Bergman projection

Remarks

The integral formula follows from the reproducing property: for $f \in L^2(\Omega)$, $Pf(z) = \ip{Pf}{K(\cdot, z)} = \ip{f}{K(\cdot, z)}$, since $K(\cdot, z) \in A^2(\Omega)$ and $f - Pf$ is orthogonal to $A^2(\Omega)$. Example: on the unit disc, the functions $z^k$, $k \ge 0$, form an orthogonal basis of $A^2$, and $\ip{\bar z}{z^k} = \int_{\mathbb{D}} \bar z \, \bar z^k \, dV = 0$ for every $k$, so $P \bar z = 0$. Similarly $P \abs{z}^2 = \frac{1}{2}$, since $\ip{\abs{z}^2}{1} = \int_{\mathbb{D}} \abs{z}^2 dV = \frac{\pi}{2}$ and $\norm{1}^2 = \pi$, and $\abs{z}^2$ is orthogonal to $z^k$ for $k \ge 1$. Non-example: the orthogonal projection of $L^2(\Omega)$ onto the harmonic functions in $L^2(\Omega)$ is a different operator, the harmonic Bergman projection; on the unit disc it fixes $\bar z$. $P$ is self-adjoint, $P^2 = P$, and $\norm{P} = 1$ when $A^2(\Omega) \ne \set{0}$. On smoothly bounded strongly pseudoconvex domains, $P$ maps $C^\infty(\bar\Omega)$ into itself.