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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.

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The answer to your first question is yes, because $f$ will be bounded entire. – Olivier Bégassat Jul 18 '11 at 17:39
for 2), take a power series rep and compare (to get $g(z)=z$) – yoyo Jul 18 '11 at 17:44
I see, we use continuity of $f$ to get a bound near $0$, and we have a bound everywhere else – RHP Jul 18 '11 at 18:14
up vote 13 down vote accepted

(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.)

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This is a very good answer. I was trying to think of how to avoid open mapping, and your remark about uncountability achieves this. Very clever! – Dylan Moreland Jul 18 '11 at 17:56
Dylan, thanks. In turn, I guess the easiest way to see that the range is uncountable is to use connectedness. – Jonas Meyer Jul 18 '11 at 18:01
@RHP: That is overkill, but to answer your question, it would follow from continuity. Continuous functions that agree on a dense subset are equal. To avoid overkill, I recommend using the identity theorem. Agreeing on an uncountable set suffices; e.g., see… – Jonas Meyer Jul 18 '11 at 18:26
Proof: Let $w\in\mathbb{C}$. If there exists $\delta>0$ such that $\left|f(z)-w\right|\geq \delta$ for all $z\in \mathbb{C}$, then the function $g:\mathbb{C}\to\mathbb{C}$ defined by the rule $g(z)=\frac{1}{f(z)-w}$ is an entire function bounded by $\frac{1}{\delta}$. Liouville's theorem shows that $g$ is constant. Hence $f$ is constant. Therefore, if $f$ is not constant, there does not exist a $\delta$ satisfying the prescribed condition. In other words, $w$ is in the closure of $f(\mathbb{C})$ and $f(\mathbb{C})$ is dense in $\mathbb{C}$. – Amitesh Datta Jul 23 '11 at 14:00
@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. – Jonas Meyer Jul 28 '11 at 4:25

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