= 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 gives
$$
\langle h,\overline g\rangle=\int_Dh(z)g(z)\,dA(z)
=\pi h(0)g(0).
$$
If $g(0)=0$, 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 \b[strict]. Set $w(z)=|z|^2-1/2$, a nonzero bounded function in $L^2(D)$. For every $h\in Y$,
$$
\langle w,h\rangle
=2\pi\overline{h(0)}\int_0^1(r^2-1/2)r\,dr=0.
$$
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
$$
\boxed{Z\ne Y^\perp.}
$$
Back to article page
