Let $D$ the unit complex disk and let $\mu$ be the Lebesgue measure on $D$. If $\nu>-1$, define $\varphi_\nu(w):=(1-|w|^2)^\nu$, define $\operatorname{d}\mu_\nu:=\varphi_\nu\operatorname{d}\mu$ and define the $\nu$-Bergman projection as: $$B_\nu :L^1(\mu_\nu)\rightarrow H(D), f\mapsto\left(z\mapsto \int_D\frac{f(w)}{(1-z\bar{w})^{2+\nu}}\operatorname{d}\mu_\nu(w)\right),$$
where $H(D)$ is the space of holomorphic functions on the disk. Following the ideas in the proof of the theorem 6, page 37 of the book Duren and Schuster – Bergman Spaces, I stumbled upon the following problem:
if $B_\nu$ is bounded from $L^1(\mu_\nu)$ to itself, then:
$$\forall f\in L^1(\mu_\nu), \forall g\in L^\infty(D), \int_D B_\nu(f)(z)\overline{g(z)}\operatorname{d}\mu_\nu(z)=\int_D f(w)\overline{B_\nu(g)(w)}\operatorname{d}\mu_\nu(w).$$
In order to prove this assertion I tried to use Fubini's theorem, but I can't see how its hypothesis are satisfied. In particular I can't prove that:
$$\int_D\int_D|f(w)||g(z)|\frac{1}{|1-z\bar{w}|^{2+\nu}}\operatorname{d}\mu_\nu(w)\operatorname{d}\mu_\nu(z)<+\infty,$$
only from the hypothesis that $B_\nu$ is bounded.
Best Answer
For $n \ge 1$, let $f_n(w) = f(w)1_{|w| \le 1-\frac{1}{n}}$. Then $$\int_D\int_D|f_n(w)||g(z)|\frac{1}{|1-z\bar{w}|^{2+\nu}}d\mu_\nu (w)d\mu_\nu(z) \le ||g||_\infty\int_D |f_n(w)|\int_D \frac{1}{|1-z\bar{w}|^{2+\nu}}d\mu_\nu(z)d\mu_\nu(w)$$ $$\le \int_D |f_n(w)| C_n d\mu_\nu(w) < \infty$$ where we used that $C_n := \sup_{|w| \le 1-\frac{1}{n}} \int_D \frac{1}{|1-z\bar{w}|^{2+\nu}}d\mu_\nu(z) < \infty$. Therefore, Fubini applies and $$\int_D B_\nu(f_n)(z)\overline{g(z)} d\mu_\nu(z) = \int_D f_n(w)\overline{B_\nu(g)(w)}d\mu_\nu(w)$$ for each $n \ge 1$. As $f_n \to f$ in $L^1(\mu_\nu)$ and $B_\nu$ is bounded, we have $B_\nu(f_n) \to B_\nu(f)$ in $L^1(\mu_\nu)$. Then since $g \in L^\infty$, we have $$\int_D B_\nu(f_n)(z)\overline{g(z)} d\mu_\nu(z) \to \int_D B_\nu(f)(z)\overline{g(z)}d\mu_\nu(z).$$ Therefore, we just need to show $$\int_D f_n(w)\overline{B_\nu(g)(w)}d\mu_\nu(w) \to \int_D f(w)\overline{B_\nu(g)(w)}d\mu_\nu(w).$$ However, $\int_D f(w)\overline{B_\nu(g)(w)}d\mu_\nu(w)-\int_D f_n(w)\overline{B_\nu(g)(w)}d\mu_\nu(w) = \int_{|w| > 1-\frac{1}{n}} f(w)\overline{B_\nu(g)(w)}d\mu_\nu(w)$ does go to $0$ as $n \to \infty$ since $f(w)\overline{B_\nu(g)(w)} \in L^1(\mu_\nu)$.