An IMO Problem Solved

17 Feb 2012

I recently encountered an IMO question that’s stated as follows: Given a function f: \mathbb{R} \rightarrow \mathbb{R} such that for any x, y \in \mathbb{R}, f(x + y) \leq yf(x) + f(f(x)), show that f(x) = 0 for all x \leq 0.

There are two solutions, and I will present the one that AM and I came up with: one can easily verify that f(x) \leq f(f(x)). Let f^2(x) = f(f(x)). We will proceed by showing the following:

1. f(x) \leq f^2(0)
2. f(x) \leq 0
3. f^2(0) = 0

At which point the result should be fairly straight-forward.

To show (1), consider f(f(x)) = f(f(x) + 0) \leq f(x)f(0) + f^2(0). From this, we have that f(x - f(0)) \leq -f(0)f(x) + f^2(x) \leq -f(0)f(x) + f(0)f(x) + f^2(0) = f^2(0). As x is arbitrary, (1) follows.

For (2), let y > 0, and consider f(x - y + y) \leq yf(x - y) + f^2(x + y). As y is positive, we have that yf(x - y) \leq -y^2f(x) + yf(f(x)), so f(x) \leq -y^2f(x) + yf^2(x) + f^2(x + y). Via (1), we have that f(f(x)) \leq f^2(0), f^2(x + y) \leq f^2(0), which in turn implies that f(x) \leq -y^2f(x) + (1 + y)f^2(0).

But this implies (1 + y^2)f(x) \leq (1 + y)f^2(0), which in turn implies that

As y is arbitrary, we have that f(x) \leq 0: this proves (2).

To prove (3), notice that for any x, f(x) = f(x - 1 + 1) \leq f(x - 1) + f(f(x - 1)), and use (1). That is 2f^2(0) \geq f(x) for any x — in particular, x = f(0). Together with (2), we have f^2(0) = 0.

Finally, as f^2(0) \leq f^3(0) = f(0), we have that f(0) = 0; now observe 0 = f(0) = f(x - x) \leq -xf(x) + f(f(x)). The main result follows.