(of Theorem 5.2) Let $0<\varepsilon<1$ be given and fixed. Let $d_{0}$ be chosen as in the proof of Corollary 5.2. For any $p\in M$, let $\eta_{p}:M\rightarrow\mathbb{R}_{+}$ be a cut-off function such that $\eta_{p}(p)=1$, $\eta_{p}\equiv 1-\theta$ outside the ball $B_{\frac{d_{0}}{2}}(p)$, with the estimate $|\nabla\eta_{p}|^{2}\leq\frac{4\theta^{2}}{d_{0}^{2}}$, $|\nabla^{2}\eta_{p}|\leq\frac{4\theta}{d_{0}^{2}}$. Here $0<\theta<1$ is to be determined later. Let $\delta>0$, $\lambda>0$ be constants to be determined. Assume the function $e^{\delta(F+\varepsilon\psi-\lambda\varphi)}$ achieves maximum at $p_{0}\in M$. We now compute

(5.9)

$$\begin{split}&\Delta_{\varphi}\big(e^{\delta(F+\varepsilon\psi-\lambda\varphi)}\eta_{p_{0}}\big)\\ &=\Delta_{\varphi}(e^{\delta(F+\varepsilon\psi-\lambda\varphi)})\eta_{p_{0}}+e^{\delta(F+\varepsilon\psi-\lambda\varphi)}\Delta_{\varphi}(\eta_{p_{0}})+e^{\delta(F+\varepsilon\psi-\lambda\varphi)}2\delta\nabla_{\varphi}(F+\varepsilon\psi-\lambda\varphi)\cdot\nabla_{\varphi}\eta_{p_{0}}\\ &=e^{\delta(F+\varepsilon\psi-\lambda\varphi)}\eta_{p_{0}}\big(\delta^{2}|\nabla_{\varphi}(F+\varepsilon\psi-\lambda\varphi)|_{\varphi}^{2}+\delta\Delta_{\varphi}(F+\varepsilon\psi-\lambda\varphi)\big)\\ &\qquad\qquad\qquad+e^{\delta(F+\varepsilon\psi-\lambda\varphi)}\Delta_{\varphi}(\eta_{p_{0}})+e^{\delta(F+\varepsilon\psi-\lambda\varphi)}2\delta\nabla_{\varphi}(F+\varepsilon\psi-\lambda\varphi)\cdot\nabla_{\varphi}\eta_{p_{0}}.\end{split}$$

First we can estimate

(5.10)

$$\begin{split}e^{\delta(F+\varepsilon\psi-\lambda\varphi)}&\Delta_{\varphi}\eta_{p_{0}}\geq-e^{\delta(F+\varepsilon\psi-\lambda\varphi)}|\nabla^{2}\eta_{p_{0}}|tr_{\varphi}g\\ &\geq-e^{\delta(F+\varepsilon\psi-\lambda\varphi)}\frac{4\theta}{d_{0}^{2}(1-\theta)}\eta_{p_{0}}tr_{\varphi}g.\end{split}$$

(5.11)

$$\begin{split}&2\delta\nabla_{\varphi}(F+\varepsilon\psi-\lambda\varphi)\cdot\nabla_{\varphi}\eta_{p_{0}}\geq-\delta^{2}\eta_{p_{0}}|\nabla_{\varphi}(F+\varepsilon\psi-\lambda\varphi)|^{2}_{\varphi}-\frac{|\nabla_{\varphi}\eta_{p_{0}}|_{\varphi}^{2}}{\eta_{p_{0}}}\\ &\geq-\delta^{2}\eta_{p_{0}}|\nabla_{\varphi}(F+\varepsilon\psi-\lambda\varphi)|_{\varphi}^{2}-\frac{|\nabla\eta_{p_{0}}|^{2}tr_{\varphi}g}{\eta_{p_{0}}}\\ &\geq-\delta^{2}\eta_{p_{0}}|\nabla_{\varphi}(F+\varepsilon\psi-\lambda\varphi)|_{\varphi}^{2}-\frac{4\theta^{2}tr_{\varphi}g}{d_{0}^{2}(1-\theta)}.\end{split}$$

Finally we compute

(5.12)

$$\begin{split}\Delta_{\varphi}(&F+\varepsilon\psi-\lambda\varphi)=-(\underline{R}+\lambda n)+tr_{\varphi}Ric+\lambda tr_{\varphi}g+\varepsilon\Delta_{\varphi}\psi\\ &\geq(-\underline{R}-\lambda n+\varepsilon nA_{\Phi}^{-\frac{1}{n}}\Phi^{\frac{1}{n}}(F))+(\lambda-\varepsilon-|Ric|)tr_{\varphi}g.\end{split}$$

Here $A_{\Phi}=\int_{M}e^{F}\Phi(F)dvol_{g}$. In the above calculation, we noticed that

$$\begin{split}\Delta_{\varphi}\psi=g_{\varphi}^{i\bar{j}}&(g_{i\bar{j}}+\psi_{i\bar{j}})-tr_{\varphi}g\geq n\big(\det(g_{\varphi}^{i\bar{j}})\det(g_{i\bar{j}}+\psi_{i\bar{j}})\big)^{\frac{1}{n}}-tr_{\varphi}g\\ &=n(e^{-F}e^{F}\Phi(F)A_{\Phi}^{-1})^{\frac{1}{n}}-tr_{\varphi}g.\end{split}$$

Plug (5.10), (5.11), (5.12) back into (5.9), we see

(5.13)

$$\begin{split}\Delta_{\varphi}&\big(e^{\delta(F+\varepsilon\psi-\lambda\varphi)}\eta_{p_{0}}\big)\geq\delta\eta_{p_{0}}e^{\delta(F+\varepsilon\psi-\lambda\varphi)}(-\underline{R}-\lambda n+\varepsilon nA_{\Phi}^{-\frac{1}{n}}\Phi^{\frac{1}{n}}(F))\\ &-e^{\delta(F+\varepsilon\psi-\lambda\varphi)}\big(\delta\eta_{p_{0}}(\lambda-\varepsilon-|Ric|)-\frac{4\theta}{d_{0}^{2}(1-\theta)}\eta_{p_{0}}-\frac{4\theta^{2}}{d_{0}^{2}(1-\theta)^{2}}\big)tr_{\varphi}g.\end{split}$$

Now we choose various constants $\delta$, $\lambda$ and $\theta$ appearing above.

Since $0<\varepsilon<1$, first we choose $\lambda=2(1+\max_{M}|Ric|)$. Then we fix $\lambda$, and choose $\delta$ to be $2n\delta\lambda=\alpha$. We need to make sure the coefficient in front of $tr_{\varphi}g$ to be positive. This can be achieved by choosing $\theta$ to be sufficiently small. Indeed, with above choice of $\delta$ and $\lambda$, we may calculate:

(5.14)

$$\begin{split}&\delta\eta_{p_{0}}(\lambda-\varepsilon-|Ric|)-\frac{4\theta\eta_{p_{0}}}{d_{0}^{2}(1-\theta)}-\frac{4\theta^{2}}{d_{0}^{2}(1-\theta)^{2}}\\ &\geq\frac{1}{2}\delta(1-\theta)\lambda-\frac{4\theta\eta_{p_{0}}}{d_{0}^{2}(1-\theta)}-\frac{4\theta^{2}}{d_{0}^{2}(1-\theta)^{2}}\geq\frac{(1-\theta)\alpha}{4n}-\frac{4\theta}{d_{0}^{2}(1-\theta)}-\frac{4\theta^{2}}{d_{0}^{2}(1-\theta)^{2}}.\end{split}$$

Hence if we choose $\theta$ small enough, above $\geq 0$. After we made all the choices of $\delta$, $\lambda$, $\theta$, we obtain from (5.13) that

(5.15)

$$\Delta_{\varphi}\big(e^{\delta(F+\varepsilon\psi-\lambda\varphi)}\eta_{p_{0}})\geq\delta\eta_{p_{0}}e^{\delta(F+\varepsilon\psi-\lambda\varphi)}(-\underline{R}-\lambda n+\varepsilon nA_{\Phi}^{-\frac{1}{n}}\Phi^{\frac{1}{n}}(F)).$$

Denote $u=e^{\delta(F+\varepsilon\psi-\lambda\varphi)}$. Now we are ready to apply Alexandroff estimate in $B_{d_{0}}(p_{0})$:

(5.16)

$$\begin{split}\sup_{B_{d_{0}}(p_{0})}&u\eta_{p_{0}}\leq\sup_{\partial B_{d_{0}}(p_{0})}u\eta_{p_{0}}\\ &+C_{n}d_{0}\bigg(\int_{B_{d_{0}}(p_{0})}\frac{u^{2n}\big((-\underline{R}-\lambda n+\varepsilon nA_{\Phi}^{-\frac{1}{n}}\Phi^{\frac{1}{n}}(F))^{-}\big)^{2n}}{e^{-2F}}dvol_{g}\bigg)^{\frac{1}{2n}}.\end{split}$$

We want to claim the integral appearing on the right hand side is bounded. Indeed, the function been integrated is nonzero only if

$$-\underline{R}-\lambda n+\varepsilon nA_{\Phi}^{-\frac{1}{n}}\Phi^{\frac{1}{n}}(F)<0.$$

By the coercivity of $\Phi$, this will imply an upper bound for $F$, say $F\leq C_{5.3}$, where the constant $C_{5.3}$ depends on $\varepsilon$, the choice of $\Phi$, the integral bound $A_{\Phi}$, and the background metric $g$. With this observation, we see

(5.17)

$$\begin{split}&\int_{B_{d_{0}}(p_{0})}\frac{u^{2n}\big((-\underline{R}-\lambda n+\varepsilon nA_{\Phi}^{-\frac{1}{n}}\Phi^{\frac{1}{n}}(F))^{-}\big)^{2n}}{e^{-2F}}dvol_{g}\\ &\leq\int_{B_{d_{0}}(p_{0})\cap\{F\leq C_{5.3}\}}e^{2n\delta(F+\varepsilon\psi-\lambda\varphi)}e^{2F}(|\underline{R}|+\lambda n)^{2n}dvol_{g}\\ &\leq(|\underline{R}|+\lambda n)^{2n}e^{(2n\delta+2)C_{5.3}}\int_{B_{d_{0}}(p_{0})}e^{2n\delta\varepsilon\psi-2n\delta\lambda\varphi}dvol_{g}.\end{split}$$

But recall $\psi\leq 0$, and $2n\delta\lambda=\alpha$, we know

(5.18)

$$\int_{B_{d_{0}}(p_{0})}e^{2n\delta\varepsilon\psi-2n\delta\lambda\varphi}dvol_{g}\leq\int_{B_{d_{0}}(p_{0})}e^{-\alpha\varphi}dvol_{g}\leq C_{5.4}.$$

Denote $I=(|\underline{R}|+\lambda n)^{2n}e^{(2n\delta+2)C_{5.3}}\int_{B_{d_{0}}(p_{0})}e^{-\alpha\varphi}dvol_{g}$. Now we go back to (5.16) and obtain:

(5.19)

$$u(p_{0})=\sup_{M}u\leq(1-\theta)\sup_{M}u+C_{n}d_{0}I^{\frac{1}{2n}}.$$

Here we recall that $\eta_{p_{0}}\equiv 1-\theta$ on $\partial B_{d_{0}}(p_{0})$. This implies $\sup_{M}u\leq\frac{C_{n}d_{0}I^{\frac{1}{2n}}}{\theta}$. ∎