TITLE: Shape with area $a$ and shortest average distance between any two points
QUESTION [17 upvotes]: Consider the collection $\mathcal{X}$ of 

compact and connected subsets of $\mathbb{R}^2$,  
that are the closure of some open subset of $\mathbb{R}^2$, and
have area $a$.

For any $\Omega \in \mathcal{X}$ and any $x,y \in \Omega$, let $d(x,y,\Omega)$ be the length of the shortest path lying entirely inside $\Omega$ and connecting $x$ to $y$.
Also, let $\mathbb{1}(x,y,\Omega)$ be the indicator function that equals $1$ if $x$ and $y$ are both in $\Omega$, and zero otherwise.
I am interested in the following problem:

$$\min_{\Omega \in \mathcal{X}} \int_{x\in \mathbb{R}^2}\int_{y\in \mathbb{R}^2} d(x,y,\Omega) \mathbb{1}(x,y,\Omega) dx dy$$


Is there a well-known solution to this problem?
Is it possible to show that a solution to this problem must be convex?
Is the problem even well-defined (e.g., does $d(x,y,\Omega)$ always exist? I tried to help guaranteeing it by including only closed and connected sets in $\mathcal{X}$, but I am not 100% sure that's enough)?
Is the problem guaranteed to have a solution without more regularity conditions?

Some related problems are described here: https://link-springer-com.dist.lib.usu.edu/content/pdf/10.1007%2Fs10958-012-0717-3.pdf. Unfortunately, these related problems either (1) focus on "networks" rather than "thick" sets, (2) focus right away on convex sets, or (3) do not fix the area and instead impose an additional "cost" to the minimization problem for increasing the area.

REPLY [5 votes]: I don't know if this is well known (I guess it is), but here is an argument showing that a disc is the unique minimizer. Since the minimizer will end up being convex, we can consider straight-line distances instead of distances within the set.
Given such a set $\Omega$ with centre of mass $(0,0),$ consider the set $\Omega^{\mathrm{new}}$ defined by
$$(x,y)\in\Omega^{\mathrm{new}} \iff |y|\leq\frac12 \int 1(x,y',\Omega)dy'.$$
So in each vertical line segment $\Omega_x=\{y\mid (x,y)\in\Omega\},$ we replace the set by a centred version.
By Fubini's theorem, the areas of $\Omega^{\mathrm{new}}$ and $\Omega$ are equal. For a fixed pair of distinct $x$-coordinates $x,x'$ consider the effect on the distance between points on those lines,
$$w(\Omega_x,\Omega_{x'})=\int_{y\in \Omega_x} \int_{y'\in \Omega_{x'}} |(x,y)-(x',y')| dy dy'$$
I claim this decreases when replacing $\Omega$ by $\Omega^{\mathrm{new}},$ except in the case where $\Omega$ and $\Omega^{\mathrm{new}}$ are a.e. equal or one is a null set.
To see this, first consider replacing $\Omega_{x'}$ by the interval (not necessarily centred at zero) of the same length but of the form $\{y\mid \Phi(y)<c\},$ where $\Phi(y)=\int_{y'\in \Omega_{x'}} |(x,y)-(x',y')| dy'.$ Note $\Phi$ is strictly convex so this does indeed produce an interval. This operation does not increase $w(\Omega_x,\Omega_{x'}).$ We then do the same the other way round. Both slices are now intervals, and a vertical translation takes them to $\Omega^{\mathrm{new}}.$ Doing this a bit more carefully would show that the weight strictly decreases for sets that actually change.
This shows that a minimizer must have reflection symmetry through every line through its centre of mass, and must therefore be rotationally invariant, and a bit of thought shows it has to be a disk.
The existence of a minimizer can be shown by compactness. The set of measures $\mu$ of total measure $a,$ contained in some fixed large disk, subject to the additional restriction $\mu(S)\leq\lambda(S)$ (bounded by Lebesgue measure), is compact in the vague (weak-*) topology. (I'm using measures here, but you can use something like $L^2$ if that's more intuitive.) Any indicator function of a compact set with area $a$ can be put into such a space, then we can invoke compactness to pick a minimizer $\mu.$
I will now argue that the minimizer has to be a set. (Actually this isn't really needed for the above argument - you could just replace sets by functions everywhere.)
 The potential $\Phi_\mu(x)=\int_{x'\in\mathbb R^2} |x-x'|d\mu(x')$ is convex, and in particular continuous. If there were sets $S_1,S_2$ with $\max \Phi_\mu(S_1)<\min \Phi_\mu(S_2)$ and $\mu(S_1)<\lambda(S_1)$ then we could move some mass from $S_2$ to $S_1$ and get a smaller total potential, a contradiction. This means $\mu$ is actually uniform distribution on a set of the form $\{\Phi_\mu<c\}.$
It is possible to have a compact connected closed set that is regular (equal to the closure of its interior) and not path-connected. Just thicken the sine part of the topologist's sine curve, and stick two of them back to back (e.g. the closure of the set of $(x,y)$ with $0<|x|<1$ and $|y-\sin 1/x|<\tfrac 1 4$), then scale to get the area correct. So the integral may not be finite. You can modify this example to get path-connected but a diverging integral.