Let \(U \subset \mathbb{C}\) be an open set and
\[A^2(U) = \{ f \text{ holomorphic on } U : \int_U |f(z)|^2 \, dx \, dy < \infty \}.\]
Define
\[\langle f, g \rangle = \int_U f(z) \overline{g(z)} \, dx \, dy, \quad \forall f, g \in A^2(U).\]
Prove that \(A^2(U)\) is a Hilbert space when equipped with this inner product.
Solution
Write \(dA=dx\,dy\) and \(\|f\|_2^2=\int_U|f|^2\,dA\).
1. The formula defines an inner product on the vector space \(A^2(U)\).
Linear combinations of holomorphic functions are holomorphic, and \(|af+bg|^2\leq2|a|^2|f|^2+2|b|^2|g|^2\) shows that they remain square integrable. Cauchy–Schwarz gives \(\int_U|f\overline g|\,dA\leq\|f\|_2\|g\|_2\), so the inner product is well defined [@Fol13]. Linearity in the first variable and conjugate symmetry follow from the integral.
If a holomorphic \(f\) is nonzero at \(a\in U\), continuity gives a disk of positive area on which \(|f|\geq|f(a)|/2\). Then \(\|f\|_2>0\). Hence \(\|f\|_2=0\) implies \(f(a)=0\) at every \(a\in U\), which is positive definiteness.
2. Every compact subset has an \(L^2\) point-evaluation bound.
For \(\overline{D(a,r)}\subset U\), Cauchy’s circle formula and Cauchy–Schwarz imply, for \(0<\rho<r\), \[ |f(a)|^2\leq\frac1{2\pi}\int_0^{2\pi}|f(a+\rho e^{it})|^2\,dt \] [@SS03; @Fol13]. Integrating this inequality against \(2\pi\rho\,d\rho\) yields \(\pi r^2|f(a)|^2\leq\int_{D(a,r)}|f|^2\,dA\). For a nonempty compact \(K\subset U\), compactness and openness give a single \(r>0\) with these closed disks contained in \(U\) for every \(a\in K\). Thus \[ \sup_{a\in K}|f(a)|\leq\frac1{\sqrt\pi r}\|f\|_2. \] The same estimate holds for differences of functions in \(A^2(U)\).
3. The inner-product space is complete.
Let \((f_n)\) be Cauchy in this norm. Completeness of \(L^2(U)\) gives an \(L^2\) limit \(F\), regarded as a measurable representative of its almost-everywhere class [@Fol13]. Step 2 makes \((f_n)\) uniformly Cauchy on every compact subset of \(U\). Pointwise completeness of \(\mathbb C\) and that same estimate give a locally uniform limit \(f\). This limit is holomorphic by the local uniform limit theorem [@SS03].
At almost every point where \(F\) is finite, \(f_n\to f\). Fatou’s lemma therefore gives \[ \int_U|f-F|^2\,dA \leq\liminf_{n\to\infty}\int_U|f_n-F|^2\,dA=0 \] [@Fol13]. Hence \(f=F\) almost everywhere. It follows that \(f\in A^2(U)\) and \(\|f_n-f\|_2=\|f_n-F\|_2\to0\). Thus every Cauchy sequence converges in \(A^2(U)\), proving that it is a Hilbert space.