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

Question

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 $0$.

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.

Answer 1

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

Answer 2

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 $n$.