We again divide into different regions and estimate separately.

For $|z(\bm{x})|\leq 1$, applying Corollary 3.24.1, we have

(4.307)

$$(\omega_{D}+T^{-1}\psi)^{n-1}=\omega_{D}^{n-1}(1+T^{-1}\Tr_{\omega_{D}}\psi+\sum_{k\geq 2}T^{-k}\Phi_{k})$$

where $\Phi_{k}=O^{\prime}(r^{k-1})$ is independent of $T$. By (4.20) we have

(4.308)

$$T^{-1}h=1+T^{-1}\Tr_{\omega_{D}}\psi+T^{-2}\underline{B}(z).$$

Using (3.316), it is easy to see that

(4.309)

$$\|\mathrm{Err}_{CY}\|_{C^{0}(\{|z(\bm{x})|\leq 1\})}=O(T^{-2}),$$

Immediately, by the definition of the weighted $C^{0}$-norm, we have

(4.310)

$$\|\mathrm{Err}_{CY}\|_{C_{\delta,\nu,\mu}^{0}(\{|z(\bm{x})|\leq 1\})}=O(T^{-2+\frac{\nu}{n}+\mu}),$$

Now consider the region $|z(\bm{x})|\geq 1$, then by (3.349) we may write

(4.311)

$\displaystyle\psi=\begin{cases}(k_{-}z)\cdot\omega_{D}+\xi,&z\leq-1,\\ (k_{+}z)\cdot\omega_{D}+\xi,&z\geq 1,\end{cases}$

where $\xi=\epsilon(z)$. So it follows that

	$\displaystyle(\omega_{D}+T^{-1}\psi)^{n-1}$	$\displaystyle=\Big((1+T^{-1}k_{\pm}z)\omega_{D}+T^{-1}\xi\Big)^{n-1}$
(4.312)			$\displaystyle=\omega_{D}^{n-1}\Big((1+T^{-1}k_{\pm}z)^{n-1}+(1+T^{-1}k_{\pm}z)^{n-2}T^{-1}\Tr_{\omega_{D}}\xi+O(T^{-2})\Big)$

By (4.16), we have

(4.313)

$$T^{-1}h=(1+T^{-1}k_{\pm}z)^{n-1}+T^{-1}\Tr_{\omega_{D}}\xi.$$

So we obtain

(4.314)

$$\mathrm{Err}_{CY}=\Big((1+T^{-1}k_{\pm}z)^{-1}-(1+T^{-1}k_{\pm}z)^{-n+1}\Big)T^{-1}\Tr_{\omega_{D}}\xi+O(T^{-2})$$

Since for $z\in[T_{-},T_{+}]$,

(4.315)

$$U_{T}(z)=T-T^{-\frac{n-2}{2}}(T+k_{\pm}z)^{\frac{n}{2}}=T(1-(1+T^{-1}k_{\pm}z)^{\frac{n}{2}})\leq-\frac{n}{2}\cdot k_{\pm}z.$$

Here we use the following elementary inequality: $(1-x)^{p}\geq 1-px$ for any $p\geq 1$ and $x\in(0,1)$. By Proposition 3.31, the asymptotics $\xi=\epsilon(z)$ has the explicit exponential decaying rate $\epsilon(z)=O(e^{-(1-\tau)\sqrt{\lambda_{1}}z})$ for any $\tau\in(0,1)$. Applying (4.315) and the the assumption

(4.316)

$$0<\delta<\delta_{e}\equiv\frac{\sqrt{\lambda_{1}}}{n(|k_{-}|+|k_{+}|)},$$

we conclude that, as $|z(\bm{x})|\to+\infty$, the growth rate of $e^{\delta U_{T}(z(\bm{x}))}$ is slower than the decaying rate of $\epsilon(z)$.

Therefore,

(4.317)

$$\|\mathrm{Err}_{CY}\|_{C^{0}(\{\|z(\bm{x})\|\geq 1\})}=O(T^{-2}).$$

By the definition of the weighted norm, we have

(4.318)

$$\|\mathrm{Err}_{CY}\|_{C_{\delta,\nu,\mu}^{0}(\{\|z(\bm{x})\|\geq 1\})}=O(T^{-2+\frac{\nu}{n}+\mu}).$$

The weighted $C^{0,\alpha}$-estimate can be obtained in a similar way. It suffices to analyze the Hölder regularity around the singular set $\mathcal{P}$. Notice that a fixed function in $O^{\prime}(r)$ has bounded $C^{0,\alpha}$ norm, so the weighted $C^{0,\alpha}$-estimate is given by

(4.319)

$$\|\mathrm{Err}_{CY}\|_{C_{\delta,\nu,\mu}^{0}(\mathcal{M}_{T})}=O(T^{-2+\frac{\nu+\alpha}{n}+\mu}).$$

∎