Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 8 / 1 / vi

For the Hurwitz proof of the planar isoperimetric inequality, take a positively oriented regular simple closed curve of length $L$ and enclosed area $A$. Write its complex position as $z(t)=x(t)+iy(t)$, with $0\le t\le 2\pi$ proportional to arc length. Then $|z^{\prime }(t)|=L/(2\pi )$. Translate the curve to make its mean position zero, and write its Fourier coefficients as $c_n$, with $c_0=0$.

Green's theorem gives the signed area, and Parseval's identity computes it:Periodic integration by parts and Parseval's identity applied to $z^{\prime }$ giveSince $n\le n^2$ for every integer $n$,The sums converge absolutely: $\sum |n||c_n{|}^2\le (\sum |c_n{|}^2{)}^{1/2}(\sum n^2|c_n{|}^2{)}^{1/2}$. Equality forces $c_n=0$ unless $n=0$ or $1$; after the mean translation, $z(t)=c_1{e}^{it}$ is a circle. Conversely a circle attains equality. Thus circles uniquely attain equality, up to translation and orientation.

The same proof applies to a rectifiable simple closed curve using its Lipschitz arc length parametrization. Its derivative exists almost everywhere and belongs to $L^2$; periodic mollification converges to the curve in the function and derivative $L^2$ norms. This justifies the derivative coefficient identity, Parseval's identity and area integral by approximation. Reversing orientation, if necessary, makes the enclosed area positive. The printed name “Hurewitz” is read as Hurwitz.