2
$\begingroup$

I wonder if anybody has tried the following kind of direct proof for the existence of an antiderivative of an analytic function on a star-shaped domain.

Theorem: Let $f:D \to \mathbb{C}$ be an analytic function on a star-shaped domain $D$. Then $f$ has an antiderivative $F$ on $D$.

"proof": For simplicity, assume that every point in $D$ is connected to $0 \in \mathbb{C}$ by a line segment. Define \begin{equation} F(z) = \int_0^1 zf(zt)dt. \end{equation} Then \begin{equation} \lim_{h \to 0} \left( \frac{ F(z+h)-F(z) }{h} \right) = \lim_{h \to 0}\int_0^1 \left(\frac{(z+h)f(zt+ht)-zf(t)}{h} \right) dt. \end{equation} It can be checked that as $h \to 0$, the integrand converges to \begin{equation} \frac{d}{dt} tf(zt). \end{equation}

Therefore, $F'(z) = f(z)$, provided that the limit and integral are interchangeable. qed.

Of course, a limit and integral cannot always be interchanged. But I wonder if anybody seriously considered the above line of proof.

Thanks. As far as I can search from several textbooks in complex variables, the above theorem is proved by using Cauchy-Goursat's theorem. More concretely, they use the equality \begin{equation} \int_{z_0}^{z+h} f(z)dz \ - \ \int_{z_0}^z f(z)dz \ = \ \int_z^{z+h} f(z)dz, \end{equation} which can be justified by Cauchy-Goursat's theorem. The point of my question is: Is it possible to directly invoke to the computation as above, without using Cauchy-Goursat's theorem? If this is possible, we get another proof of Cauchy-Goursat's theorem, at least for star-convex domains.

$\endgroup$
1
  • $\begingroup$ Welcome to Math.SE! Can you please clarify what exactly is being asked here? There doesn't appear to be a precise question. $\endgroup$ Dec 2, 2015 at 5:26

1 Answer 1

2
$\begingroup$

I'm not sure I understand exactly what you want to ask, but perhaps this is helpful:

A continuous function $f : \Omega \to \mathbb{C}$ has an anti-derivative on $\Omega$ if and only if $$ \int_\gamma f(z)\,dz = 0 $$ for every simple closed curve $\gamma$ in $\Omega$.

In other words, it seems difficult to completely circumvent some variant of Cauchy-Goursat.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .