# Functions without a fundamental period

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?

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

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.

• 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 :) – Pspl Dec 30 '19 at 12:05
• @Pspl: I added some details explaining why it's not possible to have nice example that can be written with a nice formula. – calcstudent Dec 30 '19 at 12:19
• could you please look at the edit part I added to my question? – Pspl Dec 30 '19 at 14:23
• @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. – calcstudent Dec 30 '19 at 15:10
• So IT IS an awful answer! :D I will stick to a trivial example then! Thanks for your time... – Pspl Dec 30 '19 at 15:46