The proof of the proposition is constructive. The basic strategy is to apply separation of variables to construct a solution to the equation (4.151). Given a function $v$ and for any fixed $z\geq 1$, there is an expansion over the fiber $Y^{3}$,

(4.153)

$$v(z,\bm{y})=\sum\limits_{k=1}^{\infty}v_{k}(z)\varphi_{k}(\bm{y}).$$

Separation of variables enables us to construct a formal solution

(4.154)

$$u(z,\bm{y})=\sum\limits_{k=1}^{\infty}u_{k}(z)\varphi_{k}(\bm{y})$$

to the equation (4.151), where $u_{k}$ are the particular solutions in Lemma 4.13. Since a priori the above series is defined in the $L^{2}$-topology along each fiber $Y^{3}\times\{z\}$, we need to verify the higher order convergence of the series, which will indicate that $u$ is a regular solution to (4.151).

First, we will show that the above series converges in the $C^{0}$-topology and thus $u$ is a $C^{0}$-function. The main point is to reduce the uniform convergence to the convergence of certain numerical series involving only in the eigenvalues $\{\Lambda_{k}\}_{k=1}^{\infty}$ of a definite fiber $(Y^{3},h_{0})$. Indeed, Lemma 4.12 guarantees that the solutions satisfy all the conditions in Lemma 4.13. Since we have obtained in Lemma 4.13 the uniform estimate for the ODE solutions $u_{k}$ and also in Lemma 4.14 the uniform estimate for the eigenfunctions, the $L^{2}$-expansion has the following bound,

(4.155)

$\displaystyle|u(z,\bm{y})|\leq\sum\limits_{k=1}^{\infty}|u_{k}(z)|\cdot|\varphi_{k}(\bm{y})|\leq C\sum\limits_{k=1}^{\infty}\frac{e^{\eta z}}{(\Lambda_{k})^{K_{0}-1}}.$

Since the spectrum of Laplacian $\{\Lambda_{k}\}_{k=1}^{\infty}$ obeys Weyl’s law on $(Y^{3},h_{0})$, it follows that for sufficiently large $k$,

(4.156)

$$C_{0}^{-1}k^{\frac{2}{3}}\leq|\Lambda_{k}|\leq C_{0}k^{\frac{2}{3}},$$

where $C_{0}>0$ depends only on $h_{0}$. Plugging the above asymptotics into (4.155), we have that

(4.157)

$$\sum\limits_{k=1}^{\infty}\frac{1}{(\Lambda_{k})^{K_{0}-1}}\leq C\sum\limits_{k=1}^{\infty}\frac{1}{k^{\frac{4}{3}}}<\infty$$

and hence

(4.158)

$$|u(z,\bm{y})|\leq Ce^{\eta z}.$$

Therefore, $u\in C^{0}(\mathcal{C})$ and $u$ exponentially decays.

Next, we will apply the standard elliptic regularity on the Calabi manifold to show that $u\in C^{2}$ and thus $u$ is a regular solution. For the expansions

(4.159)

$\displaystyle u(z,\bm{y})=\sum\limits_{k=1}^{\infty}u_{k}(z)\varphi_{k}(\bm{y}),\ v(z,\bm{y})=\sum\limits_{k=1}^{\infty}v_{k}(z)\varphi_{k}(\bm{y}),$

we denote by

(4.160)

$\displaystyle U_{N}(z,\bm{y})\equiv\sum\limits_{k=1}^{N}u_{k}(z)\varphi_{k}(\bm{y}),\ V_{N}(z,\bm{y})\equiv\sum\limits_{k=1}^{N}v_{k}(z)\varphi_{k}(\bm{y})$

the partial sums of $u$ and $v$ respectively. Immediately,

(4.161)

$$\Delta_{g_{\mathcal{C}}}U_{N}=V_{N}.$$

For every $\bm{x}\equiv(z,\bm{y})\in\mathcal{C}$, we will apply the elliptic regularity on the ball $B_{2}(\bm{x})\subset\mathcal{C}$ to obtain the higher regularity of $u$. For this purpose, first we prove the following claim.

The proof of the higher order convergence is exactly the same. In fact, we just need to replace $\|v\|_{C^{2K_{0}}}$ with the higher order norm $\|v\|_{C^{2K_{0}+m}}$ with $m\leq K_{0}$. Since $\Delta_{g_{\mathcal{C}}}U_{N}=V_{N}$, the standard $W^{2,p}$- implies that regularity for every $1<p<\infty$, $\|U_{N}\|_{W^{2,p}(B_{1}(\bm{x}))}\leq C_{p,\bm{x}}$. By assumption $v\in C^{3K_{0}}(\mathcal{C})$ with $K_{0}\geq 3$, we have $\|V_{N}\|_{C^{2}(B_{2}(\bm{x}))}\leq C_{\bm{x}}$. Hence the regularity of $u$ will be improved as follows, for every $1<p<\infty$,

(4.166)

$$\|U_{N}\|_{W^{4,p}(B_{1}(\bm{x}))}\leq C_{p,\bm{x}}(\|U_{N}\|_{W^{2,p}(B_{3/2}(\bm{x}))}+\|V_{N}\|_{W^{2,p}(B_{2}(\bm{x}))})\leq C_{p,\bm{x}}.$$

Now taking $p>4$ and applying the Sobolev embedding,

(4.167)

$$\|U_{N}\|_{C^{3,\alpha}(B_{1}(\bm{x}))}\leq C_{p,\bm{x}},\ \alpha\equiv 1-\frac{4}{p},$$

which implies that $U_{N}$ converges to a smooth solution $u$. Then applying the standard Schauder estimate and bootstrapping, the statement of the proposition just follows.

∎