4.5. Proof of the Liouville theorem [03I0]

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4.5. Proof of the Liouville theorem

With the above technical preparations, we complete the proof of the main result in this section, Theorem 4.3. We need the following lemma which is an immediate corollary of the Bochner formula and the maximum principle.

Lemma 4.17.

Let $(M^{n},g,p)$ be a complete non-compact manifold with $\Ric_{g}\geq 0$ . Let $\omega$ be a harmonic $1$ -form on $(M^{n},g)$ , i.e., $\Delta_{H}\omega=0$ and assume that

(4.168)

\lim\limits_{d_{g}(x,p)\to+\infty}|\omega(x)|=0,

then $\omega\equiv 0$ on $M^{n}$ .

Proof.

Since $\omega$ is harmonic, by Bochner’s formula,

(4.169)

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

then $|\omega|^{2}$ is subharmonic. Given the asymptotic property (4.168), applying the maximum principle to the above subharmonic function $|\omega|^{2}$ , we have $\omega\equiv 0$ on $M^{n}$ .

∎

Proof of Theorem 4.3.

Let $u$ satisfy $\Delta_{g}u=0$ $(X^{4},g)$ . We also assume that $u$ satisfies the asymptotic behavior

(4.170)

u=O(e^{\ell_{0}z})

for some $\ell_{0}\in(0,1)$ . The main part of the proof is to determine a positive number $\ell_{0}>0$ such that if (4.170) holds, then $u$ has at most linear growth at infinity, which enables us to apply Lemma 4.17.

By assumption, there is a diffeomorphism

(4.171)

\Phi:X^{4}\setminus K\longrightarrow[10^{2},+\infty)\times Y^{3}

such that for all $k\geq 0$

(4.172)

\|g-\Phi^{*}g_{\mathcal{C}}\|_{C^{k}}\leq Ce^{-\delta z}.

To obtain an accurate growth order of $u$ , we will study the equation of $u$ in terms of the metric $g_{\mathcal{C}}$ on the model space $[10^{2},+\infty)\times Y^{3}$ .

First, we will show that a harmonic function on $(X^{4},g)$ with exponential growth is well behaved in terms of the model metric $g_{\mathcal{C}}$ near infinity. Preciesly, we will prove the following claim.

Claim 4.18.

Assume that $(X^{4},g)$ is $\delta$ -asymptotically Calabi. Let $\hat{\delta}\in(0,\delta/10)$ such that $u$ satisfies

(4.173)

\displaystyle\begin{split}\Delta_{g}u&=0\\ u&=O(e^{\hat{\delta}z}),\end{split}

then for every fixed $k\in\mathbb{Z}_{+}$ , let $\bm{x}_{0}\in[T_{0}(k),+\infty)\times Y^{3}$ with $T_{0}(k)\geq 100^{k^{3}}>0$ and denote $z_{0}\equiv z(\bm{x}_{0})$ , we have

(4.174)

\|\nabla^{k}\Delta_{g_{\mathcal{C}}}u(\bm{x}_{0})\|\leq C(k,g)\cdot e^{-\frac{\delta z_{0}}{2}}.

Proof.

Denoting $\phi\equiv(\Delta_{g}-\Delta_{g_{\mathcal{C}}})u$ , then $\Delta_{g}u=0$ implies

(4.175)

\Delta_{g_{\mathcal{C}}}u+\phi=0.

We will show that for each $k\in\mathbb{N}$ we have

(4.176)

\|\nabla^{k}\phi(\bm{x}_{0})\|\leq C(k,g)\cdot e^{-\frac{\delta z_{0}}{2}},

where $C(k,g)>0$ depends only on $k\in\mathbb{N}$ and the curvature bound of the cutoff region $(X^{4}\setminus K,g)$ .

The higher order derivative estimate will be proved by the $W^{k,p}$ -estimate for harmonic functions on the complete space $(X^{4},g)$ . Since the metric $g$ is collapsing near the infinity, the standard elliptic estimate cannot be directly applied. To overcome this difficulty, we will scale up the metric $\tilde{g}=\lambda^{2}g$ such that $B_{1}(\bm{x_{0}})$ is non-collapsing for $\tilde{g}$ which guarantees the elliptic estimate holds in terms of the rescaled metric $\tilde{g}$ . For fixed $\bm{x}_{0}\in[T_{0}(k),+\infty)$ , we take

(4.177)

\lambda=z_{0}^{\frac{1}{2}}

and hence there is some constant $v_{0}>0$ which is independent of the $z$ -coordinate such that

(4.178)

\Vol_{\tilde{g}}(B_{1}(\bm{x_{0}}))\geq v_{0}>0.

By explicit calculation on the model space $\mathcal{C}$ using (4.4) one easily sees that curvatures are uniformly bounded in a ball of definite size of radius, i.e.

(4.179)

\sup\limits_{B_{2}(\bm{x_{0}})}\|\Rm\|_{\tilde{g}}\leq\Lambda_{0},

where $\Lambda_{0}>0$ is independent of the $z$ -coordinate. It follows that for every $k\in\mathbb{N}$ and $1<p<\infty$ , there exists $C(k,v_{0},\Lambda_{0},p)>0$ such that under the rescaled metric $\tilde{g}$ ,

(4.180)

\|u\|_{W_{\tilde{g}}^{k+2,p}(B_{1}(x_{0}))}\leq C\|u\|_{W_{\tilde{g}}^{k,p}(B_{1+\frac{1}{k^{2}}}(x_{0}))},

which implies that for every $k\in\mathbb{Z}_{+}$ ,

(4.181)

\|u\|_{W_{\tilde{g}}^{k,p}(B_{1}(x_{0}))}\leq C\sup\limits_{B_{3}(\bm{x}_{0})}|u|.

Therefore, for every $k\in\mathbb{Z}_{+}$ and sufficiently large $p\in(1,\infty)$ , applying the Sobolev embedding on $(B_{4/3}(\bm{x}_{0}),\tilde{g})$ , there exists $C(k,p,v_{0},\Lambda_{0})>0$ such that

(4.182)

\sup\limits_{B_{1}(\bm{x}_{0})}|\nabla^{k}u|_{\tilde{g}}\leq C\|\nabla^{k+1}u\|_{L^{p}(B_{4/3}(x_{0}))}.

By (4.181) and the growth assumption on $u$ , there is some constant $C>0$ such that

(4.183)

\sup\limits_{B_{1}(\bm{x}_{0})}|\nabla^{k}u|_{\tilde{g}}\leq C\sup\limits_{B_{2}(\bm{x}_{0})}|u|\leq Ce^{\hat{\delta}z_{0}}.

In terms of the original metric $g$ , we have

(4.184)

|\nabla^{k}u(\bm{x}_{0})|_{g}\leq\sup\limits_{B_{1/\lambda}(\bm{x}_{0})}|\nabla^{k}u|_{g}\leq C\cdot z_{0}^{2k}e^{\hat{\delta}z_{0}}<Ce^{\delta^{\prime}z_{0}},

where $\delta^{\prime}\in\Big(\hat{\delta},(1+10^{-3})\hat{\delta}\Big)$ .

Next, by (4.172), there is some constant $\delta>0$ such that

(4.185)

\|\Phi^{*}g_{\mathcal{C}}-g\|_{C^{k}(B_{2}(\bm{x}_{0}))}\leq C_{k}e^{-\delta z}.

then the elliptic estimate (4.184) and (4.185) imply that

(4.186)

|\phi(\bm{x}_{0})|=|(\Delta_{g}-\Delta_{g_{\mathcal{C}}})u(\bm{x}_{0})|\leq Ce^{-\frac{\delta\cdot z_{0}}{2}}

and similarly

(4.187)

|\nabla^{k}\phi(\bm{x}_{0})|\leq C_{k}e^{-\frac{\delta\cdot z_{0}}{2}}.

∎

The above error estimate enables us to construct a harmonic function with respect to the model metric $g_{\mathcal{C}}$ on $[10^{2},+\infty)\times Y^{3}$ which has at most linear growth and is exponentially close to the original function $u$ . Let

(4.188)

\ell_{0}\in\Big(0,\min\{\frac{\delta}{10^{2}},\underline{\delta}\}\Big),

where $\underline{\delta}>0$ is the constant in Proposition 4.10. By assumption the harmonic function $u$ satisfies the asymptotic behavior,

(4.189)

u=O(e^{\ell_{0}z}).

Then applying the above claim and Proposition 4.15 on $[T_{0},+\infty)\times Y^{3}$ , there exists a solution to the equation

(4.190)

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

such that

(4.191)

v=O(e^{-\ell z})

for some $\ell\in(-\delta/2,0)$ . Therefore, combine (4.175) and (4.190), we have

(4.192)

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

and $u+v=O(e^{\ell_{0}z})$ . Since $0<\ell_{0}<1$ has been specified in (4.188), now we are in a position to apply Proposition 4.10 to $u+v$ , which shows that

(4.193)

(u+v)=az+b+O(e^{-\underline{\delta}z}),

and hence in the non-compact part $[T_{0},+\infty)\times Y^{3}$ ,

(4.194)

\displaystyle u=az+b+O(e^{-\delta^{\prime\prime}z}),\ \delta^{\prime\prime}\equiv\min\{\ell,\underline{\delta}\}.

The above asymptotics immediately implies that

(4.195)

|du|_{g}\to 0\ \text{as}\ z\to\infty.

Let $\Delta_{H}$ be the Hodge-Laplacian on $(X^{4},g)$ . Since $\Delta_{g}u=0$ , it holds that

(4.196)

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

Since the complete space $(X^{4},g)$ satisfies $\Ric_{g}\geq 0$ , and $|du|$ satisfies the decay property (4.195), applying Lemma 4.17 implies that

(4.197)

|du|_{g}\equiv 0\ \text{on}\ X^{4}.

Therefore, $u$ has to be a constant. ∎