Does there always exist a unique point with minimal average distance to any curve? Given any curve $\gamma:[0,1]\to\mathbb{R}^2$, we can consider the average distance from any point $x\in\mathbb{R}^2$ to the curve, $\int_0^1d(x,\gamma(t))dt$. I am interested in the points $x\in\mathbb{R}^2$ which minimize this distance. It is easy to prove that such points exist by a compactness argument, but are they always unique?
This problem can be expressed much more generally: given a compactly supported probability measure $m$ on $\mathbb{R}^2$, is there always a unique point $x$ that minimizes $\int_{\mathbb{R}^2}d(x,y)dm(y)$? In this case the general answer is no, for example, we can consider a measure where the points $(0,0)$ and $(0,1)$ each have probability $\frac{1}{2}$: then any point between them minimizes the average distance. But are there some simple sufficient conditions that eliminate "trivial" counterexamples like that one? (e.g. the measure not being supported in two points or something more restrictive if necessary).
The case of the first paragraph with a curve $\gamma$ can be seen as a concrete case of a compactly supported probability measure, just by defining a measure $m$ in $\mathbb{R}^2$ by $m(A)=\mu(\gamma^{-1}(A))$, where $\mu$ is the usual Lebesgue measure in $[0,1]$.
If the uniqueness part is true, it would have some cool consequences: for example, if the curve is a regular polygon parametrized by arc length, then the unique point at minimal average distance has to be the center of the polygon. The same would apply to other curves whose isometry group (that is, the group of isometries $\phi:\mathbb{R}^2\to\mathbb{R}^2$ such that $\phi\circ\gamma=\gamma$) fixes just $1$ point.
 A: Okay, so I got the result I wanted:
For any curve $\gamma:[0,1]\to\mathbb{R}^2$, only one point achieves the minimal average distance to $\gamma$.
This will be deduced from the following result:
Proposition: If a compactly supported probability measure $m$ in $\mathbb{R}^2$ is not supported in a line, then the set $A$ of points at minimal average distance from $m$ contains just one point.
After proving the result I adress separately the case where the image of $\gamma$ is contained in some line.
Proof of the general result:
Note that for any point $x$, the function $z\to d(x,z)$ is convex: concretely, we have $\forall x,y,z\in\mathbb{R}^2$ and $\lambda\in(0,1)$ that
$d(\lambda x+(1-\lambda)y,z)=|z-(\lambda x+(1-\lambda)y)|=|\lambda(z-x)+(1-\lambda)(z-y)|$
$\leq \lambda|z-x|+(1-\lambda)|z-y|=\lambda d(x,z)+(1-\lambda)d(y,z).$
If $x\neq y$ and $\lambda\in(0,1)$, then equality is reached only when $|\lambda(z-x)+(1-\lambda)(z-y)|=\lambda|z-x|+(1-\lambda)|z-y|$, and this only happens when $z-x$ and $z-y$ are linearly dependent and they don't have opposite directions. So, when $z$ is in the line formed by $x$ and $y$ and not inside the "open" segment $xy$.
Now let $F(x):=\int_{\mathbb{R}^2}d(x,z)dm(z)$ be the average distance of a point $x\in\mathbb{R}^2$ to the measure $m$, and suppose there are two distinct points $x,y\in A$ where $F$ reaches its minimum value $k$ (it is not difficult to show by a compactness argument that this value is reached). Then for any $\lambda\in[0,1]$ we have
$$F(\lambda x+(1-\lambda)y)=\int_{\mathbb{R}^2}d(\lambda x+(1-\lambda)y,z)dm(z)\leq \int_{\mathbb{R}^2}\lambda d(x,z)+(1-\lambda)d(y,z)dm(z)=\int_{\mathbb{R}^2}\lambda d(x,z)+\int_{\mathbb{R}^2}(1-\lambda)d(y,z)dm(z)=\lambda k+(1-\lambda)k=k,$$
so $F(\lambda x+(1-\lambda)y)\leq k$. But as $k$ is the minimum value of $F$, we necessarily have $F(\lambda x+(1-\lambda)y)=k$. This implies that the inequalities above are all equalities, so if we let $G(z)=d(\lambda x+(1-\lambda)y,z)$ and $H(z)=\lambda d(x,z)+(1-\lambda)d(y,z)$, then we have
$$\int_{\mathbb{R}^2}G(z)dm(z)=\int_{\mathbb{R}^2}H(z)dm(z).$$
However we already know that $G(z)\leq H(z)$ $\forall z$. So for the integrals to be equal, we need to have $G(z)= H(z)$ for $m$-almost every $z$. That is, almost every $z$ has to be in the line containing $x$ and $y$ and outside the open segment $xy$. So the measure $m$ is supported in the line formed by $x$ and $y$.$\square$
Some comments:

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*This also works for compactly supported measures in $\mathbb{R}^n$ for any $n$, or more generally in any Hilbert space $H$ (where $|x|+|y|=|x+y|$ implies that $x,y$ are linearly dependent). Also, for this to work we don't need the probability measure $m$ to be compactly supported, we just need the functions $z\mapsto d(x,z)$ to be in $L^1$, or equivalently, we need $\int_H|z|dm(z)<\infty$.


*If the measure $m$ is supported in a line $l$, then it is clear that no point outside of $m$ is in $A$ (its projection onto $l$ has less average distance). Moreover, $A$ is a closed segment, and as we have seen, for any $x,y\in A$ the measure is supported outside the open segment $xy$. So we have two cases:
Case 1: $A$ is just one point.
Case 2: $A$ will be a closed segment with "interior" $A_0$ such that the
measure is supported in $l\setminus A_0$. Moreover, it is easy to
check that if $l^+$ and $l^-$ are the two components of $l\setminus
   A_0$, then $m(l^+)$ and $m(l^-)$ are both $\frac{1}{2}$: if not, the
average distance from the points of $A$ to $m$ would not be constant.

*

*If our measure is induced by a curve $\gamma:[0,1]\to\mathbb{R}^2$ contained in a line $l$, then we cannot have case 2 from above: if we had a segment $A_0$ such that $m(l^+)$ and $m(l^-)$ are both $\frac{1}{2}$, then the image of the curve would intersect both $l^+$ and $l^-$, so it would intersect $A_0$. Moreover $\gamma^{-1}(A_0)$ is open in $[0,1]$, so $A_0$ would have positive measure, contradicting $m(l^+)=m(l^-)=\frac{1}{2}$.

