prove that a non constant periodic, continuous function has a "smallest period"

Let $P$ be the set of periods of $f$. Using your definition, $P$ is non-empty and bounded below by $0$. Consider $p^*=\inf P$. Take $p_n \in P \to p^*$. Fix $x \in \mathbb R$. Then $x+p_n \to x+p^*$ and $f(x+p^*)= \lim f(x+p_n)=f(x)$.

If $p^*>0$, then $p^* \in P$ and so $p^* = \min P$.

If $p^*=0$, then we need to argue that $f$ is constant.

For a more conceptual approach, here is a roadmap:

• The set of periods of a function is an additive subgroup of $\mathbb R$.

• An additive subgroup of $\mathbb R$ is either cyclic or dense.

• The set of periods of a continuous function is a closed set.

• A continuous function with a dense set of periods is constant.

Outline of the proof: Assume by contradiction that there is no smallest period. Use first the fact that the difference between two periods is also a period to show that you can find a decreasing sequence of periods which converge to 0.

This means that for each $\delta >0$ you can find some $T$ period such that

$$0 < T <\delta$$

Now, pick $x,y$ arbitrary.

Fix $\epsilon >0$, then there exists a $\delta$ such that for all $z$ with $$|y-z| < \delta \Rightarrow |f(y)-f(z)|<\epsilon$$

Pick some $0< T < \delta$.

Show now that there exists some $n \in \mathbb Z$ such that $|(x+nT)-y|<\delta$.

Then $$|f(x)-f(y)|=|f(x+nT)-f(y)| <\epsilon$$

Since this is true for all $\epsilon$ we get $f(x)=f(y)$. As those are arbitrary, you are done.