The solutions constructed so far belong to $\mathcal E(X,\theta ,\phi )$, but in general their difference from $\phi $ is unbounded. We now show that the difference is bounded when the measure is dominated by a power $1+\delta >1$ of the Monge–Ampère capacity $\operatorname {Cap}_{\theta ,\phi }$ relative to $\phi $ (definition 16.25.5). If $\mu _ u\leq A\operatorname {Cap}_{\theta ,\phi }^{1+\delta }$, the sublevel estimate of proposition 16.25.7 bounds the capacity of the deeper sublevel set $\{ u<\phi -s-t\} $ by $At^{-n}$ times the $(1+\delta )$th power of the capacity of $\{ u<\phi -s\} $. Iterating with geometrically decreasing steps $t$, we reach a finite level beyond which the capacity vanishes. Measures with $L^ p$ densities, $p>1$, satisfy such a bound by corollary 16.27.10.
Theorem 17.5.20.F21AE677873A Let $u\in \mathcal E(X,\theta ,\phi )$ with $u\leq \phi $. Let $0\leq a<1$, $A>0$ and $\delta >0$, and let $\nu $ be a positive Borel measure on $X$ such that
2646B3091195
17.5.21
\begin{equation} \mu _ u\leq \nu +a\mu _\phi ,\qquad \nu (E)\leq A\operatorname {Cap}_{\theta ,\phi }(E)^{1+\delta } \quad \text{for every Borel set }E\subset X. \end{equation}
Then there is a finite constant $C\geq 0$ such that $\phi -C\leq u\leq \phi $ on $X$. If, in addition, $I=\int _ X(\phi -u)\, \mathrm d\nu <\infty $, then one can take
856E2E7BF89A
17.5.22
\begin{equation} C=2+\frac I{1-a}\left(2\left(\frac A{1-a}\right)^{1/n}\right)^{n/\delta } +\frac1{1-2^{-\delta }}. \end{equation}
Here $\phi -u$ and the sublevel sets $\{ u<\phi -s\} $ are taken on the set where $\phi $ is finite; its complement is pluripolar, hence of zero capacity (proposition 16.25.6) and therefore $\nu $-null.
Proof.
We derive a recursive inequality for the capacities of the sublevel sets of $u-\phi $ and iterate it along levels that converge to a finite depth. For $s>0$ put
\[ E_ s=\{ u<\phi -s\} ,\qquad g(s)=\operatorname {Cap}_{\theta ,\phi }(E_ s)^{1/n},\qquad B=\left(\frac A{1-a}\right)^{1/n}. \]
The sets $E_ s$ decrease as $s$ increases, so $g$ is nonincreasing.
We first compare the measures on $E_ s$. Both $u$ and $\phi -s$ belong to $\mathcal E(X,\theta ,\phi )$, so the comparison principle of proposition 16.25.4, applied to $u$ and $\phi -s$, gives $\mu _\phi (E_ s)\leq \mu _ u(E_ s)$. By (17.5.21),
\[ \mu _ u(E_ s)\leq \nu (E_ s)+a\mu _\phi (E_ s)\leq \nu (E_ s)+a\mu _ u(E_ s), \]
and hence $(1-a)\mu _ u(E_ s)\leq \nu (E_ s)$. Since $u$ has full mass relative to $\phi $, proposition 16.25.7 gives $t^ n\operatorname {Cap}_{\theta ,\phi }(E_{s+t})\leq \mu _ u(E_ s)$ for $0<t\leq 1$. Combining this with the last inequality and the capacity bound in (17.5.21), we obtain $t^ ng(s+t)^ n\leq B^ ng(s)^{n(1+\delta )}$, that is,
E2AC3FD3AA08
17.5.23
\begin{equation} t g(s+t)\leq B g(s)^{1+\delta },\qquad s>0,\quad 0<t\leq 1. \end{equation}
By the second assertion of proposition 16.25.7, $g(s)\to 0$ as $s\to \infty $. Choose $s_0>0$ with $2Bg(s_0)^\delta \leq 1$, and put
A41B6BAF1A25
17.5.24
\begin{equation} C=s_0+\frac{2B}{1-2^{-\delta }}g(s_0)^\delta . \end{equation}
We claim that $g(C)=0$. If $g(s_0)=0$, then $C=s_0$ and there is nothing to prove. Otherwise, as long as $g(s_ j)>0$, define inductively
\[ t_ j=2B g(s_ j)^\delta ,\qquad s_{j+1}=s_ j+t_ j. \]
Since $g$ is nonincreasing, $t_ j\leq 2Bg(s_0)^\delta \leq 1$, and (17.5.23) with $s=s_ j$ and $t=t_ j$ gives
\[ g(s_{j+1})\leq \frac{Bg(s_ j)^{1+\delta }}{t_ j}=\frac{g(s_ j)}2. \]
Hence $g(s_ j)\leq 2^{-j}g(s_0)$ and $t_ j\leq 2^{-j\delta }\cdot 2Bg(s_0)^\delta $, so that
\[ s_ j\leq s_0+2B g(s_0)^\delta \sum _{i=0}^\infty 2^{-i\delta }=C \qquad \text{for every }j. \]
If the construction stops because $g(s_ j)=0$ for some $j$, then $g(C)\leq g(s_ j)=0$. Otherwise $g(C)\leq g(s_ j)\leq 2^{-j}g(s_0)$ for every $j$, and again $g(C)=0$. This proves the claim.
Thus $\operatorname {Cap}_{\theta ,\phi }(E_ C)=0$. The capacity bound in (17.5.21) gives $\nu (E_ C)=0$. Since $\phi $ itself is a competitor for $\operatorname {Cap}_{\theta ,\phi }$ (definition 16.25.5), we also have $\mu _\phi (E_ C)\leq \operatorname {Cap}_{\theta ,\phi }(E_ C)=0$. The measure inequality in (17.5.21) therefore gives $\mu _ u(\{ u<\phi -C\} )=0$. As $\phi -C\in \mathcal E(X,\theta ,\phi )$, the domination principle, theorem 16.22.10, yields $u\geq \phi -C$ everywhere.
Now suppose that $I<\infty $. On $E_ s$ the weight $\phi -u$ exceeds $s$, so Markov's inequality gives $\nu (E_ s)\leq I/s$. Taking $t=1$ in the sublevel estimate and using the measure inequality on $E_ s$, we find
\[ g(s+1)^ n\leq \mu _ u(E_ s)\leq \frac{\nu (E_ s)}{1-a} \leq \frac I{(1-a)s}. \]
Choose $s=1+I(2B)^{n/\delta }/(1-a)$ and $s_0=s+1$. Since $(1-a)s>I(2B)^{n/\delta }$, we get $g(s_0)^ n\leq (2B)^{-n/\delta }$, that is, $2Bg(s_0)^\delta \leq 1$. Thus $s_0$ satisfies the condition imposed on it above, and the constant in (17.5.24) is at most $s_0+(1-2^{-\delta })^{-1}$. Since $(2B)^{n/\delta }=\bigl (2(A/(1-a))^{1/n}\bigr )^{n/\delta }$, this is the right side of (17.5.22). The lower bound $\phi -C\leq u$ remains true when $C$ is increased, which proves (17.5.22).
$\square$
Corollary 17.5.25.6DAA97EFFAAF Let $u\in \mathcal E^1(X,\theta ,\phi )$ with $u\leq \phi $, and let $A>0$ and $\delta >0$ be such that
\[ \mu _ u(E)\leq A\operatorname {Cap}_{\theta ,\phi }(E)^{1+\delta } \quad \text{for every Borel set }E\subset X. \]
If $L\geq 0$ and $E_\theta ^\phi (u)\geq -L$, then
\[ \phi -\left(2+(n+1)L(2A^{1/n})^{n/\delta } +\frac1{1-2^{-\delta }}\right)\leq u\leq \phi . \]
Proof.
Apply theorem 17.5.20 with $a=0$ and $\nu =\mu _ u$. By the first inequality in (17.3.15), $I=\int _ X(\phi -u)\, \mu _ u\leq (n+1)\bigl (-E_\theta ^\phi (u)\bigr )\leq (n+1)L$, and (17.5.22) with $a=0$ gives the stated bound.
$\square$
Corollary 17.5.26.DAF36530C8C1 Let $p>1$, let $0\leq f\in L^ p(X,\omega ^ n)$ and $0\leq a<1$, and let $u\in \mathcal E(X,\theta ,\phi )$ satisfy $\sup _ Xu=0$ and
\[ \mu _ u\leq f\omega ^ n+a\mu _\phi . \]
Then $\phi -C\leq u\leq \phi $, where $C$ depends only on $X,\theta ,\omega ,p,m,\| f\| _{L^ p}$ and $(1-a)^{-1}$. In particular, if $\int _ X f\omega ^ n=m$, then the equation $\mu _ u=f\omega ^ n$ has a solution $u\in \mathcal E(X,\theta ,\phi )$ with $\sup _ Xu=0$, and this solution has bounded difference from $\phi $.
Proof.
Since $\sup _ Xu=0$, lemma 16.22.2 gives $u\leq \phi $; recall also that $\sup _ X\phi =0$, by definition 16.21.1. Let $q=p/(p-1)$ be the conjugate exponent. By corollary 10.3.2, the $L^ q(X,\omega ^ n)$ norms of $u$ and $\phi $ are at most a constant $M_ q$ depending only on $X,\theta ,\omega ,q$. Hölder's inequality gives
\[ I=\int _ X(\phi -u)f\omega ^ n \leq \| f\| _{L^ p}\| \phi -u\| _{L^ q} \leq 2M_ q\| f\| _{L^ p}. \]
By corollary 16.27.10 with $r=2$, the measure $\nu =f\omega ^ n$ satisfies the capacity bound in (17.5.21) with $\delta =1$ and $A=B_2\| f\| _{L^ p}/m^2$, where $B_2$ depends only on $X,\theta ,\omega ,p$. (If $f=0$ almost everywhere, take $A=1$ instead.) With $\delta =1$, the constant (17.5.22) equals $4+2^ nAI/(1-a)^2$. Inserting the bounds for $A$ and $I$, and recalling that the lower bound $\phi -C\leq u$ survives an increase of $C$, we see that theorem 17.5.20 gives the assertion with
\[ C=4+\frac{2^{n+1}B_2M_ q\| f\| _{L^ p}^2}{m^2(1-a)^2}, \]
which has the stated dependence.
Finally, the measure $f\omega ^ n$ is non-pluripolar, as noted after definition 17.5.2. If its mass is $m$, theorem 17.5.15 gives a solution $u\in \mathcal E(X,\theta ,\phi )$ of $\mu _ u=f\omega ^ n$ with $\sup _ Xu=0$, and the bound just proved, with $a=0$, shows that $u$ has bounded difference from $\phi $.
$\square$
Pome
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