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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?

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  • $\begingroup$ @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 $(\pi R^2)^{-1}$. $\endgroup$
    – L-JS
    Commented Jun 28, 2023 at 5:58
  • $\begingroup$ Compare math.stackexchange.com/q/3266985/42969 $\endgroup$
    – Martin R
    Commented Jun 28, 2023 at 7:38
  • $\begingroup$ @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 $\Omega$. $\endgroup$
    – L-JS
    Commented Jun 28, 2023 at 8:44
  • $\begingroup$ From $$ \pi R^2 f_n(z_0) = \int_0^R \int_0^{2 \pi} f_n(z_0 + re^{i \theta}) r \, d\theta dr = \int_{B_R(z_0)} f_n(x+iy) \, dx dy \, . $$ $\endgroup$
    – Martin R
    Commented Jun 28, 2023 at 8:51

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Here is my answer following the prompts in the comments.

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.

Please do let me know if there any gaps in my proof. Thank you!

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