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Alexandre Eremenko
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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 $p(z)=p(a-z)$.

Alexandre Eremenko
  • 91.8k
  • 9
  • 259
  • 429