Uniform continuity of continuous periodic function

Question

Defn: Let $f$ be a function from $\mathbb{R}$ into a set $X$. We say that $f$ is periodic if there exists $p>0$ such that for all $x\in \mathbb{R}$, we have $f(x+p)=f(x)$.

Prove: If $f$ is a continuous periodic function from $\mathbb{R}$ into a metric space $M$, then $f$ is uniformly continuous on $\mathbb{R}$.

Attempt: I think I can use the fact that for all $x \in \mathbb{R}$, $[x,x+p]$ is a closed and bounded interval. Then $f$ is compact and hence uniformly continuous on the interval.

I also tried considering $[0,p]$. In that case, $x = np+\alpha$ and $y = mp + \beta$ for some $m,p \in \mathbb{Z}$ and $\alpha,\beta \in \mathbb{R}$. If $n<0$, then we can choose $\alpha \in \mathbb{R^\mathbf{-}}$, so that $f(np+\alpha)=f(|\alpha|)$ and not $f(1-\alpha)$. Hope that makes sense.

Then $\alpha,\beta \in [0,p]$, and I can find a $\delta$ that works for all $\alpha, \beta$ in $[0,p]$. The problem is, I need to constrain $|x-y|$ and somehow get that $|\alpha-\beta| < \delta$. So far I haven't figured out how to do this.

Been furrowing my brow at this for a while.. any hints very welcomed...

Thanks

Answer 1

For $r>0$ let $s_r\in (0,p)$ such that $\forall x,y\in [0,2p]\;(|x-y|<s_r\implies |f(x)-f(y)|<r). $

For any $x,y\in \mathbb R$ with $|x-y|<s_r$ there exists $n\in \mathbb Z$ such that $\{x,y\}\subset [np, (n+2)p].$ Because if $n=\max \{m\in \mathbb Z: mp\leq \min (x,y)\}$ then $\min (x,y)<(n+1)p,$ so $$np\leq \min (x,y)\leq\max (x,y)<\min (x,y)+s_r<(n+1)p+s_r<(n+2)p .$$

Now $\{x-np,y-np\}\subset [0,2p]$ and $|(x-np)-(y-np)|=|x-y|<s_r.\;$ Therefore $$|f(x)-f(y)|=|f(x-np)-f(y-np)|<r.$$

Answer 2

HINT:

your idea is to translate each point to the interval $[0,p]$ and hope that the translates are also close by. Better to translate both by the same multiple of p, so that the smaller one lands inside, and the larger one is still guaranteed not too far. So you consider the restriction of your function to [0, p+1]. hope you can continue from here.