×
Well done. You've clicked the tower. This would actually achieve something if you had logged in first. Use the key for that. The name takes you home. This is where all the applicables sit. And you can't apply any changes to my site unless you are logged in.

Our policy is best summarized as "we don't care about _you_, we care about _them_", no emails, so no forgetting your password. You have no rights. It's like you don't even exist. If you publish material, I reserve the right to remove it, or use it myself.

Don't impersonate. Don't name someone involuntarily. You can lose everything if you cross the line, and no, I won't cancel your automatic payments first, so you'll have to do it the hard way. See how serious this sounds? That's how serious you're meant to take these.

×
Register


Required. 150 characters or fewer. Letters, digits and @/./+/-/_ only.
  • Your password can’t be too similar to your other personal information.
  • Your password must contain at least 8 characters.
  • Your password can’t be a commonly used password.
  • Your password can’t be entirely numeric.

Enter the same password as before, for verification.
Login

Grow A Dic
Define A Word
Make Space
Set Task
Mark Post
Apply Votestyle
Create Votes
(From: saved spaces)
Exclude Votes
Apply Dic
Exclude Dic

Click here to flash read.

arXiv:2404.13616v1 Announce Type: new
Abstract: In this paper, we investigate Monge-Kantorovich problems for which the absolute continuity of marginals is relaxed. For $X,Y\subseteq\mathbb{R}^{n+1}$ let $(X,\mathcal{B}_X,\mu)$ and $(Y,\mathcal{B}_Y,\nu)$ be two Borel probability spaces, $c:X\times Y\to\mathbb{R}$ be a cost function, and consider the problem \begin{align*}\tag{MKP}\label{MKPEQ} \inf\left\{\int_{X\times Y} c(x,y)\,d\lambda\ :\ \lambda \in\Pi(\mu,\nu) \right\}. \end{align*} Inspired by the seminal paper \cite{GANGBOMCCANN2} with applications in shape recognition problem, we first consider \eqref{MKPEQ} for the cost $c(x,y)=h(x-y)$ with $h$ strictly convex defined on the multi-layers target space \begin{align*} X=\overline{X}\times\{\overline{x}\},\quad\text{and}\quad Y=\bigcup_{k=1}^K \left(\overline{Y}_{k}\times \{\overline{y}_k\}\right), \end{align*} where $\overline{X}, \overline{Y}_{k}\subseteq \mathbb{R}^{n}$ for $k\in \{1,\ldots,K\},$ $\overline{x}\in \mathbb{R}$, and $\{\overline{y}_1,..., \overline{y}_K\}\subseteq \mathbb{R}$. Here, we assume that $\mu|_\overline{X}\ll\mathcal{L}^n$ (the Lebesgue measure on $\mathbb{R}^n$), but $\mu$ is singular w.r.t. $\mathcal{L}^{n+1}$. When $K=1$, this translates to the standard \eqref{MKPEQ} for which the unique solution is concentrated on a map. We show that for $K\geq 2,$ the solution is still unique but it concentrates on the graph of several maps. Next, we study \eqref{MKPEQ} for a closed subset $X\subseteq \mathbb{R}^{n+1}$ and its $n$-dimensional submanifold $X_0$ with the first marginal of the form \begin{align*} \int_X f(x)\,d\mu(x)=\int_X f(x)\alpha(x)\,d\mathcal{L}^{n+1}(x)+\int_{X_0} f(x_0)\,d S(x_0),\ \ \forall f\in C_b(X). \end{align*} Here, $S$ is a measure on $X_0$ such that $S\ll \mathcal{L}^{n}$ on each coordinate chart of $X_0$. This can be seen as a two-layers problem as the measure $\mu$ charges both $n$- and $n+1$-dimensional subsets.

Click here to read this post out
ID: 818205; Unique Viewers: 0
Unique Voters: 0
Total Votes: 0
Votes:
Latest Change: April 23, 2024, 7:33 a.m. Changes:
Dictionaries:
Words:
Spaces:
Views: 7
CC:
No creative common's license
Comments: