Existence of a function in $C([0,1], \mathbb{R}), ||.||_{\infty}$ s.t. $2f(x) -f(x)^2 + f(x^2)=g(x)$ Let $g \in E= C([0,1], \mathbb{R}), ||.||_{\infty}$ s.t. $\|g\|_\infty< \frac{1}{4}$. I need to show that there exists $f$ in the same set such that 
$$2f(x) - f(x)^2 + f(x^2)=g(x).$$
Can anyone please provide me with a hint? Should I maybe look into $\Gamma :E \to E$, $\Gamma (f)= 2f(x)-f(x)^2+ f(x^2)$?
 A: Let $X = \{ f\in E: \|f\|_\infty \le C\}$, where $C<\frac 12$ is close to $\frac 12$ such that 
$$\left(C-\frac 12\right)^2 \le \frac 14 - \| g\|_\infty.$$
Consider the mapping 
$$L : X \to X, \ \ Lf (x) = \frac{1}{2} (g(x) - f(x^2) + f(x)^2).$$
First we check that $L$ really maps $X$ to $X$. Note that if $f\in X$, 
$$\begin{split} \|Lf\|_\infty &\le \frac{1}{2}( \| g\|_\infty + C + C^2) = \frac{1}{2} (C^2 - C + \frac 14 - \frac 14 + \|g\|_\infty + 2C)\\
&= \frac{1}{2}( (C-\frac 12)^2 - (\frac 14 - \|g\|_\infty) + 2C)\\
&\le C
\end{split}$$
Thus $Lf \in X$. Now we check that $L$ is a contractoin:
$$\begin{split}
\|Lf_1 - Lf_2\|_\infty &= \left\| \frac{1}{2} \left(f_2(x^2) - f_1(x^2) + f_1(x)^2 - f_2(x)^2 \right)\right\|_\infty\\
&\le \frac 12 \|f_1 - f_2\|_\infty + \frac 12 \left(C+C \right)\|f_1 - f_2\|_\infty \\
&= \frac{1}{2} (1+ 2C) \|f_1 - f_2\|_\infty.
\end{split}$$
Since 
$$\frac{1}{2} (1+ 2C) < \frac{1}{2}(1+ 1)  =1$$
as $C<\frac 12$, $L$ is a contraction. As $X$ is a complete metric space, there is $f\in X$ so that $Lf = f$, by the contraction mapping theorem. But this is the same as 
$$2 f(x) - f(x)^2 + f(x^2) = g(x).$$
