Let \(U\) be an open subset of \(\mathbb{C}\). We use the notation
\[\|f\|_{L^2(U)} = \left(\int_U |f|^2 \, dx \, dy\right)^{1/2}.\]
- Let \(f : U \to \mathbb{C}\) be a holomorphic function. Show that for any compact set \(K \subset U\), there is a constant \(C_K\) such that
\[\sup_{z \in K} |f(z)| \leq C_K \|f\|_{L^2(U)}.\]
Prove that \(\{f \text{ is holomorphic on } U : \|f\|_{L^2(U)} \leq 1\}\) is a normal family.
Suppose \(U\) is the punctured disc \(D(0,1) \setminus \{0\}\). If \(f\) is holomorphic on \(U\) and \(\|f\|_{L^2(U)} < \infty\), prove that \(z = 0\) is a removable singularity of \(f\).
Solution
Write \(dA=dx\,dy\). Normality here means that every sequence has a subsequence converging uniformly on compact subsets to a finite holomorphic function.
1. A disk contained in \(U\) gives a pointwise mean-square estimate.
Suppose \(\overline{D(a,r)}\subset U\). For \(0<\rho<r\), Cauchy’s formula at the center and Cauchy–Schwarz give \[ f(a)=\frac1{2\pi}\int_0^{2\pi}f(a+\rho e^{it})\,dt, \qquad |f(a)|^2\leq\frac1{2\pi}\int_0^{2\pi}|f(a+\rho e^{it})|^2\,dt \] [@SS03; @Fol13]. Multiplying by \(2\pi\rho\) and integrating over \(0<\rho<r\) yields, by polar coordinates, \[ \pi r^2|f(a)|^2\leq\int_{D(a,r)}|f|^2\,dA. \]
2. The required compact-set bound holds with a constant independent of \(f\).
For nonempty compact \(K\subset U\), choose \(r>0\) with \(\overline{D(a,r)}\subset U\) for all \(a\in K\). If \(U\ne\mathbb C\), take half the distance from \(K\) to the closed set \(\mathbb C\setminus U\); this distance is positive by compactness and openness. For \(U=\mathbb C\), take \(r=1\). Step 1 gives \[ \sup_{a\in K}|f(a)|\leq\frac1{\sqrt\pi r}\|f\|_{L^2(U)}. \] Thus \(C_K=1/(\sqrt\pi r)\) works. If the norm is infinite, the inequality holds in the extended sense. For empty \(K\) the pointwise assertion has no points to check.
3. The family with \(L^2\) norm at most one is normal on \(U\).
Step 2 makes this family uniformly bounded on every compact subset of \(U\). Montel’s theorem gives normality [@SS03]. This also applies when \(U\) is disconnected: its components are open and each contains a point with rational real and imaginary parts, so there are at most countably many components. Apply the theorem to each component and take a diagonal subsequence. Every compact subset of \(U\) meets only finitely many components, since the components form an open cover of that compact set. The diagonal subsequence therefore converges uniformly on every compact subset of \(U\), to the componentwise holomorphic limit. When \(U\) is empty the assertion is immediate.
In fact the limit remains in the family: pointwise convergence and Fatou’s lemma give \(\int_U|f|^2\,dA\leq\liminf_j\int_U|f_{n_j}|^2\,dA\leq1\) [@Fol13].
4. Square integrability makes the isolated singularity removable.
Now let \(U=\{0<|z|<1\}\) and \(\int_U|f|^2\,dA<\infty\). For \(0<|a|<1/2\), the closed disk of radius \(r=|a|/2\) centered at \(a\) lies in \(U\). Step 1 implies \[ |a|^2|f(a)|^2 \leq\frac4\pi\int_{D(a,|a|/2)}|f|^2\,dA \leq\frac4\pi\int_{0<|z|<3|a|/2}|f(z)|^2\,dA. \] The last integral tends to zero as \(a\to0\). To see this, the indicators of shrinking punctured disks tend pointwise to zero on \(U\) and are dominated by one; dominated convergence applies to the integrable function \(|f|^2\) [@Fol13]. Hence \(af(a)\to0\).
The function \(h(z)=zf(z)\) is holomorphic on the punctured disk and tends to zero at the puncture. The removable singularity theorem extends it holomorphically with \(h(0)=0\) [@SS03]. Its Taylor series is divisible by \(z\), so \(h(z)=zF(z)\) for a holomorphic \(F\) near zero. For \(z\ne0\), this \(F\) equals \(f\). It supplies the required holomorphic extension across zero.