Putnam 2009, Problem B-5

Let $f : (1, \infty) \to \mathbb{R}$ be a differentiable function such that
\[
f'(x) = \frac{x^2 - f(x)^2}{x^2 (f(x)^2 + 1)} \quad \text{for all } x > 1.
\]
Prove that $\lim_{x \to \infty} f(x) = \infty$.

Solution:
**First solution.** If \( f(x) \geq x \) for all \( x > 1 \), then the desired conclusion clearly holds. We may thus assume hereafter that there exists \( x_0 > 1 \) for which \( f(x_0) < x_0 \).

Rewrite the original differential equation as
\[
f'(x) = 1 - \frac{x^2 + 1}{x^2} \frac{f(x)^2}{1 + f(x)^2}.
\]
Put \( c_0 = \min\{0, f(x_0) - 1/x_0\} \). For all \( x \geq x_0 \), we have \( f'(x) > -1/x^2 \) and so
\[
f(x) \geq f(x_0) - \int_{x_0}^x \frac{dt}{t^2} > c_0.
\]
In the other direction, we claim that \( f(x) < x \) for all \( x \geq x_0 \). To see this, suppose the contrary; then by continuity, there is a least \( x \geq x_0 \) for which \( f(x) \geq x \), and this least value satisfies \( f(x) = x \). However, this forces \( f'(x) = 0 < 1 \) and so \( f(x - \varepsilon) > x - \varepsilon \) for \( \varepsilon > 0 \) small, contradicting the choice of \( x \).

Put \( x_1 = \max\{x_0, -c_0\} \). For \( x \geq x_1 \), we have \( |f(x)| < x \) and so \( f'(x) > 0 \). In particular, the limit \( \lim_{x \to +\infty} f(x) = L \) exists.

Suppose that \( L < +\infty \); then \( \lim_{x \to +\infty} f'(x) = 1/(1 + L^2) > 0 \). Hence for any sufficiently small \( \varepsilon > 0 \), we can choose \( x_2 \geq x_1 \) so that \( f'(x) \geq \varepsilon \) for \( x \geq x_2 \). But then \( f(x) \geq f(x_2) + \varepsilon(x - x_2) \), which contradicts \( L < +\infty \). Hence \( L = +\infty \), as desired.

**Variant.** (by Leonid Shteyman) One obtains a similar argument by writing
\[
f'(x) = \frac{1}{1 + f(x)^2} - \frac{f(x)^2}{x^2(1 + f(x)^2)},
\]
so that
\[
-\frac{1}{x^2} \leq f'(x) - \frac{1}{1 + f(x)^2} \leq 0.
\]
Hence \( f'(x) - 1/(1 + f(x)^2) \) tends to 0 as \( x \to +\infty \), so \( f(x) \) is bounded below, and tends to \( +\infty \) if and only if the improper integral \( \int dx/(1 + f(x)^2) \) diverges. However, if the integral were to converge, then as \( x \to +\infty \) we would have \( 1/(1 + f(x)^2) \to 0 \); however, since \( f \) is bounded below, this again forces \( f(x) \to +\infty \).

**Second solution.** (by Catalin Zara) The function \( g(x) = f(x) + x \) satisfies the differential equation
\[
g'(x) = 1 + \frac{1 - (g(x)/x - 1)^2}{1 + x^2(g(x)/x - 1)^2}.
\]
This implies that \( g'(x) > 0 \) for all \( x > 1 \), so the limit \( L_1 = \lim_{x \to +\infty} g(x) \) exists. In addition, we cannot have \( L_1 < +\infty \), or else we would have \( \lim_{x \to +\infty} g'(x) = 0 \) whereas the differential equation forces this limit to be 1. Hence \( g(x) \to +\infty \) as \( x \to +\infty \).

Similarly, the function \( h(x) = -f(x) + x \) satisfies the differential equation
\[
h'(x) = 1 - \frac{1 - (h(x)/x - 1)^2}{1 + x^2(h(x)/x - 1)^2}.
\]
This implies that \( h'(x) \geq 0 \) for all \( x \), so the limit \( L_2 = \lim_{x \to +\infty} h(x) \) exists. In addition, we cannot have \( L_2 < +\infty \), or else we would have \( \lim_{x \to +\infty} h'(x) = 0 \) whereas the differential equation forces this limit to be 1. Hence \( h(x) \to +\infty \) as \( x \to +\infty \).

For some \( x_1 > 1 \), we must have \( g(x), h(x) > 0 \) for all \( x \geq x_1 \). For \( x \geq x_1 \), we have \( |f(x)| < x \) and hence \( f'(x) > 0 \), so the limit \( L = \lim_{x \to +\infty} f(x) \) exists. Once again, we cannot have \( L < +\infty \), or else we would have \( \lim_{x \to +\infty} f'(x) = 0 \) whereas the original differential equation (e.g., in the form given in the first solution) forces this limit to be \( 1/(1 + L^2) > 0 \). Hence \( f(x) \to +\infty \) as \( x \to +\infty \), as desired.

**Third solution.** (by Noam Elkies) Consider the function \( g(x) = f(x) + \frac{1}{3}f(x)^3 \), for which
\[
g'(x) = f'(x)(1 + f(x)^2) = 1 - \frac{f(x)^2}{x^2}
\]
for \( x > 1 \). Since evidently \( g'(x) < 1 \), \( g(x) - x \) is bounded above for \( x \) large. As in the first solution, \( f(x) \) is bounded below for \( x \) large, so \( \frac{1}{3}f(x)^3 - x \) is bounded above by some \( c > 0 \). For \( x \geq c \), we obtain \( f(x) \leq (6x)^{1/3} \).

Since \( f(x)/x \to 0 \) as \( x \to +\infty \), \( g'(x) \to 1 \) and so \( g(x)/x \to 1 \). Since \( g(x) \) tends to \( +\infty \), so does \( f(x) \). (With a tiny bit of extra work, one shows that in fact \( f(x)/(3x)^{1/3} \to 1 \) as \( x \to +\infty \).)