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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?

math.n00b
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2 Answers2

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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

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

Daniel Fischer
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  • 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? – math.n00b Apr 30 '14 at 14:55
  • 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). – Daniel Fischer Apr 30 '14 at 15:00
  • 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? – math.n00b Apr 30 '14 at 16:15
  • I'm off to dinner now, will read and respond when I return. – Daniel Fischer Apr 30 '14 at 16:17
  • OK, thanks. Bon appetit! – math.n00b Apr 30 '14 at 16:34
  • The road leads both ways. If we have $G$, we can construct an $f$ whose period group is $G$ (with the caveat that it's usually not practically easy to determine whether a given real number lies in $G$ or in which coset it lies). If we have $f$, we obtain $G$ (although again, in practice, that can be very difficult). Here, the task was a) to determine that a continuous function whose period group contains two elements with irrational quotient is constant, and b) to construct a non-constant discontinuous function with such a period group. The key for a) is that such a group must be dense in ... – Daniel Fischer Apr 30 '14 at 17:32
  • ... $\mathbb{R}$. For b), the task is such that we need to go from the group to the function. So we choose some such group, $\mathbb{Z} + \pi\mathbb{Z}$ or $\mathbb{Z}+ \sqrt{2}\mathbb{Z}$ for example, and construct $f$ from that. – Daniel Fischer Apr 30 '14 at 17:39
  • If we have $G$, why can we construct a function $f$ whose period group $P(f)$ "is" $G$? What I understand from your argument is that for any map $h$ (not necessarily a homomorphism) a function $f$ is induced such that its period group "contains" G. But how bigger $P(f)$ is from $G$ is determied by $h$. Right? – math.n00b Apr 30 '14 at 18:34
  • Right. Any $h$ gives an $f$ with $P(f) \supset G$. But if you take an $h$ such that some fibre of $h$ is precisely one coset, it follows that in fact $P(f) = G$. You can have $P(f) = G$ also if all fibres of $h$ contain more than one coset, but picking $h$ so that some fibre contains only one coset is the simplest way to make sure that you have $P(f) = G$. – Daniel Fischer Apr 30 '14 at 18:42
  • It took me sometime to understand your idea but I'm still not sure if I've understood it correctly because this is the first time I read about fibers. I think fiber is a fancy term for the pre-image, am I right? Are you in some sense creating an inverse for $h$? – math.n00b May 02 '14 at 14:15
  • Yes, a fibre is the preimage of one point. It's not a fancy term, though, just one that isn't often introduced early (which I forgot). I'm not creating an inverse of $h$, more constructing $h$ by saying what the preimages of points are. – Daniel Fischer May 02 '14 at 15:32
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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

math.n00b
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    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. – Daniel Fischer Apr 30 '14 at 17:45