0
$\begingroup$

Let $f$ be entire such that $\forall |z|\geq1$, $|f(z)|\leq e^{-|z|}$. Then how do we prove that $f$ is constant?

Can we map the complement of unit disk conformally onto the complex plane or the unit disk, by virtue of which we may then invoke Liouville's Theorem? Any ideas. Thanks beforehand

$\endgroup$
  • $\begingroup$ Removed the wholly unnecessary tag liouville-theorem. $\endgroup$ – Cameron Williams Feb 20 '17 at 15:45
3
$\begingroup$

For $|z| \geq 1$ we have $|f(z)|\leq e^{-|z|} \leq 3 $.

Also, for $|z| \leq 1$ $|f|$ is bounded by a constant $C$, since the unit disc is compact (and $f$ is continuous).

So $|f(z)|\leq \max \{3, C \}$ on all of $\mathbb{C}$. Now Liouville's Theorem does the rest.

$\endgroup$

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.