PomeProposition 13.15.6.8B51BF8706B7 If $u\in \mathrm{PSH}(X,\omega )$ and $u\leq 0$, then $F_ a(u)$ is a bounded $\omega $-plurisubharmonic function, and
\[ \mathrm{e}^{na u}\mu _ u\leq \mathrm{MA}_\omega (F_ a(u)). \]
13.15.5 Exponential damping569E9BAF1CA2
The exponential weight can be handled directly. A convex increasing function with bounded range transforms $u$ into a bounded $\omega $-plurisubharmonic function, and the chain rule shows that the transformed current dominates an exponential multiple of $\omega +\mathrm{dd}^{\mathrm{c}}u$. The Monge–Ampère measure of the bounded transform therefore controls the damped measure.
For $a>0$ put $F_ a(t)=a^{-1}\mathrm{e}^{at}$ for $t\leq 0$, with $F_ a(-\infty )=0$. Then $0\leq F_ a\leq a^{-1}$, while $0<F_ a'\leq 1$ and $F_ a''\geq 0$ at finite arguments.
Proof. First suppose that $u$ is bounded. The chain rule (5.27.9) gives
\[ \omega +\mathrm{dd}^{\mathrm{c}}F_ a(u) =\mathrm{e}^{au}(\omega +\mathrm{dd}^{\mathrm{c}}u)+(1-\mathrm{e}^{au})\omega +a\mathrm{e}^{au}\, \mathrm{d}u\wedge \mathrm d^{\mathrm c}u. \]
All three summands are positive, and positivity of their bounded mixed products yields the claimed inequality.
For general $u$, use the truncations $u_ t=\max (u,-t)$. The bounded functions $F_ a(u_ t)$ decrease and take values between zero and $a^{-1}$, so their limit $F_ a(u)$ is a bounded $\omega $-plurisubharmonic function. On the plurifinely open set $D_ t=\{ u>-t\} $, it agrees with $F_ a(u_ t)$. Locality and the bounded case therefore give
\[ \mathrm{e}^{na u}\mathbf1_{D_ t}\mu _ u \leq \mathbf1_{D_ t}\mathrm{MA}_\omega (F_ a(u)). \]
Let $t\to +\infty $. The pole set is not charged by $\mu _ u$, and it has zero mass for the bounded product on the right. $\square$
Theorem 13.15.7.4394319FDE01 For $p>0$, the damped measures $\nu _{p,u}=\mathrm{e}^{pu}\mu _ u$ satisfy
\[ \nu _{p,u}(X)\leq C_\omega (1+p^{-1})^{r_ n}, \qquad \nu _{p,u}(E)\leq \max (1,n/p)^ n\operatorname {Cap}_\omega (E) \]
for every Borel set $E\subset X$, uniformly over all nonpositive $u\in \mathrm{PSH}(X,\omega )$.
Proof. Take $a=p/n$ in proposition 13.15.6. Since $F_{p/n}(u)$ takes values in $[0,n/p]$, proposition 13.15.2 with $A=n/p$ gives the total mass bound, after enlarging $C_\omega $. For the capacity bound put $B=\max (1,n/p)$ and $v=F_{p/n}(u)$. Then $v/B$ is a test function for the capacity, and
\[ B\bigl (\omega +\mathrm{dd}^{\mathrm{c}}(v/B)\bigr ) =(\omega +\mathrm{dd}^{\mathrm{c}}v)+(B-1)\omega \geq \omega +\mathrm{dd}^{\mathrm{c}}v. \]
Consequently $\nu _{p,u}\leq \mathrm{MA}_\omega (v)\leq B^ n\mathrm{MA}_\omega (v/B)$, which gives the second bound. The measures $\nu _{p,u}$ do not charge pluripolar sets, because $\mu _ u$ does not. $\square$
Since $\mu _ u$ does not charge the pole set, multiplication by $\mathrm{e}^{-pu}$ recovers $\mu _ u$ from $\nu _{p,u}$, including when its mass is infinite. An additive shift $u\mapsto u+c$ multiplies the damped measure by $\mathrm{e}^{pc}$.
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