Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am studying for a real analysis final tomorrow and I stumbled across this interesting problem that I am wondering how to do. It goes as follows:

"Suppose that the function $f$ is analytic on $(a,b)$, prove $f(x) = 0$, $\forall x \in (c,d) $ $\subset (a,b) \implies f(x) = 0, \forall$ $x \in (a,b)$."

I think that using induction is the way to go. We know that the infinite derivatives on the interval $(c,d)$ are all $0$. How do we use this to find out that all the infinite derivatives on $(a,b)$ are all identically $0$?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

The simple proof of this is the fact that if you have power series representation on a subball of your ball, the power series is valid everywhere on your ball. In particular, since $f$ has the zero power series on the ball $(c,d)$ it has the zero power series on the ball $(a,b)$.

A much stronger statement is true though.

Namely, assume that $f:(a,b)\to\mathbb{R}$ is real analytic (with $(a,b)$ finite) and there exists a non-discrete set $X\subseteq (a,b)$ for which $f$ vanishes, then $f=0$ on $(a,b)$.

To see why this is true, use the common fact then that $f$ extends to a holomorphic map $\widetilde{f}:D\to\mathbb{C}$ with $D$ open, from which the fact that $\widetilde{f}=0$ is zero on a set with a limit point gives you that $f$ is zero on $D$ and so zero on $(a,b)$.

The only reason I go to the complex case is because I feel like the result that fibers are discrete is better known there.

share|improve this answer

Write $f$ as a power series centered around some point $x\in(c,d)$. $f$ is equal to its power series wherever that series converges. What is the power series and what is its radius of convergence?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.