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Let $v\not= 1$ be a real number. Let $f(s)$ be real analytic on an open interval containing $v$ and $1$, with a zero of order $m\ge 1$ at $s=1$.

My question is: Can we solve this functional equation with respect to $f$:

$$f(sv)=\frac{(s-1)f'(s)}{f(s)}$$ where $s$ is in the proximity of $s=1$.

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    $\begingroup$ Something is obviously wrong or imprecise with this statement of the problem. If $v\ne1$, then we define $f(s)$ arbitrarily in a sufficiently small neighborhood of $1$ (the only requirements being that it is real analytic and has zero of order $m$) and then use your equation to define it in a neighborhood of $v$. So long as these neighborhoods have an empty intersection, we are done. This is if you require $f$ to be real analytic only near $1$ (and hence near $v$). If, however, you wish $f$ to be real analytic on $R$, then why require equation to hold only near $1$? It then holds everywhere. $\endgroup$
    – Vladimir
    Commented Jul 21, 2014 at 14:12
  • $\begingroup$ @Vladimir: Yes, Christian Remling is right. This is exactely my question. $\endgroup$
    – Safwane
    Commented Jul 21, 2014 at 20:39
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    $\begingroup$ These comments refer to an earlier version; I've edited, in order to clarify. $\endgroup$
    – Christian Remling
    Commented Jul 22, 2014 at 0:42

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