# Uniform convergence of holomorphic functions

Question: Let $$\Omega$$ be a non-empty open set in $$\mathbb{C}$$ and let $$f$$ be a continuous function on $$\Omega$$. Suppose $$\{f_n\}$$ is a sequence of holomorphic functions on $$\Omega$$ such that $$\lim_{n\to \infty} \int_D |f_n(x+iy) - f(x+iy)|\, dx \, dy =0$$ for all closed disk $$D \subset \Omega$$. Show that $$f$$ is holomorphic in $$\Omega$$ and both $$f_n \to f$$ and $$f_n' \to f'$$ uniformly on compact subsets of $$\Omega$$.

My attempt: To prove $$f$$ is holomorphic in $$\Omega$$, I aim to show that $$f$$ satisfy the Cauchy integral formula, that is $$f(z_0) = \int_{|z-z_0| = R} \frac{f(z)}{z - z_0}\, dz$$ for all $$z \in \Omega$$ where $$\{z\colon |z- z_0| \leq R\} \subset \Omega$$. Given that $$f_n$$ is holomorphic in $$\Omega$$ by the use of polar coordinates $$f_n(z_0) =\int_D\frac{f_n(z)}{z - z_0}\, dz = \frac{1}{\pi R^2} \int_0^{2\pi} \int_0^R f_n(z_0 + R e^{i\theta})r \, d r \, d \theta$$ where the compact closure of $$D = D(z_0, R)$$ is contained in $$\Omega$$ whereas $$\int_D f_n(x+iy) \, d x\, dy = \int_0^{2\pi} \int_0^R f_n(z_0 + r e^{i\theta}) r \, d r\, d \theta$$ so $$\int_D f_n(x+iy)\, dx \, dy$$ and $$f_n(z_0)$$ are not directly comparable. are there any ways to resolve this issue? Or is there an alternative method to solve this problem?

• @geetha290krm may I know how did u get that equality? I have the integral expression of $f_n(z_0)$ and $\int_D f_n(x+iy)\, dx \, dy$ to be different since the integrand in each terms are different, as indicated in the second and third equations. Furthermore, there is a missing scaling constant of $(\pi R^2)^{-1}$.
– L-JS
Jun 28 at 5:58
• Jun 28 at 7:38
• @MartinR Thanks for the resource. Do you know how did the answer obtain the following equality in the first place $m(D)|f_n(z) - f_m(z)| = |\int_D (f_n - f_m)|$ where $z \in \Omega$ is given and $D$ is ball containing $z$ that is contained in $\Omega$.
– L-JS
Jun 28 at 8:44
• From $$\pi R^2 f_n(z_0) = \int_0^R \int_0^{2 \pi} f_n(z_0 + re^{i \theta}) r \, d\theta dr = \int_{B_R(z_0)} f_n(x+iy) \, dx dy \, .$$ Jun 28 at 8:51

Let $$z_0 \in \Omega, \epsilon > 0$$ be given and choose $$R > 0$$ such that $$\overline{D_{R + \epsilon}(z_0)} \subset \Omega$$. Then, there exists $$N \in \mathbb{N}$$ so that
$$n \geq N \implies \int_{D_{R + \epsilon}(z_0)}|f_n(x+iy) - f(x+iy)|\, dx \, dy < \frac{|D_{\epsilon}(z_0)|\epsilon}{2}.$$
and in particular $$n,m \geq N \implies \int_{D_{R + \epsilon}(z_0)} |f_n(x+iy) - f_m(x+iy)|\, dx \, dy < |D_{\epsilon}(z_0)|\epsilon.$$ Also given $$f_n, f_m$$ are holomorphic in $$\Omega$$ then by mean-value property $$f_n(z_0) = \frac{1}{2\pi} f_n(z_0 + r e^{i \theta})\, d \theta, \quad f_m(z_0) =f_m(z_0 + r e^{i \theta})\, d \theta$$ for all $$0 < r \leq R$$. Then by the use of polar coordinates whenever $$n, m \geq N$$ \begin{align} |D_R(z_0)| |f_n(z_0) - f_m(z_0)| &= \pi R^2 |f_n(z_0) - f_m(z_0)|\\ &= \left| \int_{D_R(z_0)}[f_n(x+iy) - f_m(x+iy)] \, dx \, dy \right| \\ &\leq \int_{D_{R + \epsilon}(z_0)}\left| f_n(x+iy) - f_m(x+iy) \right|\, dx \, dy \\ &\leq |D_\epsilon(z_0)| \epsilon \end{align} Thus, $$n,m \geq N \implies |f_n(z_0) - f_m(z_0)| \leq \frac{|D_\epsilon(z_0)|}{|D_R(z_0)|}\epsilon < \epsilon.$$ Given any $$z_1 \in \overline{D_R(z_0)}$$ then $$\overline{D_\epsilon(z_1)} \subset \overline{D_{R + \epsilon}(z_0)}$$ so \begin{align} |D_\epsilon(z_1)| |f_n(z_1) - f_m(z_1)| &= \pi \epsilon^2 |f_n(z_1) - f_m(z_1)|\\ &= \left| \int_{D_\epsilon(z_1)}[f_n(x+iy) - f_m(x+iy)] \, dx \, dy \right| \\ &\leq \int_{D_{R + \epsilon}(z_0)}\left| f_n(x+iy) - f_m(x+iy) \right|\, dx \, dy \\ &\leq |D_\epsilon(z_0)| \epsilon \end{align} Thus, $$n,m \geq N \implies |f_n(z_1) - f_m(z_1)| \leq \epsilon.$$ Therefore, $$\{f_n\}$$ is uniformly Cauchy in $$\overline{D_R(z_0)}$$ so $$\{f_n\}$$ is uniformly convergent in $$\overline{D_R(z_0)}$$. As such closed disks are arbitrary we have $$f_n$$ converges uniformly on every compact subsets of $$\Omega$$ implying that $$f_n$$ converges uniformly to $$F$$, a holomorphic function in $$\Omega$$. Now, we are left to show that $$F(z) = f(z), \forall z \in \Omega.$$ Given any closed disk $$D \subset \Omega$$, we have $$0 = \lim_{n \to \infty} \int_D |f_n(x+iy) - f(x+iy)|\, dx \, dy = \int_D |F(x+iy) - f(x+iy)|\, dx \, dy$$ thus $$F = f$$ a.e. in $$D$$. If for a contradiction that the set $$\{z \in D \colon f(z) \neq F(z)\} = (f - F)^{-1}(D - \{0\})$$ is nonempty then it being open must have positive measure, contrary to our conclusion. Thus, $$f(z) = F(z), \forall z \in D$$. Again, $$D$$ is an arbitrary closed disk in $$\Omega$$ then $$f(z) = F(z), \forall z \in \Omega$$, this concludes the proof. As for the convergence of the derivatives, it follows a standard argument available in most complex analysis text.