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I have the following question: Let $f$ be an analytic function satisfying the functional equation: $f(z)=u(z)f(a-z)$ where $a$ is a real constant. Let $g$ be another function satisfying the same functional equation. In this case I asking if $f=g$. Thank you in advance.

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NO, a=0, u(z)=1. – Marc Palm Oct 22 '12 at 14:10
Thank you very much. – China-Hong Kong Oct 22 '12 at 14:39
A more profound example is $u(z)=1,$ but $a=2\pi.$ In that case, both $\sin$ and $\cos$ satisfy the equation. – Igor Rivin Oct 22 '12 at 15:41
You must say, WHERE is your function analytic, to obtain a non-trivial answer. – Alexandre Eremenko Oct 22 '12 at 16:21
Note that your equation is linear in $f$, so if there is one nonzero solution there is at least a one-parameter family of them. – Robert Israel Oct 22 '12 at 18:51
up vote 2 down vote accepted

In the solution by Robert Israel, the uniqueness question (which was asked) is not addressed.

Of course one cannot conclude that $f=g$, because if $f$ satisfies the equation, and $p$ is any function such that $p(z)=p(a-z)$ then $fp$ satisfies the equation. And this is the general description of all solutions because if $f$ and $g$ satisfy the equation, then their ratio has the property that $p(z)=p(a-z)$.

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See my comment to the question. $0$ is always a solution. As soon as you have a nonzero solution, you have nonuniqueness. If there are no nonzero solutions, the solution $0$ is unique. – Robert Israel Oct 22 '12 at 22:52

A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = u(a/2) = \pm 1$: if it is $1$, then $f(z) = \sqrt{u(z)}$ is a solution. If it is $-1$, then $f(z) = (z - a/2) \sqrt{u(z)}$ is a solution.

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Thank you very much. – China-Hong Kong Oct 24 '12 at 6:55

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