Functions

Note that for any \(x \in \mathbb{R}\), we have \[ f(x + 1)^2 \leq f(x) , \] which shows that \(f(x) \geq 0\). To show that \(f(x) \leq 1\) for any \(x \in \mathbb{R}\), let us establish the following claim.

For any \(x, y \in \mathbb{R}\) with \(x > y\), and for any positive integer \(n\), the inequality \[ f(x)^{2^n} \leq f(y) \] holds.

Proof of the claim. Note that the inequality holds for \(n = 1\) by hypothesis. Let us assume that \(n\) is a positive integer such that the inequality holds for any \(x, y \in \mathbb{R}\) with \(x > y\). Then, for any \(x, y \in \mathbb{R}\) with \(x > y\), we have \[ x > \frac{x + y}{2} > y , \] which implies that \[ f(x)^{2^n} \leq f\left(\frac{x + y}{2}\right) \quad \text{and} \quad f\left(\frac{x + y}{2}\right)^{2} \leq f(y) . \] Combining these two inequalities and using that \(f(x)\) is nonnegative, we obtain \[ f(x)^{2^{n + 1}} \leq f(y) . \] By the principle of mathematical induction, the claim follows.

Let \(x\) be an arbitrary real number. Using the above claim, we obtain \[ f(x)^{2^n} \leq f(x - 1) \] for any positive integer \(n\). If \(f(x) \leq 1\), then we are done. It remains to consider the case that \(f(x) > 1\), which we assume from now on. It follows that \[ f(x - 1) \geq 1 + 2^n \left(f(x) - 1\right) \] for any positive integer \(n\). This is impossible since \(f(x) - 1\) is positive. Combining the above, we conclude that \[ 0 \leq f(x) \leq 1 \] holds for any \(x \in \mathbb{R}\).