Ngô Quốc Anh

July 31, 2009

Picard’s Theorem + Hadamard’s Theorem = ?

Question. Let $f$ be an entire non-constant function that satisfies the functional equation $\displaystyle f(1 - z) = 1 - f(z)$

for all $z \in \mathbb C$. Show that $f(\mathbb C) = \mathbb C$.

Solution. Assume by contradiction, then W.L.O.G. by the Picard’s Theorem we can assume that $f$ misses $a \in \mathbb C$. By the Hadamard’s Theorem, $\displaystyle f(z)-a = e^{p(z)}$

for some polynomial $p$. Therefore, $\displaystyle f(z) = a +e^{p(z)}$

for all $z \in \mathbb C$. From the fact that $\displaystyle f(1-z)=1-f(z)$

we get $\displaystyle \underbrace{a+{e^{p\left({1-z}\right)}}}_{f\left({1-z}\right)}=\underbrace{1-\left({a+{e^{p\left( z\right)}}}\right)}_{1-f\left( z\right)}$

which yields $\displaystyle {e^{p\left( z \right)}} = 2a - 1 + {e^{p\left( {1 - z} \right)}}$.

Put $z=0$ and $z=1$, we obtain $\displaystyle {e^{p\left( 0 \right)}} = 2a - 1 + {e^{p\left( 1 \right)}}, \quad {e^{p\left( 1 \right)}} = 2a - 1 + {e^{p\left( 0 \right)}}$.

Hence $\displaystyle {e^{p\left( 0 \right)}} = 2\left( {2a - 1} \right) + {e^{p\left( 0 \right)}}$

which implies $a=\frac{1}{2}$. From the identity $f(1-z)=1-f(z)$ put $z=\frac{1}{2}$ we then deduce that $\displaystyle f\left( {\frac{1} {2}} \right) = \underbrace {\frac{1} {2}}_a$

Note: I think I should post some applications of the Hadamard’s Theorem.

Corrigendum to the previous proof. Assume by contradiction, then $f$ takes all values except possibly some $a$. Also notethat $f(1/2) = 1/2$. So we can assume $a \neq 1/2$ which means that $1-a \ne a$. It then does take the value $1-a$, and hence takes the value $a$. The proof is now complete.

I thank Xu Wei Biao for pointing out a mistake in the previous solution.

1. By the Hadamard’s Theorem, how can I say f(z)-a= e^p(z)????

Comment by scsdxc — May 4, 2016 @ 1:15

2. I think there is a big calculation mistake after the line “which yields” and after calculation I can’t show a=1/2..please correct this mistake

Comment by RUHUL ALI KHAN — May 4, 2016 @ 2:27

• Thank you. To save our time, please let me know the mistake so that we can fix asap :).

Comment by Ngô Quốc Anh — May 4, 2016 @ 2:36

• You are right, there was missing a minus sign in front of $e^{p(z)}$ :(.

Comment by Ngô Quốc Anh — May 4, 2016 @ 2:43

• inspite of that I can’t show a=1/2

Comment by RUHUL ALI KHAN — May 4, 2016 @ 15:57

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