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Let $\mu(dx)=\sum_{i=1}^np_i\delta_{x_i}(dx)$ and $\nu(dy)=\rho(y)dy$ be two probability measures on $\mathbb R^d$. Consider the $2-$Wasserstein distance below:

$$W_2(\mu,\nu)^2 \quad := \quad \inf_{\pi\in\Pi(\mu,\nu)}~ \int_{\mathbb R^d\times\mathbb R^d}~ |x-y|^2\pi(dx,dy),$$

where $\Pi(\mu,\nu)$ denotes the collection of couplings $\pi$ of $\mu$ and $\nu$. Obviously $W_2(\mu,\nu)$ depends on the points $x_1,\ldots, x_n$, and we denote $\mu\equiv \mu(x_1,\ldots,x_n)$ and define the energy function

$$E(x_1,\ldots, x_n): = W_2^2\big(\mu(x_1,\ldots, x_n),\nu\big).$$

I am interested in the minimisation of $(x_1,\ldots, x_n)\mapsto E(x_1,\ldots, x_n)$. My first intuition is to compute the gradient of $E$ and to use Newton's gradient method (the domain of definition of $E$ is not convex). I would like to know whether there is any related reference on this problem, and all answers or comments are highly appreciated!

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  • $\begingroup$ Let $\Omega \subset \mathbb R^d \times \cdots \times \mathbb R^d$ be the subspace containing all elements $(x_1,\ldots, x_n)$ s.t. $x_i\neq x_j$ for all $I\neq j$. Notice that $E$ is defined on $\Omega$ which is open and convex (although $E$ can be extended continuously on $\mathbb R^d \times \cdots \times \mathbb R^d$). $\endgroup$
    – user111097
    Commented Aug 3, 2018 at 20:33

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