I have been stuck for a few days in a seemingly harmless question.


Given a simply connected open set $\Sigma\subset\mathbb{R}^2$, with smooth boundary $\partial\Sigma$, I am interested in estimating
$$
\int\_{\Sigma}d(x,\partial\Sigma)\;dx,
$$
where $d(x,\partial\Sigma)=\inf\_{y\in\partial\Sigma}|x-y|$.


I would love to be able to prove that
$$
\int\_{\Sigma}d(x,\partial\Sigma)\;dx\ge\frac{2}{3} \frac{|\Sigma|^2}{|\partial\Sigma|}.
$$
Equality holds for $\Sigma$ being a disk and violating the smoothness assumption for $\Sigma$ being a square.


The only thing that seems to be doing on the right hand side of the inequality is using the Isoperimetric Inequality to say
$$
\frac{2}{3} \frac{|\Sigma|^2}{|\partial\Sigma|}\le \frac{1}{3\sqrt{\pi}}|\Sigma|^{\frac{3}{2}},
$$
with equality, if and only if, $\Sigma$ is a disk.


Ideas or counter-examples are very much welcomed.



Integral of the distance function to the boundary of a planar set
The correct constant is $1/2$. Let $L(s)$ be the perimeter of the set of points whose distance to the boundary is $\ge s$. For simply-connected domains $L(s)$ is non-increasing, the integral of the distance to the boundary is $\int\_0^\infty sL(s)\,ds$ and the area is $\int\_0^\infty L(s)\,ds$. Now we want to minimize the first integral given the second one and $L(0)=\max\_s L(s)$. Clearly, the worst case is when we push all the mass to the left as much as we can, so $L(s)=L(0)$ up to some point and then $0$. That gives you $1/2$ for the ratio in question and that scenario is almost realized by a long narrow rectangle, as Nate pointed out.


