Solution to an ODE with every $n$-moment finite. Let $f$ be a differentiable function from $[0, \infty]$ to $\mathbb{R}$ satisfying the following:
$$
f'(x)=f(2x)-f(x),
$$
$$
M_n=\int_0^\infty x^n f(x) \ dx < \infty.
$$
Show that there exists a non null function that satisfies the hypothesis and then enumerate all the sequences $(a_n)_{n \in \mathbb{N}}$ such that $a_n=M_n, \ \forall n \in \mathbb{N}$. 
My approach was to search a function of the type 
$$
f(x) = \sum_n a_n  e^{b_n x}
$$
with $a_n \ne 0, b_n <0$, since such a function converges to a $\mathcal{C}^1$ function as the series converges uniformly. However doing some computation I imagine that $b_n$ as to be something like $-2^n$, but I do not have many arguments to state that formally, let us say that I only hope it has to be something like that since no polynomial satisfies the hypothesis.
Any suggestions? And moreover in such a problem how do I establish a candidate to such condition instead of strongly hoping to have at least a function that is very similar to its derivative? 
 A: Here are some thoughts. 
No nonconstant analytic solutions of the ODE can exist. If $f(x)=\sum_{n=0}^\infty a_nx^n$ is a solution, then $na^n=(2^{n−1}−1)a_{n−1}$ which implies that if $a_{n−1}=0$ then $a_n=0$. Since $a_1=f′(0)=f(0)−f(0)=0$, then $a_n=0$ for all $n \geq 1$ and so constant functions are the only analytic solutions of the ODE. So if a nonzero solution of the problem exists, it must be $C^\infty$ smooth but not analytic. 
However, the second part of the problem can be done as follows. Suppose we have found $f(x)$ satisfying the conditions. In particular, $x^n f(x) \rightarrow 0$ as $x\rightarrow \infty$ for all $n\geq 0$. Integration by parts gives
$$\int_0^\infty x^{n+1}f'(x)dx=x^{n+1}f(x)|_0^\infty -\int_0^\infty (n+1)x^nf(x)dx.$$
$$\int_0^\infty x^{n+1}f'(x)dx=\int_0^\infty x^{n+1}f(2x)dx -M_{n+1}=2^{-n}\int_0^\infty u^{n+1}f(u)du-M_{n+1}=(2^{-n}-1)M_{n+1}.$$
Substituting this into the first equation gives
$$(2^{-n}-1)M_{n+1}=-(n+1)M_n \Rightarrow M_{n+1}=\frac{n+1}{1-2^{-n}}M_n.$$
by induction, we have
$$M_n=\frac{n!M_1}{(1-2^{-n-1})\ldots(1-2^{-1})}=\alpha_n M_1.$$
If $f(x)$ is a solution of the ODE, then $kf(x)$ is also a solution and $M_1(kf)=kM_1(f)$. So the sequence $(M_1,M_2,\ldots)$ is a real multiple of the sequence $(\alpha_1,\alpha_2,\ldots)$, hence unique up to a multiplicative constant. 
A: I will suggest a solution to the problem that does not require any prior knowledge about the correct functional form.
First take the Laplace transform of the equation, defining $\tilde{f}(s)=\int^{\infty}_0f(t)e^{-st}dt$, and also using some properties of the Laplace transform we end up with the equation:
$$(s+1)\tilde{f}(s)=\tilde{f}(\frac{s}{2})+f(0)$$
Notice that we can solve for $\tilde{f}(s)$ and we could iterate the above equation, halving s at every iteration:
$$\tilde{f}(s)=f(0)\sum\limits^{N}_{i=1}\prod\limits^{i}_{k=0}\frac{1}{1+\frac{s}{2^k}}+f\big(\frac{s}{2^N}\big)\prod\limits^{N}_{k=1}\frac{1}{1+\frac{s}{2^k}}$$
Then we could take the limit $N\rightarrow \infty$ here and obtain the general solution in terms of its Laplace transform and with $f(0)$ being an arbitrary real value:
$$\tilde{f}(s)=f(0)\Big(\sum\limits^{\infty}_{i=1}\prod\limits^{i}_{k=0}\frac{1}{1+\frac{s}{2^k}}+\prod\limits^{\infty}_{k=1}\frac{1}{1+\frac{s}{2^k}}\Big)$$
In principle we would need to invert the Laplace transform to obtain the solution, but this looks like an arduous task. Let us instead make the observation that the structure of the LT is fairly simple: it has poles at $s=-2^k, k\in\mathbb{N}$ which upon inversion will produce terms of the form $e^{-2^kx}$. Therefore we can try the ansatz:
$$f(x)=\sum\limits^{\infty}_{n=0}a_n e^{-2^nx}$$
Upon plugging in to the original equation we obtain:
$$a_n= \frac{1}{1-2^n}{a_{n-1}}, n>0$$
or equivalently:
$$a_n=(-1)^na_0\prod\limits^{n}_{k=1}\frac{1}{2^k-1}$$
It can be readily shown that the series converges for any $x>0$ since $\lim\limits_{n\rightarrow \infty}|\frac{a_{n+1}}{a_n}|=0, x>0$ and therefore f is truly the solution to the equation. Finally we compute the moments of this function which are finite thanks to the quick convergence of the coefficients (which can be easily verified again by taking the ratio test):
$$M_k=\frac{1}{k!}\sum\limits^{\infty}_{k=0}a_n (2^{k+1})^n$$
