Bonnesen's inequality for non-simple curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:19:57Z http://mathoverflow.net/feeds/question/100941 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100941/bonnesens-inequality-for-non-simple-curves Bonnesen's inequality for non-simple curves Dorian 2012-06-29T15:10:54Z 2012-06-30T13:43:59Z <p>Given a closed curve in the plane $\mathbb{R}^2$, it is well known that $L^2 \geq 4\pi A$ where $L$ is the length of the curve and $A$ is the area of the interior of the curve.</p> <p>For a <em>simple</em> closed curve $\gamma$, the stronger inequality due to Bonnesen holds: $L^2 - 4\pi A \geq \pi^2 (R_{out}- R_{in})^2$, where, setting $\Omega =$ Int $(\gamma)$, $R_{in}$ and $R_{out}$ denote the inner and outer radii of the sets:</p> <p>$R_{in} = \sup_{B_r \subset \Omega} r$</p> <p>$R_{out} = \inf_{\Omega \subset B_R} R$</p> <p><strong>Question:</strong> Does this inequality continue to hold <strong>without</strong> the assumption that the curve is <strong>simple</strong>? In particular, does it hold for any <em>connected, rectifiable set</em>?</p> <p>If one adds components to the interior of a simple curve, it is clear that this increases the isoperimetric defecit. However while $R_{out}$ will remain unchanged, $R_{in}$ will necessarily decrease, making it not immediately clear that the inequality would continue to hold. </p> <p><strong>Update:</strong> I just realized this is <em>false</em>. Take a countable union of points in the interior of the unit ball. Around the kth point, make a circle of radius $\epsilon/2^k$. Distribute the points well enough and $R_{in}$ can be made arbitrarily small. Then if you make $\epsilon$ small enough, you don't change the length or area too much and so the desired inequality will be violated. </p>