Consider the fiberwise average

(2.32)

$$\mathcal{V}_{\infty}(z)\equiv\frac{1}{{\rm Area}_{g_{0}}(\mathbb{T}^{2})}\int_{\mathbb{T}^{2}\times\{z\}}V_{\infty}(x,y,z)\dvol_{g_{0}}(x,y).$$

This is well-defined, smooth in $z$ for $z\neq 0$, and continuous at $z=0$. For $z\neq 0$ we have

(2.33)

$$\mathcal{V}_{\infty}^{\prime\prime}(z)=\int_{\mathbb{T}^{2}\times\{z\}}\frac{d^{2}}{dz^{2}}V_{\infty}=-\int_{\mathbb{T}^{2}\times\{z\}}\Delta_{\mathbb{T}^{2}}V_{\infty}=0.$$

This implies that $\mathcal{V}_{\infty}(z)$ is a piecewise linear function. Since $V_{R}(x,y,z)\leq-C_{R}$ for $|z|\geq R$ with $C_{R}\to\infty$ as $R\to\infty$, it follows that $\lim\limits_{z\to+\infty}V_{\infty}(x,y,z)=-\infty$, and hence for all $z>0$ that

(2.34)

$$\mathcal{V}_{\infty}^{\prime}(z)=const\equiv k_{+}<0.$$

Denote $D_{0}\equiv\diam_{g_{0}}(\mathbb{T}^{2})$. Choose $R_{0}>10D_{0}$ large enough so that $V_{\infty}(x,y,z)\leq-1$ in $(\mathbb{T}^{2}\times\mathbb{R})\setminus B_{R_{0}}(p)$. Then for any fixed $q\in(\mathbb{T}^{2}\times\mathbb{R})\setminus B_{2R_{0}}(p)$ and $r\in(2D_{0},4D_{0})$, we can apply the Harnack inequality to the harmonic function $V_{\infty}$, which is negative in the geodesic ball $B_{r}(q)\subset(\mathbb{T}^{2}\times\mathbb{R})\setminus B_{R_{0}}(p)$. More precisely, passing to the universal cover and applying the standard Harnack inequality for positive harmonic functions on a fixed ball in $\mathbb{R}^{3}$, we see that there is a uniform constant $C_{0}>0$ depending only on $D_{0}>0$ such that for all $w_{1},w_{2}\in B_{r}(q)$,

(2.35)

$$\frac{1}{C_{0}}\leq\frac{V_{\infty}(w_{1})}{V_{\infty}(w_{2})}\leq C_{0}.$$

Since the fiber average $\mathcal{V}_{\infty}(z)$ of $V_{\infty}$ is linear in $z$ with slope $k_{+}<0$, (2.35) yields that

(2.36)

$$-C_{2}z\leq V_{\infty}(x,y,z)\leq-C_{1}z$$

for $z\gg 1$, where the constants $C_{1}$ and $C_{2}$ depend only on the constants $C_{0}$ and $k_{+}$.

We denote by $\Lambda_{\mathbb{T}^{2}}=\{\lambda_{j}\}_{j=1}^{\infty}$ the positive spectrum of $-\Delta_{\mathbb{T}^{2}}$ and expand $V_{\infty}$ according to the eigenfunctions of $\Delta_{\mathbb{T}^{2}}$ along the torus fiber $\mathbb{T}^{2}\times\{z\}$ for each fixed $z>0$. This yields

(2.37)

$$V_{\infty}(x,y,z)=(k_{+}z+\beta_{+})+\sum_{j=1}^{\infty}f_{j}(z)h_{j}(x,y),$$

where $k_{+}<0$ is the constant of (2.34) and where

(2.38)

$\displaystyle f_{j}^{\prime\prime}(z)=\lambda_{j}f_{j}(z),\ -\Delta_{\mathbb{T}^{2}}h_{j}=\lambda_{j}h_{j},\ \int_{\mathbb{T}^{2}}|h_{j}|^{2}=1.$

Immediately,

(2.39)

$$f_{j}(z)=c_{j}e^{-\sqrt{\lambda_{j}}z}+c_{j}^{*}e^{\sqrt{\lambda_{j}}z}.$$

Notice that

(2.40)

$$\int_{\mathbb{T}^{2}\times\{z\}}|V_{\infty}|^{2}=(k_{+}z+\beta_{+})^{2}+\sum_{j=1}^{\infty}|f_{j}(z)|^{2}.$$

By the linear growth property (2.36), we obtain that $c_{j}^{*}=0$ for all $j\in\mathbb{Z}_{+}$. Therefore,

(2.41)

$$\int_{\mathbb{T}^{2}\times\{z\}}|V_{\infty}-(k_{+}z+\beta_{+})|^{2}=\sum_{j=1}^{\infty}|c_{j}|^{2}e^{-2\sqrt{\lambda_{j}}z}=O(e^{-2\sqrt{\lambda_{1}}z})\ \text{as}\ z\to+\infty,$$

where $\lambda_{1}>0$ is the minimum of $\Lambda_{\mathbb{T}^{2}}$. To see the $O(e^{-2\sqrt{\lambda_{1}}z})$ estimate, note that the series converges for $z=1$. Applying elliptic regularity to the harmonic function $\widehat{V}_{\infty}\equiv V_{\infty}-(k_{+}z+\beta_{+})$,

(2.42)

$$\|\widehat{V}_{\infty}\|_{W^{2,2}(B_{r/2}(q))}\leq C\|\widehat{V}_{\infty}\|_{L^{2}(B_{r}(q))}\leq Ce^{-2\sqrt{\lambda_{1}}z}$$

for all balls $B_{r}(q)$ as above, where $C$ depends only on the diameter and on the injectivity radius of $\mathbb{T}^{2}$. By the $3$-dimensional Sobolev embedding $W^{2,2}\hookrightarrow C^{0,\frac{1}{2}}$,

(2.43)

$$|V_{\infty}(x,y,z)-(k_{+}z+\beta_{+})|=O(e^{-\sqrt{\lambda_{1}}z})\ \text{as}\ z\to+\infty.$$

Standard elliptic regularity then shows that for any $k\in\mathbb{N}$,

(2.44)

$$|\nabla_{g_{0}}^{k}(V_{\infty}(x,y,z)-(k_{+}z+\beta_{+}))|=O(e^{-\sqrt{\lambda_{1}}z})\ \text{as}\ z\to+\infty.$$

The same argument applies in the case $z\rightarrow-\infty$.

Now we prove the slope relation (2.30). Fix $R>0$. Then by Green’s formula,

(2.45)

$$\mathcal{V}_{\infty}^{\prime}(R)-\mathcal{V}_{\infty}^{\prime}(-R)=\int_{\mathbb{T}^{2}\times[-R,R]}\Delta V_{\infty}.$$

Thus, by the definition of $k_{+}$ and the analogous definition of $k_{-}$,

(2.46)

$$k_{+}\Area_{g_{0}}(\mathbb{T}^{2})-k_{-}\Area_{g_{0}}(\mathbb{T}^{2})=-2\pi.$$

It follows that

(2.47)

$$k_{-}-k_{+}=\frac{2\pi}{\Area_{g_{0}}(\mathbb{T}^{2})}.$$

Since $V_{\infty}$ is symmetric in $z$, it holds that $k_{-}=-k_{+}$ and the claim follows. ∎