Given $K$, for all $t>0$ small enough the sets $\lambda_{t}^{-1}(K)$ are all contained in a fixed compact set $K^{\prime}\subset B\times\mathbb{C}^{n-m}$. We wish to deduce (4.16) from (4.11). To see this, write on $B\times\mathbb{C}^{n-m}$

$$\begin{split}\lambda_{t}^{*}p^{*}T_{-\sigma}^{*}\tilde{\omega}_{t}=&\sqrt{-1}\bigg(\sum_{i,j}A_{i\overline{j}}(t,y,z)dz^{i}\wedge d\overline{z}^{j}+\sum_{i,j}B_{i\overline{j}}(t,y,z)dy^{i}\wedge d\overline{y}^{j}\\ &+\sum_{i,j}C_{i\overline{j}}(t,y,z)dy^{i}\wedge d\overline{z}^{j}+\sum_{i,j}D_{i\overline{j}}(t,y,z)dz^{i}\wedge d\overline{y}^{j}\bigg).\end{split}$$

Thanks to (4.11), on $K^{\prime}$ the coefficents $A,B,C,D$ satisfy uniform $C^{k}$ estimates in the variables $(y,z)$ independent of $t$. We then pull back this equation via the map $\lambda_{1/t}$ (the inverse of $\lambda_{t}$) and get

$$\begin{split}p^{*}T_{-\sigma}^{*}\tilde{\omega}_{t}&=\sqrt{-1}\bigg(t\sum_{i,j}A_{i\overline{j}}(t,y,z\sqrt{t})dz^{i}\wedge d\overline{z}^{j}+\sum_{i,j}B_{i\overline{j}}(t,y,z\sqrt{t})dy^{i}\wedge d\overline{y}^{j}\\ &+\sqrt{t}\sum_{i,j}C_{i\overline{j}}(t,y,z\sqrt{t})dy^{i}\wedge d\overline{z}^{j}+\sqrt{t}\sum_{i,j}D_{i\overline{j}}(t,y,z\sqrt{t})dz^{i}\wedge d\overline{y}^{j}\bigg),\end{split}$$

and the new coefficients are uniformly bounded in $C^{k}$ on $K$, thus proving (4.16). ∎