Home > Algebra, Problem Solving > IMO 2011 Problem 3

## 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 $y$. Prove that $f(x) = 0$ for all $x \leq 0$.

IMO 2011 Problem 3

Solution: First, suppose there exists $c$ with $f(c)>0$. This means that $f(c+y) \leq y f(c)+f(f(c)),\ \forall y \in \Bbb{R}$. Taking $y \to -\infty$ we get $\lim_{x\to -\infty} f(x)=-\infty$. For some $k \in \Bbb{R}$, take $y=k-x$. Then $f(k) \leq (k-x)f(x)+f(f(x))$. For $x \to -\infty$ we have $f(k) \leq -\infty$. Contradiction. Therefore, for all $x \in \Bbb{R}$ we have $f(x)\leq 0$.

For $y=f(x)-x$ we get $0 \leq f(x)(f(x)-x)$. For $y=0$ we get $f(x)\leq f(f(x))$. Assume that $f(x)<0,\ \forall x \in \Bbb{R}$. Then it follows by the inequality above that $f(x)\leq x,\ \forall x \in \Bbb{R}$ and furthermore $-xf(x) \leq -x^2,\ \forall x< 0$.

From the hypothesys $f(0) \leq -xf(x)+f(f(x)) \leq -xf(x)\leq -x^2,\ \forall x <0$. Contradiction. Therefore there is a value $x_0$ with $f(x_0)=0$. Then $0=f(x_0)\leq f(f(x_0))=f(0)\leq 0$. This proves that $f(0)=0$.

It is easy now to see that $f(0)=0 \leq -xf(x)\leq 0 ,\ \forall x <0$, which means that $f(x)=0,\ \forall x<0$, and we are done.

For more proofs and informations see the discussion in the following link: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=729&t=418798&&start=20

1. October 28, 2011 at 3:31 pm

When $x\to -\infty, -x\to \infty$ and you claimed $f(x)\to -\infty$, so $(k-x)f(x)\to -\infty$. What about $f(f(x))$? Is that possible $f(f(x))\to \infty$ and cancel out $(k-x)f(x)\to- \infty$?

2. October 28, 2011 at 3:34 pm

Well, when $x\to -\infty$, $f(x)\to -\infty$, in turn, $f(f(x))\to -\infty$.

• October 28, 2011 at 4:13 pm

Ok, you’ve figured it out. 🙂

3. November 23, 2011 at 3:05 am

This was an amazingly elegant solution.

4. November 23, 2011 at 4:03 am

Doesn’t this assume a special value for f(0) though? that it can’t be infinite?

• November 23, 2011 at 7:06 am

Why it assumes a special value for $f(0)$? Of course $f(0)$ can’t be infinite, since $f: \Bbb{R} \to \Bbb{R}$.

5. November 23, 2011 at 10:55 am

Thanks! Doy.

Thank you for this very elegant solution. I learnt alot here.

• November 23, 2011 at 10:56 am

I’m glad to hear that 🙂