Let us first calculate $\Delta_{\varphi}(|\nabla_{\varphi}f|_{\varphi}^{2})$ for any smooth function $f$ in $M.\;$ First we do the calculation under an orthonormal frame $g_{\varphi}$.

$$\begin{array}[]{lcl}\Delta_{\varphi}|\nabla_{\varphi}f|^{2}&=&(f_{i}f_{\bar{i}})_{,j\bar{j}}\\ &=&f_{,ij\bar{j}}f_{\bar{i}}+f_{i}f_{,\bar{i}j\bar{j}}+|f_{,ij}|_{\varphi}^{2}+|f_{,i\bar{j}}|_{\varphi}^{2}\\ &=&f_{,ji\bar{j}}f_{\bar{i}}+f_{i}f_{,j\bar{i}\bar{j}}+|f_{,ij}|_{\varphi}^{2}+|f_{,i\bar{j}}|_{\varphi}^{2}\\ &=&(\Delta_{\varphi}f)_{i}f_{\bar{i}}+f_{i}(\Delta_{\varphi}f)_{\bar{i}}+Ric_{\varphi,i\bar{j}}f_{j}f_{\bar{i}}+|f_{,ij}|_{\varphi}^{2}+|f_{,i\bar{j}}|_{\varphi}^{2}.\end{array}$$

In the above, $f_{,ij\cdots}$ denote covariant derivatives under the metric $g_{\varphi}$. Let $B(\lambda):\mathbb{R}\rightarrow\mathbb{R}$ be a smooth function, now we calculate $\Delta_{\varphi}(e^{B(f)}|\nabla_{\varphi}f|_{\varphi}^{2})$.

(4.3)

$$\begin{split}&e^{-B(f)}\cdot\Delta_{\varphi}(e^{B(f)}|\nabla_{\varphi}f|_{\varphi}^{2})\\ &=\Delta_{\varphi}(|\nabla_{\varphi}f|_{\varphi}^{2})+B^{\prime}(f_{i}(|\nabla_{\varphi}f|_{\varphi}^{2})_{\bar{i}}+f_{\bar{i}}(|\nabla_{\varphi}f|_{\varphi}^{2})_{i})\\ &\quad\quad\quad\quad+\left((B^{\prime 2}+B^{\prime\prime})|\nabla_{\varphi}f|^{2}+B^{\prime}\Delta_{\varphi}f\right)|\nabla_{\varphi}f|_{\varphi}^{2}\\ &=(\Delta_{\varphi}f)_{i}f_{\bar{i}}+f_{i}(\Delta_{\varphi}f)_{\bar{i}}+Ric_{\varphi,i\bar{j}}f_{j}f_{\bar{i}}+|f_{,ij}|_{\varphi}^{2}+|f_{,i\bar{j}}|_{\varphi}^{2}\\ &+B^{\prime}\left(f_{i}f_{j}f_{,\bar{j}\bar{i}}+f_{i}f_{,j\bar{i}}f_{\bar{j}}+f_{\bar{i}}f_{j}f_{,\bar{j}i}+f_{\bar{i}}f_{,ji}f_{\bar{j}}\right)\\ &\quad\quad\quad\quad\quad\quad+\left((B^{\prime 2}+B^{\prime\prime})|\nabla_{\varphi}f|_{\varphi}^{2}+B^{\prime}\Delta_{\varphi}f\right)|\nabla_{\varphi}f|_{\varphi}^{2}\\ &\geq(\Delta_{\varphi}f)_{i}f_{\bar{i}}+f_{i}(\Delta_{\varphi}f)_{\bar{i}}+Ric_{\varphi,i\bar{j}}f_{j}f_{\bar{i}}+|f_{,i\bar{j}}|_{\varphi}^{2}\\ &+B^{\prime}\left(f_{i}f_{,j\bar{i}}f_{\bar{j}}+f_{\bar{i}}f_{j}f_{,\bar{j}i}\right)+\left(B^{\prime\prime}|\nabla_{\varphi}f|_{\varphi}^{2}+B^{\prime}\Delta_{\varphi}f\right)|\nabla_{\varphi}f|_{\varphi}^{2}.\end{split}$$

In the inequality above, we noticed and dropped the following complete square:

$$B^{\prime 2}|\nabla_{\varphi}f|_{\varphi}^{4}+B^{\prime}f_{i}f_{j}f_{,\bar{j}\bar{i}}+B^{\prime}f_{\bar{i}}f_{\bar{j}}f_{,ij}+|f_{,ij}|_{\varphi}^{2}=|f_{,ij}+B^{\prime}f_{i}f_{j}|_{\varphi}^{2}.$$

We apply above calculation to $F$. Notice that

$$Ric_{\varphi,i\bar{j}}=R_{i\bar{j}}-F_{i\bar{j}}.$$

Set $B^{\prime}={1\over 2},\;$and we switch to normal coordinate of $g$ (c.f. (2.1)), then we have

(4.4)

$$\begin{split}e^{-{F\over 2}}\Delta_{\varphi}(e^{F\over 2}|&\nabla_{\varphi}F|^{2})\geq{{(\Delta_{\varphi}F)_{i}F_{\bar{i}}+(\Delta_{\varphi}F)_{\bar{i}}F_{i}}\over{1+\varphi_{i\bar{i}}}}+\frac{R_{j\bar{i}}F_{i}F_{\bar{j}}}{(1+\varphi_{i\bar{i}})(1+\varphi_{j\bar{j}})}\\ &\qquad\qquad+\frac{|F_{i\bar{\alpha}}|^{2}}{(1+\varphi_{i\bar{i}})(1+\varphi_{\alpha\bar{\alpha}})}+{1\over 2}\Delta_{\varphi}F|\nabla_{\varphi}F|^{2}_{\varphi}.\end{split}$$

Next we wish to use the equation satisfied by $F$:

$$\Delta_{\varphi}F=-\underline{R}+tr_{\varphi}Ric.$$

Take derivative with respect to $z_{i}$ on both sides, we obtain:

$$\begin{array}[]{lcl}(\Delta_{\varphi}F)_{i}&=&-g_{\varphi}^{k\bar{p}}g_{\varphi,\bar{p}qi}g_{\varphi}^{q\bar{l}}R_{k\bar{l}}+g_{\varphi}^{k\bar{l}}R_{k\bar{l},i}\\ &=&-{{\varphi_{\bar{k}li}R_{k\bar{l}}}\over{(1+\varphi_{k\bar{k}})(1+\varphi_{l\bar{l}})}}+{R_{k\bar{k},i}\over{1+\varphi_{k\bar{k}}}}.\end{array}$$

Plugging this into equation (4.4), we have

(4.5)

$$\begin{split}e^{-{F\over 2}}\Delta_{\varphi}(e^{F\over 2}|&\nabla_{\varphi}F|^{2})\geq-\frac{F_{\bar{i}}\varphi_{\beta\bar{\alpha}i}R_{\alpha\bar{\beta}}+F_{i}\varphi_{\beta\bar{\alpha}\bar{i}}R_{\alpha\bar{\beta}}}{(1+\varphi_{i\bar{i}})(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}+\frac{F_{\bar{i}}R_{\alpha\bar{\alpha},i}+F_{i}R_{\alpha\bar{\alpha},\bar{i}}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{i\bar{i}})}\\ &+\frac{R_{j\bar{i}}F_{i}F_{\bar{j}}}{(1+\varphi_{i\bar{i}})(1+\varphi_{j\bar{j}})}+\frac{|F_{i\bar{\alpha}}|^{2}}{(1+\varphi_{i\bar{i}})(1+\varphi_{\alpha\bar{\alpha}})}+{1\over 2}(-\underline{R}+\frac{R_{i\bar{i}}}{1+\varphi_{i\bar{i}}})|\nabla_{\varphi}F|^{2}_{\varphi}.\end{split}$$

In the above, $\varphi_{\beta\bar{\alpha}i}$, $R_{\alpha\bar{\alpha},i}$ etc are just usual derivatives taken under the coordinate as specified above. Notice that there will be no more terms like $\frac{F_{j\bar{i}}F_{i}F_{\bar{j}}}{(1+\varphi_{i\bar{i}})(1+\varphi_{j\bar{j}})}$, because the choice $B^{\prime}\equiv\frac{1}{2}$ makes such terms exactly cancel out. Now we proceed further from (4.5). As preparation, we observe that for any $1\leq i\leq n$:

(4.6)

$$\frac{1}{1+\varphi_{i\bar{i}}}=e^{-F}\Pi_{j\neq i}(1+\varphi_{j\bar{j}})\leq e^{-F}(n+\Delta\varphi)^{n-1}.$$

First we can estimate as follows, with various constants $C_{i}$ depending only on $n$, $||F||_{0}$, and the curvature bound of the original metric $g$.

(4.7)

$$|{1\over 2}e^{{F\over 2}}(-\underline{R}+\frac{R_{i\bar{i}}}{1+\varphi_{i\bar{i}}})|\leq C_{4.3}(1+\sum_{i}\frac{1}{1+\varphi_{i\bar{i}}})\leq C_{4.3}(1+ne^{-F}(n+\Delta\varphi)^{n-1}).$$

(4.8)

$$\begin{split}\frac{|F_{\bar{i}}\varphi_{\beta\bar{\alpha}i}R_{\alpha\bar{\beta}}|}{(1+\varphi_{i\bar{i}})(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}&\leq\frac{1}{2}\frac{|\varphi_{\beta\bar{\alpha}i}|^{2}|R_{\alpha\bar{\beta}}|^{2}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}+\frac{1}{2}\frac{|F_{\bar{i}}|^{2}}{(1+\varphi_{i\bar{i}})^{2}(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}\\ &\leq C_{4.4}\frac{|\varphi_{\beta\bar{\alpha}i}|^{2}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}+\frac{C_{4.4}|F_{i}|^{2}}{1+\varphi_{i\bar{i}}}(n+\Delta\varphi)^{3n-3}.\end{split}$$

In the second line of above estimate, we used (4.6) to estimate the extra powers of $\frac{1}{1+\varphi_{\alpha\bar{\alpha}}}$. The conjugate term will satisfy the same estimate as above.

(4.9)

$$\begin{split}|\frac{F_{\bar{i}}R_{\alpha\bar{\alpha},i}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{i\bar{i}})}|&\leq\frac{1}{2}\frac{|F_{\bar{i}}|^{2}|R_{\alpha\bar{\alpha},i}|^{2}}{1+\varphi_{i\bar{i}}}+\frac{1}{2}\frac{1}{(1+\varphi_{\alpha\bar{\alpha}})^{2}(1+\varphi_{i\bar{i}})}\\ &\leq C_{4.5}\frac{|F_{i}|^{2}}{1+\varphi_{i\bar{i}}}+C_{4.5}(n+\Delta\varphi)^{3n-3}.\end{split}$$

Finally

(4.10)

$$\begin{split}|\frac{R_{j\bar{i}}F_{i}F_{\bar{j}}}{(1+\varphi_{i\bar{i}})(1+\varphi_{j\bar{j}})}|&\leq\frac{1}{2}\frac{|R_{j\bar{i}}|^{2}|F_{i}|^{2}}{(1+\varphi_{i\bar{i}})(1+\varphi_{j\bar{j}})}+\frac{1}{2}\frac{|R_{j\bar{i}}|^{2}|F_{\bar{j}}|^{2}}{(1+\varphi_{i\bar{i}})(1+\varphi_{j\bar{j}})}\\ &\leq C_{4.6}\frac{|F_{i}|^{2}}{1+\varphi_{i\bar{i}}}(n+\Delta\varphi)^{n-1}.\end{split}$$

Now combining the estimates in (4.7), (4.8), (4.9), (4.10), we obtain from (4.5):

(4.11)

$$\begin{split}\Delta_{\varphi}(e^{\frac{1}{2}F}|\nabla_{\varphi}F|_{\varphi}^{2})&\geq-C_{4.7}(n+\Delta\varphi)^{3n-3}|\nabla_{\varphi}F|_{\varphi}^{2}-C_{4.7}\frac{|\varphi_{\beta\bar{\alpha}i}|^{2}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}\\ &-C_{4.7}(n+\Delta\varphi)^{3n-3}+\frac{1}{C_{4.7}}\frac{|F_{i\bar{\alpha}}|^{2}}{(1+\varphi_{i\bar{i}})(1+\varphi_{\alpha\bar{\alpha}})}.\end{split}$$

Note that $n+\Delta\varphi\geq ne^{\frac{F}{n}}$, hence has a positive uniform lower bound since $F$ is bounded from below. Here $C_{4.7}$ is some constant depending only on $n$, $||F||_{0}$, and the curvature bound of the original metric $g$. In order to handle the second term on the right hand side, we need to consider $\Delta_{\varphi}(n+\Delta\varphi)$. For this we can recall our calculation in (3.6):

(4.12)

$$\begin{split}\Delta_{\varphi}(n+\Delta\varphi)&=\frac{R_{i\bar{i}\alpha\bar{\alpha}}(1+\varphi_{i\bar{i}})}{1+\varphi_{\alpha\bar{\alpha}}}+\frac{|\varphi_{\alpha\bar{\beta}i}|^{2}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}+F_{i\bar{i}}-R_{i\bar{i}}\\ &\geq\frac{-C_{4.8}(1+\varphi_{i\bar{i}})}{1+\varphi_{\alpha\bar{\alpha}}}+\frac{|\varphi_{\alpha\bar{\beta}i}|^{2}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}+F_{i\bar{i}}-C_{4.8}\\ &\geq-C_{4.9}(n+\Delta\varphi)^{n}+\frac{|\varphi_{\alpha\bar{\beta}i}|^{2}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}+F_{i\bar{i}}-C_{4.8}.\end{split}$$

Let $K>0$ be a constant, we combine (4.11), (4.12), and conclude:

(4.13)

$$\begin{split}\Delta_{\varphi}(e^{\frac{1}{2}F}&|\nabla_{\varphi}F|_{\varphi}^{2}+K(n+\Delta\varphi))\geq-C_{4.7}(n+\Delta\varphi)^{3n-3}|\nabla_{\varphi}F|_{\varphi}^{2}+(K-C_{4.7})\frac{|\varphi_{\alpha\bar{\beta}i}|^{2}}{(1+\varphi_{\alpha\bar{\alpha}})(1+\varphi_{\beta\bar{\beta}})}\\ &-C_{4.7}(n+\Delta\varphi)^{3n-3}-KC_{4.9}(n+\Delta\varphi)^{n}+KF_{i\bar{i}}-KC_{4.8}+\frac{1}{C_{4.7}}\frac{|F_{i\bar{\alpha}}|^{2}}{(1+\varphi_{i\bar{i}})(1+\varphi_{\alpha\bar{\alpha}})}.\end{split}$$

First we choose $K=C_{4.7}+1$, and calculate:

(4.14)

$$|KF_{i\bar{i}}|\leq\frac{1}{2C_{4.7}}\frac{|F_{i\bar{i}}|^{2}}{(1+\varphi_{i\bar{i}})^{2}}+\frac{K^{2}C_{4.7}}{2}(1+\varphi_{i\bar{i}})^{2}\leq\frac{1}{2C_{4.7}}\frac{|F_{i\bar{i}}|^{2}}{(1+\varphi_{i\bar{i}})^{2}}+\frac{nK^{2}C_{4.7}}{2}(n+\Delta\varphi)^{2}.$$

Hence there exists a constant $C_{4.91}$, with the same dependence as said above, such that

(4.15)

$$\begin{split}\Delta_{\varphi}&(e^{\frac{1}{2}F}|\nabla_{\varphi}F|_{\varphi}^{2}+K(n+\Delta\varphi))\geq-C_{4.91}(n+\Delta\varphi)^{3n-3}|\nabla_{\varphi}F|_{\varphi}^{2}-C_{4.91}(n+\Delta\varphi)^{3n-3}\\ &\geq-C_{4.91}e^{-\frac{1}{2}F}(n+\Delta\varphi)^{3n-3}(e^{\frac{1}{2}F}|\nabla_{\varphi}F|_{\varphi}^{2}+K(n+\Delta\varphi))-C_{4.91}(n+\Delta\varphi)^{3n-3}.\end{split}$$

Set

$$u=e^{\frac{1}{2}F}|\nabla_{\varphi}F|_{\varphi}^{2}+K(n+\Delta\varphi),$$

we obtain the key estimate from here:

(4.16)

$$\Delta_{\varphi}u\geq-C_{4.92}(n+\Delta\varphi)^{3n-3}u-C_{4.92}(n+\Delta\varphi)^{3n-3}.$$

Next we plan to do iteration, using (4.16). Notice that for any $p>0$:

(4.17)

$${1\over{2p+1}}\Delta_{\varphi}(u^{2p+1})=u^{2p}\Delta_{\varphi}u+2pu^{2p-1}|\nabla_{\varphi}u|_{\varphi}^{2}.$$

Integrate over $M$, we obtain:

(4.18)

$$\int_{M}2pu^{2p-1}|\nabla_{\varphi}u|_{\varphi}^{2}dvol_{\varphi}=-\int_{M}u^{2p}\Delta_{\varphi}udvol_{\varphi}.$$

Plug in the key estimate (4.16), we get:

(4.19)

$$\begin{split}\int_{M}2pu^{2p-1}|\nabla_{\varphi}u|_{\varphi}^{2}dvol_{\varphi}&\leq C_{4.92}\int_{M}(n+\Delta\varphi)^{3n-3}u^{2p+1}dvol_{\varphi}\\ &+C_{4.92}\int_{M}(n+\Delta\varphi)^{3n-3}u^{2p}dvol_{\varphi}.\end{split}$$

Or equivalently:

(4.20)

$$\int_{M}|\nabla_{\varphi}(u^{p+\frac{1}{2}})|^{2}_{\varphi}dvol_{\varphi}\leq\frac{C_{4.92}(p+\frac{1}{2})^{2}}{2p}\int_{M}(n+\Delta\varphi)^{3n-3}(u^{2p+1}+u^{2p})dvol_{\varphi}.$$

Observe that $u\geq K(n+\Delta\varphi)\geq e^{\frac{F}{n}}$, $u^{2p}\leq u^{2p+1}e^{-\frac{F}{n}}\leq C\cdot u^{2p+1}$. Hence for some constant $C_{3.91}$, we have

(4.21)

$$\int_{M}|\nabla_{\varphi}(u^{p+\frac{1}{2}})|_{\varphi}^{2}dvol_{\varphi}\leq\frac{C_{4.93}(p+\frac{1}{2})^{2}}{2p}\int_{M}(n+\Delta\varphi)^{3n-3}u^{2p+1}dvol_{\varphi}.$$

Since $dvol_{\varphi}=e^{F}dvol_{g}$, and $F$ is bounded, we see

(4.22)

$$\int_{M}|\nabla_{\varphi}(u^{p+\frac{1}{2}})|_{\varphi}^{2}dvol_{g}\leq\frac{C_{4.94}(p+\frac{1}{2})^{2}}{2p}\int_{M}(n+\Delta\varphi)^{3n-3}u^{2p+1}dvol_{g}.$$

Fix $\varepsilon\in(0,2)$ to be determined, we estimate the right hand side of (4.22):

(4.23)

$$\int_{M}(n+\Delta\varphi)^{3n-3}u^{2p+1}dvol_{g}\leq\bigg(\int_{M}u^{(p+\frac{1}{2})(2+\varepsilon)}dvol_{g}\bigg)^{\frac{2}{2+\varepsilon}}\bigg(\int_{M}(n+\Delta\varphi)^{\frac{(3n-3)(2+\varepsilon)}{\varepsilon}}\bigg)^{\frac{\varepsilon}{2+\varepsilon}}.$$

Denote $v=u^{p+\frac{1}{2}}$, then (4.22) now becomes:

(4.24)

$$\int_{M}|\nabla_{\varphi}v|_{\varphi}^{2}dvol_{g}\leq\frac{C_{4.94}(p+\frac{1}{2})^{2}}{2p}\bigg(\int_{M}(n+\Delta\varphi)^{\frac{(3n-3)(2+\varepsilon)}{\varepsilon}}\bigg)^{\frac{\varepsilon}{2+\varepsilon}}\cdot\bigg(\int_{M}v^{2+\varepsilon}dvol_{g}\bigg)^{\frac{2}{2+\varepsilon}}.$$

We estimate the left hand side of (4.24) from below:

(4.25)

$$\begin{split}|\nabla v|^{2-\varepsilon}&\leq\sum_{i}|v_{i}|^{2-\varepsilon}=\sum_{i}\frac{|v_{i}|^{2-\varepsilon}}{(1+\varphi_{i\bar{i}})^{\frac{2-\varepsilon}{2}}}\cdot(1+\varphi_{i\bar{i}})^{\frac{2-\varepsilon}{2}}\\ &\leq\bigg(\sum_{i}\frac{|v_{i}|^{2}}{1+\varphi_{i\bar{i}}}\bigg)^{\frac{2-\varepsilon}{2}}(\sum_{i}(1+\varphi_{i\bar{i}})^{\frac{2-\varepsilon}{\varepsilon}}\bigg)^{\frac{\varepsilon}{2}}\leq|\nabla_{\varphi}v|_{\varphi}^{2-\varepsilon}n^{\frac{\varepsilon}{2}}(n+\Delta\varphi)^{\frac{2-\varepsilon}{2}}.\end{split}$$

Integrate and use Holder inequality, we get:

(4.26)

$$\begin{split}\int_{M}&|\nabla v|^{2-\varepsilon}dvol_{g}\leq n^{\frac{\varepsilon}{2}}\int_{M}|\nabla_{\varphi}v|_{\varphi}^{2-\varepsilon}(n+\Delta\varphi)^{\frac{2-\varepsilon}{2}}dvol_{g}\\ &\leq n^{\frac{\varepsilon}{2}}\bigg(\int_{M}|\nabla_{\varphi}v|_{\varphi}^{2}\bigg)^{\frac{2-\varepsilon}{2}}\bigg(\int_{M}(n+\Delta\varphi)^{\frac{2-\varepsilon}{\varepsilon}}dvol_{g}\bigg)^{\frac{\varepsilon}{2}}.\end{split}$$

Therefore, for $p\geq\frac{1}{2}$, we may apply (4.24) to get:

(4.27)

$$\begin{split}\bigg(\int_{M}|\nabla v|^{2-\varepsilon}dvol_{g}\bigg)^{\frac{2}{2-\varepsilon}}&\leq n^{\frac{\varepsilon}{2-\varepsilon}}\bigg(\int_{M}(n+\Delta\varphi)^{\frac{2-\varepsilon}{\varepsilon}}dvol_{g}\bigg)^{\frac{\varepsilon}{2-\varepsilon}}\int_{M}|\nabla_{\varphi}v|_{\varphi}^{2}dvol_{g}\\ &\leq C_{4.95}pK_{\varepsilon}\bigg(\int_{M}v^{2+\varepsilon}dvol_{g}\bigg)^{\frac{2}{2+\varepsilon}}.\end{split}$$

Here

(4.28)

$$K_{\varepsilon}=n^{\frac{\varepsilon}{2-\varepsilon}}\cdot\bigg(\int_{M}(n+\Delta\varphi)^{\frac{2-\varepsilon}{\varepsilon}}dvol_{g}\bigg)^{\frac{\varepsilon}{2-\varepsilon}}\cdot\bigg(\int_{M}(n+\Delta\varphi)^{\frac{(3n-3)(2+\varepsilon)}{\varepsilon}}\bigg)^{\frac{\varepsilon}{2+\varepsilon}}.$$

Apply the Sobolev embedding with exponent $2-\varepsilon$, and denote $\theta=\frac{2n(2-\varepsilon)}{2n-2+\varepsilon}$ to be the improved integrability, we get

$$||v||_{L^{\theta}(dvol_{g})}\leq C_{sob}(||\nabla v||_{L^{2-\varepsilon}(dvol_{g})}+||v||_{L^{2-\varepsilon}(dvol_{g})}).$$

Recall that $v=u^{p+\frac{1}{2}}$, this means:

(4.29)

$$\begin{split}\bigg(\int_{M}u^{(p+\frac{1}{2})\theta}dvol_{g}&\bigg)^{\frac{2}{\theta}}\leq C_{sob}\bigg(\big(\int_{M}|\nabla(u^{p+\frac{1}{2}})|^{2-\varepsilon}dvol_{g}\big)^{\frac{2}{2-\varepsilon}}+\big(\int_{M}u^{(p+\frac{1}{2})(2-\varepsilon)}dvol_{g}\big)^{\frac{2}{2-\varepsilon}}\bigg)\\ &\leq C_{sob}\bigg(C_{4.95}pK_{\varepsilon}\big(\int_{M}u^{(p+\frac{1}{2})(2+\varepsilon)}dvol_{g}\big)^{\frac{2}{2+\varepsilon}}+\big(\int_{M}u^{(p+\frac{1}{2})(2-\varepsilon)}dvol_{g}\big)^{\frac{2}{2-\varepsilon}}\bigg)\\ &\leq C_{4.96,\varepsilon}p\big(\int_{M}u^{(p+\frac{1}{2})(2+\varepsilon)}dvol_{g}\big)^{\frac{2}{2+\varepsilon}}.\end{split}$$

Here $C_{4.96,\varepsilon}$ has the same dependence as $C_{i}$’s above, but with additional dependence on $\varepsilon$. From the 1st line to 2nd line, we used (4.27). Now choose $\varepsilon>0$ small so that $\theta>2+\varepsilon$, then above estimate indeed improves integrability, namely we need

(4.30)

$$\frac{2n(2-\varepsilon)}{2n-2+\varepsilon}>2+\varepsilon.$$

We fix $\varepsilon$ and (4.29) gives for $p\geq\frac{1}{2}$:

(4.31)

$$||u||_{L^{(p+\frac{1}{2})\theta}}\leq\big(C_{4.97}p\big)^{\frac{1}{p+\frac{1}{2}}}||u||_{L^{(p+\frac{1}{2})(2+\varepsilon)}}.$$

Denote $\chi=\frac{\theta}{2+\varepsilon}>1$, and choose $p+\frac{1}{2}=\chi^{i}$, for $i\geq 0$. Then we obtain:

(4.32)

$$||u||_{L^{(2+\varepsilon)\chi^{i+1}}}\leq\big(C_{4.97}\chi^{i}\big)^{\frac{1}{\chi^{i}}}||u||_{L^{(2+\varepsilon)\chi^{i}}}.$$

It follows that

(4.33)

$$||u||_{L^{\infty}}\leq C_{4.97}^{\sum_{i\geq 0}\frac{1}{\chi^{i}}}\cdot\chi^{\sum_{i\geq 0}\frac{i}{\chi^{i}}}||u||_{L^{2+\varepsilon}}\leq C_{4.97}^{\sum_{i\geq 0}\frac{1}{\chi^{i}}}\cdot\chi^{\sum_{i\geq 0}\frac{i}{\chi^{i}}}||u||_{L^{1}}^{\frac{1}{2+\varepsilon}}||u||_{L^{\infty}}^{\frac{1+\varepsilon}{2+\varepsilon}}.$$

From above we get estimate of $||u||_{L^{\infty}}$ in terms of $||u||_{L^{1}}$. But recall $u=e^{\frac{1}{2}F}|\nabla_{\varphi}F|^{2}+K(n+\Delta\varphi)$, so $L^{1}$ estimate is available.

Indeed, it is clear that $n+\Delta\varphi\in L^{1}$. To see $e^{\frac{1}{2}F}|\nabla_{\varphi}F|_{\varphi}^{2}\in L^{1}$, we just need to show $|\nabla_{\varphi}F|_{\varphi}^{2}\in L^{1}$ since $F$ is now assumed to be bounded. Then we can calculate:

(4.34)

$$\Delta_{\varphi}(F^{2})=2|\nabla_{\varphi}F|_{\varphi}^{2}+2F\Delta_{\varphi}F=2|\nabla_{\varphi}F|_{\varphi}^{2}+2F(-\underline{R}+tr_{\varphi}Ric).$$

Integrate with respect to $dvol_{\varphi}=e^{F}dvol_{g}$, we see

(4.35)

$$\begin{split}\int_{M}e^{F}&|\nabla_{\varphi}F|_{\varphi}^{2}dvol_{g}=\int_{M}e^{F}F(\underline{R}-tr_{\varphi}Ric)dvol_{g}\leq C_{4.98}\int_{M}(1+tr_{\varphi}g)dvol_{g}\\ &\leq C_{3.96}(n+1)vol(M).\end{split}$$

Here $C_{4.98}$ may depend on $||F||_{0}$. To see the range of $p_{n}$ asserted in the Remark 4.3, we notice the choice of $\varepsilon=\frac{1}{2n}$ verifies the requirement in (4.30). With this choice, the highest power of $n+\Delta\varphi$ appearing in (4.28) is exactly $(3n-3)(4n+1)$. Once we have control over $K_{\varepsilon}$, the rest of the proof goes through. ∎