Let $(X, d)$ be a metric space, $(E, || \cdot ||)$ a Banach space, $(AE(X), || \cdot ||)$ - as described below.
$AE_0(X) = \{ u : X \rightarrow \mathbb{R} \ : \ u^{-1} (\mathbb{R} \setminus \{0 \}) \ \ \text{is finite}, \ \sum_{x \in X} u(x)=0 \}$,
for $x,y,z \in X, \ x \neq y, \ m_{xy} \in AE_0(X), \ \ m_{xy} (x)=1, \ m_{xy}(y)=-1, \ m_{xy}(z)=0$ for $z \neq x, y$ and $m_{xx} \equiv 0$
for $u \in AE_0(X), \ \\ ||u||_d = \inf _{n \ge 1} \{ \sum_{k=1}^n |a_k| d(x_k, y_k) \ : \ u= \sum_{k=1}^n a_km_{x_k, y_k}, a_k \in \mathbb{R}, x_k, y_k \in X\}$
I have trouble proving that $||m_{xy}|| = d(x,y)$, the inequality $\le$ is trivial - we just take $x_1 = x, \ y_1 = y, a_1 = 1$.
The other direction seems very difficult to me.
Here's what I've tried:
One of $x_k$ must equal x. For all those $x_k \neq x$ there must exist $y_j$ such that $x_k = y_j$ and $a_k = a_j$. (Because $m_{xy}(x_k) = 0$)
I think it's a good idea to consider $m_{x x_1}, \ m_{x_1 x_2} $ etc where $x \neq x_1 \neq x_2 ...$
This way if we are given a sum $m_{xy} = a_1 m_{xx_1} + a_2 m_{x_1x_2} +a_3 m_{x_2 x_3} + ... + a_{n}m_{x_{n-1} y}$ we have $a_2 = ... = a_{n-1}=a$, and $a_1 = a_n = 1$, because $m_{xy}(x) = 1$ and so $a=1$.
Also we can get rid of repetitions like $m_{x_1x_2}, \ m_{x_1x_4}$ by noting that $m_{x_1y_4} = m_{x_1x_2} + m_{x_2 x_3} + m_{x_3x_4}$
Is this a good approach? Could you help me a bit?