[proofplan] We compute the first variation of the numerator and denominator of the CR Yamabe quotient at the positive function $u$. The stated integration-by-parts convention converts the horizontal gradient variation into the sub-Laplacian term, and criticality forces the resulting weak identity against every smooth [test function](/page/Test%20Function). Since the test function is arbitrary and all functions are smooth, the weak identity is the pointwise Euler-Lagrange equation. Finally, the standard pseudohermitian conformal scalar-curvature and volume transformation formulae identify this equation with constant Webster scalar curvature for $\hat\theta=u^{2/n}\theta$ and give the normalization consequences. [/proofplan]

[step:Define the energy and constraint functionals]Set \begin{align*} a:=\frac{2(n+1)}{n}. \end{align*} Let $\mathcal P$ denote the positive cone \begin{align*} \mathcal P:=\{v\in C^\infty(M;\mathbb R):v>0\text{ on }M\}. \end{align*} Define the energy functional \begin{align*} N:\mathcal P&\to\mathbb R \end{align*} by \begin{align*} N(v):=\int_M \left(a|\nabla_b v|_\theta^2+R_\theta v^2\right)\,dV_\theta \end{align*} and define the constraint functional \begin{align*} I:\mathcal P&\to(0,\infty) \end{align*} by \begin{align*} I(v):=\int_M v^p\,dV_\theta. \end{align*} Then \begin{align*} Q_\theta[v]=N(v)I(v)^{-2/p}. \end{align*} Since $M$ is compact and $u>0$ is continuous, for each $\varphi\in C^\infty(M;\mathbb R)$ there exists $\varepsilon>0$ such that $u+s\varphi>0$ on $M$ whenever $|s|<\varepsilon$. Thus the curve \begin{align*} \gamma_\varphi:(-\varepsilon,\varepsilon)&\to C^\infty(M;\mathbb R) \end{align*} \begin{align*} s&\mapsto u+s\varphi \end{align*} is an admissible curve in the positive cone.[/step]

[guided]We isolate the two pieces of the quotient because the Euler-Lagrange equation comes from differentiating a ratio. Define \begin{align*} a:=\frac{2(n+1)}{n}. \end{align*} Let $\mathcal P$ denote the positive cone \begin{align*} \mathcal P:=\{v\in C^\infty(M;\mathbb R):v>0\text{ on }M\}. \end{align*} Define \begin{align*} N:\mathcal P&\to\mathbb R \end{align*} by \begin{align*} N(v):=\int_M \left(a|\nabla_b v|_\theta^2+R_\theta v^2\right)\,dV_\theta \end{align*} and define \begin{align*} I:\mathcal P&\to(0,\infty) \end{align*} by \begin{align*} I(v):=\int_M v^p\,dV_\theta. \end{align*} With this notation, the CR Yamabe quotient is \begin{align*} Q_\theta[v]=N(v)I(v)^{-2/p}. \end{align*} The variation must stay inside the domain of $Q_\theta$, namely the positive smooth functions. Fix an arbitrary test direction $\varphi\in C^\infty(M;\mathbb R)$. Because $M$ is compact and $u>0$ is continuous, $u$ has a positive minimum: \begin{align*} m:=\min_{x\in M}u(x)>0. \end{align*} If $\varphi=0$, any $\varepsilon>0$ works. If $\varphi\ne 0$, define the supremum norm \begin{align*} \|\varphi\|_{C^0(M)}:=\max_{x\in M}|\varphi(x)| \end{align*} and choose \begin{align*} \varepsilon:=\frac{m}{2\|\varphi\|_{C^0(M)}}. \end{align*} Then for $|s|<\varepsilon$ and every $x\in M$, \begin{align*} u(x)+s\varphi(x)\ge m-|s|\|\varphi\|_{C^0(M)}> \frac{m}{2}>0. \end{align*} Therefore the map \begin{align*} \gamma_\varphi:(-\varepsilon,\varepsilon)&\to C^\infty(M;\mathbb R) \end{align*} \begin{align*} s&\mapsto u+s\varphi \end{align*} is an admissible variation through positive smooth functions. This justifies differentiating $Q_\theta[u+s\varphi]$ at $s=0$ for an arbitrary real smooth direction $\varphi$.[/guided]

[step:Differentiate the numerator and move the gradient term by integration by parts] Fix $\varphi\in C^\infty(M;\mathbb R)$. Differentiating under the integral sign, which is valid because the integrand is a smooth polynomial expression in $s$ on the compact manifold $M$, gives \begin{align*} \frac{d}{ds}\bigg|_{s=0}N(u+s\varphi) =2\int_M \left(a\langle\nabla_b u,\nabla_b\varphi\rangle_\theta+R_\theta u\varphi\right)\,dV_\theta. \end{align*} Using the defining integration-by-parts convention with $f=u$ and $g=\varphi$, \begin{align*} \int_M \langle\nabla_b u,\nabla_b\varphi\rangle_\theta\,dV_\theta=-\int_M \varphi\,\Delta_b u\,dV_\theta. \end{align*} Hence \begin{align*} \frac{d}{ds}\bigg|_{s=0}N(u+s\varphi) =2\int_M \varphi\left(-a\Delta_b u+R_\theta u\right)\,dV_\theta. \end{align*} [/step]

[step:Differentiate the denominator and impose criticality] Since $u>0$ and \begin{align*} p=2+\frac{2}{n}, \end{align*} differentiating the constraint gives \begin{align*} \frac{d}{ds}\bigg|_{s=0}I(u+s\varphi) =p\int_M u^{p-1}\varphi\,dV_\theta. \end{align*} The real function \begin{align*} J:(0,\infty)&\to \mathbb R \end{align*} \begin{align*} t&\mapsto t^{-2/p} \end{align*} has derivative $J'(t)=-\frac{2}{p}t^{-2/p-1}$. Therefore the chain rule gives \begin{align*} 0=\frac{d}{ds}\bigg|_{s=0}Q_\theta[u+s\varphi] =2I(u)^{-2/p}\int_M \varphi\left(-a\Delta_b u+R_\theta u\right)\,dV_\theta-2N(u)I(u)^{-2/p-1}\int_M u^{p-1}\varphi\,dV_\theta. \end{align*} Multiplying by $\frac{1}{2}I(u)^{2/p}$ gives \begin{align*} \int_M \varphi\left(-a\Delta_b u+R_\theta u-\frac{N(u)}{I(u)}u^{p-1}\right)\,dV_\theta=0 \end{align*} for every $\varphi\in C^\infty(M;\mathbb R)$. [/step]

[step:Convert the weak identity into the pointwise Euler-Lagrange equation]Define \begin{align*} \lambda:=\frac{N(u)}{I(u)}\in\mathbb R. \end{align*} Also define the smooth real function \begin{align*} F:M&\to\mathbb R \end{align*} \begin{align*} x&\mapsto -a\Delta_b u(x)+R_\theta(x)u(x)-\lambda u(x)^{p-1}. \end{align*} The preceding step says \begin{align*} \int_M \varphi F\,dV_\theta=0 \end{align*} for every $\varphi\in C^\infty(M;\mathbb R)$. Taking $\varphi=F$ is allowed because $F$ is smooth and real-valued, and gives \begin{align*} \int_M F^2\,dV_\theta=0. \end{align*} Since $dV_\theta$ is a positive smooth volume form and $F^2$ is continuous and non-negative, this implies $F=0$ on $M$. Thus \begin{align*} -\frac{2(n+1)}{n}\Delta_b u+R_\theta u=\lambda u^{p-1}. \end{align*} Because \begin{align*} p-1=1+\frac{2}{n}, \end{align*} this is exactly \begin{align*} -\frac{2(n+1)}{n}\Delta_b u+R_\theta u=\lambda u^{1+2/n}. \end{align*}[/step]

[guided]The integral identity from the first variation is a weak form of the Euler-Lagrange equation. We now show that, because every object is smooth, the weak form is equivalent to a pointwise identity. Define the real number \begin{align*} \lambda:=\frac{N(u)}{I(u)}. \end{align*} This is well-defined because $u>0$ on $M$ and $dV_\theta$ is a positive volume form, so \begin{align*} I(u)=\int_M u^p\,dV_\theta>0. \end{align*} Define the smooth real function \begin{align*} F:M&\to\mathbb R \end{align*} \begin{align*} x&\mapsto -a\Delta_b u(x)+R_\theta(x)u(x)-\lambda u(x)^{p-1}. \end{align*} The critical point computation gives \begin{align*} \int_M \varphi F\,dV_\theta=0 \end{align*} for every $\varphi\in C^\infty(M;\mathbb R)$. The decisive point is that the test direction $\varphi$ is arbitrary. Since $u$, $R_\theta$, and $\Delta_b u$ are smooth, the function $F$ is itself a legitimate smooth real test direction. Choosing $\varphi=F$ gives \begin{align*} \int_M F^2\,dV_\theta=0. \end{align*} The integrand $F^2$ is continuous and non-negative, and $dV_\theta$ is a positive smooth volume form. If $F$ were nonzero at some point $x_0\in M$, continuity would give an open neighbourhood $U\subset M$ of $x_0$ and a real number $c>0$ such that $F^2\ge c$ on $U$. Positivity of the volume form would then imply \begin{align*} \int_M F^2\,dV_\theta\ge \int_U F^2\,dV_\theta>0, \end{align*} contradicting the integral identity. Hence $F=0$ on $M$. Therefore \begin{align*} -a\Delta_b u+R_\theta u=\lambda u^{p-1}. \end{align*} Substituting \begin{align*} a=\frac{2(n+1)}{n} \end{align*} and \begin{align*} p-1=1+\frac{2}{n} \end{align*} yields \begin{align*} -\frac{2(n+1)}{n}\Delta_b u+R_\theta u=\lambda u^{1+2/n}. \end{align*} This is the desired Euler-Lagrange equation.[/guided]

[step:Use the conformal Webster curvature law to identify the constant curvature form] We use the pseudohermitian conformal scalar-curvature transformation formula, with the normalization stated in the theorem, for the convention \begin{align*} \int_M \langle \nabla_b f,\nabla_b g\rangle_\theta\,dV_\theta=-\int_M g\,\Delta_b f\,dV_\theta. \end{align*} For a positive smooth function $u\in C^\infty(M;\mathbb R)$ and \begin{align*} \hat\theta:=u^{2/n}\theta, \end{align*} the formula is \begin{align*} -\frac{2(n+1)}{n}\Delta_b u+R_\theta u=R_{\hat\theta}u^{1+2/n}. \end{align*} Comparing this identity with the Euler-Lagrange equation gives \begin{align*} R_{\hat\theta}u^{1+2/n}=\lambda u^{1+2/n}. \end{align*} Since $u>0$ on $M$, division by $u^{1+2/n}$ gives \begin{align*} R_{\hat\theta}=\lambda \end{align*} on $M$. [/step]

[step:Compute the volume scaling and the normalized multiplier] For $\hat\theta=u^{2/n}\theta$, the [exterior derivative](/theorems/1525) is \begin{align*} d\hat\theta=d(u^{2/n})\wedge\theta+u^{2/n}d\theta. \end{align*} In the wedge product \begin{align*} \hat\theta\wedge(d\hat\theta)^n, \end{align*} every term containing $d(u^{2/n})\wedge\theta$ vanishes because it also contains the factor $\hat\theta=u^{2/n}\theta$ and hence contains two copies of $\theta$. Therefore \begin{align*} dV_{\hat\theta}=\hat\theta\wedge(d\hat\theta)^n=u^{2/n}\theta\wedge\left(u^{2/n}d\theta\right)^n=u^{2+2/n}\theta\wedge(d\theta)^n=u^p\,dV_\theta. \end{align*} Thus, if \begin{align*} \int_M u^p\,dV_\theta=1, \end{align*} then \begin{align*} \operatorname{Vol}(M,\hat\theta)=\int_M dV_{\hat\theta}=\int_M u^p\,dV_\theta=1. \end{align*} It remains to identify $\lambda$ under this normalization. Multiplying the Euler-Lagrange equation by $u$ and integrating gives \begin{align*} \int_M u\left(-a\Delta_b u+R_\theta u\right)\,dV_\theta=\lambda\int_M u^p\,dV_\theta. \end{align*} Using the integration-by-parts convention with $f=u$ and $g=u$, \begin{align*} -\int_M u\,\Delta_b u\,dV_\theta=\int_M |\nabla_b u|_\theta^2\,dV_\theta. \end{align*} Hence \begin{align*} \lambda\int_M u^p\,dV_\theta=\int_M \left(a|\nabla_b u|_\theta^2+R_\theta u^2\right)\,dV_\theta=N(u). \end{align*} If $\int_M u^p\,dV_\theta=1$, then $\lambda=N(u)$. Since \begin{align*} Q_\theta[u]=N(u)\left(\int_M u^p\,dV_\theta\right)^{-2/p}, \end{align*} the same normalization gives \begin{align*} \lambda=Q_\theta[u]. \end{align*} This proves all asserted conclusions. [/step]