$d$ is a metric. Under which conditions $f\circ d$ is also a metric? Given a set $\mathscr{M}$ such that $\# \mathscr{M}\geq 2$, we know that there is, at least, an uncountable number of metrics defined on $\mathscr{M}$:
$$\begin{array}{cccc}d_a:&\mathscr{M}\times \mathscr{M}&\longrightarrow &\mathbb{R}\\
& (x,y)&\longmapsto & a\cdot \delta_{x,y}\end{array}$$
with $a\in \mathbb{R}_+$ and $\delta_{x,y}=0$ if $x=y$ and $\delta_{x,y}=1$ otherwise.  
Now, my question is: given a metric $d$ on a set $\mathscr{M}$ is there some sufficient and necessary conditions over a function $f:A\subset \mathbb{R} \longrightarrow \mathbb{R}$ in order to
$$D_f=f\circ d:\mathscr{M}\times \mathscr{M} \longrightarrow \mathbb{R}$$
also be a metric over $\mathscr{M}$?  
I found that


*

*$I\supset d(\mathscr{M}\times \mathscr{M})$;

*$f(0)=0$ and $f(x)> 0,\forall x \in d(\mathscr{M}\times \mathscr{M})-\{0\}$.  


are obviously sufficient and necessary to be possible to compose and to have $D_f(x,y)\geq 0$ with $D_f(x,y)=0\iff x=y$. We don't need to request anything from $f$ in order to gain $D_f(x,y)=D_f(y,x)$, since $d$ is symmetric and $f$ is a function. The trouble arises at the triangular inequality. Obviously,
$$D_f(x,y)\leq D_f(x,z)+D_f(z,y)$$
is a sufficient and necessary condition, but I don't want to suppose anything about $D_f$, only about $f$ itself.  
Well, I found that


*$f$ is monotonically increasing, i.e. $x\leq y \implies f(x)\leq f(y)$, when restricted to $d(\mathscr{M}\times \mathscr{M})$;

*$f(x+y)\leq f(x)+f(y), \forall x,y\in d(\mathscr{M}\times \mathscr{M})$  


are SUFFICIENT conditions for this, but seems far from being necessary (at least, I couldn't show it).  
Note, for example, that given $a>0$ a real number and $f:\mathbb{R}\longrightarrow \mathbb{R}$, $f(x)=\min \{a,x\}$, then $f$ satisfies 1, 2, 3 and 4.  
Note that my question is not duplicate to


*

*What operations is a metric closed under?, since here are not asked about NECESSARY conditions; and

*If $d$ is a metric and $f$ a function when is $d\circ f$ a metric?, since the author of this question supposed $f$ to be decreasing and, furthermore, there is not an answer yet.

 A: My teacher at university was able to figure out that, under some adaptation, 4 is also necessary, as follows.  
We see that if $d$ is any metric on $\mathscr{M}$ then, if $f$ satisfies 1, 2, 3 and 4, we have that $f\circ d$ is also a metric on $\mathscr{M}$.  
Suppose now that $D_f=f\circ d$ is a metric for ANY metric $d$ on ANY set $\scr{M}$. Then, consider the following example:  
$$\mathscr{M}=\mathbb{R}, \hspace{5mm}d(x,y)=|x-y|$$ 
and suppose that there are $x_0,y_0\in d(\mathscr{M}\times \mathscr{M})=\mathbb{R}_+\cup \{0\}$ such that
$$f(x_0+y_0)>f(x_0)+f(y_0).$$
And, in this case, we have
$$f(|x_0-(-y_0)|)>f(|x_0-0|)+f(|0-y_0|)$$
$$\implies D_f(x_0,-y_0)>D_f(x_0,0)+D_f(0,-y_0)$$
and this contradicts the fact that $D_f$ is supposed to be a metric.  
Thus, we must have $f(x+y)\leq f(x)+f(y),\forall x,y \in d(\mathscr{M}\times \mathscr{M})$ ($f$ must be sub-additive in $d(\mathscr{M}\times \mathscr{M})$), if we request $f\circ d$ to be a metric for any $d$.  
This is not a complete converse to my statement, but it is a n important improvement in the direction to build a class of functions $\mathscr{F}$ such that if $f\in \mathscr{F}$ then $f\circ d$ is a metric, for any metric $d$ on any set $\mathscr{M}$.  
After this little adaptation, what do you guys think about the item 3?? We weren't able to find nothing yet...
A: And, curiously (found by my teacher again) there is a simple example that shows that 3 IS NOT necessary (now if we consider again, as in the beginning, $d$ a fixed metric).  
Let $\mathscr{M}$ be $\{0,1,2\}$ and $d:\mathscr{M}\times \mathscr{M} \longrightarrow \mathbb{R}$ be given by $d(x,y)=|x-y|$ and $f:\{0,1,2\}\to \mathbb{R}$ given by
$$f(0)=0, \hspace{5mm} f(1)=2, \hspace{5mm} f(2)=1.$$
$f$ is clearly decreasing, although $D_f=f\circ d$ defines a metric on $\mathscr{M}$ (we verify for example the triangular inequality case by case, since the set is finite).  
Note that, curiously, in this case, $f$ is sub-additive! (Note that this is not necessary, at least not with the proof below where we allow $d$ to be ANY metric, not a fixed one).
In fact, $f(1+1)=f(2)=1<2+2=f(1)=f(1)$, the cases with $0$ are trivial and $1+2\notin d(\mathscr{M}\times \mathscr{M})$, in such a way that there are not $x,y\in d(\mathscr{M}\times \mathscr{M})$ that contradict $f(x+y)\leq f(x)+f(y)$. 
