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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?

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  • $\begingroup$ What are radii? $\endgroup$ Apr 12, 2012 at 17:50
  • $\begingroup$ There is a mistake in your first inequlity. Bonnesen's inequality reads π^2(R_out−R_in)^2≤(L^2−4πA) $\endgroup$
    – user54265
    Jun 22, 2014 at 10:43

1 Answer 1

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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.

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  • $\begingroup$ Very nice counter example. Thank you Rbega. $\endgroup$
    – Dorian
    Apr 13, 2012 at 9:54

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