To start with, we choose a closed Sakaki manifold $Y^{3}$ which is given by the level set $\{\varrho=r_{0}\}$ in the Calabi space $\mathcal{C}$. Denote by $\{\Lambda_{k}\}_{k=1}^{\infty}$ be the spectrum of $(Y^{3},h_{0})$, where $h_{0}$ is the induced Riemannian metric from $g_{\mathcal{C}}$. Let $\varphi_{k}\in C^{\infty}(Y^{3})$ be the eigenfunctions satisfying

(4.95)

$$\begin{split}-\Delta_{h_{0}}\varphi_{k}&=\Lambda_{k}\varphi_{k}\\ \mathcal{L}_{\partial_{\theta}}\varphi_{k}&=\sqrt{-1}j_{k}\varphi_{k},\ j_{k}\in\mathbb{N}.\end{split}$$

As computed in Section 4.1, separation of variables gives the following expansion,

(4.96)

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

where $\bm{y}\in Y^{3}$, and $u_{k}$ satisfies the equation

(4.97)

$$\frac{d^{2}u_{k}(z)}{dz^{2}}-(j_{k}^{2}z^{2}+\lambda_{k})u_{k}(z)=0,\ z\geq 1,$$

for some $j_{k}\in\mathbb{N}$ and $\lambda_{k}\geq 1$. Note that, in Section (4.1), we have shown the relations

(4.98)

$$\Lambda_{k}=\frac{\lambda_{k}}{z_{0}}+\frac{2z_{0}\cdot j_{k}^{2}}{r_{0}^{2}}$$

and $\lambda_{k}\geq j_{k}$, where $z_{0}\equiv(-\log r_{0}^{2})^{\frac{1}{2}}$. So all the estimates obtained in the previous sections directly apply here.

Since the harmonic function $u$ is smooth, so the convergence (4.96) is in the $C^{\infty}$ topology in any compact subset of $\mathcal{C}$. Immediately, for $k\in\mathbb{Z}_{+}$ and for some fixed $z_{0}>1$,

	$\displaystyle|u_{k}(z_{0})|$	$\displaystyle=\Big|\int_{Y^{3}}u\cdot\varphi_{k}\dvol_{h_{0}}\Big|=\Big|\int_{Y^{3}}u\cdot\frac{(-\Delta_{h_{0}})^{K_{0}}\varphi_{k}}{(\Lambda_{k})^{K_{0}}}\dvol_{h_{0}}\Big|$
(4.99)			$\displaystyle\leq\frac{1}{(\Lambda_{k})^{K_{0}}}\Big|\int_{Y^{3}}|(-\Delta_{h_{0}})^{K_{0}}u|\cdot\varphi_{k}\dvol_{h_{0}}\Big|\leq\frac{C_{1}}{(\Lambda_{k})^{K_{0}}},$

where

(4.100)

$$C_{1}\equiv\|u\|_{C^{2K_{0}}(Y^{3}\times\{z_{0}\})}\cdot(\Vol_{h_{0}}(Y^{3}))^{1/2}.$$

Before discussing the asymptotic behavior of the harmonic function $u$, let us give a more precise expression for each ODE solution $u_{k}$ under the growth condition $u=O(e^{\delta z})$ for $0<\delta<\underline{\delta}$. First, for every $k\in\mathbb{Z}_{+}$, there exist constants $C_{k}$ and $C_{k}^{*}$ such that

(4.101)

$$u_{k}(z)=C_{k}\cdot\mathcal{U}_{k}(z)+C_{k}^{*}\cdot\mathcal{F}_{k}(z).$$

The growth condition on $u$ gives the growth of $u_{k}$. Indeed, by assumption for any sufficiently large $z\in(2z_{0},+\infty)$ with $z_{0}>10^{6}$, it holds that

(4.102)

$$|u(z)|\leq C_{0}\cdot e^{\delta z},$$

which implies that

(4.103)

$$|u_{k}(z)|=\Big|\int_{Y^{3}}u\cdot\varphi_{k}\Big|\leq C_{0}\cdot(\Vol_{h_{0}}(Y^{3}))^{1/2}e^{\delta_{0}z}.$$

There are two cases to analyze:

First, we consider the case $j_{k}=0$, then $u_{k}$ satisfies the linear equation

(4.104)

$$u_{k}^{\prime\prime}(z)-\lambda_{k}\cdot u_{k}(z)=0.$$

We only consider $\lambda_{k}>0$. Otherwise, the solution is just a linear function. In this case, we pick the fundamental solutions

(4.105)

$$\mathcal{F}_{k}(z)\equiv e^{\sqrt{\lambda_{k}}\cdot z}\ \text{and}\ \mathcal{U}_{k}(z)\equiv e^{-\sqrt{\lambda_{k}}\cdot z},\$$

We define

(4.106)

$$\underline{\delta}\equiv\min\Big\{\sqrt{\lambda_{k}}\Big|k\in\mathbb{Z}_{+}\Big\}>0.$$

If we choose $\delta\in(0,\underline{\delta})$, then (4.103) implies

(4.107)

$$C_{k}^{*}=0$$

for each $k\in\mathbb{Z}_{+}$ which satisfies $j_{k}=0$, and hence

(4.108)

$$u_{k}(z)=C_{k}e^{-\sqrt{\lambda_{k}}\cdot z}.$$

Next, we consider the case $k\in\mathbb{Z}_{+}$ such that $j_{k}\neq 0$. Lemma 4.7 implies that $\mathcal{F}_{k}$ is growing and $\mathcal{U}_{k}$ is decaying. Therefore, apply (4.103) again, we have

(4.109)

$$C_{k}^{*}=0$$

for every $k\in\mathbb{Z}_{+}$ which satisfies $j_{k}\neq 0$.

Combining the above two cases, we conclude that if $\delta\in(0,\underline{\delta})$, then for every $k\in\mathbb{Z}_{+}$, there exists some constant $C_{k}\in\mathbb{R}$ such that

(4.110)

$$u_{k}(z)=C_{k}\cdot\mathcal{U}_{k}(z)$$

and hence

(4.111)

$$u(z,\bm{y})=\sum\limits_{k=1}^{\infty}C_{k}\cdot\mathcal{U}_{k}(z)\cdot\varphi_{k}(\bm{y}).$$

By definition, in our context $\mathcal{U}_{m,h}(z)>0$, so there is no harm to assume $u_{k}(z_{0})\neq 0$.

Now we are in a position to estimate the upper bound of the harmonic function $u$ which satisfies $u=O(e^{\delta z})$ with $0<\delta<\underline{\delta}$. We still separate in two cases. First, we consider $k\in\mathbb{Z}_{+}$ with $j_{k}=0$. For fixed $z_{0}>10^{6}$, we apply (4.108), then for every sufficiently large $z\in(2z_{0},+\infty)$,

(4.112)

$$\Big|\frac{u_{k}(z)}{u_{k}(z_{0})}\Big|=\Big|\frac{\mathcal{U}_{k}(z)}{\mathcal{U}_{k}(z_{0})}\Big|=e^{-\sqrt{\lambda_{k}}\cdot(z-z_{0})}\leq e^{-\underline{\delta}(z-z_{0})},$$

and hence

	$\displaystyle\Big|\sum_{\begin{subarray}{c}k>0\\ j_{k}\geq 1\end{subarray}}u_{k}(z)\cdot\varphi_{k}(\bm{y})\Big|$	$\displaystyle=\Big|\sum_{\begin{subarray}{c}k>0\\ j_{k}=0\end{subarray}}\frac{u_{k}(z)}{u_{k}(z_{0})}\cdot u_{k}(z_{0})\cdot\varphi_{k}(\bm{y})\Big|$
(4.113)			$\displaystyle\leq\sum_{\begin{subarray}{c}k>0\\ j_{k}=0\end{subarray}}\Big|\frac{u_{k}(z)}{u_{k}(z_{0})}\Big|\cdot|u_{k}(z_{0})|\cdot|\varphi_{k}(\bm{y})|\leq Ce^{-\underline{\delta}z/2}\cdot\sum_{\begin{subarray}{c}k>0\\ j_{k}=0\end{subarray}}\frac{1}{(\Lambda_{k})^{K_{0}-1}}.$

Next, let $k\in\mathbb{Z}_{+}$ satisfy $j_{k}\geq 1$. For fixed $z_{0}>10^{6}$ and take $z\in(2z_{0},\infty)$, then we have

(4.114)

$$\Big|\frac{u_{k}(z)}{u_{k}(z_{0})}\Big|=\Big|\frac{\mathcal{U}_{k}(z)}{\mathcal{U}_{k}(z_{0})}\Big|\leq e^{-\frac{j_{k}(z^{2}-z_{0}^{2})}{2}}\leq Ce^{-\frac{3z^{2}}{8}}.$$

Taking the sum,

	$\displaystyle\Big|\sum_{\begin{subarray}{c}k>0\\ j_{k}\geq 1\end{subarray}}u_{k}(z)\cdot\varphi_{k}(\bm{y})\Big|$	$\displaystyle=\Big|\sum_{\begin{subarray}{c}k>0\\ j_{k}\geq 1\end{subarray}}\frac{u_{k}(z)}{u_{k}(z_{0})}\cdot u_{k}(z_{0})\cdot\varphi_{k}(\bm{y})\Big|$
(4.115)			$\displaystyle\leq\sum_{\begin{subarray}{c}k>0\\ j_{k}\geq 1\end{subarray}}\Big|\frac{u_{k}(z)}{u_{k}(z_{0})}\Big|\cdot|u_{k}(z_{0})|\cdot|\varphi_{k}(\bm{y})|\leq Ce^{-\frac{3z^{2}}{8}}\sum_{\begin{subarray}{c}k>0\\ j_{k}\geq 1\end{subarray}}\frac{1}{(\Lambda_{k})^{K_{0}-1}}.$

The estimates in the above two cases imply that

	$\displaystyle|u(z,\bm{y})|$	$\displaystyle=\Big|\sum_{k=1}^{\infty}u_{k}(z)\cdot\varphi_{k}(\bm{y})\Big|$
(4.116)			$\displaystyle\leq C\Big(e^{-\underline{\delta}z/2}\sum_{\begin{subarray}{c}k>0\\ j_{k}=0\end{subarray}}\frac{1}{(\Lambda_{k})^{K_{0}-1}}+e^{-\frac{3z^{2}}{8}}\sum_{\begin{subarray}{c}k>0\\ j_{k}\geq 1\end{subarray}}\frac{1}{(\Lambda_{k})^{K_{0}-1}}\Big)\leq Ce^{-\underline{\delta}z/2}\sum_{k=1}^{\infty}\frac{1}{(\Lambda_{k})^{K_{0}-1}}.$

We can choose $K_{0}\geq 3$, applying Weyl’s law, then the above numerical series converges and hence

(4.117)

$$|u(z,\bm{y})|\leq C.$$

Therefore, there exists sufficiently large $N_{0}>10^{6}$ such that for all $z\in(N_{0},\infty)$

(4.118)

$$|u(z)-(a_{0}z+b_{0})|\leq Ce^{-2\pi z},$$

where $C$ depends on $\Vol_{h_{0}}(Y^{3})$ and $\|u\|_{C^{K_{0}}(Y^{3}\times\{z_{0}\})}$ for some fixed $z_{0}>10^{6}$ and $K_{0}\geq 3$.

∎

The proof of the lemma is identical to the convergence arguments in Proposition 4.10.

∎