Let \(U \subseteq \mathbb{C}\) be an open set and
\[A^2(U) = \{ f \text{ holomorphic on } U : \int_U |f(z)|^2 \, dx \, dy < \infty \}.\]
Define
\[(f, g) = \int_U f(z) \overline{g(z)} \, dx \, dy, \quad \forall f, g \in A^2(U).\]
Prove that \(A^2(U)\) is a Hilbert space when equipped with this inner product.
Solution
Write \(dA=dx\,dy\) and \(\|f\|_2=(\int_U|f|^2\,dA)^{1/2}\). If \(U\) is empty, the space consists of the unique empty function and is the zero Hilbert space. Assume \(U\ne\varnothing\).
1. The displayed formula is an inner product on \(A^2(U)\).
Linear combinations remain holomorphic, and \(|af+bg|^2\leq2|a|^2|f|^2+2|b|^2|g|^2\) proves their square integrability. Thus \(A^2(U)\) is a complex vector space. Cauchy–Schwarz gives \(\int_U|f\overline g|\,dA\leq\|f\|_2\|g\|_2\), so the inner product is finite [@Fol13]. Linearity in the first variable and conjugate symmetry follow from the integral.
If \(f(a)\ne0\) at some \(a\in U\), continuity gives a disk inside \(U\) on which \(|f|\geq|f(a)|/2>0\). Its positive area makes \(\int_U|f|^2\,dA>0\). Hence zero norm forces the holomorphic function itself to be zero everywhere, not just almost everywhere. This proves positive definiteness.
2. The \(L^2\) norm controls uniform convergence on each compact subset.
Let \(h\) be holomorphic on \(U\), and suppose \(\overline{D(a,r)}\subset U\). Cauchy’s formula and Cauchy–Schwarz on each circle of radius \(0<\rho<r\) give \[ |h(a)|^2\leq\frac1{2\pi}\int_0^{2\pi} |h(a+\rho e^{it})|^2\,dt \] [@SS03]. Multiply by \(2\pi\rho\) and integrate from zero to \(r\). Polar coordinates yield \[ \pi r^2|h(a)|^2\leq\int_{D(a,r)}|h|^2\,dA. \] For a nonempty compact \(K\subset U\), choose one \(r>0\) such that \(\overline{D(a,r)}\subset U\) for every \(a\in K\). Such an \(r\) exists by compactness and openness; when \(U\ne\mathbb C\), take less than the positive distance from \(K\) to the closed complement, and otherwise take \(r=1\). It follows that \[ \sup_{a\in K}|h(a)|\leq\frac1{\sqrt\pi r}\|h\|_2. \] Apply this to differences of elements of \(A^2(U)\).
3. Every Cauchy sequence in \(A^2(U)\) converges in its norm.
Let \((f_n)\) be such a sequence. Step 2 makes it uniformly Cauchy on each compact subset of \(U\). Completeness of \(\mathbb C\) gives a pointwise limit \(f\), and the same Cauchy estimates show that convergence is uniform on each compact subset. The local uniform limit of holomorphic functions is holomorphic [@SS03]. Thus \(f\) is holomorphic on \(U\).
The norms \(\|f_n\|_2\) are bounded because the sequence is Cauchy. Fatou’s lemma gives \[ \int_U|f|^2\,dA\leq\liminf_{n\to\infty}\|f_n\|_2^2<\infty, \] so \(f\in A^2(U)\) [@Fol13]. Given \(\varepsilon>0\), choose \(N\) such that \(\|f_n-f_m\|_2<\varepsilon\) for \(n,m\geq N\). For fixed \(n\geq N\), apply Fatou again: \[ \|f_n-f\|_2^2\leq\liminf_{m\to\infty}\|f_n-f_m\|_2^2 \leq\varepsilon^2. \] Therefore \(f_n\to f\) in norm. The inner-product space is complete, hence is a Hilbert space.