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I came up with the following problem myself.

Prove or disprove:

Let $f_n$ be a sequence of real-valued continuous periodic functions defined on $\mathbb{R}$. Suppose that there exists a $M$ such that for each $n$, there exists a period $T_n$ of $f_n$ and $T_n<M$. If $f_n$ converges poinwise to $f$, then $f$ is periodic. If it is not true, then does the conclusion hold if the convergence is uniform?

If the existence of such an $M$ is not required, then we have the following counterexample

$f_n(x)= \begin{cases} x\text{$\ \ \ $if $0\leq x\leq n$,}\\ 2n-x\text{$\ \ \ $if $n< x< 2n$,}\\ f_n(x-2nm)\text{$\ \ \ $ where $m$ is the integer satisfying $m\leq \frac{x}{2n}< m+1$ otherwise.} \end{cases}$

Then each $f_n$ has a period $2n$ and $f_n$ converges to $|x|$ which is not periodic.

If the $f_n$'s are not required to be continuous, then we have the following counterexample

Let $\{p_n\}$ be a strictly increasing sequence of primes. Define $f_n$ as follows

$f_n(x)= \begin{cases} 1 \text{$\ \ \ $ if $x=\frac{m}{\sqrt{p_n}}$ for some $m\in\mathbb{Z}$,}\\ 0 \text{$\ \ \ $ otherwise.} \end{cases}$

Then each $f_n$ has period $\frac{1}{\sqrt{p_n}}<1$ and $f_n$ converges to $f$ defined by $f(0)=1$ and $f(x)=0$ for $x\neq0$, which is not periodic.

Now, when both of the conditions are required, I can't prove or come up with a counterexample.

19021605
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    Neither convergence in your examples is uniform. Are you intereseted in uniform or pointwise convergence? – mihaild Feb 27 '25 at 14:40
  • "If $f_n$ converges to $f$ ..." can mean many things because when dealing with functions there are several types of convergence. – jjagmath Feb 27 '25 at 14:49
  • @mihalid ${\sin(x/n)}_{n\geq 1}$ should converge uniformly to $0$. And the periods are not bounded ($T_n = 2\pi n$). – XavierO Feb 27 '25 at 14:52
  • @XavierO yes, so it is sequence of periodic functions uniformly converging to periodic function, thus not a counterexample. – mihaild Feb 27 '25 at 14:54
  • Oh, yes, of course. Sorry – XavierO Feb 27 '25 at 15:03
  • @mihaild You're right. I was careless. I am fine with either convergence. But I would appreciate if we could solve the problem for the case of uniform convergence also. – 19021605 Feb 27 '25 at 15:26
  • @mihaild I edited so that the question asks for both pointwise and uniform convergence which is different since that question only asks for uniform convergence. – 19021605 Feb 28 '25 at 00:56

2 Answers2

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Here is a counter-example without uniform convergence.

Let $f_n$ be the continuous, even and $\frac{n+2}{n}$-periodic function such that $f_n(x)=1-nx$ on $[0,\frac 1n]$ and $f_n(x)=0$ on $[\frac 1n,\frac{n+1}{n}]$.

Then $f_n(0)\to 1$ and we prove that for all $x\neq 0$, $f_n(x)\to 0$. That means that the limit $f$ is not periodic (and not continuous).

Assume WLOG $x>0$. With $k=\lceil x\rceil -1$ the last integer strictly before $x$, and $x=k+r$, then for all $n>\frac{2k+1}{r}$, we get $x>\frac{n+2}{n}k+\frac 1n$ so the last bump before $x$ is passed and $\frac{n+2}n(k+1)-\frac 1n\ge k+1>x$ so the next one is still over, and $f_n(x)=0$.

Christophe Boilley
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  • Thank you for answering. I do not quite understand how you get $r>\frac{n+2}{n}k$ from $n>\frac{2k+1}{r}$. At best I can get $r>\frac{2k+1}{n}=\frac{2+1/k}{n}k$. – 19021605 Feb 28 '25 at 01:13
  • Correct me if I'm wrong but if we set $x=2$, then $r=k=1$, then $r>\frac{n+2}{n}k$ implies $0>\frac{2}{n}$. – 19021605 Feb 28 '25 at 02:35
  • Nevermind, I proved the convergence myself. For any $x>0$, let $k$ be the integer as you have defined. Since $\lim\left(\frac{n+2}{n}k+\frac{1}{n}\right)=k<x$, there exists some $N$ such that $n\geq N$ implies $x>\frac{n+2}{n}k+\frac{1}{n}$. We have $\frac{n+2}{n}k+\frac{n+1}{n}>x>\frac{n+2}{n}k+\frac{1}{n}$ for all $n\geq N$. We have $\lim f_n(x)=\lim f_n(x-\frac{n}{n+2}k)=0$. – 19021605 Feb 28 '25 at 03:33
  • @19021605 Curiously, I didn’t save my last edit. Now I corrected the last inequality. – Christophe Boilley Feb 28 '25 at 07:10
  • There's a typo in the last inequality. – 19021605 Feb 28 '25 at 08:00
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If $f_n$ (with the properties given) converges uniformly to $f$ then $f$ must be periodic. The proof is actually quite simple as $T_n$ bounded means it has a convergent subsequence so passing to a subsequence we can assume $T_n \to T$ and it is easily shown then that $f$ is periodic with period $T$.

See for example this question The limit of periodic function sequence is periodic

Note that if we remove the boundness condition on $T_n$ it is very easy to construct $f_n$ that converge uniformly to a nonperiodic continuous $f$, though $f$ will be uniformly almost periodic, so will have lots of properties similar to the ones of periodic function (eg generalized Fourier expansion with countably many exponents).

For example take $f(x)=\sum_{n \ge 1}\frac{\cos (2\pi x/n)}{n^2}$ and $f_n$ the partial sums

Conrad
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  • But what if there is another subsequence that converges to $T'$, such that $T'/T\in\mathbb{R}\backslash\mathbb{Q}$? – chirico Feb 28 '25 at 10:15
  • @Chirico The function is constant then which is periodic with any period - the point is that once a function has a period $T$ it is periodic though of course $T$ may not be the least period – Conrad Feb 28 '25 at 13:38