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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.)

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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.

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  • $\begingroup$ 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

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