that $\gamma_{1,\overline{1}}=0$, we see that$$\begin{array}
[c]{ccl}\int_{M}\gamma_{1,0}u_{\overline{1}}d\mu & = & \int_{M}\gamma_{1}u_{\overline{1}0}d\mu \\
& = & -\int_{M}\gamma_{1}\left(  u_{0\overline{1}}-A_{\overline{1}\overline
{1}}u_{1}\right)  d\mu \\
& = & \int_{M}A_{\overline{1}\overline{1}}u_{1}\gamma_{1}d\mu.
\end{array}
\label{31A}$$ It follows from $\left(  \ref{30A}\right)  \ $and$\  \left(  \ref{31A}\right)
$ that$$\begin{array}
[c]{l}\  \  \ 2\int_{M}Qud\mu+2\int_{M}\left(  P_{0}u\right)  ud\mu \\
=i\int_{M}\left[  \left(  A_{11}u_{\overline{1}}\gamma_{\overline{1}}-A_{\overline{1}\overline{1}}u_{1}\gamma_{1}\right)  -conj\right]  d\mu \\
=-2\int_{M}Tor\left(  d_{b}u,\gamma \right)  d\mu.
\end{array}$$ Thus by (\[34\]) $$\int_{M}\left(  R-\frac{1}{2}Tor-\frac{1}{2}Tor^{\prime}\right)  \left(
\gamma,\gamma \right)  d\mu+\int_{M}\left \vert \gamma_{1,1}\right \vert ^{2}d\mu+\int_{M}Qud\mu+\int_{M}\left(  P_{0}u^{\bot}\right)  u^{\bot}d\mu=0.$$

(proof of Theorem \[t3\] and Corollary \[c4\])  If we assume that $$\ker P_{1}=\ker P_{0}.$$ Then we also have $$0=\int_{M}\left(  R-\frac{1}{2}Tor-\frac{1}{2}Tor^{\prime}\right)  \left(
\gamma,\gamma \right)  d\mu+\int_{M}\left \vert \gamma_{1,1}\right \vert ^{2}d\mu+\int_{M}Qu^{\bot}d\mu+\int_{M}\left(  P_{0}u^{\bot}\right)  u^{\bot}d\mu.\label{2018F}$$ Here we have used the fact that $P_{0}$ is self-adjoint and$$\int_{M}Tor\left(  d_{b}u_{\ker},\gamma \right)  d\mu=0.$$

Now if $\widetilde{\theta}=e^{\frac{\left(  f+2u\right)  }{3}}\theta$ is a pseudo-Einstein contact form for any CR-pluriharmonic function $f,$ it follows from (\[26A\]) that $$\gamma=0$$ and then $$0=\int_{M}Qu^{\bot}d\mu+\int_{M}\left(  P_{0}u^{\bot}\right)  u^{\bot}d\mu=\int_{M}Q^{\bot}u^{\bot}d\mu+\int_{M}\left(  P_{0}u^{\bot}\right)
u^{\bot}d\mu.$$ By  the Hőrder inequality and essentially positivity of the CR Paneitz operator, we have$$\begin{array}
[c]{ccl}\int_{M}Q^{\bot}u^{\bot}d\mu+\int_{M}\left(  P_{0}u^{\bot}\right)  u^{\bot
}d\mu & \geq & \Lambda \int_{M}(u^{\bot})^{2}d\mu-(\int_{M}(Q^{\bot})^{2}d\mu)^{\frac{1}{2}})(\int_{M}(u^{\bot})^{2}d\mu)^{\frac{1}{2}}\\
& \geq & [\Lambda(\int_{M}(u^{\bot})^{2}d\mu)^{\frac{1}{2}}-(\int_{M}(Q^{\bot
})^{2}d\mu)^{\frac{1}{2}}](\int_{M}(u^{\bot})^{2}d\mu)^{\frac{1}{2}}\text{ }\end{array}$$ and then $$0\geq \Lambda(\int_{M}(u^{\bot})^{2}d\mu)^{\frac{1}{2}}-(\int_{M}(Q^{\bot})^{2}d\mu)^{\frac{1}{2}}.$$ Hence $$\int_{M}(Q^{\bot})^{2}d\mu \geq \Lambda^{2}\int_{M}(u^{\bot})^{2}d\mu.$$

Furthermore, if he CR  $Q$-curvature is pluriharmonic (i.e. $Q^{\bot}=0),$ then $$u^{\bot}=0$$ and by (\[0\]) $$W_{1}=0.$$ Hence $\theta$ is also a globally defined 