Stronger version of the isoperimetric inequality - MathOverflow

Stronger version of the isoperimetric inequality

2011-02-14T12:44:06Z

I have been searching for a version of the isoperimetric inequality which is something like:

$P(\Omega) - 2\sqrt{\pi} Vol(\Omega)^{1/2} \geq \pi (r_{out}^2 - r_{in}^2)$ where $r_{out}$ and $r_{in}$ are the inner and outer radius of a given set. There are of course details which I am missing such as what kind of sets this applies to (clearly connected and possibly simply connected). I was hoping somebody may recognize this inequality and be able to direct me to a source for it along with a proof.

Update: I'm curious if anyone can direct me to a some papers which relate the isoperimetric deficit to Hausdorff distance. Such as: $P(\Omega)^2 - 4\pi Vol(\Omega) \geq C d_H(\Omega,B)^2$ whre $B$ is a sphere in $\mathbb{R}^2$ which may be the inner or outer circle.

Update April 12: I would like to know if the first Bonnesen inequality written below is strictly stronger than the one in higher dimensions? In particular, if one considers the Fraenkel assymetry $\alpha(\Omega) = \min_B |\Omega \Delta B|$ where $|B|=|\Omega|$, does it hold on a bounded domain that

$ r_{out}^2 - r_{in}^2 \leq C \alpha(\Omega)$,

for some constant $C>0$? This seems like it should be true but I can't seem to find a concise proof of it.

Answer by Andrey Rekalo for Stronger version of the isoperimetric inequality

2011-02-14T13:44:08Z

There is a sharpened version of the plane isoperimetric inequality due to Benson which involves the inner and outer radii. Let $$\Gamma=\{(r,\theta):\ r=r(s),\theta=\theta(s)\}$$ be a simple closed rectifiable curve on the plane, parametrized by the arc length $s$, and let $$r_1=\sup\{r:\ (r,\theta)\in\Gamma\},\qquad r_2= \inf\{r:\ (r,\theta)\in\Gamma\}.$$ Assume that $\Gamma$ winds once arround the inner circle. Then

$$L^2-4\pi A\geq\frac{(2FA-2\pi E-\pi/(2F))^2}{1+4EF},$$

where $L$ is the perimeter of $\Gamma$, $A$ is the area of the enclosed region, and $$F=\frac{1}{r_1-r_2},\qquad E=\frac{r_1r_2(r_1+r_2)}{(r_1-r_2)^2}.$$

The reference is: D.Benson, "Sharpened Forms of the Plane Isoperimetric Inequality", The American Mathematical Monthly, Vol. 77 (1970), pp. 29-34.

Answer by Mark Meckes for Stronger version of the isoperimetric inequality

2011-02-14T13:54:32Z

A classical result along these lines is Bonnesen's inequality, which states $$ L^2 - 4\pi A \ge \pi^2 (r_{out} - r_{in})^2, $$ where $L$ is the length and $A$ is the enclosed area of a simple planar closed curve. There are many other results along these lines, called "stability estimates" for the isoperimetric inequality. There are several pointers to the literature in Note 6 following section 6.2 of Schneider's book Convex Bodies: the Brunn-Minkowski Theory.

Added: Bonnesen's inequality also suffices for the updated question. If $B_{in}$ and $B_{out}$ are the inner and outer disks, respectively, then since $B_{in} \subseteq \Omega \subseteq B_{out}$, $$ d_H(\Omega, B) \le d_H(B_{in},B_{out}) \le 2r_{out} - 2r_{in} $$ (the extreme case in the latter inequality being when the circles are tangent), so you get your desired result with $C = \pi^2/4$.

Answer by Gil Kalai for Stronger version of the isoperimetric inequality

2011-02-14T20:41:29Z

A very good source of Bonnesen type inequalties is the paper by Rovert Osserman entitled Bonnesen style isoprimetric inequalities, Americam Math Monthly 86(1979) 1-29. Here is another link for the same paper through this page. Osserman's 1978 Bulletin AMS paper on the ioperimetric inequality is also a good related source.