Solving the functional equation $f(x)=\frac{2-x^2}{2} f(\frac{x^2}{2-x^2})$

Question

Let $f:[-1,1]\to\Bbb R$ be a continuous function such that $$f(x)=\frac{2-x^2}{2} f(\frac{x^2}{2-x^2})$$ for every x in $[-1,1]$, $f(0)=1$, find $f(x)$. Seeing the expression I think that a recurrence relation may be created by some suitable substitution but couldn’t proceed like that. I tried substitution $x\to\frac{x^2}{2-x^2} $ but then expression became worse. Or a trigonometric substitution like $x=\sqrt2$ $cos\theta$ but then LHS becomes unworkable. Please provide an approach to continue.

Answer 1

Using the hint that it 'can' be made into a recurrence relation,$$f(x)=\frac{2-x^2}{2} f\left(\frac{x^2}{2-x^2}\right)$$

Putting $x=1$, we get one initial condition $f(1)=0$.

Substituting trigonometric functions did not help me either so I tried to invert the expression inside $f$ be substituting $x=\frac{2y}{1+y^2}$.

$$f\left(\frac{2y}{1+y^2}\right)=\frac{1+y^4}{y^4+2y^2+1} f\left(\frac{2y^2}{1+y^4}\right)$$

Substituting $h(y)=f\left(\frac{2y}{1+y^2}\right)$, we can notice a pattern;

$$h(y)=\frac{1+y^4}{y^4+2y^2+1}h(y^2)$$

On rearranging,

$$h(y^2)=\frac{y^4+2y^2+1}{y^4+1}h(y)$$

Replace $y^2 \to x$,

$$h(x)=\frac{x^2+2x+1}{x^2+1}h(\sqrt[2]x)$$

I think you can continue from here.

Note - As Gonçalo found, you should be getting $f(x)=\sqrt{1-x^2}$

Answer 2

By trial and error I found the following solution: $$ f(x)=\sqrt{1-x^2}. $$