Let us consider the unit disc $\mathcal{D}=\{z: |z|\leq 1\}$ and an arbitrary continuous function $f:\partial \mathcal{D} \rightarrow \mathbb{C}$.

I want to find an extension of $f$ that is continuous on $\mathcal{D}$ and holomorphic on $int(\mathcal{D})$, by applying the Cauchy-Riemann integral formula: $f(z_0)=\int_{\partial \mathcal{D}}\frac{f(z)}{z-z_0}dz$.

Moreover, by Morera's theorem, a continuous, complex-valued function ƒ defined on an open set D in the complex plane that satisfies $\int_{\gamma}f(z)dz=0$ for every closed piecewise $\mathcal{C^1}$ curve $\gamma$ in $\mathcal{D}$ must be holomorphic on $\mathcal{D}$.

How can I apply the Morera's theorem to find out whether there is an extension of $f$ that is holomorphic on $int(\mathcal{D})$?

  • $\begingroup$ You have given a formula for $f$, so what then is the question? $\endgroup$ – Lord Shark the Unknown Oct 13 '17 at 21:33
  • $\begingroup$ I would like to understand Morera's theorem. Does it suffice to show that the integral on the unit disc is zero? $\endgroup$ – FunnyBuzer Oct 13 '17 at 21:39
  • 2
    $\begingroup$ You want to show that the integral of $f$ over any nice curve inside $D$ is zero, that is $$\int_\gamma\int_{\partial D}\frac{f(w)}{w-z}\,dw\,dz.$$ Faced with this integral, I'd try changing the order of integration. $\endgroup$ – Lord Shark the Unknown Oct 13 '17 at 21:42
  • $\begingroup$ If there is a holomorphic extension of $f$ to the unit disk, then by Cauchy's integral theorem, one must have $$\int_{\partial \mathcal{D}} z^k f(z)\,dz = 0\tag{1}$$ for every integer $k \geqslant 0$. This condition is also sufficient, if $f\in C(\partial \mathcal{D})$ satisfies $(1)$ (for all $k \geqslant 0$), then there is an $F \in C(\mathcal{D})$, holomorphic in the interior, with $F\lvert_{\partial \mathcal{D}} = f$. $\endgroup$ – Daniel Fischer Oct 13 '17 at 21:57

There is no such a thing as the “Cauchy-Riemann integral equation”; what you have in mind is Cauchy's integral formula.

See what happens if $f(z)=\frac1z$ when $|z|=1$. Then, if $0<|z_0|<1$,\begin{align}\int_{\partial D}\frac{f(z)}{z-z_0}\mathrm dz&=\int_{\partial D}\frac1{z_0}\left(\frac1{z-z_0}-\frac1z\right)\,\mathrm dz\\&=\frac{2\pi i}{z_0}\bigl(\operatorname{Ind}_{\partial D}(z_0)-\operatorname{Ind}_{\partial D}(0)\bigr)\\&=0,\end{align}and it is easy to see that $\int_{\partial D}\frac{f(z)}z\mathrm dz=0$ too. So, you get the null function in $\operatorname{Int}D$.

  • $\begingroup$ You are definitely right, sorry for the descriptive mistake. $\endgroup$ – FunnyBuzer Oct 13 '17 at 21:40
  • $\begingroup$ In your example, you chose a function that allows an extension that is continuous on $\mathcal{D}$ and holomorphic on $int(\mathcal{D})$. Suppose $\mathcal{Im}(z)=0$. Then, computing the integral on the unit disc, one gets that the integral is $\pi i \neq 0$. Isn't it a contradiction with the fact that $f$ is holomorphic on $\mathbb{C}$? $\endgroup$ – FunnyBuzer Oct 14 '17 at 17:26
  • $\begingroup$ @FunnyBuzer Are you saying that there is a continuous function $F\colon D\longrightarrow\mathbb C$ which is continuous on $\mathring{D}$ and such that $|z|=1\implies f(z)=\frac1z$? $\endgroup$ – José Carlos Santos Oct 14 '17 at 17:37
  • $\begingroup$ Not exactly. Rather consider a function on the disc of radius $1-\varepsilon$ (which is still a compact set) and $f(z)=\mathcal{Re}(z)$. $\endgroup$ – FunnyBuzer Oct 14 '17 at 17:55
  • $\begingroup$ @FunnyBuzer I'm sorry, but I really don't know what your asking. $\endgroup$ – José Carlos Santos Oct 14 '17 at 17:56

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