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So I'm new to uniform continuity and this is just an exercise that was scribbled on the board in my real analysis class. But it's is a tricky question for me:

if $f:R\to R$ is periodic with period P (so $f(x+P)=f(x)$) and $f$ is continuous on $(0,p+a)$ for some $\epsilon >0$, then $f$ is uniformly continuous

how can we go about proving this?

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There is a well-known theorem about uniform continuity of a real-valued function -- which does most of the work in answering this question.

Theorem: A continuous function with a compact domain is uniformly continuous.

So the restriction of the given function $f(x)$ to the interval $[0,P]$ is uniformly continuous. Now since $f(x)$ is periodic we can always relate its "rate of change" $\Delta_\varepsilon(x) = \displaystyle \frac{f(x+\varepsilon)-f(x)}{\varepsilon}$ to $\Delta_\varepsilon(y)$ for some $y \in [0,P]$.

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  • $\begingroup$ thanks so much! Is that Heine's theorem that Bernard above mentioned? I'm familiar with Heine-Borel and applying that to find compactness, but I didn't know I could apply any of that to continuity. $\endgroup$
    – Tye
    Oct 27, 2015 at 17:45
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    $\begingroup$ Wikipedia lists it as the Heine-Cantor theorem. Note the function has to be continuous in the first place to conclude it is uniformly continuous. Think of the theorem as a generalisation of how if the function is continuously differentiable, then the derivative is a function on a compact space, hence bounded, so the "rate of change" is bounded -- which is essentially uniform continuity. $\endgroup$
    – Daron
    Oct 27, 2015 at 22:50

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