Planar isoperimetric inequality

This is the planar case of the isoperimetric inequality. Every bounded planar domain with a sufficiently regular boundary of length $L$ and area $A$ satisfiesEquality holds exactly for circular domains.

Fix two attachments a distance $2a$ apart on a straight line and a simple fence of length $2l$ in one half-plane, with $a<l<\pi a/2$. There is a unique $\theta \in (0,\pi /2)$ satisfying $l/a=\theta /\sin \theta$: its derivative has numerator $\sin \theta -\theta \cos \theta >0$, since that numerator starts at zero and has derivative $\theta \sin \theta >0$. A radius-$R$ minor circular arc with half-angle $\theta$ then has the required chord and length. Its graph is $y=\sqrt{R^2-x^2}-\sqrt{R^2-a^2}$, for $-a\le x\le a$, and sector-minus-triangle geometry gives the displayed area.

To prove global maximality, attach the complementary major arc of this circle, lying in the opposite half-plane, to any competing fence. The resulting closed curve has perimeter $2l+(2\pi R-2l)=2\pi R$. Its enclosed area is the competitor's area plus a fixed major-segment area. The planar isoperimetric inequality bounds the total by $\pi R^2$, with equality only for a circle. Subtracting that fixed added area proves the minor arc is the unique maximizing shape on the chosen side. The same arc follows from the Beltrami identity for the area-minus-length functional.

Parametrize a positively oriented closed plane curve at constant arc length speed on $[0,2\pi ]$, and let $c_n$ be its complex-position Fourier coefficients. Green's theorem and Parseval's identity give $A=\pi {\sum }_{n}n|c_n{|}^2$ and $L^2/(4{\pi }^2)={\sum }_n{n}^2|c_n{|}^2$. Since $n\le n^2$ for integers, the planar isoperimetric inequality follows. Equality leaves only the constant and positive first harmonic, a translated circle.

Equality in the Cauchy-Schwarz and periodic Wirtinger bounds makes the arclength boundary parametrization a first harmonic with orthogonal equal-length coefficient vectors. The boundary is therefore a circle, and the enclosed domain is a disc.