Let $\varphi$ solve the equation (1.1), where $F$ is a given smooth function and $f(\lambda[\cdot])$ is the nonlinear operator as introduced in Section 1. The setting of Theorem 1 with a fixed background $\omega_{X}$ can be viewed as a special case of the setting of Theorem 2 with $\chi=0$, and $t$ taken to be $1$, $\omega_{1}=\omega_{X}$, and in the notations (1.7) and (1.1) for the two settings,

$\displaystyle e^{F}=c_{1}e^{F_{1}},\quad V=V_{1}.$

(3.1)

We observe that, in view of the normalization ${1\over V}\int_{X}e^{nF_{1}}\omega_{X}^{n}=1$ for $F_{1}$,

$\displaystyle c_{1}^{n}={1\over V}\int_{X}e^{nF}\omega_{X}^{n}\leq{\rm Ent}_{p}(F)+e^{n}$

(3.2)

for any $p\geq 1$. Thus, applying Theorem 2 and assuming that ${\rm Ent}_{p}$ is bounded, we find that $L^{\infty}$ bounds for $\varphi$ would follow if we can control the energy $E=E_{t=1}$. However, an easy application of Hölder’s inequality gives

$$E(\varphi)=\frac{1}{V}\int_{X}(-\varphi)e^{nF}\omega_{X}^{n}\leq\Big(\frac{1}{V}\int_{X}(-\varphi)^{\frac{n}{n-1}}e^{nF}\omega_{X}^{n}\Big)^{(n-1)/n}\leq C$$

(3.3)

so it suffices to control the right hand side. This is done in the following theorem, part (c), which completes the proof of Theorem 1:

We observe that, in the special case of the Monge-Ampère equation and when $p=1$, these estimates have been established in [10], using pluripotential theory. Further if $p>n$, then Theorem 1 implies the solutions are $L^{\infty}$ bounded. Theorem 3 gives a more complete integral estimate of such solutions for the full range of $p\in(0,\infty)$, using pure PDE methods.

For the proof of Theorem 1, we only need part (c) of Theorem 3, but we give the proofs of the other parts as well, as they are of independent interest. As mentioned in the Introduction, for this we need the ABP method developed in [5] and [7] in order to get better integral bounds for $\varphi$ as stated in (3.4) and (3.5). In particular, we have fix a background metric $\omega_{X}$, as this method is not effective for handling degenerating families.

Suppose $p\in(0,n]$ and write

$$\Psi_{p}:=\frac{1}{V}\int_{X}(F^{2}+1)^{p/2}e^{nF}\omega_{X}^{n}\sim\frac{1}{V}\int_{X}|F|^{p}e^{nF}\omega_{X}^{n}={\mathrm{Ent}}_{p}(F).$$

In the proof below, for a smooth function $u$ on $X$, we denote

$$\Box u:=G^{i\bar{j}}u_{\bar{j}i},\quad|\nabla u|_{G}^{2}:=G^{i\bar{j}}\nabla_{\bar{j}}u\nabla_{i}u$$

(3.8)

and

$${\rm tr}_{G}\alpha:=G^{i\bar{j}}\alpha_{\bar{j}i},\quad\mbox{ for a smooth $(1,1)$-form $\alpha=\alpha_{\bar{j}i}\sqrt{-1}dz^{i}\wedge d\bar{z}^{j}$},$$

where as in Lemma 1, $G^{i\bar{j}}=\frac{\partial\,{\rm log}\,f(\lambda[h])}{\partial h_{ij}}$ is the coefficient matrix of the linearized operator of $\,{\rm log}\,f(\lambda[\cdot])$ at $h=\omega_{X}^{-1}\cdot\omega_{\varphi}$, and by the structure condition (1.4) on $f$, we have ${\rm det}(G^{i\bar{j}})\geq\gamma f^{-n}$ and $G^{i\bar{j}}$ is positive definite.

For notation convenience, we set $\phi_{\delta}(t)=t+\sqrt{t^{2}+\delta}>0$, which is a smoothing of $2\max(t,0)$ and converges to it as $\delta\to 0$. Here we will first fix a $\delta>0$ small and later on $\delta$ will be sent to zero. All constants appearing in the proof are independent of $\delta$, unless stated otherwise. From now on we will consider $\phi_{\delta}(\rho)$ which is monotone decreasing and converges to $2\rho_{+}$ as $\delta\to 0$. We will omit the $\delta$ in $\phi_{\delta}(\rho)$ and simply write $\phi(\rho)$.

We can now calculate,

$$\Box(H\eta)=\eta\Box H+H\Box\eta+2Re\big(G^{i\bar{j}}\nabla_{\bar{j}}H\nabla_{i}\eta\big).$$

(3.13)

We observe that the middle term in (3.13) satisfies

$$H\Box\eta=H{\rm tr}_{G}i\partial\bar{\partial}\eta\geq-H\frac{10\theta}{r^{2}}{\rm tr}_{G}\omega_{X}.$$

The last term in (3.13) satisfies

	$\displaystyle 2Re\big(G^{i\bar{j}}\nabla_{\bar{j}}H\nabla_{i}\eta\big)$	$\displaystyle=$	$\displaystyle 2b\phi(\rho)^{b-1}Re\big(G^{i\bar{j}}\nabla_{\bar{j}}\phi(\rho)\nabla_{i}\eta\big)$
		$\displaystyle\geq$	$\displaystyle-\frac{b(b-1)}{2}\phi(\rho)^{b-2}|\nabla\phi(\rho)|^{2}_{G}-\frac{2b}{b-1}\phi(\rho)^{b}|\nabla\eta|^{2}_{G}$
		$\displaystyle\geq$	$\displaystyle-\frac{b(b-1)}{2}\phi(\rho)^{b-2}|\nabla\phi(\rho)|^{2}_{G}-\frac{2b}{b-1}\phi(\rho)^{b}\frac{10\theta^{2}}{r^{2}}{\rm tr}_{G}\omega_{X}$

where in the first inequality we applied the Cauchy-Schwarz inequality. The first term in (3.13) is

	$\displaystyle\eta\Box H$	$\displaystyle=$	$\displaystyle b\eta\phi(\rho)^{b-1}\Box\phi(\rho)+b(b-1)\eta\phi(\rho)^{b-2}|\nabla\phi(\rho)|_{G}^{2}$		(3.14)
		$\displaystyle=$	$\displaystyle b\eta\phi(\rho)^{b-1}\phi^{\prime}(\rho)\Box\rho+b\eta\phi(\rho)^{b-1}\phi^{\prime\prime}(\rho)|\nabla\rho|^{2}_{G}+b(b-1)\eta\phi(\rho)^{b-2}|\nabla\phi(\rho)|_{G}^{2}.$

We note that the middle term in (3.14) is nonnegative due to the fact that

$$\phi^{\prime\prime}(t)=\frac{1}{\sqrt{t^{2}+\delta}}-\frac{t^{2}}{(\sqrt{t^{2}+\delta})^{3}}=\frac{\delta}{(t^{2}+\delta)^{3/2}}>0.$$

To deal with the first term in (3.14) we note by the homogeneity of degree one assumption on $f$ that

$$\Box\varphi={\rm tr}_{G}i\partial\bar{\partial}\varphi={\rm tr}_{G}\omega_{\varphi}-{\rm tr}_{G}\omega_{X}=1-{\rm tr}_{G}\omega_{X}.$$

Then we calculate

	$\displaystyle\Box\rho$	$\displaystyle=$	$\displaystyle\Box(-(-\psi+\Lambda)^{\beta}-\lambda\varphi)$		(3.15)
		$\displaystyle=$	$\displaystyle\beta(-\psi+\Lambda)^{\beta-1}\Box\psi+\beta(1-\beta)(-\psi+\Lambda)^{\alpha-2}|\nabla\psi|_{G}^{2}-\lambda\Box\varphi$
		$\displaystyle\geq$	$\displaystyle\beta(-\psi+\Lambda)^{\beta-1}{\rm tr}_{G}\omega_{\psi}-\beta(-\psi+\Lambda)^{\beta-1}{\rm tr}_{G}\omega_{X}-\lambda+\lambda{\rm tr}_{G}\omega_{X}$
		$\displaystyle\geq$	$\displaystyle n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{(F^{2}+1)^{p/2}}{\Psi_{p}}\big)^{1/n}+(\lambda-\beta(-\psi+\Lambda)^{\beta-1}){\rm tr}_{G}\omega_{X}-\lambda$
		$\displaystyle\geq$	$\displaystyle n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{|F|^{p}}{\Psi_{p}}\big)^{1/n}+(\lambda-\beta\Lambda^{\beta-1}){\rm tr}_{G}\omega_{X}-\lambda,$

where in the second inequality we used the arithmetic-geometric inequality and the equations for $\varphi$ and $\psi$. Plugging these inequalities into (3.13), we obtain

	$\displaystyle\Box(H\eta)$	$\displaystyle\geq$	$\displaystyle-\frac{10\theta}{r^{2}}\phi(\rho)^{b}{\rm tr}_{G}\omega_{X}-\frac{2b}{b-1}\phi(\rho)^{b}\frac{10\theta^{2}}{r^{2}}{\rm tr}_{G}\omega_{X}$		(3.16)
			$\displaystyle+b\eta\phi(\rho)^{b-1}\phi^{\prime}(\rho)\Big(n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{|F|^{p}}{\Psi_{p}}\big)^{1/n}+(\lambda-\beta\Lambda^{\beta-1}){\rm tr}_{G}\omega_{X}-\lambda\Big)$
		$\displaystyle\geq$	$\displaystyle b\phi(\rho)^{b-1}\Big\{\frac{9}{10}(\lambda-\beta\Lambda^{\beta-1})\phi^{\prime}(\rho){\rm tr}_{G}\omega_{X}-\frac{20\theta}{r^{2}b}\phi(\rho){\rm tr}_{G}\omega_{X}$
			$\displaystyle-\frac{20\theta^{2}}{(b-1)r^{2}}\phi(\rho){\rm tr}_{G}\omega_{X}+n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{|F|^{p}}{\Psi_{p}}\big)^{1/n}-\lambda\phi^{\prime}(\rho)\Big\},$

where we ignored $\eta$ in the last inequality since $\eta>9/10$. To deal with the right hand side in the equation (3.16), we note that on the set $\{\rho\leq 0\}$

$$0\leq\phi(\rho)=\rho+\sqrt{\rho^{2}+\delta}=\frac{\delta}{\sqrt{\rho^{2}+\delta}-\rho}\leq\sqrt{\delta},$$

and on this same set the function $\phi^{\prime}(\cdot)$ satisfies

$$1\geq\phi^{\prime}(\rho)=1+\frac{\rho}{\sqrt{\rho^{2}+\delta}}=\frac{\phi(\rho)}{\sqrt{\rho^{2}+\delta}}\geq 0.$$

So on the set $\{\rho\leq 0\}$ the right hand side of (3.16) is greater or equal to

$\displaystyle b\phi(\rho)^{b-1}\Big(-\frac{20\theta}{r^{2}b}\sqrt{\delta}{\rm tr}_{G}\omega_{X}-\frac{20\theta^{2}}{(b-1)r^{2}}\sqrt{\delta}{\rm tr}_{G}\omega_{X}-\lambda\Big)$

On the other hand, on the set $\{\rho>0\}$, we know $2\geq\phi^{\prime}(\rho)>1$, so the first three terms on the right hand side of (3.16) are positive due to the choice of $\theta$ in (3.11), and $\Lambda,\lambda$ in (3.9) and the fact that $\phi(\rho)\leq M^{1/b}$. So on the set $\{\rho>0\}$ the right hand side of (3.16) is greater or equal to

$$b\phi(\rho)^{b-1}\Big(n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{|F|^{p}}{\Psi_{p}}\big)^{1/n}-\lambda\Big).$$

Combining the above two cases, we obtain

	$\displaystyle\Box(H\eta)$	$\displaystyle\geq$	$\displaystyle b\phi(\rho)^{b-1}\Big(n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{|F|^{p}}{\Psi_{p}}\big)^{1/n}-\lambda\Big)\cdot\chi_{\{\rho>0\}}$
			$\displaystyle-b\phi(\rho)^{b-1}\Big(\frac{20\theta}{r^{2}b}\sqrt{\delta}{\rm tr}_{G}\omega_{X}+\frac{20\theta^{2}}{(b-1)r^{2}}\sqrt{\delta}{\rm tr}_{G}\omega_{X}+\lambda\Big)\cdot\chi_{\{\rho\leq 0\}},$

where $\chi_{E}$ denotes the characteristic function of a set $E$.

We now apply the ABP maximum principle to the function $H\eta$ on the domain $B_{r}(x_{0})=:B_{0}$ in ${\bf R}^{2n}$. It follows that

	$\displaystyle\sup_{B_{0}}(H\eta)$	$\displaystyle\leq$	$\displaystyle\sup_{\partial B_{0}}(H\eta)+C(n)r\Big\{\int_{B_{0}\cap\{\rho>0\}}\frac{\phi(\rho)^{2n(b-1)}(n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{|F|^{p}}{\Psi_{p}}\big)^{1/n}-\lambda\big)_{-}^{2n}}{({\rm det}G^{i\bar{j}})^{2}}\omega_{X}^{n}$		(3.17)
			$\displaystyle+\int_{B_{0}\cap\{\rho\leq 0\}}\frac{\phi(\rho)^{2n(b-1)}(\frac{20\theta}{r^{2}b}\sqrt{\delta}{\rm tr}_{G}\omega_{X}+\frac{20\theta^{2}}{(b-1)r^{2}}\sqrt{\delta}{\rm tr}_{G}\omega_{X}+\lambda)^{2n}}{({\rm det}G^{i\bar{j}})^{2}}\omega_{X}^{n}\Big\}^{1/2n}$
		$\displaystyle\leq$	$\displaystyle\sup_{\partial B_{0}}(H\eta)+C(n)\Big\{\int_{B_{0}\cap\{\rho>0\}}\frac{\sqrt{\phi(\rho)}\big(n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{|F|^{p}}{\Psi_{p}}\big)^{1/n}-\lambda\big)_{-}^{2n}}{e^{-2nF}}\omega_{X}^{n}$
			$\displaystyle+C^{\prime}\delta^{n(b-1)}\Big\}^{1/2n}$

where the constant $C^{\prime}=C^{\prime}(n,F,G,\omega_{X})$ in the last term may not be uniformly bounded, but this is not a concern, since later on we will let $\delta\to 0$. We observe that the integral involved in the last inequality is in fact integrated over the set where $n\gamma^{1/n}\beta(-\psi+\Lambda)^{\beta-1}\big(\frac{|F|^{p}}{\Psi_{p}}\big)^{1/n}-\lambda<0$ and $\rho>0$, and over this set we have from the choice of constants in (3.9)

$$|F|\leq(\Psi_{p})^{1/p}\Big(\frac{\lambda}{n\gamma^{1/n}\beta}\Big)^{n/p}(-\psi+\Lambda)^{(1-\beta)n/p}=\frac{\alpha_{0}}{2}(-\psi+\Lambda)^{(1-\beta)n/p}.$$

At the same time, on the same set, we have $0<\rho\leq-\lambda\varphi$ and $\phi(\rho)\leq 2\rho+\sqrt{\delta}$. Therefore, we obtain from (3.17) that

	$\displaystyle M$	$\displaystyle\leq$	$\displaystyle(1-\theta)\sup_{\partial B_{0}}H+C\Big(\int_{B_{0}}(-\varphi+\sqrt{\delta})^{1/2}\,{\rm exp}\,\Big(\frac{\alpha_{0}}{2}(-\psi+\Lambda)^{(1-\beta)n/p}\Big)\omega_{X}^{n}+C^{\prime}\delta^{n(b-1)}\Big)^{1/2n}$		(3.18)
		$\displaystyle\leq$	$\displaystyle(1-\theta)M+C\Big(\int_{B_{0}}\left(-\varphi+\,{\rm exp}\,\Big({\alpha_{0}}(-\psi+\Lambda)^{(1-\beta)n/p}\Big)\right)\omega_{X}^{n}+C^{\prime}\delta^{n(b-1)}\Big)^{1/2n}$
		$\displaystyle\leq$	$\displaystyle(1-\theta)M+C_{0}+C^{\prime}\delta^{(b-1)/2}$

where $C_{0}=C_{0}(n,\omega_{X},\gamma,\Psi_{p},p)>0$ is independent of $\delta$ and in the last inequality we have used the following inequalities:

(1) $\int_{X}(-\varphi)\omega_{X}^{n}\leq C(n,\omega_{X})$ which follows from the Green’s formula, $n+\Delta_{\omega_{X}}\varphi>0$ (since $\lambda[h_{\varphi}]\in\Gamma\subset\{\lambda_{1}+\cdots+\lambda_{n}>0\}$) and the normalization condition $\sup_{X}\varphi=0$.

(2) $\int_{X}\,{\rm exp}\,\Big({\alpha_{0}}(-\psi+\Lambda)^{(1-\beta)n/p}\Big)\omega_{X}^{n}\leq C(n,\omega_{X},\Psi_{p},p)$. By the choice of $\beta$, if $p<n$, $(1-\beta)n/p=1$; and if $p=n$, then $(1-\beta)n/p=1-N^{-1}<1$, Young’s inequality gives the desired estimate.

Hence we conclude that with the choice of $\theta$ in (3.11)

$$\min(M^{1-\frac{1}{b}},M)\leq C_{0}+C^{\prime}\delta^{(b-1)/2}\,\,\Rightarrow\,M\leq C_{0}+C^{\prime}\delta^{2b},$$

which implies that

$$\sup_{X}2\rho_{+}\leq\sup_{X}\phi(\rho)=M\leq C_{0}+C^{\prime}\delta^{2b}.$$

Finally letting $\delta\to 0$ yields the desired estimate $\sup_{X}\rho_{+}\leq C_{0}$ for some positive constant $C_{0}=C_{0}(n,\omega_{X},\gamma,\Psi_{p})$ (which may be different from the $C_{0}$ in (3.18)). The proof of Lemma 3 is complete.

The inequality (3.6) in (c) is an immediate consequence of the estimates in (a) and (b), and Jensen’s inequality. More precisely, we have

$$\frac{1}{V}\int_{X}e^{-nF+c_{0}(-\varphi)^{N}}e^{nF}\omega_{X}^{n}=\frac{1}{V}\int_{X}e^{c_{0}(-\varphi)^{N}}\omega_{X}^{n}\leq C(n,\omega_{X},\gamma,\Psi_{p},N).$$

Taking the logarithms of both sides and applying Jensen’s inequality yields

$$\frac{1}{V}\int_{X}\big(c_{0}(-\varphi)^{N}-nF\big)e^{nF}\omega_{X}^{n}\leq C(n,\omega_{X},\gamma,\Psi_{p},N),$$

from which (3.6) follows after noting that $\Psi_{p}$ is equivalent to the entropy ${\mathrm{Ent}}_{p}$ and if $p\geq 1$

$$\frac{1}{V}\int_{X}Fe^{nF}\omega_{X}^{n}\leq{\mathrm{Ent}}_{p}.$$

The proof of Theorem 3 is complete.