Find all differentiable functions with $f'(0)=1$, such that they satisfy the following functional equation:
$$ f(2x) = 2f^2(x) -4f(x) + 3$$
So, here's what I tried:
And now, I'm stuck because I have no clue how to simplify the final limit. Plus, I think there must be a more general method to solve functional equations of the kind $f(nx) = g(f(x))$ where $g(x)$ is a polynomial of degree $n$, or at least the case $f(2x) = quadratic in f(x)$.
So, what can I do ahead of this?
First note that differentiating the relation $$ f(2x) = 2f(x)^2 - 4f(x)+3 $$ at $x=0$ we obtain $$ 2f'(0) = 4f'(0)f(0) - 4f'(0) $$ and since $f'(0)=1$ this gives $f(0)=3/2$.
Instead of your choice of $g$, take $$ g(x) = 2(f(x)-1), $$ and thus $g(0)=1$. Then you obtain the functional equation $$ g(2x) = g(x)^2, $$ where $g'(0)=2$. Now, we have that $$ g(x) = g\left( \frac{x}{2} \right)^2 \geq 0 $$ for all $x\in\mathbb{R}$. If $g(x_0)=0$ for some $x_0\in\mathbb{R}$, then by induction we can prove that $g(2^{-n}x_0)=0$ for all positive integer $n$, so taking the limit as $n\to \infty$ we obtain that $g(0)=0$ (note that $g$ is continuous at $0$ since it is differentiable at $0$), but this contradicts $g(0)=1$. Thus we have that $g(x)>0$ for all $x\in\mathbb{R}$.
Now, define $$ h(x) = \log(g(x)). $$ Here $\log$ is the natural logarithm. Then you obtain the functional equation $$ h(2x) = 2h(x) $$ with $h(0)=0$ and $h$ differentiable at $0$. This equation is well known and is solved for example here. Note that by the first answer to that question, you only need the differentiability at $0$, not on $\mathbb{R}$.
Thus by the answer to the above mentioned question, we have that $h(x)=ax$ for some $a\in\mathbb{R}$. Thus $g(x)=e^{ax}$ and hence $$ f(x) = 1 + \frac{e^{ax}}{2}. $$ From the condition $f'(0)=1$ you can show that $a=2$ so $$ f(x) = 1 + \frac{e^{2x}}{2} $$ is the only solution to the given functional equation.
