Putnam 2012, Problem A-3

Let $f : [-1,1] \to \mathbb{R}$ be a continuous function such that  
(i) $f(x) = \frac{2 - x^2}{2} f\left( \frac{x^2}{2 - x^2} \right)$ for every $x$ in $[-1,1]$,  
(ii) $f(0) = 1$, and  
(iii) $\lim_{x \to 1^-} \frac{f(x)}{\sqrt{1 - x}}$ exists and is finite.  
Prove that $f$ is unique, and express $f(x)$ in closed form.

Solution:
We will prove that $f(x) = \sqrt{1-x^2}$ for all $x \in [-1,1]$. Define $g : (-1,1) \to \mathbb{R}$ by $g(x) = f(x)/\sqrt{1-x^2}$. Plugging $f(x) = g(x)\sqrt{1-x^2}$ into equation (i) and simplifying yields
$$g(x) = g\left(\frac{x^2}{2-x^2}\right)$$
for all $x \in (-1,1)$. Now fix $x \in (-1,1)$ and define a sequence $\{a_n\}_{n=1}^{\infty}$ by $a_1 = x$ and $a_{n+1} = \frac{a_n^2}{2-a_n^2}$. Then $a_n \in (-1,1)$ and thus $|a_{n+1}| \leq |a_n|^2$ for all $n$. It follows that $\{|a_n|\}$ is a decreasing sequence with $|a_n| \leq |x|^n$ for all $n$, and so $\lim_{n \to \infty} a_n = 0$. Since $g(a_n) = g(x)$ for all $n$ by (1) and $g$ is continuous at $0$, we conclude that $g(x) = g(0) = f(0) = 1$. This holds for all $x \in (-1,1)$ and thus for $x = \pm 1$ as well by continuity. The result follows.

**Remark.** As pointed out by Noam Elkies, condition (iii) is unnecessary. However, one can use it to derive a slightly different solution by running the recursion in the opposite direction.