I'm trying to show that there exists only one solution to the problem :
$$ f\in C([0,1]) $$
$$ s.t :f(x)=x+\frac{xf(\sqrt{x})+(1-x)f(x^2)}{2} $$
Now I understand I need to use the contraction mapping theorem, since $C([0,1])$ is a complete metric space, all I need to show is that the mapping $T:C([0,1])\rightarrow C([0,1])$ defined by :
$$ T(f)(x)=x+\frac{xf(\sqrt{x})+(1-x)f(x^2)}{2} $$
is contractive. That is for every $f,g\in C([0,1])$ I show :
$$\left\Vert T\left(f\right)-T\left(g\right)\right\Vert _{\infty} =\left\Vert x+\frac{xf(\sqrt{x})+(1-x)f(x^{2})}{2}-\left(x+\frac{xg(\sqrt{x})+(1-x)g(x^{2})}{2}\right)\right\Vert _{\infty} =\left\Vert \frac{xf(\sqrt{x})+(1-x)f(x^{2})}{2}-\frac{xg(\sqrt{x})+(1-x)g(x^{2})}{2}\right\Vert _{\infty} =\left\Vert \frac{xh(\sqrt{x})+(1-x)h(x^{2})}{2}\right\Vert _{\infty}$$
When $h(x)=f(x)-g(x)$. My intuition tells me that : $$\left\Vert \frac{xh(\sqrt{x})+(1-x)h(x^{2})}{2}\right\Vert _{\infty}\leq\left\Vert \frac{h(x)}{2}\right\Vert _{\infty}$$ But I'm not able to prove this formally.
$|h(\sqrt x)|\leq \|h\|_{\infty}$ and $|h(x^{2})|\leq \|h\|_{\infty}$. Since $xh(\sqrt x)+(1-x)h(x^2)$ is a convex combination of these two we get $|xh(\sqrt x)+(1-x)h(x^2)|\leq \|h\|_{\infty}$ for every $x$.