An asymptotic version of the Isoperimetric inequality

Question

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?

Answer 1

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.