Inequality between area and boundary length, $4\pi A \leq L^2 $

Question

Suppose we have a simply connected region $R$ in $\mathbb{R}^2$ with area $A$ and the boundary of $R$ is a curve sufficiently well behaved (say piecewise $C^1$) that we can say it has length $L$. Then

$$4\pi A \leq L^2 $$

I read the inequality in an article which offered no proof and suggested it was 'well known'. It's a lovely result and equality is clearly attained for a circle.

How does one show the inequality in general? Feel free to refine the set up if you need different hypotheses.

Thanks.

Answer 1

The isoparametric inequality is covered in depth in other sources, so I will just leave a quick bit of intuition.

If you have a closed region in the plane bounded by a curve $\gamma(s)$, two facts turn out to be true about the region:

1. The "gradient" of the enclosed area, i.e. the direction the boundary should flow to most steeply increase the enclosed area, is the boundary normal $\hat{n}(s)$ (this notion of gradient can be made precise using the calculus of variations.)
2. The "gradient" of the perimeter is the curvature times the normal $\kappa(s)\hat{n}(s)$.

Now suppose $\gamma$ isn't optimal. To improve $\gamma$, you want to flow it in the gradient direction of the area, while projecting out the part of that gradient that would change perimeter, i.e. in the language of the method of Lagrange multipliers you want to flow the surface in the direction $$\hat{n}(s) + \lambda \kappa(s)\hat{n}(s)$$ where the scalar $\lambda$ is whatever it needs to be so that the flow keeps the perimeter constant. When do you stop flowing? When the two gradients are scalar multiples of each other and the projection cancels everything out, i.e. when $\kappa$ is constant. This means that curves $\gamma$ with constant curvature are extrema of area while keeping perimeter constant; and of course such curves are precisely circles.

The above generalizes in a very beautiful way to a variety of settings; in 3D the role of curvature is played by the mean curvature $2H$ and you get constant mean curvature surfaces as solutions to isoperimetric problems (and indeed, water in space will arrange itself into droplets of constant mean curvature.) You can also relax the regularity requirements on $\gamma$, and allow it to be e.g. piecewise-$C^1$ and much of the above still carries through (in fact one of the ways to define the curvature of a piecewise-linear curve at the vertices is as the magnitude of the gradient of perimeter!)