Problem: Find all functions $f:[0,2]\to (0,1]$ such that $f'(0)$ exists and $f$ satisfies $$ \forall x \in [0,1],\quad f(2x)=2f(x)^2 -1. $$
My attempt: Motivated from the identity $\cos 2\theta = 2\cos^2 \theta -1,$ let us define $g(x)=\cos^{-1} f(x)$ for $x\in [0, 2].$ We can show from the given functional equation that $g(x)=2g(x/2)$ holds for every $x\in [0, 2].$ On the other hand, we find that $f(0)=1$ and $f'(0)=0.$ Now, if we somehow show that $g'(0)$ exists, then letting $n\to\infty$ in the identity $$ \forall n\ge 1,\quad \frac{g(x)}{x}=\frac{ g(x/2^n)}{x/2^n} $$ would give us $g(x) = g'(0) x$ for every $x.$ In other words, $f(x) = \cos kx$ for some constant $k.$ Note that the range is given to be $(0,1],$ which restricts $k$ to lie in the interval $(-\frac{\pi}{4}, \frac{\pi}{4}).$
Now my question is, how to show that $g'(0)$ exists? The usual chain rule does not apply here, because $\cos^{-1}$ does not have a finite derivative at $f(0)=1$.
There are more such functions than you think.
See, the differentiability of $f$ at $0$ has very weak implications on $g$. Like OscarRascal suggested in the comments, we may define $g$ arbitrarily on $(1,2]$, then extend it everywhere else using the equation (2) and $g(0)=0$, then go back to $f$. Such a function, however, would not necessarily satisfy the differentiability condition. Now what we put just one really weak limitation on $g$?
We will define $g$ on $(1,2]$ as an arbitrary bounded function. It can still be very bad in all other regards (like nowhere continuous). Then we extend it using (2). It follows that $g(x)\over x$ is also bounded by some constant $C$, and moreover, remains bounded by the same constant as we extend the function's domain all the way to $0$. Now we complete the construction with $g(0)=0$ and $f(x)=\cos g(x)$. The equation (1) is obviously satisfied. Now what about the derivative? $$|f'(0)|=\left|\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}\right|=\left|\lim\limits_{x\to0}\dfrac{\cos g(x)-1}{x}\right|\leqslant\left|\lim\limits_{x\to0}\dfrac{g(x)^2}{2x}\right|\leqslant\left|\lim\limits_{x\to0}\dfrac{(Cx)^2}{2x}\right|=0$$
So $f'(0)$ does exist, and moreover, is known. Except that, little good can be said of $f$. It can be non-differentiable (and, for that matter, non-continuous) anywhere else. The choice remains pretty wide.
So it goes.