IMO 2011 Q3

f: \mathbb {R} \to \mathbb {R} satisfies f(x+y) \leq yf(x)+f(f(x)) for all real numbers x and y. Show that f(x)=0 for all x \leq 0.

First, set y=0 gives us inequality f(f(x)) \geq f(x) for all x\in \mathbb R.

Let the inequality be denoted by P(x,y).

 P(x, f(y)-x) gives us f(f(y)) \leq (f(y)-x)f(x) + f(f (x)) for all x,\,y \in \mathbb R. Similarly, we have f(f(x)) \leq (f(x)-y)f(y) + f(f(y)), by switching x and y.

Adding the two inequalities up, we get 2f(x)f(y) \geq xf(x) + yf(y),\, \forall x,\,y \in \mathbb R. Let y=2f(x) in this inequality, we see that xf(x) \leq 0 \, \forall x \in \mathbb R.

If x<0, we see that from xf(x) \geq 0, we get either f(x)=0 or f(x) >0. We now try to rule out the latter case, by observing that replace x in the inequality xf(x) \leq 0 with f(x) and we get f(x)f(f(x)) \leq 0. This demands f(f(x)) to be non-positive.

Now, since f(x) is positive, f(f(x)) is non-positive, we have f(f(x)) < f(x) for that certain x<0, which contradicts our initial inequality f(f(x)) \geq f(x). Thus this cannot happen and f(x)=0,\, \forall x<0.

It remains to show f(0)=0. Now, we re-use the inequality f(f(x)) \geq f(x) by letting x be any negative number, which gives off f(x)=0 and thus f(0) \geq 0.

P(0,y) gives f(y) \leq yf(0) + f(f(0)). Assume now that f(0) > 0, and let y \to - \infty, which requires yf(0)+ f(f(0)) \to - \infty since f(0) is positive, and thus we see for y<0,\, |y| large enough, f(y) <0, which is impossible as now y is negative and f(y)=0. Contradiction. Thus f(0)>0 has been wrong which combined with f(0) \geq 0 gives f(0)=0.


Comment: The functional inequality does not lead to simple non-trivial solution, and we thus hope to exploit the “badly-behavedness” of the condition by reducing it to something simple yet non-trivial. (In this problem’s case, the inequaity xf(x) \leq 0.) Since the condition is an inequality, it is not easy to derive an equality out of it. In this solution we established equality from both sides, to thus force the equality to be true.

The motivation for the substitution y \to f(y)-x is to yield a f(f(y)) term on one side, being symmetric to the f(f(x)) term on the other side. Since on the other side y only appears once and appears outside f, we can afford to substitute a relatively complicated form of y. Once we reach 2f(x)f(y) \geq xf(x) + yf(y), substituting y \to 2f(x) is simply giving up the strength of that inequality while preserving its non-triviality. It should be noted that substituting x=y leads to weaker inequality f(x)^2 \geq xf(x).

The difficulty of the problem lies in the difficulty in dealing with the iterated application of f, given that it is not easy to find a non-trivial solution to this equation–under the pressure of IMO and the anxiety of the windmill problem. (I personally do not know of any elementary non-trivial solution to the inequality, which nonetheless makes it a great problem.)  Also, functional inequality is much less common than functional equations. I reckon that if it appeared on the IMO as Q2 instead with the windmill problem being Q3, it would be better done–people would likely spend more time on it. And it proves itself not as challenging as the windmill problem after all!

This solution is probably very close to the official solution.

The problem is proposed by Igor Voronovich, Belarus, and appeared in IMO 2011 as Q3.


One thought on “IMO 2011 Q3

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s