# Given reals $p_1,p_2$ and a function $f_1$ with minimal period $p_1$.Existence a function $f_2$ with minimal period $p_2$,$f_1-f_2$ periodic?

Given any two positive real numbers $p_1,p_2$ and a function $f_1 : \mathbb{R} \to \mathbb{R}$ such that $f_1$ has a minimal positive period of $p_1$ . Then is it true that whatever be the choice of $f_1$ there is always a function $f_2 : \mathbb{R} \to \mathbb{R}$ having smallest positive period of $p_2$ such that the function $f_1-f_2$ is also periodic ?

A Miklos Schweitzer problem says :

Let $p_1,p_2$ be any two positive real numbers. Then prove that there is always functions $f_1,f_2 : \mathbb{R} \to \mathbb{R}$ such that $f_i$ has minimal positive period of $p_i$ and the function $f_1-f_2$ is also periodic.

In the proof the function $f_1$ was constructed as $f_1(x) = x$ for every real number $x \in [0,p_1)$ and then extended periodically on $\mathbb{R}$

• your $f_1(x)=x$ is not periodic, right? – Fedor Petrov May 8 '17 at 12:29
• No I mean define $f_1(x)=x$ for $x \in [0,p_1)$ and then extend it periodically. – Aditya Guha Roy May 8 '17 at 15:05
• I also wonder if we could explicitly construct $f_1,f_2$ in course of the Miklos Schweitzer problem. – Aditya Guha Roy May 11 '17 at 15:32

If $p_3$ is a period of $f_1-f_2$, we see that $f_1$ must satisfy an equation $f_1(x)+f_1(x+p_2+p_3)=f_1(x+p_2)+f_1(x+p_3)$ for all $x$ (because both $f_2$ and $f_1-f_2$ satisfy this equation). If, say, $f_1(x)=\sin x$, this rewrites as $\sin(x+\frac{p_2+p_3}2)\cos(\frac{p_2+p_3}2)=\sin(x+\frac{p_2+p_3}2)\cos(\frac{p_2-p_3}2)$, $\cos(\frac{p_2+p_3}2)=\cos(\frac{p_2-p_3}2)$, $\sin\frac{p_2}2\sin\frac{p_3}2=0$. If $p_2$ is not divisible by $2\pi$, we see that $p_3$ should be divisible by $2\pi$, but in this case $f_2=f_1-(f_1-f_2)$ is $p_3$-periodic, that in general is not so (for example, it is not so for $f_2=\sin(x\sqrt{2})$).