# A periodic function with no fundamental period and continous at one point is constant.

Theorm : Let $$f:\mathbb{R} \to \mathbb{R}$$ be a periodic function and suppose $$f$$ is continous at some $$\zeta \in \mathbb{R}$$ and that $$f$$ has no fundamental period then prove that $$f$$ is constant .

My trial proof using sequences

Let $$\{p_n\}$$ be a decreasing sequence of periods of $$f$$ converging to $$0$$.

If $$f$$ is not constant then $$\exists$$ a point $$a$$ such that $$f(a) \neq f(\zeta)$$.

Let $$a\gt \zeta$$.

There exist $$m\in \mathbb{N}$$ such that $$0\lt p_n \lt a-\zeta, \forall n \gt m$$

We choose $$x_1, x_2 , ..., x_m$$ as the same real number $$a$$

For $$n\gt m$$, we select $$x_n \in (\zeta, \zeta+p_n)$$ such that $$f(x_n)=f(a)$$ which is possible by the periodicity of $$f$$

Clearly $$x_n \to \zeta$$ as $$n\to \infty$$ but the corresponding functional sequence $$f(x_n)=f(a)\to f(a)\neq f(\zeta)$$ as $$n\to \infty$$ thus contradicting that $$f$$ is continous at $$\zeta$$

Similar technique for $$a\lt \zeta$$

Thus there is no such $$a$$ and so the result follows.

I know there are several questions like this posted here but as far as I have seen none of them use sequences.

My proof looks too simple . Is everything correct or am I overlooking something?

• – Invisible Aug 24 '20 at 14:56

Your proof seems correct, but there is a much simpler way to prove the statement.

Fix $$x \in \mathbb{R}$$, and let $$\varepsilon > 0$$.

Because $$f$$ is continuous at $$\zeta$$, there exists $$\eta > 0$$ such that for all $$y \in [\zeta-\eta, \zeta + \eta]$$, $$|f(y)-f(\zeta)| \leq \varepsilon$$. Let $$T$$ be a period of $$f$$ such that $$0< T < 2\eta$$. There exists $$N \in \mathbb{Z}$$ such that $$x + NT \in [\zeta-\eta, \zeta + \eta]$$, so you deduce that $$|f(x)-f(\zeta)| = |f(x+NT)-f(\zeta)| \leq \varepsilon$$. Because this has to be true for all $$\varepsilon > 0$$, you deduce that $$f(x)=f(\zeta)$$.

Hence $$f$$ is constant.

You can further simplify by showing that if $$P$$ is the set of periods of $$f$$ and if $$a\in \Bbb R$$ then the set $$S(a)=\{a+mp: m\in \Bbb Z\land p\in P\}$$ is dense in $$\Bbb R.$$ And if $$x\in S(a)$$ then (obviously) $$f(x)=f(a).$$ By the denseness of $$S(a)$$ there exists a sequence $$(x_j)_{j\in \Bbb N}$$ of members of $$S(a)$$ that converges to $$\zeta.$$ Hence $$f(\zeta)=\lim_{j\to \infty}f(x_j)=\lim_{j\to \infty}f(a)=f(a).$$

Addendum: To show that $$S(a)$$ is dense: Suppose $$b Take $$p\in P\cap (0,c-b).$$ There exist $$n_1,n_2\in \Bbb N$$ with $$a-n_1p\le b$$ and $$a+n_2p\ge c.$$

Let $$m_0 =\max \{m\in \Bbb Z: -n_1\le m

Then $$a+m_0p\le b (because $$0). So $$a+(1+m_0)p\in S(a)\cap (b,c).$$