This particular question was asked to me by a junior as a part of his real analysis quiz and I was unable to solve it.
Question: Let a function to be called "analytic" if for each $a\in\Bbb R$ there exists $\delta_a >0 $ such that on $(a-\delta_a, a+\delta_a)$ , $f$ has a power series expansion. Then show that zeroes of an "analytic" function on $\Bbb R$ have no limit points.
In complex analysis $f$ is said to be analytic at $a$ if $f$ is differentiable in each neighborhood of $a$ and alternatively it has a power series expansion which is convergent in some disc about $a$. So, it's analytic in complex analysis terms. So, I can use the well known result( proof) that zeroes of analytic functions are isolated in $\Bbb C$ hence also in $\Bbb R$ .
Is this approach right?
Best Answer
Yes, you can use it. Not because it works in $\Bbb C$ (many true statements about complex analytic functions are false in the case of real analytic functions), but because that proof works both for real and for complex analytic functions.