I am trying to find a proof of Green's formula for rectifiable Jordan curves $\Ga$ (and the corresponding interior regions $R$). There is a proof by Ridder, followed by very similar proofs by Verblunsky and Potts. The idea in those three papers, which looks quite natural, is to subdivide $R\cup \Ga$ into small regions $R_i\cup \Ga_i$ by vertical and horizontal lines. The problem is how to deal with the "partial" regions $R_i\cup \Ga_i$ that are not rectangles; more precisely, with the line integrals over the corresponding $\Ga_i$'s. To deal with such line integrals rigorously, one first of all needs to parametrize the boundaries $\Ga_i$. However, none of the mentioned three authors (or any other ones known to me) even mentions such a parameterization.

In Mathematical Analysis: a Modern Approach to Advanced Calculus, 1957, by Apostol, an apparent attempt is made to make Ridder's approach rigorous. The Green formula in question is stated there as Theorem 10--43. The crucial part in the proof of Apostol's Theorem 10--43 is Theorem 10--42, which tries to deal with the mentioned "partial" regions $R_i\cup \Ga_i$. However, there are a few places in the proof of Apostol's Theorem 10--42 that I don't understand.

Let $\Ga$ be a rectifiable Jordan curve bounding the corresponding interior region $R$. First, Apostol takes a lowest and highest points, $p_1$ and $p_2$ on $\Ga$. These two points are then separated by a horizontal line, say $\ell$. Then $p_1$ and $p_2$ are joined by "arcs" $C_1$ and $C_2$ to points $q_1$ and $q_2$ in $R$ close to $p_1$ and $p_2$, respectively, so that $q_1$ is below $\ell$ and $q_2$ is above $\ell$.

Question 1. Apostol says such "arcs" (possibly non-rectifiable) exist, without proof. I do not immediately see how to prove this, even though it's probably simple. (The $C_j$'s are never mentioned again in the proof, after they are introduced.)

Then the points $q_1$ and $q_2$ are connected by a polygonal line $C\subset R$, which must have an intersection point with $\ell$. Without loss of generality, there are finitely many such points. For each such point $p$, let $[u,v]$ be the maximal segment on $\ell$ such that the open interval $(u,v)$ is contained in $R$. Then Apostol says there must be at least one such maximal segment $[u,v]$ that has an odd number of intersections with the polygonal line $C$; I guess this follows because the total number of the intersections of $C$ with $\ell$ can be proved to be odd, by induction. But then Apostol just says: "Using such a segment, it is clear that $L[u,v]$[$=[u,v]$ -- I.P.] forms, with $\Ga$, two rectifiable Jordan curves $\Ga_1$ and $\Ga_2$, one of which contains $p_1$ (call this one $\Ga_1$) and the other contains $p_2$. Moreover, these curves form the boundaries of two regions $R_1$ and $R_2$, whose union is $R$. The positively oriented boundaries $\Ga(R_1)$ and $\Ga(R_2)$ of $R_1$ and $R_2$ such that $\Ga(R)=\Ga(R_1)+\Ga(R_2)$."

There is a picture there, making the above claims plausible. Yet, here is my main question:

Question 2. Nothing in that passage is clear to me. How does this "forming" of curves $\Ga_1$ and $\Ga_2$ occur? Are they Jordan curves? If so, why? How are $\Ga_1$ and $\Ga_2$ parameterized? Why is the union of $R_1$ and $R_2$ equal to $R$? What precisely are the roles of the points $q_j$ and the "arcs" $C_j$ in this proof?

Can one help me decipher this proof? Or, perhaps even better, is there a completely rigorous, fully detailed proof of this obviously important and natural result? (Of course, there is a general Stokes theorem for manifolds, but it does not seem at all obvious how to build a bridge from there to Jordan curves.)


One can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\Gamma$ and any $p>2$. Given a function $f\in L^p(K)$, we can define its Cauchy transform $$ \left(\mathcal{C}f \right)(z)=\frac{1}{\pi}\int_K\frac{f(w)}{z-w}. $$ By Hölder inequality, this is a continuous operator from $L^p(K)$ to $C(K)$. This operator inverts the $\bar{\partial}$ operator, namely, $\mathcal{C}\bar{\partial}\varphi\equiv\varphi$ for any $\varphi \in C^1_0(K)$; see the computation after the equation (8) in Chapter 5 of "Lectures on quasiconformal mappings" by Ahlfors (1966).

Now, the composition of $\mathcal C$ and integration over $\Gamma$ is a continuous linear functional on $L^p(K)$: $$ L(f):= \frac{1}{2i}\int_\Gamma \left(\mathcal{C}f\right)(z)dz=\int_K f(z)\psi(z), $$ for some $\psi \in L^{p'}(K)$. It follows that for any $\varphi \in C^1_0(K)$, we have $$ \frac{1}{2i}\int_\Gamma \varphi(z) dz = \int_K\psi(z)\bar{\partial}\varphi(z). $$Hence, it remains to check that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside; since $\Gamma$ itself is rectifiable, its area is zero. To this end, it suffices to check that $L(f)=\int_{\text{inside }\Gamma} f$ for any $f\in C_0(K\setminus\Gamma)$. For such an $f$, we can exchange the integrals: $$ L(f)=\frac{1}{2\pi i}\int_\Gamma dz\int_K \frac{f(w)}{z-w}=\int_K f(w)\frac{1}{2\pi i} \int_\Gamma \frac{dz}{z-w}. $$ The inner integral can be interpreted as the increment of $\log (z-w)$ along $\Gamma$, and thus it is equal to the winding number of $\Gamma$ w. r. t. $w$. Therefore, the result follows from the topological statement that any Jordan curve has winding number $\pm 1$ w. r. t. a point inside it, and $0$ w. r. t. a point outside. Some references for this statement are discussed here.

| cite | improve this answer | |
  • $\begingroup$ Thank you for this answer. I have some further questions, though. (i) Is this proof yours or does it appear elsewhere? (ii) The Cauchy transform appears to be usually defined as a line integral. Can your version of it be found elsewhere? (iii) Can you fill in details, at least concerning (a) $\mathcal{C}\bar{\partial}\varphi\equiv \varphi$ for $\varphi\in C_0^\infty$ and (b) "this boils down to checking that ..."? $\endgroup$ – Iosif Pinelis Aug 7 '18 at 14:22
  • $\begingroup$ Previous comment, continued: It would also be helpful if you added details on the limit transition (?) from $\varphi\in C_0^\infty$ to a (what?) specific class of more general functions, concerning "it remains to show that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside". $\endgroup$ – Iosif Pinelis Aug 7 '18 at 14:22
  • $\begingroup$ @Iosif, (i) I have not seen this proof elsewhere; (ii) I am no longer sure about the terminology, but I believe what I wrote is also called the Cauchy transform. Naturally, being the inverse of $\bar{\partial}$, it is a very standard tool. A version of it is the operator P from Chapter 5 of Ahlfors, Lectures on quasiconformal mappings; (iii) I will add the details. $\endgroup$ – Kostya_I Aug 7 '18 at 15:17
  • $\begingroup$ I will be looking forward to the details. $\endgroup$ – Iosif Pinelis Aug 7 '18 at 17:45
  • $\begingroup$ Thank you very much! This is a very nice way to reduce Green's theorem for general rectifiable Jordan curves to that for circles (say). It can apparently be generalized to higher dimensions, too. $\endgroup$ – Iosif Pinelis Aug 7 '18 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.