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$f$ be an entire function such that $|f(z)|\le c|z|^3\forall |z|\ge 3,f(z)=f(iz)\forall z$, my question is : is there such $f$ exists?

$f(z)=a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+a_5z^5+\dots=a_0+a_1iz-a_2z^2-ia_3z^3+a_4z^4+ia_5z^5\dots$

$|g(z)|=|{f(z)\over z^3}|\le c\forall |z|\ge 3$ so by Liuvilles $f(z)=kz^3\forall |z|\ge 3$ also $kz^3=-ki z^3$ so $k=0$, so $f(z)=0\forall |z|\ge 3$? so $f\equiv 0$? am I wrong anywhere?

Myshkin
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  • This is hard to read. Could you put some more efforts in formatting the question please? :-) – BIS HD Oct 22 '13 at 14:49
  • What does $f(z)\le cz^3$ mean? Unless we know that $f(z)$ and $cz^3$ are both real, there's no natural way to talk about them being ordered. Should it perhaps read $|f(z)|\le c|z|^3$? – Cameron Buie Oct 22 '13 at 14:54
  • ha[][][][][][][[ – Myshkin Oct 22 '13 at 14:57
  • too many $\forall$ symbols, not enough words :( – Tyler Oct 22 '13 at 14:59
  • Your constraint ($\forall |z|\geq 3$) doesn't imply the Liouville result you use; rather, you have the generic $f(z) = a_0+a_1z+a_2z^2+a_3z^3$. Now just equate coefficients of the power series/polynomial. – Steven Stadnicki Oct 22 '13 at 15:00

1 Answers1

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Since you only have the estimate

$$\lvert f(z)\rvert \leqslant c\lvert z\rvert^3\tag{1}$$

for $\lvert z\rvert \geqslant 3$, you can't assume that

$$g(z) = \frac{f(z)}{z^3}$$

is an entire function. It may have a pole in $0$.

However, $(1)$ says, by the Cauchy-estimates, that $f$ is a polynomial of degree at most $3$. Then

$$a_0 + a_1 z + a_2 z^2 + a_3 z^3 \equiv a_0 + a_1 iz - a_2 z^2 - a_3 i z^3$$

tells you that $a_1 = a_2 = a_3 = 0$, so the only functions with these properties are the constant functions.

Daniel Fischer
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