If p>q>2, the maximal t for which the t-snowflake of L_q admits a bi-Lipschitz embedding into L_p is t=q/p

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 13th January 2016, Assaf Naor submitted to arxiv a paper in which he proved that, for every 2<q<p, the maximal \theta\in(0,1] for which the \theta-snowflake of L_q admits a bi-Lipschitz embedding into L_p is equal to \frac{q}{p}.

Short context: It is known that, if 2<q<p, then L_q does not embed into L_p. Hence, if \theta^*=\theta^*(p,q) denotes the maximal \theta\in(0,1] for which the \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. In 2004, Mendel and Naor proved that \frac{q}{p}\leq \theta^*. In a paper submitted in 2014, Naor and Schechtman proved that \theta^* \leq 1-\frac{(p-q)(q-2)}{2p^3} and conjectured that \theta^*=\frac{q}{p}. The Theorem confirms 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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