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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?

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2 Answers

up vote 11 down vote accepted

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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Didn't Robert just say that? –  Fly by Night Aug 28 '12 at 22:52
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@FlybyNight: I was trying to solve the problem by myself too, since I just took my first Complex Analysis class last semester, and really wanted to answer the question . If it is in breach of M.SE etiquette to answer a question a little late using strikingly similar methods, I would willingly delete my reply. I meant no offense. Please let me know. –  Ravi Donepudi Aug 28 '12 at 23:00
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@FortuonPaendrag: Don't delete your answer, it is on topic and correct. Quite often two people will write out similar answers, and post them at around the same time. I see no problem with this. –  Eric Naslund Aug 29 '12 at 3:49
    
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. –  Fly by Night Aug 29 '12 at 19:16
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