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Proof.
Since by construction
(4.338)
T n − 2 n ω = T π ∗ ω D + d d c ϕ . T^{\frac{n-2}{n}}\omega=T\pi^{*}\omega_{D}+dd^{c}\phi.
We have
(4.339)
Δ T 2 n − 2 n ω ϕ = n − T ⋅ Tr T 2 n − 2 n ω π ∗ ω D . \Delta_{T^{\frac{2n-2}{n}}\omega}\phi=n-T\cdot\Tr_{T^{\frac{2n-2}{n}}\omega}\pi^{*}\omega_{D}.
Since π ∗ ω D \pi^{*}\omega_{D} is smooth on 𝒱 \mathcal{V} and ω \omega is parallel, again the above discussion gives that
(4.340)
| Tr T 2 n − 2 n ω π ∗ ω D | C δ , ν , μ 1 , α ( 𝒱 ) ≤ e C T . |\Tr_{T^{\frac{2n-2}{n}}\omega}\pi^{*}\omega_{D}|_{C^{1,\alpha}_{\delta,\nu,\mu}(\mathcal{V})}\leq e^{CT}.
So by Proposition 4.22 we get that
(4.341)
| ϕ | C δ , ν , μ 3 , α ( 𝒱 ) ≤ e C T + C | ϕ | C 0 δ , ν , μ ( π − 1 ( U β ) ∩ { | z | ≤ 1 } ) . |\phi|_{C^{3,\alpha}_{\delta,\nu,\mu}(\mathcal{V})}\leq e^{CT}+C|\phi|_{C^{0}_{\delta,\nu,\mu}(\pi^{-1}(U_{\beta})\cap\{|z|\leq 1\})}.
To bound the right hand side we use the formula
(4.342)
ϕ = ∫ T + z u h ( u ) 𝑑 u + ϕ ( T + ) = ∫ T + 0 u h ( u ) 𝑑 u + ϕ ( T + ) + ∫ 0 z u h ( u ) 𝑑 u . \phi=\int_{T_{+}}^{z}uh(u)du+\phi(T_{+})=\int_{T_{+}}^{0}uh(u)du+\phi(T_{+})+\int_{0}^{z}uh(u)du.
Hence
(4.343)
ϕ = T 2 z 2 + 1 2 r + B T + O ( 1 ) , \phi=\frac{T}{2}z^{2}+\frac{1}{2}r+B_{T}+O(1),
which gives
(4.344)
| ϕ | C 0 δ , ν , μ ( π − 1 ( U β ) ∩ { | z | ≤ 1 } ) ≤ O ( T m ) |\phi|_{C^{0}_{\delta,\nu,\mu}(\pi^{-1}(U_{\beta})\cap\{|z|\leq 1\})}\leq O(T^{m})
for some m > 0 m>0 . The conclusion then follows.
∎