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Let $(X,d,x)$ be a separable pointed metric space and let $\mathcal{F}(X)$ be its Arens-Eells (also called its Lipschitz-Free space; in the case where $X$ is Banach) space. We view the $1$-Wasserstein space $\mathcal{W}(X)$ as a convex subset of $\mathcal{F}(X)$ via the usual embedding: $$ \mathcal{W}(X)\ni \mu \mapsto \left[f\mapsto \int_{u\in X} f(u) [\mu-\delta_{x}](du)\right] \in \operatorname{Lip}_0(X)' =:\mathcal{F}(X); \qquad \boldsymbol{(1)} $$ where the functions $f$ on the right-hand side belong to $\operatorname{Lip}_0(X)$ (so $f(x)=0$, but for generality I have defined them on all of $\operatorname{Lip}(X)$).

THE ASSUMPTION:

We assume that $(X,d,x)$ is such that $\mathcal{F}(X)$ has the bounded approximation property (BAP) (extensive research has been conducted over the last decade on this question, and many "interesting/typical" pointed metric spaces satisfy this assumption). Furthermore, in a broad range of "tame cases" explicit estimates on the operator-norm of the sequence of operators realizing the bounded-approximation property are known (for instance for any closed-subset of an $N$-dimensional Euclidean space the rate is $C\sqrt{N}$ for some universal constant $C$.

THE QUESTION(s): Let $K\subseteq \mathcal{W}(X)$ be given and let $(T_n)_{n=1}^{\infty}$ be a sequence of finite-rank operators converging to the identity on $K$ and whose operator norm is uniformly bounded by some constant $\lambda>0$.

Then, can we perturb the build a sequence of affine maps $(A_n)_{n=1}^{\infty}$ which are finite-rank, approximate the identity, and such that $A_n(K)\subseteq K$ for all $n$?

Comment: I expect that is possible and that one can construct this map by exploiting the fact that $\mathcal{W}(X)$ is a convex subset of a co-dimension $0$ affine subset of $\mathcal{F}(X)$.

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  • $\begingroup$ As far as I know, exact projections onto nearest points of convex subsets requires reflexivity, which you never have for $\mathcal{F}(X)$ when $X$ is infinite. What you're asking for is weaker, but still seems unlikely to me. $\endgroup$
    – Nik Weaver
    Commented Dec 1, 2021 at 2:32
  • $\begingroup$ BTW if $X$ isn't compact then you don't get $\mathcal{W}(X) \subset \mathcal{F}(X)$, but you can slightly modify the definition of $\mathcal{W}(X)$ so that this works. This is probably well-known, but anyway it's done in this paper. $\endgroup$
    – Nik Weaver
    Commented Dec 1, 2021 at 2:34
  • $\begingroup$ @NikWeaver Ins't the modification just $\mu \mapsto \mu - \delta_{x}$ (for the distinguishe point $x$)? $\endgroup$
    – ABIM
    Commented Dec 1, 2021 at 2:39
  • $\begingroup$ @NikWeaver Fair, initially I was thinking of $\epsilon$-best projections (like in Respov's book; Chapter 6) but this also seemed difficult and a likely less explored research question by others. $\endgroup$
    – ABIM
    Commented Dec 1, 2021 at 2:40
  • $\begingroup$ As a final thought, on my end, if we equip $\mathcal{F}(X)$ with a Borel probability measure $\nu$, then since $K$ is compact then one can combine the measurable maximum theorem (18.19 in Charalambos et al.'s book) and Lusin's Theorem to conclude that a metric projection must exist on a closed subset of $\mathcal{F}(X)$ of arbitrarily-high $\nu$-probability. (Though, this is kind of cheating). $\endgroup$
    – ABIM
    Commented Dec 1, 2021 at 2:54

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