Let $f: \mathbb{R} \to \mathbb{R}$ be a smooth function. Suppose $f^{(n)}(a) \neq 0$ for some $n \geq 2$.
I was wondering is it possible to prove that there exists $\epsilon > 0$ such that
$f$ only has finitely many zeros in the interval $[a-\epsilon, a+ \epsilon]$?
If $n=1$ this is clear.
The function given in the answer in Smooth function with all derivatives zero is an example of a non-constant smooth function which has infinitely many zeros
in $[0-\epsilon, 0 + \epsilon]$ for all $\epsilon > 0$. But this example satisfies $f^{(n)}(a) = 0$ for all $n$. I was wondering if we can make the number of zeros to be finite with some kind of non-zero derivative assumption. Thank you.
Best Answer
Yes, this follows the usual way. There is some minimal $k$ so that $f^{(k)}(a)\ne 0$, then there is by the continuity of that derivative some $ϵ>0$ so that $|f^{(k+1)}(x)(x-a)|<\frac12|f^{(k)}(a)|$ for $x\in(a−ϵ,a+ϵ)$. From that it follows that $f$ is close enough to its most trivial non-trivial Taylor polynomial (monomial) to conclude that there are no roots in that interval for $k=0$ or a $k$-fold root at $x=a$ for $k>0$ and no other roots.
It is perhaps easier to just divide out the linear factors, $f(x)=(x-a)^kg(x)$ with $g(a)=\frac1{k!}f^{(k)}(a)$, so that roots of $f$ other than $a$ are also roots of $g$. Then argue by the continuity of $g$ and $g(a)\ne 0$ that $a$ is an isolated root (or that there are no roots at all close to $a$).