# Another functional inequality

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

• You might note that all nondecreasing nonnegative functions on $[0,1]$ satisfy the inequality. – Robert Israel May 19 '19 at 4:05
• 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. – Gianfranco OLDANI May 19 '19 at 16:13
• 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 – Pietro Majer May 21 '19 at 21:27