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?
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 $(X,x_0)$".
