Q: Cauchy integral formula proof relating to Laurent series existence proof.

195 Views Asked by At

THIS POST HAS BEEN EDITED

I am studying a proof for Cauchy's integral formula using Walter Rudin's ''Real and Complex analysis''.

Rudin uses a cicle $\Gamma$, but I am going to use a contour $\gamma \subset \Omega$, where $\Omega$ is an open set where $f$ is holomorphic, such that $Ind_{\gamma}(a)=0$ for every $a$ not in $\Omega$, and $Ind_{\gamma}(a)=1 \, \forall a \in \Omega $.

So.. I am going to skip some steps and define $\Omega_1=\{z \in \mathbb{C} : Ind_{\gamma}(z)=0\}$ and define $h_1: (\Omega_1 - \gamma *) \to \mathbb{C}$ such that:

$h_1(z)= \int_{\gamma}\frac{f(w)}{w-z}dw$, now they assert that $h_1 \in H(\Omega_1 - \gamma *)$.

My question is: How do you prove this? And, What hypothesis do you use? I get that $h_1(z)-h_1(z_0)= \int_{\gamma}\frac{f(w)(z-z_0)}{(w-z)(w-z_0)}dw$ and the idea is to take the limit when $z$ approaches $z_0$.

So for now I buy that you can insert the limit into the integral. But that would mean (since you can define $h_1$ in ($\mathbb{C}-\gamma*$), that $h_1$ is analytic in the whole plane (except the countour). But if this statement is true, considering the proof of existence of Laurent series for holomorphic functions in a ring, where they define a similar $h_1$, there is something wrong.

To prove existence of Laurent series, the proof I have seen is that taking the cycle:

$\Gamma=\gamma_1 + (-\gamma_2)$ where $\gamma_1$ and $\gamma_2$ are the lower and upper contour of the ring respectively.

So by Cauchy, you have that (since $\Gamma$ satisfies the hypothesis):

$f(z)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(w)}{w-z}dw=\frac{1}{2\pi i}\int_{\gamma_2}\frac{f(w)}{w-z}dw - \frac{1}{2\pi i}\int_{\gamma_1}\frac{f(w)}{w-z}dw$.

Now if the statement of $h_1$ being holomorphic is true, then the second and the first term of the sum are analytic, then f also is, which is false.

If anyone can help me out with this it would be greatly appreciated.

P.S. Sorry if I have some syntax errors, or use different notations for some elements, English is not my first language.

From here I edited this:

So I am going to answer my own question, maybe it will help somebody in the future.

The problem is that $h_1$ has a natural boundary of non isolated singularities at $\gamma*$, the proof still holds, but $h_1$ will have 2 power series, one when the domain is the disc contained inside $\gamma*$ the other outside it.

So for the Laurent series, since $f$ is defined in the ring (lets say of center $a$), i.e. in the intersection of the "interior" (pardon the abuse of notation here) of $\gamma_2$ and the complement of the "interior" of $\gamma_1$ the fact that the second summand of

$f(z)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(w)}{w-z}dw=\frac{1}{2\pi i}\int_{\gamma_2}\frac{f(w)}{w-z}dw - \frac{1}{2\pi i}\int_{\gamma_1}\frac{f(w)}{w-z}dw$ (**)

can be expressed as a power series of center $a$ "inside" $\gamma_1$ is of no concern, since $f$ is defined "outside" it. Hence you cannot express $f$ as a power series because of the fact that this equality (**) holds, only inside the ring.