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Given a non-negative, smooth function $f: [a,b] \rightarrow \mathbb{R}_{\geq 0}$. If there exists a sequence (of pairwise disjoint points) $x_n \in [a,b]$ such that $f(x_n)=0$ for all $n \in \mathbb{N}$, does it mean that $f(x)=0$ for all $x \in [a,b]$?

Motivation: Consider smooth curves in polar coordinates that is a map $c: [0,2\pi) \rightarrow \mathbb{R}^2 , t \mapsto r(t) \cdot (\cos(t),\sin(t))$. If we restrict this curve to a compact intervall, is it regular (i.e. $r'(t)^2 +r^2(t) \neq 0$ for all $t \in [0,2\pi)$) up to finitely many points or finitely many closed intervalls? This would follow, if the above conclusion holds.

Thoughts: By compactness we can find (by abuse of notation) a subsequence $x_n \rightarrow x \in [a,b]$ such that $f(x)=0$. From the taylor expansion we see that all derivatives in $x$ must vanish, that is $f^{(n)}(x)=0$ for all $n \in \mathbb{N}$.

Remarks: Smooth function on a closed intervall means there is an extension to an open intervall containing it.

user21820
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ELDG
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1 Answers1

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Let's start with $f_1(x) = \sin(1/x)$, which has infinitely many zeroes, but it's clearly not smooth (not even continuous). You can square it to get a nonnegative function $f_2(x) = \sin^2(1/x)$.

As you observed, if $f$ is to be smooth, a flat function has to appear, e.g. $e^{-1/x^2}$. So consider $$f_3(x) = e^{-1/x^2} \sin^2(1/x).$$ You should be able to check that the zero of $e^{-1/x^2}$ at $x=0$ is so strong that the product is smooth, and in fact $f_3^{(n)}(0) = 0$ for each natural $n$.

Michał Miśkiewicz
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