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I came across with an exercise that asks for an example of a real periodic function without a fundamental period. Since it's an exercise to be valued by the teacher, I would like to give him a non trivial example. I know that constant functions $f(x)=C$, $C\in\mathbb R$ and Dirichlet functions such as

$$g(x) = \left\{ \begin{array}{cc} a& x \in \mathbb{Q}\\ b & x \in \mathbb{R}\setminus\mathbb{Q}\end{array}\right., a\neq b$$

are periodic but don't have a fundamental period:

$$T_f \in \mathbb{R}^+:f(x+T_f)=f(x) \Rightarrow T_f \in \mathbb{R}^+$$ $$T_g \in \mathbb{R}^+:g(x+T_g)=g(x) \Rightarrow T_g \in \mathbb{Q}^+$$

I wonder if there is another kind of example. Does anyone know? Thanks.

EDIT:

Please, take a look at this function:

$$h(x)=\lim_{n\to \infty}{\sin(nx)}$$

I know that $\lim_{x\to \infty}{\sin x}$ doesn't exist, but if we, for a moment, think that it does, doesn't $h$ is periodic without any fundamental period?

function h

I'm not sure if this answer is great or awful...

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Pick a list of real numbers $C$, and consider the set $S$ of all finite linear combinations of these numbers with integer coefficients (that is, numbers of the form $\sum_{i=1}^{n}k_{i}c_{i}$ where $n$ is finite, $c_{i}$ are the chosen number from $C$, and $k_{i}$ are integers). Make sure that $S$ has nonzero number arbitrarily close to $0$. This condition can easily be satisfied by many ways, for example by explicitly choose numbers from a sequence going to $0$, or by making sure there are $2$ numbers in $C$ that are incommensurable.

Then the function $f(x)=1$ if $x\in S$, $0$ if not, is a periodic function with set of periods $S$, and so lack a fundamental period.

EDIT:

Let me explain why there won't be any hopes of finding any "nice" function with this property other than constant. Let $f$ be a function with set of periods $S$. Then $S$ has the property that if $x,y\in S$ then $-x,-y,x+y\in S$, which is obvious to prove. Now we assuming that $S$ has nonzero number arbitrarily close to $0$, and that $f$ is continuous. Let $f(p)=v$ for some point $p$ and value $v$. Consider an arbitrary point $q$. For any $n$ pick $s_{n}$ be a positive number in $S$ less than $\frac{1}{n}$, let $k_{n}$ be the largest integer such that $q-p>=k_{n}s_{n}$. Then $f(p+k_{n}s_{n})=f(p)=v$ and also $|q-p+k_{n}s_{n}|<=s_{n}<\frac{1}{n}$. Then $p+k_{n}s_{n}$ converge to $q$ in the limit, and since $f$ is continuous, $f(q)=v$. So $f$ must be constant.

So there shouldn't be any hope of being able to write a function with a nice formula that satisfy a claim beside constant function.

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  • $\begingroup$ Your answer is quite clever and I think it works. However, this is "another version" of a Dirichlet function (at least it's generated by an analogical mechanism). I was looking for a completely not related example! But I will upvote your answer, non the less :) $\endgroup$ – Pspl Dec 30 '19 at 12:05
  • $\begingroup$ @Pspl: I added some details explaining why it's not possible to have nice example that can be written with a nice formula. $\endgroup$ – calcstudent Dec 30 '19 at 12:19
  • $\begingroup$ could you please look at the edit part I added to my question? $\endgroup$ – Pspl Dec 30 '19 at 14:23
  • $\begingroup$ @Pspl: I think what your idea is to just take the graph of the sine function then just squash it down along the left-right direction. What you should get is a thick solid horizontal bar. Which honestly, is just as boring as a constant function, except it also fail to be a function. $\endgroup$ – calcstudent Dec 30 '19 at 15:10
  • $\begingroup$ So IT IS an awful answer! :D I will stick to a trivial example then! Thanks for your time... $\endgroup$ – Pspl Dec 30 '19 at 15:46

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