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Find all functions $f:\mathbb{R}_+ \to \mathbb{R}_+$ (not necessarily continues function) where
$\mathbb{R}_+ = ${$r \in \mathbb{R} : r \geq 0$}, such that $$f(x+y) \geq f(x) + y f(f(x)) \quad\forall x,y \in \mathbb{R}_+$$

I tend to believe that the only solution will be the trivial one ($f = 0 $).

My ideas so far: for $x \neq y: \frac{f(x+y)-f(x)}{y} \geq f(f(x)) \geq 0$ so in particular $f$ is weak monotonous. also if $f$ is bounded then it must be the zero function, because by now $f$ is monotonic and bounded there for converges and since $0 \leq f(f(x)) \leq \frac{f(x+y)-f(x)}{y} \to 0 $ where x tends to 0 we get that the monotonic $x \to f(f(x))$ function converges to 0 in infinity and there for $f(f(x)) = 0$ from which we easly deduce that $f=0$. we still need to deal with the unbounded case.

thanks ahead

Dr. Mathva
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ned grekerzberg
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  • It's not true that $ f \circ f \equiv 0 $ implies $ f \equiv 0 $, and it's not true that the constant zero function is the only solution. For a counterexample, let $ f ( x ) = 0 $ for $ 0 \le x \le 1 $ and $ f ( x ) = 1 $ for $ x > 1 $. – Mohsen Shahriari Mar 14 '19 at 17:25

1 Answers1

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From the given inequality, $f$ is increasing. Now, if $f(x)\geq x+1$ for some $x$, then from the given inequality for $y=1$, $$f(x+1)\geq f(x)+f(f(x))\geq f(x)+f(x+1),$$ which implies that $f(x)\leq 0$, a contradiction. Therefore $f(x)<x+1$ for all $x>0$. Hence, for all $x,y>0$, $$f(f(x))\leq\frac{f(x)}{y}+f(f(x))\leq\frac{f(x+y)}{y}\leq\frac{x+y+1}{y},$$ and letting $y\to\infty$ we obtain that $f(f(x))\leq 1$ for all $x>0$.

Now, if $f$ is not bounded, then there exists $x_1>0$ such that $f(x_1)>2$. Then, there exists $x_2$ such that $f(x_2)>x_1$, and since $f$ is increasing, we obtain that $$f(f(x_2))\geq f(x_1)>2,$$ which is a contradiction with $f(f(x)\leq 1$ for all $x$. Hence $f$ is bounded, and we can then proceed as in your answer.

detnvvp
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