# Thread: Why holomorphic ?

1. ## Why holomorphic ?

I cannot understand the following statement in the proof of Lemma 8.7 in "Lectures on Riemann Surfaces" by Otto Forster.

$\phi(z)=\frac{1}{2 \pi i}\int_{|w-w_0|=\epsilon}w\frac{F_w(z,w)}{F(z,w)}dw, \;\;\; \left( F_w := \frac{\partial F}{\partial w}\right)$
Since the integral depends holomorphically on $z$, the function $z\mapsto \phi(z)$is holomorphic on $D(r)$.
where
$D(R)=\{z\in C : |z|0,$
$F(z,w)=w^n+c_1(z)w^{n-1}+\cdots+c_n(z),$
$c_1(z)\cdots c_n(z)$are holomorphic functions on $D(r)$.

In the equation
$\phi(z)-\phi(z_0)=\frac{1}{2 \pi i}\int_{|w-w_0|=\epsilon}w\left(\frac{F_w(z,w)}{F(z,w)}-\frac{F_w(z_0,w)}{F(z_0,w)}\right)dw$
the integrand can be written as $(z-z_0)P(z,w)$ where $P(z,w)$ is holomorphic with respect to $z$, to prove $\phi(z)$ to be holomorphic, it seems enough to show that $\int_{|w-w_0|=\epsilon}P(z,w) dw$ is continuous with respenct to $z$. But, I don't know how to show it.

Any help would be appreciated.

2. ## Re: Why holomorphic ?

This lemma is just the implicit function theorem. Holomorphicity of $\phi(z)$ is established by differentiating under the integral sign wrt to $z$.

3. ## Re: Why holomorphic ?

Thank you for your help.

I only know that for real functions the differentiation under the integral sign is possible only if the derivative of the integrand is continuous as a function of both variables.
Is this condition applicable to complex functions?
If so, how do we know the derivative is continuous as a function of both variables.

4. ## Re: Why holomorphic ?

A function $f(z)$ will be analytic if $\partial f/\partial \bar z=0$. Taking $\frac{\partial}{\partial \bar z}=\frac{1}{2}\frac{\partial}{\partial x} +\frac{i}{2} \frac{\partial}{\partial y}$ under the integral sign only requires the real result. In addition to continuity, you also need local uniform convergence of the integral. Do you have a complex analysis text addressing analyticity of integrals?

5. ## Re: Why holomorphic ?

I have "COMPLEX ANALYSIS" by Ahlfors, and "Elementary Theory of Analytic Functions of One or Several Complex Variables" by H. Cartan, although I don't understand what you mean by analyticity of integrals.

6. ## Re: Why holomorphic ?

What I mean is the analyticity of functions like $\phi(z)$ which are defined by integrals. There are theorems on this. Take a look in the texts you've got to see if they address it.

7. ## Re: Why holomorphic ?

The only theorem I could find is

Suppose that $\phi(\zeta)$is continuous on $\gamma$. Then,
$F_n(z)=\int_\gamma \frac{\phi(\zeta)}{(\zeta-z)^n}d\zeta$
is analytic on each domain determined by $\gamma$. Moreover,
its derivative is $F_n'(z)=nF_{n+1}(z)$.
from Ahlfors's book.

But, the integral that I mentioned is not in this form.
Could you suggest some references ?

8. ## Re: Why holomorphic ?

I think the result you need is Leibnitz's rule. It's covered in Nevanlinna and Paatero's book "Intro to complex analysis", 9.7.

9. ## Re: Why holomorphic ?

I read the section 9.7. of the book you suggested.
That is exactly what I need.
Thank you.