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The Dunford-Pettis property for Lipschitz-free spaces over ℝn
Abstract
We show that for every n∈ ℕ, the Lipschitz-free space ℱ(ℝn) has the Dunford-Pettis property. As a consequence, for every n∈ ℕ and every nonempty M⊂ ℝn, the space ℱ(M) has the Dunford-Pettis property. This implies that ℱ(M) is not isomorphic to ℱ(X) for any separable Banach space X that fails the Dunford-Pettis property. In particular, this applies to X=ℓ2, and more generally to every infinite-dimensional separable reflexive space. We also deduce that if 1<p<∞, then ℱ(ℓp) is not isomorphic to (⨁n=1∞ ℱ(ℓpn) )ℓ1.
Related
- The Ball-Covering Property in Lipschitz-Free Spaces
- Kwapień's property (K) in Lipschitz-free spaces
- Structural consequences of the Schur p-property for Lipschitz-free p-spaces
- Isometric copies of ℓ∞n in Lipschitz-free spaces over finite metric spaces
- Lipschitz-free spaces over separable Banach spaces and their complementability in the bidual
BibTeX
@misc{mason2026dunfordpettis,
title = {{The Dunford-Pettis property for Lipschitz-free spaces over $\mathbbR^n$}},
author = {Fraser Mason},
howpublished = {arXiv},
year = {2026},
month = sep,
doi = {10.48550/arxiv.2609.28842},
eprint = {2609.28842},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2609.28842},
}
RIS
TY - UNPB
TI - The Dunford-Pettis property for Lipschitz-free spaces over ℝ^n
AU - Fraser Mason
PY - 2026
DA - 2026/09/23
PB - arXiv
DO - 10.48550/arxiv.2609.28842
UR - https://arxiv.org/abs/2609.28842
AB - We show that for every n∈ ℕ, the Lipschitz-free space ℱ(ℝ^n) has the Dunford-Pettis property. As a consequence, for every n∈ ℕ and every nonempty M⊂ ℝ^n, the space ℱ(M) has the Dunford-Pettis property. This implies that ℱ(M) is not isomorphic to ℱ(X) for any separable Banach space X that fails the Dunford-Pettis property. In particular, this applies to X=ℓ_2, and more generally to every infinite-dimensional separable reflexive space. We also deduce that if 1<p<∞, then ℱ(ℓ_p) is not isomorphic to (⨁_n=1^∞ ℱ(ℓ_p^n) )_ℓ_1.
ER -