# Building the Wasserstein space by pushforwards

Let $$\mathbb{R}^d$$ denote the $$d$$-dimensional Euclidean space, $$\mathcal{W}_2(\mathbb{R}^d)$$ denote the $$2$$-Wasserstein space with respect to the $$d$$-dimensional Euclidean space $$\mathbb{R}^d$$. Let $$L^2(\mathbb{R}^d)$$ denote the Bochner space of all Borel functions $$f:\mathbb{R}^d\rightarrow \mathbb{R}^d$$ satisfying $$\int \, \lVert f(x)\rVert^2dx<\infty$$.

Let $$\mathcal{X}\subseteq \mathcal{W}_2(\mathbb{R}^d)$$ consist of all measures $${\nu}$$ for which there is some $$f\in L^2(\lambda,\mathbb{R}^d)$$ satisfying: $${\nu}=f_{\#}\lambda$$ where $$\lambda$$ is the uniform measure on $$[0,1]^d$$.

How is $$\mathcal{X}$$ related to $$\mathcal{W}_2(\mathbb{R}^d)$$? Is $$\mathcal{X}$$ a dense subset of $$\mathcal{W}_2(\mathbb{R}^d)$$?

• I think the way the question is written now $X$ is empty, as you will always have $\tilde{\nu}(\mathbb{R}^d) = \infty$. You may have more luck by replacing $\lambda$ with a finite measure $\nu$. In that case I think the answer is equality if $\nu$ is absolutely continuous and proper subset if it has atoms, but I am not sure about the inbetween.
– mlk
Jan 10, 2022 at 11:23
• @mlk is right, as written the space $\mathcal{X}$ is not included in the Wasserstein space as it is not made of probability measures. If you replace $\lambda$ by the restriction of Lebesgue measure to a cube for example, then every probability measure is a push-forward of that one (and the map $f$ will be in $L^2(\lambda)$ if the given measure has finite second moment). This can be done by the Knothe-Rosenblatt rearrangement, or in many other ways. Jan 10, 2022 at 12:08
• @BenoîtKloeckner Perfect! Do you happen to have a reference for this?
– ABIM
Jan 10, 2022 at 13:21
• Your question as currently written doesn't seem to make sense: What does "Let $\mathbb R^d$ let $\mathcal W_2(\mathbb R^d)$ denote the 2-Wasserstein space with respect to $X$ mean"? Is the "Let $\mathbb R^d$" superfluous? What is $X$? Is it the same as $\mathcal X$ later? Is $\tilde\nu \in \mathcal X$ the same as $\nu$ with $f \in L^2(\nu, \mathbb R^d)$? Finally, your last sentence asks if $\mathcal W_2$, which contains $\mathcal X$ by the definition of $\mathcal X$, is a proper subset of $\mathcal X$; did you mean it the other way around? Jan 10, 2022 at 15:58
• @TomTheQuant The KNothe-Rosenblatt rearrangement is briefly discussed in Villani's "Old and New", and some references are given. You can also look the keyword "Gromov's proof of isopermietric inequality"; Brenier's map (i.e. the solution to Monge's problem for the quadratic cost) would also work, although it is easier when the starting measure has density bounded below (but you can first map $\lambda$ to a Gaussian measure and proceed from there. Jan 11, 2022 at 14:08

Yes, your set is dense. Indeed, $$\mathcal X$$ contains finitely atomic measures, which are obviously weakly-$$\ast$$ dense in the Wasserstein space and therefore also dense in distance (it is well known that the Wasserstein distance $$W_2$$ metrize the weak-$$\ast$$ convergence, modulo minor decay conditions at infinity given by convergence of the second moments).
To see that $$\mathcal X$$ contains finitely atomic measures, fix an arbitrary integer $$N$$ and take any $$\nu=\sum_{i=1}^N \nu_i\delta_{x_i}\in \mathcal W_2(\mathbb R^d)$$. Choose now any partition of your unit cube $$[0,1]^d=\bigcup_{i=1}^1 C_i$$ into $$N$$ disjoint sets with corresponding measures $$|C_i|=\nu_i$$. Then the piecewise constant map $$f$$ defined as $$f(x)=x_i$$ for $$x\in C_i$$ pushes forward $$\lambda$$ to $$\nu$$.
Perhaps another simple argument that $$\mathcal{X}$$ is indeed equal to $$\mathcal{W}_2(\mathbb{R}^d$$): Starting from the fact that there is a bimeasurable bijection $$b : \mathbb{R}^d \rightarrow \mathbb{R}$$. (c.f. here).
For any $$\nu \in \mathcal{W}_2(\mathbb{R}^d)$$, let $$\tilde\nu := b_{\#}\nu$$, and thus for the inverse $$a$$ of $$b$$ it holds $$\nu = a_{\#} \tilde\nu$$. Let further $$\mathcal{U}$$ be the uniform measure on $$(0, 1)$$ and note that $$\mathcal{U} = (\pi_1)_{\#} \lambda$$, where $$\pi_1$$ is the projection onto the first coordinate. Let $$Q$$ be the quantile function of $$\tilde\nu$$ and thus $$\tilde\nu = Q_{\#} \mathcal{U}$$. Putting things together, $$\nu = (a \circ Q \circ \pi_1)_{\#} \lambda$$, where $$a \circ Q \circ \pi_1 \in L^2(\lambda)$$ follows through this equality.