The easy(?) part of IMO 2011 Problem 3

Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$.

How can I prove that $f(0) = 0$?

First, we show that $f(x) \leq 0$ for all $x$.

Suppose that $f(z) > 0$ for some $z$. The functional inequality implies that $$\lim_{x \to -\infty} f(x) = -\infty,$$ since $f$ is bounded by a strictly increasing linear function. We also have $$ f(y) = f(0+y) \leq f(0)y + f(f(0)). $$ Using this for $y = f(x)$, $$ f(0) = f(x-x) \leq -xf(x) + f(f(x)) \leq (f(0) - x)f(x) + f(f(0)).$$ As $x \to -\infty$, the right-hand side tends to $-\infty$, leading to a contradiction.

Second, let $x > 0$. Note that $$ f(0) \leq x f(-x) + f(f(-x)) \leq xf(-x). $$ Thus $f(-x) \geq f(0)/x$. As $x\to\infty$, the righthand side tends to zero. Since $$ f(-x) = f(-x+0) \leq f(f(-x)), $$ we get that there is a sequence of points $x_n = -f(-n)$ tending to zero such that $f(-x_n) \to 0$.

Suppose that $f(f(0)) < 0$. Then $$f(-x_n) = f(0-x_n) \leq -f(0)x_n + f(f(0)).$$ Since $x_n\to 0$, the righthand side tends to $f(f(0)) < 0$, contradiction.

We conclude that $f(f(0)) = 0$. This implies that $$0 = f(f(0)) \leq f(f(f(0))) = f(0), $$ and so $f(0) = 0$.

Third, as noted above, for $x > 0$ we have $$xf(-x) \geq f(0) = 0.$$ Thus $f(-x) = 0$ for all $x \geq 0$.

Addendum: Putting $g(x) = -f(x)$, the functional inequality reduces to $$ g(x+y) \geq yg(x), $$ where now $g\colon \mathbb{R}_+ \to \mathbb{R}_+$. An example is the function $g(x) = \exp(x)$.

This is my solution: $$f(f(x)) = f(y+f(x)-y) \le (f(x)-y).f(y) + f(f(y)),\forall x, y (1)$$ swap $x, y$, we have: $$f(f(y))\le (f(y)-x)f(x)+f(f(x)), \forall x, y (2)$$ (1), (2) $\Rightarrow 0 \le 2f(x)f(y)-xf(x)-yf(y), \forall x, y.$

$\Rightarrow -xf(x) \ge (y-2f(x))f(y), \forall x, y \Rightarrow -xf(x)\ge 0, \forall x (*)$ (using $y=2f(x)$)

In the other hand: $f(y)=f(x+y-x)\le (y-x)f(x)+f(f(x)), \forall x, y$. Suppose that there exist $x: f(x)>0$, then $\lim\limits_{y\to-\infty}(y-x)f(x)=-\infty \Rightarrow \lim\limits_{y\to-\infty}f(y)=-\infty \Rightarrow \lim\limits_{x\to-\infty}(-xf(x))=-\infty$! (absurd from $(*)$).

Therefore $f(x) \le 0, \forall x (**)$

From $(*), (**) \Rightarrow\forall x<0: -xf(x)\ge 0 \Rightarrow f(x)\ge 0 \Rightarrow f(x)=0.$

The last one: $0=f(-1)\le f(f(-1))=f(0)\Rightarrow f(0)=0.$ [End of proof]

koreagerman.