Putnam 2005, Problem B-3

Find all differentiable functions $f : (0, \infty) \to (0, \infty)$ for which there is a positive real number $a$ such that $f'\left( \frac{a}{x} \right) = \frac{x}{f(x)}$ for all $x > 0$.

Solution:
First solution: The functions are precisely \(f(x) = cx^d\) for \(c,d > 0\) arbitrary except that we must take \(c = 1\) in case \(d = 1\). To see that these work, note that \(f'(a/x) = dc(a/x)^{d-1}\) and \(x/f(x) = 1/(cx^{d-1})\), so the given equation holds if and only if \(dc^2 a^{d-1} = 1\). If \(d \neq 1\), we may solve for \(a\) no matter what \(c\) is; if \(d = 1\), we must have \(c = 1\). (Thanks to Brad Rodgers for pointing out the \(d = 1\) restriction.)

To check that these are all solutions, put \(b = \log(a)\) and \(y = \log(a/x)\); rewrite the given equation as
\[
f(e^{b - y}) f'(e^y) = e^{b - y}.
\]
Put
\[
g(y) = \log f(e^y);
\]
then the given equation rewrites as
\[
g(b - y) + \log g'(y) + g(y) - y = b - y,
\]
or
\[
\log g'(y) = b - g(y) - g(b - y).
\]
By the symmetry of the right side, we have \(g'(b - y) = g'(y)\). Hence the function \(g(y) + g(b - y)\) has zero derivative and so is constant, as then is \(g'(y)\). From this we deduce that \(f(x) = cx^d\) for some \(c,d\), both necessarily positive since \(f'(x) > 0\) for all \(x\).

Second solution: (suggested by several people) Substitute \(a/x\) for \(x\) in the given equation:
\[
f'(x) = \frac{a}{x f(a/x)}.
\]
Differentiate:
\[
f''(x) = -\frac{a}{x^2 f(a/x)} + \frac{a^2 f'(a/x)}{x^3 f(a/x)^2}.
\]
Now substitute to eliminate evaluations at \(a/x\):
\[
f''(x) = -\frac{f'(x)}{x} + \frac{f'(x)^2}{f(x)}.
\]
Clear denominators:
\[
x f(x) f''(x) + f(x) f'(x) = x f'(x)^2.
\]
Divide through by \(f(x)^2\) and rearrange:
\[
0 = \frac{f'(x)}{f(x)} + \frac{x f''(x)}{f(x)} - \frac{x f'(x)^2}{f(x)^2}.
\]
The right side is the derivative of \(x f'(x)/f(x)\), so that quantity is constant. That is, for some \(d\),
\[
\frac{f'(x)}{f(x)} = \frac{d}{x}.
\]
Integrating yields \(f(x) = cx^d\), as desired.