This lemma is just the implicit function theorem. Holomorphicity of is established by differentiating under the integral sign wrt to .
I cannot understand the following statement in the proof of Lemma 8.7 in "Lectures on Riemann Surfaces" by Otto Forster.
Since the integral depends holomorphically on , the function is holomorphic on .
are holomorphic functions on .
In the equation
the integrand can be written as where is holomorphic with respect to , to prove to be holomorphic, it seems enough to show that is continuous with respenct to . But, I don't know how to show it.
Any help would be appreciated.
Thanks in advance.
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.
A function will be analytic if . Taking 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?
Thank you for your quick reply.
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.
The only theorem I could find is
from Ahlfors's book.Suppose that is continuous on . Then,
is analytic on each domain determined by . Moreover,
its derivative is .
But, the integral that I mentioned is not in this form.
Could you suggest some references ?