Putnam 2000, Problem B-4

Let $f(x)$ be a continuous function such that $f(2x^2 - 1) = 2x f(x)$ for all $x$. Show that $f(x) = 0$ for $-1 \leq x \leq 1$.

Solution:
For \(t\) real and not a multiple of \(\pi\), write \(g(t) = \frac{f(\cos t)}{\sin t}\). Then \(g(t + \pi) = g(t)\); furthermore, the given equation implies that  
\[
g(2t) = \frac{f(2\cos^2 t - 1)}{\sin(2t)} = \frac{2(\cos t)f(\cos t)}{\sin(2t)} = g(t).
\]  

In particular, for any integer \(n\) and \(k\), we have  
\[
g(1 + n\pi/2^k) = g(2^k + n\pi) = g(2^k) = g(1).
\]  

Since \(f\) is continuous, \(g\) is continuous where it is defined; but the set \(\{1 + n\pi/2^k \mid n,k \in \mathbb{Z}\}\) is dense in the reals, and so \(g\) must be constant on its domain. Since \(g(-t) = -g(t)\) for all \(t\), we must have \(g(t) = 0\) when \(t\) is not a multiple of \(\pi\). Hence \(f(x) = 0\) for \(x \in (-1,1)\). Finally, setting \(x = 0\) and \(x = 1\) in the given equation yields \(f(-1) = f(1) = 0\).