1
$\begingroup$

Let $U$ be a simply connected bounded open set in $\mathbb{R}^2$. The area of $U$ is denoted by $A$.

(We do not assume any thing about its boundary).

Assume that $\gamma_n$,s are smooth simple closed curves which lie in $U$. The perimiter and area of $\gamma_n$ are denoted by $l_n$ and $A_n$, respectively. We assume that $\gamma_n$,s eventually leave compact subsets of $U$. That is for every compact subset $K\subset U$, there is a natural number $N$ such that $\gamma_n$ has empty intersection with $K$, for every $n>N$. Assume that $A_n$ converges to $A$ and $l_n$ converges to a real number $l$ and we have$ 4\pi A=l^2$.

Is $U$ necessarily the interior of a circle?

$\endgroup$
4
$\begingroup$

I may have missed something but it should follow from Bonnesen's inequality, which states that every domain $\Omega\subset\mathbb{R}^2$ satisfies : $$\mathcal{L}(\partial\Omega)^2-4\pi\mathcal{A}(\Omega)\geq \pi^2(r_\text{ex}(\Omega)-r_\text{in}(\Omega))^2 $$ where $r_\text{in}(\Omega)$ (resp. $r_\text{ex}(\Omega)$) is the biggest( resp. smallest) possible radius of disk contained in $\Omega$ (resp. which contains $\bar\Omega$).

If one denotes by $\Omega_n$ the domain bounded by your $\gamma_n$. Then your hypothesis imply that each of the $\Omega_n$ are sandwiched between two disks of closer and closer radius, which should be enough to conclude.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your very interesting answer. $\endgroup$ – Ali Taghavi Feb 2 '19 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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