# Math Help - prove uniform continuity given a periodic and continuous function

1. ## prove uniform continuity given a periodic and continuous function

I'm stuck on how to do this proof.

Originally Posted by Gaugan. Introduction To Analysis, 5th ed, p105
A function $f:R\rightarrow R$ is periodic iff there is a real number $h\neq0$ such that $f(x+h)=f(x)$ for all $x\in R$. Prove that if $f:R\rightarrow R$ is periodic and continuous, then $f$ is uniformly continuous.
I thought I might start like this...

Choose $\epsilon>0$. By definition of continuity, for each $z_{\alpha}\in R$, there exists $\delta_{\alpha}>0$ such that if $|z-z_{\alpha}|<\delta_{\alpha}$ and $z\in R$, then $|f(z)-f(z_{\alpha})|<\epsilon$.

That way, if I can somehow show that $\inf\{\delta_{\alpha}\}>0$, then I could just finish thusly:

Choose $\delta=\inf\{\delta_{\alpha}\}$. Then for each $x\in R$ and $y\in\{z_{\alpha}\}=R$, if $|x-y|<\delta$ and $x\in R$, then $|f(x)-f(y)|<\epsilon$.

But how do I show that $\inf\{\delta_{\alpha}\}>0$? Or, if that's impossible, then perhaps instead of setting $\delta_{\alpha}>0$, I could set it to $\delta_{\alpha}\in(0,h/4)$, and then somehow show a relationship between $|x-y|$ and $|z-z_0|$. But I wouldn't know how to do that, either.

2. Do you know that if a function is continuous on a closed interval then it is informally continuous on that interval?
If $h>0$ in the definition of periodic, then is the function continuous on $[0,h]$?
What can you do with that?

3. Originally Posted by Plato
Do you know that if a function is continuous on a closed interval then it is informally continuous on that interval?
If $h>0$ in the definition of periodic, then is the function continuous on $[0,h]$?
What can you do with that?
Well that would be a simple matter.

Choose $\alpha_n,\beta_n\in R$ with $\alpha_n<\beta_n$ and $\cup_{n=1}^{\infty}[\alpha_n,\beta_n]=R$ $\; \Rightarrow \;$ $[\alpha_n,\beta_n]$ is closed and bounded* $\; \Rightarrow \;$ $[\alpha_n,\beta_n]$ is compact and $f$ is continuous over $[\alpha_n,\beta_n]$ $\; \Rightarrow \;$ $f$ is uniformly continuous over $R$.

EDIT: Or am I missing something, here? The textbook obviously wants me to incorporate $h$ somehow into my proof, which I have not done.

*- Does "bounded" here mean that $f$ is bounded over the interval, or that the interval itself is bounded? My textbook indicates the latter, but that doesn't make sense to me. Aren't all closed intervals bounded?

4. Is it true that on any interval $[a,b]$ such that $b-a<\delta the graph of $f$ is identical to its graph an equivalent sub-interval of $[0,2h]$?

5. Originally Posted by Plato
Is it true that on any interval $[a,b]$ such that $b-a<\delta the graph of $f$ is identical to its graph an equivalent sub-interval of $[0,2h]$?
Of course. But I do not see how this helps. In fact, I do not see what if anything is wrong with the proof from my last post.

By the way, thanks for the help so far. I really appreciate it.

6. Originally Posted by hatsoff
Of course. But I do not see how this helps. In fact, I do not see what if anything is wrong with the proof from my last post.
By the way, thanks for the help so far. I really appreciate it.

7. Originally Posted by Plato
Choose $n\in \mathbb{Z}$. Then $z\in[0,h]\Rightarrow(z+hn)\in[hn,hn+h]$ with $f(z)=f(z+hn)$. Since $f$ is uniformly continuous over every interval $[hn,hn+h]$ and $\cup_{n\in \mathbb{Z}}[hn,hn+h]=R$, $f$ is uniformly continuous over $R$.

But I still don't see why $h$ is significant. Why not any real number?

8. Originally Posted by hatsoff
But I still don't see why $h$ is significant. Why not any real number?
This is my last comment on this topic!
Any periodic function in essence copies the graph on intervals over and over again.
Thus if it is uniformly continuous of an interval what does that mean?

9. Originally Posted by Plato
This is my last comment on this topic!
Any periodic function in essence copies the graph on intervals over and over again.
Thus if it is uniformly continuous of an interval what does that mean?
For any $x,y\in R$ with $|x-y|<|h|$, then $\exists\;a,b\in[0,|h|]$ with $f(x)=f(a)$ and $f(y)=f(b)$, and $|x-y|=|a-b|$. Since $f$ is continuous over the compact interval $[0,|h|]$, then $\exists\;\delta\in(0,|h|)$ so that $|x-y|<\delta\;\Rightarrow\;|a-b|<\delta\;\Rightarrow\;|f(a)-f(b)|<\epsilon\;\Rightarrow\;|f(x)-f(y)|<\epsilon$. So $f$ is uniformly continuous.
Choose $\alpha_n,\beta_n\in R$ with $\alpha_n<\beta_n$ and $\cup_{n=1}^{\infty}[\alpha_n,\beta_n]=R$ $\; \Rightarrow \;$ $[\alpha_n,\beta_n]$ is closed and bounded $\; \Rightarrow \;$ $[\alpha_n,\beta_n]$ is compact and $f$ is continuous over $[\alpha_n,\beta_n]$ $\; \Rightarrow \;$ $f$ is uniformly continuous over $R$.