5. A Liouville Theorem on asymptotically Calabi spaces [053E]

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5. A Liouville Theorem on asymptotically Calabi spaces

Our main goal in this section is prove a Liouville type theorem for harmonic functions on a class of complete Riemannian manifolds. This is a crucial technical component in proving the uniform injectivity estimate in Section 6 and Section 7.

First we introduce a definition.

Definition 5.1 ( $\delta$ -aysmptotically Calabi space).

Given some constant $\delta>0$ , a complete Riemannian manifold $(X^{2n},g)$ of dimension $2n$ is said to be $\delta$ -asymptotically Calabi if there exist a compact subset $K\subset X^{2n}$ , a Calabi model space $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ (as in Section 2.2) with $\dim_{\mathbb{C}}(\mathcal{C}^{n})=n$ , and a diffeomorphism

(5.1)

\Phi:\mathcal{C}^{n}\setminus K^{\prime}\rightarrow X^{2n}\setminus K

with $K^{\prime}=\{|\xi|\geq C\}\subset\mathcal{C}^{n}$ (for some $C>0$ ) such that for all $k\geq 0$ ,

(5.2)

|\nabla_{g_{\mathcal{C}^{n}}}^{k}(\Phi^{*}g-g_{\mathcal{C}^{n}})|_{g_{\mathcal{C}^{n}}}=O(e^{-\delta z^{\frac{n}{2}}})\ \text{as}\ z\to+\infty,

where $z\equiv(-{\log|\xi|^{2}})^{\frac{1}{n}}$ denotes the natural moment map coordinate on $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ .

Now we state the main theorem to be proved in this section

Theorem 5.2 (Liouville Theorem).

Let $(X^{2n},g)$ be a complete Riemannian manifold of dimension $2n$ which has non-negative Ricci curvature and is $\delta$ -asymptotically Calabi for some $\delta>0$ . Then there exists a $\epsilon_{X}>0$ depending on $(X^{2n},g)$ such that if $u$ is a harmonic function on $(X^{2n},g)$ satisfying

(5.3)

|u|=O(e^{\epsilon_{X}\cdot z^{\frac{n}{2}}}),\ z\to+\infty,

then $u$ is a constant.

Remark 5.2.1.

In the gluing construction in Section 7.3 we shall apply Theorem 5.2 to the Tian-Yau spaces (c.f. Section 7.2).

Remark 5.2.2.

The special case $n=2$ of Theorem 5.2 was proved in [HSVZ18].

This section is organized as follows. In Section 5.1 we recall the separation of variables in [HSVZ18] and write down the ODE for the Laplace equation on the Calabi model space. This ODE is not familiar at first sight, which leads us to perform the change of variables to transform the ODE to known ones. Depending on whether the Fourier mode with respect to the natural $S^{1}$ -action vanishes or not, we shall get either modified Bessel equations, or confluent hypergeometric equations. Solutions to these equations have known asymptotics, but for our analysis we need uniform estimates. These will be done in Section 5.2 and 5.3. The key technical ingredients involve estimating exponential integrals using Laplace’s method. With these preparations, in Section 5.4, we show a harmonic function on the Calabi model space which has slowly exponential growth at infinity must decompose as the sum of the linear function in $z$ (the moment coordinate in the Calabi model space, c.f. Section 2.2) and an exponentially decaying terms. In Section 5.5, we show the Poisson equation on the Calabi model space can be solved using separation of variables for a function with certain growth control at infinity. Section 5.6 is dedicated to the proof of Theorem 5.2. First transplanting the harmonic function to an approximately harmonic function on the Calabi model space, then correct this to a harmonic function by solving a Poisson equation. These imply the function $u$ grows at most linearly in $z$ . The later then implies $du$ is a decaying harmonic 1-form, and must vanish by applying the Bochner technique (which uses the assumption $\Ric(g)\geq 0$ ) and maximum principle.

Now we list some notations and make basic conventions for the convenience of later discussions in this section:

•

The Laplace-Beltrami operator $\Delta_{g}$ acting on functions is given by

(5.4) $\Delta_{g}u\equiv\Tr_{g}(\nabla^{2}u).$

For example, $\Delta_{\mathbb{R}^{n}}\equiv\sum\limits_{j=1}^{n}\frac{\partial^{2}}{\partial x_{j}^{2}}$ . Notice this is different from the Laplacian $\Delta$ used in Section 2 to 4 which is the Hodge Laplacian.
•

Let $k\in\mathbb{Z}_{+}$ and $x\in\mathbb{R}$ , we define

(5.5) $(x)_{k}\equiv\prod\limits_{m=1}^{k}(x+m-1)\ and\ (x)_{0}\equiv 1.$
•
Given two positive functions $f(z)$ and $g(z)$ defined on $\mathbb{R}_{+}$ , then
1. (1)
  
  We say $f(z)\sim g(z)$ if
  
  (5.6) $\lim\limits_{z\to+\infty}\frac{f(z)}{g(z)}=1.$
2. (2)
  
  Given two $C^{1}$ -functions $f(y)$ and $g(y)$ , their Wronskian is denoted by
  
  (5.7) $\mathcal{W}(f,g)(y)\equiv f(y)g^{\prime}(y)-f^{\prime}(y)g(y).$

5.1. Separation of variables and ODE reduction

Let $\mathcal{C}^{n}$ be a Calabi model space applied to an ample line bundle $L$ over an $n-1$ dimensional compact Calabi-Yau manifold $(D,\omega_{D},\Omega_{D})$ , then the Kähler form of the Calabi metric is given by

(5.8)

\omega_{\mathcal{C}^{n}}=\frac{n}{n+1}\sqrt{-1}\partial\bar{\partial}(-\log|\xi|^{2})^{\frac{n+1}{n}},

which is well-defined for $|\xi|<1$ . In order to carry out separation of variables, we will study the local representation of the Laplace operator $\Delta_{\mathcal{C}^{n}}$ on $\mathcal{C}^{n}$ . The authors have developed separation of variables in Section 4.1 of [HSVZ18], so we just briefly review the computations and basic estimates obtained there.

Let $\{w_{i}\}_{i=1}^{n-1}$ be some local holomorphic coordinates on $D$ , and fix a local holomorphic trivialization $e_{0}$ of the line bundle $L$ with $|e_{0}|^{2}=e^{-\psi}$ , where $\psi:D\to\mathbb{R}$ is a smooth function. So we get local holomorphic coordinates $(\underline{w},\zeta)\equiv(w_{1},\ldots,w_{n-1},\sigma)$ on $\mathcal{C}^{n}$ by writing a point $\xi\in\mathcal{C}^{n}$ as $\xi=\eta e_{0}(\underline{w})$ , then $|\xi|^{2}=|\eta|^{2}e^{-\psi}$ . We may assume $\psi(0)=1$ , $d\psi(0)=0$ and $\sqrt{-1}\partial\bar{\partial}\psi=\omega_{D}$ . Let $\pi:\mathcal{C}^{n}\rightarrow D$ be the obvious projection map. Denote

(5.9)

\rho\equiv|\xi|,

then we can write

(5.10)

w=\rho e^{\frac{\psi}{2}+\sqrt{-1}\theta},

where $\partial_{\theta}$ generates the natural $S^{1}$ -rotation on the total space of $L$ .

Now we fix some $\rho_{0}\in(0,1)$ , and define $(Y^{2n-1},h_{0})$ to be the level set $\{\rho=\rho_{0}\}$ endowed with the induced Riemannian metric $h_{0}$ . We denote by $\{\Lambda_{k}\}_{k=0}^{\infty}$ the spectrum of $\Delta_{h_{0}}$ with $\Lambda_{0}\equiv 0$ , and let $\{\varphi_{k}\}_{k=0}^{\infty}$ be an orthonormal basis of (complex-valued) eigenfunctions which are homogeneous under the $S^{1}$ action and with

(5.11)

-\Delta_{h_{0}}\varphi_{k}=\Lambda_{k}\cdot\varphi_{k}.

From [HSVZ18], Section 4.1 we know that $\Lambda_{k}$ can be always represented as follows,

(5.12)

\Lambda_{k}=\frac{\lambda_{k}}{z_{0}}+\frac{nz_{0}^{n-1}\cdot j_{k}^{2}}{r_{0}^{2}}

such that $j_{k}\in\mathbb{N}$ and

(5.13)

\lambda_{k}\geq\frac{(n-1)\cdot j_{k}}{2}.

Notice $j_{k}$ and $\lambda_{k}$ have geometric meanings as explained in [HSVZ18], Section 4.1. Namely, $\varphi_{k}$ has weight $\pm j_{k}$ with respect to the $S^{1}$ -action (notice the weight of $\bar{\varphi}_{k}$ is negative the weight of $\varphi_{k}$ ), and $\varphi_{k}$ corresponds to a smooth section of the induced complex line bundle over $D$ , which is an eigenfunction of the $\bar{\partial}$ -Hodge Laplacian with eigenvalue $\lambda_{k}$ . In particular $\lambda_{0}=j_{0}=0$ and $\varphi_{0}$ is a constant. Moreover when $j_{k}=0$ , $\varphi_{k}$ corresponds to an eigenfunction on $D$ and

(5.14)

\underline{\lambda}\equiv\inf\{\lambda_{k}>0|j_{k}=0,k\in\mathbb{Z}_{+}\}>0.

Now we carry out separation of variables for the Laplace equation on $\Delta_{\mathcal{C}^{n}}$ . Let $u$ be a harmonic function on the model space $\mathcal{C}^{n}$ , namely,

(5.15)

\Delta_{\mathcal{C}^{n}}u=0

For every fixed $z$ , we can write the $L^{2}$ -expansion along the fiber $Y^{2n-1}$ ,

(5.16)

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

The computations in [HSVZ18] tell us that for each $k\in\mathbb{N}$ , $u_{k}(z)$ satisfies the differential equation

(5.17)

\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=0,\ z\geq 1.

We also consider the Poisson equation

(5.18)

\Delta_{\mathcal{C}^{n}}u=v.

Take the $L^{2}$ -expansion of $v$ in the direction of the cross section $Y^{2n-1}$ ,

(5.19)

v(z,\bm{y})=\sum\limits_{k=1}^{\infty}\xi_{k}(z)\cdot\varphi_{k}(\bm{y}),

then the same procedure of separation of variables leads to an ordinary differential equation

(5.20)

\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=z^{n-1}\cdot\xi_{k}(z),\ z\geq 1.

Since we will study the solutions (5.16) and (5.19) in terms of the fiber-wise $L^{2}$ -expansions, so there are two fundamental ingredients to analyze: First, in order to show the $L^{2}$ -expansions in fact converge, we need to obtain some uniform estimates for the ODE solutions which are independent of the subscript $k\in\mathbb{N}$ . The other basic aspect is to understand the asymptotics of the linearly independent solutions $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ as $z\to+\infty$ , which in turn gives the asymptotics of the solutions (5.16) and (5.19).

Technically speaking, we will study the solutions to (5.17) and (5.20) in two different cases: $j_{k}=0$ and $j_{k}\neq 0$ . The first step is to understand the solutions to homogeneous equation (5.17). Notice that, by using the change of variables $\zeta\equiv z^{n}$ , (5.17) will become a homogeneous equation with linear coefficients, so that we can apply the theory of special functions to obtain some effective estimates for the solutions. Now letting

(5.21)

\displaystyle\begin{cases}\zeta=-\log r^{2}=z^{n}\\ w_{k}(\zeta)\equiv u_{k}(z)=u_{k}(\zeta^{\frac{1}{n}}),\end{cases}

we have

(5.22)

\zeta\cdot\frac{d^{2}w_{k}(\zeta)}{d\zeta^{2}}+(1-\frac{1}{n})\frac{dw_{k}(\zeta)}{d\zeta}-(\frac{j_{k}^{2}}{4}\cdot\zeta+\frac{\lambda_{k}}{n})w_{k}(\zeta)=0.

In the first case $j_{k}=0$ , we make the transformation of the above solution $w(\zeta)$ as follows,

(5.23)

\begin{cases}y=2\sqrt{\frac{\lambda}{n}}\cdot\zeta^{\frac{1}{2}}\geq 0\\ w_{k}(\zeta)=\zeta^{\frac{1}{2n}}\cdot\mathcal{B}\Big(2\sqrt{\frac{\lambda}{n}}\cdot\zeta^{\frac{1}{2}}\Big),\end{cases}

then the function $\mathcal{B}(y)$ satisfies the modified Bessel equation,

(5.24)

y^{2}\cdot\frac{d^{2}\mathcal{B}(y)}{dy^{2}}+y\cdot\frac{d\mathcal{B}(y)}{dy}-(y^{2}+\frac{1}{n^{2}})\cdot\mathcal{B}(y)=0.

In the latter case $j_{k}\neq 0$ , we make the following transformation

(5.25)

\displaystyle\begin{cases}y=-j_{k}\cdot\zeta\leq 0\\ w_{k}(\zeta)=e^{\frac{j_{k}\cdot\zeta}{2}}\cdot\mathcal{J}(-j_{k}\cdot\zeta),\end{cases}

then $\mathcal{J}(y)$ satisfies the confluent hypergeometric equation,

(5.26)

y\cdot\frac{d^{2}\mathcal{J}(y)}{dy^{2}}+(\fa-y)\cdot\frac{d\mathcal{J}(y)}{dy}-\fb\cdot\mathcal{J}(y)=0,

where

(5.27)

\displaystyle\begin{cases}\fa=1-\frac{1}{n}\\ \fb=\frac{1}{2}(1-\frac{1}{n})-\frac{\lambda_{k}}{j_{k}\cdot n}.\end{cases}

It is straightforward to see that $\alpha\in(0,1)$ and $\beta\in(-\infty,0]$ .

Remark 5.2.3.

The above ODE transformations were first used by [KK10].

Remark 5.2.4.

The homogeneous equation (5.17) was studied by the authors in the special case $n=\dim_{\mathbb{C}}(\mathcal{C}^{n})=2$ . When $j_{k}=0$ , (5.17) has standard solutions given by exponential functions. When $j_{k}>0$ , the transformation was chosen as

(5.28)

\begin{cases}y=j_{k}^{\frac{1}{2}}\cdot z^{\frac{n}{2}}\\ u_{k}(z)=e^{-\frac{j_{k}z^{n}}{2}}\cdot Q(j_{k}^{\frac{1}{2}}\cdot z^{\frac{n}{2}}).\end{cases}

We refer the readers to Section 4 of [HSVZ18] for more details. In the special case $n=2$ , $Q(y)$ is an Hermite function which satisfies the Hermite differential equation

(5.29)

\frac{d^{2}Q(y)}{dy^{2}}-2y\frac{dQ(y)}{dy}-2(h+1)Q(y)=0.

The key tool to prove the estimates for $Q$ essentially relies on its integral representation formula. However, when $n>2$ , if we perform the transformation as (5.28) then the resulting equation for $Q$ is more complicated to study. It turns out the transformation (5.25) is a more suitable choice.

5.2. The case $j_{k}=0$ : uniform estimates and asymptotics

In this subsection, we consider the case $j_{k}=0$ so (5.17) reduces to the homogeneous ODE

(5.30)

\frac{d^{2}u_{k}(z)}{dz^{2}}-n\lambda_{k}\cdot z^{n-2}u_{k}(z)=0,z\geq 1.

When $\lambda_{k}=0$ the equation has trivial solutions given by linear functions. In this subsection we always assume $\lambda_{k}\neq 0$ . As discussed in Section 5.1 under the change of variables given by (5.21) and (5.23), we are lead to study the modified Bessel equation.

(5.31)

y^{2}\cdot\frac{d^{2}\mathcal{B}(y)}{dy^{2}}+y\cdot\frac{d\mathcal{B}(y)}{dy}-(y^{2}+\nu^{2})\cdot\mathcal{B}(y)=0,\ \nu\in\mathbb{R}.

There are two linearly independent solutions $I_{\nu}(y)$ and $K_{\nu}(y)$ called the modified Bessel functions, whose definition is given in Appendix A. These yield two linearly independent solutions to the original equation (5.17), given by

(5.32)

\begin{cases}\mathcal{G}_{k}(z)\equiv z^{\frac{1}{2}}\cdot I_{\frac{1}{n}}\Big(2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}\Big),\\ \mathcal{D}_{k}(z)\equiv z^{\frac{1}{2}}\cdot K_{\frac{1}{n}}\Big(2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}\Big).\end{cases}

First by the definition of $I_{\nu}$ and $K_{\nu}$ we can compute its Wronskian

Proposition 5.3.

Let $\nu>0$ and $y>0$ , then

(5.33)

\mathcal{W}(I_{\nu}(y),K_{\nu}(y))=-\frac{1}{y}.

Proof.

Since $I_{\nu}$ and $K_{\nu}$ satisfy

(5.34)		$\displaystyle\frac{d}{dy}(y\cdot I_{\nu}^{\prime}(y))-(y+\frac{\nu^{2}}{y})I_{\nu}(y)=0,$
(5.35)		$\displaystyle\frac{d}{dy}(y\cdot K_{\nu}^{\prime}(y))-(y+\frac{\nu^{2}}{y})K_{\nu}(y)=0.$

This implies that

(5.36)

K_{\nu}(y)\cdot\frac{d}{dy}(y\cdot I_{\nu}^{\prime}(y))-I_{\nu}(y)\cdot\frac{d}{dy}(y\cdot K_{\nu}^{\prime}(y))=0,

and hence

(5.37)

\frac{d}{dy}\Big(y\cdot\mathcal{W}(I_{\nu}(y),K_{\nu}(y))\Big)=0.

Therefore, $y\cdot\mathcal{W}(I_{\nu}(y),K_{\nu}(y))$ is a constant.

Next, we will compute this constant which equals the limit of $y\cdot\mathcal{W}(I_{\nu}(y),K_{\nu}(y))$ as $y\to 0$ . By definition,

(5.38)

\lim\limits_{y\to 0}I_{\nu}(y)\Big/\Big(\frac{y^{\nu}}{\Gamma(\nu+1)\cdot 2^{\nu}}\Big)=1,\ \lim\limits_{y\to 0}K_{\nu}(y)\Big/\Big(\frac{\pi}{2\sin(\nu\pi)}\cdot\frac{2^{\nu}\cdot y^{-\nu}}{\Gamma(1-\nu)}\Big)=1.

Notice that

(5.39)

\Gamma(\nu+1)\Gamma(1-\nu)=\nu\Gamma(\nu)\Gamma(1-\nu)=\frac{\nu\pi}{\sin(\nu\pi)},

then it is straightforward that

(5.40)

\lim\limits_{y\to 0}y\cdot(I_{\nu}(y)K_{\nu}^{\prime}(y)-K_{\nu}(y)I_{\nu}^{\prime}(y))=-1.

This completes the proof. ∎

Corollary 5.3.1.

For any $z>0$ , we have

(5.41)

\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=-\frac{n}{2}.

Proof.

Applying Lemma 5.3 and the chain rule,

(5.42)

\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=-z\cdot n(\frac{\lambda_{k}}{n})^{\frac{1}{2}}\cdot z^{\frac{n}{2}-1}\cdot\frac{1}{2(\frac{\lambda_{k}}{n})^{\frac{1}{2}}z^{\frac{n}{2}}}=-\frac{n}{2}.

∎

By Corollary A.8.1, we also have the asymptotics of the solutions for each fixed $k$ .

Lemma 5.4.

As $z\rightarrow\infty$ we have

(5.43)		$\displaystyle\mathcal{G}_{k}(z)$	$\displaystyle\sim\frac{1}{2\sqrt{\pi}\cdot(\frac{\lambda_{k}}{n})^{\frac{1}{4}}}\cdot\frac{e^{2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}}}{z^{\frac{n-2}{4}}},$
(5.44)		$\displaystyle\mathcal{D}_{k}(z)$	$\displaystyle\sim\frac{\sqrt{\pi}}{2(\frac{\lambda_{k}}{n})^{\frac{1}{4}}}\cdot\frac{e^{-2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}}}{z^{\frac{n-2}{4}}}.$

In our proof of Theorem 5.2, we need uniform estimates (with respect to $k$ and $z$ ) on $\mathcal{G}_{k}$ and $\mathcal{D}_{k}$ . So in the following, we will prove uniform estimates for $I_{\nu}(y)$ and $K_{\nu}(y)$ for all $y\geq 1$ . Notice that, in this subsection we are interested in the case $j_{k}=0$ which corresponds to $\nu=\frac{1}{n}$ . However, the following formulae and estimates work for general $\nu\in\mathbb{R}$ , and we shall need the case $\nu=-\frac{1}{n}$ in Section 5.3. We will apply appropriate integral representations of $I_{\nu}(y)$ and $K_{\nu}(y)$ to study their upper bounds and asymptotic behaviors. The following integral formulae will play a fundamental role in our estimates: Let $y>0$ , then by Lemma A.1, we have

(5.45)

I_{\nu}(y)=\frac{1}{\pi}\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta-\frac{\sin(\nu\pi)}{\pi}\int_{0}^{\infty}e^{-y\cosh t-\nu t}dt

and

(5.46)

K_{\nu}(y)=\int_{0}^{\infty}e^{-y\cosh t}\cosh(\nu t)dt.

Proposition 5.5.

The following hold

(1)

For all $\nu\in\mathbb{R}$ , there is a constant $C(\nu)>1$ such that

(5.47)		$\displaystyle C^{-1}(\nu)\cdot\frac{e^{-y}}{\sqrt{y}}\leq K_{\nu}(y)\leq C(\nu)\cdot\frac{e^{-y}}{\sqrt{y}},\qquad y\geq 1;$
(5.48)		$\displaystyle I_{\nu}(y)\leq\begin{cases}C(\nu)\cdot\frac{e^{y}}{\sqrt{y}},&y\geq 1,\\ C(\nu)\cdot y^{\nu},&0<y\leq 1.\end{cases}$

(2)

For all $\nu>-1$ , we have

(5.49)

\displaystyle I_{\nu}(y)\geq\begin{cases}C(\nu)^{-1}\cdot\frac{e^{y}}{\sqrt{y}},&y\geq 1,\\ C(\nu)^{-1}\cdot y^{\nu},&0<y\leq 1.\end{cases}

Proof.

In the proof the constant $C(\nu)$ may vary from line to line. First we prove Item (1). To start with, we prove the upper bound estimate for the solution $K_{\nu}(y)$ . Notice that $\cosh(t)\geq 1+\frac{t^{2}}{2}$ for every $t\geq 0$ , then

(5.50)	$\displaystyle K_{\nu}(y)$	$\displaystyle=$	$\displaystyle\int_{0}^{\infty}e^{-y\cosh t}\cosh(\nu t)dt$
		$\displaystyle\leq$	$\displaystyle\int_{0}^{\infty}e^{-y(1+\frac{t^{2}}{2})}\cosh(\nu t)dt$
		$\displaystyle=$	$\displaystyle\frac{e^{-y}}{2}\Big(\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}+\nu t}dt+\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}-\nu t}dt\Big).$

Now we prove that, for $y\geq 1$ and $\nu\in\mathbb{R}$ ,

(5.51)

\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}+\nu t}dt\leq C(\nu)\cdot\frac{1}{\sqrt{y}}.

It is by straightforward computation that

(5.52)		$\displaystyle\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}+\nu t}dt$	$\displaystyle=$	$\displaystyle\int_{0}^{\infty}e^{-(\sqrt{\frac{y}{2}}t-\frac{\nu}{2}\sqrt{\frac{2}{y}})^{2}+\frac{\nu^{2}}{2y}}dt$
(5.52)			$\displaystyle=$	$\displaystyle\sqrt{\frac{2}{y}}\cdot e^{\frac{\nu^{2}}{2y}}\int_{-\frac{\nu}{2}\sqrt{\frac{2}{y}}}^{\infty}e^{-\tau^{2}}d\tau,$

where $\tau=\sqrt{\frac{y}{2}}t-\frac{\nu}{2}\sqrt{\frac{2}{y}}$ . Notice that

(5.53)

\int_{-\frac{\nu}{2}\sqrt{\frac{2}{y}}}^{\infty}e^{-\tau^{2}}d\tau\leq\int_{-\infty}^{\infty}e^{-\tau^{2}}d\tau=\sqrt{\pi}.

Moreover, the assumption $y\geq 1$ implies $e^{\frac{\nu^{2}}{2y}}\leq e^{\frac{\nu^{2}}{2}}$ , so it holds that

(5.54)

\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}+\nu t}dt\leq C(\nu)\cdot\frac{1}{\sqrt{y}}.

Similarly,

(5.55)

\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}-\nu t}dt\leq C(\nu)\cdot\frac{1}{\sqrt{y}}.

Therefore, we have

(5.56)

K_{\nu}(y)\leq C(\nu)\cdot\frac{e^{-y}}{\sqrt{y}},

where $C(\nu)>0$ depends only on $\nu$ .

Next we prove the lower bound estimate for $K_{\nu}(y)$ . The integral representation of $K_{\nu}(y)$ can be written as follows,

(5.57)

\displaystyle K_{\nu}(y)

\displaystyle=\frac{e^{-y}}{2}\Big(\int_{0}^{\infty}e^{-y(\cosh t-1)+\nu t}dt+\int_{0}^{\infty}e^{-y(\cosh t-1)-\nu t}dt\Big).

We will give lower bound estimates for the above two integrals respectively. It is straightforward that

(5.58)

\int_{0}^{\infty}e^{-y(\cosh t-1)+\nu t}dt\geq\int_{0}^{1}e^{-y(\cosh t-1)+\nu t}dt=\int_{0}^{1}e^{-\frac{y\cdot\cosh(\theta_{t})t^{2}}{2}+\nu t}dt

for some $0\leq\theta_{t}\leq 1$ , which implies that

(5.59)

\int_{0}^{\infty}e^{-y(\cosh t-1)+\nu t}dt\geq\int_{0}^{1}e^{-2yt^{2}+\nu t}dt.

The calculations in the last step imply that for $y\geq 1$ ,

(5.60)

\frac{C^{-1}(\nu)}{\sqrt{y}}\leq\int_{0}^{1}e^{-2yt^{2}+\nu t}\leq\frac{C(\nu)}{\sqrt{y}}.

Therefore,

(5.61)

\int_{0}^{1}e^{-y(\cosh t-1)+\nu t}dt\geq\frac{C^{-1}(\nu)}{\sqrt{y}}.

By the same calculations,

(5.62)

\int_{0}^{1}e^{-y(\cosh t-1)-\nu t}dt\geq\frac{C^{-1}(\nu)}{\sqrt{y}}.

This completes the proof of (5.47).

To see (5.48) we first assume $y\geq 1$ . We use the integral representation

(5.63)

I_{\nu}(y)=\frac{1}{\pi}\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta-\frac{\sin(\nu\pi)}{\pi}\int_{0}^{\infty}e^{-y\cosh t-\nu t}dt.

To estimate the second term, we use the integral estimate

(5.64)

\int_{0}^{\infty}e^{-y\cosh t-\nu t}dt\leq e^{-y}\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}-\nu t}dt\leq C(\nu)\cdot\frac{e^{-y}}{\sqrt{y}}.

Next, we estimate the first term of $I_{\nu}(y)$ . Since for every $\theta\in[0,\frac{\pi}{3}]$ ,

(5.65)

\cos\theta\leq 1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{24}\leq 1-\frac{\theta^{2}}{4},

then

(5.66)

\displaystyle\Big|\frac{1}{\pi}\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta\Big|

\displaystyle\leq

\displaystyle\frac{1}{\pi}\int_{0}^{\frac{\pi}{3}}e^{y\cos\theta}d\theta+\frac{1}{\pi}\int_{\frac{\pi}{3}}^{\pi}e^{y\cos\theta}d\theta

Estimating the right hand side separately, we get

\displaystyle\Big|\frac{1}{\pi}\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta\Big|

\displaystyle\leq

\displaystyle\frac{e^{y}}{\pi}\int_{0}^{\frac{\pi}{3}}e^{-\frac{y\cdot\theta^{2}}{4}}d\theta+\frac{2e^{\frac{y}{2}}}{3}\leq\frac{2e^{y}}{\sqrt{\pi}\cdot\sqrt{y}}+\frac{2e^{\frac{y}{2}}}{3}\leq\frac{10e^{y}}{\sqrt{y}}.

Therefore,

(5.67)

I_{\nu}(y)\leq\frac{10e^{y}}{\sqrt{y}}+\frac{C(\nu)\cdot e^{-y}}{\sqrt{y}}\leq\frac{C(\nu)\cdot e^{y}}{\sqrt{y}}.

Now we assume $y\in(0,1]$ . Since $I_{\nu}$ is smooth, we only need to analyze the behavior of $I_{\nu}(y)$ as $y\to 0$ . By the definition of $I_{\nu}(y)$ we see if $\nu\geq 0$ or $\nu$ is a negative integer, $\lim\limits_{y\rightarrow 0}I_{\nu}(y)=0$ . For any $\nu<0$ , we have

(5.68)

\lim\limits_{y\to 0}I_{\nu}(y)\Big/\frac{(\frac{y}{2})^{\nu}}{\Gamma(\nu+1)}=1.

Therefore, for any $y\in(0,1]$ ,

(5.69)

I_{\nu}(y)\leq C(\nu)\cdot y^{\nu}.

Now we prove Item (2). First we observe that by the definition of $I_{\nu}$ using power series, when $\nu\in(-1,0)$ , $I_{\nu}(y)$ is positive for all $y\in(0,\infty)$ . So the lower bound of $I_{\nu}$ for $y\in(0,1]$ follows just as before. Now we assume $y\geq 1$ . To get the lower bound on $I_{\nu}$ , it suffices to get the lower bound on the first term of (5.63). Suppose $\nu\neq 0$ , denote $\eta_{\nu}=\min(\pi,\frac{\pi}{3|\nu|})$ , then we divide the integral into two parts

(5.70)

\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta=\int_{0}^{\eta_{\nu}}e^{y\cos\theta}\cos(\nu\theta)d\theta+\int_{\eta_{\nu}}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta.

Since $\cos\theta\geq 1-\frac{\theta^{2}}{2}$ we get

(5.71)

\int_{0}^{\eta_{\nu}}e^{y\cos\theta}\cos(\nu\theta)d\theta\geq\frac{1}{2}e^{y}\int_{0}^{\eta_{\nu}}e^{-\frac{\theta^{2}}{2}y}d\theta\geq C(\nu)\frac{e^{y}}{\sqrt{y}},

and for the second term we have

(5.72)

\Big|\int_{\eta_{\nu}}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta\Big|\leq\int_{\eta_{\nu}}^{\pi}e^{y\cos\theta}d\theta\leq(\pi-\eta_{\nu})e^{\cos(\eta_{\nu})y}.

So we get

(5.73)

I_{\nu}(y)\geq C^{-1}(\nu)\frac{e^{y}}{\sqrt{y}}.

For $\nu=0$ the argument is similar. This completes the proof of Item (1).

∎

Converting the above back to $\mathcal{G}_{k}$ and $\mathcal{D}_{k}$ , we obtain

Corollary 5.5.1.

There is a dimensional constant $C(n)>0$ such that $z\geq 2^{-\frac{2}{n}}n^{\frac{1}{n}}\underline{\lambda}^{-\frac{1}{n}}$ , we have

(5.74)		$\displaystyle\frac{C^{-1}(n)}{\lambda_{k}^{\frac{1}{4}}}\cdot\frac{e^{-2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}}}{z^{\frac{n-2}{4}}}$	$\displaystyle\leq\mathcal{D}_{k}(z)\leq\frac{C(n)}{\lambda_{k}^{\frac{1}{4}}}\cdot\frac{e^{-2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}}}{z^{\frac{n-2}{4}}},$
(5.75)		$\displaystyle\frac{C^{-1}(n)}{\lambda_{k}^{\frac{1}{4}}}\cdot\frac{e^{2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}}}{z^{\frac{n-2}{4}}}$	$\displaystyle\leq\mathcal{G}_{k}(z)\leq\frac{C(n)}{\lambda_{k}^{\frac{1}{4}}}\cdot\frac{e^{2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}}}{z^{\frac{n-2}{4}}}.$

5.3. The case $j_{k}\neq 0$ : uniform estimates and asymptotics

In this subsection, we consider the case $j_{k}\neq 0$ of the homogeneous equation

(5.76)

\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=0.\ z\geq 1,

Under the change of variables given by (5.21) and (5.25), the above equation is transformed into the confluent hypergeometric equation,

(5.77)

y\cdot\frac{d^{2}\mathcal{J}(y)}{dy^{2}}+(\fa-y)\cdot\frac{d\mathcal{J}(y)}{dy}-\fb\cdot\mathcal{J}(y)=0,\ y<0,

where

(5.78)

\displaystyle\begin{cases}\fa=1-\frac{1}{n}\\ \fb=\frac{1}{2}(1-\frac{1}{n})-\frac{\lambda_{k}}{j_{k}\cdot n}.\end{cases}

Since we have shown in Section 5.1 that $\lambda_{k}\geq\frac{j_{k}(n-1)}{2}$ , we have that

(5.79)

\beta\leq 0\ \text{and}\ \ \alpha-\beta\geq 1-\frac{1}{n}>0.

According to the discussion in Appendix A, in our case $y<0$ , the confluent hypergeometric equation (5.77) has two linearly independent solutions

(5.80)

\Ku(\beta,\alpha,y)\equiv\sum\limits_{k=0}^{\infty}\frac{(\beta)_{k}}{(\alpha)_{k}}\cdot\frac{y^{k}}{k!}

and

(5.81)

\Tri(\beta,\alpha,y)\equiv\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt.

By Item (3) of Lemma A.3, as $y\to-\infty$ , $\Tri(y)$ is a decaying solution to (5.77) for every $\alpha>\beta$ , while Lemma A.5 shows that, in the case $\beta<0$ , the solution $\Ku(y)$ is growing of certain polynomial rate as $y\to-\infty$ . These then yield two linearly independent solutions to the homogeneous equation (5.76),

(5.82)

\begin{cases}\mathcal{G}_{k}(z)=e^{\frac{j_{k}z^{n}}{2}}\cdot\Ku(\beta,\alpha,-j_{k}z^{n}),\\ \mathcal{D}_{k}(z)=e^{\frac{j_{k}z^{n}}{2}}\cdot\Tri(\beta,\alpha,-j_{k}z^{n}).\end{cases}

First we can compute the Wronskian

Proposition 5.6.

For every $k\in\mathbb{N}$ , the Wronskian of $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ is a constant given by

(5.83)

\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot j_{k}^{\frac{1}{n}}.

Proof.

Since $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ solve the homogeneous equation

(5.84)

\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=0

which misses the first order term. Immediately, for all $z\geq 0$ ,

(5.85)

\frac{d}{dz}\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=0,

which implies that the Wronskian $\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))$ is a constant. So it suffices to calculate it at $z=0$ . By the definition of the Wronskian,

	$\displaystyle\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=$	$\displaystyle e^{j_{k}z^{n}}\cdot\Big(\Ku(\beta,\alpha,-j_{k}z^{n})\cdot\frac{d}{dz}\Tri(\beta,\alpha,-j_{k}z^{n})$
(5.86)			$\displaystyle-\frac{d}{dz}\Ku(\beta,\alpha,-j_{k}z^{n})\cdot\Tri(\beta,\alpha,-j_{k}z^{n})\Big).$

To calculate $\frac{d}{dz}\Tri(\beta,\alpha,-j_{k}z^{n})$ , we will apply Kummer’s transformation law to relate $\Tri$ and $\Ku$ , that is,

(5.87)		$\displaystyle\Tri(\beta,\alpha,-j_{k}z^{n})$
	$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\mathcal{U}(\alpha-\beta,\alpha,{j_{k}}z^{n})$
	$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\Big(\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\alpha-\beta,\alpha,{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot({j_{k}}z^{n})^{1-\alpha}\Ku(1-\beta,2-\alpha,{j_{k}}z^{n})\Big)$
	$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\Big(\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\alpha-\beta,\alpha,{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}z\cdot\Ku(1-\beta,2-\alpha,{j_{k}}z^{n})\Big)$
	$\displaystyle=$	$\displaystyle\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\beta,\alpha,-{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}z\cdot\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n}).$

So it follows that

(5.88)		$\displaystyle\frac{d}{dz}\Tri(\beta,\alpha,-{j_{k}}z^{n})$
	$\displaystyle=$	$\displaystyle\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\frac{d}{dz}\Ku(\beta,\alpha,-{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}\cdot\Big(\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})$
	$\displaystyle+$	$\displaystyle z\cdot\frac{d}{dz}\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})\Big).$

Since $n\geq 2$ , it directly follows from the definition of $\Ku$ that

(5.89)		$\displaystyle\frac{d}{dz}\Big\|_{z=0}\Ku(\beta,\alpha,-{j_{k}}z^{n})=0,$
(5.90)		$\displaystyle\frac{d}{dz}\Big\|_{z=0}\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})=0.$

Therefore,

(5.91)		$\displaystyle\frac{d}{dz}\Big\|_{z=0}\Tri(\beta,\alpha,-{j_{k}}z^{n})$	$\displaystyle=$	$\displaystyle\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}\cdot\Ku(1-\alpha+\beta,2-\alpha,0)$
(5.91)			$\displaystyle=$	$\displaystyle\frac{\Gamma(\alpha-1)\cdot{j_{k}}^{\frac{1}{n}}}{\Gamma(\alpha-\beta)}.$

Now evaluate (5.86) at $z=0$ , we have

(5.92)

\mathcal{W}(\mathcal{G}_{k},\mathcal{D}_{k})(z)=\mathcal{W}(\mathcal{G}_{k},\mathcal{D}_{k})(0)=\frac{d}{dz}\Big|_{z=0}\Tri(\beta,\alpha,-{j_{k}}z^{n})=\frac{\Gamma(\alpha-1)\cdot{j_{k}}^{\frac{1}{n}}}{\Gamma(\alpha-\beta)}.

∎

Applying Lemma A.3 and Lemma A.5, immediately we have the following asymptotics for the solutions $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ for fixed $k$ .

Lemma 5.7.

For each fixed $k$ , as $z\to+\infty$ , we have

(5.93)		$\displaystyle\mathcal{G}_{k}(z)$	$\displaystyle\sim\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(j_{k}z^{n})^{-\beta}\cdot e^{\frac{j_{k}z^{n}}{2}},$
(5.94)		$\displaystyle\mathcal{D}_{k}(z)$	$\displaystyle\sim(j_{k}z^{n})^{\beta-\alpha}\cdot e^{-\frac{j_{k}z^{n}}{2}}.$

Again we need to derive uniform estimates and asymptotic behavior for $\Ku$ and $\Tri$ . The idea is to first estimate them in terms of certain integrals and then apply Laplace’s method. To start with, we need some preliminary calculations for $\Ku$ and $\Tri$ .

By definition,

(5.95)		$\displaystyle\Tri(\fb,\fa,y)$	$\displaystyle=$	$\displaystyle\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{0}^{\infty}e^{yt+(\alpha-\beta-1)\log t+(\beta-1)\log(t+1)}dt$
(5.95)			$\displaystyle=$	$\displaystyle\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{0}^{\infty}e^{yt+(\alpha-\beta-1)\log\frac{t}{t+1}}\cdot\frac{1}{(1+t)^{1+\frac{1}{n}}}dt.$

For simplicity, we denote

(5.96)

F(t)\equiv yt+(\fa-\fb-1)\log\frac{t}{t+1},

then

(5.97)

\Tri(\fb,\fa,y)=\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{0}^{\infty}e^{F(t)}\cdot\frac{1}{(1+t)^{1+\frac{1}{n}}}dt.

Now we give both upper and lower bounds for $\Ku(\fb,\fa,y)$ by simpler exponential integrals.

Lemma 5.8.

Let $y\leq-1$ , then following holds,

(5.98)

C_{n}^{-1}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{\frac{1}{\sqrt{-y}}}^{\infty}e^{G(u)}du\leq\Ku(\fb,\fa,y)\leq C_{n}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{0}^{\infty}e^{G(u)}du,

where

(5.99)

G(u)\equiv-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u.

Proof.

To prove this estimate, we need the following integral representation formula for $\Ku(\fb,\fa,y)$ ,

(5.100)

\Ku(\fb,\fa,y)=\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot e^{y}(-y)^{\frac{1-\fa}{2}}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt.

The proof is included in Lemma A.2 of Appendix A.

The key point in the proof of (5.98) is to apply the estimate of $I_{\alpha-1}$ in Proposition 5.5. By definition, $\alpha=1-\frac{1}{n}$ and hence $\alpha-1=-\frac{1}{n}\geq-\frac{1}{2}$ . Applying the upper bound estimate of $I_{\alpha-1}$ in (5.48) of Proposition 5.5,

(5.101)		$\displaystyle I_{\alpha-1}(2\sqrt{-yt})=I_{-\frac{1}{n}}(2\sqrt{-yt})$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot\max\Big\{(2\sqrt{-yt})^{-\frac{1}{n}},(2\sqrt{-yt})^{-\frac{1}{2}}\cdot e^{2\sqrt{-yt}}\Big\}$
(5.101)			$\displaystyle\leq$	$\displaystyle C_{n}\cdot(-yt)^{-\frac{1}{4}}\cdot e^{2\sqrt{-yt}}.$

Substituting the above in (5.100),

(5.102)	$\displaystyle\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-t+2\sqrt{-yt}}\cdot t^{\frac{2\alpha-3}{4}-\beta}dt$
		$\displaystyle=$	$\displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-t+2\sqrt{-yt}+(\frac{2\alpha-3}{4}-\beta)\log t}dt$
		$\displaystyle=$	$\displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.$

Therefore,

(5.103)

\displaystyle\Ku(\fb,\fa,y)

\displaystyle\leq

\displaystyle C_{n}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{0}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.

Next, $\Ku$ can be also bounded below in a similar way. In fact, we consider the integral domain $t\geq\frac{1}{-y}$ with $y\leq-1$ , then

(5.104)

I_{\alpha-1}(2\sqrt{-yt})\geq C_{n}^{-1}\cdot\frac{e^{2\sqrt{-yt}}}{(-yt)^{\frac{1}{4}}},

and hence

	$\displaystyle\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt$	$\displaystyle\geq\int_{\frac{1}{-y}}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt$
		$\displaystyle\geq C_{n}^{-1}\cdot\int_{\frac{1}{-y}}^{\infty}e^{-t+2\sqrt{-yt}+(\frac{2\alpha-3}{4}-\beta)\log t}dt$
(5.105)			$\displaystyle=C_{n}^{-1}\cdot\int_{\frac{1}{\sqrt{-y}}}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.$

Therefore,

(5.106)

\Ku(\fb,\fa,y)\geq C_{n}^{-1}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{\frac{1}{\sqrt{-y}}}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.

∎

Now we set up a few notations for convenience. Let

(5.107)

Q\equiv\alpha-\beta-1\geq-\frac{1}{n},\ \gamma_{n}\equiv\frac{1}{2}+\frac{1}{n},

and recall the notations (5.96) and (5.99),

(5.108)		$\displaystyle F(t)$	$\displaystyle=yt+Q\log\frac{t}{t+1},$
(5.109)		$\displaystyle G(u)$	$\displaystyle=-u^{2}+2(-y)^{\frac{1}{2}}\cdot u+(2Q+\gamma_{n})\cdot\log u.$

By direct calculation

(5.110)		$\displaystyle F^{\prime\prime}(t)$	$\displaystyle=Q(-\frac{1}{t^{2}}+\frac{1}{(t+1)^{2}}),$
(5.111)		$\displaystyle G^{\prime\prime}(u)$	$\displaystyle=-2-\frac{2Q+\gamma_{n}}{u}.$

Notice that $2Q+\gamma_{n}\geq\frac{1}{2}-\frac{1}{n}\geq 0$ . Therefore, $G(u)$ is strictly concave in $\mathbb{R}_{+}$ , and $F$ is strictly concave in $\mathbb{R}$ if $Q>0$ .

We will split our analysis in two different cases:

Case (A): $Q\geq 1$ .

Case (B): $Q\leq 1$ .

Our main focus is Case (A) which is more difficult. The upper bound estimates in Case (B) follows from elementary integral calculations (see Lemma 5.12).

Case (A)

Let $t_{0}>0$ be the unique critical point of $F(t)$ and let $u_{0}>0$ be the unique critical point of $G(u)$ , then $t_{0}$ and $u_{0}$ satisfy the equations

(5.112)		$\displaystyle t_{0}^{2}+t_{0}+\frac{Q}{y}=0,$
(5.113)		$\displaystyle u_{0}^{2}-(-y)^{\frac{1}{2}}\cdot u_{0}-\frac{2Q+\gamma_{n}}{2}=0.$

Immediately we have

(5.114)		$\displaystyle t_{0}$	$\displaystyle=\frac{-1+\sqrt{1+\frac{4Q}{-y}}}{2},$
(5.115)		$\displaystyle u_{0}$	$\displaystyle=\frac{(-y)^{\frac{1}{2}}}{2}\cdot\Big(1+\sqrt{1+\frac{4Q}{-y}+\frac{2\gamma_{n}}{-y}}\Big).$

Now prove the following effective estimates on $\Ku$ and $\Tri$ . The difference from Lemma 5.7 is here the estimates holds uniformly for all $\beta\leq 0$ (recall $\alpha$ is the fixed number $1-\frac{1}{n}$ ).

Proposition 5.9.

There exists some dimensional constant $C_{n}>0$ such that for every $y\leq-1$ , the following estimates hold:

(5.116)		$\displaystyle C_{n}^{-1}\cdot Q^{-\frac{1}{4}-\frac{1}{2n}}\cdot\frac{(-y)^{-1}\cdot e^{y+F(t_{0})}}{\Gamma(Q+1)}$	$\displaystyle\leq\Tri(\beta,\alpha,y)\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot\frac{e^{y+F(t_{0})}}{\Gamma(Q+1)},$
(5.117)		$\displaystyle C_{n}^{-1}\cdot Q^{-\frac{1}{4}}\cdot\frac{(-y)^{\frac{1-2\alpha}{4}}\cdot e^{y+G(u_{0})}}{\Gamma(Q+1)}$	$\displaystyle\leq\Ku(\beta,\alpha,y)\leq C_{n}\cdot\frac{(-y)^{\frac{1-2\alpha}{4}}\cdot e^{y+G(u_{0})}}{\Gamma(Q+1)}.$

Proof.

Our main strategy is to apply Laplace’s method. The basic idea is that the above exponential integrals are concentrated at the critical values $t_{0}$ and $u_{0}$ .

First, we prove the uniform estimate for $\Tri(\fb,\fa,y)$ . By (5.97),

(5.118)

\Tri(\fb,\fa,y)\leq\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{F(t)}dt.

Clearly, the upper bound of $\Tri(\fb,\fa,y)$ follows from the upper bound estimate of $\int_{0}^{\infty}e^{F(t)}dt$ . Write

(5.119)

\int_{0}^{\infty}e^{F(t)}dt=\int_{0}^{2t_{0}}e^{F(t)}dt+\int_{2t_{0}}^{\infty}e^{F(t)}dt.

We will estimate the two terms separately.

To estimate the first term in (5.119), we make a change of variable

(5.120)

t=t_{0}\cdot(1+\xi),\ \xi\in(-1,1),

then Taylor’s theorem gives that

(5.121)	$\displaystyle F(t)-F(t_{0})$	$\displaystyle=$	$\displaystyle F(t_{0}(1+\xi))-F(t_{0})$
		$\displaystyle=$	$\displaystyle F^{\prime}(t_{0})\cdot t_{0}\cdot\xi+\frac{F^{\prime\prime}(\theta)}{2}\cdot t_{0}^{2}\cdot\xi^{2}$
		$\displaystyle=$	$\displaystyle\frac{F^{\prime\prime}(\theta)}{2}\cdot t_{0}^{2}\cdot\xi^{2},$

where $\theta$ is between $t$ and $t_{0}$ . Now we need to estimate the quadratic error term. It is straightforward calculation that

(5.122)

\displaystyle F^{\prime\prime\prime}(t)

\displaystyle=2(\frac{Q}{t^{3}}-\frac{Q}{(t+1)^{3}})>0,

then $F^{\prime\prime}(t)$ is increasing in $t$ . Since $\theta$ is between $t_{0}$ and $t\in[0,2t_{0}]$ , the above monotonicity of $F^{\prime\prime}$ implies $F^{\prime\prime}(\theta)\leq F^{\prime\prime}(2t_{0})<0$ . So the first term of (5.119) becomes

(5.123)		$\displaystyle\int_{0}^{2t_{0}}e^{F(t)}dt$	$\displaystyle=$	$\displaystyle e^{F(t_{0})}\int_{0}^{2t_{0}}e^{F(t)-F(t_{0})}dt$
(5.123)			$\displaystyle\leq$	$\displaystyle e^{F(t_{0})}\cdot t_{0}\cdot\int_{-1}^{1}e^{\frac{F^{\prime\prime}(2t_{0})}{2}\cdot t_{0}^{2}\cdot\xi^{2}}d\xi$

By direct computations, $F^{\prime\prime}(2t_{0})=-\frac{(4t_{0}+1)}{4t_{0}^{2}(2t_{0}+1)^{2}}\cdot Q$ . So we have,

(5.124)	$\displaystyle\int_{0}^{2t_{0}}e^{F(t)}dt$	$\displaystyle\leq$	$\displaystyle e^{F(t_{0})}\cdot t_{0}\cdot\int_{-1}^{1}e^{-\frac{4t_{0}+1}{8(2t_{0}+1)^{2}}\cdot Q\cdot\xi^{2}}d\xi$
		$\displaystyle\leq$	$\displaystyle C_{n}\cdot\frac{t_{0}(2t_{0}+1)}{\sqrt{4t_{0}+1}\cdot\sqrt{Q}}\cdot e^{F(t_{0})}$
		$\displaystyle\leq$	$\displaystyle C_{n}\cdot Q^{\frac{1}{4}}\cdot e^{F(t_{0})},$

where we used that $t_{0}\leq C_{n}\cdot Q^{1/2}$ (since $y\leq-1$ and $Q\geq 1$ ). Immediately, we have

(5.125)

\Tri(\beta,\alpha,y)\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot\frac{e^{y+F(t_{0})}}{\Gamma(\alpha-\beta)}.

Next, we estimate the second term in (5.119). Since we have proved $F^{\prime\prime}(t)<0$ , so this implies that $F^{\prime}(t)$ is decreasing and hence $F^{\prime}(t)\leq F^{\prime}(2t_{0})$ for any $t\geq 2t_{0}$ . Now Taylor’s theorem gives that

(5.126)

F(t)\leq F(2t_{0})+F^{\prime}(2t_{0})\cdot(t-2t_{0}),

which implies that

(5.127)

\int_{2t_{0}}^{\infty}e^{F(t)}dt\leq e^{F(2t_{0})}\int_{2t_{0}}^{\infty}e^{F^{\prime}(2t_{0})\cdot(t-2t_{0})}dt=\frac{e^{F(2t_{0})}}{-F^{\prime}(2t_{0})}.

One can check that $F^{\prime}(2t_{0})=\frac{y(3t_{0}+1)}{2(2t_{0}+1)}<0$ with $0<t_{0}<+\infty$ . Since $F^{\prime}(t)<0$ for all $t>t_{0}$ , so $F(2t_{0})\leq F(t_{0})$ and hence for $y\leq-1$ we have

(5.128)

\int_{2t_{0}}^{\infty}e^{F(t)}dt\leq C_{n}e^{F(t_{0})}.

Combining the above, we have

(5.129)

\int_{0}^{\infty}e^{F(t)}dt\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot e^{F(t_{0})}.

Therefore,

(5.130)

\displaystyle\Tri(\beta,\alpha,y)

\displaystyle\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot\frac{e^{y+F(t_{0})}}{\Gamma(\alpha-\beta)}.

The lower bound estimate for $\Tri(\fb,\fa,y)$ also follows from Laplace’s method and we just sketch the computations.

	$\displaystyle\Tri(\fb,\fa,y)$	$\displaystyle=\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{F(t)}\cdot\frac{1}{(t+1)^{1+\frac{1}{n}}}dt$
		$\displaystyle\geq\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}\cdot\frac{1}{(t+1)^{1+\frac{1}{n}}}dt$
(5.131)			$\displaystyle\geq\frac{e^{y}}{\Gamma(\alpha-\beta)\cdot(1+2t_{0})^{1+\frac{1}{n}}}\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}dt.$

By the concavity of $F(t)$ and the monotonicity of $F^{\prime\prime}(t)$ in the domain $t_{0}\leq t\leq 2t_{0}$ , we have

(5.132)

\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}dt\geq e^{F(t_{0})}\int_{t_{0}(y)}^{2t_{0}(y)}e^{\frac{F^{\prime\prime}(t_{0})}{2}(t-t_{0})^{2}}dt\geq C_{n}\cdot e^{F(t_{0})}\frac{t_{0}(t_{0}+1)}{\sqrt{2t_{0}+1}\cdot\sqrt{Q}}

It is elementary to see that

(5.133)

C_{n}Q^{\frac{1}{2}}(-y)^{-1}\leq t_{0}\leq C_{n}\cdot{Q^{\frac{1}{2}}}

Therefore,

(5.134)

\Tri(\fb,\fa,y)\geq C_{n}\cdot Q^{-\frac{1}{4}-\frac{1}{2n}}\cdot\frac{e^{y}\cdot(-y)^{-1}}{\Gamma(\alpha-\beta)}\cdot e^{F(t_{0})}.

The uniform estimate for $\Ku(\fb,\fa,y)$ stated in (5.117) can be proved in the same way. One just needs to apply Laplace’s method to the integral estimate formula in Lemma 5.8. We can eventually obtain

(5.135)

C_{n}^{-1}\cdot Q^{-\frac{1}{4}}\cdot e^{G(u_{0})}\leq\int_{0}^{\infty}e^{G(u)}du\leq C_{n}\cdot e^{G(u_{0})}.

We omit the computations here.

∎

Converting into the variables $z$ , we obtain

Corollary 5.9.1.

There exists $C_{n}>0$ such that for all $z\geq 1$ , we have

(5.136)		$\displaystyle C_{n}^{-1}\cdot\frac{Q^{-\frac{1}{4}-\frac{1}{2n}}}{\Gamma(Q+1)}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}+F(t_{0}(z))}\cdot(j_{k}z^{n})^{-1}$	$\displaystyle\leq\mathcal{D}_{k}(z)\leq C_{n}\cdot\frac{Q^{\frac{1}{4}}}{\Gamma(Q+1)}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}+F(t_{0}(z))},$
(5.137)		$\displaystyle C_{n}^{-1}\cdot Q^{-\frac{1}{4}}\cdot\frac{(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma(Q+1)}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}+G(u_{0}(z))}$	$\displaystyle\leq\mathcal{G}_{k}(z)\leq C_{n}\cdot\frac{(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma(Q+1)}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}+G(u_{0}(z))},$

where $Q\equiv\alpha-\beta-1\geq 1$ .

The next Proposition essentially gives an estimate of the product of $\Ku$ and $\Tri$ .

Proposition 5.10.

There exists some dimensional constant $C_{n}>0$ such that for any $y\leq-1$ , we have

(5.138)

e^{F(t_{0})+G(u_{0})}\leq C_{n}(-y)^{\frac{\gamma_{n}}{2}}e^{-y}e^{-Q}Q^{Q+\frac{\gamma_{n}}{2}}.

In particular we have

(5.139)

\Tri\cdot\Ku\leq C_{n}\cdot\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)^{2}}(-y)^{\frac{1}{n}}Q^{Q}e^{-Q}e^{y}.

Proof.

The calculation in the proof is purely elementary. The order estimate involving the parameter $Q$ will be used at crucial places for our later estimates, so we include the detailed proof. Plugging the critical points formulae (5.114) and (5.115) into the expression of $F$ and $G$ ,

(5.140)

F(t_{0})+G(u_{0})=yt_{0}+(-y)^{\frac{1}{2}}u_{0}-\frac{2Q+\gamma_{n}}{2}+Q\log\frac{t_{0}}{t_{0}+1}+(2Q+\gamma_{n})\log u_{0},

where $Q\equiv\alpha-\beta-1$ and $\gamma_{n}\equiv\frac{1}{2}+\frac{1}{n}$ as before.

First, it is straightforward that

(5.141)

yt_{0}+(-y)^{\frac{1}{2}}u_{0}\leq(-y)+\frac{\gamma_{n}}{2}.

So this implies that

(5.142)	$\displaystyle e^{F(t_{0})+G(u_{0})}$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot e^{-y}\cdot e^{-Q}\cdot\Big(\frac{t_{0}}{1+t_{0}}\Big)^{Q}\cdot u_{0}^{2Q+\gamma_{n}}$
		$\displaystyle=$	$\displaystyle C_{n}\cdot e^{-y}\cdot e^{-Q}\cdot\Big(\frac{t_{0}^{2}}{t_{0}(1+t_{0})}\Big)^{Q}\cdot u_{0}^{2Q+\gamma_{n}}$
		$\displaystyle=$	$\displaystyle C_{n}\cdot e^{-y}\cdot u_{0}^{\gamma_{n}}\cdot e^{-Q}\cdot\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}},$

where the last equality follows from (5.112).

Now we claim

(5.143)

u_{0}t_{0}\leq(-y)^{-\frac{1}{2}}\cdot(Q+\frac{\gamma_{n}}{2}).

To prove this, we denote $\tau\equiv\frac{2\gamma_{n}}{-y}>0$ and $\widehat{Q}\equiv\frac{4Q}{-y}>0$ . Then using the critical point formulae of $u_{0}$ and $t_{0}$ given by (5.114) and (5.115), we obtain

(5.144)		$\displaystyle u_{0}t_{0}$
	$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(1+\sqrt{1+\widehat{Q}}\Big)\cdot\Big(-1+\sqrt{1+\widehat{Q}+\tau}\Big)$
	$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(-1+\sqrt{1+\widehat{Q}}\cdot\sqrt{1+\widehat{Q}+\tau}+\sqrt{1+\widehat{Q}}-\sqrt{1+\widehat{Q}+\tau}\Big)$
	$\displaystyle\leq$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(-1+\sqrt{1+\widehat{Q}+\tau}\cdot\sqrt{1+\widehat{Q}+\tau}+\sqrt{1+\widehat{Q}+\tau}-\sqrt{1+\widehat{Q}+\tau}\Big)$
	$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot(\widehat{Q}+\tau)$
	$\displaystyle=$	$\displaystyle(-y)^{-\frac{1}{2}}\cdot(Q+\frac{\gamma_{n}}{2}).$

Then it follows that

(5.145)

\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}}\leq\frac{(Q+\frac{\gamma_{n}}{2})^{2Q}}{Q^{Q}}=Q^{Q}\cdot(1+\frac{\gamma_{n}}{2Q})^{2Q}\leq e^{\gamma_{n}}\cdot Q^{Q}.

Moreover, we notice that

(5.146)

u_{0}^{\gamma_{n}}\leq C_{n}\cdot Q^{\frac{\gamma_{n}}{2}}\cdot(-y)^{\frac{\gamma_{n}}{2}}.

Therefore, combining all the above, we have

(5.147)		$\displaystyle e^{F(t_{0})+G(u_{0})}$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot e^{-y}\cdot u_{0}^{\gamma_{n}}\cdot e^{-Q}\cdot\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}}$
(5.147)			$\displaystyle\leq$	$\displaystyle C_{n}\cdot(-y)^{\frac{\gamma_{n}}{2}}\cdot e^{-y}\cdot e^{-Q}\cdot Q^{Q+\frac{\gamma_{n}}{2}}.$

∎

In the next subsections, we will also need the following monotonicity formula to study the integral estimates for the above fundamental solutions $\mathcal{G}_{k}$ and $\mathcal{D}_{k}$ .

Lemma 5.11.

Let

(5.148)		$\displaystyle\widehat{F}(z)$	$\displaystyle\equiv-\frac{jz^{n}}{2}+F(t_{0}(z)),$
(5.149)		$\displaystyle\widehat{G}(z)$	$\displaystyle\equiv-\frac{jz^{n}}{2}+G(u_{0}(z)),$

then for all $\eta\geq 0$ , when $z\geq\eta^{\frac{2}{n}}$ , $\widehat{F}(z)+\eta\cdot z^{\frac{n}{2}}$ is decreasing and $\widehat{G}(z)-\eta\cdot z^{\frac{n}{2}}$ is increasing.

Proof.

Let $y=-jz^{n}$ , then it is straightforward that

(5.150)

\displaystyle\frac{d\widehat{F}(y)}{dy}=\frac{1}{2}+t_{0}(y)+F^{\prime}(t_{0}(y))\cdot\frac{dt_{0}(y)}{dy}=\frac{1}{2}+t_{0}(y)=\frac{1}{2}\sqrt{1+\frac{4Q}{-y}}\geq\frac{1}{2}.

This implies that, as $z\geq\eta^{\frac{2}{n}}$ ,

(5.151)

\displaystyle\frac{d(\widehat{F}(z)+\eta z^{\frac{n}{2}})}{dz}=\frac{d\widehat{F}(y)}{dy}\cdot(-nj\cdot z^{n-1})+\frac{n\cdot\eta}{2}\cdot z^{\frac{n}{2}-1}\leq-\frac{n}{2}\cdot z^{\frac{n}{2}-1}(j\cdot z^{\frac{n}{2}}-\eta)\leq 0.

By similar calculations, one can also obtain that $\widehat{G}(z)-\eta\cdot z^{\frac{n}{2}}$ is increasing as $z\geq\eta^{\frac{2}{n}}$ .

∎

Case (B): Now we consider the case when $Q\leq 1$ . As mentioned in the above, this case is easier.

Lemma 5.12.

Let $Q\leq 1$ , then there is some dimensional constant $C_{n}>0$ such that

(5.152)		$\displaystyle C_{n}^{-1}\cdot e^{y}\cdot(-y)^{\beta-\alpha}$	$\displaystyle\leq\Tri(\fb,\fa,y)\leq e^{y}\cdot(-y)^{\beta-\alpha},$
(5.153)		$\displaystyle C_{n}^{-1}\cdot(-y)^{-\beta}$	$\displaystyle\leq\Ku(\fb,\fa,y)\leq C_{n}\cdot(-y)^{-\beta}.$

for all $y\leq-1$ .

Remark 5.12.1.

In the case $Q\leq 1$ , the estimate is optimal in the sense that it coincides with the asymptotic behavior of $\Tri$ and $\Ku$ for fixed $\alpha$ and $\beta$ , as given in Lemma A.3 and Lemma A.5.

Proof.

First, we prove (5.152). Both the upper bound and lower bound estimates can be proved in the similar way:

	$\displaystyle\Tri(\beta,\alpha,y)$	$\displaystyle=\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt$
		$\displaystyle\leq\frac{e^{y}}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}dt$
		$\displaystyle=\frac{e^{y}\cdot(-y)^{\beta-\alpha}}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{-u}u^{\fa-\fb-1}du$
(5.154)			$\displaystyle=e^{y}\cdot(-y)^{\beta-\alpha}.$

Similarly,

	$\displaystyle\Tri(\fb,\fa,y)$	$\displaystyle\geq\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{1}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt$
		$\displaystyle\geq C_{n}\cdot e^{y}\int_{0}^{1}e^{yt}t^{\fa-\fb-1}dt$
(5.155)			$\displaystyle\geq C_{n}\cdot e^{y}\cdot(-y)^{\fb-\fa}.$

Next, we prove the upper bound estimate for $\Ku$ . Notice in the proof of Lemma 5.9 we do not need the condition $Q\leq 1$ for the upper bound on $\Ku$ . So we have

(5.156)

\Ku(\beta,\alpha,y)\leq C_{n}\cdot\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(-y)^{\frac{1-2\alpha}{4}}\cdot e^{y+G(u_{0})}.

To prove (5.153), we need an upper bound estimate for $e^{y+G(u_{0})}$ . This follows from elementary computations. In fact,

e^{y+G(u_{0})}=e^{y-u_{0}^{2}+2\sqrt{-y}u_{0}}\cdot(u_{0})^{2Q+\gamma_{n}}\\ \leq C_{n}\cdot e^{y-u_{0}^{2}+2\sqrt{-y}u_{0}}\cdot(-y)^{Q+\frac{\gamma_{n}}{2}}.

Notice that $u_{0}$ satisfies $G^{\prime}(u_{0})=0$ , i.e.,

(5.157)

u_{0}^{2}-\sqrt{-y}\cdot u_{0}-\frac{2Q+\gamma_{n}}{2}=0,

so we have

(5.158)

e^{y+G(u_{0})}\leq C_{n}\cdot e^{y+\sqrt{-y}u_{0}}\cdot(-y)^{Q+\frac{\gamma_{n}}{2}}.

By (5.115), it is straightforward that

(5.159)

\displaystyle y+\sqrt{-y}u_{0}=\frac{y}{2}\Big(1-\sqrt{1+\frac{4Q+2\gamma_{n}}{-y}}\Big)=\frac{2Q+\gamma_{n}}{1+\sqrt{1+\frac{4Q+2\gamma_{n}}{-y}}}\in[C_{n}^{-1},C_{n}],

for some dimensional constant $C_{n}>0$ . Therefore,

(5.160)

\displaystyle e^{y+G(u_{0})}\leq C_{n}(-y)^{Q+\frac{1}{4}+\frac{1}{2n}},

and hence

(5.161)

\displaystyle\Ku(\fb,\fa,y)\leq C_{n}(-y)^{Q+\frac{1}{n}}=C_{n}(-y)^{-\beta}.

This completes the proof. ∎

Converting into the variables $z$ we obtain

Corollary 5.12.1.

There exists $C_{n}>0$ such that for all $z\geq 1$ , we have

(5.162)		$\displaystyle C_{n}^{-1}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}}\cdot(j_{k}z^{n})^{\beta-\alpha}$	$\displaystyle\leq\mathcal{D}_{k}(z)\leq C_{n}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}}\cdot(j_{k}z^{n})^{\beta-\alpha},$
(5.163)		$\displaystyle C_{n}^{-1}\cdot e^{\frac{j_{k}\cdot z^{n}}{2}}\cdot(j_{k}z^{n})^{-\beta}$	$\displaystyle\leq\mathcal{G}_{k}(z)\leq C_{n}\cdot e^{\frac{j_{k}\cdot z^{n}}{2}}\cdot(j_{k}z^{n})^{-\beta}.$

We end this subsection by making some remarks regarding the above estimates on $\Ku$ and $\Tri$ . Notice that in the case $Q\equiv\fa-\fb-1\leq 1$ we applied Laplace’s method to turn the problem into estimates on exponential integrals. One may wonder how far the uniform estimates in Lemma 5.9 is from optimal comparing to the non-uniform estimate with the optimal order in Lemma 5.12. We can consider two extreme cases depending on the size of $Q$ compared with $-y$ .

First we assume $\frac{Q^{2}}{-y}\ll 1$ , which obviously includes the case when we fix $Q$ and let $y\rightarrow-\infty$ . Then by definition we see that

(5.164)

t_{0}=\frac{Q}{-y}+O\Big((\frac{Q}{-y})^{2}\Big),

and we get

(5.165)

F(t_{0})=yt_{0}+Q\log\frac{t_{0}}{t_{0}+1}=-Q+Q\log Q-Q\log(-y)+O(\frac{Q}{-y}).

So by Lemma 5.9 we get

(5.166)

C_{n}^{-1}Q^{-\frac{1}{4}-\frac{1}{2n}}e^{y}(-y)^{-Q-1}Q^{Q}e^{-Q}\leq\Tri\leq C_{n}\frac{1}{\Gamma(\alpha-\beta)}e^{y}(-y)^{-Q}Q^{Q}e^{-Q}Q^{\frac{1}{4}}.

Notice by Stirling’s formula for $Q$ large $\Gamma(\alpha-\beta)=Q\Gamma(Q)$ is comparable to $C_{n}Q^{\frac{3}{2}}Q^{Q}e^{-Q}$ . So up to polynomial errors in $Q$ this estimate is optimal comparing with (A.27). Similarly, we have

(5.167)

u_{0}=(-y)^{\frac{1}{2}}\Big(1+\frac{Q+\frac{1}{2}\gamma_{n}}{-y}+O((\frac{Q}{-y})^{2})\Big),

and

(5.168)

G(u_{0})=-u_{0}^{2}+2(-y)^{\frac{1}{2}}u_{0}+(2Q+\gamma_{n})\log u_{0}\leq C_{n}e^{-y}(-y)^{Q+\frac{\gamma_{n}}{2}}.

(5.169)

\Ku\leq C_{n}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}(-y)^{\frac{1-2\alpha}{4}}(-y)^{Q+\frac{\gamma_{n}}{2}}=C_{n}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}(-y)^{-\beta},

which is again optimal comparing with (A.34).

Secondly we assume the other extreme $\frac{Q}{(-y)^{3}}\gg 1$ . In this case we have

(5.170)

t_{0}=\sqrt{\frac{Q}{-y}}-\frac{1}{2}+O(\sqrt{\frac{-y}{Q}}).

Then we get

(5.171)

F(t_{0})=-2\sqrt{-Qy}-\frac{1}{2}y+O(1),

and

(5.172)

C_{n}^{-1}Q^{-\frac{1}{4}-\frac{1}{2n}}e^{\frac{1}{2}y-\sqrt{-Qy}}(-y)^{-Q-1}\leq\Tri(y)\leq C_{n}\frac{1}{\Gamma(\alpha-\beta)}Q^{\frac{1}{4}}(-y)^{-Q}e^{\frac{1}{2}y-2\sqrt{-Qy}}.

Similarly, we get

(5.173)

G(u_{0})=2\sqrt{-Qy}-\frac{y}{2}+(Q+\frac{1}{2}\gamma_{n})\log Q-Q.

(5.174)

\Ku(y)\leq C_{n}\cdot\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}(-y)^{\frac{1-2\alpha}{4}}e^{\frac{1}{2}y+2\sqrt{-Qy}}e^{-Q}Q^{Q+\frac{1}{2}\gamma_{n}}.

In this case even though in the produce $\Tri\cdot\Ku$ there is a good cancellation each of them does behave quite differently from the previous case. This also gives a reason why we do get an optimal estimate (up to polynomial errors in $y$ and $Q$ ) for the product $\Tri\cdot\Ku$ , comparing with (A.27) and (A.34).

5.4. Asymptotics of harmonic functions on the Calabi model space

As Section 5.1, we fix $r_{0}\in(0,1)$ , and view the Calabi model space $\mathcal{C}^{n}$ as the product of a fixed cross section $Y^{2n-1}\cong\{\rho=\rho_{0}\}$ with the restricted metric $h_{0}=g_{\mathcal{C}^{n}}|_{\{\rho=\rho_{0}\}}$ with a ray $\mathbb{R}^{+}$ . The spectrum of the Laplacian operator on $Y$ is given by $\{\Lambda_{k}\}_{k=0}^{\infty}$ , with $\Lambda_{0}=0$ , and we have chosen an orthonormal basis of complex valued eigenfunctions of the form $\{\varphi_{k}\}_{k=0}^{\infty}$ such that

(5.175)

\displaystyle\begin{cases}\Delta_{Y^{2n-1}}\varphi_{k}=\Lambda_{k}\cdot\varphi_{k},\\ \|\varphi_{k}\|_{L^{2}(Y^{2n-1})}=1.\end{cases}

We need a basic lemma on the decay of Fourier coefficients of the expansion of a sufficiently smooth function in terms of eigenfunctions.

Lemma 5.13.

Let $K_{0}\geq 1$ and let $\xi\in C^{2K_{0}}(Y^{2n-1})$ satisfy the $L^{2}$ -expansion

(5.176)

\xi(\bm{y})=\sum\limits_{k=1}^{\infty}\xi_{k}\cdot\varphi_{k}(\bm{y}),

then for all $k\in\mathbb{Z}_{+}$ ,

(5.177)

|\xi_{k}|\leq\frac{C|\xi|_{C^{2K_{0}}(Y^{2n-1})}}{(\Lambda_{k})^{K_{0}}},

where the constant $C>0$ is independent of $k$ .

Proof.

The estimate is proved by the standard integration by parts. Since the eigenfunctions $\varphi_{k}$ satisfy

(5.178)

-\Delta_{h_{0}}\varphi_{k}=\Lambda_{k}\cdot\varphi_{k}

and $\|\varphi_{k}\|_{L^{2}(Y^{2n-1})}=1$ , we have that

	$\displaystyle\|\xi_{k}(z)\|$	$\displaystyle=\Big\|\int_{Y^{2n-1}}\xi\cdot\varphi_{k}\Big\|=\Big\|\int_{Y^{2n-1}}\xi\cdot\frac{(-\Delta_{h_{0}})^{K_{0}}\varphi_{k}}{(\Lambda_{k})^{K_{0}}}\Big\|$
(5.179)			$\displaystyle\leq\frac{1}{(\Lambda_{k})^{K_{0}}}\int_{Y^{2n-1}}\|\Delta_{h_{0}}^{K_{0}}\xi\|\cdot\|\varphi_{k}\|$
(5.180)			$\displaystyle\leq\frac{C\|\xi\|_{C^{2K_{0}}(Y^{2n-1})}}{(\Lambda_{k})^{K_{0}}},$

where $C>0$ depends only on the geometry of $Y^{2n-1}$ .

∎

Proposition 5.14 (Asymptotics of harmonic functions).

Let $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ be a Calabi model space with $\dim_{\mathbb{C}}(\mathcal{C}^{n})=n$ . Define a constant

(5.181)

\delta_{b}\equiv 2\underline{\lambda}^{\frac{1}{2}}n^{-\frac{1}{2}}>0

where $\underline{\lambda}>0$ is given by (5.14). If $u$ is a harmonic function outside a compact set in $\mathcal{C}^{n}$ satisfying

(5.182)

|u(z,\bm{y})|=O(e^{\delta\cdot z^{\frac{n}{2}}})

for some $\delta\in(0,\delta_{b})$ as $z\rightarrow\infty$ . Then $u$ can be decomposed as

(5.183)

u(z,\bm{y})=L(z)+h(z,\bm{y})

with the following properties:

(1)

$L(z)=\kappa_{0}\cdot z+c_{0}$ for some $\kappa_{0},c_{0}\in\mathbb{R}$ .
(2)

$h(z,\bm{y})$ is harmonic and for any $k\in\mathbb{N}$ , there is some $C_{k}>0$ such that

(5.184) $|\nabla^{k}h(z,\bm{y})|\leq C_{k}\cdot e^{-\underline{\delta}\cdot z^{\frac{n}{2}}}$

for all $\underline{\delta}\in(0,\delta_{b})$ , as $z\to+\infty$ .

Proof.

The proof consists of two steps.

In the first step, we will apply separation of variables to show that if a harmonic function $u$ satisfies (5.182), then $u(z,\bm{y})=k_{0}\cdot z+c_{0}+h(z,\bm{y})$ for some $k_{0},c_{0}\in\mathbb{R}$ and $h(z,\bm{y})$ has some exponential decaying rate.

Since $u$ is smooth, for any fixed $z\geq 1$ , we have the fiber-wise $L^{2}$ -expansion of $u$ as follows,

(5.185)

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

where $\bm{y}\in Y^{2n-1}$ and $u_{k}$ satisfies the equation

(5.186)

\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=0,\ z\geq 1,

for some $j_{k}\in\mathbb{N}$ and $\lambda_{k}\geq 0$ . Notice that the expansion (5.185) converges in the $C^{\infty}$ -topology. This follows from Lemma 5.13, Lemma 3.32 and the Weyl law for spectrum asymptotics.

For $k=0$ we have $j_{k}=\lambda_{k}=0$ , and $u_{k}$ is a linear function of the form $\kappa_{0}\cdot z+c_{0}$ . For $k\geq 1$ , we can write $u_{k}$ as a linear combination of the two linearly independent solutions discussed in Section 5.2 and 5.3.

(5.187)

u_{k}(z)=C_{k}\cdot\mathcal{D}_{k}(z)+C_{k}^{*}\cdot\mathcal{G}_{k}(z),

where $\mathcal{G}_{k}$ is a growing and $\mathcal{D}_{k}$ is decaying.

We claim $C_{k}^{*}=0$ for all $k\in\mathbb{Z}_{+}$ . To see this, we apply Lemma 5.13 to $u(z,\bm{y})$ , then for all $k\in\mathbb{Z}_{+}$

(5.188)

|u_{k}(z)|=O(e^{\delta z^{\frac{n}{2}}}).

So the claim follows from the asymptotics of $\mathcal{G}_{k}(z)$ in Lemma 5.4 and 5.7 which corresponds to $j_{k}=0$ and $j_{k}\in\mathbb{Z}_{+}$ respectively.

Now we define

(5.189)

h(z,\bm{y})\equiv u(z,\bm{y})-(\kappa_{0}\cdot z+c_{0})=\sum_{k=1}^{\infty}u_{k}(z)\cdot\varphi_{k}(\bm{y}).

It suffices to show $h(z,\bm{y})$ decays at the desired rate. Let $z_{0}>(2\delta_{b})^{2/n}$ be sufficiently big so that $u$ is defined on $\{z\geq z_{0}\}$ . Now we fix $K_{0}\equiv 2n+1$ . Applying Lemma 5.13 to $u(z_{0},\bm{y})$ we get for all $k\in\mathbb{Z}_{+}$ ,

(5.190)

|u_{k}(z_{0})|\leq C_{1}(\Lambda_{k})^{-K_{0}}.

We separate in several cases. First, we consider $k\in\mathbb{Z}_{+}$ with $j_{k}=0$ . Applying (5.74), then for any $\epsilon>0$ with $\underline{\delta}=(1-\epsilon)\delta_{b}<\delta_{b}$ , if $z\geq\frac{1}{\epsilon^{\frac{2}{n}}}\cdot z_{0}$ ,

(5.191)

\Big|\frac{u_{k}(z)}{u_{k}(z_{0})}\Big|=\Big|\frac{\mathcal{D}_{k}(z)}{\mathcal{D}_{k}(z_{0})}\Big|\leq Ce^{-2\sqrt{\frac{\lambda_{k}}{n}}\cdot(z^{\frac{n}{2}}-z_{0}^{\frac{n}{2}})}\leq Ce^{-(1-\epsilon)\delta_{b}\cdot z^{\frac{n}{2}}}=Ce^{-\underline{\delta}\cdot z^{\frac{n}{2}}}.

This implies that

(5.192)		$\displaystyle\Big\|\sum_{\begin{subarray}{c}k>0\\ j_{k}=0\end{subarray}}\frac{u_{k}(z)}{u_{k}(z_{0})}\cdot u_{k}(z_{0})\cdot\varphi_{k}(\bm{y})\Big\|$	$\displaystyle\leq$	$\displaystyle\sum_{\begin{subarray}{c}k>0\\ j_{k}=0\end{subarray}}\Big\|\frac{u_{k}(z)}{u_{k}(z_{0})}\Big\|\cdot\|u_{k}(z_{0})\|\cdot\|\varphi_{k}(\bm{y})\|$
(5.192)			$\displaystyle\leq$	$\displaystyle Ce^{-\delta_{b}\cdot z^{\frac{n}{2}}}\cdot\sum_{\begin{subarray}{c}k>0\\ j_{k}=0\end{subarray}}\frac{1}{(\Lambda_{k})^{K_{0}-\frac{n}{2}}},$

where the eigenfunction estimate

(5.193)

\|\varphi_{k}\|_{L^{\infty}(Y^{2n-1})}\leq C\cdot(\Lambda_{k})^{\frac{n}{2}}.

follows from Lemma 3.32.

When $j_{k}\in\mathbb{Z}_{+}$ we divide into two cases. When $Q\geq 1$ we apply Corollary 5.9.1 and Lemma 5.11 (with $\eta=2\delta_{b}$ ) to get

(5.194)		$\displaystyle\Big\|\sum_{\begin{subarray}{c}j_{k}\geq 1\\ Q\geq 1\end{subarray}}\frac{u_{k}(z)}{u_{k}(z_{0})}\cdot u_{k}(z_{0})\cdot\varphi_{k}(\bm{y})\Big\|$	$\displaystyle\leq$	$\displaystyle\sum_{\begin{subarray}{c}j_{k}\geq 1\\ Q\geq 1\end{subarray}}\Big\|\frac{u_{k}(z)}{u_{k}(z_{0})}\Big\|\cdot\|u_{k}(z_{0})\|\cdot\|\varphi_{k}(\bm{y})\|$
(5.194)			$\displaystyle\leq$	$\displaystyle Ce^{-\delta_{b}\cdot z^{\frac{n}{2}}}\sum_{\begin{subarray}{c}j_{k}\geq 1\\ Q\geq 1\end{subarray}}\frac{1}{(\Lambda_{k})^{K_{0}-\frac{n}{2}-1}}.$

Now when $Q\leq 1$ we apply instead Corollary 5.12.1 to get

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

Summing up all the above we get

(5.196)

|h(z,\bm{y})|\leq Ce^{-\delta_{b}\cdot z^{\frac{n}{2}}}\sum_{k=1}^{\infty}\frac{1}{(\Lambda_{k})^{K_{0}-\frac{n}{2}-1}}.

Since $K_{0}=2n+1$ we see the series converges. So the proof of the first step is done.

The second step is to prove the higher decaying estimate for the error function $h(z,\bm{y})$ , which follows from the uniform Schauder estimate. We have proved that the error function $h(z,\bm{y})$ as a harmonic function satisfies

(5.197)

|h(z,\bm{y})|\leq C_{0}\cdot e^{-\underline{\delta}\cdot z^{\frac{n}{2}}}.

By explicit and straightforward computations, a Calabi space $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ is collapsing with bounded curvatures as $z\to+\infty$ . We just lift the harmonic function $h$ to the local universal cover which is non-collapsed with uniformly bounded geometry. So the following Schauder estimate holds for any $k\in\mathbb{Z}_{+}$ and $\alpha\in(0,1)$ on the local universal cover,

(5.198)

|h|_{C^{k,\alpha}(B_{r_{0}}(\bm{x}))}\leq C_{k}\cdot|h|_{C^{0}(B_{2r_{0}}(\bm{x}))}\leq C_{k}\cdot e^{-\underline{\delta}\cdot z^{\frac{n}{2}}}.

where $r_{0}>0$ is some fixed constant of some definite size which is independent of $\bm{x}\in\mathcal{C}^{n}$ . In particular, at the center $\bm{x}=(z,\bm{y})$ , we have

(5.199)

|\nabla^{k}h(z,\bm{y})|\leq C_{k}\cdot|h|_{C^{0}(B_{2r_{0}}(\bm{x}))}\leq C_{k}\cdot e^{-\underline{\delta}\cdot z^{\frac{n}{2}}}.

This completes the proof of (5.184).

∎

5.5. The Poisson equation with prescribed asymptotics

In this subsection, we will construct solutions to the Poisson equation on the Calabi space $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ ,

(5.200)

\Delta_{g_{\mathcal{C}^{n}}}u=v

with controlled asymptotic behavior. As in Section 5.1, we carry out separation of variables. Suppose $v$ is a smooth function defined on $\{z\geq 1\}$ . We write

(5.201)

\displaystyle u(z,\bm{y})=\sum\limits_{k=1}^{\infty}u_{k}(z)\cdot\varphi_{k}(\bm{y}),\quad v(z,\bm{y})=\sum\limits_{k=1}^{\infty}\xi_{k}(z)\cdot\varphi_{k}(\bm{y}).

So the Poisson equation

(5.202)

\Delta_{g_{\mathcal{C}^{n}}}u=v

is reduced to the following inhomogeneous ODE

(5.203)

\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=z^{n-1}\cdot\xi_{k}(z),\quad z\geq z_{1}.

Let $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ be the growing solution and decaying solution to the corresponding homogeneous equation, which were analyzed in Section 5.2 and 5.3. So applying standard Liouville’ formula, Equation (5.203) has a particular solution

(5.204)

u_{k}(z)\equiv\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\cdot\Big(\xi_{k}(r)\cdot r^{n-1}\Big)dr+\frac{\mathcal{D}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z_{1}}^{z}\mathcal{G}_{k}(r)\cdot\Big(\xi_{k}(r)\cdot r^{n-1}\Big)dr,

where $\mathcal{W}_{k}$ is the Wronskian

(5.205)

\mathcal{W}_{k}(z)\equiv\mathcal{W}\Big(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z)\Big).

Lemma 5.15.

Assume that the function $\xi_{k}(z)$ satisfies the following property: there are $\eta_{0}\in(-\delta_{b}/2,\delta_{b}/2)$ , a sequence of positive constants $\mathfrak{B}_{k}>0$ such that

(5.206)

|\xi_{k}(z)|\leq\mathfrak{B}_{k}\cdot e^{\eta_{0}\cdot z^{\frac{n}{2}}}.

Let $u_{k}(z)$ be the particular solution (5.204), then there exists some constant $C_{0}>0$ such that the particular solution $u_{k}$ satisfies the uniform estimate

(5.207)

|u_{k}(z)|\leq C_{0}\cdot\mathfrak{B}_{k}\cdot(\Lambda_{k})^{\frac{1}{2n}}\cdot e^{\eta\cdot z^{\frac{n}{2}}}

for any $\eta>\eta_{0}$ .

Proof.

We will estimate the two terms in (5.204) individually, and we also divide into several cases.

First consider $j_{k}=0$ and $k=0$ . In this case the solutions $u_{k}$ is given by simple integrals of $\xi_{k}$ and the conclusion is easy to see.

The second case is that $k\in\mathbb{Z}_{+}$ and $j_{k}=0$ . Applying Proposition 5.5, the fundamental solutions $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ satisfy the uniform estimates

(5.208)		$\displaystyle\mathcal{G}_{k}(z)$	$\displaystyle\leq\frac{C}{\lambda_{k}^{\frac{1}{4}}}\cdot z^{\frac{2-n}{4}}\cdot e^{2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}},$
(5.209)		$\displaystyle\mathcal{D}_{k}(z)$	$\displaystyle\leq\frac{C}{\lambda_{k}^{\frac{1}{4}}}\cdot z^{\frac{2-n}{4}}\cdot e^{-2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}}.$

By Lemma 5.3.1, $\mathcal{W}_{k}(z)=\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=\frac{n}{2}$ . Let us denote $\tilde{\lambda}_{k}\equiv 2\sqrt{\frac{\lambda_{k}}{n}}$ , then $\tilde{\lambda}_{k}\geq 2\sqrt{\frac{\lambda_{1}}{n}}=\delta_{b}$ . Now the first integral term in (5.204) has the following bound,

(5.210)				$\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\|\xi_{k}(r)\cdot r^{n-1}\|dr$
(5.210)			$\displaystyle\leq$	$\displaystyle\frac{C\cdot\mathfrak{B}_{k}}{\lambda_{k}^{\frac{1}{2}}}\cdot z^{\frac{2-n}{4}}\cdot e^{\tilde{\lambda}_{k}\cdot z^{\frac{n}{2}}}\cdot\int_{z}^{\infty}r^{\frac{3n}{4}-\frac{1}{2}}\cdot e^{(-\tilde{\lambda}_{k}+\eta_{0})\cdot r^{\frac{n}{2}}}dr.$

By assumption, $|\eta_{0}|<\frac{\delta_{b}}{2}\leq\frac{\tilde{\lambda}_{k}}{2}$ , then

(5.211)		$\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\|\xi_{k}(r)\cdot r^{n-1}\|dr$	$\displaystyle\leq$	$\displaystyle\frac{C\cdot\mathfrak{B}_{k}}{\lambda_{k}^{\frac{1}{2}}}\cdot z^{\frac{2-n}{4}}\cdot e^{\tilde{\lambda}_{k}\cdot z^{\frac{n}{2}}}\cdot e^{(-\tilde{\lambda}_{k}+\eta^{\prime})\cdot z^{\frac{n}{2}}}$
(5.211)			$\displaystyle\leq$	$\displaystyle C\cdot\mathfrak{B}_{k}\cdot e^{\eta\cdot z^{\frac{n}{2}}},$

where $\eta>\eta^{\prime}>\eta_{0}>0$ . Similarly,

(5.212)

\displaystyle\frac{\mathcal{D}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z_{0}}^{z}\mathcal{G}_{k}(r)|\xi_{k}(r)\cdot r^{n-1}|dr\leq C\cdot\mathfrak{B}_{k}\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

In the third case $j_{k}\in\mathbb{Z}_{+}$ and $Q\geq 1$ , we need to apply Lemma 5.11. In fact,

(5.213)		$\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\xi_{k}(r)\cdot r^{n-1}dr$
	$\displaystyle\leq$	$\displaystyle C_{n}\cdot\frac{Q^{\frac{1}{4}}\cdot(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma^{2}(Q+1)}\cdot\frac{e^{\widehat{G}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{F}_{k}(r)}\xi_{k}(r)\cdot r^{n-1}dr$
	$\displaystyle\leq$	$\displaystyle C_{n}\cdot\mathfrak{B}_{k}\cdot\frac{Q^{\frac{1}{4}}\cdot(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma^{2}(Q+1)}\cdot\frac{e^{\widehat{G}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{F}_{k}(r)+\eta^{\prime}\cdot r^{\frac{n}{2}}}dr,$

where $\eta^{\prime}>\eta_{0}$ . We choose any $\epsilon\in(\delta_{b}/100,\delta_{b}/10)$ and denote $\eta^{\prime\prime}\equiv\eta^{\prime}+\epsilon$ , then by Lemma 5.11,

(5.214)		$\displaystyle\frac{e^{\widehat{G}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{F}_{k}(r)+\eta^{\prime}\cdot r^{\frac{n}{2}}}dr$
	$\displaystyle=$	$\displaystyle\frac{e^{\widehat{G}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{F}_{k}(r)+\eta^{\prime\prime}\cdot r^{\frac{n}{2}}}\cdot e^{-\epsilon r^{\frac{n}{2}}}dr$
	$\displaystyle\leq$	$\displaystyle\frac{e^{\widehat{F}_{k}(z)+\widehat{G}_{k}(z)+\eta^{\prime\prime}\cdot z^{\frac{n}{2}}}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{-\epsilon\cdot r^{\frac{n}{2}}}dr$
	$\displaystyle\leq$	$\displaystyle C_{n}\cdot\frac{e^{\widehat{F}_{k}(z)+\widehat{G}_{k}(z)+\eta^{\prime\prime}\cdot z^{\frac{n}{2}}}}{\mathcal{W}_{k}(z)}.$

Therefore,

(5.215)

\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\xi_{k}(r)\cdot r^{n-1}dr\leq C_{n}\cdot\mathfrak{B}_{k}\cdot\frac{Q^{\frac{1}{4}}\cdot(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma^{2}(Q+1)}\cdot\frac{e^{\widehat{F}_{k}(z)+\widehat{G}_{k}(z)+\eta^{\prime\prime}\cdot z^{\frac{n}{2}}}}{\mathcal{W}_{k}(z)}.

Plugging Lemma 5.10 and Proposition 5.6 into the above inequality,

(5.216)	$\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\xi_{k}(r)\cdot r^{n-1}dr$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot\mathfrak{B}_{k}\cdot\frac{j_{k}^{\frac{1}{n}}\cdot e^{-Q}\cdot Q^{Q+1}}{\Gamma(Q+1)}\cdot z\cdot e^{\eta^{\prime\prime}\cdot z^{\frac{n}{2}}}.$
		$\displaystyle\leq$	$\displaystyle C_{n}\cdot\mathfrak{B}_{k}\cdot j_{k}^{\frac{1}{n}}\cdot e^{\eta\cdot z^{\frac{n}{2}}}$
		$\displaystyle\leq$	$\displaystyle C_{n}\cdot\mathfrak{B}_{k}\cdot(\Lambda_{k})^{\frac{1}{2n}}\cdot e^{\eta\cdot z^{\frac{n}{2}}}$

for any $\eta\in(\eta^{\prime\prime},\eta^{\prime\prime}+\frac{\delta_{b}}{100})$ , where we used Stirling’s formula for estimating $\Gamma(Q+1)$ . Similarly we get the bound for the other term of (5.204).

The fourth case is when $j_{k}\geq 1$ and $Q\leq 1$ . This case is simpler and follows from Corollary 5.12.1 and the argument in the second case.

This completes the proof of the proposition.

∎

Based on the above ODE estimate, we prove the following $C^{0}$ and $C^{1}$ estimate for the equation to the Poisson equation.

Proposition 5.16.

Let $\{z\geq 1\}\subset\mathcal{C}^{n}$ be a subset and let $K_{0}\geq 2n+1$ be a positive integer. Given any $\eta_{0}\in(-\delta_{b}/2,\delta_{b}/2)\setminus\{0\}$ , if $v\in C^{3K_{0},\alpha}(\{z\geq 1\})$ for and

(5.217)

|v|=O(e^{\eta_{0}\cdot z(\bm{x})^{\frac{n}{2}}}),

then the Poisson equation

(5.218)

\Delta_{g_{\mathcal{C}^{n}}}u=v

has a solution $u\in C^{3K_{0}+2,\alpha}(\{z\geq 1\})$ such that for any $\eta>\eta_{0}$

(5.219)

|u(\bm{x})|+|\nabla u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}},

as $z(\bm{x})\to+\infty$ , where $C>0$ is independent of $\bm{x}\in\mathcal{C}^{n}$ .

Proof.

The proof is constructive, which will be done in two steps.

The first step, as the main part, is to find a solution $u$ with the prescribed growth (or decay) rate. We will use the method of separation of variables described as follows.

For a fixed slice $Y^{2n-1}\subset\mathcal{C}^{n}$ , let $\{\Lambda_{k}\}_{k=0}^{\infty}$ with $\Lambda_{0}=0$ be the spectrum of $\Delta_{\mathcal{C}^{n}}$ acting on functions. Let $\{\varphi_{k}\}_{k=0}^{\infty}$ be the eigenfunctions satisfying

(5.220)

\displaystyle\begin{cases}-\Delta_{\mathcal{C}^{n}}\varphi_{k}=\Lambda_{k}\varphi_{k},\\ \|\varphi_{k}\|_{L^{2}(Y^{2n-1})}=1.\end{cases}

Given a function $v$ and for any fixed $z\geq 1$ , we have the fiberwise $L^{2}$ -expansion on $Y^{2n-1}$ ,

(5.221)

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

Then we can first construct a formal solution

(5.222)

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

to (5.218), which holds in the $L^{2}$ -sense for each fixed $z\geq 1$ . Here the coefficient functions $u_{k}(z)$ are the particular solutions constructed in Lemma 5.15. The main part is to prove that the above series $u(z,\bm{y})$ converges with higher regularity and hence $u(z,\bm{y})$ is a regular solution to (5.218).

To begin with, we will prove that the series $u(z,\bm{y})$ converges in the $C^{0}$ -norm and hence gives a $C^{0}$ -function. Combining Lemma 5.13, Lemma 5.15 and the eigenfunction estimate in Lemma 3.32, we have

(5.223)

\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\cdot z^{\frac{n}{2}}}}{(\Lambda_{k})^{K_{0}-\frac{n}{2}-\frac{1}{2n}}}.

Applying Weyl’s law to the spectrum $\{\Lambda_{k}\}_{k=1}^{\infty}$ ,

(5.224)

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

where $C_{0}>0$ depends only on $Y^{2n-1}$ and $k$ is sufficiently large. Let $K_{0}\geq 2n+1$ , then

(5.225)

|u(z,\bm{y})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}\cdot\sum\limits_{k=1}^{\infty}\frac{1}{(\Lambda_{k})^{\frac{3n}{2}}}\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}\cdot\sum\limits_{k=1}^{\infty}\frac{1}{k^{\frac{3n}{2n-1}}}\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

Therefore, $u\in C^{0}(\mathcal{C}^{n})$ and $u$ satisfies the $C^{0}$ -asymptotic estimate in (5.219).

Based on the above $C^{0}$ -regularity, we will apply the standard elliptic regularity on $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ to show that $u\in C^{2}(\mathcal{C}^{n})$ is a regular solution to $\Delta_{g_{\mathcal{C}^{n}}}u=v$ . We take the partial sums

(5.226)

\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})

of the expansions

(5.227)

\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}).

It is obvious that,

(5.228)

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

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

As a starter, by the same arguments as the above, we have $\|V_{N}-v\|_{C^{0}(B_{2}(\bm{x}))}\to 0$ as $N\to\infty$ . The proof of the higher order convergence is almost verbatim. In fact, we just need to use $\|v\|_{C^{2K_{0}+m}}$ with $m\leq K_{0}$ . Since $\Delta_{g_{\mathcal{C}}}U_{N}=V_{N}$ , the standard $W^{2,p}$ - implies that regularity for every $1<p<\infty$ ,

(5.229)

\|U_{N}\|_{W^{2,p}(B_{1}(\bm{x}))}\leq C_{p,\bm{x}}\cdot(\|V_{N}\|_{C^{0}(B_{2}(\bm{x}))}+(\|U_{N}\|_{C^{0}(B_{2}(\bm{x}))}).

By assumption $v\in C^{3K_{0}}(\mathcal{C}^{n})$ for $K_{0}\geq 2n+1$ , so it follows that $\|V_{N}\|_{C^{2}(B_{2}(\bm{x}))}\leq C_{\bm{x}}$ . Therefore, for every $1<p<\infty$ ,

(5.230)

\|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 it suffices to choose $p>2n$ , so the Sobolev embedding implies

(5.231)

\|U_{N}\|_{C^{3,\alpha}(B_{1}(\bm{x}))}\leq C_{p,\bm{x}},\ \alpha\equiv 1-\frac{2n}{p},

which implies that $U_{N}\to u$ in the $C^{3}$ -norm with respect to $g_{\mathcal{C}^{n}}$ . The proof of the first step is done.

We have constructed a solution $u$ satisfying $|u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}$ . Now we are ready to show that

(5.232)

|\nabla u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

This can be accomplished by the elliptic $W^{2,p}$ -estimate. Since a Calabi space $(\mathcal{C}^{n},g_{\mathcal{C}^{n}})$ is collapsed with bounded curvatures as $z\to+\infty$ , so there is some constant $r_{0}>0$ such that for each $\bm{x}\in\mathcal{C}^{n}$ satisfying $z(\bm{x})\geq 1$ , the universal cover $(\widetilde{B_{2r_{0}}(\bm{x})},\tilde{\bm{x}})$ is non-collapsing. Now we lift the solution $u$ to this non-collapsing local universal cover, then for any $p>1$ , there exists $C_{p}>0$ such that

(5.233)

|u|_{W^{2,p}(B_{r_{0}}(\tilde{\bm{x}}))}\leq C_{p}\cdot(|u|_{L^{\infty}(B_{2r_{0}}(\tilde{\bm{x}}))}+|v|_{L^{\infty}(B_{2r_{0}}(\tilde{\bm{x}}))})\leq C_{p}\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

We can choose any $p>2n$ , then Sobolev embedding gives

(5.234)

|u|_{C^{1,\alpha}(B_{r_{0}}(\tilde{\bm{x}}))}\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

In particular,

(5.235)

|\nabla u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}},

where $\alpha\equiv 1-\frac{2n}{p}$ . So the proof of the proposition is done.

∎

5.6. Proof of the Liouville theorem

In this subsection, we will complete the proof of Theorem 5.2.

To begin with, we prove the following lemma, which states that any harmonic function with slow exponential growth rate on a $\delta$ -asymptotically Calabi space is in fact almost harmonic with repsect to the Calabi model metric.

Lemma 5.17.

Let $(X^{2n},g)$ be a complete non-compact Riemannian manifold which is $\delta$ -asymptotically Calabi space in the sense of Definition 5.1. Let $\underline{\delta}\in(0,\delta/10)$ be a constant such that $u$ satisfies

(5.236)

\displaystyle\begin{split}\Delta_{g}u&=0\\ u&=O(e^{\underline{\delta}\cdot z^{\frac{n}{2}}}),\end{split}

then there exists $z_{0}>0$ , such that for every fixed $k\in\mathbb{Z}_{+}$ , we have for all $z\geq z_{0}$ ,

(5.237)

\|\nabla^{k}_{g_{\mathcal{C}^{n}}}\Delta_{g_{\mathcal{C}^{n}}}u(z,\bm{y})\|\leq C_{k}\cdot e^{-\frac{\delta}{2}\cdot z^{\frac{n}{2}}},

where $C_{k}$ is a constant depending only on $X$ and $k$ .

The proof of this is essentially the same as the proof of Claim 4.18 in [HSVZ18]. We omit the details here. By quite explicit computations, the curvatures of the Calabi model space are uniformly bounded as $z\to+\infty$ , which allows us to use the local elliptic estimate even though the geometry is collapsing at infinity.

Proof of Theorem 5.2.

We let

(5.238)

\underline{\delta}_{0}\equiv\min(\frac{\delta}{10},\delta_{b}).

Let $u$ be a harmonic function on the $\delta$ -asymptotically Calabi space $(X^{2n},g)$ , which satisfies

(5.239)

u=O(e^{\underline{\delta}_{0}\cdot z^{\frac{n}{2}}}).

By assumption, there exists some large constant $z_{1}\gg 1$ , and a diffeomorphism

(5.240)

\Phi:[z_{1},+\infty)\times Y^{2n-1}\to X^{2n}\setminus K

such that for all $k\in\mathbb{N}$

(5.241)

\|\nabla^{k}(\Phi^{*}g-g_{\mathcal{C}^{n}})\|_{g_{\mathcal{C}^{n}}}\leq Ce^{-\delta\cdot z^{\frac{n}{2}}}.

By the Lemma 5.17, there is some large constant $z_{0}\gg 1$ such that

(5.242)		$\displaystyle\Delta_{g_{\mathcal{C}^{n}}}u$	$\displaystyle=\phi,$
(5.243)		$\displaystyle\|\nabla^{k}_{g_{\mathcal{C}^{n}}}\phi\|$	$\displaystyle=O(e^{-\delta z^{\frac{n}{2}}})$

for all $z\geq z_{0}$ and $k\in\mathbb{N}$ .

Then applying Proposition 5.16 on $[z_{0},+\infty)\times Y^{2n-1}$ , there exists a solution to the equation

(5.244)

\Delta_{g_{\mathcal{C}^{n}}}v=\phi

such that

(5.245)

|v|+|\nabla v|=O(e^{-\ell\cdot z^{\frac{n}{2}}})

for any $\ell\in(0,\delta/2)$ . Notice that, as $z\to+\infty$ , curvatures are uniformly bounded in the Calabi space. Therefore, we have

(5.246)

\displaystyle 0=\Delta_{g}(u)=\Delta_{g_{\mathcal{C}^{n}}}(u-v),

and $u-v=O(e^{\underline{\delta}_{0}\cdot z^{\frac{n}{2}}})$ . Now we are in a position to apply Proposition 5.14 to $u-v$ , which shows that there is some harmonic function $h$ on the Calabi space such that

(5.247)

u-v=\kappa_{0}\cdot z+c_{0}+h,

where $|h|+|\nabla h|=O(e^{-\underline{\delta}\cdot z^{\frac{n}{2}}})$ for all $\underline{\delta}\in(0,\delta_{b})$ . Also $|dz|_{g_{\mathcal{C}^{n}}}\rightarrow 0$ as $z\rightarrow\infty$ , then

(5.248)

|du|_{g}\leq C|du|_{g_{\mathcal{C}^{n}}}\leq C(|dv|_{g_{\mathcal{C}^{n}}}+|dz|_{g_{\mathcal{C}^{n}}}+|dw|_{g_{\mathcal{C}^{n}}})\rightarrow 0,\ \ \ \ z\rightarrow\infty.

Since $\Delta_{g}u=0$ , so it holds that

(5.249)

\Delta(du)=dd^{*}(du)=-d\Delta_{g}u=0.

By assumption, $(X^{2n},g)$ satisfies $\Ric_{g}\geq 0$ , then Bochner’s formula implies that

(5.250)

\frac{1}{2}\Delta_{g}|du|^{2}=|\nabla du|^{2}+\Ric_{g}(du,du)\geq 0.

Applying the decay property of $|du|$ in (5.248) and the maximum principle,

(5.251)

|du|_{g}\equiv 0\ \text{on}\ X^{2n}.

Therefore, $u$ is a constant.

∎

	$\displaystyle\|\xi_{k}(z)\|$	$\displaystyle=\Big\|\int_{Y^{2n-1}}\xi\cdot\varphi_{k}\Big\|=\Big\|\int_{Y^{2n-1}}\xi\cdot\frac{(-\Delta_{h_{0}})^{K_{0}}\varphi_{k}}{(\Lambda_{k})^{K_{0}}}\Big\|$
(5.179)			$\displaystyle\leq\frac{1}{(\Lambda_{k})^{K_{0}}}\int_{Y^{2n-1}}\|\Delta_{h_{0}}^{K_{0}}\xi\|\cdot\|\varphi_{k}\|$
(5.180)			$\displaystyle\leq\frac{C\|\xi\|_{C^{2K_{0}}(Y^{2n-1})}}{(\Lambda_{k})^{K_{0}}},$

5. A Liouville Theorem on asymptotically Calabi spaces [053E]

5. A Liouville Theorem on asymptotically Calabi spaces

Definition 5.1 (δ\delta-aysmptotically Calabi space).

Theorem 5.2 (Liouville Theorem).

Remark 5.2.1.

Remark 5.2.2.

5.1. Separation of variables and ODE reduction

Remark 5.2.3.

Remark 5.2.4.

5.2. The case jk=0j_{k}=0: uniform estimates and asymptotics

Proposition 5.3.

Proof.

Corollary 5.3.1.

Proof.

Lemma 5.4.

Proposition 5.5.

Proof.

Corollary 5.5.1.

5.3. The case jk≠0j_{k}\neq 0: uniform estimates and asymptotics

Proposition 5.6.

Proof.

Lemma 5.7.

Lemma 5.8.

Proof.

Proposition 5.9.

Proof.

Corollary 5.9.1.

Proposition 5.10.

Proof.

Lemma 5.11.

Proof.

Lemma 5.12.

Remark 5.12.1.

Proof.

Corollary 5.12.1.

5.4. Asymptotics of harmonic functions on the Calabi model space

Lemma 5.13.

Proof.

Proposition 5.14 (Asymptotics of harmonic functions).

Proof.

5.5. The Poisson equation with prescribed asymptotics

Lemma 5.15.

Proof.

Proposition 5.16.

Proof.

5.6. Proof of the Liouville theorem

Lemma 5.17.

Proof of Theorem 5.2.

Definition 5.1 ( $\delta$ -aysmptotically Calabi space).

5.2. The case $j_{k}=0$ : uniform estimates and asymptotics

5.3. The case $j_{k}\neq 0$ : uniform estimates and asymptotics