Proposition 14.19.40.21F8F4F18D0A Fix $1<p<\infty $ and $M\geq 1$. Suppose $f_ j\geq 0$ satisfy
\[ \int _ X f_ j\, \mathrm{d}\nu =1,\qquad \| f_ j\| _{L^ p(X,\nu )}\leq M,\qquad c_{f_ j}\longrightarrow 1. \]
For each $j$, let $u_ j$ be any continuous solution of $\omega _{u_ j}^3=c_{f_ j}f_ j\omega ^3$ normalized by $\sup _ Xu_ j=0$. Then
\[ \| u_ j-Au_ j\| _{C^0(X)}\longrightarrow 0,\qquad f_ j-Af_ j\rightharpoonup 0\quad \hbox{weakly in }L^ p(X,\nu ). \]
Every subsequence has a further subsequence with $u_ j\to \pi ^*q$ uniformly and $f_ j\rightharpoonup \pi ^*g$ weakly in $L^ p$, where $q$ is a continuous $\omega _{\mathrm{FS}}$-plurisubharmonic function, $g\geq 0$, $\int _{\mathbb P^2}g\omega _{\mathrm{FS}}^2=1$, and
\[ (\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}q)^2=g\omega _{\mathrm{FS}}^2. \]
Proof.
Since the constants $c_{f_ j}$ converge, the densities $c_{f_ j}f_ j$ have a common $L^ p$ bound. By theorem 14.19.21, every subsequence has a further subsequence on which $u_ j\to u$ uniformly. Since $L^ p$ is reflexive, a further extraction gives $f_ j\rightharpoonup f$ weakly in $L^ p$. Positivity, the unit integral and the bound $\| f\| _ p\leq M$ pass to the limit, and continuity of bounded products under uniform convergence gives
\[ (\omega +\mathrm{dd}^{\mathrm{c}}u)^3=f\omega ^3. \]
Thus $c_ f=1$. Since also $I_ f=1$, proposition 14.19.38 gives $u=\pi ^*q$ and $f=\pi ^*g$. The bounded cube identity proved there, tested against pullbacks of continuous functions on the base, gives the asserted base equation. In particular, both limits are fixed by $A$.
Haar averaging is a contraction on $C^0(X)$ and on $L^ p(X,\nu )$. It is also self-adjoint with respect to $\nu $: Fubini's theorem and the invariance of $\nu $ under the fiber translations give $\int \chi Af\, \mathrm{d}\nu =\int A\chi f\, \mathrm{d}\nu $ for $\chi \in L^{p'}$. Along the extracted subsequence, therefore, $u_ j-Au_ j\to 0$ uniformly and $f_ j-Af_ j\rightharpoonup 0$ in $L^ p$. If either convergence failed for the original sequence, some subsequence would stay away from the limit (for the weak convergence, when tested against one fixed function in $L^{p'}$), and a further subsequence of the kind just constructed would give a contradiction. This proves both conclusions for the whole sequence.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.