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Assume we have $n$ points $p_0\ldots p_{n-1}$ which form a discrete metric space $V$ with metric $d$. Can we define a function $f:V\rightarrow \mathbb{R}$ with $f(p_0) = 0$, $f(p_1) = d(p_0,p_1)$ and $d(p_i,p_j) \geq |f(p_i) - f(p_j)|$ for all $i,j$?

If the $p_i$ are points in a vector space with inner product, $f$ could be just the orthogonal projection onto the ray of direction $p_j - p_i$, but for general metrics, I do not know whether such a function exists.

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Take $f(x)=d(x,p_0)/2-d(x,p_1)/2+d(p_0,p_1)/2$.

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    $\begingroup$ IMO, as a simple answer one should give $f(x) = d(x,p_0)$. $\endgroup$
    – Bill Johnson
    Commented Oct 26, 2014 at 2:15
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    $\begingroup$ I was blind, but now I see... $\endgroup$
    – J Fabian Meier
    Commented Oct 26, 2014 at 9:38
  • $\begingroup$ We've all been there, J. Fabian. I hope you did not my "joke" answer, which was the most complicated proof I saw off the top of my head. I was hoping that someone would suggest eliminating the axiom of choice by quoting the non linear Hahn-Banach theorem instead of the Hahn-Banach theorem, but apparently no one else was in a jovial mood. $\endgroup$
    – Bill Johnson
    Commented Oct 26, 2014 at 11:31
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Sure. Think about $V$ being contained in a Banach space $X$ in such a way that $0=p_0$ and let $f$ be a norm one linear functional s.t. $f(p_1)= \|p_1\|$.

It is of course elementary that a pointed metric space $(M,p_0)$ embeds isometrically into the Banach space $\ell_\infty(M)$ in such a way that $p_0$ is sent to $0$.

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