Picard’s Theorem + Hadamard’s Theorem = ?

July 31, 2009

Picard’s Theorem + Hadamard’s Theorem = ?

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 2:51

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

$\displaystyle f\left( {\frac{1} {2}} \right) = \underbrace {\frac{1} {2}}_a$

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.

Comments (5)

5 Comments »

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

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

Reply
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

Reply
- 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
  
  Reply
- 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
  
  Reply
  - inspite of that I can’t show a=1/2
    
    Comment by RUHUL ALI KHAN — May 4, 2016 @ 15:57

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Ngô Quốc Anh

July 31, 2009