# Stronger version of the isoperimetric inequality

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.

• I took me more than one second to be sure that you focus on the planar case. Just to mention that in higher dimension, there is no such refined inequality in terms of the outer and inner radii, because you can add to your body an arbitrarily long spike, with arbitrarily small area. Instead, stability estimates involve the volume of the minimum ${\mathcal A}(\Omega)$ of the symmetric difference between the body and the balls of volume $|\Omega|$. For instance, one has $P(\Omega)-P(B)\ge C{\mathcal A}(\Omega)^2$ for some positive constant $C$. – Denis Serre Feb 14 '11 at 14:47
• Doesn't Bonnesen's inequality give exactly what you ask for in your update? – Mark Meckes Apr 27 '11 at 13:20
• No since that measures the difference between inner and outer radii of the set. This doesn't say anything about hausdorff distance to a disk. – Dorian Apr 27 '11 at 14:20
• Let $B_{in}$ and $B_{out}$ be the inner and outer disks. Since $B_{in} \subseteq \Omega$, $d_H(\Omega,B_{in}) = \sup_{x\in \Omega} d(x,B_{in})$. Since $\Omega \subseteq B_{out}$, for every $x \in \Omega$, $d(x,B_{in}) \le r_{out} - r_{in}$, and equality is achieved if $x \in \partial B_{out}$. Therefore $d_H(\Omega,B_{in}) = r_{out} - r_{in}$. – Mark Meckes Apr 27 '11 at 14:35
• Sorry, I was assuming that $B_{in}$ and $B_{out}$ are concentric (which is true if $\Omega = - \Omega$ at least). I'm not sure about the general case offhand. – Mark Meckes Apr 27 '11 at 14:37

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 $C = \pi^2/4$.

• I proved the same inequality but using a much more complicated proof. Thanks! – Dorian Apr 28 '11 at 13:29

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.

• I was just thinking about this inequality. As written I think it is incorrect. Take a circle which is not centered at the origin. Then r_1-r_2 is not zero but the left side is zero.. – Dorian Jun 2 '12 at 16:05

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.