Let $\varphi_{t}$ solve the equation (1.7). Fix $p>n$, and any upper bound ${\overline{E}}_{t}>0$ for $E_{t}(\varphi_{t})$. We shall actually show that

$$\sup_{X}|\varphi_{t}|\leq C_{0}\Big\{\frac{c_{t}^{n}}{V_{t}}\big({\rm Ent}_{p}+1+{\rm exp}(C_{1}{\overline{E}}_{t})\big)\Big\}^{\frac{n}{p-n}}{\overline{E}}_{t}+C(n,p),$$

for constants $C_{0},C_{1}$ depending only on $n,\omega_{X},\chi,\gamma$, and $C(n,p)$ depending only on $n,p$. Throughout the proof we will fix $t\in(0,1]$, but the constants will be independent of $t$, unless stated explicitly otherwise.

For any $s>0$, we let $\Omega_{s}:=\{\varphi_{t}\leq-s\}$ be the sub-level set of $\varphi_{t}$.

We solve an auxiliary complex Monge-Ampère equation on $X$

$$(\omega_{t}+i\partial\bar{\partial}\psi_{t,k})^{n}=\frac{\tau_{k}(-\varphi_{t}-s)}{A_{s,k}}f(\lambda[h_{\varphi_{t}}])^{n}\omega_{X}^{n}=\frac{\tau_{k}(-\varphi_{t}-s)}{A_{s,k}}c_{t}^{n}e^{nF_{t}}\omega_{X}^{n},$$

(2.2)

with $\sup\psi_{t,k}=0$ where $A_{s,k}:=\frac{c_{t}^{n}}{V_{t}}\int_{X}\tau_{k}(-\varphi_{t}-s)e^{nF_{t}}\omega_{X}^{n}$ is chosen so that the integrals of both sides of (2.2) are equal. Note that (2.2) admits a unique smooth solution by Yau’s theorem [25]. We also observe that as $k\to\infty$

$$A_{s,k}\to A_{s}=\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}(-\varphi_{t}-s)e^{nF_{t}}\omega_{X}^{n},$$

(2.3)

which follows from Lebesgue’s dominated convergence theorem. The limit $A_{s}$ satisfies $A_{s}\leq{\overline{E}}_{t}$, the assumed upper bound of $E_{t}(\varphi_{t})$.

Denote $\Phi$ to be the smooth function

$\displaystyle\Phi:=-\varepsilon(-\psi_{t,k}+\Lambda)^{\frac{n}{n+1}}-(\varphi_{t}+s)$

(2.4)

where

$$0<\varepsilon:=(\frac{n+1}{n^{2}})^{\frac{n}{n+1}}A_{s,k}^{\frac{1}{n+1}}\gamma^{{-\frac{1}{n+1}}},\quad 0<\Lambda:=\frac{1}{(n+1)n^{n-1}}\frac{A_{s,k}}{\gamma}$$

(2.5)

where $\gamma>0$ is the constant in the structure condition (1.4) of $f$.

Since $X$ is compact without boundary, the maximum of $\Phi$ must be attained at some point, say, $x_{0}\in X$. If $x_{0}\in X\backslash\Omega_{s}^{\circ}$, then

$$\sup_{X}\Phi=\Phi(x_{0})=-\varepsilon(-\psi_{t,k}(x_{0})+\Lambda)^{\frac{n}{n+1}}-(\varphi_{t}(x_{0})+s)<-\varphi_{t}(x_{0})-s\leq 0.$$

Otherwise $x_{0}\in\Omega_{s}^{\circ}$. Then at $x_{0}$, $i\partial\bar{\partial}\Phi(x_{0})\leq 0$, and on the right hand side of (2.2) we have

$$\tau_{k}(-\varphi_{t}-s)(x_{0})=-(\varphi_{t}(x_{0})+s)+1/k>0$$

by the definition of $\tau_{k}$ in (2.1).

We denote by $G^{i\bar{j}}=\frac{\partial\,{\rm log}\,f(\lambda[h])}{\partial h_{ij}}=\frac{1}{f}\frac{\partial f(\lambda[h])}{\partial h_{ij}}$ the coefficients of the linearization of the operator $\,{\rm log}\,f(\lambda[h])$ with $h=\omega_{X}^{-1}\cdot\omega_{t,\varphi_{t}}$. By the ellipticity assumption of $f(\lambda[\cdot])$, $(G^{i\bar{j}})$ is positive definite. Moreover, by the structure condition (1.4) on $f$, we have

$${\rm det}\,G^{i\bar{j}}=f^{-n}{\rm det}\big(\frac{\partial f(\lambda[h])}{\partial h_{ij}}\big)\geq\frac{\gamma}{f(\lambda)^{n}}.$$

Recall that the eigenvalues of $h$ are by definition $\lambda=(\lambda_{1},\ldots,\lambda_{n})$. Working in a basis where $h$ is diagonal and $\omega_{X}$ the identity, we find using the definition of $G^{i\bar{j}}$ that

$$\sum_{i,j}G^{i\bar{j}}(\omega_{t,\varphi_{t}})_{\bar{j}i}=\frac{1}{f(\lambda)}\sum_{j}\frac{\partial f(\lambda)}{\partial\lambda_{j}}\lambda_{j}=1$$

where we have used the assumption that $f$ is homogeneous of degree one, so that $\sum_{i}\lambda_{i}\frac{\partial f(\lambda)}{\partial\lambda_{i}}=f(\lambda)$. At the maximum point $x_{0}$ of $\Phi$, we can write

	$\displaystyle 0$	$\displaystyle\geq$	$\displaystyle G^{i\bar{j}}\Phi_{\bar{j}i}(x_{0})$
		$\displaystyle=$	$\displaystyle\frac{n\varepsilon}{n+1}(-\psi_{t,k}+\Lambda)^{-\frac{1}{n+1}}G^{i\bar{j}}(\psi_{t,k})_{\bar{j}i}+\frac{n\varepsilon}{(n+1)^{2}}(-\psi_{t,k}+\Lambda)^{-\frac{n+2}{n+1}}G^{i\bar{j}}(\psi_{t,k})_{\bar{j}}(\psi_{t,k})_{i}-G^{i\bar{j}}(\varphi_{t})_{\bar{j}i}$
		$\displaystyle\geq$	$\displaystyle\frac{n\varepsilon}{n+1}(-\psi_{t,k}+\Lambda)^{-\frac{1}{n+1}}G^{i\bar{j}}(\omega_{t,\psi_{t,k}})_{\bar{j}i}-G^{i\bar{j}}(\omega_{t,\varphi_{t}})_{\bar{j}i}+\big(1-\frac{\varepsilon n}{n+1}(-\psi_{t,k}+\Lambda)^{-\frac{1}{n+1}}\big)G^{i\bar{j}}(\omega_{t})_{\bar{j}i}$
		$\displaystyle\geq$	$\displaystyle\frac{\varepsilon n}{n+1}(-\psi_{t,k}+\Lambda)^{-\frac{1}{n+1}}n\Big({\rm det}G^{i\bar{j}}\cdot{\rm det}(\omega_{t,\psi_{t,k}})_{\bar{j}i}\Big)^{1/n}-1+\big(1-\frac{\varepsilon n}{n+1}(-\psi_{t,k}+\Lambda)^{-\frac{1}{n+1}}\big)G^{i\bar{j}}(\omega_{t})_{\bar{j}i}$
		$\displaystyle\geq$	$\displaystyle\frac{\varepsilon n^{2}}{n+1}(-\psi_{t,k}+\Lambda)^{-\frac{1}{n+1}}\gamma^{1/n}\Big(\frac{\tau_{k}(-\varphi_{t}-s)}{A_{s,k}}\Big)^{1/n}-1+\big(1-\frac{\varepsilon n}{n+1}\Lambda^{-\frac{1}{n+1}}\big)G^{i\bar{j}}(\omega_{t})_{\bar{j}i}$
		$\displaystyle\geq$	$\displaystyle\frac{\varepsilon n^{2}\gamma^{1/n}}{n+1}(-\psi_{t,k}+\Lambda)^{-\frac{1}{n+1}}\Big(\frac{-\varphi_{t}-s+1/k}{A_{s,k}}\Big)^{1/n}-1$

where in the third inequality we used the arithmetic-geometric inequality and in the last one we used the choice of $\varepsilon$ and $\Lambda$ in (2.5). Thus at $x_{0}$ we have

$$-(\varphi_{t}+s)(x_{0})<A_{s,k}\Big(\frac{n+1}{n^{2}\varepsilon\gamma^{1/n}}\Big)^{n}(-\psi_{t,k}(x_{0})+\Lambda)^{n/(n+1)}=\varepsilon(-\psi_{t,k}(x_{0})+\Lambda)^{n/(n+1)}$$

which implies that $\Phi(x_{0})\leq 0$. Hence we can conclude that $\sup_{X}\Phi\leq 0$, that is, on $X$

$$\frac{-(\varphi_{t}+s)}{A_{s,k}^{1/(n+1)}}\leq(\frac{n+1}{n^{2}})^{\frac{n}{n+1}}\gamma^{-\frac{1}{n+1}}\big(-\psi_{t,k}+\frac{1}{(n+1)n^{n-1}}\frac{A_{s,k}}{\gamma}\big)^{\frac{n}{n+1}}\leq C_{n}\big(-\psi_{t,k}+\frac{A_{s,k}}{\gamma}\big)^{\frac{n}{n+1}},$$

(2.6)

for some constant $C_{n}$ depending only on $n$ and $\gamma$. Taking the $\big(\frac{n+1}{n}\big)$-th power of both sides of the previous equation, multiplying it by some small $\beta_{0}>0$, taking the exponential on both sides and then integrating the resulting inequality over $\Omega_{s}$, we obtain

$$\int_{\Omega_{s}}\,{\rm exp}\,\Big\{\beta_{0}\big(\frac{-(\varphi_{t}+s)}{A_{s,k}^{1/(n+1)}}\big)^{\frac{n+1}{n}}\Big\}\omega_{X}^{n}\leq\,{\rm exp}\,(C_{n}\beta_{0}A_{s,k})\int_{\Omega_{s}}\,{\rm exp}\,(-C_{n}\beta_{0}\psi_{t,k})\omega_{X}^{n}.$$

(2.7)

Recall that $\omega_{t}+i\partial\bar{\partial}\psi_{t,k}>0$ and $\omega_{t}=\chi+t\omega_{X}$. We may assume $\chi\leq(a_{0}-1)\omega_{X}$ for some $a_{0}=a_{0}(\chi,\omega)>1$, so $\psi_{t,k}$ is also $(a_{0}\omega_{X})$-plurisubharmonic. Now it is a basic fact in Kähler geometry that, for any Kähler class $\hat{\chi}$ on $X$, there is a constant $\alpha=\alpha(X,\hat{\chi})$ so that

$\displaystyle\int_{X}e^{-\alpha_{0}\psi}\omega_{X}^{n}\leq C(\alpha_{0},n,\hat{\chi},\omega_{X})$

(2.8)

for any $\alpha_{0}<\alpha$ and any $\hat{\chi}$-plurisubharmonic function $\psi$ with $\sup_{X}\psi=0$. The local version of this statement is in [13], and the above global version in [20]. We apply this statement with $\hat{\chi}=a_{0}\omega_{X}$, and fix $\alpha_{0}$ with $0<\alpha_{0}<\alpha(X,\hat{\chi})$. Then we choose $\beta_{0}=\beta_{0}(n,\omega_{X},\chi,\gamma)>0$ in (2.7) such that $\beta_{0}C_{n}=\alpha_{0}$, and from (2.7) we can then deduce that

$$\int_{\Omega_{s}}\,{\rm exp}\,\Big\{{\beta_{0}\big(\frac{-(\varphi_{t}+s)}{A_{s,k}^{1/(n+1)}}\big)^{\frac{n+1}{n}}}\Big\}\omega_{X}^{n}\leq Ce^{CA_{s,k}},$$

(2.9)

for some constant $C=C(n,\omega_{X},\chi,\gamma)>0$. Letting $k\to\infty$ in (2.9) we obtain from (2.3)

$$\int_{\Omega_{s}}\,{\rm exp}\,\Big\{{\beta_{0}\big(\frac{-(\varphi_{t}+s)}{A_{s}^{1/(n+1)}}\big)^{\frac{n+1}{n}}}\Big\}\omega_{X}^{n}\leq Ce^{CA_{s}}\leq Ce^{C{\overline{E}}_{t}},$$

(2.10)

for some constant $C=C(n,\omega_{X},\chi,\gamma)>0$. The proof of Lemma 1 is complete.

We come now to the proof of Theorem 2 proper. Fix $p>n$, and define $\eta:{\bf R}_{+}\to{\bf R}_{+}$ by $\eta(x)=(\,{\rm log}\,(1+x))^{p}$. Note that $\eta$ is a strictly increasing function with $\eta(0)=0$, and let $\eta^{-1}$ be its inverse function. If we let

$\displaystyle v:=\frac{\beta_{0}}{2}\big(\frac{-\varphi_{t}-s}{A_{s}^{1/(n+1)}}\big)^{(n+1)/n}$

(2.11)

then we have for any $z\in\Omega_{s}$, by the generalized Young’s inequality with respect to $\eta$,

	$\displaystyle v(z)^{p}e^{nF_{t}(z)}$	$\displaystyle\leq$	$\displaystyle\int_{0}^{\,{\rm exp}\,({nF_{t}(z)})}\eta(x)dx+\int_{0}^{v(z)^{p}}\eta^{-1}(y)dy$
		$\displaystyle\leq$	$\displaystyle\,{\rm exp}\,({nF_{t}(z)})(\,{\rm log}\,(1+\,{\rm exp}\,({nF_{t}(z)})))^{p}+\int_{0}^{\,{\rm exp}\,({v(z)-1})}x\eta^{\prime}(x)dx$
		$\displaystyle\leq$	$\displaystyle\,{\rm exp}\,({nF_{t}(z)})(1+n|F_{t}(z)|)^{p}+v(z)^{p}\,{\rm exp}\,({v(z)})$
		$\displaystyle\leq$	$\displaystyle\,{\rm exp}\,({nF_{t}(z)})(1+n|F_{t}(z)|)^{p}+C(p)\,{\rm exp}\,({2v(z)})$

We integrate both sides in the inequality above over $z\in\Omega_{s}$, and get by Lemma 1 that

	$\displaystyle\int_{\Omega_{s}}v(z)^{p}e^{nF_{t}(z)}\omega_{X}^{n}$	$\displaystyle\leq$	$\displaystyle\int_{\Omega_{s}}e^{nF_{t}}(1+n|F_{t}(z)|)^{p}\omega_{X}^{n}+\int_{\Omega_{s}}e^{2v(z)}\omega_{X}^{n}$
		$\displaystyle\leq$	$\displaystyle\|e^{nF_{t}}\|_{L^{1}(\,{\rm log}\,L)^{p}}+C+Ce^{C{\overline{E}}_{t}},$

where the constant $C>0$ depends only on $n,\omega_{X},\chi,\gamma,p$. In view of the definition of $v$, this implies

$$\int_{\Omega_{s}}(-\varphi_{t}-s)^{\frac{(n+1)p}{n}}e^{nF_{t}(z)}\omega_{X}^{n}\leq 2^{p}\beta_{0}^{-p}A_{s}^{\frac{p}{n}}\big(\|e^{nF_{t}}\|_{L^{1}(\,{\rm log}\,L)^{p}}+C+Ce^{C{\overline{E}}_{t}}\big).$$

(2.12)

From the definition of $A_{s}$ in (2.3), it follows from Hölder inequality that

	$\displaystyle A_{s}$	$\displaystyle=$	$\displaystyle\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}(-\varphi_{t}-s)e^{nF_{t}}\omega_{X}^{n}$
		$\displaystyle\leq$	$\displaystyle\Big(\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}(-\varphi_{t}-s)^{\frac{(n+1)p}{n}}e^{nF_{t}}\omega^{n}_{X}\Big)^{\frac{n}{(n+1)p}}\cdot\Big(\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}e^{nF_{t}}\omega_{X}^{n}\Big)^{1/q}$
		$\displaystyle\leq$	$\displaystyle A_{s}^{\frac{1}{n+1}}\Big(\frac{c_{t}^{n}}{V_{t}}2^{p}\beta_{0}^{-p}\big(\|e^{nF_{t}}\|_{L^{1}(\,{\rm log}\,L)^{p}}+C+Ce^{C{\overline{E}}_{t}}\big)\Big)^{\frac{n}{(n+1)p}}\cdot\Big(\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}e^{nF_{t}}\omega_{X}^{n}\Big)^{1/q}$

where $q>1$ satisfies $\frac{n}{p(n+1)}+\frac{1}{q}=1$, i.e. $q=\frac{p(n+1)}{p(n+1)-n}$. The inequality above yields

$$A_{s}\leq\Big(\frac{c_{t}^{n}}{V_{t}}2^{p}\beta_{0}^{-p}\big(\|e^{nF_{t}}\|_{L^{1}(\,{\rm log}\,L)^{p}}+C+Ce^{C{\overline{E}}_{t}}\big)\Big)^{1/p}\cdot\Big(\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}e^{nF_{t}}\omega_{X}^{n}\Big)^{\frac{1+n}{qn}}.$$

(2.13)

Observe that the exponent of the integral on the right hand of (2.13) satisfies

$$\frac{1+n}{qn}=\frac{pn+p-n}{pn}=1+\delta_{0}>1,$$

for $\delta_{0}:=\frac{p-n}{pn}>0$. For notation convenience, set

$$B_{0}:=\Big(\frac{c_{t}^{n}}{V_{t}}2^{p}\beta_{0}^{-p}\big(\|e^{nF_{t}}\|_{L^{1}(\,{\rm log}\,L)^{p}}+C+Ce^{C{\overline{E}}_{t}}\big)\Big)^{1/p}.$$

(2.14)

From (2.13) we then get

$$A_{s}\leq B_{0}\Big(\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}e^{nF_{t}}\omega_{X}^{n}\Big)^{1+\delta_{0}}.$$

(2.15)

For any $r\in[0,1]$, we note that $-\varphi_{t}-s\geq r$ on $\Omega_{s+r}=\{\varphi_{t}\leq-s-r\}$. Thus

$$A_{s}=\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}(-\varphi_{t}-s)e^{nF_{t}}\omega_{X}^{n}\geq r\cdot\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s+r}}e^{nF_{t}}\omega_{X}^{n}.$$

(2.16)

If we define $\phi:{\bf R}_{+}\to{\bf R}_{+}$ by

$$\phi(s):=\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}e^{nF_{t}}\omega_{X}^{n}$$

then (2.15) and (2.16) imply that

$$r\phi(s+r)\leq B_{0}\phi(s)^{1+\delta_{0}},\quad\forall r\in[0,1]\mbox{ and }s\geq 0.$$

(2.17)

$\phi$ is clearly nonincreasing and continuous, so the lemma below applies to $\phi$. It is a classic lemma due to De Giorgi, which was also used in [14, 12]. We include a sketch of the proof for the readers’ convenience, and to exhibit the dependence of $\|\varphi_{t}\|_{L^{\infty}}$ on the given data.

We return now to the proof of Theorem 2. By Chebyshev’s inequality, we have

$$\phi(s)=\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}e^{nF_{t}}\omega_{X}^{n}\leq\frac{1}{s}\frac{c_{t}^{n}}{V_{t}}\int_{\Omega_{s}}(-\varphi_{t})e^{nF_{t}}\omega_{X}^{n}\leq\frac{{\overline{E}}_{t}}{s}\to 0\mbox{ as }s\to\infty.$$

Thus we may choose $s_{0}=(2B_{0})^{1/\delta_{0}}{\overline{E}}_{t}$ in the proof of Lemma 2. By (2.17) and Lemma 2, we deduce that

$$\Omega_{S_{\infty}}=\{\varphi_{t}\leq-S_{\infty}\}=\emptyset,$$

so hence

$$\inf_{X}\varphi_{t}\geq-S_{\infty}=-(2B_{0})^{1/\delta_{0}}{\overline{E}}_{t}-\frac{1}{1-2^{-\delta_{0}}},$$

(2.18)

where $B_{0}$ is the constant in (2.14) and $\delta_{0}:=\frac{p-n}{pn}>0$ depends only on $n$ and $p$. The proof of Theorem 2 is complete.