Let $\mathbb{P}$ be a Borel probability measure on $\mathbb{R}^n$ with $\int\_{x \in \mathbb{R}^n} \|x\|^p\mathbb{P}(dx)<\infty$. For which $p\in [1,\infty)$ (other than $p=2$) is the following optimality set non-empty:
$$
X(\mathbb{P}):=\left\{
x\in \mathbb{R}^n:\, \int\_{u \in \mathbb{R}^n}\|u-x\|^p\mathbb{P}(du)=\inf\_{x'\in \mathbb{R}^n}
\int\_{u \in \mathbb{R}^n}\|u-x'\|^p\mathbb{P}(du)
\right\}.
$$


More importantly, for which such $p$ does there exist a continuous selector $S:\mathcal{P}\_p(\mathbb{R}^n)\rightarrow \mathbb{R}^n$ with:
$$
S(\mathbb{P})\in X(\mathbb{P})
?
$$
Note, here, we equip $\mathcal{P}\_p(\mathbb{R}^n)$ with the $L^p$-Wasserstein distance.


This question is related to [this post](https://mathoverflow.net/questions/397420/holder-continuous-barycenter-maps).



$L^p$-barycenters via continuous selectors
There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that
$$
\int\_{B\_R(0)}\|u\|^pd\mathbb{P}(u) > \frac{1}{2}\int\_{\mathbb{R}^n}\|u\|^pd\mathbb{P}(u).
$$
Then if $x > B\_{3R}(0)$, we see that
$$
\int\_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u) > \int\_{\mathbb{R}^n}\|u\|^pd\mathbb{P}(u)
$$
and hence is
$$
\inf\_{x\in\mathbb{R}^n}\int\_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u) = \inf\_{x\in B\_{3R}(0)}\int\_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u).
$$
Now, just note that $x\mapsto \int\_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B\_{3R}(0)$. Moreover, we can say that the minimizer is unique since the function is convex (that follows from the convexity of $x\mapsto \|x\|^p$).


