# Cauchy's theorem: what about non-smooth homotopies?

This morning I realized I have never understood a technical issue about Cauchy's theorem (homotopy form) of complex analysis. To illustrate, let me first give a definition.

(In what follows $\Omega$ will always denote an open subset of the complex plane.)

Definition Let $\gamma, \eta\colon [0, 1]_t \to \Omega$ be two piecewise smooth closed curves (or circuits). We say that $\gamma, \eta$ are $\Omega$-homotopic if there exists a continuous mapping $H \colon [0, 1]_\lambda \times [0, 1]_t \to \Omega$ s.t.

1. $H(0, t)=\gamma(t)$ and $H(1, t)=\eta(t), \quad \forall t \in [0, 1]$;
2. $H(\lambda, 0)=H(\lambda, 1), \quad \forall \lambda \in [0, 1]$.

Theorem (Cauchy) Let $f\colon \Omega \to \mathbb{C}$ be holomorphic. If $\gamma, \eta$ are $\Omega$-homotopic circuits, then

$$\int_{\gamma} f(z)\, dz= \int_{\eta}f(z)\, dz.$$

Problem The function $H$ above is only continuous and need not be smooth. So for $0< \lambda < 1$, the closed curve $H(\lambda, \cdot)$ may be pretty much everything (a Peano curve, for example). Does this void the validity of theorem as it is stated above? How can integration be defined over such a pathological object?

The proof of Cauchy's theorem that I have in mind goes as follows. To begin, one observes that for a sufficiently small value of $\lambda_1$, the circuits $\gamma=H(0, \cdot)$ and $H(\lambda_1, \cdot)$ are close toghether; that is, they can be covered by a finite sequence of disks not leaving $\Omega$ like in the following figure:

Since $f$ is locally exact, its integrals over every single disk depend only on the local primitive. Playing a bit with this, one arrives at

$$\int_\gamma f(z)\, dz= \int_{H(\lambda_1, \cdot)} f(z)\, dz.$$

Then one repeats this process, yielding a $\lambda_2$ greater than $\lambda_1$ and such that

$$\int_{H(\lambda_1,\cdot)} f(z)\, dz= \int_{H(\lambda_2, \cdot)} f(z)\, dz.$$

And so on. A compactness argument finally shows that this algorithm ends in a finite number of steps.

Problem is: this proof assumes implicitly that $H(\lambda_1, \cdot), H(\lambda_2, \cdot) \ldots$ are piecewise smooth, to make sense of integrals $$\int_{H(\lambda_j, \cdot)}f(z)\, dz.$$ This, however, does not follow from the definition if $H$ is only assumed to be continuous. Therefore this proof works only for smooth $H$.

Is this regularity condition necessary?

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I never thought about this deeply, but I'd have said you need a certain smoothness condition on $H$, and e.g. a Lipschitz condition on $H$ should be enough to ensure that all the paths you're considering are rectifiable Jordan curves, for which the Cauchy integral formula holds. This should be enough to push your argument through. Also, I think that this should be treated in Dieudonné's multi-volume book or Conway's book on complex analysis. – t.b. Jun 9 '11 at 12:43
Can't we use "smooth approximation theorems"? I can look this up in a couple of hours, but I'm pretty sure that two continuously homotopic piecewise smooth curves should be smoothly homotopic. – Akhil Mathew Jun 9 '11 at 12:46
I think I agree with @Akhil. His approach is certainly the better one if you're happy with piecewise smooth curves that are only continuously homotopic. The approach I'm suggesting should yield an extension for rectifiable Jordan curves though, for which I'm not so sure that there are good enough approximation theorems. – t.b. Jun 9 '11 at 12:52
@Theo: I found this article lying in my hard drive. Here the author employs a strategy similar to yours: first he uses a technique (different from the one I sketched above) to prove the theorem in case of Lipschitz-continuous homotopies, then he shows that a Lipschitz continuous homotopy can always be built from a continuous one. Still I haven't got it very well, though. – Giuseppe Negro Jun 9 '11 at 13:14

As has already been pointed out by Akhil in the comments, any two smooth curves $\gamma_0, \; \gamma_1$ which are continuously homotopic are also smoothly homotopic. The point here is to approximate a continuous homotopy, rather than the curves themselves.

More concretely, given homotopic smooth curves $\gamma_0, \gamma_1: I \to \Omega$, let $H(s,t): I\times [0,1] \to \Omega$ be a continuous homotopy between $\gamma_0$ and $\gamma_1$.

By the approximation theorem of your choice (mine is due to Whitney), we can find a smooth map $G: I\times[0,1] \to \Omega$, which coincides with $H$ on $I \times \{0\} \cup I \times \{1\}$ (since $H$ is smooth there). This means that we find a smooth homotopy $G$ between $\gamma_0$ and $\gamma_1$, so we are done in this special case.

Now, if we start out with rectifiable curves rather than smooth ones, we can find two smooth curves $\tilde \gamma_0$ and $\tilde \gamma_1$, which approximate our initial $\gamma_0$ and $\gamma_1$, respectively, and such that the linear homotopy

$$H_\lambda(s,t) := t\gamma_\lambda(s) + (1-t)\tilde \gamma_\lambda(s) \qquad \lambda = 0, 1$$

maps into $\Omega$ (choose a $\epsilon$-neighborhood of the image $\gamma_\lambda(I)$ of $I$ which is contained in $\Omega$ and take $\tilde \gamma_\lambda$ to be contained within this neighborhood).

It is not difficult to see that $H_\lambda$ will then be a "rectifiable" homotopy.

But with $\tilde \gamma_0$, $\tilde \gamma_1$ we are again in the first situation, so there is a smooth homotopy between them. Now we can build a rectifiable homotopy between $\gamma_0$ and $\gamma_1$ in three steps

1. Homotop $\gamma_0$ to $\tilde \gamma_0$ by the linear homotopy.
2. Use a smooth homotopy between $\tilde \gamma_0$ and $\tilde \gamma_1$
3. Go from $\tilde \gamma_1$ to $\gamma_1$ by the straight line homotopy.

Thus proving that all notions of "homotopic" agree.

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Ps: You are completely right that a merely continuous homotopy will in general do crazy things, and we lose our control over the value of the integrals during the homotopy. (Indeed, the main concern is that the integral might not even be defined for intermediate curves...) – Sam Jun 9 '11 at 14:53
This answer completely solves the problem and is very interesting. What do you mean by rectifiable homotopy, exactly? I guess it is a homotopy $H$ s.t. $H(\lambda, \cdot)$ is a rectifiable path for all $\lambda$, right? – Giuseppe Negro Jun 10 '11 at 10:44
Yeah, that's exactly what I meant (so that the issues you were pointing out in your post don't come up). – Sam Jun 10 '11 at 18:28
All right. Thank you. It is interesting to note that there are various ways to "circumvent this difficulty", so as to quote @Malik. – Giuseppe Negro Jun 11 '11 at 9:44

If I understand your question correctly, the problem that $H$ may be non-smooth can be solved by approximating with polygonal smooth paths, see for example Rudin's real and complex analysis (3rd edition), thm 10.40 and the remark after it.

As an interesting note, Rudin adds that another way to circumvent this difficulty is to extend the definition of index to closed curves, which is sketched in one of the exercises of the book.

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Thank you. That chapter of Rudin's book is one I'll want to read for sure. – Giuseppe Negro Jun 10 '11 at 10:42
You're welcome. I'm glad that helped! – Kalim Jun 12 '11 at 3:38