If p>q>2, and the t-snowflake of L_q admits a bi-Lipschitz embedding into L_p, then t is bounded away from 1

Analysis 1 Minute

You need to know: Metric space.

Background: We say that a metric space (X,\rho_X) admit a bi-Lipschitz embedding (or just “embeds” for short) into metric space (Y,\rho_Y) if there exist constants m,M>0 and a function f:X\to Y such that m \rho_X(x,y) \leq \rho_Y(f(x),f(y)) \leq M \rho_X(x,y), \, \forall x,y \in X. For \theta\in(0,1], a \theta-snowflake of metric space (X,\rho_X) is the metric space on the same set X with distance \rho(x,y)=(\rho_X(x,y))^\theta, \, \forall x,y \in X. By L^p space we mean, for concreteness, space of functions f:[0,1]\to{\mathbb R} with norm ||f||_p=\left(\int_0^1|f(x)|^pdx\right)^{1/p}<\infty.

The Theorem: On 25th August 2014, Assaf Naor and Gideon Schechtman submitted to arxiv a paper in which they, among other results, proved that, for every 2<q<p, if \theta\in(0,1] is such that the \theta-snowflake of L_q admits a bi-Lipschitz embedding into L_p, then necessarily \theta \leq 1-\frac{(p-q)(q-2)}{2p^3}.

Short context: It is known that, if 2<q<p, then L_q does not embed into L_p. Hence, if \theta-snowflake of L_q embeds into L_p, then \theta<1. Quantifying by “how much” \theta is bounded away from 1 gives an important quantitative refinement of non-embeddability L_q into L_p. However, before 2014, no estimate in the form \theta \leq 1-\delta(p,q) for any explicit function \delta(p,q) has been known. The Theorem provides the first such estimate. In fact, the authors conjectured that the inequality in the Theorem can be improved to \theta \leq \frac{q}{p}, which would be the best possible. In a later work, Naor proved this conjecture.

Links: Free arxiv version of the original paper is here, journal version is here.

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Published by Bogdan Grechuk

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