5. Liouville theorem for half-harmonic 1-forms [03I6]

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5. Liouville theorem for half-harmonic 1-forms

We call a $1$ -form $\gamma$ half-harmonic if it is in the kernel of the operator $\mathscr{D}\equiv d^{+}\oplus d^{*}$ . If the manifold is compact, then such a form is automatically harmonic, but this is no longer true on a non-compact manifold. The main result of this section is the following.

Theorem 5.1.

Let $(X^{4}=M\setminus D,\omega)$ be given by the Tian-Yau construction where $M$ is a smooth Fano manifold of complex dimension $2$ and $D$ is a smooth anti-canonical divisor. Then there is some positive constant $\delta_{h}>0$ which depends only on the geometry of $X^{4}$ such that if a 1-form $\gamma$ on $X$ satisfies

(5.1)

d^{+}\gamma=d^{*}\gamma=0

and

(5.2)

|\gamma|_{\omega}=O(e^{\delta_{h}r^{\frac{2}{3}}})

as $r\to\infty$ , where $r$ is the distance function on $X$ to a fixed point, then $\gamma=0$ .

Proof.

We begin with an interpretation of the equation (5.1) in terms of complex geometric data. Notice in the Tian-Yau construction we have a preferred complex structure on $X$ induced from $M$ , which we denote by $J$ . With respect to $J$ we can write $\gamma=\gamma^{1,0}+\gamma^{0,1}$ with $\gamma^{1,0}=\overline{\gamma^{0,1}}$ . Then by the Kähler identities we have

(5.3)

d^{+}\gamma=0\Longleftrightarrow\begin{cases}\bar{\partial}\gamma^{0,1}=0\\ \sqrt{-1}(\bar{\partial}^{*}\gamma^{0,1}-\partial^{*}\gamma^{1,0})=0\end{cases}

and

(5.4)

d^{*}\gamma=0\Longleftrightarrow\bar{\partial}^{*}\gamma^{0,1}+\partial^{*}\gamma^{1,0}=0.

Thus, equation (5.1) is equivalent to

(5.5)

\bar{\partial}\gamma^{0,1}=0,\ \bar{\partial}^{*}\gamma^{0,1}=0.

The theorem follows from Theorem 4.3 once we prove that there exists some small $\delta>0$ and a smooth function $f=O(e^{\delta z})$ such that $\bar{\partial}f=\gamma$ (note that $\Delta f=\bar{\partial}^{*}\gamma=0$ ).

We next give a brief outline of the proof. In Step 1, we will construct a solution $f$ to $\bar{\partial}f=\gamma$ such that $f=O(e^{\epsilon z^{2}})$ for all $\epsilon>0$ . This is done using a complex geometric argument which amounts to an application of Hörmander’s weighted $L^{2}$ estimates for the $\bar{\partial}$ -operator. Interestingly it does not seem to be possible to obtain the required improvement $f=O(e^{\delta z})$ using only this type of method, owing to the fact that the function $z^{a}$ is plurisubharmonic on the Calabi model space $(\mathcal{C},g_{\mathcal{C}})$ if and only if $a\geq 2$ .

To overcome this problem we use the elliptic theory on $(\mathcal{C},g_{\mathcal{C}})$ developed in Section 4. Thanks to the bound $f=O(e^{\epsilon z^{2}})$ for all $\epsilon>0$ from Step 1 and the $O(e^{-(\frac{1}{2}-\epsilon)z^{2}})$ complex structure asymptotics of Proposition 3.4, it follows that $\bar{\partial}_{\mathcal{C}}f=O(e^{\delta z})$ on $(\mathcal{C},g_{\mathcal{C}})$ . In particular, since $\Delta={\rm tr}(\sqrt{-1}\partial\bar{\partial})$ , the Poisson equation estimates of Proposition 4.15 imply that $f$ can be decomposed into an $O(e^{\delta z})$ part $f_{1}$ and a $g_{\mathcal{C}}$ -harmonic part $f_{2}$ which is $O(e^{\epsilon z^{2}})$ for all $\epsilon>0$ (see Step 2 for details). Observe that it would not be possible to compare $\Delta_{TY}f$ and $\Delta_{\mathcal{C}}f$ directly because $g_{TY}$ and $g_{\mathcal{C}}$ are only asymptotic at rate $O(e^{-\delta z})$ , which is too slow to beat the $O(e^{\epsilon z^{2}})$ growth of $f$ from Step 1.

Step 3 analyzes the $g_{\mathcal{C}}$ -harmonic part $f_{2}$ of $f$ . It is clear from Section 4 that $f_{2}=O(e^{Cz})$ for some large constant $C$ . The required improvement $f_{2}=O(e^{\delta z})$ comes from the first-order equation $\bar{\partial}_{\mathcal{C}}f_{2}=O(e^{\delta z})$ satisfied by $f_{2}$ (in addition to $\Delta_{g_{\mathcal{C}}}f_{2}=0$ ). Technically this is done using separation of variables for the $\bar{\partial}_{\mathcal{C}}$ -operator but the underlying idea can be easily explained: being of $O(e^{Cz})$ rather than $O(e^{Cz^{2}})$ growth, the leading terms of the harmonic function $f_{2}$ must be $S^{1}$ -invariant, but on $S^{1}$ -invariant functions the $\bar{\partial}_{\mathcal{C}}$ -operator directly controls the radial derivative $\frac{\partial}{\partial z}$ .

Step 4 concludes the proof by appealing to Theorem 4.3.

Step 1. In this step, we prove the following proposition.

Proposition 5.2.

There is a smooth function $f$ on $X$ with $\bar{\partial}f=\gamma$ and $|f|=O(e^{\epsilon z^{2}})$ for all $\epsilon>0$ .

Remark 5.3.

We also have $\Delta_{\omega}f=0$ , but at this point we cannot apply Theorem 4.3 directly to conclude that $f$ is a constant since this would require stronger control, $|f|=O(e^{\delta z})$ .

Proof of Proposition 5.2.

We work on the compact manifold $M$ . Let $S$ be a holomorphic section of $K_{M}^{-1}$ with $S^{-1}(0)=D$ , and let $h$ be a smooth hermitian metric on $K_{M}^{-1}$ whose curvature form $\omega_{h}$ is a Kähler form on $M$ with positive Ricci curvature. By Theorem 3.3 near $D$ we have

(5.6)

C^{-1}\sqrt{-1}\partial\bar{\partial}(-{\log|S|^{2}_{h}})^{3/2}\leq\omega_{TY}\leq C\sqrt{-1}\partial\bar{\partial}(-{\log|S|^{2}_{h}})^{3/2}.

By a straightforward computation this implies that

(5.7)

\omega_{TY}\leq C|S|_{h}^{-2}\omega_{h}

and hence, trivially,

(5.8)

|\gamma|_{\omega_{h}}\leq C^{-1}|S|_{h}^{-1}|\gamma|_{\omega_{TY}}=O(|S|_{h}^{-1-\epsilon})

for any $\epsilon>0$ . Define $\alpha=\gamma\otimes S$ . This is a section of $\Lambda_{M}^{0,1}\otimes K_{M}^{-1}$ which lies in $L^{p}_{\omega_{h}}(M,\Lambda^{0,1}_{M}\otimes K_{M}^{-1})$ for all $p\geq 1$ . Since $\bar{\partial}\gamma=0$ , one can directly check that $\bar{\partial}\alpha=0$ in the distributional sense. Now notice that $H^{1}(M,K_{M}^{-1})=H^{1}(M,K_{M}\otimes L)=0$ by the Kodaira vanishing theorem applied to the ample line bundle $L=K_{M}^{-2}$ . Thus, we can define $\beta=\bar{\partial}^{*}\Delta_{\bar{\partial}}^{-1}\alpha$ with respect to $\omega_{h}$ . It follows from elliptic regularity that $\beta\in W^{1,p}_{\omega_{h}}(M,K_{M}^{-1})$ for all $p\geq 1$ , so that $\beta\in C^{\alpha}_{\omega_{h}}(M,K_{M}^{-1})$ for all $\alpha<1$ . Moreover by local regularity we know $\beta$ is smooth outside $D$ and $\bar{\partial}\beta=\alpha$ . Let $f=\beta\otimes S^{-1}$ , then on $X$ we have $\bar{\partial}f=\gamma$ . The immediate estimate we get is that for some constant $C>0$ ,

(5.9)

f=O(|S|_{h}^{-1})=O(e^{Cz^{2}}).

The lemma below allows us to improve (5.9) to the growth order $e^{\epsilon z^{2}}$ for any $\epsilon>0$ . The key point is that the estimate (5.7) can be improved to almost $O(1)$ in directions tangential to $D$ .

Lemma 5.4.

Denote $\beta_{0}:=\beta|_{D}$ , then $\bar{\partial}\beta_{0}=0$ , i.e. $\beta_{0}$ is a holomorphic section of $K_{M}^{-1}|_{D}$ .

Proof.

We choose a finite cover $D=\bigcup_{k=1}^{N_{0}}O_{k}$ such that for each $k$ there exists a local holomorphic coordinate system $(z,w)$ on some domain $U_{k}\subset M$ such that $U_{k}\cap D=O_{k}=\{w=0\}$ . We will show that $\bar{\partial}\beta_{0}=0$ in every $O_{k}\subset D$ in the distributional sense. Let $\psi$ be a smooth section of $\Lambda^{0,1}_{D}\otimes(K_{M}^{-1}|_{D})$ with compact support in $O_{k}$ . It suffices to show that $\langle\beta_{0},\bar{\partial}^{*}\psi\rangle_{O_{k}}=0$ . To this end, write $\psi(z)=\sigma(z)d\overline{z}\otimes(dz\wedge dw)^{-1}$ for some smooth function $\sigma\in C^{\infty}_{0}(O_{k},\mathbb{C})$ and use this to define the trivial extension $\hat{\psi}(z,w)=\sigma(z)d\overline{z}\otimes(dz\wedge dw)^{-1}$ for all $(z,w)\in U_{k}$ . Denote by $O_{k}(\tau)$ the slice $\{w=\tau\}$ in $U_{k}$ , which is a complex submanifold of $M$ , and equip $O_{k}(\tau)$ with the restriction of the Kähler metric $\omega_{h}$ from $M$ . Notice that $\hat{\psi}$ restricts to a smooth section of $\Lambda_{O_{k}(\tau)}^{0,1}\otimes(K_{M}^{-1}|_{O_{k}(\tau)})$ with compact support in $O_{k}(\tau)$ . Since $\bar{\partial}\beta=\alpha$ and $\beta\in W^{1,p}\cap C^{\alpha}$ for any $p\geq 1$ , it follows that

(5.10)

\displaystyle\langle\beta_{0},\bar{\partial}^{*}{\psi}\rangle_{O_{k}}=\lim\limits_{\tau\to 0}\langle\beta,\bar{\partial}^{*}\hat{\psi}\rangle_{O_{k}(\tau)}=\lim\limits_{\tau\to 0}\langle\bar{\partial}\beta,\hat{\psi}\rangle_{O_{k}(\tau)}=\lim\limits_{\tau\to 0}\langle\alpha,\hat{\psi}\rangle_{O_{k}(\tau)}.

Notice that

(5.11)

|\gamma(\partial_{\bar{z}})|\leq|\gamma|_{\omega_{TY}}|\partial_{\bar{z}}|_{\omega_{TY}}\leq|\gamma|_{\omega_{TY}}(-\log|S|^{2}_{h})^{\frac{1}{4}}=O(|S|_{h}^{-\epsilon}).

Since $\alpha=\gamma\otimes S$ , it then follows that $|\alpha(\partial_{\bar{z}})|=O(|S|_{h}^{1-\epsilon})\rightarrow 0$ uniformly as $w\rightarrow 0$ . Using (5.10), it follows that

(5.12)

\langle\beta_{0},\bar{\partial}^{*}{\psi}\rangle_{O_{k}}=0,

as desired. By standard elliptic regularity, $\beta_{0}$ is a holomorphic section. ∎

Since $M$ is Fano we have $H^{1}(M,\mathcal{O}_{M})=0$ so by a standard exact sequence ([GH94, p.139]) the restriction map $H^{0}(M,K_{M}^{-1})\rightarrow H^{0}(D,K_{M}^{-1}|_{D})$ is surjective. This means we can find some $\beta_{1}$ $\in$ $H^{0}(M,K_{M}^{-1})$ such that $\beta_{1}|_{D}=\beta_{0}|_{D}$ . Let $f=(\beta-\beta_{1})\otimes S^{-1}$ . Then we still have $\bar{\partial}f=\gamma$ on $X$ but now since $\beta-\beta_{1}=0$ on $D$ and $\beta\in C^{\alpha}_{\omega_{h}}(M,\mathbb{C})$ for all $\alpha<1$ , we finally obtain Proposition 5.2.∎

Step 2. Let $f=u+\sqrt{-1}v$ be the smooth function constructed by Proposition 5.2 with $\Delta_{\omega_{TY}}u=\Delta_{\omega_{TY}}v=0$ . In this step, we reduce the problem to a question on the Calabi model space through the diffeomorphism $\Phi:(\mathcal{C}\setminus K^{\prime},\omega_{C},J_{C})\rightarrow(X\setminus K,\omega_{TY},J_{TY})$ chosen in Proposition 3.4. The main point is to obtain the decomposition $u=u_{1}+u_{2}$ and $v=v_{1}+v_{2}$ such that $u_{1}=O(e^{\delta z})$ , $v_{1}=O(e^{\delta z})$ and $\Delta_{\omega_{\mathcal{C}}}u_{2}=\Delta_{\omega_{\mathcal{C}}}v_{2}=0$ . The growth estimates for $u_{2}$ and $v_{2}$ will be shown in Step 3.

The idea of the proof of Step 2 is as follows. First, we will estimate $\Delta_{\omega_{\mathcal{C}}}u$ and $\Delta_{\omega_{\mathcal{C}}}v$ and all of their derivatives. Specifically, we will prove that they have slow exponential growth rates (as shown in (5.23)). Then applying Proposition 4.15, we can construct solutions to the Poisson equations

(5.13)

\displaystyle\Delta_{\omega_{\mathcal{C}}}u_{1}=\Delta_{\omega_{\mathcal{C}}}u,\ \Delta_{\omega_{\mathcal{C}}}v_{1}=\Delta_{\omega_{\mathcal{C}}}v,

such that $u_{1}=O(e^{\delta z})$ and $v_{1}=O(e^{\delta z})$ This completes the desired decomposition of $u$ and $v$ .

To obtain the derivative estimates for $\Delta_{\omega_{\mathcal{C}}}u$ and $\Delta_{\omega_{\mathcal{C}}}v$ , we will prove the derivative estimates for $du+J_{\mathcal{C}}dv$ . To start with, by the assumption on $\gamma$ and the first order equation given by Step 1,

(5.14)

du+J_{TY}dv=Re(\gamma)=O(e^{C\delta_{h}z}).

Applying the asymptotic estimate for $J_{TY}$ in Proposition 3.4, we can convert the above growth control to the corresponding estimate for $du+J_{\mathcal{C}}dv$ .

In fact, applying Item (b) of Proposition 3.4, for any $\epsilon>0$ and for any $k\in\mathbb{N}$ ,

(5.15)

|\nabla_{g_{\mathcal{C}}}^{k}(\Phi^{*}J_{TY}-J_{\mathcal{C}})|_{g_{\mathcal{C}}}=O(e^{(-\frac{1}{2}+\epsilon)z^{2}}).

We also need derivative estimates for $u$ and $v$ with respect to the model metric $\omega_{\mathcal{C}}$ . Notice that by Step 1, $f=u+\sqrt{-1}v=O(e^{\epsilon z^{2}})$ for any $\epsilon>0$ , which implies that

(5.16)

\displaystyle|u|+|v|

\displaystyle\leq 2|f|=O(e^{\epsilon r^{4/3}})=O(e^{\epsilon z^{2}}),

where $z$ is the natural coordinate on $\mathcal{C}$ . Since $u$ and $v$ satisfy $\Delta_{\omega_{TY}}u=\Delta_{\omega_{TY}}v=0,$ by applying the same $W^{k,p}$ -estimate as in the proof of Theorem 4.3, we have for all $\epsilon>0$ and $k\geq 1$ ,

(5.17)

|\nabla^{k}u|_{\omega_{TY}}=O(e^{\epsilon z^{2}}),\ |\nabla^{k}v|_{\omega_{TY}}=O(e^{\epsilon z^{2}}).

Since the asymptotic order of harmonic functions $u$ and $v$ is dominated by $e^{\epsilon z^{2}}$ and the asymptotic order of the metric $\omega_{TY}$ is $e^{-\underline{\delta}z}$ , so in terms of the model metric we have

(5.18)

|\nabla^{k}u|_{\omega_{\mathcal{C}}}\leq Ce^{\frac{\epsilon z^{2}}{2}},\ |\nabla^{k}v|_{\omega_{\mathcal{C}}}\leq Ce^{\frac{\epsilon z^{2}}{2}}.

Now we apply the assumption $|\gamma|_{\omega_{TY}}=O(e^{C\delta_{h}z})$ and the above elliptic regularity to (5.14), we get for $k\in\mathbb{N}$ ,

	$\displaystyle\|\nabla^{k}(du+J_{\mathcal{C}}dv)\|_{\omega_{\mathcal{C}}}$	$\displaystyle=\Big\|\nabla^{k}(du+J_{TY}dv)+\nabla^{k}\Big((J_{\mathcal{C}}-\Phi^{*}J_{TY})dv\Big)\Big\|_{\omega_{\mathcal{C}}}$
(5.19)			$\displaystyle=O(e^{C_{k}\delta_{h}z})+O(e^{-\frac{z^{2}}{4}})=O(e^{C_{k}\delta_{h}z}).$

Now we proceed to prove the derivative estimates for $\Delta_{\omega_{\mathcal{C}}}u$ and $\Delta_{\omega_{\mathcal{C}}}v$ by making use of the system

(5.20)

du+J_{TY}dv=Re(\gamma).

The advantage of the above equation is that $\Delta_{\omega_{TY}}=\Tr_{\omega_{\mathcal{C}}}(dJ_{TY}d)$ so that the behavior $\Delta_{\omega_{TY}}$ will follow from the asymptotics of $J_{TY}$ . In fact, taking the differential of (5.20),

(5.21)

dJ_{TY}du=dJ_{TY}Re(\gamma),\ dJ_{TY}dv=dRe(\gamma).

Then using Item $(a)$ of Proposition 3.4, similar to the above we have for all $k\in\mathbb{Z}_{+}$

(5.22)

|\nabla^{k}dJ_{\mathcal{C}}du|=O(e^{C_{k}\delta_{h}z}),\ |\nabla^{k}dJ_{\mathcal{C}}dv|=O(e^{C_{k}\delta_{h}z}).

Taking the trace, then we obtain

(5.23)

|\nabla^{k}\Delta_{\omega_{\mathcal{C}}}u|=O(e^{C_{k}\delta_{h}z}),\ |\nabla^{k}\Delta_{\omega_{\mathcal{C}}}v|=O(e^{C_{k}\delta_{h}z}).

Applying the linear theory for $\Delta_{\omega_{\mathcal{C}}}$ in Proposition 4.15, if $\delta_{h}\ll\underline{\delta}$ , then we choose two solutions $u_{1}$ and $v_{1}$ provided by Proposition 4.15

(5.24)

\Delta_{\omega_{\mathcal{C}}}u_{1}=\Delta_{\omega_{\mathcal{C}}}u\ \text{and}\ \Delta_{\omega_{\mathcal{C}}}v_{1}=\Delta_{\omega_{\mathcal{C}}}v

such that $u_{1}$ , $v_{1}$ satisfy

(5.25)

|\nabla^{k}u_{1}|_{\omega_{\mathcal{C}}}=O(e^{C_{k}\delta_{h}z}),\ |\nabla^{k}v_{1}|_{\omega_{\mathcal{C}}}=O(e^{C_{k}\delta_{h}z}).

So we have finished the proof of the decomposition $u=u_{1}+u_{2}$ and $v=v_{1}+v_{2}$ such that

(5.26)

\Delta_{\omega_{\mathcal{C}}}u_{2}=\Delta_{\omega_{\mathcal{C}}}v_{2}=0.

We also obtain that

(5.27)

|du_{2}+J_{\mathcal{C}}dv_{2}|_{\omega_{\mathcal{C}}}=O(e^{C\delta_{h}z})

and

(5.28)

|u_{2}|=O(e^{\epsilon z^{2}}),\ |v_{2}|=O(e^{\epsilon z^{2}}).

Step 3. Now we estimate the harmonic functions $u_{2}$ and $v_{2}$ with respect to the model metric $\omega_{\mathcal{C}}$ using separation of variables. The goal is to improve the growth order of $u_{2}$ and $v_{2}$ from $O(e^{\epsilon z^{2}})$ for all $\epsilon>0$ to $O(z)$ , using the fact that they also satisfy a first-order equation.

Proposition 5.5.

We have

(5.29)

\displaystyle|u_{2}|=O(z),\ |v_{2}|=O(z).

Remark 5.6.

The operator $(u,v)\mapsto du+J_{\mathcal{C}}dv$ has a kernel which consists of pairs $(u,v)$ such that $u+\sqrt{-1}v$ is holomorphic. In our case this is eliminated since we have the growth control (5.28). Notice that the smallest growth rate of a non-constant holomorphic function on $\mathcal{C}$ is $e^{\frac{1}{2}z^{2}}.$

Proof of Proposition 5.5.

We denote

\psi=du_{2}+J_{\mathcal{C}}dv_{2}=O(e^{C\delta_{h}z}),

Then we have the following expansion as in Section 4.1: let $\{\Lambda_{k}\}_{k=1}$ be the spectrum of $Y^{3}$ and $\{\varphi_{k}\}_{k=1}^{\infty}$ are the corresponding eigenfunctions on $Y$ with $\mathcal{L}_{\partial_{\theta}}\varphi_{k}=\sqrt{-1}j_{k}\varphi_{k}$ ,

(5.30)

u_{2}=\sum\limits_{k}f_{k}(z)\varphi_{k}(z_{\alpha},\theta),\ v_{2}=\sum\limits_{k}g_{k}(z)\varphi_{k}(z_{\alpha},\theta),

which implies that

(5.31)

du_{2}=\sum\limits_{k}(f_{k}^{\prime}(z)\cdot\varphi_{k}\cdot dz+f_{k}(z)\cdot d\varphi_{k}),\ dv_{2}=\sum\limits_{k}(g_{k}^{\prime}(z)\varphi_{k}\cdot dz+g_{k}(z)\cdot d\varphi_{k}).

On $\mathcal{C}$ by the definition in Section 4.1, we have $J_{\mathcal{C}}(zdz)=d\theta$ , so we have

(5.32)

\displaystyle\begin{split}\sum\limits_{k}\big(f_{k}^{\prime}(z)-\sqrt{-1}j_{k}\cdot z\cdot g_{k}(z)\big)\varphi_{k}&=\psi(\partial_{z})\\ \sum\limits_{k}\big(z^{-1}\cdot g_{k}^{\prime}(z)+\sqrt{-1}j_{k}\cdot f_{k}(z)\big)\varphi_{k}&=\psi(\partial_{\theta}).\end{split}

This implies that for each $k$ ,

(5.33)

\begin{split}f_{k}^{\prime}(z)-\sqrt{-1}j_{k}\cdot z\cdot g_{k}(z)=O(e^{C\delta_{h}z})\\ z^{-1}\cdot g_{k}^{\prime}(z)+\sqrt{-1}j_{k}\cdot f_{k}(z)=O(e^{C\delta_{h}z}).\end{split}

There are three different cases.

If $j_{k}\neq 0$ , then we can write $f_{k}$ and $g_{k}$ are given by a linear combination of one growing solution $\mathcal{F}_{k}$ and one decaying solution $\mathcal{U}_{k}$ . Using the analysis in Section 4 we know that the asymptotic order of $\mathcal{F}_{k}$ is $e^{\frac{j_{k}z^{2}}{2}}$ (see Lemma 4.7). The control (5.28) then implies both $f_{k}$ and $g_{k}$ can only be a multiple of the decaying solution $\mathcal{U}_{k}=O(e^{-\frac{j_{k}z^{2}}{2}})$ .

If $j_{k}=0$ and $\lambda_{k}\neq 0$ , then $f_{k}$ and $g_{k}$ are given by linear combinations of the exponential functions of the form $e^{\sqrt{\lambda_{k}}z}$ and $e^{-\sqrt{\lambda_{k}}z}$ . Let $\underline{\delta}>0$ be the positive constant given in Proposition 4.10, we use (5.33) and the fact that $\psi=O(e^{C\delta_{h}z})$ to conclude that, if $C\delta_{h}<\underline{\delta}$ , then both $f_{k}$ and $g_{k}$ must be proportional to the decaying solutions.

If $j_{k}=0$ and $\varphi_{k}$ is constant, then $f_{k}$ and $g_{k}$ are linear functions in $z$ . Now since $u_{2}$ and $v_{2}$ are harmonic functions on $\mathcal{C}$ , by Lemma 4.11, we conclude that

(5.34)

|u_{2}|=O(z),\ |v_{2}|=O(z).

This completes the proof of Proposition 5.5.∎

Step 4. We now complete the proof of Theorem 5.1. By Proposition 5.5 and (5.25),

(5.35)

|u|=O(e^{C\delta_{h}z}),\ |v|=O(e^{C\delta_{h}z}).

If we further choose $C\delta_{h}\leq\ell_{0}$ , where $\ell_{0}$ is the constant of Theorem 4.3, we conclude that $u$ and $v$ must be constant, hence $\gamma=0$ . ∎