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In the book that I'm reading they said that this fact is trivial , but I'm not completely sure about something and I prefer to confirm it. It's about a lot of equivalences, but I have question about two.

Let $f$ be a function defined on a domain $D \subset {\Bbb C}$. Assume that $f$ has a power series expansion at each point of $D$, and let $\zeta \in D$. Then the following are equivalent:

$\quad$ i) $f^{(n)} (\zeta) = 0,\;\text{ for }\;n = 0,1,2\ldots$

$\quad$ ii) $f \equiv 0$ in a neighborhood of $\zeta$

I know how to prove that i) implies ii); I want to see the other side of the proof. I think that I must use the fact that I can write $f$ locally in the form $f(z) = \sum_{n = 0}^\infty a_n (z - \zeta)^n = \sum_{n = 0}^\infty \frac{f^{(n)}(\zeta)} {n!} (z - \zeta)^n $ and then by the assumption we have : $\sum_{n = 0}^\infty \frac{f^{(n)}(\zeta)}{n!}(z - \zeta)^n = 0$ But now I don't know what can I do, because the sum could be zero, because the sum has some positive terms, and other negative terms, and because this the sum it's zero. I don't know how to finish

Remark: I know that there are other equivalences, even one of them asserts that $f$ vanishes not only on a neighborhood,but in the whole domain $D$, but I want to see for the moment only this two equivalence. Thanks =)!

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    @Jonas: Ah, I agree that is most likely. Sorry for the confusion, Sasha :)2011-11-27

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Suppose that $f\equiv 0$ on a neighborhood $U$ of $\zeta$. The derivative of the zero function (or any other constant function) is the zero function, so f'\equiv 0 on $U$. Similarly, f''\equiv 0 on $U$, f'''\equiv 0 on $U$, etc. If $f^{(k)}\equiv 0$ on $U$, then in particular $f^{(k)}(\zeta)=0$.

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    @August: You're welcome. I was looking for an easy problem when I opened the question. It was truth in advertising.2011-11-27