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)}}.


\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

a contradiction.

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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