How to solve the functional equation $f(2x) = (e^x+1)f(x)$?

Question

I need to solve $f(2x)=(e^x+1)f(x)$. I am thinking about Frobenius type method: $$\sum_{k=0}^{\infty}2^ka_kx^k=\left(1+\sum_{m=0}^{\infty}\frac{x^m}{m!}\right)\sum_{n=0}^{\infty}a_nx^n\\ \sum_{k=0}^{\infty}(2^k-1)a_kx^k=\left(\sum_{m=0}^{\infty}\frac{x^m}{m!}\right)\left(\sum_{n=0}^{\infty}a_nx^n\right)=\sum_{m=0}^{\infty}\left(\frac{x^m}{m!}\sum_{n=0}^{\infty}a_nx^n\right)=\sum_{m=0}^{\infty}\left(\sum_{n=0}^{\infty}\frac{a_nx^{m+n}}{m!}\right)$$ So I think $$(2^k-1)a_k=\sum_{m=0}^k\frac{a_{k-m}}{m!}.$$ Now I think about a recurrence relation: $$(2^k-1)r^k=\sum_{m=0}^k\frac{r^{k-m}}{m!}.$$ Can somebody help me? Firstly, is my Frobenius technique right? Useful? Is there another easier way? Recurrence relation?

Answer 1

As suggested by @Shivang in comments, substituting $g(x)=f(x)/(e^x-1)$ gives: $$f(2x)=(e^x+1)f(x)\implies g(2x)=g(x)$$ From a known process one can then prove that g is constant and thus f is $a(e^x-1)$ where a is a constant.

Answer 2

What about $f(x)=D(x)(e^x-1)?$

But the additional condition:

There exist $\lim\limits_{x\to 0} \frac{f(x)}{x}$

will delete such counterexamples

Answer 3

$\because$ the trivial solution is $f(x)=e^x-1$

$\therefore$ the general solution is $f(x)=\Theta(\log_2x)(e^x-1)$ , where $\Theta(x)$ is an arbitrary periodic function with unit period