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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?)

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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)

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.

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  • $\begingroup$ Thanks! I think it answers to the question. But note there is not plenty of eigenfunctions, as if we use Fourier, for example, we transform the g(t+c) in a product and then we find there is no solution accepting Fourier transform as we have a product egal to zero (as the sum starts at n=1) - (if sum starts at n=0 there is no solution as at infinity g tends to 0 on one side and not on other side) $\endgroup$
    – Bertrand
    Commented Dec 10, 2014 at 15:59
  • $\begingroup$ As I said, it depends on your space of functions. I recommend to work with Schwartz's tempered distributions. In this case you always have infinitely many eigenfunctions (trig sums). Their Fourier transforms are combinations of delta-functions. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 10, 2014 at 19:20
  • $\begingroup$ 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. $\endgroup$
    – Bertrand
    Commented Dec 11, 2014 at 8:24
  • $\begingroup$ 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. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 11, 2014 at 12:56

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