3 added 2 characters in body

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.

For a simple closed curve $\gamma$, the stronger inequality due to Bonnesen holds: $L^2 - 4\pi A \geq \pi pi^2 (R_{out}- R_{in})^2$, where, setting $\Omega =$ Int $(\gamma)$, $R_{in}$ and $R_{out}$ denote the inner and outer radii of the sets:

$R_{in} = \max_{B_r sup_{B_r \subset \Omega} r$

$R_{out} = \min_{\Omega inf_{\Omega \subset B_R} R$

Question: Does this inequality continue to hold without the assumption that the curve is simple? In particular, does it hold for any connected, rectifiable set?

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.

Update: I just realized this is false. 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.

2 point out that my original question is false

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.

For a simple closed curve $\gamma$, the stronger inequality due to Bonnesen holds: $L^2 - 4\pi A \geq \pi (R_{out}- R_{in})^2$, where, setting $\Omega =$ Int $(\gamma)$, $R_{in}$ and $R_{out}$ denote the inner and outer radii of the sets:

$R_{in} = \max_{B_r \subset \Omega} r$

$R_{out} = \min_{\Omega \subset B_R} R$

Question: Does this inequality continue to hold without the assumption that the curve is simple? In particular, does it hold for any connected, rectifiable set?

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.

Update: I just realized this is false. 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.

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# Bonnesen's inequality for non-simple curves

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.

For a simple closed curve $\gamma$, the stronger inequality due to Bonnesen holds: $L^2 - 4\pi A \geq \pi (R_{out}- R_{in})^2$, where, setting $\Omega =$ Int $(\gamma)$, $R_{in}$ and $R_{out}$ denote the inner and outer radii of the sets:

$R_{in} = \max_{B_r \subset \Omega} r$

$R_{out} = \min_{\Omega \subset B_R} R$

Question: Does this inequality continue to hold without the assumption that the curve is simple? In particular, does it hold for any connected, rectifiable set?

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.