# Eigenfunctions of an infinite summation operator

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?)

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)
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.
• Alexandre, we have $\hat{g}(z).(\lambda + a_1+\sum_{2} a_n e^{i z c_n})=0$ but how can this trig sum can be zero? I though they were linearly independant, so no solution? Am I wrong? Thanks for your help. – Bertrand Dec 11 '14 at 8:24
• The expression in parenthesis has infinitely many zeros. So $\hat{g}$ is a linear combination of delta-functions sitting at those zeros. The inverse transform $g$ of the linear combination of delta functions is a trig sum. – Alexandre Eremenko Dec 11 '14 at 12:56