Prescribed Webster Scalar Curvatures on Compact Pseudo-Hermitian Manifolds

Abstract

In this paper, we investigate the problem of prescribing Webster scalar curvatures on compact pseudo-Hermitian manifolds. In terms of the method of upper and lower solutions and the perturbation theory of self-adjoint operators, we can describe some sets of Webster scalar curvature functions which can be realized through pointwise CR conformal deformations and CR conformally equivalent deformations respectively from a given pseudo-Hermitian structure.

Appendix

In this section, we give an alternative proof of Theorem 3.4 by following the spirit of [30]. Although the expressions are different, the following theorem and Theorem 3.4 are completely equivalent.

Theorem 4.7

Let $(M^{2n+1},H,J,\theta )$ be a compact pseudo-Hermitian manifold. Then $S=\{f\in C^\infty (M):\ f<0\}\subset \text {Im}\ T$ if and only if $\lambda _1<0$, where T is as in (4.1) with $\rho \in C^\infty (M)$, and $\text {Im}\ T=\{Tu: 0<u\in S^p_2(M)\}$.

Proof

If $S\subset \text {Im}\ T$, then $-1\in \text {Im}\ T$, i.e., $T(u)=-1$ for some $C^\infty (M)$ function $u>0$. Thus, $Lu=-u^a$. Let $\psi$ be the positive eigenfunction of L with respect to $\lambda _1$, namely, $L\psi =\lambda _1\psi$. So we have the following

$$\begin{aligned} \lambda _1\langle \psi , u\rangle _{L^2}=\langle L \psi , u\rangle _{L^2}=\langle \psi , Lu\rangle _{L^2}=-\langle \psi , u^a\rangle _{L^2}<0 \end{aligned}$$

(4.19)

which implies $\lambda _1<0$.

For the converse, assume that $\lambda _1<0$. Set $K=S\cap \text {Im}\ T$. From $L\psi =\lambda _1\psi$, we have $T(\psi )=\lambda _1\psi ^{1-a}<0$, thus $\lambda _1\psi ^{1-a}\in K$. Consequently, K is nonempty. Clearly, S is connected. In order to prove $K=S$, it is sufficient to prove K is a both open and closed subset in S. For openness, we will show that for any $u>0$, $T(u)\in K$ implies $\ker T'(u)=\ker A(u)=0$, where A(u) is defined by (4.4), and thus K is open subset of S in terms of Theorem 4.2. Let $\mu _1$ be the first eigenvalue of A(u) and $\phi$ be the corresponding positive eigenfunction. By (4.4), we have

$$\begin{aligned} \mu _1\langle \phi , u\rangle _{L^2}=\langle A(u)\phi , u\rangle _{L^2}=\langle \phi , A(u)u\rangle _{L^2}=\frac{1-a}{b_n}\langle \phi , T(u)u^a \rangle _{L^2}>0 \end{aligned}$$

(4.20)

which gives $\mu _1>0$, and so $\ker A(u)=0$. For closeness, we assume that $f_j\in K$ and $f_j\xrightarrow {C^0} f\in S$, we need prove $f\in K$, i.e., there is $u\in S^p_2(M)$ such that $T(u)=f$. Since $f_j\in K$, there exists a function $0<u_j\in C^\infty (M)$ satisfying $T(u_j)=f_j$. Let $w_j=\log {\frac{u_j}{\psi }}$ where $\psi$ is the positive eigenfunction associated with the first eigenvalue $\lambda _1$ of L. Then $w_j$ satisfies

$$\begin{aligned} -b_n\Delta _\theta w_j-b_n \nabla ^H w_j\cdot \left( \nabla ^H w_j+2\frac{\nabla ^H\psi }{\psi } \right) =-\lambda _1+f_j\psi ^{a-1}e^{(a-1)w_j} \end{aligned}$$

(4.21)

where $\cdot$ is the inner product induced by the Webster metric $g_\theta$. Considering the maximum and minimum of $w_j$ and using the classical maximum principle, it is easy to show that there are two constants $m_1, m_2>0$ independent of j such that $0<m_1\le u_j\le m_2$. Hence, applying Lemma 4.1 to the operator $L_4=-\Delta _\theta + id$, we have

$$\begin{aligned} \Vert u_j\Vert _{S^p_2(M)}\le C\Vert L_4u_j\Vert _{L^p(M)}=C\left\| \frac{1}{b_n}(f_ju_j^a-\rho u_j)+u_j\right\| _{L^p(M)}\le \hat{C} \end{aligned}$$

(4.22)

where $C, \hat{C}$ are constants independent of j. Using the compactly embedding theorem $S^p_2(M)\subset W^{1,p}(M)\subset \subset C^0(M)$ ( $p>2n+1$, cf. Theorem 19.1 of [11]), there exists a subsequence $\{u_{j_k}\}$ such that $u_{j_k}\xrightarrow {C^0} u$ as $k\rightarrow +\infty$, where $u>0$ in M since $0<m_1\le u_{j_k}\le m_2$. Moreover, the subsequence $\{u_{j_k}\}$ is a Cauchy sequence in $S^p_2(M)$, because

$$\begin{aligned} \Vert u_{j_k}-u_{j_l}\Vert _{S^p_2(M)}&\le C \Vert L_4(u_{j_k}-u_{j_l})\Vert _{L^p(M)}\nonumber \\&=C\left\| (u_{j_k}-u_{j_l})+\frac{1}{b_n}(f_{j_k}u_{j_k}^a-f_{j_l}u_{j_l}^a+\rho u_{j_l}-\rho u_{j_k})\right\| _{L^p(M)}\nonumber \\&\rightarrow 0, \end{aligned}$$

(4.23)

as $k,l \rightarrow +\infty$. Therefore, $u_{j_k}\rightarrow u$ in $S^p_2(M)$ as $k\rightarrow +\infty$. Let $k\rightarrow \infty$ in $T(u_{j_k})=f_{j_k}$, by the continuity of $T: S^p_2(M)\rightarrow L^p(M)$, we obtain $T(u)=f$, so $f\in K$. $\square$