If periodic function without lowest positive period is continuous then it is constant.

Question

I am beginner at real analysis so I am doubtfull on my proof technices. Please check my proof.

Proposition: Suppose that $f$ is periodic, continuous, and that there exists no lowest positive period. Show that $f$ is constant.

My proof: By definition continiuty we have $ \vert x-0 \vert \ < \delta \Rightarrow \vert f(x)-f(0)\vert < \varepsilon $

And we know that if there is no lowest period then every open interval contains a period (however this interval is small).

Consider an increasing sequence $\delta_k < \delta$

Choose $y_k$ and a period $t_k$ from the interval $(-\frac {\delta_k}{2}, \frac {\delta_k}{2})$.

Hence $\vert f(y_k+t_k)-f(0)\vert=\vert f(y_k+nt_k)-f(0)\vert < \varepsilon$

By this way we know that as $y_k, t_k$ can be arbitary small and n ranges over integers we can obtain all domain of the function.

Answer 1

By hypothesis, there exists a sequence $p_n$ such that $f(x+p_n) = f(x)$ for all $x$ and $p_n \to 0$. Think about the image of such an $f$: $$\mathrm{im} f = f([0,p_n]) \ \forall n.$$ You should be able to finish by using continuity to deduce that $\mathrm{im} f = f(\{0\})$.

Answer 2

We have that $f$ is periodic for any $a \in (0,\infty)$ $$f(x)=f(x+a)$$ for any $x \in \mathbb{R}$. We can see that $f(x)$ is differentiable and the derivative always yields $$\lim_{b \to 0}\frac{f(x+b)-f(x)}{b}=\lim_{b \to 0}\frac{1}{b}\cdot 0=0$$ It follows that $f(x)$ is also infinitely differentiable. The Taylor series yields $f(x)=a_0,\forall x \in \mathbb{R}$ so $f(x)$ is just a constant function.