Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm sure that many readers are already familiar with the well known Bonnesen inequality in the plane for a smooth, connected curve:

$(R_{out} - R_{in})^2 \leq \pi^2 (L^2 - 4\pi A),$

where $R_{out}$ and $R_{in}$ are the outer and inner radii respectively, $L$ is the length of the curve and $A$ is it's area.

This form turns out to often be quite inconvenient for me, and I was wondering if there was a version which read as follows:

$R_{out}^2 - R_{in}^2 \leq C (L^2 - 4\pi A)$,

where now $C$ is a constant which may depend on the support of the curve or other quantities possibly. The main difference is that when $R_{in} - R_{out}$ is small, the second inequality I've written is strictly stronger if one assumes the curve is constrainted to a bounded domain. The sacrifice I make though is that I do not care what the constant is, or if it depends on the support of the curve.

This inequality seems quite reasonable but I have thus far been unable to prove it. Has anyone encountered such a version of this inequality?

share|improve this question
    
What are radii? –  Alexander Chervov Apr 12 '12 at 17:50
    
There is a mistake in your first inequlity. Bonnesen's inequality reads π^2(R_out−R_in)^2≤(L^2−4πA) –  user54265 Jun 22 at 10:43

1 Answer 1

up vote 4 down vote accepted

I seriously doubt there is any such estimate (at least one that doesn't have the constant $C$ depend in a complicated and extremely unnatural way on the geometry of the curve).

My heuristic reasoning is as follows. Let $\sigma$ be the unit circle and $f$ be a non-zero smooth function on $\sigma$ so that $\int_{\sigma} f=0$ and $\max_\sigma f =-\min_\sigma f:=m(f)>0$.

Set $$\sigma_s=\lbrace (1+sf) \mathbf{x}(p): p\in \sigma\rbrace\subset\mathbb{R}^2$$ where here $\mathbf{x}(p)=(x_1(p),x_2(p))$ are the restrictions of the coordinate functions. For $s$ small we have we have that $\sigma_s$ is a smooth Jordan curve (and is smooth and with geometry as close to that of the unit circle as desired).

If $L_s$ is the length of $\sigma_s$, $A_s$ the area of the enclosed region and $R_s^\pm$ the in and out radius we have $$L_s=L_0+o(s)=2\pi +o(s)$$ $$A_s=A_0+o(s)=\pi+o(s)$$ $$R_s^\pm=R_0^\pm \pm sm(f) +o(s)=1\pm s m(f)+o(s)$$

If your inequality were to hold we would then have a constant $C$ so that $$ 4 s m(f) +o(s)\leq C o(s) $$ which is clearly impossible after letting $s\to 0$.

It works for Bonnesen's inequality as there is a favorable cancellation.

share|improve this answer
    
Very nice counter example. Thank you Rbega. –  Dorian Apr 13 '12 at 9:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.