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Thread: Why holomorphic ?

  1. #1
    Aki
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    Why holomorphic ?

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

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

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

    Any help would be appreciated.
    Thanks in advance.
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  2. #2
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    Re: Why holomorphic ?

    This lemma is just the implicit function theorem. Holomorphicity of $\displaystyle \phi(z)$ is established by differentiating under the integral sign wrt to $\displaystyle z$.
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  3. #3
    Aki
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    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.
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    Re: Why holomorphic ?

    A function $\displaystyle f(z)$ will be analytic if $\displaystyle \partial f/\partial \bar z=0$. Taking $\displaystyle \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?
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    Aki
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    Re: Why holomorphic ?

    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.
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    Re: Why holomorphic ?

    What I mean is the analyticity of functions like $\displaystyle \phi(z)$ 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.
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  7. #7
    Aki
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    Re: Why holomorphic ?

    The only theorem I could find is

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

    But, the integral that I mentioned is not in this form.
    Could you suggest some references ?
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    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.
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  9. #9
    Aki
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    Re: Why holomorphic ?

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