I began this way
Assume we have a periodic continuous function $f$. That is, there exists a $\delta$ such that if $0<|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\varepsilon$ and $f(x+h)=f(x)$ for all $x$ and some $h\neq 0$. We wish to show that $f$ is uniformly continuous. That is, there exists a $\delta$ such that $0<|x-y|<\delta$ implies that $|f(x)-f(x+y)|<\varepsilon$ where $\varepsilon>0$.
I know this seems like a duplicate but I have looked at the solutions and hints already posted and it is not. I need to do this without HB or anything past it. Thanks for your help!
Best Answer
Let $T$ be the period of $f$. On $[0,T+\zeta], \:\zeta>0$, $f$ is continuous and thus uniform continuous. So given $\epsilon>0$, there is a $0<\delta<\zeta$ such that for any $x,y\in[0,T+\zeta],\:|x-y|<\delta$, there is $$|f(x)-f(y)|<\epsilon\tag1$$
Now consider any $x',y'\in\Bbb{R}$ that $|x'-y'|<\delta$. WLOG, assume $x'<y'$.
If $nT\leqslant x'<(n+1)T$ and $nT\leqslant y'<(n+1)T$, then there are $x,y\in[0,T]$ that $f(x)=f(x')$ and $f(y)=f(y')$. So by $(1)$ $$ |f(x')-f(y')|=|f(x)-f(y)|<\epsilon $$ If $nT\leqslant x'<(n+1)T$ and $(n+1)T\leqslant y'<(n+2)T$, then there are $x,y\in[0,T+\zeta]$ that $f(x)=f(x')$ and $f(y)=f(y')$ for $\delta<\zeta$. So by $(1)$ again $$ |f(x')-f(y')|=|f(x)-f(y)|<\epsilon $$ So $f$ is uniform continuous on $\Bbb{R}$.