1 \end{align*} $

其中, $ m_h $是$ \mu_{h} $的中位数.

注   $ m_0 = \pi/2 $, 从上面结论可以看出$ \lim_{h\rightarrow\infty}m_h = 0 $, 而且不难得到对任意$ h>0 $时, 都有$ 0<m_h<\pi/2 $.反设$ m_h\geq\pi/2 $, $ \int_{0}^{m_h}e^{h\cos\theta}d\theta = \int_{0}^{\frac{\pi}{2}}e^{h\cos\theta}d\theta+\int_{\frac{\pi}{2}}^{m_h}e^{h\cos\theta}d\theta>\frac{\pi}{2} $而$ \int_{m_h}^{\pi}e^{h\cos\theta}d\theta<\left( \pi-m_h\right)\leq\frac{\pi}{2} $与中位数定义矛盾.

证   a) 当$ h<1 $时, 已知$ C_2(h)\geq 2 $, 且$ \frac{1}{2}C_2(h) = \int_{0}^{m_h}e^{h\cos\theta}d\theta\leq\frac{2e^h}{\sqrt h}\int_{0}^{\sqrt{h(1-\cos m_h)}}e^{-t^2}dt\leq\frac{2e}{\sqrt h}\sqrt{h(1-\cos m_h)}. $因此可以得出结论

$ \frac{h}{4e^2}\leq h(1-\cos m_h)\leq h. $

b) 当$ h\geq 1 $时, 一方面, 由引理2.4知$ \ C_2(h)\geq\frac{e^h}{2\sqrt{h}} $, 且

$ \int_{0}^{m_h}e^{h\cos\theta}d\theta\leq\frac{2e^h}{\sqrt h}\int_{0}^{\sqrt{h(1-\cos m_h)}}e^{-t^2}dt\leq\frac{2e^h}{\sqrt h}\sqrt{h(1-\cos m_h)}, $

从而可得

$ h(1-\cos m_h)\geq\frac{1}{64}; $

另一方面, $ \int_{0}^{m_h}e^{h\cos\theta}d\theta\geq\frac{\sqrt{2}e^h}{\sqrt {h}}\int_{0}^{\sqrt{h(1-\cos m_h)}}e^{-t^2}dt, $利用极坐标变换, 有

$ \begin{align*} ( \int_{0}^{\sqrt{h(1-\cos m_h)}}e^{-x^2}dx) ^2 = &( \int_{0}^{\sqrt{h(1-\cos m_h)}}e^{-x^2}dx) ( \int_{0}^{\sqrt{h(1-\cos m_h)}}e^{-y^2}dy) \\ = &\int_{0}^{\sqrt{h(1-\cos m_h)}}\int_{0}^{\sqrt{h(1-\cos m_h)}}e^{-(x^2+y^2)}dxdy\\ \geq&\int_{0}^{\sqrt{h(1-\cos m_h)}}\int_{0}^{\pi/2}e^{-r^2}rdrd\theta\\ = &\frac{1}{4}\pi( 1-e^{-h(1-\cos m_h)}), \end{align*} $

于是

$ \frac{C_2(h)}{2} = \int_{0}^{m_h}e^{h\cos\theta}d\theta\geq\frac{e^h}{\sqrt {2h}}\sqrt{\pi(1-e^{-h(1-\cos m_h)})}. $

而对于$ C_2(h) $, 类似引理2.4的证明可得

$ \begin{align*} C_2(h) = & \int_{0}^{\frac{\pi}{2}}e^{h\cos\theta}d\theta+\int_{\frac{\pi}{2}}^{\pi}e^{h\cos\theta}d\theta \leq\frac{\pi}{2}+\int_{0}^{1}\frac{e^{ht}}{\sqrt{1-t^2}}dt\\ = &\frac{\pi}{2}+\frac{2e^{h}}{\sqrt{h}}\int_{0}^{\sqrt{h}}e^{-v^2}dv \leq\frac{\pi}{2}+e^h\sqrt{\frac{\pi}{h}}. \mathrm{} \end{align*} $

故而有,

$ \begin{align*} &\frac{\sqrt{2}e^h}{\sqrt {h}}\sqrt{\pi(1-e^{-h(1-\cos m_h)})} \leq \frac{\pi}{2}+e^h\sqrt{\frac{\pi}{h}}\\ \Longrightarrow\, &\sqrt{1-e^{-h(1-\cos m_h)}}\leq \frac{\sqrt{\pi h}}{2\sqrt{2}e^h}+\frac{1}{\sqrt{2}}\leq \frac{1}{\sqrt{2}}\left( 1+\frac{\sqrt{\pi}}{2e}\right)\\ \Longrightarrow\, &e^{-h(1-\cos m_h)}\geq \frac{4e^2-\pi-4e\sqrt{\pi}}{8e^2}\geq \frac{1}{2e^2}\\ \Longrightarrow\, &h(1-\cos m_h)\leq \log 2+2<3. \end{align*} $

证毕！

定理3.1   对于任意$ h>0 $, $ \nu_{h} $为Boltzmann测度$ \mu_{h} $在映射$ x\rightarrow d(e_1, x) $下的像测度, 则$ \nu_h $满足一致的对数Sobolev不等式, 即对于任意光滑函数$ f:S\rightarrow\mathbb{R} $, 有$ {\rm Ent}_{\nu_h}(f^2)\leq2C_{LS}(\nu_h)\int_{0}^{\pi} (f')^2 d\nu_{h}, $其中对数Sobolev最佳常数$ C_{LS}(\nu_h) $一致有界, 且满足

$ \begin{eqnarray*} &&\frac{{\pi}^2}{32e^2}\log(1+\frac{2}{\pi} ) \leq C_{LS}(\nu_h)\leq 8{\pi}^2\log2, \quad 0<h<1;\\ &&\frac{1}{4e}\leq C_{LS}(\nu_h)\leq 32e\sqrt{\pi}\bigvee\frac{24e^3}{h}. \qquad \qquad \qquad h\geq1. \end{eqnarray*} $

进而有, 环$ S $上的Boltzmann测度$ \mu_{h} $满足一致的对数Sobolev不等式, 且最佳常数$ C_{LS}(\mu_h) $满足

$ \begin{eqnarray*} &&\frac{{\pi}^2}{32e^2}\log(1+\frac{2}{\pi} ) \leq C_{LS}(\mu_h)\leq 8{\pi}^2\log2+1, \quad 0<h<1, \\ &&\frac{1}{4e}\leq C_{LS}(\mu_h)\leq 32e\sqrt{\pi}\bigvee\frac{24e^3}{h}+1. \qquad \qquad \qquad h\geq1. \end{eqnarray*} $

注   从定理结论中我们可以看出, 对于任意的$ h>0 $, $ C_{LS}(\mu_h) = O(1) $.而在[6]中, 作者给出了该测度谱系的比较精确的刻画$ 1\vee \frac{|h|}{7} \le\lambda_1(\mu_h)\le \sqrt{3}|h|+\frac{2h^2+3}{h^2+3}. $此即说明Poincaré常数$ C_{P}(\mu_h) = 1/\lambda_1(\mu_h) = O(1/h) $, 再通过[10]中的结论, 我们还可知道Talagrand传输不等式常数$ C_{W_2H}(\mu_h)\leq O(1/\sqrt{h}) $, 此也可佐证对数Sobolev不等式严格强于Talagrand传输不等式.

根据降维定理2.1, 我们仅需考虑一维测度$ \nu_h $, 再由定理2.3, 考虑如下的$ b_+, b_- $

$ \begin{align*} &b_+ = \sup\limits_{m_h<\alpha<\pi}\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{\alpha}^{\pi}e^{h\cos\theta }d\theta})\int_{m_h}^{\alpha}\frac{d\theta}{e^{h\cos\theta}}, \\ &b_- = \sup\limits_{0<\alpha<m_h}\int_{0}^{\alpha}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{0}^{\alpha}e^{h\cos\theta }d\theta})\int_{m_h}^{\alpha}\frac{d\theta}{e^{h\cos\theta}}. \end{align*} $

讨论将分两种情形展开.

情形1   $ 0<h<1 $.先看$ b_+ $的下界:由$ 0<m_h<\pi/2 $, 有

$ \begin{align*} b_+\geq&\sup\limits_{\frac{\pi}{2}<\alpha<\pi}\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{\alpha}^{\pi}e^{h\cos\theta }d\theta} )\int_{\frac{\pi}{2}}^{\alpha}\frac{d\theta}{e^{h\cos\theta}}\\ \geq&\frac{1}{e}(\pi-\frac{3\pi}{4} ) \log(1+\frac{C_2(h)}{\pi-\frac{3\pi}{4}}) e^{-\frac{\sqrt{2}}{2}}\frac{\pi}{4} = \frac{1}{e^{1+\frac{\sqrt{2}}{2}}}\frac{{\pi}^2}{16}\log(1+\frac{C_2(h)}{\pi} )\geq \frac{1}{e^2}\frac{{\pi}^2}{16}\log(1+\frac{2}{\pi} ). \end{align*} $

其次, 考虑$ b_+ $的上界.因对任意$ C>0 $，$ x\log(1+\frac{C}{x}) $在$ x>0 $上都是单调递增的, 故有

$ \begin{align*} b_+\leq&\sup\limits_{0<\alpha<\pi}\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{\alpha}^{\pi}e^{h\cos\theta }d\theta} )\int_0^{\alpha}\frac{d\theta}{e^{h\cos\theta}} \\ \leq&\sup\limits_{0<\alpha<\pi}(\pi-\alpha)\alpha \log(1+\frac{\pi}{\pi-\alpha}) = \sup\limits_{0<\alpha<\pi}\alpha \log(1+\frac{\pi}{\alpha} ) (\pi-\alpha) \end{align*} $

$ \begin{align*} \leq&\sup\limits_{0<\alpha<\pi}\pi\alpha \log(1+\frac{\pi}{\alpha} ) = {\pi}^2\log2. \end{align*} $

最后, 考虑$ b_- $的上界

$ \begin{align*} b_-\leq&\sup\limits_{0<\alpha<\pi/2}\int_{0}^{\alpha}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{0}^{\alpha}e^{h\cos\theta }d\theta} )\int_{\alpha}^{m_h}\frac{d\theta}{e^{h\cos\theta}}\\ \leq&\sup\limits_{0<\alpha<\frac{\pi}{2}}e\alpha \log\left(1+\frac{\pi}{e\alpha} \right)\left( \frac{\pi}{2}-\alpha\right) \leq\frac{{\pi}^2}{4}e\log\left(1+\frac{2}{e} \right). \end{align*} $

故当$ 0<h<1 $时, 有$ \frac{{\pi}^2}{32e^2}\log\left(1+\frac{2}{\pi} \right) \leq C_{LS}(\nu_h)\leq 8{\pi}^2\log2. $

情形2    $ h\geq1 $.同情形1, 首先考虑$ b_+ $的下界

$ \begin{align*} b_+&\geq\sup\limits_{\pi/2<\alpha<\pi}\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta \log(1+\frac{C_2(h)}{2\int_{\alpha}^{\pi}e^{h\cos\theta }d\theta})\int_{\pi/2}^{\alpha}\frac{d\theta}{e^{h\cos\theta}} \end{align*} $

由$ \int_{\alpha}^{\pi}e^{h\cos\theta}d\theta\geq\frac{\sqrt{2}e^{-h}}{\sqrt{h}}\int^{\sqrt{h(1+\cos\alpha)}}_0e^{t^2}dt, \quad \int_{\pi/2}^{\alpha}\frac{d\theta}{e^{h\cos\theta}}\geq\frac{\sqrt{2}e^h}{\sqrt{h}}\int_{\sqrt{h(1+\cos\alpha)}}^{\sqrt{h}}e^{-t^2}dt, $及$ x \log(1 + \frac{C}{x}), C>0 $在$ x >0 $上的递增性, 有

$ \begin{align*} b_+&\geq\frac{2}{h}\sup\limits_{0<x<\sqrt{h}}\int_0^xe^{t^2}dt \log(1+\frac{\sqrt{h}e^hC_2(h)}{2\sqrt{2}\int_0^xe^{t^2}dt}) \int_x^{\sqrt{h}}e^{-t^2}dt\\ &\geq\frac{2}{h}\sup\limits_{0<x<1}\int_0^xe^{t^2}dt \log(1+\frac{e^{2h}}{4\sqrt{2}\int_0^xe^{t^2}dt}) \int_x^1e^{-t^2}dt\\ &\geq\frac{2}{h}\int_0^{1/2}e^{t^2}dt \log(1+\frac{e^{2h}}{4\sqrt{2}\int_0^{1/2}e^{t^2}dt}) \int_{1/2}^1e^{-t^2}dt\\ &\geq\frac{2}{h}\frac{1}{2} \log(1+\frac{e^{2h}}{2\sqrt{2}})\frac{1}{2e} \geq\frac{1}{2e}. \end{align*} $

其次, 对$ b_+ $的上界进行估计.

$ \begin{align*} b_+ = &\sup\limits_{m_h<\alpha<\pi}\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta})\int_{m_h}^{\alpha}e^{-h\cos\theta}d\theta\\ = &\sup\limits_{\pi/2\leq\alpha<\pi}\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta})\int_{m_h}^{\alpha}e^{-h\cos\theta}d\theta\\ &\bigvee\sup\limits_{m_h<\alpha<\pi/2}\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{\alpha}^{\pi}e^{h\cos\theta}d\theta})\int_{m_h}^{\alpha}e^{-h\cos\theta}d\theta \\ = &\sup\limits_{-1<x\leq 0}\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt\log(1+\frac{C_2(h)}{2\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt}) \int_x^{\cos m_h}\frac{e^{-ht}}{\sqrt{1-t^2}}dt \\ &\bigvee\sup\limits_{0<x<\cos m_h}\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt\log(1+\frac{C_2(h)}{2\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt}) \int_x^{\cos m_h}\frac{e^{-ht}}{\sqrt{1-t^2}}dt. \end{align*} $

记

$ \begin{align*} b_{+}^{1}:& = \sup\limits_{-1<x\leq 0}\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt\log(1+\frac{C_2(h)}{2\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt}) \int_x^{\cos m_h}\frac{e^{-ht}}{\sqrt{1-t^2}}dt\\ b_{+}^{2}:& = \sup\limits_{0<x<\cos m_h}\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt\log(1+\frac{C_2(h)}{2\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt}) \int_x^{\cos m_h}\frac{e^{-ht}}{\sqrt{1-t^2}}dt. \end{align*} $

1) 先考虑$ b_+^1 $的上界:通过一个简单的放缩以及之前的变换, 易得

$ \begin{align*} b_+^1&\leq\sup\limits_{-1<x\leq 0}\int_{-1}^x\frac{e^{ht}}{\sqrt{1+t}}dt\log(1+\frac{C_2(h)}{2\int_{-1}^x\frac{e^{ht}}{\sqrt{1+t}}dt})\int_x^{\cos m_h}\frac{e^{-ht}}{\sqrt{1+t}}dt\\ & = \frac{4}{h}\sup\limits_{0<x<\sqrt{h}}\int_{0}^{x}e^{t^2}dt\log(1+\frac{C_2(h)e^h\sqrt{h}}{4\int_{0}^{x}e^{t^2}dt})\int_{x}^{\sqrt{h(1+\cos m_h)}}e^{-t^2}dt\\ &\leq\frac{4}{h}\sup\limits_{0<x<\sqrt{h}}\int_{0}^{x}e^{t^2}dt\log(1+\frac{3e^{2h}}{4\int_{0}^{x}e^{t^2}dt})\int_x^{\infty}e^{-t^2}dt. \end{align*} $

若上确界在$ x<1 $时取得, 则由$ \int_{0}^{x}e^{u^2}du\leq\int_{0}^{1}e^{u^2}du\leq e, \quad \int_{x}^{\infty}e^{-u^2}du\leq \int_{0}^{\infty}e^{-u^2}du = \frac{\sqrt{\pi}}{2} $可得

$ \begin{align*} b_+^1\leq&\frac{4}{h}e\log(1+\frac{3e^{2h}}{4e})\frac{\sqrt{\pi}}{2}<4e\sqrt{\pi}; \end{align*} $

若上确界在$ x\geq1 $时取得, 则由

$ \begin{align*} &\int_{0}^{x}e^{u^2}du\leq\int_{0}^{1}e^{u^2}du+\int_{ 1}^{x}ue^{u^2}du\leq \frac{e^{x^2}+e}{2}\leq e^{x^2}\\ & \int_{x}^{\infty}e^{-u^2}du\leq \int_{x}^{\infty}ue^{-u^2}du\leq \frac{e^{-x^2}}{2}, \end{align*} $

可得

$ \begin{align*} b_+^1\leq&\frac{4}{h}\sup\limits_{0<x<\sqrt{h}}e^{x^2}\log(1+\frac{3e^{2h}}{4e^{x^2}})\frac{1}{2e^{x^2}} = \frac{2}{h}\log(1+\frac{3e^{2h}}{4})\leq 4. \end{align*} $

综上所述, 当$ h\geq1 $时, 有

$ b_+^1\leq 4e\sqrt{\pi}. $

2) 再考虑$ b_+^2 $的上界:注意$ 0<x<\cos m_h<1 $, 由$ \int_0^xe^{t^2}dt\leq e^{x^2} $, 可得

$ \begin{align*} \int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt& = \int_{-1}^{0}\frac{e^{ht}}{\sqrt{1-t^2}}dt+\int_{0}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt \end{align*} $

$ \begin{align*} &\leq\frac{2e^{-h}}{\sqrt{h}}\int_0^{\sqrt{h}}e^{t^2}dt+\frac{2e^{h}}{\sqrt{h}}\int_{\sqrt{h(1-x)}}^{\sqrt{h}}e^{-t^2}dt \leq\frac{2}{\sqrt{h}}+\frac{2e^{h}}{\sqrt{h}}\int_{\sqrt{h(1-x)}}^{\sqrt{h}}e^{-t^2}dt. \end{align*} $

又$ \int_x^{\cos m_h}\frac{e^{-ht}}{\sqrt{1-t^2}}dt \leq\int_x^1 \frac{e^{-ht}}{\sqrt{1-t}}dt \leq\frac{2e^{-h}}{\sqrt{h}}\int_0^{\sqrt{h(1-x)}}e^{t^2}dt, $令$ y = \sqrt{h(1-x)} $, 从而有

$ \begin{align*} b_+^2& = \sup\limits_{0<x<\cos m_h}\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt\log(1+\frac{C_2(h)}{2\int_{-1}^x\frac{e^{ht}}{\sqrt{1-t^2}}dt}) \int_x^{\cos m_h}\frac{e^{ht}}{\sqrt{1-t^2}}dt\\ &\leq \sup\limits_{0<y<\sqrt{h}}( \frac{4}{he^h}+\frac{4}{h}\int_y^{\sqrt{h}}e^{-t^2}dt) \log(1+\frac{C_2(h)}{\frac{4}{\sqrt{h}}+\frac{4e^h}{\sqrt{h}}\int_y^{\sqrt{h}}e^{-t^2}dt })(\int_0^ye^{t^2}dt)\\ &\leq\sup\limits_{0<y<\sqrt{h}}\frac{4}{he^h}(1+\frac{e^{h-y^2}}{2y}) \log(1+\frac{\sqrt{h}C_2(h)}{4(1+\frac{e^{h-y^2}}{2y})})(\int_0^ye^{t^2}dt)\\ &\leq\sup\limits_{0<y<1}\frac{4e}{he^h}\frac{\sqrt{h}C_2(h)}{4} \bigvee\sup\limits_{1\leq y\leq\sqrt{h}}\frac{4}{he^{h-y^2}}(1+\frac{e^{h-y^2}}{2}) \log(1+\frac{\sqrt{h}C_2(h)}{4(1+\frac{e^{h-y^2}}{2})})\\ &\leq\frac{3e}{h} \bigvee\sup\limits_{1\leq y\leq\sqrt{h}}\frac{4}{h}(\frac{1}{e^{h-y^2}}+\frac{1}{2}) \log(1+\frac{\sqrt{h}C_2(h)}{4})\\ &\leq\frac{3e}{h} \bigvee\frac{6}{h}\log(1+\frac{3e^h}{4})\leq 12. \end{align*} $

上面第二个不等式分别应用了$ \int_y^{\sqrt{h}}e^{-t^2}dt\leq\int_y^{\infty}e^{-t^2}dt\leq\frac{e^{-y^2}}{2y} $, 而第三个不等式则用到了$ \log(1+x)\leq x\ \text{与} \int_0^xe^{t^2}dt\leq e^{x^2}, x\geq 0 $.

最后, 对$ b_- $的上界进行估计:注意$ C_2(h)\leq\frac{3e^h}{\sqrt{h}}, \ h(1-\cos m_h)<3, \ \int_0^xe^{t^2}dt\leq e^{x^2}. $

$ \begin{align*} b_-& = \sup\limits_{0<\alpha<m_h}\int_{0}^{\alpha}e^{h\cos\theta}d\theta\log(1+\frac{C_2(h)}{2\int_{0}^{\alpha}e^{h\cos\theta}d\theta})\int_{\alpha}^{m}e^{-h\cos\theta}dt\\ &\leq\sup\limits_{\cos m_h<x<1}\int_x^{1}\frac{e^{ht}}{\sqrt{1-t}}dt\log( 1+\frac{C_2(h)}{2\int_x^{1}\frac{e^{ht}}{\sqrt{1-t}}dt}) \int_{\cos m_h}^x\frac{e^{-ht}}{\sqrt{1-t}}dt\\ &\leq\sup\limits_{0<y<\sqrt{h(1-\cos m_h)}}\frac{4}{h}\int_{0}^ye^{-t^2}dt\log(1+\frac{C_2(h)e^{-h}\sqrt{h}}{4\int_{0}^ye^{-t^2}dt})\int_y^{\sqrt{h(1-\cos m_h)}}e^{t^2}dt \\ &\leq\frac{4}{h}\sup\limits_{0<y<\sqrt{3}}\int_{0}^ye^{-t^2}dt\log(1+\frac{3}{4\int_{0}^ye^{-t^2}dt})\int_y^{\sqrt{3}}e^{t^2}dt \\ &\leq\frac{4}{h}\int_{0}^{\infty}e^{-t^2}dt\log(1+\frac{3}{4\int_{0}^{\infty}e^{-t^2}dt})\int_0^{\sqrt{3}}e^{t^2}dt \leq\frac{2\sqrt{\pi}e^3}{h}\log(1+\frac{3}{2\sqrt{\pi}}) \leq\frac{3e^3}{h}. \end{align*} $

综合可得, $ b_+\bigvee b_- \leq 4e\sqrt{\pi}\bigvee\frac{3e^3}{h} $, 进而可得定理3.1中的结论.