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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) <= || Ku - Kv || <= 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

Conjecture: Let K: l_1 --> c_0 be a Lipschitz embedding. Then K cannot be monotone 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) < 0.

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Can you supply a reference? Is Aharoni's proof constructive? Can you give an example of such an embedding? –  Dave Penneys Nov 29 '09 at 5:29
    
The original reference is: I. Aharoni, Every separable metric space is Lipschitz equivalent to a subset of c_0, Israel J. Math. 19 (1974), 284-291. Yes, Aharoni's proof is constructive. Another reference is the Kalton & Lancien preprint from 2007 arxiv.org/abs/0708.3924 –  Ady Nov 29 '09 at 6:18
    
I just thought it might be useful to strip the construction to the bones (at the cost of relaxing the Lipschitz constant). Let $X$ be a separable metric space and let $x_j$ be a countable dense set. Let $d_j(x)$ be the distance from $x$ to $x_j$. Put $(Kx)_n=\min(4d_1(x),4d_2(x),\dots,4d_{n-1}(x),d_n(x))$. Then $\frac 13d(x,y)\le \|Kx-Ky\|\le 4d(x,y)$. Unfortunately, this is highly non-monotone. What I wonder though is if there is any monotone Lipshitz (not necessarily bi-Lipschitz) mapping $K$ with $K_1(x)=\|x\|_1$ –  fedja Jan 23 '10 at 0:59
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2 Answers 2

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 $c_0$.

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Thanks, Fedja; I did not see that. –  Bill Johnson Jan 18 '10 at 15:44
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Ady, this could be a hard problem. Why are you interested in the answer?

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You have more than 50 reputation, which means that you can leave comments by clicking "add comment" below the question. –  Jonas Meyer Dec 27 '09 at 23:21
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@Professor Johnson My deep motivation relies on the following (IMHO) abnormal situation. It looks like there are two distinct, and distant things, namely: (A) the nonlinear functional analysis [monotone operators, Leray-Schauder degree, Browder-Petryshyn degree, and so on] and (B) the so-called "Geometric nonlinear functional analysis", that is nonlinear Banach space theory, in fact [uniform homeo, Lipschitz homeo, your works, Kalton's works, Lindenstrauss' works, and so on]. –  Ady Dec 28 '09 at 6:18
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@Professor Johnson Simply I think these are two branches of the same tree and thus they must be somehow connected. Because it's a bit frustrating to see, e.g., that even the most general form of the topological degree cannot "touch", e.g., the uniform embeddings. Investigating the "common border" of (A) and (B), and their interrelations, that would be interesting, I think. E.g., the Mazur homeomorphisms, although simple, are elements of this "edge". P.S. I am aware that this problem may be difficult. Unfortunately, I think that the answer would be affirmative. –  Ady Dec 28 '09 at 6:22
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