Monotone Lipschitz embedding ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:21:50Z http://mathoverflow.net/feeds/question/7110 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7110/monotone-lipschitz-embedding Monotone Lipschitz embedding ? Ady 2009-11-29T04:52:00Z 2010-01-18T14:25:03Z <p>In 1974, Aharoni proved that every separable metric space (X, d) is Lipschitz isomorphic to a subset of the Banach space c_0. Thus, for some constant L, there is a map K: X --> c_0 that satisfies the inequality d(u,v) &lt;= || Ku - Kv || &lt;= Ld(u,v) for all u and v in X. Now, suppose X = l_1 (in this case, L = 2 is best possible). I have the following</p> <p><strong>Conjecture</strong>: Let K: l_1 --> c_0 be a Lipschitz embedding. Then K cannot be <em>monotone</em> w.r.t. the natural duality pairing (.,.) between l_1 and c_0, i.e., there are some u and v in l_1 such that (u - v, Ku - Kv) &lt; 0.</p> http://mathoverflow.net/questions/7110/monotone-lipschitz-embedding/9911#9911 Answer by Bill Johnson for Monotone Lipschitz embedding ? Bill Johnson 2009-12-27T23:03:25Z 2009-12-27T23:03:25Z <p>Ady, this could be a hard problem. Why are you interested in the answer?</p> http://mathoverflow.net/questions/7110/monotone-lipschitz-embedding/12191#12191 Answer by fedja for Monotone Lipschitz embedding ? fedja 2010-01-18T14:25:03Z 2010-01-18T14:25:03Z <p>To answer Bill Johnson's question, a monotone linear bi-Lipschitz embedding (actually, an isometric one) $\ell^1\to\ell^\infty$ is very easy to construct. Just take any antisymmetric matrix $A$ of $\pm 1$s with the property that for each $n$ every combination of signs in the first $n$ positions appears in some row of $A$ (you can easily build it by induction) and take $Lx=x+Ax$. Unfortunately, I do not see how to convert it into a mapping to $c_0$. </p>