Past exam of the analysis course of the University of Oxford / 2011 / 5

The orthogonal complement isIt is a linear subspace and is closed, since it is an intersection of kernels of the continuous linear functionals $w\mapsto ⟨w,y⟩$. Continuity follows from the Cauchy-Schwarz inequality.

The minimizing point and orthogonality proved below write every $x$ as $x=y_0+(x-y_0)\in Y+Y^{\perp }$. If $w\in Y\cap Y^{\perp }$, then $\Vert w{\Vert }^2=⟨w,w⟩=0$, so the intersection is zero. ConsequentlyIn particular the minimizing point is unique: an alternative orthogonal decomposition would have a difference in this zero intersection.

Let $d={\inf }_{y\in Y}\Vert x-y\Vert$ and choose $y_n\in Y$ with $\Vert x-y_n\Vert \to d$. The parallelogram law givesThe right side tends to zero. Thus $(y_n)$ is Cauchy. The closed subspace $Y$ of a Hilbert space is complete, so $y_n\to y_0\in Y$. Continuity of the norm gives $\Vert x-y_0\Vert =d$, proving the distance is attained.

Put $w=x-y_0$. For any $y\in Y$ and any real $t$, minimality givesThe derivative at zero must vanish, so $\mathrm{Re}⟨w,y⟩=0$. In a complex Hilbert space, apply the same argument to $iy$ to get the imaginary part zero as well. In the real case the first conclusion is already enough. Hence

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