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)

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

∎