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It is well known that if $f:\mathbb R\to \mathbb R$ is real analytic then we have that either $f\equiv 0$ or $f$ has finitely many zeros on the interval $[a,b]$.

It is also known that you can have a nonnegative function $f:\mathbb R\to \mathbb R$ that is smooth with infinitely many zeros on $[a,b]$. A bump function, for example.

My question is:

Given a nonnegative function $f\in C^1([a,b],\mathbb R)$, not identically $0$, is there an additional constraint other than analytic that tells you that $f$ has finitely many zeros?

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    $\begingroup$ How about "$f$ is strictly monotone"? $\endgroup$
    – Willie Wong
    Commented Feb 17, 2021 at 1:29
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    $\begingroup$ @WillieWong Or more generally, one condition is that $f'$ has finitely many zeros. $\endgroup$
    – user168590
    Commented Feb 17, 2021 at 1:32
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    $\begingroup$ @JochenGlueck Sorry, I meant analytic $\mathbb R \to \mathbb R$ and then study the zeros on $[a,b]$. $\endgroup$
    – user168590
    Commented Feb 17, 2021 at 1:35
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    $\begingroup$ @WillieWong and user168590: And then, by induction, ... :-) $\endgroup$
    – Jochen Glueck
    Commented Feb 17, 2021 at 1:36
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    $\begingroup$ Analyticity is pretty much the largest regularity class that avoids bump functions. Gevrey classes, for instance, are the most popular class used between the smooth and analytic classes, and they support bump functions also. en.wikipedia.org/wiki/Gevrey_class $\endgroup$
    – Terry Tao
    Commented Feb 17, 2021 at 2:48

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