A real continuous periodic function with two incommensurate periods is constant.

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?

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

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

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