From: djb@koobera.math.uic.edu (D. J. Bernstein)
Newsgroups: sci.math
Subject: Re: Isoperimetric problem
Date: 15 Apr 1998 23:29:19 GMT

Teck Cheong Lim <tlim@erols.com> wrote:
> Can anyone show an easy (not too advanced) proof of this theorem?

Peter Lax published a straightforward proof in the Monthly in 1995.

Consider a closed curve of length 2 pi. Parametrize it by arc length:
the curve is (x(s),y(s)) for 0 <= s <= 2 pi. We may assume that y(0) and
y(pi) are both 0. Note that the derivative of y^2 cot s is 2yy' cot s -
y^2 csc^2 s; note also that 2yx' <= y^2 + x'^2 = y^2 + 1 - y'^2.

Write I(f) for the integral from 0 to 2 pi of f ds. Now double the area
inside the curve is I(2yx') <= I(y^2 + 1 - y'^2) =
I(y^2 + 1 - y'^2 + 2yy' cot s - y^2 csc^2 s) =
I(1 - (y cot s - y')^2) <= I(1) = 2 pi.

---Dan
Smaller, faster, safer than inetd+tcpd. http://pobox.com/~djb/ucspi-tcp.html