any answer to this question or part of it is much appreciated.
I'm working through Exercise 33 of Santambrogio's Optimal Transport for Applied Mathematicians, which reads:
Let $\gamma \in \Pi(\mu,\nu)$ be a transport plan between two probabilities on two compact spaces $X$ and $Y$, respectively. Let $\mu_n \rightharpoonup \mu$ and $\nu_n \rightharpoonup \nu$ be two weakly converging sequences of probabilities. Prove that there exists a sequence $\gamma_n \in \Pi(\mu_n,\nu_n)$ with $\gamma_n \rightharpoonup \gamma$… [also something about converging transport costs, but not important here.]
The "gluing lemma" (Lemma 5.5 in the book, for the standard proof of Wasserstein triangle inequality) lets me glue transport plans together, and Santambrogio's hint suggests to use that lemma to modify the marginals of $\gamma$.
What I can think to do with it is to build $\gamma_n$ from $\gamma$ gluing the optimal plan $\gamma^\mu_n$ from $\mu_n$ to $\mu$ and the optimal plan $\gamma^\nu_n$ from $\nu_n$ to $\nu$ appropriately so that $\gamma_n$ is indeed from $\mu_n$ to $\mu$: intuitively to get from $\mu_n$ to $\nu_n$ we go first by $\gamma^\mu_n$, then $\gamma$, then by $\gamma^\nu_n$. Since $\mu_n$ and $\nu_n$ weakly converge to $\mu$ and $\nu$ respectively, the part I'm modifying $\gamma$ by to obtain $\gamma_n$ should, loosely speaking, become negligible as $n \to \infty$.
1) Does this $\gamma_n$ work? I don't see any other way to build it so that it weakly converges to $\gamma$.
2) How do I show $\gamma_n \rightharpoonup \gamma$? I believe I should use the fact that the Wasserstein convergence metrizes weak convergence, but (a) the option to show $W_p(\gamma_n,\gamma) \to 0$ seems messy/there isn't a clear triangle inequality here and (b) I'm not even sure how to check directly by testing against $C_b(X \times Y)$.
3) In trying to test convergence against $C_b(X \times Y)$ I got the following "proof":
The separable functions, i.e. those $\phi$ satisfying $\phi(x,y) = \sum_i \psi(x)\eta(y)$, are dense in $C(X \times Y)$ by Stone-Weierstrass (using compactness here), so we can assume that our test functions are separable. Then
\begin{align*}
\int_{X \times Y} \phi(x,y) \, d \gamma_n(x,y) &= \int_{X \times Y} \psi(x)\eta(y) \, d \gamma_n(x,y) \\
&= \int_X \psi(x) \, d \mu_n(x) \int_Y \eta(y) \, d \nu_n(y) \\
& \to \int_X \psi(x) \, d \mu(x) \int_Y \eta(y) \, d \nu(y) \\
&= \int_{X \times Y} \psi(x) \eta(y) \, d\gamma(x,y) \\
&= \int_{X \times Y} \phi(x,y) \, d \gamma(x,y) \, .
\end{align*}
My issue with this argument is that it seems to work for any $\gamma_n \in \Pi(\mu_n,\nu_n)$, and this would imply that $\gamma_n \rightharpoonup \gamma$ for any $\gamma$ (i.e. there's a unique $\gamma$ since the weak limits are unique), which I know is wrong. I'm not sure where the argument is wrong.
Again any comments/insight is appreciated.
Best Answer
Let me start by declaring I do not have a solution to the problem, but I may have some answers to your questions:
1) You cannot take "optimal" plans since you do not have any cost function. You may take some plans between $\mu,\mu_n$ and $\nu,\nu_n$ but I am not sure it will work. For example if you take the product measures $$\gamma_n^{\mu}=\mu_n\otimes\mu\ ;\ \gamma_n^{\nu}=\nu\otimes\nu_n$$ Then by taking $\gamma_n$ as the gluing of $\gamma_n^{\mu},\gamma,\gamma_n^{\nu}$ will lead to $\gamma_n=\mu_n\otimes\nu_n$ (to see that, work with the definition of the gluing in the proof of lemma 5.5) which does not necessarily converge to $\gamma$. You may need to choose the plans $\gamma_n^{\mu},\gamma_n^{\nu}$ differently.
2) If you obtained $\gamma_n$ using gluing of measures, it might work to use the definition of gluing to find the limit. If the spaces are compact, you can check directly using test functions in $C(X\times Y)$ (since continuous on compact is bounded).
3) Your argument is wrong since the second equality holds only if $\gamma_n=\mu_n\otimes\nu_n$, whereas the third equality holds only if $\gamma=\mu\otimes\nu$.