I want to know whether my attempt for the following proof is correct:

Let $(f_n)_{n\in \mathbb N}$ be a sequence of entire functions converging uniformly on $\partial D$ (boundary of the unit disk, i.e. unit circle) to some function $g: \partial D \to \mathbb C$. Show that there exists a holomorphic function $f: D \to \mathbb C$ such that $f_n(z) \to f(z)$ for each $z\in D$.

My attempt: By the Cauchy integral formula, the function $\tilde g: D \to \mathbb C$, $$z\mapsto \frac{1}{2\pi i}\int_{\partial D}\frac{g(\zeta)}{\zeta -z}d\zeta$$ is holomorphic on $D$. Now, for each $z\in D$:

$$\lim_{n\to \infty} f_n(z) = \lim_{n\to \infty} \frac{1}{2\pi i} \int_{\partial D}\frac{f_n(\zeta)}{\zeta -z}d\zeta = \frac{1}{2\pi i} \int_{\partial D}\lim_{n\to \infty} \frac{f_n(\zeta)}{\zeta -z}d\zeta = \frac{1}{2\pi i}\int_{\partial D}\frac{g(\zeta)}{\zeta -z}d\zeta$$ where we exchanged limit and integral by uniform convergence. So $\tilde g$ is our desired function.

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    $\begingroup$ How do you use Cauchy intergral formula to show that $\tilde{g}$ is holomorphic? $\endgroup$ – Adayah Oct 13 '17 at 11:05
  • $\begingroup$ I don't know. I thought the so defined function is always holomorphic inside of the disk. $\endgroup$ – Staki42 Oct 13 '17 at 11:55
  • $\begingroup$ It is, but I don't see how to use the Cauchy formula to prove it. I would compute the derivative from the definition instead. $\endgroup$ – Adayah Oct 13 '17 at 12:43
  • $\begingroup$ Well, it was poorly put by me. I used the formula for the proof, but it doesn't follow from the formula that the so defined function is holomorphic. $\endgroup$ – Staki42 Oct 13 '17 at 13:11
  • $\begingroup$ The fact that it's holomorphic on the inside of the disk is a standard thing. What is not clear is that-with this definition on the inside- the function is continuous on the closed disk. That was not part of the problem, but it should be, as seen in some solution provided. $\endgroup$ – Orest Bucicovschi Oct 13 '17 at 15:33

By the maximum principle: $$ \sup_{z\in D} |f_n(z)-f_m(z)| \le \sup_{z\in \partial D} |f_n(z)-f_m(z)| $$ so $\{ f_n \}$ is uniformly Cauchy on $\bar D$. Hence $f_n$ converges uniformly to some (necessarily holomorphic on $D$ and continuous on $\bar D$) function $f$, where $f = g$ on $\partial D$.

The fact that $f$ is holomorphic on $D$ follows, for example, from Morera's theorem: If $\gamma$ is any closed curve in $D$, then $$ \int_\gamma f\,dz = \int_\gamma (\lim_{n\to\infty} f_n)\,dz = \lim_{n\to\infty} \int_\gamma f_n\,dz = 0 $$ by Cauchy's integral theorem and properties of uniform convergence. Hence $f$ is holomorphic.

  • $\begingroup$ That's a neat solution. So is my try completely false or does it have correct parts? $\endgroup$ – Staki42 Oct 13 '17 at 12:26
  • $\begingroup$ @lappen68 Your attempt can be saved, but you need to show that your $\tilde g$ is holomorphic. (Either you need to know a little bit about the Cauchy transform, or you can prove it directly.) $\endgroup$ – mrf Oct 13 '17 at 12:33
  • $\begingroup$ Alright. I thought that was an obvious thing, but I guess it's not. thanks! $\endgroup$ – Staki42 Oct 13 '17 at 12:42

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