proof of Goursat’s theorem
We argue by contradiction. Set
Figure 1: subdivision of the rectangle contour.
Now subdivide each of the four sub-rectangles, to get 16 congruent sub-sub-rectangles , and then continue ad infinitum to obtain a sequence of nested families of rectangles , with the values of integrated along the corresponding contour.
Orienting the boundary of 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.
Continuing inductively, let be such that is the maximum of . We then have
Now the sequence of nested rectangles converges to some point ; more formally
The derivative is assumed to exist, and hence for every there exists a sufficiently large, so that for all we have
Now we make use of the following.
The first of these assertions follows by the Fundamental Theorem of Calculus; after all the function has an anti-derivative. The second assertion follows from the fact that the absolute value 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 there exists a sufficiently large that
where denotes the length of perimeter of the rectangle . This contradicts the earlier estimate (1). Therefore .
|Title||proof of Goursat’s theorem|
|Date of creation||2013-03-22 12:54:37|
|Last modified on||2013-03-22 12:54:37|
|Last modified by||rmilson (146)|