proof of Goursat’s theorem

We argue by contradictionMathworldPlanetmathPlanetmath. Set


and suppose that η0. Divide R into four congruentMathworldPlanetmath rectangles R1,R2,R3,R4 (see Figure 1), and set


Figure 1: subdivision of the rectangle contour.

Now subdivide each of the four sub-rectangles, to get 16 congruent sub-sub-rectangles Ri1i2,i1,i2=14, and then continue ad infinitum to obtain a sequenceMathworldPlanetmath of nested families of rectangles Ri1ik, with ηi1ik the values of f(z) integrated along the corresponding contour.

Orienting the boundary of R and all the sub-rectangles in the usual counter-clockwise fashion we have


and more generally


In as much as the integrals along oppositely oriented line segments cancel, the contributions from the interior segments cancel, and that is why the right-hand side reduces to the integrals along the segments at the boundary of the composite rectangle.

Let j1{1,2,3,4} be such that |ηj1| is the maximum of |ηi|,i=1,,4. By the triangle inequalityMathworldMathworldPlanetmath we have


and hence


Continuing inductively, let jk+1 be such that |ηj1jkjk+1| is the maximum of |ηj1jki|,i=1,,4. We then have

|ηj1jkjk+1|4-(k+1)|η|. (1)

Now the sequence of nested rectangles Rj1jk converges to some point z0R; more formally


The derivativePlanetmathPlanetmath f(z0) is assumed to exist, and hence for every ϵ>0 there exists a k sufficiently large, so that for all zRj1jk we have


Now we make use of the following.

Lemma 1

Let QC be a rectangle, let a,bC, and let f(z) be a continuousMathworldPlanetmathPlanetmath, complex valued function defined and bounded in a domain containing Q. Then,


where M is an upper bound for |f(z)| and where P is the length of Q.

The first of these assertions follows by the Fundamental Theorem of CalculusMathworldPlanetmathPlanetmath; after all the function az+b has an anti-derivative. The second assertion follows from the fact that the absolute valueMathworldPlanetmathPlanetmathPlanetmath of an integral is smaller than the integral of the absolute value of the integrand — a standard result in integration theory.

Using the Lemma and the fact that the perimeter of a rectangle is greater than its diameter we infer that for every ϵ>0 there exists a k sufficiently large that


where |R| denotes the length of perimeter of the rectangle R. This contradicts the earlier estimate (1). Therefore η=0.

Title proof of Goursat’s theoremMathworldPlanetmath
Canonical name ProofOfGoursatsTheorem
Date of creation 2013-03-22 12:54:37
Last modified on 2013-03-22 12:54:37
Owner rmilson (146)
Last modified by rmilson (146)
Numerical id 13
Author rmilson (146)
Entry type Proof
Classification msc 30E20