Putnam 2010, Problem B-5

Is there a strictly increasing function \( f : \mathbb{R} \to \mathbb{R} \) such that \( f'(x) = f(f(x)) \) for all \( x \)?

Solution:
**First solution.** The answer is no. Suppose otherwise. For the condition to make sense, $f$ must be differentiable. Since $f$ is strictly increasing, we must have $f'(x) \geq 0$ for all $x$. Also, the function $f'(x)$ is strictly increasing: if $y > x$ then $f'(y) = f(f(y)) > f(f(x)) = f'(x)$. In particular, $f'(y) > 0$ for all $y \in \mathbb{R}$.

For any $x_0 \geq -1$, if $f(x_0) = b$ and $f'(x_0) = a > 0$, then $f'(x) > a$ for $x > x_0$ and thus $f(x) \geq a(x-x_0) + b$ for $x \geq x_0$. Then either $b < x_0$ or $a = f'(x_0) = f(f(x_0)) = f(b) \geq a(b-x_0) + b$. In the latter case, $b \leq a(x_0 + 1)/(a+1) \leq x_0 + 1$. We conclude in either case that $f(x_0) \leq x_0 + 1$ for all $x_0 \geq -1$.

It must then be the case that $f(f(x)) = f'(x) \leq 1$ for all $x$, since otherwise $f(x) > x + 1$ for large $x$. Now by the above reasoning, if $f(0) = b_0$ and $f'(0) = a_0 > 0$, then $f(x) > a_0 x + b_0$ for $x > 0$. Thus for $x > \max\{0, -b_0/a_0\}$, we have $f(x) > 0$ and $f(f(x)) > a_0 x + b_0$. But then $f(f(x)) > 1$ for sufficiently large $x$, a contradiction.

**Second solution.** (Communicated by Catalin Zara.) Suppose such a function exists. Since $f$ is strictly increasing and differentiable, so is $f \circ f = f'$. In particular, $f$ is twice differentiable; also, $f''(x) = f'(f(x))f'(x)$ is the product of two strictly increasing nonnegative functions, so it is also strictly increasing and nonnegative. In particular, we can choose $\alpha > 0$ and $M \in \mathbb{R}$ such that $f''(x) > 4\alpha$ for all $x \geq M$. Then for all $x \geq M$,
$$f(x) \geq f(M) + f'(M)(x-M) + 2\alpha(x-M)^2.$$

In particular, for some $M' > M$, we have $f(x) \geq \alpha x^2$ for all $x \geq M'$.

Pick $T > 0$ so that $\alpha T^2 > M'$. Then for $x \geq T$, $f(x) > M'$ and so $f'(x) = f(f(x)) \geq \alpha f(x)^2$. Now
$$\frac{1}{f(T)} - \frac{1}{f(2T)} = \int_T^{2T} \frac{f'(t)}{f(t)^2} dt \geq \int_T^{2T} \alpha dt;$$
however, as $T \to \infty$, the left side of this inequality tends to 0 while the right side tends to $+\infty$, a contradiction.

**Third solution.** (Communicated by Noam Elkies.) Since $f$ is strictly increasing, for some $y_0$, we can define the inverse function $g(y)$ of $f$ for $y \geq y_0$. Then $x = g(f(x))$, and we may differentiate to find that $1 = g'(f(x))f'(x) = g'(f(x))f(f(x))$. It follows that $g'(y) = 1/f(y)$ for $y \geq y_0$; since $g$ takes arbitrarily large values, the integral $\int_{y_0}^\infty dy/f(y)$ must diverge. One then gets a contradiction from any reasonable lower bound on $f(y)$ for $y$ large, e.g., the bound $f(x) \geq \alpha x^2$ from the second solution. (One can also start with a linear lower bound $f(x) \geq \beta x$, then use the integral expression for $g$ to deduce that $g(x) \leq \gamma \log x$, which in turn forces $f(x)$ to grow exponentially.)