Pseudometrics and non-expansive maps Let $X$ be any set, and $(Y,\rho)$ be a pseudometric space. For a collection of functions $F\subseteq Y^X$ define
$$
  d_{F,\rho}(x',x''):=\sup_{f\in F}\rho(f(x'),f(x'')).
$$
It is easy to show that $(X,d_{F,\rho})$ is always a pseudometric space, and any function $f\in F$ is nonexpansive with respect to $d_{F,\rho}$, that is $F\subseteq N(d_{F,\rho})$. I have several questions:


*

*Is that true that $d_{F,\rho}$ generates the smallest topology under which any $f\in F$ is continuous? In such case, can we say that $d_{F,\rho}$ is "the smallest" pseudometric which makes all $f\in F$ being non-expansive?

*Under which conditions $F$ is dense in $N(d_{F,\rho})$ with respect to the distance
$$
  R(f',f'') := \sup_{x\in X}\rho(f'(x),f''(x)).
$$

*Let $a$ be some pseudometric on $X$. Under which conditions $N(d_{N(a),\rho}) = N(A)$ or more stronger $a = d_{N(a),\rho}$?
 A: Let me start by noting that to make $d_{F,\rho}$ a pseudometric, you must either
drop the usual requirement that pseudometrics are finite, or require that $\rho$
is bounded, or something similar.
While it is clear that $d_{F,\rho}$ is the minimum of all the pseudometrics that
make every function in $F$ non-expansive, the topology it induces is usually
strictly finer than the initial topology of $F$. To see that it is always finer,
it is of course enough to note that nonexpansive  functions are continuous. A
simple example where $d_{F,\rho}$ is the discrete metric is
when $F = \mathcal{C}([0,1], [0,1])$. For every $x \ne y$ there is
a function such that $f(x) = 0$ and $f(y) = 1$, which forces 
$d_{F,\rho}(x, y) \ge 1$, even though the functions are already continuous in
the absolute value metric.
Your second question is a little vague, and no natural conditions spring to
mind.
For your last question: it is always true that $N(d_{N(a),\rho}) = N(a)$.
You have already noted that $F\subseteq N(d_{F,\rho})$, and since 
$$
N(a) = \{f \in Y^X \mid \rho(f(x'), f(x'')) \le a(x', x'') \:\text{for all}\: x', x'' \in X  \}
$$
it follows immediately from the definition that $d_{N(a),\rho} \le a$, which
implies the reverse inclusion.
A necessary and sufficient condition on $(Y, \rho)$ such that
$d_{N(a),\rho} = a$ for every $(X, a)$ is that for every $r > \epsilon > 0$ it contains a path of length $r$ where the distance between the end points is at
least $r - \epsilon$. To see this, observe that the problem can be reduced to the case $(X, a) = ([0, \infty), |\cdot|)$, since for every
$x', x'' \in X$ the function $h_{x'}: X \to [0, \infty)$ defined as 
$h_{x'}(x'') = a(x', x'')$  is a non-expansive function with $|h(x') - h(x'')| = a(x', x'')$
