5
$\begingroup$

In my undergrad complex analysis, our professor used a quick way to derive the Cauchy integral formula (without Goursat's theorem and all that) with the following setup: Let $U$ be an open set of $\mathbb{C}$. Consider a complex function $f:U\to\mathbb{C}$ as a zero form on $U$, i.e., $f\in \Lambda^0(U).$ Then analyticity of $f$ is equivalent to Cauchy-Riemann equation $\partial_{\bar{z}}f=0$, which is equivalent to the form $f(z)dz\in\Lambda^1(U)$ being closed: $$d(fdz)=df\wedge dz=\partial_{\bar{z}}fd\overline{z}\wedge dz+\partial_{z}fdz\wedge dz=0.$$ Then one can prove the Cauchy integral formula as follows:

Suppose $\Omega\subset U$ is a relatively compact set with piecewise smooth boundary and $U$ is an open subset of complex plane, and $f:U\to\mathbb{C}$ is holomorphic. Then for any $a\in \Omega$, we have $$f(a)=\frac{1}{2\pi I}\int_{\partial\Omega}\frac{f(z)}{z-a}dz.$$

Proof: Let $\Omega_\epsilon=\Omega-D_\epsilon(a).$ Then we consider the one form $f(z)/(z-a)\in\Lambda^1(\Omega_\epsilon).$ By Stoke's theorem ($\int_{\Sigma}d\omega=\int_{\partial \Sigma}\omega$) we have $$ \int_{\partial\Omega_\epsilon}\frac{f(z)}{z-a}dz=\int_{\Omega_\epsilon}d\left(\frac{f(z)}{z-a}dz\right)=0 $$ since $f/(z-a)$ is holomorphic on $\Omega_\epsilon$. Now $$ \int_{\partial\Omega_\epsilon}\frac{f(z)}{z-a}dz=\int_{\partial\Omega}\frac{f(z)}{z-a}dz-\int_{\partial D_\epsilon(a)}\frac{f(z)}{z-a}dz=0. $$ Taking $\epsilon\to 0$ the second term goes to $2\pi if(a).$ This concludes the proof.

However, my graduate complex analysis professor told us that one cannot apply Stoke's theorem directly until one verifies that $f'$ is integrable. I never learned Stoke's theorem properly so I wonder:

  • What do we need to know prior to the proof above about $f,f'$ in order for the proof to be valid? (differentiability of $f'$ or something like that, or even smoothness of $f$?) What should we have proved before this point to make the above proof work?

  • Is there a way of using differential forms and Stokes theorem to prove the Cauchy integral formula, without going into the traditional Goursat's theorem for triangles and polygons and all that? Are there any recommended references on this topic?

Thanks in advance

$\endgroup$

1 Answer 1

6
$\begingroup$

The usual proof of Stokes's theorem requires the differential form to be at least of class $C^1$ (i.e., differentiable everywhere with continuous first partial derivatives). There are versions that work with weaker hypotheses, but just assuming the existence of derivatives is definitely not enough.

Your proof of Cauchy's formula works fine as long as you assume that $f$ is of class $C^1$ and satisfies the Cauchy-Riemann equations. But the Cauchy-Goursat argument is needed if you only assume that $f'(z)$ exists for all $z\in\Omega$, because then Stokes’s theorem doesn't apply.

$\endgroup$
3
  • $\begingroup$ Thank you for this answer. It has been more 30 years that I sat for the last time in a complex analysis class. May I ask one question ? When $f$ is holomorphic it is $C^\infty$, and on $\Omega_\epsilon$ which exludes the singularity, $f/(z-a)$ is also $C^\infty$. Do we then have to assume anything more about $f$ ? $\endgroup$
    – Kurt G.
    Commented Oct 13, 2022 at 6:30
  • 2
    $\begingroup$ It is true that holomorphic functions are $C^\infty$. However, this is usually proven using the Cauchy Integral Formula, so you run into circular reasoning here. $\endgroup$
    – tolUene
    Commented Oct 13, 2022 at 11:31
  • 1
    $\begingroup$ Thank you for your answer, Professor Lee! And I am a great fan of your books! $\endgroup$ Commented Oct 13, 2022 at 23:27

You must log in to answer this question.

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