Is the set of periodic functions from $\mathbb R$ to $\mathbb R$a subspace of $\mathbb R^{\mathbb R}$?

Question

A function $f:\mathbb R→\mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $x\in\mathbb R$. Is the set of periodic functions from $\mathbb{R}$ to $\mathbb{R}$ a subspace of $\mathbb{R}^\mathbb{R}$? Explain.

The question has been asked here

I still don't quite understand why the ratio of two periods should not be irrational.

For example if $\beta/\alpha=r/s$ and $r/s = \sqrt{2}$, the following equation still works? $$h(x+s\beta)=f(x+r\alpha)+g(x+s\beta)=f(x)+g(x)=h(x)$$ $$h(x+\beta)=f(x+\sqrt{2}\alpha)+g(x+\beta)=f(x)+g(x)=h(x)$$

Answer 1

If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.

Note that you can only write $f(x+r\alpha)=f(x),g(x+s\beta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=\sin(\sqrt2 x),g(x)=\sin(\sqrt3x)$.

Answer 2

$f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $\frac r s$ rational.