Stuck with a tricky existence proof Show that there exists a continuous function $f: [-1, 1] \rightarrow \mathbb{R}$ such
$f(0) = 1$ and
$f(x) = \frac{2-x^2}{2} \cdot f(\frac{x^2}{2-x^2})$   
$\forall x \in [-1, 1]$
I tried putting in $x = 1$ and $x = -1$ in the second condition to find that $f(1) = f(-1) = 0$. 
I also took the derivative of the second equation to find that:
$f'(x) = x (f'(\frac{x^2}{2-x^2})\frac{2}{2-x^2}-f(\frac{x^2}{2-x^2}))$
This gives me
$f'(0) = f'(1) = f'(-1) = 0$ but now I'm stuck. Anybody see a way?
 A: I recall solving this problem before... was it on a Putnam? Sadly, I don't remember what leap of intuition led me to the answer.
One may take $f(x) = \sqrt{1-x^2}$. It is clear that $f$ is continuous on $[-1,1]$ and that $f(0) = 1$, and one sees that
$$\frac{2-x^2}{2} \cdot \sqrt{1-\frac{x^4}{(2-x^2)^2}} = \frac{2-x^2}{2} \cdot \sqrt{\frac{x^4-4x^2+4-x^4}{x^4-4x^2+4}} \\
= \frac{2-x^2}{2} \cdot \sqrt{\frac{4-4x^2}{(2-x^2)^2}} = \sqrt{1-x^2},$$
so that $f(x) = \frac{2-x^2}{2} f(\frac{x^2}{2-x^2})$.
A: Let's find the most general solution.
Let continuous function $f : [-1,1] \to \mathbb R$ satisfy
$$
f(x) = \frac{2-x^2}{2} \cdot f\left(\frac{x^2}{2-x^2}\right)\qquad x \in [-1,1].
\tag{$1$}$$
and $f(1)=1$.  Observe $f$ is even.  Define $G : (0,1] \to \mathbb R$ by
$$
G(y) := \frac{1}{y}\cdot f\left(\sqrt{1-y^2}\right)
$$
Put $x=\sqrt{1-y^2}$ into $(1)$ to get
$$
y G(y) = yG\left(\frac{2y}{y^2+1}\right) .
\tag{$2$}$$
Now if $\phi(y) = 2y/(y^2+1)$, observe that the iterates $\phi^{[n]}(y)$ converge to $1$ as $n \to \infty$ for any $y \in (0,1]$.  From (2) we have
$$
G(y) = G(\phi(y)) = G(\phi^{[2]}(y)) = \dots,
$$
so that
$$
G(y) = \lim_{n \to \infty} G(\phi^{[n]}(y)) = G(1) = f(0) = 1
$$
for all $y \in (0,1]$.  Thus $f(x) = \sqrt{1-x^2}$ for all $x \in (-1,1)$, and thus for all $x \in [-1,1]$.
