 Solution

For $h,g\in Y$, their product $hg$ is holomorphic and integrable on the disk by the Cauchy-Schwarz inequality. Its angular mean on every centered circle is $h(0)g(0)$ by the mean value property. Polar integration therefore givesIf $g(0)=0$, this vanishes for every $h\in Y$, proving $\boxed{Z\subseteq Y^{\perp }}$. Integrability of the product justifies integration over the full disk, even when the individual holomorphic functions grow near its boundary.

The inclusion is strict. Set $w(z)=|z{|}^2-1/2$, a nonzero bounded function in $L^2(D)$. For every $h\in Y$,Thus $w\in Y^{\perp }$. It cannot equal $\overline{g}$ for holomorphic $g$: equality almost everywhere between these continuous functions would give equality everywhere, making $g=w$ a real-valued nonconstant holomorphic function, which is impossible. Therefore