Let $f(x)$ be a continuous function from $[-1,1] \rightarrow \mathbb{R}$ having the following properties: $\boxed{1} f(x)=\dfrac{2-x^2}{2}\cdot f\!\left(\dfrac{2-x^2}{2}\right)$
$\boxed{2} f(0)=1$
$\boxed{3} \lim\limits_{x\to1^{-}}\dfrac{f(x)}{\sqrt{1-x}}$ exists and is finite.
Find $f(x)$.
Let $g(x) = (2 - x^2)/2$, so equation (1) says $f(x) = g(x) f(g(x))$. $g$ has a fixed point $p = \sqrt{3}-1$, and $0 < g(p)< 1$ so $f(p) = 0$. Moreover, for any $x_0 \in (0,1)$, we can show that the iterates $x_n = g(x_{n-1})$ converge to $p$ as $n \to \infty$, with $f(x_n) = f(g(x_{n-1})) =f(x_{n-1})/g(x_{n-1})$. Thus $$f(x_n) = f(x_0)/ \prod_{j=0}^{n-1} g(x_j)$$ and since $\prod_{j=0}^{n-1}g(x_j) \to 0$ while $f(x_n) \to f(p)$ we must have $f(x_0) = 0$. In particular $f(0)=1$ is impossible.
