$L^p$-barycenters via continuous selectors

Question

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.

Answer 1

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 $x\mapsto \|x\|^p$).