By the Jordan Curve Theorem any closed Jordan curve $\gamma$ divides the plane into two regions one of which is bounded. This region is usually called the interior of $\gamma$.
We write $\operatorname{int} ( \gamma )$ for the interior and $\mathcal{L}(\gamma)$ for the length of $\gamma$. Also we use the notation $\lambda$ for the 2-dimensional Lebesgue measure.
Then what is
$$
\sup \left\lbrace \lambda (\operatorname{int} \left( \gamma) \right) : \gamma \text{ Jordan curve with } \mathcal{L}(\gamma) = 1 \right\rbrace?
$$
In other words: What is the largest possible area that a Jordan curve of length 1 can circumscribe?
This question is no homework nor taken from some math contest. I'm just curious.
Here are my thoughts:
- I suspect that the supremum is actually finite and attained. Namely when $\gamma$ describes the unit circle (just my gut feeling).
- I thought of applying Green's Theorem to express the area as a one-dimensional integral and then maximize this expression.
However, this most likely requires methods from variational calculus and I have no clue about this topic.
Can anyone provide a reference, an idea of how to prove or disprove my guess or maybe some result under the additional condition that $\gamma$ be differentiable?
Thanks in advance.
Best Answer
This is known as the isoperimetric inequality, $4\pi A\leq L^2$, with equality attained only by circles.
Many different proofs have been developed in the 2-dimensional case (Steiner symmetrization, Fourier series, Green's theorem) but the proof I like the most is based on the Brunn-Minkowski inequality, since it proves the statement in arbitrary dimension (with the appropriate constants and exponents, of course).