Non-decreasing functions satisfying functional inequalities $f(1+ax)\leq a f(1+x)$ and $f(xy)\leq f(x)+f(y)$

Question

Can we exactly determine a class of non-decreasing functions defined on the set of non-negative real numbers satisfying: $$f(1+ax)\leq a f(1+x)$$ and $$f(xy)\leq f(x)+f(y)$$ for any $x,y\in [0,\infty)$, assuming that $a\in[0,1)$ can vary? Par example, log fulfills second inequality, but not the first one... Looking for any kind of input or advice. (or book to look for some results on this topic)

Answer 1

We establish the following claim.

CLAIM: Let $f:\mathbb R_{\ge 0}\rightarrow\mathbb R$ be a function. Then $f$ satisfies all of the following conditions if and only if $f(x)=0$ for all $x\ge 0$.

1. The function $f$ is non-decreasing.
2. For all $a\in[0,1)$ we have $f(1+ax)\le af(1+x)$.
3. for all $x,y\ge 0$ we have $f(xy)\le f(x)+f(y)$.

First, notice that the all-zero function satisfies all three conditions. Conversely, assume that all three conditions hold for $f$. By the second condition for $a=0$ we have $f(1)\le 0$, so by the first condition we have $f(x)\le 0$ for $x\le 1$. Assume that $f(0)<0$, then we get a contradiction with the third condition for $x=y=0$, so we have $f(0)=0$. By the first condition we have $f(x)=0$ for $0\le x\le 1$.

Also, by the first condition we have $f(x)\ge 0$ for $x>1$. Now, assume there exists $v>0$ such that $f(1+v)>0$. By the third condition for $x=y=1+v$ we have $f((1+v)^2)\le 2f(1+v)$. But for $x=(1+v)^2-1$ and $a=v/x$ we have $$af(1+x)\le \frac{v}{2v+v^2}\cdot 2f(1+v)<f(1+v)=f(1+ax).$$ This violates the first condition, so for all $v>0$ we have $f(1+v)=0$, which completes the proof.