The function $f:\mathbb{R} \to \mathbb{R}$ is such that $f(x+y) \geq f(x) + yf(f(x))$ for all $x,y \in \mathbb{R}$. Prove that: a) $f(f(x)) \leq 0$ for all $x \in \mathbb{R}$; b) if $f(0) \geq 0$, then $f(x) = 0$ for all $x\in \mathbb{R}$
Problem
Source: Bulgaria EGMO TST 2015 Day 1 Problem 3 (out of 3)
Tags: Functional inequality, substitutions, limit, function, inequalities
laikhanhhoang_3011
04.02.2023 13:42
a)
$\bullet$ $$f(x+1)\overset{P(x;1)}{\geq}f(x)+f(f(x))\overset{P\left ( x+1;f(x)-\left ( x+1 \right ) \right )}{\geq}f(x)+f(x+1)+\left [ f(x)-\left ( x+1 \right ) \right ].f(f(x+1)), (*)$$
$\because \, $ If $\exists x_0: f(f(x_0))>0,$ $P(x_0;x-x_0):$ $f(x) \geq ax+b$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ ,where $a=f(f(x_0))>0; b=x_0-f(f(x_0)).$
$$\Rightarrow f(f(x)) \geq a.f(x)+b \geq a^2.x+ab+b$$$\rightarrow \exists x_1: f(f(x_1))>2+\frac{1}{2}.$
$\bullet$ $ P(x_1;x-x_1):$ $f(x)\geq (x-x_1).f(f(x_1)) +f(x_1)$ $$\rightarrow f(x) \geq 2x, \forall x \geq max\left\{ 2x_1f(f(x_1))-2f(x_1);0 \right\}=a.$$$\bullet$
$$(*) \Rightarrow 0 \geq f(x)+\left [ f(x)-\left ( x+1 \right ) \right ].f(f(x+1)) \geq 2x+(x-1).2.2.(x+1), \forall x \geq a,$$adsurb.
b)
$\bullet$ $f(x+y)\geq f(x)+y.f(f(x)) \geq f(x), \, \forall \, y \leq 0 \rightarrow$ $f$ is decreasing.
$\bullet$ $f(0) \geq 0 \geq f(f(x)) \rightarrow f(x) \geq 0, \forall x. \rightarrow 0 \geq f(f(x)) \geq 0 \rightarrow f(f(x))=0, \forall x.$
$$\rightarrow f(x+y) \geq f(x), \forall x, (1).$$$\bullet$ $\,$ In $(1),$ $0 =f(f(x)) \overset{\left (f(x)-y; \, y \right )}{\geq} f(x)\overset{\left ( f(x); \, x-f(x) \right )}{\geq} f(f(x))=0.$
So, $$f(x)=0, \, \forall x.$$
Q.E.D
Assassino9931
05.02.2023 22:36
Additionally, for RMM 2018 TST the following problem was given: if $f$ satisfies the given inequality and is continous, then $f(x) \leq 0$ for all $x\geq -1$. One firstly has to think of and prove part a) of the EGMO TST problem and then do as follows: $f(x+y) \geq f(x) + yf(f(x)) \geq f(x)$ for all $y<0$, so $f$ is decreasing and $\lim_{t\to \infty} f(t) = \infty$. Hence by the definition of a limit there is a real number bigger than all $t$ with $f(t) > 0$ - let $a$ be the smallest such. With $x<a$ and $y=f(x) - x$ we get $f(f(x))(1+x-f(x)) \geq f(x) > 0$ and so $f(x) > x+1$. Taking $x\to a$ and using the continuity of $f$ yields $a\leq -1$, as desired.