Here is my guess.

Let $f$ be a entire function From $\mathbb C$ to $\mathbb C$

(1) If for $f$, there exist a sequence $z_n$ such that $|z_{n}|\rightarrow \infty$, with $|f(z_{n})|\rightarrow\infty$, Then $f$ must be polynomial.

(2) If for all $z_n$ such that $|z_{n}|\rightarrow \infty$, $|f(z_{n})|\rightarrow z<\infty$, then $f$ is bounded, so that $f$ is constant by Liouville's theorem.

Is this true?

  • $\begingroup$ Am I misunderstanding (2)? I think that, as you state it, the hypothesis is satisfied by all entire functions. $|f(z_n)|<\infty$ is true for every $z_n\in \mathbb C$ if $f$ is entire. "Bounded" doesn't mean "$|f(z)|<\infty$". It means there is an $R$ with "$|f(z)|<R$" for for all $z$ in the domain of $f$. $\endgroup$ – MPW Dec 31 '15 at 5:34
  • $\begingroup$ @MPW you are right, sorry to confuse you. I'm not familar with '$' things. My first try. I'll fix it. $\endgroup$ – nicksohn Dec 31 '15 at 5:58

(1) is not correct. Let $f(z)=e^z$. Consider sequence $1,2,3,\cdots$. It satisfies the condition. But $f$ is not polynomial. [For $f$ to be polynomial, the condition should be stronger; it should hold for all the possible sequences].

(2) Seems to be correct with your reason.

  • 1
    $\begingroup$ (1) is somewhat worse than you make it out to be here; for any non-constant function, such a sequence $z_n$ exists. $\endgroup$ – Milo Brandt Dec 31 '15 at 4:51
  • $\begingroup$ yes; just negation of his (2); provided $f$ is entire. [Are you saying that for any non-constant, even non-entire function, such a sequence exists]? $\endgroup$ – p Groups Dec 31 '15 at 4:52
  • $\begingroup$ Thanks a lot! so, is it impossible to find all entire function? $\endgroup$ – nicksohn Dec 31 '15 at 4:56
  • 1
    $\begingroup$ @nicksohn What does "find all entire functions" mean? What would you consider a successful classification? $\endgroup$ – Noah Schweber Dec 31 '15 at 6:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.