# Measurable functions with arbitrarily small periods are constant

Let $$f:\mathbb{R}\to \mathbb{R}$$ be a measurable function such that $$f$$ is $$T$$-periodic for arbitrarily small $$T$$. I need to show that $$f$$ is constant, at least almost everywhere.

Can someone find an elementary proof of this ?

One proof: I found a proof in which I used distribution theory. First replace $$f$$ by $$f_A:= f\mathbf{1}_{\vert f\vert\leq A}$$ for $$A>0$$ to get a bounded function. Therefore, $$f$$ belongs to $$L_{loc}^1(\mathbb{R})$$ and can be seen as an element of $$\mathcal{D}'(\mathbb{R})$$.

Take $$T$$ a period of $$f$$ and denote by $$\tau_Tf:= f(\cdot-T)$$ the translation operator by $$T$$. Take $$\varphi$$ as a test function, and compute \begin{aligned} \left &=\left<\tau_T f, \varphi \right> \\ &= \left< f,\tau_{-T} \varphi \right> \end{aligned} Thus, for $$T$$ arbitrarily small $$\left = 0.$$ Then, one uses the fact that if $$\underset{n\to\infty}{\lim}T_n =0$$, then $$\underset{T_n\to0}{\lim} \frac{\tau_{T_n}\varphi - \varphi}{T_n} = \varphi'\quad \text{in}\ \mathcal{D(\mathbb{R})}.$$
Passing to the limit in the above inequality one gets that $$\left = 0$$ for any test function $$\varphi$$. Thus $$f' = 0\quad\text{in}\ \mathcal{D'}(\mathbb{R}).$$ By a classical result, this implies that $$f$$ is constant almost everywhere.

Motivation: In a paper called "On the approximation of Lebesgue integrals by Riemann sums", Jessen proves the following theorem. Let $$f$$ be a $$L^1(\mathbf{T})$$ function. Denote its Riemann sum $$f_n:= \frac{1}{n}\sum_{k=1}^n f(x+\frac{k}{n}).$$ Then $$\underset{n\to\infty}{\lim} f_{2^n}(x) = \int_0^1 f\quad a.e.x$$ Jessen considers $$\phi(x):=\overline{\lim} f_{2^n}(x)$$ which is a $$2^{-n}$$ periodic function for all $$n$$, hence an almost everywhere constant function. The proof reduces to show that this constant is $$\int_0^1 f$$.

• Do you mean that for every $T>0$, $f$ is $T$-periodic, or that for every $\epsilon>0$ there is a $T<\epsilon$ such that $f$ is $T$-periodic? – md2perpe Aug 24 '19 at 20:04

I think your question is equivalent to :

Let $$f : \mathbb R \to \mathbb R$$.$$f$$ is measurable.

If $$f(ax)=f(x)$$ ae for any $$a>0$$

Then $$f(x)=c$$ for almost every $$x \in (0,\infty)$$, where $$c$$ is a constant.

You get a similar result for $$x \in (-\infty, 0)$$, when $$a <0$$

If I'm right then your answer is here and you don't need distribution theory.

• I understood the question to mean something slightly differently; having $$\text{ If for any a>0 there exists 0<T<a such that f(Tx)=f(x) for all x\in\Bbb{R}.}$$ in stead of the second line. I don't know if this is equivalent to what you wrote; probably not? – Servaes Aug 10 '19 at 8:01
• @Servaes so it is equivalent to that in the link ? – ibnAbu Aug 10 '19 at 8:03
• I don't think it is equivalent, but I don't have a proof/counterexample at hand. – Servaes Aug 10 '19 at 8:04
• My question is the one formulated by @Servaes. Basically, my problem was a bit weaker since the function is $2^{-n}$-periodic for all $n$: $$f(x+2^{-n})=f(x)\quad a.e.x$$ Both problems are equivalent since $f$ turns out to be constant in any case. But this equivalence is not clear at first glance. – Wulfenite Aug 10 '19 at 8:32

If $$f$$ is periodic with period $$p$$, then $$\int_x^{x+p} f(t) \, dt = \int_y^{y+p} f(t) \, dt$$, for all $$x$$ and $$y$$.

By the Fundamental Theorem of Calculus for Lebesgue integration (see Royden 4th edition, Chapter 6, Theorem 14), $$\lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t) \, dt = f(x)$$ almost everywhere. Let $$A$$ be a set of full measure witnessing this.

Let $$p_n$$ be a sequence of periods converging to 0. For $$x,y \in A$$, we have

$$f(x) = \lim_{n \to \infty} \frac{1}{p_n}\int_x^{x+p_n} f(t) \, dt = \lim_{n \to \infty} \frac{1}{p_n}\int_y^{y+p_n} f(t) \, dt = f(y).$$