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Let $f$ be entire and non-constant. Assuming $f$ satisfies the functional equation $f(1-z)=1-f(z)$, can one show that the image of $f$ is $\mathbb{C}$?

The values $f$ takes on the unit disc seems to determine $f$...

Any ideas?

2 Answers 2

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If $f({\mathbb C})$ misses $w$, then it also misses $1-w$. The only case where $w = 1-w$ is $1/2$, which is $f(1/2)$. Now use Picard's "little" theorem.

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Since the function is entire, it misses at most one point by Picard's little theorem. If the point is say, $y$, then $1-y$ must be in the range, unless $y=1-y$. But, in this case, $y=\frac{1}{2}$ and we can deduce that $f(\frac{1}{2})=\frac{1}{2}$. So, $y\neq 1-y$. So, if there is a $z$ such that $f(z)=1-y$, then, $f(1-z)=1-f(z)=y$. So, the range of $f$ is $\mathbb C$.

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    It is possible that they were posted at the same time. But what usually happens, and what tends to annoy me, is that people reply without bothering to read the other answers. I'm not suggesting that this is the case here, and I hope I have not upset Fortuon. I've just become very cynical.2012-08-29