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A function that is analytic in the whole plane and which vanishes along with all its derivatives at any one point in the plane is identical to $0$. Now consider a function $f(z)$, which is supposedly analytic everywhere such that $$\lim_{z\rightarrow\infty}f^{(n)}(z)=0$$ for $n=0, 1, 2...$ Is the conclusion that $f(z)$ is identical to $0$ the only possibility?

vnd
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1 Answers1

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Yes. $\lim_{z\rightarrow\infty}f(z)=0$ already implies that $f \equiv 0$.

That follows from the maximum modulus principle, or from Liouville's theorem.

Martin R
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  • That's Liouville, not maximum principle. – zhw. Dec 20 '15 at 22:45
  • Yes, but the answer did remind me that $f$ would be bounded. – vnd Dec 20 '15 at 22:47
  • @zhw.: Thanks! My impression was that the maximum modulus principle implies Liouville, but after reading http://math.stackexchange.com/questions/894304/maximum-modulus-principle-implies-liouvilles-theorem I understand that this is not completely correct. – Martin R Dec 20 '15 at 22:56
  • @zhw.: But the conclusion $\lim_{z\rightarrow\infty}f(z)=0 \Longrightarrow f \equiv 0$ would also follow from the maximum principle, without using Liouville, doesn't it? – Martin R Dec 20 '15 at 23:03
  • Yes it would ... – zhw. Dec 20 '15 at 23:04