6
$\begingroup$

A couple more questions about entire functions that I'm having difficulty with:

(1) Suppose $f$ is entire with $f(0)=0$ and $|f(z)|\leq e^{1/|z|}$ for all $z\neq0$. Must $f$ be identically $0$?

(2) Suppose that $g$ is entire with $g\circ g=g$. If $g$ is not constant, must $g$ be the identity?

Thanks again for any/all advice.

  • 1
    I see, we use continuity of $f$ to get a bound near $0$, and we have a bound everywhere else2011-07-18

1 Answers 1

13

(1) If $f$ is bounded on $\{z:|z|\leq M\}$ and on $\{z:|z|\geq M\}$, then $f$ is bounded on $\mathbb{C}$.

(2) The equation $g\circ g=g$ implies that $g$ is the identity when restricted to the range of $g$. The range of a nonconstant analytic function is always so big that if 2 analytic functions agree on this range, then they must agree everywhere. (E.g., because the range is uncountable, or because it is open.)

  • 1
    @RHP: In the previous comment Amitesh has provided a proof that the range of a nonconstant analytic function is dense, which is much easier than Picard's theorem but still enough to finish this problem without using the identity theorem.2011-07-28