14.19.25 Construction of continuous solutionsFEE6FC2B120F

of theorem 14.19.2. Choose smooth strictly positive functions $f_ j\to f$ in $L^ p(X,\mathrm{d}V)$. By theorem 14.2.1, there are smooth $u_ j$ with $\omega _{u_ j}>0$ and constants $c_ j$ such that

We first bound the constants. For $n\geq 2$, fix a Gauduchon metric $\gamma $. By proposition 11.5.4,

Since $|x^{1/n}-y^{1/n}|\leq |x-y|^{1/n}$ for $x,y\geq 0$, Hölder's inequality gives $A_ j\to A>0$, where $A=\int _ X(f\mathrm{d}V/\gamma ^ n)^{1/n}\gamma ^ n$. Thus $c_ j$ is bounded above. In dimension one, Stokes' theorem gives the same estimate, with $A_ j=\int _ Xf_ j\, \mathrm{d}V$.

The lower bound for the density norm in theorem 14.19.19 gives $c_ j\| f_ j\| _ p\geq c_ p$, so the constants are bounded away from zero. The densities $c_ jf_ j$ are therefore uniformly bounded in $L^ p$, and the same theorem gives a common bound for the $u_ j$. By theorem 5.25.2, we may pass to a subsequence with $c_ j\to c>0$ and $u_ j$ Cauchy in $L^1$. By theorem 14.19.21, $u_ j\to u$ uniformly, with $u$ continuous and $\sup _ Xu=0$.

Bounded Bedford–Taylor continuity gives $\omega _{u_ j}^ n\rightharpoonup \omega _ u^ n$, while $c_ jf_ j\to cf$ in $L^ p$. Hence $\omega _ u^ n=cf\, \mathrm{d}V$, as required. $\square$

Proof. Absorbing the fixed smooth positive ratio $\mathrm{d}V/\omega ^ n$ into $f_ j$ and $f$ preserves their $L^ p$ convergence, so we may assume $\mathrm{d}V=\omega ^ n$. Choose $\rho >0$ smooth with $\mathrm{dd}^{\mathrm{c}}(\rho \omega ^{n-1})=0$, using theorem 11.4.2; for $n=1$, take $\rho =1$. Put

Set $A=\int _ X f^{1/n}\rho \omega ^ n$. The inequality for $n$th roots used in the preceding existence proof gives $A_ j\to A>0$. Since $\mathrm{dd}^{\mathrm{c}}(\rho \omega ^{n-1})=0$, the trace $\int _ X\omega _{u_ j}\wedge \rho \omega ^{n-1}$ equals $\alpha $; the arithmetic–geometric mean inequality therefore implies

Applying Jensen's inequality to the probability measure $A_ j^{-1}f_ j^{1/n}\rho \omega ^ n$ and taking logarithms, we obtain

By lemma 5.26.2, the normalized potentials $\psi _ j$ have uniform $L^ q$ bounds for every finite $q$. With $q=np/(np-1)$, Hölder's inequality gives a uniform bound for the last integral, since $\| f_ j^{1/n}\| _{np}=\| f_ j\| _ p^{1/n}$. This proves the upper bound for $M_ j$.

Finally, the density $F_ j=\mathrm{e}^{\lambda u_ j}f_ j$ satisfies $\| F_ j\| _ p\leq \mathrm{e}^{\lambda M_ j}\| f_ j\| _ p$. The lower-density bound of theorem 14.19.19, applied to $\psi _ j$, gives $c_ p\leq \mathrm{e}^{\lambda M_ j}\| f_ j\| _ p$, which bounds $M_ j$ below. $\square$

of theorem 14.19.3. Choose smooth strictly positive functions $f_ j\to f$ in $L^ p$. By theorem 14.7.2, there are smooth $u_ j$ with $\omega _{u_ j}>0$ and $\omega _{u_ j}^ n=\mathrm{e}^{\lambda u_ j}f_ j\, \mathrm{d}V$. The preceding lemma bounds $M_ j=\sup _ Xu_ j$, and consequently bounds $F_ j=\mathrm{e}^{\lambda u_ j}f_ j$ in $L^ p$. The normalized potentials $\psi _ j=u_ j-M_ j$ are uniformly bounded by theorem 14.19.19. We may pass to a subsequence with $M_ j\to M$ and $\psi _ j$ Cauchy in $L^1$. Theorem 14.19.21 gives uniform convergence $u_ j\to u\in \mathrm{PSH}(X,\omega )\cap C(X)$.

Uniform convergence and $f_ j\to f$ in $L^ p$ give $\mathrm{e}^{\lambda u_ j}f_ j\to \mathrm{e}^{\lambda u}f$ in $L^ p$, and continuity of bounded products therefore gives $\omega _ u^ n=\mathrm{e}^{\lambda u}f\, \mathrm{d}V$. Uniqueness, even among all bounded solutions, follows from bounded exponential comparison, corollary 14.19.12. $\square$

Proof. Since $f_ a\leq f_ b$, the two equations give $\mathrm{e}^{-\lambda b}\omega _ b^ n\geq \mathrm{e}^{-\lambda a}\omega _ a^ n$. Hence $b\leq a$ by corollary 14.19.12, applied with $u=a$ and $v=b$. $\square$

Proof. Since $u$ is bounded, the density $\mathrm{e}^{-u}f$ is nonnegative, nonzero and belongs to $L^ p$. By theorem 14.19.3, applied with $\lambda =1$ to this density, there is a continuous solution $r$ of

The bounded potential $u$ solves the same exponential equation. Applying lemma 14.19.28 with $f_ a=f_ b=\mathrm{e}^{-u}f$ in both orders gives $u=r$, so $u$ is continuous. $\square$