By definition,

		$\displaystyle\Big(\mathscr{N}(\sqrt{-1}\partial\bar{\partial}\phi_{1})-\mathscr{N}(\sqrt{-1}\partial\bar{\partial}\phi_{2})\Big)\cdot\omega(t)^{n}$
(6.35)		$\displaystyle=$	$\displaystyle\sum\limits_{k=2}^{n}\begin{pmatrix}n\\ k\end{pmatrix}\cdot\omega(t)^{n-k}\wedge\Big((\sqrt{-1}\partial\bar{\partial}\phi_{1})^{k}-(\sqrt{-1}\partial\bar{\partial}\phi_{2})^{k}\Big).$

By the definition of the norm on $\mathfrak{S}_{1}$, we have

(6.36)

$$\|\sqrt{-1}\partial\bar{\partial}\phi_{1}\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{T})}\leq\varrho,\quad\|\sqrt{-1}\partial\bar{\partial}\phi_{1}\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{T})}\leq\varrho.$$

With $\mu$ specified by (6.13), by Lemma 4.20, the weight function $\rho_{\delta,\nu+2,\mu}^{(\alpha)}:\mathcal{M}_{T}\to\mathbb{R}_{+}$ satisfies for any $\bm{x}\in\mathcal{M}_{T}$,

(6.37)

$$\rho_{\delta,\nu+2,\mu}^{(\alpha)}(\bm{x})\geq 1.$$

This implies the following weight-free estimates,

(6.38)

$$\|\sqrt{-1}\partial\bar{\partial}\phi_{1}\|_{C^{0,\alpha}(\mathcal{M}_{T})}\leq C_{0}\cdot\varrho\quad\text{and}\quad\|\sqrt{-1}\partial\bar{\partial}\phi_{2}\|_{C^{0,\alpha}(\mathcal{M}_{T})}\leq C_{0}\cdot\varrho,$$

where $C_{0}>0$ is a uniform constant independent of $T\gg 1$.

Since the $L^{\infty}$-norm of the Kähler form $\omega_{T}$ is bounded by a uniform constant (independent of $T\gg 1$), so the above estimates imply the pointwise estimate for $\mathscr{N}$,

(6.39)

$$|\mathscr{N}(\sqrt{-1}\partial\bar{\partial}\phi_{1})-\mathscr{N}(\sqrt{-1}\partial\bar{\partial}\phi_{2})|\leq C_{N}\cdot\varrho\cdot|\sqrt{-1}\partial\bar{\partial}(\phi_{1}-\phi_{2})|,$$

where $C_{N}>0$ is a uniform constant independent of $T$. Write the above in terms of the weighted norms, we have

(6.40)

$$\|\mathscr{N}(\sqrt{-1}\partial\bar{\partial}\phi_{1})-\mathscr{N}(\sqrt{-1}\partial\bar{\partial}\phi_{2})\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{T})}\leq C_{N}\cdot\varrho\cdot\|\sqrt{-1}\partial\bar{\partial}(\phi_{1}-\phi_{2})\|_{C_{\delta,\nu+2,\mu}^{0,\alpha}(\mathcal{M}_{T})}.$$

The proof is done.

∎