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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!

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  • $\begingroup$ Let $T$ be the period of $f$, $f:[0,T]\rightarrow R$ is uniformly continuous since $[0,T]$ is compact, you can use that show the result $\endgroup$ – Tsemo Aristide Oct 19 '15 at 0:50
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    $\begingroup$ How do you show $[0,T]$ compact without Heine-Borel? $\endgroup$ – Aaron Zolotor Oct 19 '15 at 0:50
  • $\begingroup$ You have to be careful: Your stated continuity is really uniform continuity, and the stated uniform continuity is not correct. You have to be careful about the quantifier (for all, there exists) when stating the theorems. $\endgroup$ – user99914 Oct 19 '15 at 0:51
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    $\begingroup$ A periodic continuous function on $\mathbb Q$ need not be uniformly continuous. If Heine-Borel is prohibited, what fact about $\mathbb R$ is allowed? $\endgroup$ – GEdgar Oct 19 '15 at 0:54
  • $\begingroup$ We can gather that $f$ is bounded $\endgroup$ – Aaron Zolotor Oct 19 '15 at 0:54
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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 $\Bbb{R}$.

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    $\begingroup$ This assumes that $f$ is uniformly continuous on a closed interval but we don't have that given? $\endgroup$ – Aaron Zolotor Oct 19 '15 at 2:42
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    $\begingroup$ $f$ is continuous on a closed interval means it is uniformly continuous. Read it again. $\endgroup$ – Math Wizard Oct 19 '15 at 2:54
  • $\begingroup$ I apologize for missing the obvious but based on what? $\endgroup$ – Aaron Zolotor Oct 19 '15 at 2:55
  • $\begingroup$ It is a well known theorem in analysis, i.e. continuous function is always uniform continuous on compact set. $\endgroup$ – Math Wizard Oct 19 '15 at 2:56
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    $\begingroup$ Not quite but I spoke with my professor and he informed me it is impossible to prove this without Heine-Borel $\endgroup$ – Aaron Zolotor Oct 19 '15 at 21:56

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