Pome

Proposition 17.5.75BDA01022188

Proposition 17.5.7.5BDA01022188 Let $\lambda >0$ and let $\nu $ be a finite positive non-pluripolar measure on $X$. If $u,v\in \mathcal E(X,\theta ,\phi )$ satisfy

\[ \mu _ u=\mathrm e^{\lambda u}\nu , \qquad \mu _ v\geq \mathrm e^{\lambda v}\nu , \]

then $u\geq v$. In particular, the equation $\mu _ u=\mathrm e^{\lambda u}\nu $ has at most one solution in $\mathcal E(X,\theta ,\phi )$.

Proof. Let $A=\{ u<v\} $. The comparison principle of proposition 16.25.4 gives

\[ \int _ A\mathrm e^{\lambda v}\, \mathrm d\nu \leq \mu _ v(A)\leq \mu _ u(A) =\int _ A\mathrm e^{\lambda u}\, \mathrm d\nu \leq m. \]

Both potentials are finite $\nu $-almost everywhere, and on $A$ we have $\mathrm e^{\lambda v}>\mathrm e^{\lambda u}$. The difference $\mathrm e^{\lambda v}-\mathrm e^{\lambda u}$ is therefore positive on $A$ and has nonpositive integral over $A$, so $\nu (A)=0$. Hence $\mu _ u(A)=0$, and the domination principle (theorem 16.22.10) gives $u\geq v$ everywhere. If $u$ and $v$ are two solutions, applying this in both directions gives $u=v$. $\square$

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