# Proving $f(z)$ entire function in complex analysis

If $f\in C(\Bbb C)\cap H(\Bbb C\backslash \delta B_1(0))$ then $f\in H(\Bbb C)$ [C means continuous, $\Bbb C$ means complex plane, H means analytic and $\delta$ means boundary]

I don't even know where to start. Thank you...

• You didn't have to make a new account :P it's alright to have trouble. What have you tried so far? What are your given definitions?
– Eoin
Nov 23, 2014 at 3:19
• @Eoin I have to prove that f is entire function given that f is continuous everywhere and analytic everywhere except on the boundary of unit circle. I don't know how should I start?
– mint
Nov 23, 2014 at 3:22
• @AlexR. f(z) is not anaytic only on the boundary of the unit circle, but it is analytic inside the circle.
– mint
Nov 23, 2014 at 3:58
• @Alex: $f$ is supposed continuous on the whole plane. The result is true. Nov 23, 2014 at 4:22
• @Alex: No, he said that the function is holomorphic in the open unit disk AND in the complement of the closed unit disk. Nov 23, 2014 at 4:26

## 2 Answers

Almost certainly you can prove this with Morera's theorem.

In particular, I think this sketch works. Suppose you have some closed simple curve that cuts $\partial\mathbb D$. (Actually Morera's theorem is true if you just test on rectangles, which circumvents the problem of dealing with nasty curves.) Then adding in the bits of the circle the curve cuts out, you get two closed curves, one lying in $\overline{\mathbb D}$ and one lying in $\overline {\mathbb D}^c$, that sum to your original curve. It suffices to show that the integral over each of these two curves is zero.

For the curve $\gamma$ inside the circle, consider the corresponding curve $\gamma_t$ you get by scaling $\gamma$ by $t\in(0,1)$ around zero. Each $\gamma_t$ is zero, and the function $t\rightarrow \int_{\gamma_t}f \ dz$ should be continuous, so the integral of $f$ over $\gamma$ is $0$. A similar argument should work for the other curve.

• Very sorry, I couldn't get the last paragraph and I am following Conway's Functions of one complex variable where Morera's theorem is described using triangle. I believe both of them have the same meaning. Sorry again
– mint
Nov 23, 2014 at 5:07
• @mint Ok, so you test on triangles instead. You are right that it won't make a difference. The problem reduces to showing that the two curves you get in the first paragraph, when integrated over, give zero. Consider just the one that lies in the closed disk. Think about shrinking it a little bit so it lies entirely in the open disk. Then Cauchy's integral theorem tells you the integral over this shrunken curve is zero. Does this make sense so far? Nov 23, 2014 at 5:15

Since functions which are holomorphic everywhere except on a line are easier to deal with, let us avoid dealing with the boundary of the circle directly.

Consider the map $g=f\circ c$, where $c$ is any Mobius transformation on $\mathbb{C}\cup \lbrace \infty \rbrace$ sending $\delta B_1(0)$ to $\mathbb{R}$; e.g., $c(z)=i\frac{1+z}{1-z}$. Then $g$ is holomorphic on $\mathbb{C}-\mathbb{R}$ and continuous on $\mathbb{C}$. Apply the argument using triangles on $\mathbb{C}-\mathbb{R}$ given by the answer in Function holomorphic except for real line and continuous everywhere is entire to conclude that $g$ is entire. Since $c$ is an entire bijection on $\mathbb{C}$, so is $c^{-1}$ and thus so is the composition $g\circ c^{-1}=f$, as desired.