# If $f$ is analytic, is this true?

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 $0$?

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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.

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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?