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Find all monotonic functions $f: R^+ \rightarrow R^+$ such that :

$f(x+y) \le f(x) + f(y)$

$f(0)=0$

Background: This exercise was created by me asking myself if, given a distance function $d_1(x,y)$, a function $d_2(x,y)=f(d_1(x,y))$ could still be a proper distance for some function $f$. If such a function $f$ exists than $d_2$ would be a proper distance because:

$d_2(x,z)=f(d_1(x,z))$

$\le f(d_1(x,y)+d_1(y,z))$ ( monotonicity of f+ triangular for $d_1$)

$\le f(d_1(x,y))+f(d_1(y,z))$ (property of f)

$=d_2(x,y)+d_2(y,z)$

Thomas
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    Quick tip. The usual term in English for an increasing (or decreasing) function is monotonic rather than monotonous. Monotonous means dull or repetitive. Of course the function might be that as well ;-) – WA Don May 21 '20 at 16:43
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    The first property is called “subadditivity”. – All increasing concave functions with $f(0)=0$ are subadditive. But the converse need not hold, see for example https://math.stackexchange.com/q/2911357/42969. – Martin R May 21 '20 at 16:43
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    Also $x \mapsto f(x)/x$ decreasing implies subadditivity, but not vice versa, as the example $f(x) = 2x + |\sin(x)|$ from https://mathoverflow.net/a/35673/116247 shows. – Martin R May 21 '20 at 16:59
  • Thanks! The relation between concavity and subadittivity is nice and I had not noticed it. I had noticed that the square root respects the inequality, which falls into this category. This means than that every concave positive function of a distance is still a distance, if I am not wrong. – Thomas May 21 '20 at 17:33
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    A related question: https://math.stackexchange.com/q/4543183/42969 – Martin R Nov 27 '22 at 06:51
  • Thanks @MartinR it is nice to see that also some professionals asked themelves the same question :) – Thomas Nov 27 '22 at 16:57
  • And it is admirable that after two years still you linked the questions! – Thomas Nov 27 '22 at 16:59

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