Convergence of the CR Yamabe flow

Abstract

We consider the CR Yamabe flow on a compact strictly pseudoconvex CR manifold M of real dimension $2n+1$. We prove convergence of the CR Yamabe flow when $n=1$ or M is spherical.

Appendices

Appendix A

First we consider the case when $n=1$. We review the definition of CR normal coordinates and recall some of its properties. Give any $x\in M$, we can find a contact form ${\widehat{\theta }}_x=\varphi _x^{\frac{2}{n}}\theta _0$ conformal to $\theta _0$, where ${\widehat{\theta }}_x$ is a contact form defined in (z, t) which is the CR normal coordinates centered at x. On the other hand, ${\widehat{\theta }}_x$ satisfies the following properties: (see Theorem 3.7 in P.172 of [15] and Proposition 6.5 in [11])

$$\begin{aligned} {\widehat{\theta }}_x= & {} \Bigg (1+O(\rho _x(y))\Bigg )\theta _{{\mathbb {H}}^n},\nonumber \\ dV_{{\widehat{\theta }}_x}= & {} \Bigg (1+O(\rho _x(y))\Bigg )dV_{\theta _{{\mathbb {H}}^n}},\nonumber \\ W_{k}= & {} \Bigg (1+O(\rho _x(y)^4)\Bigg )Z_k+O(\rho _x(y)^4){\overline{Z}}_k+O(\rho _x(y)^5)\frac{\partial }{\partial t}\quad \text{ for } 1\le k\le n,\nonumber \\ \end{aligned}$$

(A.1)

in $\{(z,t):(t^2+|z|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}$ for some ${\widehat{\rho }}>0$. Here (z, t) represents the point $y\in M$ in the CR normal coordinates centered at x, $\rho _x(y)$ is the distance in the CR normal coordinates at x, which implies that

$$\begin{aligned} \rho _{x_k}(y)=(t^2+|z|^4)^{\frac{1}{4}}. \end{aligned}$$

(A.2)

Also, $\theta _{{\mathbb {H}}^n}=dt + \sqrt{-1}\sum \nolimits _{j=1}^n(z_jd{\overline{z}}_j-{\overline{z}}_jdz_j)$ is the standard contact form of the Heisenberg group ${\mathbb {H}}^n=\{(z,t)=(z_1,\ldots ,z_n,t)\in {\mathbb {C}}^n\times {\mathbb {R}}\}$. Moreover, $\{W_{k}\}$ is a pseudo-Hermitian frame, i.e. $\{W_{k}\}$ is a local frame of ${\widehat{\theta }}_x$ satisfying $-\sqrt{-1}d{\widehat{\theta }}_x(W_{k},{\overline{W}}_{l})=\delta _{kl}$ (see P.165 in [15]), and $Z_k=\frac{\partial }{\partial z_k}+\sqrt{-1}{\overline{z}}_k\frac{\partial }{\partial t}$. We also have the following expression for the CR conformal sub-Laplacian: (see P.114 in [18])

$$\begin{aligned} -\left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_x}+R_{{\widehat{\theta }}_x}=-\left( 2+\frac{2}{n}\right) \Delta _{\theta _{{\mathbb {H}}^n}}+O(\rho _x(y)^2). \end{aligned}$$

(A.3)

On the other hand, the Webster scalar curvature of ${\widehat{\theta }}_x$ satisfies (see P.38 in [11])

$$\begin{aligned} |R_{{\widehat{\theta }}_x}(y)|\le C\rho _x(y)^2, \end{aligned}$$

(A.4)

where $\rho _x(y)$ is the distance in the CR normal coordinates at x. Let $G_x$ be the Green’s function with pole at x. Then we have (see (105) in [11])

$$\begin{aligned} -\left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_x}G_x(y)+R_{{\widehat{\theta }}_x} G_x(y)=0 \quad \text{ for } y\ne x. \end{aligned}$$

(A.5)

Moreover, the Green’s function satisfies the estimates (see Propositions 5.2 and 5.3 in [11])

$$\begin{aligned} |G_x(y)-\rho _x(y)^{-2n}-A_x|\le C\rho _x(y) \end{aligned}$$

(A.6)

and

$$\begin{aligned} |\nabla _{{\widehat{\theta }}_x}(G_x(y)-\rho _x(y)^{-2n})|_{{\widehat{\theta }}_x}\le C, \end{aligned}$$

(A.7)

where $A_x$ is the CR mass. Note that $\rho _x(y)$, the distance in the CR normal coordinates at x, and d(x, y), the Carnot–Carathéodory distance on M with respect to the contact form $\theta _0$ are equivalent, i.e. there exists a uniform constant $C_0$ such that

$$\begin{aligned} \frac{1}{C_0}d(x,y)\le \rho _x(y)\le C_0 d(x,y) \quad \text{ for } \text{ all } x, y\in M. \end{aligned}$$

(A.8)

When M is spherical, give any $x\in M$, we can find a smooth function $\varphi _x$ such that $\theta _{{\mathbb {H}}^n}=\varphi _x^{\frac{2}{n}}\theta _0$ in a neighborhood of x. That is,

$$\begin{aligned} {\widehat{\theta }}_x=\theta _{{\mathbb {H}}^n} \end{aligned}$$

(A.9)

in the above notation. Note also that $R_{{\widehat{\theta }}_x}=R_{\theta _{{\mathbb {H}}^n}}\equiv 0$ in this case. Therefore, (A.1) and (A.4) are still true. On the other hand, the Green’s function $G_x(y)$ defined as in (A.5) satisfies (A.6) and (A.7) by Lemma 5.1 in [10].

Under the assumptions of Theorem 1.1, the CR mass is positive, i.e. $A_x>0$ for all $x\in M$ by the CR positive mass theorem (see Theorem 1.1 in [11] and Corollary C in [10]). Note that the function $x\mapsto A_x$ is continuous. See Proposition 3.3 in [37] for the proof when M is spherical. See also Remark I.1.2 in [21] and Proposition 3.5 in [20] for the proof of the corresponding statement in the Riemannian case. Hence, we have

$$\begin{aligned} \inf _{x\in M}A_x>0. \end{aligned}$$

Suppose that we are given a set of pairs $(x_k,\varepsilon _k)_{1\le k\le m}$. For every $1\le k\le m$, we define ${\overline{u}}_{(x_k,\varepsilon _k)}$ by

$$\begin{aligned} {\overline{u}}_{(x_k,\varepsilon _k)}(y)=\varphi _{x_k}(y)\,{\overline{U}}_{(x_k,\varepsilon _k)}(y). \end{aligned}$$

(A.10)

Here

$$\begin{aligned} {\overline{U}}_{(x_k,\varepsilon _k)}(y)= & {} \left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _k^n \left[ \frac{\chi _\delta (\rho _{x_k}(y))}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}} \right. \nonumber \\&\left. +\, (1-\chi _\delta (\rho _{x_k}(y))v)G_{x_k}(y)\right] , \end{aligned}$$

(A.11)

where $\chi _\delta (s)=\chi (\frac{s}{\delta })$ and $\chi :{\mathbb {R}}\rightarrow [0,1]$ is a cut-off function satisfying $\chi (s)=1$ for $s\le 1$ and $\chi (s)=0$ for $s\ge 2$. On the other hand, $\delta$ is a positive real number such that $\varepsilon _k\ll \delta$ for all $1\le k\le m$.

Proposition A.1

For $n=1$, we have

$$\begin{aligned}&\left| \left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}(y) -R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}(y) +r_\infty {\overline{U}}_{(x_k,\varepsilon _k)}(y)^{1+\frac{2}{n}}\right. \\&\qquad \left. +\left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _k^n A_{x_k}\left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)) \right| \\&\quad \le C\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n}{2}}\rho _{x_k}(y)^2\,1_{\{\rho _{x_k}(y)\le 2\delta \}} +C\frac{\varepsilon _k^n}{\delta }\,1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}\\&\qquad +\,C\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n+2}{2}}1_{\{\rho _{x_k}(y)\ge \delta \}}. \end{aligned}$$

Proof

By definition of ${\overline{U}}_{(x_k,\varepsilon _k)}$, we have

$$\begin{aligned}&\left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}(y) -R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}(y) +r_\infty {\overline{U}}_{(x_k,\varepsilon _k)}(y)^{1+\frac{2}{n}}\\&\qquad +\left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _k^n A_{x_k}\cdot \left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)) \\&\quad =\left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _k^n\,\Bigg (I_1+I_2+I_3+I_4+I_5+I_6+I_7+I_8\Bigg ). \end{aligned}$$

Here,

$$\begin{aligned} I_1= & {} \chi _\delta (\rho _{x_k}(y))\left[ \phantom {\left( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right) ^{1+\frac{2}{n}}} \left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}\frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right. \\&\left. +\,(2n+2)n\varepsilon _k^2 \left( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right) ^{1+\frac{2}{n}}\right] ,\\ I_2= & {} -\chi _\delta (\rho _{x_k}(y))R_{{\widehat{\theta }}_{x_k}}\frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}},\\ I_3= & {} -\left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y))\Bigg ( G_{x_k}(y)-\rho _{x_k}(y)^{-2n}-A_{x_k}\Bigg ),\\ I_4= & {} \left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y))\Bigg ( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}-\rho _{x_k}(y)^{-2n}\Bigg ),\\ I_5= & {} -2\left( 2+\frac{2}{n}\right) \langle \nabla _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)),\nabla _{{\widehat{\theta }}_{x_k}}(G_x(y)-\rho _x(y)^{-2n})\rangle _{{\widehat{\theta }}_{x_k}},\\ I_6= & {} 2\left( 2+\frac{2}{n}\right) \Bigg \langle \nabla _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)),\nabla _{{\widehat{\theta }}_{x_k}} \Bigg ( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}-\rho _{x_k}(y)^{-2n}\Bigg )\Big \rangle _{{\widehat{\theta }}_{x_k}},\\ I_7= & {} (2n+2)n\varepsilon _k^2\left[ \left( \frac{\chi _\delta (\rho _{x_k}(y))}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}+ \Bigg (1-\chi _\delta (\rho _{x_k}(y))\Bigg )G_{x_k}(y)\right) ^{1+\frac{2}{n}}\right. \\&\left. -\,\chi _\delta (\rho _{x_k}(y))\left( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right) ^{1+\frac{2}{n}}\right] ,\\ I_8= & {} 7\Bigg (1-\chi _\delta (\rho _{x_k}(y)\Bigg )\left[ \left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}G_{x_k}(y)-R_{{\widehat{\theta }}_{x_k}} G_{x_k}(y)\right] . \end{aligned}$$

Since

$$\begin{aligned} \begin{aligned} \Delta _{\theta _{{\mathbb {H}}^n}}\left( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right)&=-\frac{n^2\varepsilon _k^2}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{1+\frac{n}{2}}} \end{aligned} \end{aligned}$$

(A.12)

we have

$$\begin{aligned} |I_1|\le & {} \chi _\delta (\rho _{x_k}(y))\left| \left( 2+\frac{2}{n}\right) (\Delta _{\theta _{{\mathbb {H}}^n}}+O(\rho _{x_k}(y)^2))\left( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right) \right. \\&\left. +\,(2n+2)n\varepsilon _k^2 \left( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right) ^{1+\frac{2}{n}}\right| \\\le & {} 1_{\{\rho _{x_k}(y)\le 2\delta \}}\frac{C\rho _{x_k}(y)^2}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}} \end{aligned}$$

by (A.1) and (A.2). By (A.4), we have

$$\begin{aligned} |I_2|\le C\rho _{x_k}(y)^2 \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\,1_{\{\rho _{x_k}(y)\le 2\delta \}}. \end{aligned}$$

By (A.6), we have

$$\begin{aligned} |I_3|\le & {} \left( 2+\frac{2}{n}\right) |\Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y))|\cdot \Bigg | G_{x_k}(y)-\rho _{x_k}(y)^{-2n}-A_{x_k}\Bigg |\\\le & {} C\frac{1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}} }{\delta ^2}\rho _{x_k}(y) \le \frac{C}{\delta }1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}. \end{aligned}$$

Note that

$$\begin{aligned} |I_4|= & {} \left( 2+\frac{2}{n}\right) |\Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y))|\cdot \left| \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}-\rho _{x_k}(y)^{-2n}\right| \\\le & {} C\frac{1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}}{\delta ^2} \cdot \left| \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}-\frac{1}{(t^2+|z|^4)^{\frac{n}{2}}}\right| \\\le & {} C\frac{1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}}{\delta ^2}\cdot \frac{\varepsilon _k^{2n}}{\delta ^{4n}}\\\le & {} \frac{C}{\delta }1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}} \end{aligned}$$

by (A.2) and the assumption that $\varepsilon _k\ll \delta$. Note also that

$$\begin{aligned} |I_6|\le & {} C|\nabla _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y))|_{{\widehat{\theta }}_{x_k}}\cdot \left| \nabla _{{\widehat{\theta }}_{x_k}}\Bigg ( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}-\rho _{x_k}(y)^{-2n}\Bigg )\right| _{{\widehat{\theta }}_{x_k}}\\\le & {} C\frac{1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}}{\delta } \cdot |z|^2 \frac{\varepsilon _k^2}{(t^2+|z|^4)^{\frac{n}{2}+1}}\\\le & {} C\frac{\varepsilon _k^2}{\delta ^{2n+3}}1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}} \le C\frac{1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}}{\delta } \end{aligned}$$

by (A.1), (A.2), and the assumption that $\varepsilon _k\ll \delta$. By (A.7), we have

$$\begin{aligned} |I_5|\le C|\nabla _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y))|_{{\widehat{\theta }}_{x_k}}\Bigg | \nabla _{{\widehat{\theta }}_{x_k}}( G_{x_k}(y)-\rho _{x_k}(y)^{-2n})\Bigg |_{{\widehat{\theta }}_{x_k}}\le C\frac{1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}}{\delta }. \end{aligned}$$

We also have

$$\begin{aligned} |I_7|= & {} (2n+2)n\varepsilon _k^2\left| \left( \frac{\chi _\delta (\rho _{x_k}(y))}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}+ \Bigg (1-\chi _\delta (\rho _{x_k}(y))\Bigg )G_{x_k}(y)\right) ^{1+\frac{2}{n}}\right. \\&\left. -\,\chi _\delta (\rho _{x_k}(y))\left( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right) ^{1+\frac{2}{n}}\right| \\\le & {} C\varepsilon _k^2\left( \left( \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}\right) ^{1+\frac{2}{n}}+ G_{x_k}(y)^{1+\frac{2}{n}}\right) 1_{\{\rho _{x_k}(y)\ge \delta \}}\\\le & {} C\frac{\varepsilon _k^2}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{1+\frac{2}{n}}}1_{\{\rho _{x_k}(y)\ge \delta \}} \end{aligned}$$

by (A.6), and

$$\begin{aligned}I_8=0 \end{aligned}$$

by (A.5). From this, the assertion follows. $\square$

Proposition A.2

When M is spherical, we have

$$\begin{aligned}&\left| \left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}(y) -R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}(y) +r_\infty {\overline{U}}_{(x_k,\varepsilon _k)}(y)^{1+\frac{2}{n}}\right. \\&\qquad \left. +\left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _k^n A_{x_k}\left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)) \right| \\&\quad \le C\frac{\varepsilon _k^n}{\delta }\,1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}} +C\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n+2}{2}}1_{\{\rho _{x_k}(y)\ge \delta \}}. \end{aligned}$$

Proof

The proof is the same as the proof of Proposition A.1, except we need to prove that $I_1=0$. But this follows from (A.9) and (A.12). $\square$

Corollary A.3

We have

$$\begin{aligned}&\left| \left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}(y) -R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}(y) +r_\infty {\overline{U}}_{(x_k,\varepsilon _k)}(y)^{1+\frac{2}{n}}\right| \\&\quad \le C\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n}{2}}\rho _{x_k}(y)^2\,1_{\{\rho _{x_k}(y)\le 2\delta \}} +C\frac{\varepsilon _k^n}{\delta ^2}\,1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}\\&\qquad +\,C\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n+2}{2}}1_{\{\rho _{x_k}(y)\ge \delta \}}. \end{aligned}$$

Proof

Combining Propositions A.1, A.2 and the following estimate:

$$\begin{aligned} \left| \left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _k^n A_{x_k}\left( 2+\frac{2}{n}\right) \Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y))\right| \le C\frac{\varepsilon _k^n}{\delta ^2}\,1_{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}, \end{aligned}$$

Corollary A.3 follows. $\square$

Proposition A.4

If $\delta$ is sufficiently small, then

$$\begin{aligned} E({\overline{u}}_{(x_k,\varepsilon _k)})\le Y(S^{2n+1})-cA_{x_k}\varepsilon _k^{2n} +C\delta ^2\varepsilon _k^{2n}+C\delta \varepsilon _k^{2n}+C\delta ^{-2n-2}\varepsilon _k^{2n+2} \end{aligned}$$

for some $c>0$.

Proof

When $n=1$, it follows from Proposition A.1 and integration by parts that

$$\begin{aligned}&\int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )|\nabla _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}|^2_{{\widehat{\theta }}_{x_k}} +R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}^2-r_\infty {\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}\Bigg )dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad =-\int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )\Delta _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)} -R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}+r_\infty {\overline{U}}_{(x_k,\varepsilon _k)}^{1+\frac{2}{n}}\Bigg ) {\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad \le \left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _k^n A_{x_k}\left( 2+\frac{2}{n}\right) \int _M\Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)){\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\qquad +\, C\int _{\{\rho _{x_k}(y)\le 2\delta \}}\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n}{2}}\rho _{x_k}(y)^2 {\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\qquad +\,C\frac{\varepsilon _k^n}{\delta }\int _{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}{\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\qquad +\,C\int _{\{\rho _{x_k}(y)\ge \delta \}}\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n+2}{2}} {\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}. \end{aligned}$$

(A.13)

We are going to estimate each of the terms on the right hand side of (A.13). Since

$$\begin{aligned}&\left| \int _M\Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)) \Bigg ({\overline{U}}_{(x_k,\varepsilon _k)}-\rho _{x_k}(y)^{-2n}\Bigg )dV_{{\widehat{\theta }}_{x_k}}\right| \\&\quad \le \frac{C}{\delta ^2}\int _{\{\delta \le \rho _{x_k}(y)\le 2\delta \}} \left| \frac{1}{(t^2+(\varepsilon _k^2+|z|^2)^2)^{\frac{n}{2}}}-\frac{1}{(t^2+|z|^4)^{\frac{n}{2}}}\right| dV_{{\widehat{\theta }}_{x_k}}\\&\qquad +\, \frac{C}{\delta ^2}\int _{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}|G_{x_k}(y)-\rho _{x_k}(y)^{-2n}|dV_{{\widehat{\theta }}_{x_k}}\\&\quad \le \frac{C}{\delta ^2}\int _\delta ^{2\delta }\frac{\varepsilon _k^{2n}r^{2n-1}dr}{r^{4n}} \le C\delta ^{-2n-2}\varepsilon _k^{2n} \end{aligned}$$

by (A.6) and (A.2), we have

$$\begin{aligned}&\int _M\Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)){\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad =-\int _M\Bigg \langle \nabla _{{\widehat{\theta }}_{x_k}}\rho _{x_k}(y)^{-2n}, \nabla _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y))\Bigg \rangle _{{\widehat{\theta }}_{x_k}} dV_{{\widehat{\theta }}_{x_k}} +C\delta ^{-2n-2}\varepsilon _k^{2n}\nonumber \\&\quad =2n\int _{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}\chi '_\delta (\rho _{x_k}(y))\rho _{x_k}(y)^{-2n-1}|\nabla _{{\widehat{\theta }}_{x_k}}\rho _{x_k}(y)|^2_{{\widehat{\theta }}_{x_k}} dV_{{\widehat{\theta }}_{x_k}} +C\delta ^{-2n-2}\varepsilon _k^{2n}\nonumber \\&\quad =-\frac{C}{\delta }\int _\delta ^{2\delta } dr +C\delta ^{-2n-2}\varepsilon _k^{2n}=-C_0 +C\delta ^{-2n-2}\varepsilon _k^{2n} \end{aligned}$$

(A.14)

by integration by parts. Here $C_0$ is a positive constant depending only on n. On the other hand, it follows form the definition of ${\overline{U}}_{(x_k,\varepsilon _k)}$ that

$$\begin{aligned}&C\int _{\{\rho _{x_k}(y)\le 2\delta \}}\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n}{2}}\rho _{x_k}(y)^2 {\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad \le C\varepsilon _k^{2n}\int _{\{\rho _{x_k}(y)\le 2\delta \}}\frac{\rho _{x_k}(y)^2}{(t^2+|z|^4)^{n}} dV_{{\widehat{\theta }}_{x_k}}=C\delta ^{4-2n}\varepsilon _k^{2n}, \end{aligned}$$

(A.15)

where we have used the assumption that $n=1$. Note also that

$$\begin{aligned}&C\frac{\varepsilon _k^n}{\delta }\int _{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}{\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad \le C\frac{\varepsilon _k^{2n}}{\delta } \int _{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}\Bigg (\rho _{x_k}(y)^{-2n}+A_{x_k}+C\rho _{x_k}(y)\Bigg )dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad \le C\frac{\varepsilon _k^{2n}}{\delta }\int _\delta ^{2\delta }\frac{r^{2n+1}dr}{r^{2n}}=C\delta \varepsilon _k^{2n}, \end{aligned}$$

(A.16)

and

$$\begin{aligned}&C\int _{\{\rho _{x_k}(y)\ge \delta \}}\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n+2}{2}} {\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad \le C\varepsilon _k^{2n+2}\int _\delta ^\infty \frac{r^{2n+1}dr}{r^{4(\frac{n+2}{2})}\cdot r^{2n}} =C\varepsilon _k^{2n+2}\int _\delta ^\infty \frac{dr}{r^{2n+3}} =C\delta ^{-2n-2}\varepsilon _k^{2n+2},\qquad \qquad \end{aligned}$$

(A.17)

where we have used (A.6) and the definition of ${\overline{U}}_{(x_k,\varepsilon _k)}$. Combining (A.13)–(A.17), we have

$$\begin{aligned}&\int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )|\nabla _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}|^2_{{\widehat{\theta }}_{x_k}} +R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}^2\Bigg )dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad \le r_\infty \int _M{\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}} -\left( \frac{n(2n+2)}{r_\infty }\right) ^{n} \varepsilon _k^{2n} A_{x_k}\left( 4+\frac{4}{n}\right) C_0\nonumber \\&\qquad +\,C\delta ^2\varepsilon _k^{2n}+C\delta \varepsilon _k^{2n}+C\delta ^{-2n-2}\varepsilon _k^{2n+2}\nonumber \\&\quad \le r_\infty \left( \int _M{\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}}\right) ^{\frac{n}{n+1}} \left( \int _M{\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}}\right) ^{\frac{1}{n+1}} \nonumber \\&\qquad -\left( \frac{n(2n+2)}{r_\infty }\right) ^{n} \varepsilon _k^{2n} A_{x_k}\left( 2+\frac{2}{n}\right) C_0 +C\delta ^2\varepsilon _k^{2n}+C\delta \varepsilon _k^{2n}+C\delta ^{-2n-2}\varepsilon _k^{2n+2}\nonumber \\ \end{aligned}$$

(A.18)

where the last inequality follows from Hölder’s inequality. It follows from the definition of ${\overline{U}}_{(x_k,\varepsilon _k)}$ that

$$\begin{aligned} \int _M{\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}}\le & {} \left( \frac{n(2n+2)}{r_\infty }\right) ^{n+1}\left( \int _M\chi _\delta (\rho _{x_k}(y)) \left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{n+1}dV_{{\widehat{\theta }}_{x_k}}\right. \nonumber \\&\left. +\,C \int _M(1-\chi _\delta (\rho _{x_k}(y)))\varepsilon _k^{2n+2}G_{x_k}(y)^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}}\right) . \end{aligned}$$

(A.19)

Note that

$$\begin{aligned}&\int _M(1-\chi _\delta (\rho _{x_k}(y)))\varepsilon _k^{2n+2}G_{x_k}(y)^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad \le \varepsilon _k^{2n+2}\int _{\rho _{x_k}(y)\ge \delta }\Bigg (\rho _{x_k}(y)^{-2n}+A_{x_k}+C\rho _{x_k}(y)\Bigg )^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad =C\varepsilon _k^{2n+2}\int _\delta ^\infty \frac{r^{2n+1}dr}{r^{2n\left( 2+\frac{2}{n}\right) }} =C\varepsilon _k^{2n+2}\int _\delta ^\infty \frac{dr}{r^{2n+3}}=C\delta ^{-2n-2}\varepsilon _k^{2n+2}\qquad \qquad \end{aligned}$$

(A.20)

by (A.6). Note also that

$$\begin{aligned}&\int _M\chi _\delta (\rho _{x_k}(y)) \left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{n+1}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad =\int _{{\mathbb {H}}^n}\left( \frac{1}{(t^2+(1+|z|^2)^2)^{\frac{n}{2}}}\right) ^{2+\frac{2}{n}}dV_{\theta _{{\mathbb {H}}^n}} +O(\varepsilon _k^{2n+2})\nonumber \\&\quad =\left( \frac{Y(S^{2n+1})}{n(2n+2)}\right) ^{n+1}+O(\varepsilon _k^{2n+2}), \end{aligned}$$

(A.21)

where we have used the change of variables $(\frac{t}{\varepsilon _k^2},\frac{z}{\varepsilon _k})\mapsto (t,z)$. Combining (A.19)–(A.21), we get

$$\begin{aligned} \int _M{\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}} \le \left( \frac{Y(S^{2n+1})}{r_\infty }\right) ^{n+1}+C\delta ^{-2n-2}\varepsilon _k^{2n+2}. \end{aligned}$$

Substituting this into (A.18), we obtain

$$\begin{aligned}&\int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )|\nabla _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}|^2_{{\widehat{\theta }}_{x_k}} +R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}^2\Bigg )dV_{{\widehat{\theta }}_{x_k}}\\&\quad \le Y(S^{2n+1}) \left( \int _M{\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}}\right) ^{\frac{n}{n+1}} -\left( \frac{n(2n+2)}{r_\infty }\right) ^{n} \varepsilon _k^{2n} A_{x_k}\left( 2+\frac{2}{n}\right) C_0 \\&\qquad +\,C\delta ^2\varepsilon _k^{2n}+C\delta \varepsilon _k^{2n}+C\delta ^{-2n-2}\varepsilon _k^{2n+2}. \end{aligned}$$

This proves Proposition A.4 for the case when $n=1$.

When M is spherical, it follows from Proposition A.2 and integration by parts that

$$\begin{aligned}&\int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )|\nabla _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}|^2_{{\widehat{\theta }}_{x_k}} +R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}^2-r_\infty {\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}\Bigg )dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\quad \le \left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _k^n A_{x_k}\left( 2+\frac{2}{n}\right) \int _M\Delta _{{\widehat{\theta }}_{x_k}}\chi _\delta (\rho _{x_k}(y)){\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}} \nonumber \\&\qquad +\,C\frac{\varepsilon _k^n}{\delta }\int _{\{\delta \le \rho _{x_k}(y)\le 2\delta \}}{\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}\nonumber \\&\qquad +\,C\int _{\{\rho _{x_k}(y)\ge \delta \}}\left( \frac{\varepsilon _k^2}{t^2+(\varepsilon _k^2+|z|^2)^2}\right) ^{\frac{n+2}{2}} {\overline{U}}_{(x_k,\varepsilon _k)}dV_{{\widehat{\theta }}_{x_k}}. \end{aligned}$$

(A.22)

Combining (A.14), (A.16), (A.17), and (A.22), we also obtain (A.18). Now we can follow the same argument as above to finish the proof. $\square$

Lemma A.5

We have

$$\begin{aligned} F({\overline{u}}_{(x_{k},\varepsilon _{k})})=r_\infty +o(1). \end{aligned}$$

Proof

It follows from (A.18) that

$$\begin{aligned}&\int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )|\nabla _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}|^2_{{\widehat{\theta }}_{x_k}} +R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}^2\Bigg )dV_{{\widehat{\theta }}_{x_k}}\\&\quad \le r_\infty \int _M{\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}} -\left( \frac{n(2n+2)}{r_\infty }\right) ^{n} \varepsilon _k^{2n} A_{x_k}\left( 2+\frac{2}{n}\right) C_0\\&\qquad +\,C\delta ^2\varepsilon _k^{2n}+C\delta \varepsilon _k^{2n}+C\delta ^{-2n-2}\varepsilon _k^{2n+2}, \end{aligned}$$

which implies that

$$\begin{aligned} \int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )|\nabla _{\theta _0}{\overline{u}}_{(x_k,\varepsilon _k)}|^2_{\theta _0} +R_{\theta _0}{\overline{u}}_{(x_k,\varepsilon _k)}^2\Bigg )dV_{\theta _0}\le r_\infty \int _M{\overline{u}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{\theta _0} +o(1). \end{aligned}$$

This implies that $F({\overline{u}}_{(x_{k},\varepsilon _{k})})\le r_\infty +o(1)$. On the other hand, following the proof of Proposition A.4, one can prove that

$$\begin{aligned}&\int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )|\nabla _{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}|^2_{{\widehat{\theta }}_{x_k}} +R_{{\widehat{\theta }}_{x_k}}{\overline{U}}_{(x_k,\varepsilon _k)}^2\Bigg )dV_{{\widehat{\theta }}_{x_k}}\\&\quad \ge r_\infty \int _M{\overline{U}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{{\widehat{\theta }}_{x_k}} -\left( \frac{n(2n+2)}{r_\infty }\right) ^{n} \varepsilon _k^{2n} A_{x_k}\left( 2+\frac{2}{n}\right) C_0\\&\qquad -\,C\delta ^2\varepsilon _k^{2n}-C\delta \varepsilon _k^{2n}-C\delta ^{-2n-2}\varepsilon _k^{2n+2}, \end{aligned}$$

by using the same arguments to obtain (A.18). This implies that

$$\begin{aligned} \int _M\Bigg (\Bigg (2+\frac{2}{n}\Bigg )|\nabla _{\theta _0}{\overline{u}}_{(x_k,\varepsilon _k)}|^2_{\theta _0} +R_{\theta _0}{\overline{u}}_{(x_k,\varepsilon _k)}^2\Bigg )dV_{\theta _0}\ge r_\infty \int _M{\overline{u}}_{(x_k,\varepsilon _k)}^{2+\frac{2}{n}}dV_{\theta _0} +o(1), \end{aligned}$$

which gives $F({\overline{u}}_{(x_{k},\varepsilon _{k})})\ge r_\infty +o(1)$. This proves the assertion. $\square$

Lemma A.6

We have the estimate

$$\begin{aligned} \int _M{\overline{u}}_{(x_i,\varepsilon _i)}{\overline{u}}_{(x_j,\varepsilon _j)}^{1+\frac{2}{n}}dV_{\theta _0} \le C\left( \frac{\varepsilon _i^2\varepsilon _j^2}{\varepsilon _j^4+d(x_i,x_j)^4}\right) ^{\frac{n}{2}}. \end{aligned}$$

Proof

It follows from the definition of ${\overline{u}}_{(x_i,\varepsilon _i)}$ that ${\overline{u}}_{(x_i,\varepsilon _i)}(y)=\varphi _{x_i}(y){\overline{U}}_{(x_i,\varepsilon _i)}(y)$ and

$$\begin{aligned} {\overline{U}}_{(x_i,\varepsilon _i)}(y)\le \left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} \varepsilon _i^n \left[ \frac{\chi _\delta (\rho _{x_i}(y))}{(\varepsilon _i^4+d(x_i,y)^4)^{\frac{n}{2}}}+ \Bigg (1-\chi _\delta (\rho _{x_i}(y))\Bigg )G_{x_i}(y)\right] . \end{aligned}$$

On the set $U=:\{2d(x_i,y)\le \varepsilon _j+d(x_i,x_j)\}$, we have

$$\begin{aligned} \varepsilon _j+d(y,x_j)\ge \varepsilon _j+d(x_i,x_j)-d(x_i,y) \ge \frac{1}{2}(\varepsilon _j+d(x_i,x_j)). \end{aligned}$$

(A.23)

Therefore, we have

$$\begin{aligned}&\int _M\left( \frac{\varepsilon _i^2}{\varepsilon _i^4+d(x_i,y)^4}\right) ^{\frac{n}{2}} \left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(y,x_j)^4}\right) ^{1+\frac{n}{2}}dV_{\theta _0}\\&\quad \le \left( \int _{U}+\int _{M{\setminus } U}\right) \left( \frac{\varepsilon _i^2}{\varepsilon _i^4+d(x_i,y)^4}\right) ^{\frac{n}{2}} \left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(y,x_j)^4}\right) ^{1+\frac{n}{2}}dV_{\theta _0}\\&\quad \le C\frac{\varepsilon _i^n\varepsilon _j^n}{(\varepsilon _j^4+d(x_i,x_j)^4)^{\frac{n+1}{2}}} \int _0^{\frac{\varepsilon _j+d(x_i,x_j)}{2}}\frac{r^{2n+1}dr}{(\varepsilon _i^4+r^4)^{\frac{n}{2}}}\\&\qquad +\,C\frac{\varepsilon _i^n\varepsilon _j^{n+2}}{(\varepsilon _j^4+d(x_i,x_j)^4)^{\frac{n}{2}}} \int _0^{\infty }\frac{r^{2n+1}dr}{(\varepsilon _j^4+r^4)^{1+\frac{n}{2}}}\\&\quad \le C\left( \frac{\varepsilon _i^2\varepsilon _j^2}{\varepsilon _j^4+d(x_i,x_j)^4}\right) ^{\frac{n}{2}} \end{aligned}$$

because

$$\begin{aligned} \int _0^{\frac{\varepsilon _j+d(x_i,x_j)}{2}}\frac{r^{2n+1}dr}{(\varepsilon _i^4+r^4)^{\frac{n}{2}}}&\le \int _0^{\frac{\varepsilon _j+d(x_i,x_j)}{2}}rdr =\frac{(\varepsilon _j+d(x_i,x_j))^2}{8} \\&\le (\varepsilon _j^4+d(x_i,x_j)^4)^\frac{1}{2} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \int _0^{\infty }\frac{r^{2n+1}dr}{(\varepsilon _j^4+r^4)^{1+\frac{n}{2}}}&\le \int _0^{\varepsilon _j}\frac{r^{2n+1}dr}{\varepsilon _j^4r^{4\cdot \frac{n}{2}}} +\int _{\varepsilon _j}^{\infty }\frac{r^{2n+1}dr}{r^{4(1+\frac{n}{2}})}=\frac{C}{\varepsilon _j^2}. \end{aligned} \end{aligned}$$

This proves the assertion. $\square$

Lemma A.7

We have the estimate

$$\begin{aligned}&\int _M{\overline{u}}_{(x_i,\varepsilon _i)}\left| \left( 2+\frac{2}{n}\right) \Delta _{\theta _0}{\overline{u}}_{(x_j,\varepsilon _j)} -R_{\theta _0}{\overline{u}}_{(x_j,\varepsilon _j)}+r_\infty {\overline{u}}_{(x_j,\varepsilon _j)}^{1+\frac{2}{n}}\right| dV_{\theta _0}\\&\quad \le C\left( \delta ^4+\delta ^{2n}+\frac{\varepsilon _j^2}{\delta ^2}\right) \left( \frac{\varepsilon _i^2\varepsilon _j^2}{\varepsilon _j^4+d(x_i,x_j)^4}\right) ^{\frac{n}{2}}. \end{aligned}$$

Proof

Recall ${\widehat{\theta }}_x=\varphi _x^{\frac{2}{n}}\theta _0$ for some smooth function $\varphi _x$ on M. Therefore, we can find a positive constant $c_0$ such that $c_0^{-n}\le \varphi _x\le c_0^{n},$ which implies $\frac{1}{c_0}d(x,y)\le \rho _x(y)\le c_0d(x,y)$ for all $x, y\in M$. Hence, it follows from Corollary A.3 that

$$\begin{aligned}&\left| \left( 2+\frac{2}{n}\right) \Delta _{\theta _0}{\overline{u}}_{(x_j,\varepsilon _j)} -R_{\theta _0}{\overline{u}}_{(x_j,\varepsilon _j)}+r_\infty {\overline{u}}_{(x_j,\varepsilon _j)}^{1+\frac{2}{n}}\right| \\&\quad \le C(\delta ^2+\delta ^{2n-2})\left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(x_j,y)^4}\right) ^{\frac{n}{2}}1_{\{d(x_j,y)\le 2c_0\delta \}}\\&\qquad +\,C\left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(x_j,y)^4}\right) ^{\frac{n+2}{2}}1_{\left\{ d(x_j,y)\ge \frac{\delta }{c_0}\right\} }. \end{aligned}$$

On the set $U:=\{2d(x_i,y)\le \varepsilon _j+d(x_i,x_j)\}\cap \{d(y,x_j)\le 2c_0\delta \}$, we have

$$\begin{aligned} d(x_i,y)\le \frac{1}{2}(\varepsilon _j+d(x_i,x_j))\le \varepsilon _j+d(y,x_j)\le 4c_0\delta . \end{aligned}$$

From this, it follows that

$$\begin{aligned}&\int _{\{d(y,x_j)\le 2c_0\delta \}} \left( \frac{\varepsilon _i^2}{\varepsilon _i^4+d(x_i,y)^4}\right) ^{\frac{n}{2}} \left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(y,x_j)^4}\right) ^{\frac{n}{2}}dV_{\theta _0}\\&\quad \le \left( \int _{U} + \int _{\{d(y,x_j)\le 2c_0\delta \}{\setminus } U}\right) \left( \frac{\varepsilon _i^2}{\varepsilon _i^4+d(x_i,y)^4}\right) ^{\frac{n}{2}} \left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(y,x_j)^4}\right) ^{\frac{n}{2}}dV_{\theta _0}\\&\quad \le C\int _{\{d(x_i,y)\le 4c_0\delta \}} \left( \frac{\varepsilon _i^2}{\varepsilon _i^4+d(x_i,y)^4}\right) ^{\frac{n}{2}} \left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(x_i,x_j)^4}\right) ^{\frac{n}{2}}dV_{\theta _0}\\&\qquad +\,C\int _{\{d(y,x_j)\le 2c_0\delta \}} \left( \frac{\varepsilon _i^2}{\varepsilon _j^4+d(x_i,x_j)^4}\right) ^{\frac{n}{2}} \left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(y,x_j)^4}\right) ^{\frac{n}{2}}dV_{\theta _0}\\&\quad \le C\delta ^2\left( \frac{\varepsilon _i^2\varepsilon _j^2}{\varepsilon _j^4+d(x_i,x_j)^4}\right) ^{\frac{n}{2}}. \end{aligned}$$

Similarly, on the set $V:=\{2d(x_i,y)\le \varepsilon _j+d(x_i,x_j)\}\cap \{d(y,x_j)\ge \frac{\delta }{c_0}\}$, we have

$$\begin{aligned} \varepsilon _j+d(y,x_j)\ge \varepsilon _j+d(x_i,x_j)-d(x_i,y)\ge \frac{1}{2}(\varepsilon _j+d(x_i,x_j)). \end{aligned}$$

This implies

$$\begin{aligned}&\int _{\left\{ d(y,x_j)\ge \frac{\delta }{c_0}\right\} } \left( \frac{\varepsilon _i^2}{\varepsilon _i^4+d(x_i,y)^4}\right) ^{\frac{n}{2}} \left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(y,x_j)^4}\right) ^{\frac{n+2}{2}}dV_{\theta _0}\\&\quad \le \left( \int _{V} + \int _{\left\{ d(y,x_j)\ge \frac{\delta }{c_0}\right\} {\setminus } V}\right) \left( \frac{\varepsilon _i^2}{\varepsilon _i^4+d(x_i,y)^4}\right) ^{\frac{n}{2}} \left( \frac{\varepsilon _j^2}{\varepsilon _j^4+d(y,x_j)^4}\right) ^{\frac{n+2}{2}}dV_{\theta _0}\\&\quad \le C\int _{\left\{ 2d(x_i,y)\le \varepsilon _j+d(x_i,x_j)\right\} } \left( \frac{\varepsilon _i^2}{\varepsilon _i^4+d(x_i,y)^4}\right) ^{\frac{n}{2}} \frac{\varepsilon _j^{n+2}}{\delta ^2(\varepsilon _i^4+d(x_i,x_j)^4)^{\frac{n+1}{2}}}dV_{\theta _0}\\&\qquad +\,C\int _{\left\{ d(y,x_j)\ge \frac{\delta }{c_0}\right\} } \left( \frac{\varepsilon _i^2}{\varepsilon _j^4+d(x_i,x_j)^4}\right) ^{\frac{n}{2}} \frac{\varepsilon _j^{n+2}}{(\varepsilon _j^4+d(y,x_j)^4)^{1+\frac{n}{2}}}dV_{\theta _0}\\&\quad \le C\frac{\varepsilon _j^2}{\delta ^2}\left( \frac{\varepsilon _i^2\varepsilon _j^2}{\varepsilon _j^4+d(x_i,x_j)^4}\right) ^{\frac{n}{2}} \end{aligned}$$

because

$$\begin{aligned} \begin{aligned} \int _0^{\frac{\varepsilon _j+d(x_i,x_j)}{2}}\frac{r^{2n+1}dr}{(\varepsilon _j^4+r^4)^{\frac{n}{2}}} \le \int _0^{\frac{\varepsilon _j+d(x_i,x_j)}{2}}\frac{r^{2n+1}dr}{r^{4\cdot \frac{n}{2}}} \le (\varepsilon _j^4+d(x_i,x_j)^4)^{\frac{1}{2}} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \int _{\frac{\delta }{c_0}}^{\infty }\frac{r^{2n+1}dr}{(\varepsilon _j^4+r^4)^{1+\frac{n}{2}}} \le \int _{\frac{\delta }{c_0}}^{\infty }\frac{r^{2n+1}dr}{r^{4(1+\frac{n}{2})}}=\frac{C}{\delta ^2}. \end{aligned} \end{aligned}$$

This proves the assertion. $\square$

Appendix B

Here we prove Theorem 5.1. We discuss both of the cases when $n=1$ or M is spherical. For convenience, we denote the CR conformal sub-Laplacian of a contact form $\theta$ by $L_\theta$, i.e.

$$\begin{aligned} L_\theta =-\left( 2+\frac{2}{n}\right) \Delta _{\theta }+R_{\theta }. \end{aligned}$$

Let $\{u_\nu \}$ be a sequence of positive functions satisfying (5.1) and (5.2). Note that $u_\nu$ is uniformly bounded in $S_1^2(M)$. To see this, it follows from integration by parts and Hölder’s inequality that

$$\begin{aligned}&\int _M\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}u_\nu |^2_{\theta _0}+R_{\theta _0}u_\nu ^2\right) dV_{\theta _0}\\&\quad \le \left( \int _M|L_{\theta _0}u_\nu -r_\infty u_\nu ^{1+\frac{2}{n}}|^{\frac{2n+2}{n+2}} dV_{\theta _0} \right) ^{\frac{n+2}{2n+2}}\left( \int _Mu_\nu ^{2+\frac{2}{n}}\right) ^{\frac{n}{2n+2}} +r_\infty \int _Mu_\nu ^{2+\frac{2}{n}} dV_{\theta _0}, \end{aligned}$$

which is uniformly bounded by (5.1) and (5.2). Thus, by passing to a subsequence if necessary, we assume that $u_\nu$ converges to $u_\infty$ weakly in $S_1^2(M)$. Since the Folland–Stein embedding $S_1^2(M)\hookrightarrow L^s(M)$ is compact for $1<s<2+\displaystyle \frac{2}{n}$ (see Proposition 5.6 in [28]), $u_\nu$ converges to $u_\infty$ in $L^s(M)$ for $1<s<2+\displaystyle \frac{2}{n}$.

Proposition B.1

The function $u_\infty$ is a smooth nonnegative function satisfying (5.3).

Proof

Note that for any $\varphi \in C^\infty (M)$

$$\begin{aligned} \int _M \varphi u_\nu ^{1+\frac{2}{n}}dV_{\theta _0}\rightarrow \int _M \varphi u_\infty ^{1+\frac{2}{n}}dV_{\theta _0} \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

(B.1)

Indeed, there exists a constant c such that

$$\begin{aligned} \left| u_\nu ^{1+\frac{2}{n}}-u_\infty ^{1+\frac{2}{n}}\right| \le c|u_\nu -u_\infty |\left( |u_\nu |^{\frac{2}{n}}+|u_\infty |^{\frac{2}{n}}\right) \end{aligned}$$

Hence, it follows from Hölder’s inequality that

$$\begin{aligned}&\left| \int _M \varphi u_\nu ^{1+\frac{2}{n}}dV_{\theta _0}-\int _M \varphi u_\infty ^{1+\frac{2}{n}}dV_{\theta _0}\right| \nonumber \\&\quad \le \int _M \left| u_\nu ^{1+\frac{2}{n}}-u_\infty ^{1+\frac{2}{n}}\right| |\varphi |dV_{\theta _0}\nonumber \\&\quad \le c\Vert \varphi \Vert _{L^\infty (M)}\Vert u_\nu -u_\infty \Vert _{L^\beta (M)} \left\| |u_\nu |^{\frac{2}{n}}+|u_\infty |^{\frac{2}{n}}\right\| _{L^{\beta '}(M)} \end{aligned}$$

(B.2)

where $\beta$ are chosen such that $1+\displaystyle \frac{2}{n}<\beta <2+\frac{2}{n}$ and $\displaystyle \frac{1}{\beta }+\frac{1}{\beta '}=1$. Therefore, $\displaystyle \frac{2}{n}\beta '<2+\frac{2}{n}$. Combining all these, we can conclude that the right hand side of (B.2) tends to 0 as $\nu \rightarrow \infty$. This proves (B.1). On the other hand, for any $\varphi \in C^\infty (M)$, we have

$$\begin{aligned} \begin{aligned} \int _M\varphi L_{\theta _0}u_\nu dV_{\theta _0} =\int _M u_\nu L_{\theta _0}\varphi dV_{\theta _0}\rightarrow \int _M u_\infty L_{\theta _0}\varphi dV_{\theta _0}=\int _M\varphi L_{\theta _0}u_\infty dV_{\theta _0} \end{aligned}\nonumber \\ \end{aligned}$$

(B.3)

as $\nu \rightarrow \infty$. Therefore, it follows from (B.1) and (B.3) that for any $\varphi \in C^\infty (M)$

$$\begin{aligned} \int _M\left( L_{\theta _0}u_\nu -r_\infty u_\nu ^{1+\frac{2}{n}}\right) \varphi dV_{\theta _0} \rightarrow \int _M\left( L_{\theta _0}u_\infty -r_\infty u_\infty ^{1+\frac{2}{n}}\right) \varphi dV_{\theta _0} \end{aligned}$$

as $\nu \rightarrow \infty$. Combining this with (5.2), we can conclude that $u_\infty$ satisfies (5.3). Since $u_\nu$ is nonnegative, $u_\infty$ is also nonnegative. It follows from Theorem 3.22 in [15] that $u_\infty$ is smooth. This proves the assertion. $\square$

Proposition B.2

There holds

$$\begin{aligned} u_\nu ^{1+\frac{2}{n}}-u_\infty ^{1+\frac{2}{n}}-|u_\nu -u_\infty |^{\frac{2}{n}}(u_\nu -u_\infty ) \rightarrow 0 \quad \text{ in } L^{\frac{2n+2}{n+2}}(M) \text{ as } \nu \rightarrow \infty . \end{aligned}$$

(B.4)

Proof

We use the following inequality:

$$\begin{aligned} \left| |a+b|^{\frac{2}{n}}-|a|^{\frac{2}{n}}\right| \le c\left( |b|^{\frac{2}{n}}+\max \{|a|,|b|\}^{\frac{2}{n}-1}|b|\right) \quad \text{ for } \text{ all } a,b. \end{aligned}$$

(B.5)

Applying (B.5), we get

$$\begin{aligned} u_\nu ^{1+\frac{2}{n}}= & {} u_\nu ^{\frac{2}{n}}u_\infty +(u_\nu -u_\infty +u_\infty )^{\frac{2}{n}}(u_\nu -u_\infty ) \nonumber \\= & {} u_\nu ^{\frac{2}{n}}u_\infty +|u_\nu -u_\infty |^{\frac{2}{n}}(u_\nu -u_\infty ) \nonumber \\&+\,O\left( u_\infty ^{\frac{2}{n}}|u_\nu -u_\infty |+\max \{|u_\nu -u_\infty |,|u_\infty |\}^{\frac{2}{n}-1}|u_\nu -u_\infty |u_\infty \right) \nonumber \\= & {} u_\nu ^{\frac{2}{n}}u_\infty +|u_\nu -u_\infty |^{\frac{2}{n}}(u_\nu -u_\infty ) +O\left( u_\infty ^{\frac{2}{n}}|u_\nu -u_\infty |+|u_\nu -u_\infty |^{\frac{2}{n}}u_\infty \right) .\nonumber \\ \end{aligned}$$

(B.6)

Applying (B.5) again, we obtain

$$\begin{aligned} u_\nu ^{\frac{2}{n}}u_\infty= & {} (u_\nu -u_\infty +u_\infty )^{\frac{2}{n}}u_\infty \nonumber \\= & {} u_\infty ^{1+\frac{2}{n}}+O\left( u_\infty |u_\nu -u_\infty |^{\frac{2}{n}}+\max \{|u_\nu -u_\infty |,|u_\infty |\}^{\frac{2}{n}-1}|u_\nu -u_\infty |u_\infty \right) \nonumber \\= & {} u_\infty ^{1+\frac{2}{n}}+O\left( u_\infty ^{\frac{2}{n}}|u_\nu -u_\infty |+|u_\nu -u_\infty |^{\frac{2}{n}}u_\infty \right) . \end{aligned}$$

(B.7)

Combining (B.6) and (B.7), we get

$$\begin{aligned} u_\nu ^{1+\frac{2}{n}}-u_\infty ^{1+\frac{2}{n}}-|u_\nu -u_\infty |^{\frac{2}{n}}(u_\nu -u_\infty ) =O\left( u_\infty ^{\frac{2}{n}}|u_\nu -u_\infty |+|u_\nu -u_\infty |^{\frac{2}{n}}u_\infty \right) .\nonumber \\ \end{aligned}$$

(B.8)

Since $u_\infty$ is smooth and $u_\nu$ converges to $u_\infty$ in $L^s(M)$ for $1<s<2+\displaystyle \frac{2}{n}$, (B.4) follows from (B.8). $\square$

If $u_\nu$ converges to $u_\infty$ strongly in $S_1^2(M)$, then Theorem 5.1 follows. Therefore, we assume that $u_\nu$ converges to $u_\infty$ weakly in $S_1^2(M)$, but not strongly in $S_1^2(M)$. On the other hand, it follows from (5.2), (5.3) and (B.4) that

$$\begin{aligned}&\int _M\Bigg |L_{\theta _0}(u_\nu -u_\infty )-r_\infty |u_\nu -u_\infty |^{\frac{2}{n}}(u_\nu -u_\infty )\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0}\\&\quad \le C\int _M\Bigg |L_{\theta _0}u_\nu -r_\infty u_\nu ^{1+\frac{2}{n}}\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0}\\&\qquad +\,C\, r_\infty \int _M\Bigg |u_\nu ^{1+\frac{2}{n}}-u_\infty ^{1+\frac{2}{n}}-|u_\nu -u_\infty |^{\frac{2}{n}}(u_\nu -u_\infty )\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0} \rightarrow 0 \end{aligned}$$

as $\nu \rightarrow \infty$. That is to say, if we let $v_\nu :=u_\nu -u_\infty$, then $\{v_\nu \}$ is a sequence of functions such that $v_\nu$ converges to 0 weakly in $S_1^2(M)$, but not strongly in $S_1^2(M)$, and satisfies

$$\begin{aligned} \int _M\Bigg |L_{\theta _0}v_\nu -r_\infty |v_\nu |^{\frac{2}{n}}v_\nu \Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

(B.9)

We are going to extract bubbles from $v_\nu$. To do this, we mainly follow the proof of Proposition 8 in [19]. The argument is almost the same except with some small modifications. However, there are some parts in the argument of [19] which are not very precise. So we are going to provide all the details of the proof.

As before, we denote

$$\begin{aligned} B_r(x)=\{y\in M:d(x,y)<r\}, \end{aligned}$$

where d is the Carnot–Carathéodory distance on M with respect to the contact form $\theta _0$.

Lemma B.3

There exists $x\in M$ such that for every $\rho >0$, there exists $\delta (\rho )>0$ such that

$$\begin{aligned} \liminf _{\nu \rightarrow \infty }\int _{B_\rho (x)}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0} \ge \delta (\rho ). \end{aligned}$$

(B.10)

Proof

Suppose that for all $x\in M$, there is $\rho (x)>0$ such that

$$\begin{aligned} \liminf _{\nu \rightarrow \infty }\int _{B_{\rho (x)}(x)}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0} =0. \end{aligned}$$

Since M is compact, we can cover M with finite number of $B_{\rho (x_i)}(x_i)$ with $i=1, \ldots , L$. As a result, there exists a subsequence of $\{v_\nu \}$, which is still denoted by $\{v_\nu \}$, such that

$$\begin{aligned} \int _{M}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0} \rightarrow 0 \end{aligned}$$

as $\nu \rightarrow \infty$. This contradicts to the assumption that $v_\nu$ does not converge to 0 strongly in $S_1^2(M)$. This proves the assertion. $\square$

Lemma B.4

The constant $\delta (\rho )$ in Lemma B.3 can be taken to be $a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1}$, where $a_0$ is any positive real number strictly less than 1.

Before we give the proof of Lemma B.4, we remark that our constant in Lemma B.4 is different from that of [19]. But one can see from the following arguments that a uniform constant will be sufficient for our purpose.

Proof of Lemma B.4

The proof is similar to the proof of Lemma 10 in [19]. Let $\psi _\nu :{\mathbb {R}}\rightarrow [0,1]$ be a $C^1$ cut-off function such that

$$\begin{aligned} \psi _\nu (s)=\left\{ \begin{array}{ll} 1, &{} \hbox { for }s\le \rho _\nu ; \\ 0, &{} \hbox { for }s\ge \rho _\nu +\delta _\nu , \end{array} \right. \end{aligned}$$

where $\rho<\rho _\nu <2\rho$ and $0<\delta _\nu <\rho$. Fix a point $x\in M$ where (B.10) holds for x. Define the function $\phi _\nu$ by $\phi _\nu (y)=\psi _\nu (d(x,y))$ where d is the Carnot–Carathéodory distance on M with respect to the contact form $\theta _0$. Then we have

$$\begin{aligned} |\nabla _{\theta _0}\phi _\nu |_{\theta _0}\le \frac{C}{\delta _\nu }. \end{aligned}$$

(B.11)

By Hölder’s inequality, we have

$$\begin{aligned} \int _M\Bigg (L_{\theta _0}v_\nu -r_\infty |v_\nu |^{\frac{2}{n}}v_\nu \Bigg )v_\nu \phi _\nu dV_{\theta _0} \le \Vert \alpha _\nu \Vert _{L^{\frac{2n+2}{n+2}}(M)}\left( \int _M |v_\nu \phi _\nu |^{2+\frac{2}{n}}dV_{\theta _0}\right) ^{\frac{n}{2n+2}}\nonumber \\ \end{aligned}$$

(B.12)

where $\alpha _\nu =L_{\theta _0}v_\nu -r_\infty |v_\nu |^{\frac{2}{n}}v_\nu .$ By integration by parts, the left hand side of (B.12) can be written as

$$\begin{aligned}&\int _M\Bigg (L_{\theta _0}v_\nu -r_\infty |v_\nu |^{\frac{2}{n}}v_\nu \Bigg )v_\nu \phi _\nu dV_{\theta _0}\nonumber \\&\quad =\left( 2+\frac{2}{n}\right) \int _M|\nabla _{\theta _0} v_\nu |_{\theta _0}^2\phi _\nu dV_{\theta _0} +\left( 2+\frac{2}{n}\right) \int _Mv_\nu \langle \nabla _{\theta _0} v_\nu ,\nabla _{\theta _0} \phi _\nu \rangle _{\theta _0} dV_{\theta _0}\nonumber \\&\qquad +\int _MR_{\theta _0}v_\nu ^2\phi _\nu dV_{\theta _0}-r_\infty \int _M|v_\nu |^{2+\frac{2}{n}}\phi _\nu dV_{\theta _0}. \end{aligned}$$

(B.13)

It follows from (B.12) and (B.13) that

$$\begin{aligned}&\left( 2+\frac{2}{n}\right) \int _M|\nabla _{\theta _0} v_\nu |_{\theta _0}^2\phi _\nu dV_{\theta _0} +\left( 2+\frac{2}{n}\right) \int _Mv_\nu \langle \nabla _{\theta _0} v_\nu ,\nabla _{\theta _0} \phi _\nu \rangle _{\theta _0} dV_{\theta _0}\nonumber \\&\qquad +\int _MR_{\theta _0}v_\nu ^2\phi _\nu dV_{\theta _0}\nonumber \\&\quad \le r_\infty \int _M|v_\nu |^{2+\frac{2}{n}}\phi _\nu dV_{\theta _0} +\Vert \alpha _\nu \Vert _{L^{\frac{2n+2}{n+2}}(M)}\left( \int _M |v_\nu \phi _\nu |^{2+\frac{2}{n}}dV_{\theta _0}\right) ^{\frac{n}{2n+2}}.\qquad \qquad \end{aligned}$$

(B.14)

Also, by Folland–Stein embedding theorem in Proposition 2.1, we have

$$\begin{aligned} \left( \int _M |v_\nu \phi _\nu |^{2+\frac{2}{n}}dV_{\theta _0}\right) ^{\frac{n}{2n+2}} \le C\left( \int _M\left( |\nabla _{\theta _0}(v_\nu \phi _\nu )|^2_{\theta _0}+R_{\theta _0}(v_\nu \phi _\nu )^2\right) dV_{\theta _0} \right) ^{\frac{1}{2}}.\nonumber \\ \end{aligned}$$

(B.15)

By Cauchy–Schwarz inequality, we have

$$\begin{aligned} \left( \int _M|\nabla _{\theta _0}(v_\nu \phi _\nu )|^2_{\theta _0}dV_{\theta _0}\right) ^{\frac{1}{2}}\le & {} \left( \int _M|\nabla _{\theta _0}v_\nu |^2_{\theta _0}\phi _\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}}\nonumber \\&+\,\left( \int _M|\nabla _{\theta _0}\phi _\nu |^2_{\theta _0}v_\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}}. \end{aligned}$$

(B.16)

Substituting (B.15) and (B.16) into (B.14), we get

$$\begin{aligned}&\left( 2+\frac{2}{n}\right) \int _M|\nabla _{\theta _0} v_\nu |_{\theta _0}^2\phi _\nu dV_{\theta _0} +\left( 2+\frac{2}{n}\right) \int _Mv_\nu \langle \nabla _{\theta _0} v_\nu ,\nabla _{\theta _0} \phi _\nu \rangle _{\theta _0} dV_{\theta _0}\nonumber \\&\qquad +\int _MR_{\theta _0}v_\nu ^2\phi _\nu dV_{\theta _0}\nonumber \\&\quad \le r_\infty \int _M|v_\nu |^{2+\frac{2}{n}}\phi _\nu dV_{\theta _0} +C\,\Vert \alpha _\nu \Vert _{L^{\frac{2n+2}{n+2}}(M)}\left[ \left( \int _M|\nabla _{\theta _0}v_\nu |^2_{\theta _0}\phi _\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}}\right. \nonumber \\&\qquad \left. +\left( \int _M|\nabla _{\theta _0}\phi _\nu |^2_{\theta _0}v_\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}} +\left( \int _MR_{\theta _0}(v_\nu \phi _\nu )^2dV_{\theta _0}\right) ^{\frac{1}{2}}\right] . \end{aligned}$$

(B.17)

By passing to a subsequence, we obtain from (B.10) that

$$\begin{aligned} \int _{M}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2\phi _\nu +R_{\theta _0}v_\nu ^2\phi _\nu \right) dV_{\theta _0} \ge \frac{\delta (\rho )}{2}. \end{aligned}$$

(B.18)

Since $\{v_\nu \}$ is uniformly bounded in $S_1^2(M)$, it follows from (B.9) that

$$\begin{aligned} \Vert \alpha _\nu \Vert _{L^{\frac{2n+2}{n+2}}(M)}\left[ \left( \int _M|\nabla _{\theta _0}v_\nu |^2_{\theta _0}\phi _\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}} +\left( \int _MR_{\theta _0}(v_\nu \phi _\nu )^2dV_{\theta _0}\right) ^{\frac{1}{2}}\right] \rightarrow 0\nonumber \\ \end{aligned}$$

(B.19)

as $\nu \rightarrow \infty$. By (B.18) and (B.19), we can rewrite (B.17) as follows:

$$\begin{aligned}&(1+o(1))\int _{M}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2\phi _\nu +R_{\theta _0}v_\nu ^2\phi _\nu \right) dV_{\theta _0}\nonumber \\&\qquad +\left( 2+\frac{2}{n}\right) \int _Mv_\nu \langle \nabla _{\theta _0} v_\nu ,\nabla _{\theta _0} \phi _\nu \rangle _{\theta _0} dV_{\theta _0}\nonumber \\&\quad \le r_\infty \int _M|v_\nu |^{2+\frac{2}{n}}\phi _\nu dV_{\theta _0} +C\,\Vert \alpha _\nu \Vert _{L^{\frac{2n+2}{n+2}}(M)}\left( \int _M|\nabla _{\theta _0}\phi _\nu |^2_{\theta _0}v_\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}}.\qquad \qquad \end{aligned}$$

(B.20)

Note that

$$\begin{aligned} \int _M|\nabla _{\theta _0}\phi _\nu |^2_{\theta _0}v_\nu ^2dV_{\theta _0} \le \frac{C}{\delta _\nu ^2}\int _Mv_\nu ^2dV_{\theta _0} \end{aligned}$$

(B.21)

by (B.11). Hence,

$$\begin{aligned}&\left| \int _Mv_\nu \langle \nabla _{\theta _0} v_\nu ,\nabla _{\theta _0} \phi _\nu \rangle _{\theta _0} dV_{\theta _0}\right| \nonumber \\&\quad \le \left( \int _M|\nabla _{\theta _0}v_\nu |^2_{\theta _0}dV_{\theta _0}\right) ^{\frac{1}{2}} \left( \int _M|\nabla _{\theta _0}\phi _\nu |^2_{\theta _0}v_\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}} \le \frac{C}{\delta _\nu }\left( \int _Mv_\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}}\qquad \qquad \end{aligned}$$

(B.22)

by Hölder’s inequality and the fact that $\{v_\nu \}$ is uniformly bounded in $S_1^2(M)$.

If we choose $\delta _\nu =\displaystyle \left( \int _M v_\nu ^2 dV_{\theta _0}\right) ^{\frac{1}{4}}$, then

$$\begin{aligned} \frac{C}{\delta _\nu }\left( \int _Mv_\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty , \end{aligned}$$

(B.23)

since $v_\nu \rightarrow 0$ in $L^2(M)$ as $\nu \rightarrow \infty$. Using (B.21)–(B.23), we can rewrite (B.20) as

$$\begin{aligned}&(1+o(1))\int _{M}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2\phi _\nu +R_{\theta _0}v_\nu ^2\phi _\nu \right) dV_{\theta _0}\nonumber \\&\quad \le r_\infty \int _M|v_\nu |^{2+\frac{2}{n}}\phi _\nu dV_{\theta _0}+o(1). \end{aligned}$$

(B.24)

In view of (B.18), we deduce from (B.24) that

$$\begin{aligned} (1+o(1))\int _{M}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2\phi _\nu +R_{\theta _0}v_\nu ^2\phi _\nu \right) dV_{\theta _0} \le r_\infty \int _M|v_\nu |^{2+\frac{2}{n}}\phi _\nu dV_{\theta _0}.\nonumber \\ \end{aligned}$$

(B.25)

We introduce $\gamma _\nu >0$ which will be chosen latter. If $\gamma _\nu$ is sufficiently small, it follows from (B.25) that

$$\begin{aligned}&(1+o(1))\int _{M}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2\phi _\nu +R_{\theta _0}v_\nu ^2\phi _\nu \right) dV_{\theta _0}\nonumber \\&\quad \le r_\infty \int _{M}|v_\nu |^{2+\frac{2}{n}}(\phi _\nu +\gamma _\nu ) dV_{\theta _0}. \end{aligned}$$

(B.26)

By the definition of the CR Yamabe constant in (3.1), we have

$$\begin{aligned}\int _{M} \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}w|_{\theta _0}^2 +R_{\theta _0}w^2\right) dV_{\theta _0}\ge Y(M,\theta _0)\Vert w\Vert _{L^{2+\frac{2}{n}}(M)}^2. \end{aligned}$$

Applying this to the function $w=|v_\nu |(\phi _\nu +\gamma _\nu )^{\frac{n}{2n+2}}$, we obtain

$$\begin{aligned}&Y(M,\theta _0) \left( \int _{M}|v_\nu |^{2+\frac{2}{n}}(\phi _\nu +\gamma _\nu ) dV_{\theta _0}\right) ^{\frac{n}{n+1}}\nonumber \\&\quad \le \left( 2+\frac{2}{n}\right) \int _{M} \Bigg (|\nabla _{\theta _0}v_\nu |_{\theta _0}^2(\phi _\nu +\gamma _\nu )^{\frac{n}{n+1}} +v_\nu ^2\Bigg |\nabla _{\theta _0}\Bigg ((\phi _\nu +\gamma _\nu )^{\frac{n}{2n+2}}\Bigg )\Bigg |_{\theta _0}^2\nonumber \\&\qquad +\,2|v_\nu |(\phi _\nu +\gamma _\nu )^{\frac{n}{2n+2}}\langle \nabla _{\theta _0}v_\nu ,\nabla _{\theta _0}\Bigg ((\phi _\nu +\gamma _\nu )^{\frac{n}{2n+2}}\Bigg )\rangle _{\theta _0} \Bigg )dV_{\theta _0}\nonumber \\&\qquad +\int _{M}R_{\theta _0}v_\nu ^2(\phi _\nu +\gamma _\nu )^{\frac{n}{n+1}}dV_{\theta _0}. \end{aligned}$$

(B.27)

By (B.21), (B.23) and the fact that $\displaystyle \frac{n}{n+1}-2<0$, we can choose $\gamma _k$ sufficiently small such that

$$\begin{aligned} \int _{M}v_\nu ^2\gamma _\nu ^{\frac{n}{n+1}-2}|\nabla _{\theta _0}\phi _\nu |_{\theta _0}^2=o(1), \end{aligned}$$

which implies that

$$\begin{aligned} \int _{M} v_\nu ^2\Bigg |\nabla _{\theta _0}\Bigg ((\phi _\nu +\gamma _\nu )^{\frac{n}{2n+2}}\Bigg )\big |_{\theta _0}^2dV_{\theta _0}= & {} \int _{M}v_\nu ^2(\phi _\nu +\gamma _\nu )^{\frac{n}{n+1}-2}|\nabla _{\theta _0}\phi _\nu |_{\theta _0}^2dV_{\theta _0} \nonumber \\\le & {} \int _{M}v_\nu ^2\gamma _\nu ^{\frac{n}{n+1}-2}|\nabla _{\theta _0}\phi _\nu |_{\theta _0}^2dV_{\theta _0}=o(1).\nonumber \\ \end{aligned}$$

(B.28)

It follows from (B.28), Hölder’s inequality and the fact that $\{v_\nu \}$ is uniformly bounded in $S_1^2(M)$ that

$$\begin{aligned}&\left| \int _{M} |v_\nu |(\phi _\nu +\gamma _\nu )^{\frac{n}{2n+2}}\langle \nabla _{\theta _0}v_\nu ,\nabla _{\theta _0}\Bigg ((\phi _\nu +\gamma _\nu )^{\frac{n}{2n+2}}\Bigg )\rangle _{\theta _0}dV_{\theta _0}\right| \nonumber \\&\quad \le \left( \int _{M} |\nabla _{\theta _0}v_\nu |_{\theta _0}^2(\phi _\nu +\gamma _\nu )^{\frac{n}{n+1}}dV_{\theta _0}\right) ^{\frac{1}{2}}\left( \int _{M} v_\nu ^2\Bigg |\nabla _{\theta _0}\Bigg ((\phi _\nu +\gamma _\nu )^{\frac{n}{2n+2}}\Bigg )\big |_{\theta _0}^2dV_{\theta _0}\right) ^{\frac{1}{2}}\nonumber \\&\quad =o(1). \end{aligned}$$

(B.29)

Putting (B.28) and (B.29) into (B.27), we get

$$\begin{aligned}&Y(M,\theta _0) \left( \int _{M}|v_\nu |^{2+\frac{2}{n}}(\phi _\nu +\gamma _\nu ) dV_{\theta _0}\right) ^{\frac{n}{n+1}}\\&\quad \le \int _{M} \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2(\phi _\nu +\gamma _\nu )^{\frac{n}{n+1}}+ R_{\theta _0}v_\nu ^2(\phi _\nu +\gamma _\nu )^{\frac{n}{n+1}}\right) dV_{\theta _0}+o(1). \end{aligned}$$

Combining this with (B.26) and using (B.18), we obtain

$$\begin{aligned}&(1+o(1))\int _{M}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2\phi _\nu +R_{\theta _0}v_\nu ^2\phi _\nu \right) dV_{\theta _0}\nonumber \\&\quad \le r_\infty Y(M,\theta _0)^{-\frac{n+1}{n}} \left( \int _{M} \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2\phi _\nu ^{\frac{n}{n+1}}+ R_{\theta _0}v_\nu ^2\phi _\nu ^{\frac{n}{n+1}}\right) dV_{\theta _0}\right) ^{\frac{n+1}{n}}.\nonumber \\ \end{aligned}$$

(B.30)

It follows from (B.30) and the definition of $\phi _\nu$ that

$$\begin{aligned}&(1+o(1))\int _{B_{\rho _\nu }(x)}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0}\nonumber \\&\quad \le r_\infty Y(M,\theta _0)^{-\frac{n+1}{n}} \left( \int _{B_{\rho _\nu }(x)} \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+ R_{\theta _0}v_\nu ^2\right) dV_{\theta _0}\right) ^{\frac{n+1}{n}}\nonumber \\&\qquad +\,r_\infty Y(M,\theta _0)^{-\frac{n+1}{n}}\left( \int _{B_{\rho _\nu +\delta _\nu }(x)-B_{\rho _\nu }(x)} \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+ R_{\theta _0}v_\nu ^2\right) dV_{\theta _0}\right) ^{\frac{n+1}{n}}\nonumber \\&\quad :=r_\infty Y(M,\theta _0)^{-\frac{n+1}{n}}(I+II). \end{aligned}$$

(B.31)

It follows from (B.18) that $I\ge C\delta (\rho )^{\frac{n+1}{n}}$. We have the following two cases:

Case (i). If $II=o(I)$, then it follows from (B.31) that

$$\begin{aligned}&(1+o(1))\int _{B_{\rho _\nu }(x)}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0}\\&\quad \le r_\infty Y(M,\theta _0)^{-\frac{n+1}{n}} \left( \int _{B_{\rho _\nu }(x)} \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+ R_{\theta _0}v_\nu ^2\right) dV_{\theta _0}\right) ^{\frac{n+1}{n}}, \end{aligned}$$

which implies that

$$\begin{aligned} \int _{B_{\rho _\nu }(x)}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0}\\ \ge a_0r_\infty ^{-n} Y(M,\theta _0)^{n+1} \end{aligned}$$

where $a_0$ is any positive real number strictly less than 1.

Case (ii). Suppose that there exists a fixed constant C such that $II\ge C\delta (\rho )^{\frac{n+1}{n}}$ for all choices of $\rho _\nu \in [\rho , 2\rho ]$ with $\delta _\nu =\displaystyle \left( \int _M v_\nu ^2 dV_{\theta _0}\right) ^{\frac{1}{4}}$. In fact, it can not occur since $\delta _\nu \rightarrow 0$ as $\nu \rightarrow 0$, and $\int _M\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0}$ is uniformly bounded. This proves Lemma B.4. $\square$

It follows from Lemma B.4 that for any $x\in M$ satisfying (B.10) and for any given $a_0<1$, and $\nu$ sufficiently large, there exists a $\rho _\nu (x)$ such that

$$\begin{aligned} \int _{B_{\rho _\nu (x)}(x)}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0} = a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1}. \end{aligned}$$

Which means for and $\rho >\rho _\nu (x)$, we have

$$\begin{aligned} \int _{B_{\rho _\nu (x)}(x)}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0} >a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1}. \end{aligned}$$

Then for every $\nu$ sufficiently large, we define:

$$\begin{aligned} \rho _{1,\nu }=\inf \rho _\nu (x), \end{aligned}$$

(B.32)

where the infimum is taken among all $x\in M$ satisfying (B.10), which is a closed set. Thus the infimum is attained. That is, there exists $x^*_{1,\nu }\in M$ such that

$$\begin{aligned} \rho _{1,\nu }=\rho _\nu (x^*_{1,\nu }). \end{aligned}$$

(B.33)

For any $\rho _0>0$, by the proof of Lemma B.4, we have

$$\begin{aligned} \int _{B_{{\widetilde{\rho }}}(x)}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0} \ge a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1} \end{aligned}$$

for all $\nu$ sufficiently large. Since $\rho _0$ is arbitrary, and $\rho _\nu (x)\le \rho _0$, then we have the following lemma:

Lemma B.5

The sequence $\{\rho _{1,\nu }\}$ converges to zero as $\nu \rightarrow \infty$.

As explained in Appendix A, we can find $\rho >0$ which is independent of x (since M is compact) with ${\widehat{\theta }}_x=\varphi _x^{\frac{2}{n}}\theta _0$ in a neighborhood $B_{3\delta }(x)$ of x such that (A.1) and (A.4) are satisfied when $n=1$ and (A.9) is satisfied when M is spherical. We define a sequence of functions $\{ {\widetilde{v}}_\nu \}$ in $B_{2\delta }(x)$ as follows:

$$\begin{aligned} {\widetilde{v}}_\nu =\varphi _x^{-1}v_\nu . \end{aligned}$$

Since ${\widehat{\theta }}_x=\varphi _x^{\frac{2}{n}}\theta _0$, by the CR transformation law, we have

$$\begin{aligned} L_{\theta _0}v_\nu =\varphi _x^{1+\frac{2}{n}}L_{{\widehat{\theta }}_x}{\widetilde{v}}_\nu , \end{aligned}$$

(B.34)

which holds in $B_{2\delta }(x)\subset M$. It follows from (A.8) that (B.34) holds in the CR normal coordinates $\{(z,t):(t^2+|z|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}\subset {\mathbb {H}}^n$, where ${\widehat{\rho }}>0$ is independent of x. By (B.9) and (B.34), we have

$$\begin{aligned} \int _{\{(t^2+|z|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\Bigg |L_{{\widehat{\theta }}_x}{\widetilde{v}}_\nu -r_\infty |{\widetilde{v}}_\nu |^{\frac{2}{n}}{\widetilde{v}}_\nu \Bigg |^{\frac{2n+2}{n+2}}dV_{{\widehat{\theta }}_x} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

(B.35)

By the properties of $v_\nu$, we know that ${\widetilde{v}}_\nu$ is bounded in $L^{2+\frac{2}{n}}$ and ${\widetilde{v}}_\nu \rightarrow 0$ in $L^s$ for all $s<2+\displaystyle \frac{2}{n}$ as $\nu \rightarrow \infty$.

When $n=1$, it follows from (A.1), (A.3), (A.4) and (B.35) that

$$\begin{aligned}&\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\left| L_{\theta _{{\mathbb {H}}^n}}{\widetilde{v}}_\nu -r_\infty |{\widetilde{v}}_\nu |^{\frac{2}{n}}{\widetilde{v}}_\nu \right| ^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\quad \le C\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\left| L_{\theta _{{\mathbb {H}}^n}} {\widetilde{v}}_\nu -L_{{\widehat{\theta }}_x}{\widetilde{v}}_\nu \right| ^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\qquad +\,C\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\left| L_{{\widehat{\theta }}_x}{\widetilde{v}}_\nu -r_\infty {\widetilde{v}}_\nu ^{1+\frac{2}{n}}\right| ^{\frac{2n+2}{n+2}}\Bigg |dV_{\theta _{{\mathbb {H}}^n}}-dV_{{\widehat{\theta }}_x}\Bigg |\nonumber \\&\qquad +\,C\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\left| L_{{\widehat{\theta }}_x}{\widetilde{v}}_\nu -r_\infty |{\widetilde{v}}_\nu |^{\frac{2}{n}}{\widetilde{v}}_\nu \right| ^{\frac{2n+2}{n+2}}dV_{{\widehat{\theta }}_x} \rightarrow 0 \end{aligned}$$

(B.36)

as $\nu \rightarrow \infty$. When M is spherical, then it follows from (A.9) and (B.35) that

$$\begin{aligned} \int _{\{(t^2+|z|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\Bigg |L_{\theta _{{\mathbb {H}}^n}}{\widetilde{v}}_\nu -r_\infty |{\widetilde{v}}_\nu |^{\frac{2}{n}}{\widetilde{v}}_\nu \Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty , \end{aligned}$$

(B.37)

since $R_{\theta _{{\mathbb {H}}^n}}=0$.

Let ${\widehat{\chi }}$ be a cut off function such that ${\widehat{\chi }}(s)=1$ if $0\le s\le \displaystyle \frac{{\widehat{\rho }}}{2}$ and 0 if $s\ge {\widehat{\rho }}$. Let $\{{\widetilde{V}}_\nu \}$ be a sequence of functions in ${\mathbb {H}}^n$ defined by

$$\begin{aligned} {\widetilde{V}}_\nu (z,t)= & {} \left\{ \begin{array}{ll} (\rho _{1,\nu })^n\,{\widehat{\chi }}\Bigg (\rho _{1,\nu }(t^2+|z|^4)^{\frac{1}{4}}\Bigg )\,{\widetilde{v}}_\nu \Bigg (\rho _{1,\nu }z,(\rho _{1,\nu })^2t\Bigg ), &{} \hbox { for } (t^2+|z|^4)^{\frac{1}{4}}<\displaystyle \frac{{\widehat{\rho }}}{\rho _{1,\nu }}; \\ 0, &{} \hbox { otherwise,} \end{array} \right. \nonumber \\ \end{aligned}$$

(B.38)

where $\rho _{1,\nu }$ is defined as in (B.32). Then we have the following:

Proposition B.6

• (i) For any fixed ball B of ${\mathbb {H}}^n$, we have

  $$\begin{aligned} \int _{B}\left| L_{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}_\nu -r_\infty |{\widetilde{V}}_\nu |^{\frac{2}{n}}{\widetilde{V}}_\nu \right| ^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}} \rightarrow 0\quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

• (ii) There exists a constant C such that

  $$\begin{aligned} \int _{{\mathbb {H}}^n}\left( |\nabla _{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}_\nu |^2_{\theta _{{\mathbb {H}}^n}}+ |{\widetilde{V}}_\nu |^{2+\frac{2}{n}}\right) dV_{\theta _{{\mathbb {H}}^n}}\le C \quad \text{ for } \text{ all } \nu . \end{aligned}$$

Proof

To prove (i), we fix $\rho >0$. Since $\rho _{1,\nu }\rightarrow 0$ as $\nu \rightarrow \infty$ by Lemma B.5, there exists N such that $\displaystyle \rho \le \frac{{\widehat{\rho }}}{2\rho _{1,\nu }}$ when $\nu \ge N$. Hence, if $\nu \ge N$, we have

$$\begin{aligned}&\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<\rho \}}\left| L_{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}_\nu (z,t)-r_\infty |{\widetilde{V}}_\nu (z,t)|^{\frac{2}{n}}{\widetilde{V}}_\nu (z,t)\right| ^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}}\\&\quad =\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<\rho \}}\left| L_{\theta _{{\mathbb {H}}^n}} \Bigg ((\rho _{1,\nu })^n\,{\widetilde{v}}_\nu \Bigg (\rho _{1,\nu }z,(\rho _{1,\nu })^2t\Bigg )\Bigg )\right. \\&\qquad \left. -\,r_\infty \Bigg |(\rho _{1,\nu })^n\,{\widetilde{v}}_\nu \Bigg (\rho _{1,\nu }z,(\rho _{1,\nu })^2t\Bigg )\Bigg |^{\frac{2}{n}} (\rho _{1,\nu })^n\,{\widetilde{v}}_\nu \Bigg (\rho _{1,\nu }z,(\rho _{1,\nu })^2t\Bigg )\right| ^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}}\\&\quad =\int _{\{({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}<\rho \rho _{1,\nu }\}} \left| L_{\theta _{{\mathbb {H}}^n}} {\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}}) -r_\infty |{\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})|^{\frac{2}{n}}{\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})\right| ^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}} =o(1), \end{aligned}$$

where the first equality follows from (B.38) and the fact that $\displaystyle \rho \le \frac{{\widehat{\rho }}}{2\rho _{1,\nu }}$, the second equality follows from the change of variables $({\tilde{z}},{\tilde{t}})=(\rho _{1,\nu }z,(\rho _{1,\nu })^2t)$, and the last equality follows from (B.36), (B.37) and the fact that $\displaystyle \rho \,\rho _{1,\nu }\le \frac{{\widehat{\rho }}}{2}$. This proves (i). For (ii), we have

$$\begin{aligned}&\int _{{\mathbb {H}}^n}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}_\nu |^2_{\theta _{{\mathbb {H}}^n}}+ r_\infty |{\widetilde{V}}_\nu |^{2+\frac{2}{n}}\right) dV_{\theta _{{\mathbb {H}}^n}}\\&\quad =\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<\frac{{\widehat{\rho }}}{\rho _{1,\nu }}\}}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}_\nu |^2_{\theta _{{\mathbb {H}}^n}}+ r_\infty |{\widetilde{V}}_\nu |^{2+\frac{2}{n}}\right) dV_{\theta _{{\mathbb {H}}^n}}\\&\quad =\int _{\{({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\left( \left( 2+\frac{2}{n}\right) \left| \nabla _{\theta _{{\mathbb {H}}^n}}\Bigg ({\widehat{\chi }}\Bigg (({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}\Bigg ) {\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})\Bigg )\right| ^2_{\theta _{{\mathbb {H}}^n}}\right. \\&\qquad \left. +\, r_\infty \Bigg |{\widehat{\chi }}\Bigg (({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}\Bigg ) {\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})\Bigg |^{2+\frac{2}{n}}\right) dV_{\theta _{{\mathbb {H}}^n}}\\&\quad \le \int _{\{({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}} \left( 2+\frac{2}{n}\right) {\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})^2 \left| \nabla _{\theta _{{\mathbb {H}}^n}}{\widehat{\chi }}\Bigg (({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}\Bigg ) \right| ^2_{\theta _{{\mathbb {H}}^n}}dV_{\theta _{{\mathbb {H}}^n}}\\&\qquad + \int _{\{({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\left( \left( 2+\frac{2}{n}\right) \Bigg ({\widehat{\chi }}\Bigg (({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}\Bigg )\Bigg )^2 |\nabla _{\theta _{{\mathbb {H}}^n}} {\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})|^2_{\theta _{{\mathbb {H}}^n}}\right. \\&\qquad \left. +\, r_\infty \Bigg |{\widehat{\chi }}\Bigg (({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}\Bigg ) {\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})\Bigg |^{2+\frac{2}{n}}\right) dV_{\theta _{{\mathbb {H}}^n}}\\&\quad \le \int _{\{({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}} \left( 2+\frac{2}{n}\right) {\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})^2dV_{\theta _{{\mathbb {H}}^n}}\\&\qquad +\, \int _{\{({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}<{\widehat{\rho }}\}}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _{{\mathbb {H}}^n}} {\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})|^2_{\theta _{{\mathbb {H}}^n}}+ r_\infty |{\widetilde{v}}_\nu ({\tilde{z}},{\tilde{t}})|^{2+\frac{2}{n}}\right) dV_{\theta _{{\mathbb {H}}^n}}\\&\quad \le C, \end{aligned}$$

where the first equality follows from (B.38), the second equality follows from the change of variables $({\tilde{z}},{\tilde{t}})=(\rho _{1,\nu }z,(\rho _{1,\nu })^2t)$, the third inequality follows from the property of the cut-off function ${\widehat{\chi }}$, and the last inequality follows from the fact that ${\widetilde{v}}_\nu$ is uniformly bounded in $S_1^2({\mathbb {H}}^n)$. This proves (ii). $\square$

It follows from Proposition B.6(ii) that, by passing to subsequence if necessary, ${\widetilde{V}}_\nu$ converges to ${\widetilde{V}}$ weakly in $S_1^2(B)$ as $\nu \rightarrow \infty$ on each ball B of ${\mathbb {H}}^n$. Since the Folland–Stein embedding $S_1^2(B)\hookrightarrow L^s(B)$ is compact for $1<s<2+\displaystyle \frac{2}{n}$ on each ball B of ${\mathbb {H}}^n$, ${\widetilde{V}}_\nu$ converges to ${\widetilde{V}}$ in $L^s(B)$ for $1<s<2+\displaystyle \frac{2}{n}$. On the other hand, it follows from Proposition B.6(i) that ${\widetilde{V}}$ satisfies

$$\begin{aligned} \left( 2+\frac{2}{n}\right) \Delta _{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}=r_\infty |{\widetilde{V}}|^{\frac{2}{n}}{\widetilde{V}}\quad \text{ in } {\mathbb {H}}^n. \end{aligned}$$

(B.39)

Lemma B.7

• (i) For every ball B in ${\mathbb {H}}^n$, there holds

  $$\begin{aligned} \int _B\Bigg |L_{\theta _{{\mathbb {H}}^n}}({\widetilde{V}}_\nu -{\widetilde{V}})-r_\infty |{\widetilde{V}}_\nu -{\widetilde{V}}|^{\frac{2}{n}}({\widetilde{V}}_\nu -{\widetilde{V}})\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}} \rightarrow 0 \text{ as } \nu \rightarrow \infty . \end{aligned}$$

• (ii) There exists a constant C such that

  $$\begin{aligned} \int _{{\mathbb {H}}^n}\left( |\nabla _{\theta _{{\mathbb {H}}^n}}({\widetilde{V}}_\nu -{\widetilde{V}})|^2_{\theta _{{\mathbb {H}}^n}}+ |{\widetilde{V}}_\nu -{\widetilde{V}}|^{2+\frac{2}{n}}\right) dV_{\theta _{{\mathbb {H}}^n}}\le C \quad \text{ for } \text{ all } \nu . \end{aligned}$$

Proof

By the same proof of Proposition B.2, we have

$$\begin{aligned} {\widetilde{V}}_\nu ^{1+\frac{2}{n}}-{\widetilde{V}}^{1+\frac{2}{n}}-|{\widetilde{V}}_\nu -{\widetilde{V}}|^{\frac{2}{n}}({\widetilde{V}}_\nu -{\widetilde{V}}) \rightarrow 0 \quad \text{ in } L^{\frac{2n+2}{n+2}}(B) \text{ as } \nu \rightarrow \infty . \end{aligned}$$

This together with (B.39) and Proposition B.6(i) implies that

$$\begin{aligned}&\int _B\Bigg |L_{\theta _{{\mathbb {H}}^n}}({\widetilde{V}}_\nu -{\widetilde{V}})-r_\infty |{\widetilde{V}}_\nu -{\widetilde{V}}|^{\frac{2}{n}}({\widetilde{V}}_\nu -{\widetilde{V}})\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}}\\&\quad \le C\int _B\Bigg |L_{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}_\nu -r_\infty |{\widetilde{V}}_\nu |^{\frac{2}{n}}{\widetilde{V}}_\nu \Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _{{\mathbb {H}}^n}}\\&\qquad +\,C\, r_\infty \int _B\Bigg | {\widetilde{V}}_\nu ^{1+\frac{2}{n}}-{\widetilde{V}}^{1+\frac{2}{n}}-|{\widetilde{V}}_\nu -{\widetilde{V}}|^{\frac{2}{n}}({\widetilde{V}}_\nu -{\widetilde{V}})\Bigg |^{\frac{2n+2}{n+2}} dV_{\theta _{{\mathbb {H}}^n}} \rightarrow 0 \end{aligned}$$

as $\nu \rightarrow \infty$. This proves (i). On the other hand, (ii) follows from (B.39) and Proposition B.6(ii). This proves the assertion. $\square$

Lemma B.8

For every ball B in ${\mathbb {H}}^n$, ${\widetilde{V}}_\nu$ converges to ${\widetilde{V}}$ strongly in $S_1^2(B)$ as $\nu \rightarrow \infty$.

Proof

By contradiction, we assume that ${\widetilde{V}}_\nu$ does not converge to ${\widetilde{V}}$ strongly in $S_1^2(B)$ as $\nu \rightarrow \infty$ for some B in ${\mathbb {H}}^n$. Therefore, it follows from Lemma B.7 that the sequence $\{{\widetilde{V}}_\nu -{\widetilde{V}}\}$ satisfied the same properties of the sequence $\{v_\nu \}$. In particular, it follows from Lemma B.4 that there exists a sequence $({\widetilde{x}}_\nu ,{\widetilde{\rho }}_\nu )$ with ${\widetilde{x}}_\nu \in B_{\rho _{1,\nu }}(x^*_{1,\nu })$ and ${\widetilde{\rho }}_\nu \rightarrow 0$ as $\nu \rightarrow \infty$ such that

$$\begin{aligned} \left( 2+\frac{2}{n}\right) \int _{\exp _{x^*_{1,\nu }}^{-1}\Bigg (B_{{\widetilde{\rho }}_\nu }({\widetilde{x}}_\nu )\Bigg )}|\nabla _{\theta _{{\mathbb {H}}^n}}({\widetilde{V}}_\nu -{\widetilde{V}})|^2_{\theta _{{\mathbb {H}}^n}} dV_{\theta _{{\mathbb {H}}^n}}\ge \frac{1+a_0}{2} r_\infty ^{-n} Y(B,\theta _{{\mathbb {H}}^n})^{n+1}\nonumber \\ \end{aligned}$$

(B.40)

for $\nu$ sufficiently large. It follows from Lemma B.7 that

$$\begin{aligned} \int _{\exp _{x^*_{1,\nu }}^{-1}\Bigg (B_{{\widetilde{\rho }}_\nu }({\widetilde{x}}_\nu )\Bigg )} \langle \nabla _{\theta _{{\mathbb {H}}^n}}{\widetilde{V}},\nabla _{\theta _{{\mathbb {H}}^n}}({\widetilde{V}}_\nu -{\widetilde{V}})\rangle _{\theta _{{\mathbb {H}}^n}} dV_{\theta _{{\mathbb {H}}^n}}\rightarrow 0 \end{aligned}$$

(B.41)

as $\nu \rightarrow \infty$. Combining (B.40) and (B.41), we get

$$\begin{aligned}&\left( 2+\frac{2}{n}\right) \int _{\exp _{x^*_{1,\nu }}^{-1}\Bigg (B_{{\widetilde{\rho }}_\nu }({\widetilde{x}}_\nu )\Bigg )}|\nabla _{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}_\nu |^2_{\theta _{{\mathbb {H}}^n}} dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\quad =\left( 2+\frac{2}{n}\right) \int _{\exp _{x^*_{1,\nu }}^{-1}\Bigg (B_{{\widetilde{\rho }}_\nu }({\widetilde{x}}_\nu )\Bigg )}|\nabla _{\theta _{{\mathbb {H}}^n}}({\widetilde{V}}_\nu -{\widetilde{V}})|^2_{\theta _{{\mathbb {H}}^n}} dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\qquad +\left( 2+\frac{2}{n}\right) \int _{\exp _{x^*_{1,\nu }}^{-1}\Bigg (B_{{\widetilde{\rho }}_\nu }({\widetilde{x}}_\nu )\Bigg )}|\nabla _{\theta _{{\mathbb {H}}^n}}{\widetilde{V}}|^2_{\theta _{{\mathbb {H}}^n}} dV_{\theta _{{\mathbb {H}}^n}} \nonumber \\&\qquad -\,2\left( 2+\frac{2}{n}\right) \int _{\exp _{x^*_{1,\nu }}^{-1}\Bigg (B_{{\widetilde{\rho }}_\nu }({\widetilde{x}}_\nu )\Bigg )} \langle \nabla _{\theta _{{\mathbb {H}}^n}}{\widetilde{V}},\nabla _{\theta _{{\mathbb {H}}^n}}({\widetilde{V}}_\nu -{\widetilde{V}})\rangle _{\theta _{{\mathbb {H}}^n}} dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\quad \ge \frac{1+a_0}{2} r_\infty ^{-n} Y(B,\theta _{{\mathbb {H}}^n})^{n+1}+o(1) \nonumber \\&\quad \ge a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1} \end{aligned}$$

(B.42)

for $\nu$ sufficiently large, where the last inequality follows from

$$\begin{aligned} Y(B,\theta _{{\mathbb {H}}^n})=Y({\mathbb {H}}^n,\theta _{{\mathbb {H}}^n})=Y(S^{2n+1},\theta _{S^{2n+1}})\ge Y(M,\theta _0). \end{aligned}$$

Since ${\widetilde{V}}_\nu$ has support in $\{(t^2+|z|^4)^{\frac{1}{4}}<\frac{{\widehat{\rho }}}{\rho _{1,\nu }}\}$, it follows from (B.42) that there exists $\widetilde{{\widetilde{x}}}_\nu \in B_{\rho _{1,\nu }}(x^*_{1,\nu })$ such that

$$\begin{aligned} \int _{B_{{\widetilde{\rho }}_\nu \rho _{1,\nu }}(\widetilde{{\widetilde{x}}}_\nu )}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0} \ge a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1} \end{aligned}$$

for $\nu$ sufficiently large. This implies that

$$\begin{aligned} \rho _{\nu }(\widetilde{{\widetilde{x}}}_\nu )\le {\widetilde{\rho }}_\nu \rho _{1,\nu } <\rho _{1,\nu }, \end{aligned}$$

where we have used the fact that ${\widetilde{\rho }}_\nu \rightarrow 0$ as $\nu \rightarrow \infty$. But this contradicts to (B.32). This proves Lemma B.8. $\square$

Since ${\widetilde{V}}$ satisfies (B.39), it follows from the result of Jerison and Lee in [26] that there exists $(z_0,t_0)\in {\mathbb {H}}^n$ and $\gamma _1>0$ such that

$$\begin{aligned} {\widetilde{V}}(z,t)=W\circ T_{(z_0,t_0)}(z,t), \end{aligned}$$

where

$$\begin{aligned} W(z,t)=\left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}}\left( \frac{\gamma _1^2}{\gamma _1^4t^2+(\gamma _1^2|z|^2+1)^2}\right) ^{\frac{n}{2}} \end{aligned}$$

and

$$\begin{aligned} T_{(z_0,t_0)}(z,t)=(z+z_0,t+t_0+2Im(z\cdot z_0)) \text{ for } (z,t)\in {\mathbb {H}}^n \end{aligned}$$

is the translation in ${\mathbb {H}}^n$. By the optimality of $(x^*_{1,\nu }, \rho _{1,\nu })$, we can conclude that $(z_0,t_0)=(0,0)$; for if $(z_0,t_0)\ne (0,0)$, we can find $({\widetilde{x}}^*_{1,\nu }, {\widetilde{\rho }}_{1,\nu })$ with ${\widetilde{\rho }}_{1,\nu }<\rho _{1,\nu }$ such that

$$\begin{aligned} \int _{B_{{\widetilde{\rho }}_{1,\nu }}({\widetilde{x}}^*_{1,\nu })}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+R_{\theta _0}v_\nu ^2\right) dV_{\theta _0} \ge a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} {\widetilde{V}}(z,t)=\left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}}\left( \frac{\gamma _1^2}{\gamma _1^4t^2+(\gamma _1^2|z|^2+1)^2}\right) ^{\frac{n}{2}}. \end{aligned}$$

(B.43)

We remark that it was claimed in [19] that $\gamma _1=1$ (see the last line in P.146 in [19]), which does not seem to be true. In fact, we will show that $\gamma _1\ge C_1$. Here $C_1$ is a positive constant depending only on $a_0$, $r_{\infty }$, and M. We need the following:

Proposition B.9

For any $x\in M$ and for any $r>0$, there holds

$$\begin{aligned} \int _{B_r(x)}v_\nu \Delta _{\theta _0}v_\nu dV_{\theta _0} =\int _{B_r(x)}|\nabla _{\theta _0}v_\nu |^2_{\theta _0}dV_{\theta _0}+o(1). \end{aligned}$$

Proof

We consider the following sequence of cut-off functions:

$$\begin{aligned} \chi _\nu (y)=\left\{ \begin{array}{ll} 1, &{} \hbox { if }d(x,y)\le r; \\ 0, &{} \hbox { if }d(x,y)\ge r+r_\nu , \end{array} \right. \end{aligned}$$

(B.44)

such that

$$\begin{aligned} 0\le \chi _\nu \le 1,\quad |\nabla _{\theta _0}\chi _\nu |_{\theta _0}\le \frac{C}{r_\nu } \quad \text{ and } \quad |\Delta _{\theta _0}\chi _\nu |\le \frac{C}{r_\nu ^2}, \end{aligned}$$

(B.45)

where $r_\nu$ will be chosen later. Since the function $\chi _\nu v_\nu$ has support in $B_{r+2r_\nu }(x)$, it follows from integration by parts that

$$\begin{aligned} \begin{aligned} \int _{B_{r+2r_\nu }(x)}\chi _\nu v_\nu \Delta _{\theta _0}(\chi _\nu v_\nu ) dV_{\theta _0} =\int _{B_{r+2r_\nu }(x)}|\nabla _{\theta _0}(\chi _\nu v_\nu )|^2_{\theta _0} dV_{\theta _0}. \end{aligned} \end{aligned}$$

(B.46)

We are going to expand the left and right hand side of (B.46). By (B.44), the left hand side of (B.46) can be written as

$$\begin{aligned}&\int _{B_{r+2r_\nu }(x)}\chi _\nu v_\nu \Delta _{\theta _0}(\chi _\nu v_\nu ) dV_{\theta _0}\\&\quad =\int _{B_{r}(x)}v_\nu \Delta _{\theta _0}v_\nu dV_{\theta _0} +\int _{B_{r+2r_\nu }(x)-B_{r}(x)}\chi _\nu ^2 v_\nu \Delta _{\theta _0} v_\nu dV_{\theta _0}\\&\qquad +\int _{B_{r+2r_\nu }(x)-B_{r}(x)}\chi _\nu v_\nu ^2 \Delta _{\theta _0} \chi _\nu dV_{\theta _0}\\&\qquad +\,2\int _{B_{r+2r_\nu }(x)-B_{r}(x)}\chi _\nu v_\nu \langle \nabla _{\theta _0} \chi _\nu , \nabla _{\theta _0} v_\nu \rangle _{\theta _0} dV_{\theta _0}\\&\quad :=\int _{B_{r}(x)}v_\nu \Delta _{\theta _0}v_\nu dV_{\theta _0} +I+II+III. \end{aligned}$$

By (B.44) and (B.45), we have

$$\begin{aligned} |II|\le \int _{B_{r+2r_\nu }(x)-B_{r}(x)}\left| \chi _\nu v_\nu ^2 \Delta _{\theta _0} \chi _\nu \right| dV_{\theta _0}\le \frac{C}{r_\nu ^2}\int _M v_\nu ^2dV_{\theta _0}. \end{aligned}$$

Since $v_\nu$ converges to 0 in $L^2(M)$ as $\nu \rightarrow \infty$, if we choose

$$\begin{aligned} r_\nu =\left( \int _M v_\nu ^2dV_{\theta _0}\right) ^{\frac{1}{4}}\rightarrow 0\quad \text{ as } \nu \rightarrow \infty , \end{aligned}$$

(B.47)

then $|II|\rightarrow 0$ as $\nu \rightarrow \infty$. Since $v_\nu$ converges to 0 in $L^2(M)$ as $\nu \rightarrow \infty$, we have

$$\begin{aligned} \int _M \left| R_{\theta _0}v_\nu ^2 \right| dV_{\theta _0}\rightarrow 0 \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

(B.48)

Since $r_\nu \rightarrow 0$ as $\nu \rightarrow \infty$ by (B.47) and $v_\nu$ is uniformly bounded in $S_1^2(M)\hookrightarrow L^{2+\frac{2}{n}}(M)$, we have

$$\begin{aligned} \int _{B_{r+2r_\nu }(x)-B_{r}(x)} |v_\nu |^{2+\frac{2}{n}} dV_{\theta _0}\rightarrow 0 \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

(B.49)

Therefore, we have

$$\begin{aligned}&\left| \left( 2+\frac{2}{n}\right) I\right| \\&\quad \le \left( \int _{B_{r+2r_\nu }(x)-B_{r}(x)} v_\nu ^{2+\frac{2}{n}}dV_{\theta _0}\right) ^{\frac{n}{2n+2}} \left( \int _M\left| L_{\theta _0} v_\nu -r_\infty |v_\nu |^{\frac{2}{n}}v_\nu \right| ^{\frac{2n+2}{n+2}}dV_{\theta _0}\right) ^{\frac{n+2}{2n+2}}\\&\qquad +\int _M \left| R_{\theta _0}v_\nu ^2 \right| dV_{\theta _0}+r_\infty \int _{B_{r+2r_\nu }(x)-B_{r}(x)} |v_\nu |^{2+\frac{2}{n}} dV_{\theta _0} \\&\quad =o(1), \end{aligned}$$

where we have used (B.9), (B.48) and (B.49) in the last equality. By (B.47), Cauchy–Schwarz inequality and the fact that $v_\nu$ is bounded in $S_1^2(M)$, we have

$$\begin{aligned} |III|\le & {} 2\int _{B_{r+2r_\nu }(x)-B_{r}(x)}\left| \chi _\nu v_\nu \langle \nabla _{\theta _0} \chi _\nu , \nabla _{\theta _0} v_\nu \rangle _{\theta _0}\right| dV_{\theta _0}\nonumber \\\le & {} \frac{C}{r_\nu }\left( \int _M v_\nu ^2dV_{\theta _0}\right) ^{\frac{1}{2}} \left( \int _M |\nabla _{\theta _0} v_\nu |^2_{\theta _0}dV_{\theta _0}\right) ^{\frac{1}{2}} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty .\qquad \end{aligned}$$

(B.50)

Combining all these, we can see that the left hand side of (B.46) is equal to

$$\begin{aligned} \int _{B_{r+2r_\nu }(x)}\chi _\nu v_\nu \Delta _{\theta _0}(\chi _\nu v_\nu ) dV_{\theta _0}=\int _{B_{r}(x)}v_\nu \Delta _{\theta _0}v_\nu dV_{\theta _0}+o(1). \end{aligned}$$

(B.51)

On the other hand, the right hand side of (B.46) can be written as

$$\begin{aligned}&\int _{B_{r+2r_\nu }(x)}|\nabla _{\theta _0}(\chi _\nu v_\nu )|^2_{\theta _0} dV_{\theta _0}\\&\quad =\int _{B_{r}(x)}|\nabla _{\theta _0} v_\nu |^2_{\theta _0} dV_{\theta _0} +\int _{B_{r+2r_\nu }(x)-B_r(x)}\chi _\nu ^2|\nabla _{\theta _0}v_\nu |^2_{\theta _0} dV_{\theta _0}\\&\qquad +\int _{B_{r+2r_\nu }(x)-B_r(x)}v_\nu ^2|\nabla _{\theta _0}\chi _\nu |^2_{\theta _0} dV_{\theta _0} +\int _{B_{r+2r_\nu }(x)-B_r(x)}\chi _\nu v_\nu \langle \nabla _{\theta _0}\chi _\nu ,\nabla _{\theta _0}v_\nu \rangle _{\theta _0} dV_{\theta _0}\\&\quad :=\int _{B_{r}(x)}|\nabla _{\theta _0} v_\nu |^2_{\theta _0} dV_{\theta _0} +I'+II'+III'. \end{aligned}$$

Since $r_\nu \rightarrow 0$ as $\nu \rightarrow \infty$ by (B.47) and $v_\nu$ is uniformly bounded in $S_1^2(M)$, we have

$$\begin{aligned} |I'|\le \int _{B_{r+2r_\nu }(x)-B_r(x)}|\nabla _{\theta _0}v_\nu |^2_{\theta _0} dV_{\theta _0} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

By (B.45) and (B.47), we have

$$\begin{aligned} |II'|=\int _{B_{r+2r_\nu }(x)-B_r(x)}\chi _\nu ^2|\nabla _{\theta _0}v_\nu |^2_{\theta _0} dV_{\theta _0} \le \frac{C}{r_\nu ^2}\int _M v_\nu ^2dV_{\theta _0} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

Since $III=III'$, it follows from (B.50) that $|III'|\rightarrow 0$ as $\nu \rightarrow \infty$. Combining all these, we can see that the right hand side of (B.46) is equal to

$$\begin{aligned} \int _{B_{r+2r_\nu }(x)}|\nabla _{\theta _0}(\chi _\nu v_\nu )|^2_{\theta _0} dV_{\theta _0} =\int _{B_{r}(x)}|\nabla _{\theta _0} v_\nu |^2_{\theta _0} dV_{\theta _0}+o(1). \end{aligned}$$

(B.52)

The assertion follows from combining (B.46), (B.51) and (B.52). $\square$

Proposition B.10

There exists a positive constant $C_1$ depending only on $a_0$, $r_\infty$ and M such that

$$\begin{aligned} \gamma _1\ge C_1. \end{aligned}$$

Proof

Since $\rho _{1,\nu }\rightarrow 0$ as $\nu \rightarrow \infty$ by Lemma B.5, we can find N such that

$$\begin{aligned} C_0\le \frac{{\widehat{\rho }}}{2\rho _{1,\nu }} \quad \text{ whenever } \nu \ge N, \end{aligned}$$

(B.53)

where $C_0$ is the uniform constant given in (A.8). Note that

$$\begin{aligned}&\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<C_0\}}|{\widetilde{V}}_\nu (z,t)|^{2+\frac{2}{n}} dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\quad =\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<C_0\}} \Bigg |(\rho _{1,\nu })^n\,{\widetilde{v}}_\nu \Bigg (\rho _{1,\nu }z,(\rho _{1,\nu })^2t\Bigg )\Bigg |^{2+\frac{2}{n}} dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\quad =\int _{\{\rho _{x^*_{1,\nu }}(y)<C_0\rho _{1,\nu }\}}|{\widetilde{v}}_\nu (y)|^{2+\frac{2}{n}} dV_{{\widehat{\theta }}_{x^*_{1,\nu }}}+o(1)\nonumber \\&\quad \ge \int _{\{d(x^*_{1,\nu },y)<\rho _{1,\nu }\}}|v_\nu (y)|^{2+\frac{2}{n}} dV_{\theta _0}+o(1), \end{aligned}$$

(B.54)

where the first equality follows from (B.53) and the definition of ${\widetilde{V}}_\nu$ in (B.38), and the second equality follows from (A.1) and the change of variables $({\tilde{z}},{\tilde{t}})=(\rho _{1,\nu }z,(\rho _{1,\nu })^2t)$, and the last inequality follows from (A.8) and the definition of ${\widetilde{v}}_\nu$. For sufficiently large $\nu$, we have

$$\begin{aligned}&r_\infty \int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })}|v_\nu |^{2+\frac{2}{n}} dV_{\theta _0}\nonumber \\&\quad \ge \int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })}v_\nu L_{\theta _0}v_\nu dV_{\theta _0}\nonumber \\&\qquad - \left( \int _{M}\left| L_{\theta _0}v_\nu -r_\infty |v_\nu |^{\frac{2}{n}}v_\nu \right| ^{\frac{2n+2}{n+2}}dV_{\theta _0}\right) ^{\frac{n+2}{2n+2}} \left( \int _M|v_\nu |^{2+\frac{2}{n}} dV_{\theta _0}\right) ^{\frac{n}{2n+2}}\nonumber \\&\quad =\int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })}\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu |^2_{\theta _0}+R_{\theta _0}v_\nu ^{2+\frac{2}{n}} \right) dV_{\theta _0}+o(1)\nonumber \\&\quad \ge a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1} +o(1), \end{aligned}$$

(B.55)

where the first inequality follows from Hölder’s inequality, the last equality follows from Proposition B.9, (B.9) and the fact that $v_\nu$ is bounded in $L^{2+\frac{2}{n}}(M)$, and the last inequality follows from the definition of $(x^*_{1,\nu },\rho _{1,\nu })$ in (B.33).

Combining (B.54) and (B.55), we obtain

$$\begin{aligned} \int _{\{(t^2+|z|^4)^{\frac{1}{4}}<C_0\}}|{\widetilde{V}}_\nu (z,t)|^{2+\frac{2}{n}} dV_{\theta _{{\mathbb {H}}^n}} \ge a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1} +o(1). \end{aligned}$$

Combining this with Lemma B.8, we get

$$\begin{aligned} \int _{\{(t^2+|z|^4)^{\frac{1}{4}}<C_0\}}{\widetilde{V}}(z,t)^{2+\frac{2}{n}} dV_{\theta _{{\mathbb {H}}^n}} \ge a_0 r_\infty ^{-n} Y(M,\theta _0)^{n+1} +o(1). \end{aligned}$$

(B.56)

We compute

$$\begin{aligned}&\int _{\{(t^2+|z|^4)^{\frac{1}{4}}<C_0\}}{\widetilde{V}}(z,t)^{2+\frac{2}{n}} dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\quad = \left( \frac{n(2n+2)}{r_\infty }\right) ^{n+1} \int _{\{(t^2+|z|^4)^{\frac{1}{4}}<C_0\}}\left( \frac{\gamma _1^2}{\gamma _1^4t^2+(\gamma _1^2|z|^2+1)^2}\right) ^{n+1}dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\quad = \left( \frac{n(2n+2)}{r_\infty }\right) ^{n+1} \int _{\{({\tilde{t}}^2+|{\tilde{z}}|^4)^{\frac{1}{4}}<\gamma _1 C_0\}}\left( \frac{1}{{\tilde{t}}^2+(|{\tilde{z}}|^2+1)^2}\right) ^{n+1}dV_{\theta _{{\mathbb {H}}^n}} \end{aligned}$$

(B.57)

where the first equality follows from (B.43) and the second equality follows from change of variables $({\tilde{z}},{\tilde{t}})=(\gamma _1z,\gamma _1^2t)$. Note that the last term in (B.57) tends to zero as $\gamma _1\rightarrow 0^+$. Hence, combining (B.56) and (B.57), we can conclude that $\gamma _1$ is bounded below by a positive constant $C_1$ depending only on $a_0$, $r_\infty$, and M. This proves the assertion. $\square$

For any $(x,\lambda )\in M\times (0,\infty )$, we can find a unique solution ${\widehat{\omega }}(x,\lambda )$ of the following equation:

$$\begin{aligned} L_{\theta _0}{\widehat{\omega }}(x,\lambda )=r_\infty \omega '(x,\lambda )^{1+\frac{2}{n}} \quad \text{ in } M, \end{aligned}$$

(B.58)

where $\omega '(x,\lambda )$ is defined as

$$\begin{aligned} \omega '(x,\lambda )(y)= \left\{ \begin{array}{ll} \chi _\delta (\rho _x(y))\varphi _x(y)\omega (x,\lambda )(y), &{} \hbox { for } y\in B_{2\delta }(x); \\ 0, &{} \hbox { otherwise.} \end{array} \right. \end{aligned}$$

Here $\chi _\delta$ is the cut-off function defined in (A.11), $\varphi _x$ is the conformal factor such that ${\widehat{\theta }}_x=\varphi _x^{\frac{2}{n}}\theta _0$ in a neighborhood $B_{3\delta }(x)$ of x. Moreover, $\omega (x,\lambda )(y)$ is given by

$$\begin{aligned} \omega (x,\lambda )(y)=\left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}}\left( \frac{\lambda ^2}{\lambda ^4t^2+(\lambda ^2|z|^2+1)^2}\right) ^{\frac{n}{2}}, \end{aligned}$$

(B.59)

where (z, t) is CR normal coordinates of y centered at x. It follows from the definition of ${\widehat{\omega }}(x,\lambda )$ that

$$\begin{aligned} {\widehat{\omega }}(x,\lambda )(y)=\omega '(x,\lambda )(y)=0 \quad \hbox { for }y\in M-B_{2\delta }(x). \end{aligned}$$

(B.60)

When $n=1$, there holds (see Proposition 1 in [18])

$$\begin{aligned} |{\widehat{\omega }}(x,\lambda )(y)-\omega '(x,\lambda )(y)|\le C\lambda ^{-1}(1+|\log (\lambda ^{-2}+\rho _x(y)^2)|) \quad \text{ for } y\in B_{2\delta }(x)\nonumber \\ \end{aligned}$$

(B.61)

and (see (3.6) in [18])

$$\begin{aligned} \Bigg |L_{\theta _0}\Bigg ({\widehat{\omega }}(x,\lambda )(y)-\omega '(x,\lambda )(y)\Bigg )\Bigg |\le \inf \Bigg \{1,\frac{C}{\rho _x(y)^2+\lambda ^{-2}}\Bigg \} \quad \text{ for } y\in B_{2\delta }(x).\nonumber \\ \end{aligned}$$

(B.62)

It follows from (B.60)–(B.62) that

$$\begin{aligned} \Vert {\widehat{\omega }}(x,\lambda )-\omega '(x,\lambda )\Vert _{S_1^2(M)}\rightarrow 0 \quad \text{ as } \lambda \rightarrow \infty . \end{aligned}$$

(B.63)

Similarly, when M is spherical, it follows from Lemmas 3 and 4 in [19] that

$$\begin{aligned} |{\widehat{\omega }}(x,\lambda )(y)-\omega '(x,\lambda )(y)|\le \frac{C}{\lambda ^n} \quad \text{ for } y\in B_{2\delta }(x) \end{aligned}$$

(B.64)

and

$$\begin{aligned} \Vert {\widehat{\omega }}(x,\lambda )-\omega '(x,\lambda )\Vert _{S_1^2(M)}=O\left( \frac{1}{\lambda }\right) . \end{aligned}$$

(B.65)

We have the following:

Proposition B.11

There holds

$$\begin{aligned} \int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })} \Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{2+\frac{2}{n}}dV_{\theta _0} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

Proof

By Lemma B.5, we can find N such that

$$\begin{aligned} C_0 \rho _{1,\nu }\le \delta \quad \text{ for } \nu \ge N, \end{aligned}$$

where $C_0$ is the constant in (A.8). Therefore, for $\nu \ge N$, we have

$$\begin{aligned}&\int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })} \Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{2+\frac{2}{n}}dV_{\theta _0}\nonumber \\&\quad =\int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })} \Bigg |{\widetilde{v}}_\nu (x)- \chi _\delta (\rho _{x^*_{1,\nu }}(x))\omega \Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )(x) \Bigg |^{2+\frac{2}{n}} dV_{{\widehat{\theta }}_{x^*_{1,\nu }}}\nonumber \\&\quad \le \int _{\{(|z|^4+t^2)^{\frac{1}{4}}\le C_0\rho _{1,\nu }\}} \Bigg |{\widetilde{v}}_\nu (z,t)- \omega \Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )(z,t)\Bigg |^{2+\frac{2}{n}} dV_{\theta _{{\mathbb {H}}^n}}\nonumber \\&\quad =\int _{\{(|z|^4+t^2)^{\frac{1}{4}}\le C_0\}} \Bigg |{\widetilde{V}}_\nu ({\tilde{z}},{\tilde{t}})-{\widetilde{V}}_\nu ({\tilde{z}},{\tilde{t}})\Bigg |^{2+\frac{2}{n}} dV_{\theta _{{\mathbb {H}}^n}}, \end{aligned}$$

(B.66)

where the first inequality follows from (A.8), the first equality from the definition of $\omega '(x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }})$ in (B.58), and the last equality follows from (B.43), (B.59) and the change of variables $({\tilde{z}},{\tilde{t}})=(\frac{z}{\rho _{1,\nu }},\frac{t}{(\rho _{1,\nu })^2})$. Thanks to Lemma B.8, the last expression in (B.66) tends to zero as $\nu \rightarrow \infty$. This proves the assertion. $\square$

We can now extract from $\{v_\nu \}$ the first bubble and consider the following new sequence of functions:

$$\begin{aligned} v_\nu ^1(x)=v_\nu (x)-{\widehat{\omega }}\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) (x). \end{aligned}$$

Lemma B.12

• (i) The sequence $\{v_\nu ^1\}$ satisfies

  $$\begin{aligned} \int _M\Bigg |L_{\theta _0}v_\nu ^1-r_\infty |v_\nu ^1|^{\frac{2}{n}}v_\nu ^1\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty \end{aligned}$$

• (ii) There exists a constant C such that

  $$\begin{aligned} \int _{M}\left( |\nabla _{\theta _0}v^1_\nu |_{\theta _0}^2+(v^1_\nu )^2\right) dV_{\theta _0}\le C \quad \text{ for } \text{ all } \nu . \end{aligned}$$

Proof

We follow the proof of Lemma 14 in [19]. By Lemma B.5, we can choose N such that $\rho _{1,\nu }\le 2\delta$ for $\nu \ge N.$ For $\nu \ge N$, it follows from (B.60) that

$$\begin{aligned}&\int _M\Bigg |L_{\theta _0}v_\nu ^1-r_\infty |v_\nu ^1|^{\frac{2}{n}}v_\nu ^1\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0}\nonumber \\&\quad \le C\int _{M}\Bigg |L_{\theta _0}v_\nu -r_\infty |v_\nu |^{\frac{2}{n}}v_\nu \Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0}\nonumber \\&\qquad +\,C\int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })}\Bigg |L_{\theta _0}{\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )-r_\infty {\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )^{1+\frac{2}{n}}\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0}\nonumber \\&\qquad +\,C\,r_\infty \int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })}\left| |v_\nu |^{\frac{2}{n}}v_\nu -{\widehat{\omega }}\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) ^{1+\frac{2}{n}}\right. \nonumber \\&\qquad \left. -\Bigg |v_\nu -{\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} \Bigg (v_\nu -{\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg )\right| ^{\frac{2n+2}{n+2}}dV_{\theta _0}. \end{aligned}$$

(B.67)

It follows from (B.60), (B.61), (B.64), Lemma B.5 and the definition of ${\widehat{\omega }}(x,\lambda )$ that

$$\begin{aligned}&\int _M\Bigg |L_{\theta _0}{\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )-r_\infty {\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )^{1+\frac{2}{n}}\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0}\nonumber \\&\quad =r_\infty \int _M\Bigg |\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )^{1+\frac{2}{n}}- {\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )^{1+\frac{2}{n}}\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0} =O(\rho _{1,\nu })=o(1).\nonumber \\ \end{aligned}$$

(B.68)

On the other hand,

$$\begin{aligned}&\left| |v_\nu |^{\frac{2}{n}}v_\nu -{\widehat{\omega }}\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) ^{1+\frac{2}{n}}-\Bigg |v_\nu -{\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} \Bigg (v_\nu -{\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg )\right| \\&\quad \le \left| |v_\nu |^{\frac{2}{n}}v_\nu -\omega '\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) ^{1+\frac{2}{n}}\right. \\&\qquad \left. -\Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} \Bigg (v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg )\right| \\&\qquad +\left| \omega '\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) ^{1+\frac{2}{n}} -{\widehat{\omega }}\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) ^{1+\frac{2}{n}}\right| \\&\qquad +\left| \omega '\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) -{\widetilde{\omega }}\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) \right| ^{1+\frac{2}{n}}\\&\qquad +\left| \Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} \Bigg ({\widetilde{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )-\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg )\right| \\&\qquad +\left| \Bigg |{\widetilde{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )-\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} \Bigg (v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg )\right| \\&\quad :=I+II+III+IV+V. \end{aligned}$$

It follows from the proof of (B.8) that

$$\begin{aligned} I=O\left( |v_\nu |^{\frac{2}{n}} \Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg | +\Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} |v_\nu |\right) . \end{aligned}$$

This together with Hölder’s inequality implies that

$$\begin{aligned}&\int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })} \left| |v_\nu |^{\frac{2}{n}}v_\nu -\omega '\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) ^{1+\frac{2}{n}}\right. \\&\qquad \left. -\Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} \Bigg (v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg )\right| ^{\frac{2n+2}{n+2}}dV_{\theta _0}\\&\quad \le C\left( \int _M|v_\nu |^{2+\frac{2}{n}}\right) ^{\frac{2}{n+2}}\left( \int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })}\Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{2+\frac{2}{n}}dV_{\theta _0}\right) ^{\frac{n}{n+2}}\\&\qquad +\,C\left( \int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })}\Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{2+\frac{2}{n}}dV_{\theta _0}\right) ^{\frac{2}{n+2}} \left( \int _M|v_\nu |^{2+\frac{2}{n}}\right) ^{\frac{n}{n+2}}\\&\quad =o(1), \end{aligned}$$

where the last equality follows from Proposition B.11 and the fact that $v_\nu$ is bounded in $L^{2+\frac{2}{n}}(M)$. Also, it follows from (B.60), (B.61) and (B.64) that

$$\begin{aligned} II\le C\frac{\rho _{1,\nu }}{\gamma _1} \quad \text{ and } \quad III\le C\frac{\rho _{1,\nu }}{\gamma _1}. \end{aligned}$$

It follows from (B.60), (B.61), (B.64) and Hölder’s inequality that

$$\begin{aligned} IV\le C\frac{\rho _{1,\nu }}{\gamma _1}\Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\varepsilon ^*_{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} \quad \text{ and } \quad V\le C\Bigg (\frac{\varepsilon ^*_{1,\nu }}{\gamma _1}\Bigg )^{\frac{2}{n}} \Bigg |v_\nu -\omega '\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |. \end{aligned}$$

Combining all these with Lemma B.5, Propositions B.10 and B.11, we can conclude that

$$\begin{aligned}&\int _{B_{\rho _{1,\nu }}(x^*_{1,\nu })}\left| |v_\nu |^{\frac{2}{n}}v_\nu -{\widehat{\omega }}\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) ^{1+\frac{2}{n}}\right. \nonumber \\&\quad \left. -\Bigg |v_\nu -{\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg |^{\frac{2}{n}} \Bigg (v_\nu -{\widehat{\omega }}\Bigg (x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\Bigg )\Bigg )\right| ^{\frac{2n+2}{n+2}}dV_{\theta _0} =o(1).\qquad \qquad \end{aligned}$$

(B.69)

Now (i) follows from (B.9), (B.67), (B.68) and (B.69).

For (ii), note that

$$\begin{aligned}&\int _M\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}{\widehat{\omega }}(x,\lambda )|_{\theta _0}^2+R_{\theta _0}{\widehat{\omega }}(x,\lambda )^2\right) dV_{\theta _0}\nonumber \\&\quad =r_\infty \int _M\omega '(x,\lambda )^{1+\frac{2}{n}}{\widehat{\omega }}(x,\lambda )dV_{\theta _0}\nonumber \\&\quad =r_\infty \int _M\omega '(x,\lambda )^{2+\frac{2}{n}}dV_{\theta _0}+O\left( \frac{1}{\lambda }\right) \int _M\omega '(x,\lambda )^{1+\frac{2}{n}}dV_{\theta _0}\nonumber \\&\quad \le r_\infty \int _M\omega '(x,\lambda )^{2+\frac{2}{n}}dV_{\theta _0}+O\left( \frac{1}{\lambda }\right) \left( \int _M\omega '(x,\lambda )^{2+\frac{2}{n}}dV_{\theta _0}\right) ^{\frac{n+2}{2n+2}}\qquad \qquad \end{aligned}$$

(B.70)

where the first equality follows from the definition of ${\widehat{\omega }}(x,\lambda )$, the second equality follows from (B.60), (B.61) and (B.64), the last inequality follows from the Hölder’s inequality. Note also that it follows from the definition of $\omega '(x,\lambda )$ that

$$\begin{aligned} \begin{aligned} \int _M\omega '(x,\lambda )^{2+\frac{2}{n}}dV_{\theta _0} \le \int _{B_{2\delta }(x)}\omega (x,\lambda )^{2+\frac{2}{n}}dV_{\theta _0} \le \int _M\omega (x,1)^{2+\frac{2}{n}}dV_{\theta _0}, \end{aligned}\qquad \qquad \end{aligned}$$

(B.71)

where the last inequality follows from the change of variables $({\tilde{z}},{\tilde{t}})=(\lambda z,\lambda ^2t)$ in the CR normal coordinates. We derive from (B.70) and (B.71) that there exists a uniform constant C such that

$$\begin{aligned} \int _M\left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}{\widehat{\omega }}(x,\lambda )|_{\theta _0}^2+R_{\theta _0}{\widehat{\omega }}(x,\lambda )^2\right) dV_{\theta _0}\le C \end{aligned}$$

(B.72)

when $\lambda$ is sufficiently large. By Cauchy–Schwarz inequality, we have

$$\begin{aligned}&\int _{M}\left( |\nabla _{\theta _0}v^1_\nu |_{\theta _0}^2+(v^1_\nu )^2\right) dV_{\theta _0}\nonumber \\&\quad \le 2\int _{M}\left( |\nabla _{\theta _0}v_\nu |_{\theta _0}^2+v_\nu ^2\right) dV_{\theta _0}\nonumber \\&\qquad +\,2\int _{M}\left( \left| \nabla _{\theta _0}{\widehat{\omega }}\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) \right| _{\theta _0}^2+ {\widehat{\omega }}\left( x^*_{1,\nu },\frac{\gamma _1}{\rho _{1,\nu }}\right) ^2\right) dV_{\theta _0}. \end{aligned}$$

(B.73)

Now (ii) follows from combining (B.72), (B.73) and the assumption that $\{v_\nu \}$ is uniformly bounded in $S_1^2(M)$. This proves Lemma B.12. $\square$

Iterating the above procedure, either $v_\nu ^1$ converges strongly to 0 in $S_1^2(M)$ as $\nu \rightarrow \infty$, or we can find a new sequence $(x^*_{2,\nu },\rho _{2,\nu })$ and extract another bubble by defining

$$\begin{aligned} v_\nu ^2(x)=v_\nu ^1(x)-{\widehat{\omega }}\left( x^*_{2,\nu },\frac{\gamma _2}{\rho _{2,\nu }}\right) (x) \end{aligned}$$

and show that

$$\begin{aligned} \int _M\Bigg |L_{\theta _0}v_\nu ^2-r_\infty |v_\nu ^2|^{\frac{2}{n}}v_\nu ^2\Bigg |^{\frac{2n+2}{n+2}}dV_{\theta _0} \rightarrow 0 \quad \text{ as } \nu \rightarrow \infty . \end{aligned}$$

On the other hand, it can be shown that (see Lemmas 15 and 16 in [19])

$$\begin{aligned} \rho _{2,\nu }\ge \frac{1}{2}\rho _{1,\nu } \quad \text{ and } \quad \frac{\rho _{2,\nu }}{\rho _{1,\nu }}+\frac{d(x^*_{1,\nu },x^*_{2,\nu })^2}{\rho _{1,\nu }\,\rho _{2,\nu }} \rightarrow \infty \end{aligned}$$

(B.74)

as $\nu \rightarrow \infty$. Here d is Carnot–Carathéodory distance on M with respect to the contact form $\theta _0$. This argument can be iterated as long as the new sequence $\{v_\nu ^l\}$ does not coverage strongly to 0 in $S_1^2(M)$. And we claim that the iteration must terminate in finite steps. To see this, note that

$$\begin{aligned}&\int _M \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu ^{l}|^2_{\theta _0} +R_{\theta _0}(v_\nu ^{l})^2\right) dV_{\theta _0}\\&\quad =\int _M \left( \left( 2+\frac{2}{n}\right) \Bigg |\nabla _{\theta _0}\Bigg (v_\nu ^{l-1}-{\widehat{\omega }}\Bigg (x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }}\Bigg )\Bigg )\Bigg |^2_{\theta _0}\right. \\&\qquad \left. +\,R_{\theta _0}\Bigg (v_\nu ^{l-1}-{\widehat{\omega }}\Bigg (x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }}\Bigg )\Bigg )^2\right) dV_{\theta _0}\\&\quad =\int _M \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu ^{l-1}|^2_{\theta _0} +R_{\theta _0}(v_\nu ^{l-1})^2\right) dV_{\theta _0}\\&\qquad +\,r_\infty \int _M{\widehat{\omega }}\left( x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }}\right) \omega '\left( x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }}\right) ^{1+\frac{2}{n}}dV_{\theta _0} +o(1), \end{aligned}$$

where the first equality follows from the fact that $v_\nu ^{l-1}$ converges to 0 weakly in $S_1^2(M)$ as $\nu \rightarrow \infty$, and the last equality follows from (B.58). We compute

$$\begin{aligned}&\int _M{\widehat{\omega }}\left( x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }}\right) \omega '\left( x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }}\right) ^{1+\frac{2}{n}}dV_{\theta _0}\\&\quad \ge \int _{B_{\delta }(x^*_{l,\nu })}\Bigg (\varphi _{x^*_{l,\nu }}(y)\omega \Bigg (x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }}\Bigg )(y)\Bigg )^{2+\frac{2}{n}}dV_{\theta _0}+o(1)\\&\quad \ge \int _{\left\{ (|z|^4+t^2)^{\frac{1}{4}}\le \frac{\delta }{C_0}\right\} }\omega \left( x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }}\right) (z,t)^{2+\frac{2}{n}}dV_{\theta _{{\mathbb {H}}^n}}+o(1)\\&\quad =\left( \frac{n(2n+2)}{r_\infty }\right) ^{n+1} \int _{\left\{ (|{\tilde{z}}|^4+{\tilde{t}}^2)^{\frac{1}{4}}\le \frac{\delta }{C_0\rho _{l,\nu }}\right\} } \left( \frac{\gamma _{l}^2}{\gamma _{l}^4{\tilde{t}}^2+(\gamma _{l}^2|{\tilde{z}}|^2+1)^2}\right) ^{n+1}dV_{\theta _{{\mathbb {H}}^n}}+o(1)\\&\quad =\left( \frac{n(2n+2)}{r_\infty }\right) ^{n+1} \int _{\left\{ (|{\widehat{z}}|^4+{\widehat{t}}^2)^{\frac{1}{4}}\le \frac{\gamma _{l}\delta }{C_0\rho _{l,\nu }}\right\} } \left( \frac{1}{{\widehat{t}}^2+(|{\widehat{z}}|^2+1)^2}\right) ^{n+1}dV_{\theta _{{\mathbb {H}}^n}}+o(1)\\&\quad \ge \left( \frac{n(2n+2)}{r_\infty }\right) ^{n+1} \int _{\left\{ (|{\widehat{z}}|^4+{\widehat{t}}^2)^{\frac{1}{4}}\le \frac{C_1\delta }{C_0}\right\} } \left( \frac{1}{{\widehat{t}}^2+(|{\widehat{z}}|^2+1)^2}\right) ^{n+1}dV_{\theta _{{\mathbb {H}}^n}}+o(1), \end{aligned}$$

where the first inequality follows from (B.60), (B.61), (B.64) and the definition of $\omega '(x^*_{l,\nu },\frac{\gamma _{l}}{\rho _{l,\nu }})$, the second inequality follows from (A.1) and (A.8), the first equality follows from the change of variables $({\tilde{z}},{\tilde{t}})=(\frac{z}{\rho _{l,\nu }},\frac{t}{(\rho _{l,\nu })^2})$, the second equality follows from the change of variables $({\widehat{z}},{\widehat{t}})=(\gamma _l{\tilde{z}},\gamma _{l}^2{\tilde{t}})$, and the last inequality follows from Proposition B.10 and Lemma B.5. Hence, if we let

$$\begin{aligned} C_2=r_\infty \left( \frac{n(2n+2)}{r_\infty }\right) ^{n+1} \int _{\{(|{\widehat{z}}|^4+{\widehat{t}}^2)^{\frac{1}{4}}\le \frac{C_1\delta }{C_0}\}} \left( \frac{1}{{\widehat{t}}^2+(|{\widehat{z}}|^2+1)^2}\right) ^{n+1}dV_{\theta _{{\mathbb {H}}^n}}, \end{aligned}$$

then it follows from the above computation that

$$\begin{aligned}&\int _M \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu ^{l}|^2_{\theta _0} +R_{\theta _0}(v_\nu ^{l})^2\right) dV_{\theta _0}\\&\quad \ge \int _M \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu ^{l-1}|^2_{\theta _0} +R_{\theta _0}(v_\nu ^{l-1})^2\right) dV_{\theta _0}+C_2+o(1). \end{aligned}$$

That is to say, the quantity $\int _M \left( \left( 2+\frac{2}{n}\right) |\nabla _{\theta _0}v_\nu ^{l-1}|^2_{\theta _0} +R_{\theta _0}(v_\nu ^{l-1})^2\right) dV_{\theta _0}$ at the l-th step decreases by at least $C_2$ after extraction of a bubble. Therefore, the iteration must stop after finite steps.

Therefore, there exists an integer m and a sequence of m-tuples $(x^*_{k,\nu },\varepsilon ^*_{k,\nu })_{1\le k\le m}$ where $\varepsilon ^*_{k,\nu }=\displaystyle \frac{\rho _{k,\nu }}{\gamma _k}$ such that

$$\begin{aligned} \varepsilon ^*_{k,\nu }\rightarrow 0 \quad \text{ as } \nu \rightarrow \infty \text{ for } \text{ all } 1\le k\le m, \end{aligned}$$

by Lemma B.5 and Proposition B.10. Also, we have

$$\begin{aligned} \Bigg \Vert v_\nu -\sum _{k=1}^m{\widehat{\omega }}\left( x^*_{k,\nu },\frac{1}{\varepsilon ^*_{k,\nu }}\right) \Bigg \Vert _{S^2_1(M)}\rightarrow 0 \quad \text{ as } \quad \nu \rightarrow \infty . \end{aligned}$$

(B.75)

Now (5.4) follows from (B.74) and Proposition B.10. On the other hand, we have

$$\begin{aligned}&\Bigg \Vert u_\nu -u_\infty -\sum _{k=1}^m{\overline{u}}_{\Bigg (x^*_{k,\nu },\varepsilon ^*_{k,\nu }\Bigg )}\Bigg \Vert _{S^2_1(M)}\\&\quad =\left\| v_\nu -\sum _{k=1}^m{\widehat{\omega }}\left( x^*_{k,\nu },\frac{1}{\varepsilon ^*_{k,\nu }}\right) -\sum _{k=1}^m\left( \omega '\left( x^*_{k,\nu },\frac{1}{\varepsilon ^*_{k,\nu }}\right) -{\widehat{\omega }}\left( x^*_{k,\nu },\frac{1}{\varepsilon ^*_{k,\nu }}\right) \right) \right. \\&\qquad \left. -\sum _{k=1}^m \varphi _{x^*_{k,\nu }}(y) \left( \frac{n(2n+2)}{r_\infty }\right) ^{\frac{n}{2}} (\varepsilon ^*_{k,\nu })^n \Bigg (1-\chi _\delta (\rho _{x^*_{k,\nu }}(y))\Bigg )G_{x^*_{k,\nu }}(y)\right\| _{S^2_1(M)}=o(1) \end{aligned}$$

where the first equality follows from (A.10) and (A.11), and the last equality follows from (B.59), (B.63), (B.65), (B.75), Lemma B.5 and the fact that the Green’s function $G_{x^*_{k,\nu }}(y)$ is bounded in $S_1^2(K)$ for any compact set $K\subset M-\{x^*_{k,\nu }\}$ (see (A.6) and (A.7)). This proves (5.5) and this completes the proof of Theorem 5.1.