Solving functional differential equation $f'(x)=2f(2x)-f(x)$ 
Show that there is at least a nonzero function $f$, differentiable on $[0,+\infty)$, satisfying $$f'(x)=2f(2x)-f(x) \qquad \forall x>0 $$
  $$M_n:=\int_{0}^{\infty}x^nf(x)dx<\infty \qquad \forall n\in
 \mathbb{N} $$

My best idea so far is to assume that the solution is a power series, i.e. 
$$ f(x)=\sum_{n=0}^{\infty}a_nx^n\qquad \forall x>0$$
Then the equation becomes
$$ \sum_{n=0}^{\infty}na_nx^{n-1}=2\sum_{n=0}^{\infty}a_n2^nx^n-\sum_{n=0}^{\infty}a_nx^n$$
equating all the coefficients of the same degree I get
$$na_n=(2^{n}-1)a_{n-1}\qquad \forall n\geq 1$$
So setting $a_0=1$, I get
$$a_{n}=\frac{1}{n!}\prod_{k=1}^{n}(2^k-1) \qquad \forall n$$
But does the power series actually converge? Using Hadamard's formula, and that $2^{k}-1\geq 2^{k-1}$,
$$ |a_n|^{1/n}\geq\frac{1}{(n!)^{1/n}}\left[2^{n(n-1)/2}\right]^{1/n}\sim\frac{e}{n(2\pi n)^{1/2n}}2^{(n-1)/2}\to \infty$$
so the radius of converge of the series is $0$, so it doesn't actually define a solution on $[0,+\infty)$.
 A: Well, it looks like your best idea (which by itself was not bad at all) fails precisely for the reasons you so comprehensively described. So what shall we do?
Guess what?
We turn it the other way around. Let's look for a function in the form of
$$f(x)=\sum_{n=0}^\infty a_ne^{-2^nx},\quad x\geqslant0$$
You might wonder what's with $2^n\cdot x$ up there in the exponent. Well, it's simple: I was going to write $e^{-nx}$ when I realized we won't be needing most of those terms, so let it be this way.
Now, the equation becomes
$$-\sum_{n=0}^{\infty}2^na_ne^{-2^nx}=2\sum_{n=0}^{\infty}a_ne^{-2^{n+1}x}-\sum_{n=0}^{\infty}a_ne^{-2^nx}$$
or
$$\sum_{n=0}^{\infty}2^na_ne^{-2^nx} = -2\sum_{n=1}^{\infty}a_{n-1}e^{-2^nx}+\sum_{n=0}^{\infty}a_ne^{-2^nx}$$
which gives
$$(2^n-1)a_n=-2a_{n-1},\quad n\geqslant1$$
So continuing in your footsteps, I set $a_0=1$ and get
$$a_n=\frac{2^n}{\prod_{k=1}^{n}(2^k-1)}, \quad n\geqslant1$$
which makes my series converge pretty fast, for the same reason why your series fails to do so.
Now that's our solution.
