Let $ (X,d) $ be a metric space and consider the function $ T:X \to \mathbb{R}^X$ such that $ T(x)(y) = 1$ if $ y = x $ and $ 0 $ for all other $ y $. Is there a norm on $ \mathbb{R}^X$ such that $ T $ is an isometry? That is, $ ||T(a) - T(b)|| = d(a,b)$ for all $ a,b \in X $.

I'm at a loss to know how to approach this. I didn't come up with any good ideas on how to define a proper norm, and I have absolutely no clue how to begin trying to prove such a norm could not exist. Any ideas?

  • 4
    $\begingroup$ It seems to me like a far more natural choice of $T$ would be $T(x)(y)=d(x,y)$. If $X$ is bounded, this is an isometry with respect to the sup norm on the space of bounded functions from $X$ to $\mathbb{R}$. In general, if $X$ is infinite, I would not expect there to be any natural norm that is well-defined on all of $\mathbb{R}^X$ (in particular, there does not exist a norm that makes every projection continuous). $\endgroup$ Nov 4, 2014 at 17:51
  • $\begingroup$ Yes, d(x,y) is a much better embedding of X in in $ R^X $, but my initial idea was the embedding described above and even though it wasn't the best one, I found the question of existence of a suitable norm(no matter how "weird" it could potentially be) quite interesting in itself. $\endgroup$
    – Ormi
    Nov 4, 2014 at 18:17
  • $\begingroup$ The answer to your question is "almost yes" but I'm curious to know in what context this question arose. Were you asked to find such a norm, or did you read somewhere that such a norm exists? $\endgroup$
    – Yemon Choi
    Nov 4, 2014 at 18:47
  • 1
    $\begingroup$ I was asked to prove that every metric space can be isometrically embedded in a Banach space, so that the image is linearly independent. My first idea was the one I described, since the linear independence is obvious there, and if a good norm could be found, any normed space be embedded in a Banach space, so that would give the desired result. Kuratowski's embedding seems to do the job much better(though I'm still not sure about the linear independence), but I found it curious to see if it was possible to find the right norm here and what the technique for doing it would be. $\endgroup$
    – Ormi
    Nov 4, 2014 at 19:03
  • 2
    $\begingroup$ I'm not asking this question with a "please solve my homework" intention, because I'm going to just keep trying to work out the linear independence for Kuratowski's embedding. I'm just genuinely curious about what can be achieved with my initial idea, so if you could give me some hints for that, or refer me to somewhere where I can read up about it, I'd be grateful. I also thought about defining the norm on the subspace of functions with finite supports, could this be the subspace you mentioned? $\endgroup$
    – Ormi
    Nov 4, 2014 at 19:47

1 Answer 1


Note that your embedding map $T$ actually takes values in the subspace $\newcommand{\R}{{\mathbb R}}$ $c_{00}(X;\R)$ of finitely supported functions $X\to\R$. If you merely want a norm on this subspace which makes $T$ an embedding, then this is possible via the Arens–Eells construction:

R. Arens, J. Eells, On embedding uniform and topological spaces. Pacific J. Math. 6 (1956) no. 3, 397-403.

(Arens and Eells proved a more general result: if you just want the embedding theorem for metric spaces then it is in Weaver's book Lipschitz spaces and also in some more recent work of e.g. Godefroy and Kalton. Google should provide links to various downloadable papers/preprints.)

The embedding is usually phrased in terms of sending $x\in X$ to $\delta_x \in c_{00}(X;\R)$, which is just another way of describing your map $T$. Of course the problem is defining the norm! One can either define it as an inf over various representations or a sup when paired with another more familiar Banach space. Let me choose the second way.

Start by fixing a basepoint $x_0\in X$. Given $f\in \R^X$ with $f(x_0)=0$ define its Lipschitz norm to be $$ \Vert f\Vert_L = \sup_{x,y\in X; x\neq y} \frac{|f(x)-f(y)|}{d(x,y)} \in [0,\infty] .$$ Then, given $c=\sum_{x\in X} c_x \delta_x$ where only finitely many of the $c_x$ are non-zero, define $$ \Vert c \Vert_{\bf AE} = \sup\left\{ \sum_{x\in X} c_x f(x) \;\colon\; f\in\R^X, \Vert f\Vert_L\leq 1, f(x_0)=0 \right\}$$.

The completion of $c_{00}(X;{\mathbb R})$ with respect to the norm $\Vert\cdot\Vert_{\bf AE}$ is the Arens–Eells space of $X$ (I'm using the terminology and borrowing the definition from Weaver's book.)

Let's check that $x\mapsto\delta_x$ is an isometry. Let $x,y\in X$ with $x\neq y$. If $f(x_0)=0$ and $\Vert f\Vert_L\leq 1$ then pairing $x$ with $\delta_x-\delta_y$ gives $f(x)-f(y)$, which is bounded in modulus by $d(x,y)$ owing to the Lipschitz condition. So $\Vert \delta_x - \delta_y \Vert_{\bf AE} \leq d(x,y)$. On the other hand, consider the function $$ h(z)=d(z,y)- d(x_0,y) \quad(z\in X).$$ Clearly $h(x_0)=0$, and the triangle inequality for $d$ shows us that $\Vert h\Vert_L\leq 1$. Hence $$ \Vert \delta_x -\delta_y \Vert_{\bf AE} \geq \vert h(x)-h(y)\vert = d(x,y). $$ Putting these together gives $\Vert \delta_x - \delta_y \Vert_{\bf AE} =d(x,y)$ as required.

For those who like the category-theoretic perspetive: the Arens–Eells space can be viewed as a left adjoint to the functor ${\bf U}: {\sf Ban} \to {\sf Met}_0$ where:

  • the first category has Banach spaces as objects and bounded linear maps as the morphisms;

  • the second category has pointed metric spaces as objects, and basepoint-preserving Lipschitz maps as the morphisms;

  • and given a Banach space $E$, ${\bf U}(E)$ is defined to be the underlying metric space of $E$, with $0_E$ as the basepoint.

Then the Arens–Eells embedding can be regarded as the unit of this adjunction.

In more "down-to-earth" language: given a pointed metric space $(X,x_0)$ let ${\bf AE}(X,x_0)$ be the Arens–Eells space as defined above. Then for any Banach space $E$ and any Lipschitz map $f: X \to E$ satisfying $f(x_0)=0$, there is a unique extension of $f$ to a continuous linear map $F: {\bf AE}(X,x_0) \to E$. Thus ${\bf AE}(X,x_0)$ can be viewed as the "free Banach space generated by $(X,x_0)$".

  • $\begingroup$ Thank you for the answer. Unfortunately I can't quite find those books you suggested available anywhere. I guess I'll just familiarise myself with uniform spaces and then will be able to read the original Aren and Eell's paper. $\endgroup$
    – Ormi
    Nov 4, 2014 at 22:20
  • 2
    $\begingroup$ The Godefroy-Kalton paper is available at kaltonmemorial.missouri.edu/docs/sm2003c.pdf $\endgroup$ Nov 5, 2014 at 14:02
  • $\begingroup$ But is it a norm? It seems that $\delta_{x_0}$ has norm 0. $\endgroup$
    – Fan Zheng
    Mar 19, 2015 at 21:37
  • $\begingroup$ In my definition I require $f(x_0)=0$ $\endgroup$
    – Yemon Choi
    Mar 19, 2015 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.