# Non-existence of an entire function

I'm having a bit of trouble coming up with a solid argument for this question.

"Show that there is no entire function $f(z)$ that satisfies $f(0) = 0$ and $|f(z)| \leq 1$ for all $z \in \mathbb{C}$ with $|z| > 1$"

I feel as if assuming there is a function, then getting a contradiction is the correct way to go - And somehow using Liouville's theorem to show that $f$ is bounded but that $f(0) \neq 0$ , or something along those lines would be useful, but I'm not sure how to formulate it.

Any help would be appreciated!

• Hmmm. $f(z) = 0$ for all $z$ satisfies the conditions and is entire. – Daniel Fischer Nov 5 '15 at 20:17
• Your function is bounded outside the unit disc by assumption and bounded inside the unit disc by continuity and compactness, so it's bounded everywhere. Now when you get past Daniel Fischer's objection, you'll be done. – Rob Arthan Nov 5 '15 at 20:19
• I understand that f is bounded outside the unit disc, but I don't really get how f is bounded inside? (I haven't taken topology so I'm unsure of the notion of compactness...) But either way, if f is bounded everywhere how does that show that f cannot be entire? – Sorey Nov 5 '15 at 20:33
• Compact = closed and bounded in $\Bbb C$. The image of a compact set under a continuous function is itself compact. To see the bounded part, assume not and choose for each $n$ a point $z_n$ in the unit disk with $|f(z_n)| > n$. Because the sequence is bounded it has a convergent subsequence with limit $z$, but this requires that $f$ not be continuous at $z$. – Paul Sinclair Nov 5 '15 at 20:45
• And to answer your other question, a bounded entire function is constant. Since your function has $f(0) = 0$, that means that $f(z) = 0$ for all $z$. As Daniel Fischer has pointed out, this $f$ satisfies the conditions. (The statement as given is wrong.) But Rob Arthan's point is that $f = 0$ is the only counter-example. – Paul Sinclair Nov 5 '15 at 20:55