For the proof, it is convenient to restate the theorem as follows. Under the above assumptions, if in each case we have

$$\|e^{f}-e^{h}\|_{L^{1}(\omega^{n})}\leq\gamma_{a}(r)r^{n+3},$$

(3.1)

then there is a small $r_{0}>0$ such that for all $0<r\leq r_{0}$

$$\sup_{X}|u_{t}-v_{t}|\leq Cr,$$

for all $t$ listed in each case in (2) and $C$ depends only on $n,k,\omega,\chi$, $K>0$ and the corresponding $p_{a}$ in each case. This is the version which we shall prove.

We begin with a lemma due to Kolodziej [12]. The proof is almost identical to that in [12], but since we would like to avoid the use of pluripotential theory, some additional smoothing is needed in the proof, and we provide a full proof of this lemma.

Choose a small $\bar{r}_{0}\in(0,\frac{1}{10})$ such that $\gamma_{a}(\bar{r}_{0})\bar{r}_{0}^{n}\leq\frac{1}{5}$. We then fix an $0<r<\bar{r}_{0}$. We remark that all relevant constants are independent of $r$. Later on we will choose an even smaller $r_{0}>0$.

By switching the roles of $u_{t}$ and $v_{t}$ if necessary, we may assume

$$\int_{\{v_{t}\leq u_{t}\}}(e^{f}+e^{h})\omega^{n}\leq 1.$$

Denote $E_{j}:=\{v_{t}\leq u_{t}-j\beta_{0}r\}$. The next lemma states that over the set $E_{2}$, the integral of $e^{h}$ is small.

$\bullet$ in case I, we have $\|e^{h_{j}}\|_{L^{1}(\,{\rm log}\,L)^{p_{1}}}\leq 5K$,

$\bullet$ in case II, we have $\|e^{h_{j}}\|_{L^{p_{2}}}\leq 5K$,

$\bullet$ in case III, we have $\|e^{h_{j}}\|_{L^{p_{3}}}\leq 5K$.

We solve the following Hessian equations which admit known to admit unique smooth solutions [15, 5],

$$(\omega_{t}+i\partial\bar{\partial}\rho_{j})^{k}\wedge\omega^{n-k}=c_{t}e^{h_{j}}\omega^{n},\quad\sup_{X}\rho_{j}=0\mbox{ and }\omega_{t}+i\partial\bar{\partial}\rho_{j}\in\Gamma_{k},$$

where $\Gamma_{k}$ is the usual open convex cone in $k$-th Hessian equations. By the choice of $\beta_{0}$, we have $-\beta_{0}\leq\rho_{j}\leq 0$ (see [9]). The following Newton inequality holds pointwise for any $1\leq l\leq k$

$$\omega_{t,u_{t}}^{l}\wedge\omega_{t,\rho_{j}}^{k-l}\wedge\omega^{n-k}\geq\Big(\frac{\omega_{t,u_{t}}^{k}\wedge\omega^{n-k}}{\omega^{n}}\Big)^{l/k}\Big(\frac{\omega^{k}_{t,\rho_{j}}\wedge\omega^{n-k}}{\omega^{n}}\Big)^{(k-l)/k}\omega^{n}.$$

Then on the set $E_{0}\backslash G$ where $G=\{e^{f}\leq(1-r^{2})e^{h}\}$, we have

$$\omega_{t,u_{t}}^{l}\wedge\omega_{t,\rho_{j}}^{k-l}\wedge\omega^{n-k}\geq c_{t}(1-r^{2})^{l/k}(\frac{3}{2})^{(k-l)/k}e^{h}\omega^{n}.$$

It follows that on $E_{0}\backslash G$

	$\displaystyle\omega_{t,r\rho_{j}+(1-r)u_{t}}^{k}\wedge\omega^{n-k}$	$\displaystyle=$	$\displaystyle\sum_{l=0}^{k}\frac{k!}{l!(k-l)!}r^{k-l}(1-r)^{l}\omega_{t,u_{t}}^{l}\wedge\omega_{t,\rho_{j}}^{k-l}\wedge\omega^{n-k}$		(3.3)
		$\displaystyle\geq$	$\displaystyle c_{t}\sum_{l=0}^{k}\frac{k!}{l!(k-l)!}r^{k-l}(1-r)^{l}(1-r^{2})^{l/k}(\frac{3}{2})^{(k-l)/k}e^{h}\omega^{n}$
		$\displaystyle=$	$\displaystyle c_{t}\big((1-r)(1-r^{2})^{1/k}+r(\frac{3}{2})^{1/k}\big)^{k}e^{h}\omega^{n}\geq c_{t}(1+b_{0}r)e^{h}\omega^{n}$

where $b_{0}=\frac{1}{2}((\frac{3}{2})^{1/k}-1)>0$, since $r$ is chosen to be small.

Note that by (3.1)

$$r^{2}\int_{G}e^{h}\omega^{n}\leq\int_{G}(e^{h}-e^{f})\omega^{n}\leq\gamma_{a}(r)r^{n+3},$$

which implies

$$\int_{G}e^{h}\omega^{n}\leq\gamma_{a}(r)r^{n+1}.$$

(3.4)

Adding the same constant to $u_{t}$ and $v_{t}$, we may assume without loss of generality $-\beta_{0}\leq u_{t}\leq 0$. The following inclusion relation holds from the definition

$$E_{2}\subset E:=\{v_{t}\leq r\rho_{j}+(1-r)u_{t}-\beta_{0}r\}\subset E_{0}.$$

All functions involved are smooth so by the comparison principle and (3.3),

	$\displaystyle c_{t}(1+b_{0}r)\int_{E\backslash G}e^{h}\omega^{n}$	$\displaystyle\leq$	$\displaystyle\int_{E}\omega_{t,r\rho_{j}+(1-r)u_{t}}^{k}\wedge\omega^{n-k}$
		$\displaystyle\leq$	$\displaystyle\int_{E}\omega_{v_{t}}^{k}\wedge\omega^{n-k}=c_{t}\int_{E\backslash G}e^{h}\omega^{n}+c_{t}\int_{G}e^{h}\omega^{n}$

Combined with (3.4) this implies

$$\int_{E\backslash G}e^{h}\omega^{n}\leq\frac{1}{b_{0}}\gamma_{a}(r)r^{n}$$

It follows that

$$\int_{E_{2}}e^{h}\omega^{n}\leq\int_{E\backslash G}e^{h}\omega^{n}+\int_{G}e^{h}\omega^{n}\leq(1+\frac{1}{b_{0}})\gamma_{a}(r)r^{n}.$$

The Lemma is proved.

We now come to the proof of Theorem 1 proper. We normalize $u_{t}$ as in the proof of Lemma 1. For $s\geq 0$, we set $\Omega_{s}=\{v_{t}\leq(1-r)u_{t}-3\beta_{0}r-s\}$. Note that $\Omega_{s}\subset E_{2}$ for any $s\geq 0$.

We follow the same strategy as in [9]. We choose a sequence of smooth positive functions $\eta_{j}:{{\bf R}}\to{{{\bf R}}}_{+}$ such that

$$\eta_{j}(x)=x+\frac{1}{j},\quad\mbox{ when }x\geq 0,$$

and

$$\eta_{j}(x)=\frac{1}{2j},\quad\mbox{ when }x\leq-\frac{1}{j},$$

and $\eta_{j}(x)$ lies between $1/2j$ and $1/j$ for $x\in[-1/j,0]$. Clearly $\eta_{j}\to\eta_{\infty}(x)=x\cdot\chi_{{\bf R}_{+}}(x)$ pointwise as $j\to\infty$.

We solve the complex Monge-Ampère equations

$$(\omega_{t}+i\partial\bar{\partial}\psi_{j})^{n}=c_{t}^{{n/}{k}}\frac{\eta_{j}(-v_{t}+(1-r)u_{t}-3\beta_{0}r-s)}{A_{s,j}}e^{\frac{n}{k}h}\omega^{n},\quad\sup\psi_{j}=0.$$

As $j\to\infty$ we have by the dominated convergence theorem

$$A_{s,j}=\frac{c_{t}^{\frac{n}{k}}}{V_{t}}\int_{X}\big(\eta_{j}(-v_{t}+(1-r)u_{t}-3\beta_{0}r-s)\big)e^{\frac{n}{k}h}\omega^{n}\to A_{s}$$

where the constant $A_{s}$ is defined by

$$A_{s}:=\frac{c_{t}^{\frac{n}{k}}}{V_{t}}\int_{\Omega_{s}}(-v_{t}+(1-r)u_{t}-3\beta_{0}r-s)e^{\frac{n}{k}h}\omega^{n}.$$

Consider $\Phi=-\varepsilon\big(-\psi_{j}+\Lambda\big)^{\frac{n}{n+1}}+\big(-v_{t}+(1-r)u_{t}-3\beta_{0}r-s\big)$ where

$$\varepsilon=\Big(\frac{k(n+1)}{n^{2}}\Big)^{\frac{n}{n+1}}c(n,k)^{-\frac{1}{n+1}}A_{s,j}^{\frac{1}{n+1}},\quad\mbox{where $c(n,k)$ is the one in (\ref{eqn:str})}$$

$$\Lambda=\Big(\frac{n}{n+1}\frac{\varepsilon}{r}\Big)^{n+1}=C(n,k)\frac{A_{s,j}}{r^{n+1}}$$

Suppose $\sup\Phi=\Phi(x_{0})$ for some point $x_{0}$ in $X$. If $x_{0}\not\in\Omega_{s}^{\circ}$, then by definition $\Phi(x_{0})<0$. Otherwise $x_{0}\in\Omega_{s}^{\circ}$. We calculate as in [9]. First note that for $G^{i\bar{j}}=\frac{\partial}{\partial(\omega_{t,v_{t}})_{i\bar{j}}}\,{\rm log}\,\sigma_{k}(\omega_{t,v_{t}})$

$${\rm det}G^{i\bar{j}}\geq c(n,k)c_{t}^{-\frac{n}{k}}e^{-\frac{n}{k}h}$$

(3.5)

for some computable constant $c(n,k)>0$. It follows that at $x_{0}$

	$\displaystyle 0$	$\displaystyle\geq$	$\displaystyle G^{i\bar{j}}(\Phi)_{i\bar{j}}(x_{0})$
		$\displaystyle\geq$	$\displaystyle\frac{n^{2}\varepsilon}{n+\delta}(-\psi_{j}+\Lambda)^{-\frac{1}{n+1}}\Big({\rm det}G\cdot{\rm det}\omega_{t,\psi_{j}}\Big)^{1/n}-k+\Big(r-\frac{n\varepsilon}{n+1}\Lambda^{-\frac{1}{n+1}}\Big)G^{i\bar{j}}(\omega_{t})_{i\bar{j}}$
		$\displaystyle\geq$	$\displaystyle\frac{n^{2}\varepsilon}{n+1}(-\psi_{j}+\Lambda)^{-\frac{1}{n+1}}c(n,k)^{1/n}\Big(\frac{-v_{t}+(1-r)u_{t}-3\beta_{0}r-s}{A_{s,j}}\Big)^{1/n}-k.$

By the choice of $\varepsilon$ and $\Lambda$, we deduce that $\Phi(x_{0})\leq 0$. Thus $\Phi\leq 0$ on $X$ and this implies

	$\displaystyle\int_{\Omega_{s}}\,{\rm exp}\,\Big\{c_{0}\Big(\frac{-v_{t}+(1-r)u_{t}-3\beta_{0}r-s}{A_{s,j}^{1/(n+1)}}\Big)^{\frac{n+1}{n}}\Big\}\omega^{n}$	$\displaystyle\leq$	$\displaystyle\int_{\Omega_{s}}\,{\rm exp}\,\Big(-\alpha_{0}\psi_{j}+\alpha_{0}\Lambda\Big)\omega^{n}$		(3.6)
		$\displaystyle\leq$	$\displaystyle C\,{\rm exp}\,\Big\{\frac{A_{s,j}}{r^{n+1}}\Big\}$

for some small $c_{0}=c_{0}(n,k,\omega,\chi)>0$, $C=C(n,k,\omega,\chi)>0$, and $\alpha_{0}$ a fixed number satisfying $0<\alpha_{0}<\alpha(X,\omega)$, where $\alpha(X,\omega)$ is the $\alpha$-invariant of $(X,\omega)$. Letting $j\to\infty$ gives

$$\int_{\Omega_{s}}\,{\rm exp}\,\Big\{c_{0}\Big(\frac{(-v_{t}+(1-r)u_{t}-3\beta_{0}r-s)}{A_{s}^{1/(n+1)}}\Big)^{\frac{n+1}{n}}\Big\}\omega^{n}\leq C\,{\rm exp}\,\Big(\frac{A_{s}}{r^{n+1}}\Big).$$

(3.7)

To estimate $\frac{A_{s}}{r^{n+1}}$ in (3.7), we need to consider separately the three cases I, II and III.

So for all cases I, II and III, we get from (3.7) that

$$\int_{\Omega_{s}}\,{\rm exp}\,\Big\{c_{0}\Big(\frac{-v_{t}+(1-r)u_{t}-3\beta_{0}r-s}{A_{s}^{1/(n+1)}}\Big)^{\frac{n+1}{n}}\Big\}\omega^{n}\leq C,$$

(3.8)

for some constant $C>0$ depending on $n,k,\chi,\omega,K$ and the exponents $p_{1},p_{2},p_{3}$ in each case, respectively. In particular this $C$ is independent of the choice of $r\in(0,\bar{r}_{0})$.

We choose $p>n$ as $p=p_{1}$ in case I, and arbitrary and large $p>n$ in cases II and III.

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

$\displaystyle\Psi:=\frac{c_{0}}{2}\Big(\frac{-v_{t}+(1-r)u_{t}-3\beta_{0}r-s}{A_{s}^{1/(n+1)}}\Big)^{\frac{n+1}{n}}$

(3.9)

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

	$\displaystyle\Psi(z)^{p}e^{\frac{n}{k}h(z)}$	$\displaystyle\leq$	$\displaystyle\int_{0}^{\,{\rm exp}\,({\frac{n}{k}h(z)})}\eta(x)dx+\int_{0}^{\Psi(z)^{p}}\eta^{-1}(y)dy$
		$\displaystyle\leq$	$\displaystyle e^{\frac{n}{k}h(z)}(1+|h(z)|)^{p}+C(p)e^{2\Psi(z)}$

We integrate both sides in the inequality above over $z\in\Omega_{s}$, and get by (3.8) that

$\displaystyle\int_{\Omega_{s}}\Psi(z)^{p}e^{\frac{n}{k}h(z)}\omega^{n}\leq\|e^{h}\|_{L^{n/k}(\,{\rm log}\,L)^{p}}+C,$

where the constant $C>0$ depends only on $n,k,\omega,\chi,p,K$. In view of the definition of $\Psi$, this implies

$$\int_{\Omega_{s}}(-v_{t}+(1-r)u_{t}-3\beta_{0}r-s)^{\frac{(n+1)p}{n}}e^{\frac{n}{k}h}\omega^{n}\leq CA_{s}^{\frac{p}{n}}\big(\|e^{h}\|_{L^{n/k}(\,{\rm log}\,L)^{p}}+1\big).$$

(3.10)

It follows from the Hölder inequality that

	$\displaystyle A_{s}$	$\displaystyle=$	$\displaystyle\frac{c_{t}^{n/k}}{V_{t}}\int_{\Omega_{s}}(-v_{t}+(1-r)u_{t}-3\beta_{0}r-s)e^{\frac{n}{k}h}\omega_{X}^{n}$
		$\displaystyle\leq$	$\displaystyle\Big(\frac{c_{t}^{n/k}}{V_{t}}\int_{\Omega_{s}}(-v_{t}+(1-r)u_{t}-3\beta_{0}r-s)^{\frac{(n+1)p}{n}}e^{\frac{n}{k}h}\omega^{n}\Big)^{\frac{n}{(n+1)p}}\cdot\Big(\frac{c_{t}^{n/k}}{V_{t}}\int_{\Omega_{s}}e^{\frac{n}{k}h}\omega^{n}\Big)^{1/q}$
		$\displaystyle\leq$	$\displaystyle CA_{s}^{\frac{1}{n+1}}\Big(\|e^{h}\|_{L^{n/k}(\,{\rm log}\,L)^{p}}+1)\Big)^{\frac{n}{(n+1)p}}\cdot\Big(\frac{c_{t}^{n/k}}{V_{t}}\int_{\Omega_{s}}e^{\frac{n}{k}h}\omega^{n}\Big)^{1/q}$

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

$$A_{s}\leq C\Big(\|e^{h}\|_{L^{n/k}(\,{\rm log}\,L)^{p}}+1\Big)^{1/p}\cdot\Big(\frac{c_{t}^{n/k}}{V_{t}}\int_{\Omega_{s}}e^{\frac{n}{k}h}\omega^{n}\Big)^{\frac{1+n}{qn}}=B_{0}\Big(\frac{c_{t}^{n/k}}{V_{t}}\int_{\Omega_{s}}e^{\frac{n}{k}h}\omega^{n}\Big)^{1+\delta_{0}}.$$

(3.11)

Observe that the exponent of the integral on the right hand of (3.11) satisfies

$$\frac{1+n}{qn}=\frac{pn+p-n}{pn}=1+\delta_{0}>1,\quad\mbox{for $\delta_{0}:=\frac{p-n}{pn}>0$.}$$

We remark that $\delta_{0}$ can be chosen to be close to $1/n$ in cases II and III by picking $p$ large enough. Furthermore, we note that

$$B_{0}:=C\Big(\|e^{h}\|_{L^{n/k}(\,{\rm log}\,L)^{p}}+1\Big)^{1/p}$$

(3.12)

is a constant depending only on $n,k,\omega,\chi,K$, and the exponents $p_{1},p_{2},p_{3}$ in each case, respectively, and in particular, it is independent of $r$ with $r\in(0,\bar{r}_{0})$.

If we define

$$\phi(s)=\frac{c_{t}^{n/k}}{V_{t}}\int_{\Omega_{s}}e^{\frac{n}{k}h}\omega^{n},$$

then (3.11) shows that if $\Omega_{s+s^{\prime}}\not=\emptyset$ then

$$s^{\prime}\phi(s+s^{\prime})\leq B_{0}\phi(s)^{1+\delta_{0}},\quad\mbox{for all }s^{\prime}\geq 0\mbox{ and }s\geq 0.$$

(3.13)

We now choose $r_{0}<\bar{r}_{0}$ small in each case as follows.

It is clear that in all cases, $r_{0}$ and $\bar{C}$ depend only on the given data, namely, $n,k,\omega,\chi,K$ and $p_{a}$, and we have $B_{0}\phi(0)^{\delta_{0}}\leq\frac{1}{2}$ and $\phi(0)\leq\bar{C}r^{1/\delta_{0}}$.