I think I have a proof for the statement, but I can't think of a counter-example when $f: \mathbb{R} \to \mathbb{R}$ is not continous. Here's the problem:

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous periodic function with two incommensurate periods $T_1$ and $T_2$; that is $\displaystyle \frac{T_1}{T_2}$ is irrational. Prove that $f$ is a constant function. Give an example of a nonconstant periodic function with two incommensurate periods.

Consider the set $G= \{n_1T_1+n_2T_2 : n_1,n_2 \in \mathbb{Z} \}$.

It's straight forward to verify that this set forms a subgroup of $\mathbb{R}$ under addition.

Since $\displaystyle \frac{T_1}{T_2}$ is irrational, $G$ will not be cyclic, because if it's cyclic, then there exists an element $m_1T_1+m_2T_2 \in G$ such that for any $n_1T_1+n_2T_2 \in G$ there exists a $p \in \mathbb{Z}$ such that:

$$n_1T_1 + n_2T_2 = p (m_1T_1+m_2T_2)$$

Rearranging the terms we obtain:

$$\frac{T_1}{T_2} = \frac{n_2-pm_2}{pm_1 -n_1} \in \mathbb{Q}$$ Which is contradiction. Therefore $G$ is not cyclic. Since $G$ is a subgroup of $\mathbb{R}$ which is not cylic, we conclude that $G$ is dense in $\mathbb{R}$.

Assume that $f(0)=C$. I'm going to show that $f(x)=C$.

Since $G$ is dense in $\mathbb{R}$, every point $x \in \mathbb{R}$ can be approached by elements of $G$ of the form $n_1T_1+n_2T_2$. In particular, for any $\delta>0$ there exists $c_1,c_2 \in \mathbb{Z}$ such that:

$$ |x - (c_1T_1+c_2T2)| < \delta$$

Since $f$ is assumed to be continuous, this means that for any $\epsilon>0$ we have:

$$|f(x) - f(c_1T_1+c_2T_2)|< \epsilon$$

but $f(c_1T_1+c_2T_2)=f(c_2T_2)=f(0)=C$.

Therefore, for any $x \in \mathbb{R}$ we have shown that $\forall \epsilon>0: |f(x)-C|< \epsilon$ which implies $f(x)=C$.

I can't think of a counter-example for when $f$ is not continuous. Can someone suggest a counter-example?


2 Answers 2


The set $P(f)$ of periods of a function $f\colon \mathbb{R}\to \mathbb{R}$ is always a subgroup of (the additive group of) $\mathbb{R}$. (Aside: if $f$ is continuous, it is a closed subgroup, so from $G$ being dense, it then follows that $P(f) \supset \overline{G} = \mathbb{R}$.)

So we have the factor group $\mathbb{R}/G$, and any map $h\colon \mathbb{R}/G \to \mathbb{R}$ induces a periodic function $f\colon \mathbb{R}\to \mathbb{R}$ with period group containing $G$ per $f = h \circ \pi$, where $\pi \colon \mathbb{R}\to \mathbb{R}/G$ is the canonical projection. $f$ will be nonconstant if and only if $h$ is non-constant. A simple non-constant function $f$ with period group containing (equal to, actually) $G$ would be

$$f(x) = \begin{cases} 0 &, x \in G\\ 1 &, x \notin G. \end{cases}$$

  • $\begingroup$ This is a great answer. Is this how you define $P(f)$: $P(f) = \{ T \in \mathbb{R}: f(x+T)=f(x) \text{ for any } x \in \mathbb{R} \}$? I guess everything we have here can be extended to any topological vector space. Am I right? $\endgroup$
    – math.n00b
    Apr 30, 2014 at 14:55
  • $\begingroup$ I would write "for all" instead of "for any" ;) Yes, with the exception of the characterisation of closed subgroups of $\mathbb{R}$ (trivial, infinite cyclic, or all), everything extends to the more general situation (as long as we demand the TVS to be Hausdorff, anyway). $\endgroup$ Apr 30, 2014 at 15:00
  • $\begingroup$ Yes,"for all". Daniel, I'm still having some troubles understanding this. The problem is that your approach induces $f$ from some $G$ and $h$. Am I right? Well, in most cases we've been given $f$ and we must look for the right $G$ and $h$ that induces $f$ with the properties we want. Right? Btw, to check my understanding, I have tried to prove some other problems in the book by using your approach. It seems like your approach is very powerful for solving this kind of problems. Would you please check my answer to see if I'm using your argument correctly or not? $\endgroup$
    – math.n00b
    Apr 30, 2014 at 16:15
  • $\begingroup$ I'm off to dinner now, will read and respond when I return. $\endgroup$ Apr 30, 2014 at 16:17
  • $\begingroup$ OK, thanks. Bon appetit! $\endgroup$
    – math.n00b
    Apr 30, 2014 at 16:34

Here are some other properties of periodic functions that I'm trying to apply Daniel Fischer's approach to them:

Show that if $f: \mathbb{R} \to \mathbb{R}$ is nonconstant, periodic and continuous, then it has a smallest positive period, the so-called fundamental period.

In other words, I have to show that $P(f)$ is cyclic! Since $P(f)$ is an additive subgroup of $\mathbb{R}$ it suffices to show that $P(f)$ is not dense in $\mathbb{R}$. Since $P(f)$ is closed for a continuous function, $P(f)$ is dense only when $P(f)=\mathbb{R}$. If $P(f)=\mathbb{R}$, then by taking $G=P(f)$ we see that any map $h: \mathbb{R}/G \to \mathbb{R}$ is constant! Contradicting the assumption that $f$ is non-constant. Therefore, $P(f)$ is cyclic and its generator is $\alpha = \inf \{x \in P(f): x>0 \}$.

Give an example of a nonconstant periodic function without a fundamental period.

Daniel Fischer's example works! Since $T_1$ and $T_2$ are incommensurate, the group $G=<T_1,T_2>$ won't be cyclic. Hence, $P(f)$ won't be cyclic either.

Prove that if $f: \mathbb{R} \to \mathbb{R}$ is a periodic function without a fundamental period, then the set of all periods of $f$ is dense in $\mathbb{R}$.

Again, this is very simple. Since the set of all periods of $f$, i.e. $P(f)$ is not cyclic, it must be dense in $\mathbb{R}$! :D

  • 1
    $\begingroup$ Depending on what is known about subgroups of $\mathbb{R}$, it may be necessary to explain why a non-dense (nontrivial) subgroup is cyclic. And for the last point, why non-cyclic subgroup must be dense. (Neither of the two are hard to prove, once one knows that that's what one wants to prove.) If those are known (not only by you, also by your audience), it works as written. $\endgroup$ Apr 30, 2014 at 17:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .