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8 votes
4 answers
681 views

Uniform density of Lipschitz maps is space of continuous function — for general metric spaces

Let $X$ and $Y$ be metric space, $X$ be compact, $C(X,Y)$ denote the set of continuous functions from $X$ to $Y$ with uniform convergence on compacts topology, and $\operatorname{Lip}(X,Y)$ denote the ...
Math_Newbie's user avatar
4 votes
0 answers
197 views

Compactness of the unit ball in the space of Radon measures w.r.t. the Kantorovich-Rubinstein norm

This question was posted previously but has not attracted any responses so I am repharising it in a slightly different language hoping to reach a wider community Let $(X,d)$ be a pointed metric space ...
Yury Korolev's user avatar
3 votes
0 answers
171 views

Covering number $C^k$-balls in $C(\mathbb{R}^n)$

Fix a positive integer $n$ and and an non-negative integer $k$. The Arzela-Ascoli theorem guarantees that for a given positive integer $k$ and a given $L>0$ the set $$ Ball_{C^{k,1}([0,1]^n)}(0,L) ...
ABIM's user avatar
  • 5,405
2 votes
0 answers
71 views

Perturbing the approximation property from the Lipschitz-free space to stay in the Wasserstein space

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 $1$-...
ABIM's user avatar
  • 5,405
1 vote
0 answers
449 views

Bound on covering number of Lipschitz functions – missing part in proofs of Kolmogorov et al

Given a metric space $(\mathcal{X},\rho)$ and $\mathcal{A}\subset\mathcal{X}$ totally bounded, i.e. $\mathcal{A}$ has a finite $\varepsilon$-covering for any $\varepsilon>0$. Consider $\...
samuel's user avatar
  • 11