I would like to find ALL eigenfunctions to the operator, for $f$ a real function on R+\*:


$f \rightarrow \sum\_{1}^{\infty} f(nx)$


So to find $f$ such that: $\sum\_{1}^{\infty} f(nx) = \lambda f(x)$


It is obvious that $f(x)=x^{a}$ is a solution but are they other?


May be you can advise me a reference on this subject ? 


Same question if the operator is : 


$f \rightarrow \sum\_{1}^{\infty} a\_n f(nx)$


(what are the condition on the $a\_i$ to have more than the obvious solution mentionned above?)



Eigenfunctions of an infinite summation operator
You do not state any conditions on the functions on which your operator is defined (it involves an infinite series which must converge). So my answer will also be purely formal.


Setting $g(t)=f(e^t)$, we obtain the equation
$$\sum\_n a\_n g(t+c\_n)=\lambda g(t),$$
where $c\_n=\log n$. This is a linear equation and such equations are solved with Fourier transform. Or Laplace transform whichever is more appropriate after you choose your space of functions. See, for example,
[On equation f(z+1)-f(z)=f'(z)](https://mathoverflow.net/questions/114875/on-equation-fz1-fz-fz/114878#114878)


You will get plenty of eigenfunctions. Indeed, taking FT, gives
$$(\sum\_n a\_n e^{ic\_nz}-\lambda)\hat{g}(z)=0.$$
The expression in parentheses is an entire function which has infinitely many zeros. For every such zero $z\_k$, we have a solution $\hat{g}(z)=\delta(z-z\_k)$. The inverse Fourier transform of delta is an exponential. So the general solution is an exponential sum.


Once we know that, we can forget about Fourier, and just plug $g(t)=e^{pt}$ to the equation and determine $p$, as we do with linear ODE with constant coefficients in our undergraduate classes.


