I have a question to the following problem:
Let $(X,d)$ be a metrical space and let be $f:[0,\infty)\to[0,\infty)$ twice differentiable with $f(0)=0$, $f(x)>0$ for $x>0$, $f'\geq 0$ and $f''\leq 0$
Proof that $f(a+b)\leq f(a)+f(b)$ for $a,b\in\mathbb{R}$ and $a,b\geq 0$
Conclude that
$d_f:X\times X\to \mathbb{R}$
$(x,y)\mapsto f(d(x,y))$
defines a metric space on X.
__
My problem ist to show that $f(a+b)\leq f(a)+f(b)$. Could someone help me with that?
I tried it with the mean value theorem and the inequation for convex functions, but nothing worked for me. :(
Thanks.
The property $f(a+b)\le f(a)+f(b)$ is called subadditivity. The problem is to show that an increasing concave function with $f(0)=0$ is subadditive.
If we rewrite the inequality as $$ f(a+b)-f(a) \le f(b)-f(0) $$ it takes the form of "the function $g(x) = f(x+b)-f(x)$ is decreasing". Which is easy to check by looking at its derivative: $$ g'(x) = f'(x+b)-f'(x) \le 0 $$ since $f''\le 0$.
The above proof does not actually need $f$ to be twice differentiable. Knowing that $f'$ is decreasing is enough.