Let $n\ge 2$ and let $S^{n-1}$ be the unit sphere on $\mathbb{R}^n$ equipped with geodesic distance $d$ and the uniform probability measure $\mu$. For $x\in\mathbb{R}^n$ with $|x| <1, $ we consider the probability measure on $S^{n-1}$ given by

$ d\mu_x^n(y)=\dfrac{(1-|x|^2)^{\frac{n-1}{2}}}{(1-(x, y))^{n-1}}d\mu(y), \quad y\in S^{n-1}. $

It is the so-called Moebius measure we are working on. In fact, this probability is the image of $\mu$ under the Moebius transformation. The factor $\dfrac{(1-|x|^2)^{\frac{n-1}{2}}}{(1-(x, y))^{n-1}}$ is known as the invariant Poisson kernel $P(x, y)$: as a function of $x, $ it is not harmonic but satisfies the equation $\tilde{\Delta} P(\cdot, y)=0, $ where $\tilde{\Delta}$ denotes the invariant Laplacian operator (the reader is referred to [3] for further information on this measure).

Let $M$ be a connected complete Riemannian manifold with Riemannian metric $d$ and $\nabla$ is the gradient on $M.$ Let $\mathcal{M}_1(M)$ be the space of all probabilities on $M.$ Given any $\mu\in \mathcal{M}_1(M), $ we say that

1. $\mu$ satisfies a Poincaré inequality with a non-negative constant $C$ if for any smooth function $f: M\to\mathbb{R}$, there exists a constant $C\ge 0$ such that

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

(1.1)

The optimal constant above is denoted by $C_{\rm P}(\mu)$.

2. $\mu$ satisfies a logarithmic Sobolev inequality with a constant $C\ge 0$ if for any smooth function $f: M\to\mathbb{R}$ with $\mu(f^2)=1, $

$ \begin{equation} {\rm Ent}_{\mu}(f^2)\le 2 C \displaystyle\int_M |\nabla f|^2 d\mu. \end{equation} $

(1.2)

We denote by $C_{\rm LS}(\mu)$ the optimal logarithmic Sobolev constant.

3. $\mu$ satisfies a Sobolev inequality with exponent $p\ge 1, $ if there exists one positive constant $C$ such that for any $f: M \to\mathbb{R}$ smooth enough,

$ \begin{equation}\label{defofp} \frac{p}{2-p}\left(\displaystyle\int_M f^2 d\mu-(\displaystyle\int |f|^p d\mu_M)^\frac{2}{p}\right)\leq C \displaystyle\int_M |\nabla f|^2 d\mu, \end{equation} $

(1.3)

In fact, the classical Poincaré inequality corresponds to the case $p=1$ and the logarithmic Sobolev inequality turns out to the limit case when $p$ tends to $2$ since

$ \lim\limits_{p\to 2^{-}} \frac{p(\displaystyle\int f^2 d\mu_x^n-(\displaystyle\int |f|^p d\mu_x^n)^\frac{2}{p})}{2-p}={\rm Ent}_{\mu_x^n}(f^2), $

where

$ {\rm Ent}_{\mu_x^n}(f^2):=\mu_x^n(f^2\log f^2)-\mu_x^n (f^2) \log (\mu_x^n (f^2)) $

is the relative entropy of $f^2$ under $\mu_x^n.$ It was proved in [4] that

$ \Phi_p(\mu_x^n, f)=\frac{p(\displaystyle\int f^2 d\mu_x^n-(\displaystyle\int |f|^p d\mu_x^n)^\frac{2}{p})}{2-p} $

is increasing on $p$ for given $f.$

In this paper, we consider the Poincaré inequality, logarithmic Sobolev inequality and Sobolev inequality for Moebius measures on the unit circle.

In [3], Schechtman and Schmuckenschläger proved that $\mu_x^n$ with any $|x| <1$ has a uniform Gaussian concentration property, which is similar to the one of $\mu^n_0$. In [5], they obtained logarithmic Sobolev and Poincaré inequalities for harmonic measures on unit sphere $S^{n-1}$ for $n\ge 3$ and in [2] they had similar results for harmonic measures when $n=2$. And then in [1], they obtained Sobolev inequalities for harmonic measures when $n\ge 2.$

Following the idea in [5], they obtained in [6] similar results for Moebius measures on unit sphere for $n\ge 3.$ In this paper, we will work on the Moebius measures on unit circle

$ \mu_x(dy)=\frac{\sqrt{1-|x|^2}}{1-(x, y)}\mu(dy), \quad y\in S $

(1.4)

with $x\in\mathbb{R}^2, |x| <1$ and $\mu$ the uniform probability on the unit circle.

The main result of this paper is the following.

Theorem 1    Let $\mu_x$ be the Moebius measure on the unit circle. We have

a) the optimal Poincaré constant $C_{\rm P}(\mu_x)$ satisfies

$ 1\le C_{\rm P}(\mu_x)\le\dfrac{2\sqrt{1+|x|}}{\sqrt{1+|x|}+\sqrt{1-|x|}}\le 2; $

b) the optimal logarithmic Sobolev constant $C_{\rm LS}(\mu_x)$ satisfies

$ \frac{\sqrt{3}}{18}\pi\log(1+\frac{\pi}{2\sqrt{1-|x|}})\le C_{\rm LS}(\mu_x)\le 8\pi\log(1+\frac{e^2\pi}{\sqrt{1-|x|}})+\log 4; $

c) the optimal Sobolev constant $C_{p}(\mu_{x})$ satisfies

$ \frac{p(1-(1+\frac{1}{2\sqrt{1-|x|}})^{\frac{p-2}{p}})}{3(2-p)}\leq C_{p}(\mu_{x})\leq \frac{8\pi p}{2-p}\left(1-\left(1+\frac{\pi(p-1)^{\frac{p}{p-2}}}{\sqrt{1-|x|}}\right)^{\frac{p-2}{p}}\right)+\log 4 $

for $1 <p <2.$

We first present a crucial lemma, which combines a particular case of Lemma 1.1 in [2] and a lemma in [1].

Lemma 2.1 Define

$ \nu_{a}(d\theta)=\frac{\sqrt{1-a^2}}{\pi}\frac{1}{1-a\cos\theta}d\theta, \; \theta\in [0, \pi] $

(2.1)

for $0 <a <1.$ We have, respectively,

(1) the corresponding Poincaré constant satisfies

$ C_{\rm P}(\nu_{|x|})\le C_{\rm P}(\mu_x)\le \max\{C_{\rm P}(\nu_{|x|}), \frac{1}{\lambda^{DD}(\nu_{|x|})}\}; $

(2) similarly, the optimal logarithmic Sobolev constants satisfy

$ C_{\rm LS}(\nu_{|x|})\le C_{\rm LS}(\mu_x)\le C_{\rm LS}(\nu_{|x|})+\frac{1}{\lambda^{DD}(\nu_{|x|})}, $

here $\lambda^{DD}(\nu_{|x|})$ is defined as

$ \lambda^{DD}(\nu_{|x|}):=\inf\left\{\frac{\displaystyle\int (f')^2 d\nu_{|x|}}{\nu(f^2)}: \; f(0)=f(\pi)=0, \quad f \quad\text{non constant}\right\}; $

(3) the optimal Sobolev constant satisfies

$ C_p(\nu_{|x|})\le C_{p}(\mu_x)\le C_p(\nu_{|x|})+\frac{1}{\lambda^{DD}(\nu_{|x|})}. $

Define the diffusion operator $\mathcal{L}_a$ as

$ \mathcal{L}_af(\theta)=f''(\theta)-\dfrac{a\sin\theta}{1-a\cos\theta}f'(\theta) $

for any smooth function $f: [0, \pi]\to\mathbb{R}.$ The corresponding Dirichlet form is

$ \mathcal{E}_a(f, f)=\displaystyle\int_0^\pi(f^{'})^2d\nu_a=\displaystyle\int_0^\pi f(-\mathcal{L}_a f)d\nu_a. $

The optimal Poincaré constant $C_{\rm P}(\nu_a)=\frac{1}{\lambda_1(\nu_{a})}, $ where $\lambda_1(\nu_a)$ has classic variational formula

$ \lambda_1(\nu_a)=\left\{\frac{\mathcal{E}_a(f, f)}{{\rm Var}_{a}(f)}, \; f \; \mbox{non constant}\right\}. $

Put $f(\theta)=1-a\cos\theta.$ We get

$ \nu_a(f)=\nu_a(f^2)=\sqrt{1-a^2} $

and

$ \mathcal{E}_a(f, f) =\frac{\sqrt{1-a^2}}{\pi}\displaystyle\int_0^{\pi}\frac{a^2\sin^2\theta}{1-a\cos\theta}d\theta\\ =\frac{\sqrt{1-a^2}}{\pi}\displaystyle\int_0^{\pi}\big(\frac{a^2-1}{1-a\cos\theta}+1-a\cos\theta\big) d\theta\\ =\sqrt{1-a^2}-(1-a^2). $

So by the variational formula

$ \lambda_1(\nu_a)\le\dfrac{\mathcal{E}_a(f, f)}{{\rm Var}_a(f)}=1. $

Therefore we have $C_{\rm P}(\nu_a)\ge 1.$ Now we work on the upper bound for $C_{\rm P}(\nu_a).$ The variational formula for $\lambda_1(\nu_a)$ by Chen in [7] could be understood as

$ C_{\rm P}(\nu_a)=\inf\limits_{f\in\mathcal{F}}\sup\limits_{\theta\in[0, \pi]}\left[\dfrac{1-a\cos\theta}{f'(\theta)}\displaystyle\int_\theta^\pi\dfrac{f(y)}{1-a\cos y}dy\right], $

where

$ \mathcal{F}=\left\{f: {f'>0 \, , x\in [0, \pi]}; \; \nu_a(f)=0\right\}. $

Set $\rho(\theta)=1-a\cos \theta-\sqrt{1-a^2}$, it is easy to see that $\rho$ is strictly increasing on $[0, \pi]$ and $ \nu_a(\rho) =0. $ Thus

$ \lambda_1(\nu_a)^{-1} \le \sup\limits_{\theta\in(0, \pi)}\dfrac{(1-a\cos\theta)}{a \sin\theta} \displaystyle\int_{\theta}^{\pi}\dfrac{1-a\cos y-\sqrt{1-a^2}}{1-a\cos y}dy\\ =\sup\limits_{\theta\in(0, \pi)}\dfrac{2(1-a\cos\theta)}{a \sin\theta}\left(\arctan (\cot\frac{\theta} 2)-\arctan \left(\sqrt{\dfrac{1-a}{1+a}} \cot\frac{\theta}2\right)\right)\\ \le \sup\limits_{\theta\in(0, \pi)}\dfrac{2(1-a\cos\theta)}{a \sin\theta}\dfrac{(1-\sqrt{\frac{1-a}{1+a}})\cot\frac{\theta}{2}}{1+\left(\sqrt{\frac{1-a}{1+a}} \cot\frac{\theta}2\right)^2}\\ =\frac{1+a}{a}(1-\sqrt{\frac{1-a}{1+a}}) \\ = \frac{2}{1+\sqrt{\frac{1-a}{1+a}}}, $

(2.2)

where the first equality comes true by the fact that for $\alpha\in [0, \pi], $

$ \begin{equation}\label{buzhidao}\displaystyle\int_{\alpha}^{\pi}\dfrac{d\theta}{1-a\cos\theta} =\dfrac{2}{\sqrt{1-a^2}}\arctan(\sqrt{\dfrac{1-a}{1+a}}\cot\dfrac{\alpha}{2}).\end{equation} $

(2.3)

And the last but second equality holds by the fact

$ \dfrac{2\cot(\frac{\theta}{2})}{1+\left(\sqrt{\frac{1-a}{1+a}} \cot\frac{\theta}2\right)^2}=\frac{(1+a)\sin\theta}{( 1+a)\sin^2\frac\theta 2+(1-a)\cos^2\frac{\theta}2}=\frac{(1+a)\sin\theta}{1-a\cos\theta}. $

Step 1    Lower bound for $\lambda^{DD}(\nu_a)$. Choose $f$ as $f(\theta)=\sin\theta$ for $\theta\in [0, \pi].$ Clearly, $f$ satisfies

$ f(0)=f(\pi)=0; f'(x)|_{0 <x <\pi/2}>0 ; f'(x)|_{\pi/2 <x <\pi} <0. $

So by Theorem 1.1 in [8],

$ \frac{1}{\lambda^{DD}(\nu_a)} \le \sup\limits_{x\in (0, \pi/2)}\frac1{\sin x}\displaystyle\int_0^x (1-a\cos y)dy\displaystyle\int_y^{\pi/2}\frac{\sin u}{1-a\cos u}du \\ \vee\sup\limits_{x\in (\pi/2, \pi)}\frac{1}{\sin x}\displaystyle\int_x^{\pi} (1-a\cos y)dy\displaystyle\int^y_{\pi/2}\frac{\sin u}{1-a\cos u}du\\ \le\sup\limits_{x\in(0, \pi), x\neq \pi/2}\frac{1-a\cos x}{\cos x}\displaystyle\int_x^{\pi/2}\frac{\sin u}{1-a\cos u}du\\ =\sup\limits_{x\in(0, \pi), x\neq \pi/2}\frac{1-a\cos x}{a\cos x}\ln \frac{1}{1-a\cos x}\\ =\sup\limits_{|t|\le a}(1-\frac1t)\ln (1-t)=(1+\frac1a)\ln (1+a). $

(2.4)

In fact, for $0\le a\le 1, $

$ (1+\frac1a)\ln(1+a)\le \frac{2}{1+\sqrt{\frac{1-a}{1+a}}}. $

Now, combining the upper and lower bound for $\lambda_1(\nu_a)$ as well as the lower bound for $\lambda^{DD}(\nu_a)$, we have by Lemma 2.1,

$ \frac{1}{2}\le\frac{\sqrt{1+|x|}+\sqrt{1-|x|}}{2\sqrt{1+|x|}}\le\lambda_1(\mu_x)\le 1. $

The part a) of Theorem 1 follows.

By (2.3), it is clear that the median of $\nu_a$ is $\theta_a=\arccos a.$ Define

$ S_{\rm LS}^{-}(a, r):=\sup\limits_{\alpha\in(0, \theta_a)}\displaystyle\int_0^{\alpha}\dfrac{d\theta}{1-a\cos\theta} \log(1+\dfrac{ r\pi}{\displaystyle\int_0^{\alpha}\frac{\sqrt{1-a^2}}{1-a\cos\theta}d\theta}) \cdot\displaystyle\int_{\alpha}^{\theta_a}(1-a\cos\theta)d\theta $

and

$ S_{\rm LS}^{+}(a, r):=\sup\limits_{\alpha\in(\theta_a, \pi)}\displaystyle\int_{\alpha}^{\pi}\dfrac{d\theta}{1-a\cos\theta} \log\left(1+\dfrac{r\pi}{\displaystyle\int_{\alpha}^{\pi}\frac{\sqrt{1-a^2}}{1-a\cos\theta}d\theta}\right) \cdot\displaystyle\int_{\theta_a}^{\alpha}(1-a\cos\theta)d\theta $

with $r=\frac12$ or $r=e^2.$ It is trivial to check that

$ \begin{equation}\label{jie}\frac{\sin \alpha}{1-a \cos \alpha}\le\displaystyle\int_{\alpha}^{\pi}\frac{1}{1-a\cos \theta}d\theta\le\frac{2}{(1+a)\sin\frac{\alpha}{2}}\end{equation} $

(3.1)

and

$ \begin{equation}\label{jie2}\displaystyle\int^{\alpha}_{\beta}\frac{1}{1-a\cos \theta}d\theta\le \frac{\pi}{\sqrt{1-a^2}}\end{equation} $

(3.2)

for any $0 <\beta <\alpha <\pi.$

Step 1   Upper bound for $S_{\rm LS}^+(a, e^2)$ and $S_{\rm LS}^-(a, e^2)$. For given $b>0$, $x\log(1+b/x)$ is increasing on $x>0.$ We have by (3.1),

$ S_{\rm LS}^+(a, e^2) \leq \frac2{1+a}\sup\limits_{\alpha\in(\theta_a, \pi)} \frac{1}{\sin\frac{\alpha}2} \log\left(1+\dfrac{e^2\pi\sqrt{1+a}\sin\frac{\alpha}2}{2\sqrt{1-a}}\right)\left(\alpha-a\sin\alpha\right)\\ \leq \frac{2}{1+a}\sup\limits_{\alpha\in(\theta_a, \pi)} \log\left(1+\dfrac{e^2\pi}{\sqrt{2(1-a)}}\right) \dfrac{\alpha-a\sin\alpha}{\sin\frac{\alpha}{2}}\\ \leq 2\pi\log\left(1+\dfrac{e^2\pi}{\sqrt{1-a}}\right), $

(3.3)

where the last inequality is true since $\frac{2}{\pi}\le \frac{\sin x}{x}\le 1$ for any $x\in (0, \frac{\pi}{2}).$

Similarly, by (3.2) and the fact $\sin\theta_a=\sqrt{1-a^2}, $ we have

$ S_{\rm LS}^-(a, e^2) \leq \sup\limits_{\alpha\in(0, \theta_a)} \left(\theta_a-a\sin\theta_a\right)\times \frac{\pi}{\sqrt{1-a^2}} \log\big(1+e^2\sqrt{1-a^2}\big)\\ \leq \frac{\pi\theta_a}{\sin\theta_a}\log(1+e^2) \leq \frac{\pi^2}{2} \log(1+e^2). $

(3.4)

Step 2 Lower bound for $S_{\rm LS}^+(a, \frac12)$. Choosing $\alpha=\frac{2\pi}3, $ we have by (3.1) that

$ \displaystyle\int_{\alpha}^{\pi} \frac{1}{1-a\cos\theta}d\theta\ge \frac{\sqrt{3}}{2+a}\ge \frac{\sqrt{3}}{3} $

and

$ \displaystyle\int_{\theta_a}^{\alpha}(1-a\cos\theta)d\theta\ge\displaystyle\int_{\pi/2}^{2\pi/3}(1-a\cos\theta)d\theta\ge \frac{\pi}{6}. $

Therefore from the monotonicity of $x\log(1+b/x)$ for $x>0$ when $b>0, $ it holds

$ S_{\rm LS}^+(a, \frac12) \geq \frac{\sqrt{3}\pi}{18}\log(1+\frac{\sqrt{3}\pi}{2\sqrt{1-a^2}}) \geq\frac{\sqrt{3}\pi}{18}\log(1+\frac{\pi}{2\sqrt{1-a}}). $

(3.5)

Barthe-Roberto's characterization for logarithmic Sobolev constants tells (see [9])

$ \max\{S_{\rm LS}^-(a, \frac12), S_{\rm LS}^+(a, \frac12)\}\le C_{\rm LS}(\nu_a)\le 4\max\{S_{\rm LS}^-(a, e^2), S_{\rm LS}^+(a, e^2)\}. $

(3.6)

Therefore, it follows from (3.3), (3.4) and (3.5) that

$ \frac{\sqrt{3}\pi}{18}\log(1+\frac{\pi}{2\sqrt{1-a}})\le C_{\rm LS}(\nu_a)\le 8\pi\log\left(1+\dfrac{e^2\pi}{\sqrt{1-a}}\right). $

By (2.4), we have

$ \frac{1}{\lambda^{DD}(\nu_a)}\le(1+\frac{1}{a})\ln (1+a)\le \log 4. $

Thereby by Lemma 2.1, we get

$ \frac{\sqrt{3}}{18}\pi\log(1+\frac{\pi}{2\sqrt{1-a}})\le C_{\rm LS}(\mu_x)\le 8\pi\log(1+\frac{e^2\pi}{\sqrt{1-a}})+\log 4, $

which completes the proof of b) of Theorem 1.

Define

$ B_+(a, p):= \sup\limits_{x\in(\theta_{a}, \pi)}\displaystyle\int_{x}^{\pi}\frac{1}{1-a\cos \theta}d\theta \displaystyle\int_{\theta_{a}}^{x}(1-a\cos\theta)d\theta\\ \times\left(1-\left(1+\frac{\pi(p-1)^\frac{p}{p-2}}{\sqrt{1-a^2}\displaystyle\int_x^{\pi}\frac{1}{1-a\cos\theta}d\theta}\right)^\frac{p-2}{p}\right), \\ B_-(a, p):= \sup\limits_{x\in(0, \theta_{a})}\displaystyle\int_{0}^{x}\frac{d\theta}{1-a\cos\theta}\displaystyle\int_{x}^{\theta_{a}}(1-a\cos\theta)d\theta\\ \times\left(1-\left(1+\frac{\pi(p-1)^\frac{p}{p-2}}{\sqrt{1-a^{2}}\displaystyle\int_{0}^{x}\frac{1}{1-a\cos\theta}}\right)^\frac{p-2}{p}\right). $

It is easy to check that both $x(1-(1+\frac{C}{x})^\frac{p-2}{p})$ and $1-(1+x)^{(p-2)/p}$ are increasing on $\mathbb{R}_+$ for $C>0.$ Recalling the estimates

$ \begin{equation}\label{jie1}\frac{\sin \alpha}{1-a \cos \alpha}\le\displaystyle\int_{\alpha}^{\pi}\frac{1}{1-a\cos \theta}d\theta\le\frac{2}{(1+a)\sin\frac{\alpha}{2}}\end{equation} $

(4.1)

and

$ \begin{equation}\label{jie3}\displaystyle\int^{\alpha}_{\beta}\frac{1}{1-a\cos \theta}d\theta\le \frac{\pi}{\sqrt{1-a^2}}\end{equation} $

(4.2)

for any $0 <\beta <\alpha <\pi.$ We get

$ B_+(a, p) \leq \sup\limits_{x\in(\theta_{a}, \pi)}\frac{2x}{(1+a)\sin\frac{x}{2}}\left(1-\left(1+\frac{(p-1)^\frac{p}{p-2}}{\frac{\sqrt{1-a^{2}}}{\pi}\frac{2}{(1+a)\sin\frac{x}{2}}}\right)^\frac{p-2}{p}\right) \\ \leq \frac{2\pi}{1+a}\left(1-\left(1+\frac{\pi(p-1)^\frac{p}{p-2}}{2} \sqrt{\frac{1+a}{1-a}}\right)^\frac{p-2}{p}\right)\\ \le 2\pi \left(1-\left(1+\frac{\pi(p-1)^\frac{p}{p-2}}{\sqrt{1-a}}\right)^\frac{p-2}{p}\right), $

(4.3)

where the second inequality holds by $\frac{x}{\sin x}\leq\frac{\pi}{2}, \; \forall x\in(0, \pi/2)$.

Similarly, we get

$ B_{-}(a, p) \le \sup\limits_{x\in(0, \theta_{a})}\frac{\theta_a\pi}{\sqrt{1-a^{2}}}\left(1-\left(1+(p-1)^\frac{p}{p-2}\right)^\frac{p-2}{p}\right)\\ = \sup\limits_{x\in(0, \theta_{a})}\frac{\pi\theta_{a}}{\sin\theta_{a}}\left(1-\left(1+(p-1)^\frac{p}{p-2}\right)^\frac{p-2}{p}\right) \le \frac{\pi^{2}}{2}. $

(4.4)

Finally, Barthe-Roberto's characterization for Sobolev constant guarantees that

$ C_{p}(\nu_{a})\leq\frac{4p}{2-p}\max\{B_{+}(a, p), B_{-}(a, p)\}\leq \dfrac{8\pi p}{2-p}\left(1-\left(1+\frac{\pi(p-1)^\frac{p}{p-2}}{\sqrt{1-a}}\right)^\frac{p-2}{p}\right). $

(4.5)

Set $f=\frac{1-a\cos\theta}{\sqrt{1-a^2}}, $ then

$ \nu_{a}(f)=1, \; \nu_{a}(f^{2})=\frac{1}{\sqrt{1-a^2}}, \; \mathcal{E}_{a}(f, f)=\frac{1}{\sqrt{1-a^2}}-1. $

By Lemma 2.4 in [1], we know

$ C_{p}(\nu_{a}) \geq\frac{p}{2-p}\frac{\nu_{a}(f^{2})-\nu_{a}(f^{p})^\frac{2}{p}}{\mathcal{E}_{a}(f, f)} \geq \frac{p}{3(2-p)}\frac{\nu_a(f^{2})}{\mathcal{E}_{a}(f, f)}(1-\nu_a(f^{2})^{\frac{p-2}{p}})\\ =\frac{p}{3(2-p)}\frac{1}{1-\sqrt{1-a^2}}\left(1-\left(1+\frac{1-\sqrt{1-a^2}}{\sqrt{1-a^2}}\right)^{\frac{p-2}{p}}\right)\\ \geq\frac{p}{3(2-p)}\left(1-(1+\frac{1}{2\sqrt{1-|x|}})^{\frac{p-2}{p}}\right), $

(4.6)

where the last inequality holds by the facts that $r(1-(1+\frac{C}{r})^\frac{p-2}{p})$ is increasing and $\frac{1}{1-\sqrt{1-a^2}}\ge 1$. Combining (4.5), (4.6), $\frac{1}{\lambda^{DD}(\nu_{a})}\leq\log4$ and Lemma 2.1 together, we have

$ \frac{p(1-(1+\frac{1}{2\sqrt{1-|x|}})^{\frac{p-2}{p}})}{3(2-p)}\leq C_{p}(\mu_{x})\leq \frac{8\pi p}{2-p}\left(1-\left(1+\frac{\pi(p-1)^{\frac{p}{p-2}}}{\sqrt{1-|x|}}\right)^{\frac{p-2}{p}}\right)+\log 4, $

which completes the proof of theorem.