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

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  1. Test
    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. Test
    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. Samuel
    November 23, 2011 at 3:05 am

    This was an amazingly elegant solution.

  4. Samuel
    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. Samuel
    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 🙂

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