English

Lipschitz-free spaces over products of sequences

Functional Analysis 2026-07-30 v1

Abstract

We answer positively a question of Aliaga and show that for any nonconstant real polynomial pp, the Lipschitz-free space over {(p(n),p(m)):n,mN}\{(p(n), p(m)):n, m\in \mathbb{N}\} is isomorphic to F(Z2)\mathcal{F}(\mathbb{Z}^2). We in fact show more generally that if dNd\in \mathbb{N}, qZ0q\in \mathbb{Z}_{\geq 0}, and ((an(i))n=1)i=1d((a_n^{(i)})_{n=1}^\infty)_{i=1}^d, ((bm(j))m=1)j=1q((b_m^{(j)})_{m=1}^\infty)_{j=1}^q are sequences with 0<a1(i)<a2(i)<0<a_1^{(i)}<a_2^{(i)}<\cdots, 0<b1(j)<b2(j)<0<b_1^{(j)}<b_2^{(j)}<\cdots, an(i)a_n^{(i)}\to \infty as nn\to \infty, an+1(i)an(i)1\frac{a_{n+1}^{(i)}}{a_n^{(i)}}\to 1 as nn\to \infty and lim infmbm+1(j)bm(j)>1\underset{m\to \infty}{\liminf}{\frac{b_{m+1}^{(j)}}{b_m^{(j)}}}>1 for all i,ji, j, then the Lipschitz-free space over the product of these d+qd+q sequences is isomorphic to F(Zd)\mathcal{F}(\mathbb{Z}^d).

Keywords

Cite

@article{arxiv.2607.28440,
  title  = {Lipschitz-free spaces over products of sequences},
  author = {Fraser Mason},
  journal= {arXiv preprint arXiv:2607.28440},
  year   = {2026}
}

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15 pages