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If there is a sequence of zeros of a function and this sequence tends to 0, then is $f$ not periodic, true or false?

I think it is false, for example, if we consider the function $\cos(\frac1{\sqrt x})$ for $x>0$, in intervals of amplitude equal to the largest zero of the function, we will obtain a periodic function, but however , the distance between its zeros also converges to $0$.

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But I would like to know what hypotheses would have to be added to be true, if it is always false or not,...

Thanks in advance.

CPMP
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2 Answers2

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The statement is false, as you correctly said. Apart from the “trivial” case that $f$ is identically zero, there are many continuous periodic functions on $\Bbb R$ with zeros accumulating at zero.

Your example does not work however, because $\cos(\frac1{\sqrt x})$ is not periodic.

A simple example (inspired by geetha290krm's comment) is $f(x) = \min(\sin(x), 0)$, which is $2\pi$-periodic and identically zero on $[0, \pi]$.

But you can start with any continuous function on an interval $[0, T]$, having zeros accumulating at $x=0$, and $f(0) = f(T) = 0$. Then extend $f$ to a periodic function on $\Bbb R $ by setting $f(x+nT) = f(x)$ for all integers $n$.

Martin R
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  • And if instead of having a succession of zeros, we have a function that gives us the difference between two consecutive zeros, and this is decreasing and has a limit of 0, can I conclude that $f$ is not periodic?

    Look: https://math.stackexchange.com/a/1454070/1049667

     – CPMP Oct 30 '22 at 10:14
  • @CPMP: No, you can't. There is already a comment at that answer explaining why the argument is wrong. – Martin R Oct 30 '22 at 10:20
  • ok, thank you for help – CPMP Oct 30 '22 at 12:57
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Take $f(x)=\sin(x)\sin\left(\frac{1}{\sin(x)}\right)$ (and define it as $0$ where that is undefined) for a continuous counterexample.

Here's how it looks

jjagmath
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