A unique continuous function satisfying $f(x) + \frac{1}{2} \ f(x^2) = x$ This is a question from a real analysis qual:

Show that there exists a unique continuous function $f$ on $[0, 1]$
satisfying the equation $f(x) + \frac{1}{2} \  f(x^2) = x$ and that this function is strictly increasing on $[0, 1]$.

So, we have $f(0)=0$. I'm not really sure how to solve this problem but I feel that it's related to  Banach fixed point theorem.
Any possible solutions would be greatly appreciated.
 A: To show existence and uniqueness you can use indeed Banach fixed point theorem. Let $X= C^0([0;1])$ be the space of continuous functions on $[0;1]$ with the usual max norm.
You can define $T : X \to X$
$$Tf(x)=x- \frac{1}{2}f(x^2)$$
A fixed point of $T$ would be a continuous function satisfying $f=Tf$, or in other words
$$f(x)=x- \frac{1}{2}f(x^2)$$
which solves the functional equation. You have to check that $T$ is well defined, and $X$ is a complete metric space.
Then we have to show that $T$ is a contraction.
$$||Tf-Tg||= \max_{x \in [0;1]} \left|x- \frac{1}{2}f(x^2)- x + \frac{1}{2}g(x^2) \right| = \frac{1}{2} \max_{x \in [0;1]} |f(x^2) - g(x^2)| = \frac{1}{2}||f-g||$$
Since $\frac{1}{2}<1$ this is a contraction.
For the part where it's askeed to show that $f$ is strictly increasing, I tried some stuff but found no solution.
A: To see $f$ is strictly increasing, first observe that
$$Tg(x)=x-g(x^2)/2$$
preserves the closed set $L \subset C[0,1]$ of nondecreasing functions that are Lipschitz-1, i.e.,
$$L=\{g \in C[0,1] : \forall y<x, \quad 0 \le g(x)-g(y) \le x-y\}\,,$$
since for $g \in L$ and $y<x$ in $[0,1]$,
$$  0 \le g(x^2)-g(y^2) \le x^2-y^2 \le 2(x-y) \,.$$
Thus the fixed point $f$ of $T$ must be in $L$, so for $y<x$ in $[0,1]$ we have
$$f(x)-f(y)= x-y-\!\frac{f(x^2)-f(y^2)}{2} \ge x-y-\frac{x^2-y^2}{2}=(x-y)\cdot\frac{2-x-y}{2}$$
A: To show $f$ is strictly increasing, note $f(x)=x-1/2(x^2-1/2(x^4-\dots)) = x -x^2/2 + x^4/4 - x^8/8 + \dots$ on $[0,1)$ and this series is exact since $f(0)=0, f$ is continuous, and $x^{2^n} \to 0.$
We have $f'(x) = 1-x+x^3-x^7+\dots$ on $[0,1),$ and we can see this is positive by the argument in the proof of the Alternating Series test.
