By (5),

	$\displaystyle(D_{\sigma}\mathfrak{F}_{k}(y,\sigma)-D_{\sigma}\mathfrak{F}_{k}(0,0))\dot{\sigma}$	$\displaystyle=$	$\displaystyle((D_{\sigma}(\Pi^{*}d\alpha_{k}|_{L(y,\sigma)})-D_{\sigma}(d\alpha_{k}|_{L}))\dot{\sigma},$
			$\displaystyle*_{h}(D_{\sigma}(\Pi^{*}d{\rm Im}\beta_{k}|_{L(y,\sigma)})-D_{\sigma}(d{\rm Im}\beta_{k}|_{L}))\dot{\sigma}).$

We can take a $k_{0}\gg 1$ such that, for $k>k_{0}$,

	$\displaystyle\|D_{\sigma}\mathfrak{F}_{k}(y,\sigma)-D_{\sigma}\mathfrak{F}_{k}(0,0)\|$	$\displaystyle\leq$	$\displaystyle 2C\|(d\alpha_{k},d\beta_{k})\|_{C^{1,\alpha}(Y_{2r},g)}(\sum_{l=0,1,\cdots,n}\|\sigma\|_{C^{1,\alpha}(L,h)}^{l})$
		$\displaystyle\leq$	$\displaystyle 2C\|(d\alpha_{k},d\beta_{k})\|_{C^{1,\alpha}(Y_{2r},g)}n\delta_{0}$
		$\displaystyle\leq$	$\displaystyle\frac{1}{4\overline{C}},$

by (2), (7) and (8). We obtain the first formula in the conclusion.

Note that $D_{\sigma}\mathfrak{F}_{k}(y,\sigma)=D_{\sigma}\mathfrak{F}_{k}(0,0)+(D_{\sigma}\mathfrak{F}_{k}(y,\sigma)-D_{\sigma}\mathfrak{F}_{k}(0,0))$, $D_{\sigma}\mathfrak{F}_{k}(0,0)$ is invertible, and $\|D_{\sigma}\mathfrak{F}_{k}(0,0)^{-1}\|\leq\overline{C}$. By the same arguments as in the proof of Lemma 4.3, and

$$\|D_{\sigma}\mathfrak{F}_{k}(0,0)^{-1}(D_{\sigma}\mathfrak{F}_{k}(y,\sigma)-D_{\sigma}\mathfrak{F}_{k}(0,0))\|\leq\frac{1}{2},$$

$D_{\sigma}\mathfrak{F}_{k}(y,\sigma)$ is also invertible, and

$$\|D_{\sigma}\mathfrak{F}_{k}(y,\sigma)^{-1}\|\leq(\sum_{j=0}^{\infty}2^{-j})\|D_{\sigma}\mathfrak{F}_{k}(0,0)^{-1}\|\leq 2\overline{C}.$$

∎