Let $(M,d)$ be a metric space and consider the new distance
$$\delta (x,y) = \sup \{ |f(x) - f(y)| \mid f \in \operatorname {Lip}(\le 1) \}$$
where $\operatorname {Lip} (\le L)$ is the space of Lipschitz functions of Lipschitz constant at most $L$.
If $f \in \operatorname {Lip} (\le 1)$ then $|f(x) - f(y)| \le d(x,y)$, whence $\delta (x,y) \le d(x,y)$.
Is the opposite inequality $d(x,y) \le \delta (x,y)$ true, in order for the two distances to coincide? If not, are they metrically equivalent (i.e. can we find $A,B > 0$ such that $Ad \le \delta \le Bd$)?
Since we always have
$$ |d(x,z) - d(y,z)| \leq d(x,y) \quad (\forall \ x,y,z \in M) $$
The function $f_z(x) = d(x,z)$ is in $\operatorname{Lip}(\leq 1)$, for any $z \in M$. Now, in particular,
$$ \delta(x,y) \geq |f_x(x) - f_x(y)| = |d(x,x) - d(x,y)| = d(x,y) $$