IMC 2012 · Day 1 · P4

Lemma 1. Either $\lim\limits_{t \to +\infty} f(t)$ does not exist or $\lim\limits_{t \to +\infty} f(t) \ne +\infty$.

Proof. Assume that the limit is $+\infty$. Then there exists $T_1 > 0$ such that for all $t > T_1$ we have $f(t) > 2$. There exists $T_2 > 0$ such that $f(t) > T_1$ for all $t > T_2$. Hence, $f'(t) > f(f(t)) > 2$ for $t > T_2$. Hence, there exists $T_3$ such that $f(t) > t$ for $t > T_3$. Then $f'(t) > f(f(t)) > f(t)$, $f'(t)/f(t) > 1$, after integration $\ln f(t) - \ln T_3 > t - T_3$, i.e. $f(t) > T_3 e^{t - T_3}$ for all $t > T_3$. Then $f'(t) > f(f(t)) > T_3 e^{f(t) - T_3}$ and $f'(t) e^{-f(t)} > T_3 e^{-T_3}$. Integrating from $T_3$ to $t$ yields $e^{-f(T_3)} - e^{-f(t)} > (t - T_3) T_3 e^{-T_3}$. The right-hand side tends to infinity, but the left-hand side is bounded from above, a contradiction.

Lemma 2. For all $t > 0$ we have $f(t) < t$.

Proof. By Lemma 1, there are some positive real numbers $t$ with $f(t) < t$. Hence, if the statement is false then there is some $t_0 > 0$ with $f(t_0) = t_0$.

Case I: There exist some value $t \ge t_0$ with $f(t) < t_0$. Let $T = \inf \{ t \ge t_0 : f(t) < t_0 \}$. By the continuity of $f$, $f(T) = t_0$. Then $f'(T) > f(f(T)) = f(t_0) = t_0 > 0$. This implies $f > f(T) = t_0$ in a right neighbourhood, contradicting the definition of $T$.

Case II: $f(t) \ge t_0$ for all $t \ge t_0$. Now we have $f'(t) > f(f(t)) \ge t_0 > 0$. So, $f'$ has a positive lower bound over $(t_0, \infty)$, which contradicts Lemma 1.

Lemma 3. (a) If $f(s_1) > 0$ and $f(s_2) \ge s_1$, then $f(s) > s_1$ for all $s > s_2$.

(b) In particular, if $s_1 \le 0$ and $f(s_1) > 0$, then $f(s) > s_1$ for all $s > s_1$.

Proof. Suppose that there are values $s > s_2$ with $f(s) \le s_1$ and let $S = \inf \{ s > s_2 : f(s) \le s_1 \}$. By the continuity we have $f(S) = s_1$. Similarly to Lemma 2, we have $f'(S) > f(f(S)) = f(s_1) > 0$. If $S > s_2$ then in a left neighbourhood of $S$ we have $f < s_1$, contradicting the definition of $S$. Otherwise, if $S = s_2$ then we have $f > s_1$ in a right neighbourhood of $s_2$, contradiction again.

Part (b) follows if we take $s_2 = s_1$.

With the help of these lemmas the proof goes as follows. Assume for contradiction that there exists some $t_0 > 0$ with $f(f(f(t_0))) > 0$. Let $t_1 = f(t_0)$, $t_2 = f(t_1)$ and $t_3 = f(t_2) > 0$. We show that $0 < t_3 < t_2 < t_1 < t_0$. By lemma 2 it is sufficient to prove that $t_1$ and $t_2$ are positive. If $t_1 < 0$, then $f(t_1) \le 0$ (if $f(t_1) > 0$ then taking $s_1 = t_1$ in Lemma 3(b) yields $f(t_0) > t_1$, contradiction). If $t_1 = 0$ then $f(t_1) \le 0$ by lemma 2 and the continuity of $f$. Hence, if $t_1 \le 0$, then also $t_2 \le 0$. If $t_2 = 0$ then $f(t_2) \le 0$ by lemma 2 and the continuity of $f$ (contradiction, $f(t_2) = t_3 > 0$). If $t_2 < 0$, then by lemma 3(b), $f(t_0) > t_2$, so $t_1 > t_2$. Applying lemma 3(a) we obtain $f(t_1) > t_2$, contradiction. We have proved $0 < t_3 < t_2 < t_1 < t_0$.

By lemma 3(a) ($f(t_1) > 0$, $f(t_0) \ge t_1$) we have $f(t) > t_1$ for all $t > t_0$ and similarly $f(t) > t_2$ for all $t > t_1$. It follows that for $t > t_0$ we have $f'(t) > f(f(t)) > t_2 > 0$. Hence, $\lim\limits_{t \to +\infty} f(t) = +\infty$, which is a contradiction. This contradiction proves that $f(f(f(t))) \le 0$ for all $t > 0$. For $t = 0$ the inequality follows from the continuity of $f$.