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Is there some general solution to the functional inequality:

$$ f(xy) \leq y f(x) + x f(y)$$

Where $x,y\in[0,1]$?

I can find many particular solutions but I just wonder if there is a more general description of functions f satisfying such functional inequality. I have the following conditions

1) $f : [0,1]\rightarrow \mathbb R$

2) $f$ is non-negative on $[0,1]$

3) $f(0) = 0$ and $f(1)$ is positive and finite, let say normalised to $1$

4) $f$ is one time continuously differentiable on $(0,1)$

Thanks for any suggestions

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    $\begingroup$ You might note that all nondecreasing nonnegative functions on $[0,1]$ satisfy the inequality. $\endgroup$ – Robert Israel May 19 at 4:05
  • $\begingroup$ Thanks a lot for your answers. I am very sorry I did a really stupid mistake in specifying the inequality. Now it is the right one I am interested in. $\endgroup$ – Gianfranco OLDANI May 19 at 16:13
  • $\begingroup$ Take a subadditive function $g$, then the funcion $f(x):=xg(\log x)$ satisfies your inequality. Conversely, if $f$ verifies your inequality, $g(x):=e^{-x}f(e^x)$ is subadditive. en.wikipedia.org/wiki/Subadditivity#Definitions $\endgroup$ – Pietro Majer May 21 at 21:27
  • $\begingroup$ Great very interesting, thanks Pietro. $\endgroup$ – Gianfranco OLDANI May 22 at 20:11
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Robert gave me an idea with his comment "...nondecreasing nonnegative functions....".
In fact if I divide my original inequality by the product x*y my inequality become:
f(xy) <= f(x) + f(y) and then a general answer to my question can be:

Take a non decreasing nonnegative function g(x). Then the function f(x) = xg(x) satisfies the inequality f(xy) <= xf(y) + yf(x)

Thanks for your help Gianfranco

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