$f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $f(x+1)=f(x)\forall x$ we need to show $f$ is uniformly continuous function.
please give me some hint:
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2Hint: Note that $f$ is uniformly continuous on $[0,1]$. – Tomás Jan 14 '13 at 22:39
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well, I knew that $f$ is uniformly continous on any $[x,x+1]$, can I say $f$ is uniformly continuos on $\mathbb{R}$ directly? – Myshkin Jan 14 '13 at 22:42
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Use the periodicity.... – user108903 Jan 14 '13 at 22:43
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@Panu It's not enough without your other assumption. $f(x)=x^2$ is uniformly continuous on any closed interval but is not uniformly continuous on $\mathbb{R}$. – Gyu Eun Lee Jan 14 '13 at 22:44
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1I really not getting, please explain a bit more – Myshkin Jan 14 '13 at 22:45
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You have uniform continuity on any $[x,x+1]$ this doesn't give uniform continuity on $\mathbb{R}$; $f(x)=x^2$ is an example that is uniformly continuous on any $[x,x+1]$ but not uniformly continuous on $\mathbb{R}$. Therefore you need to apply $f(x)=f(x+1)$ in some way. What Tomas and user108903 are suggesting is that since $f$ is uniformly continuous on $[0,1]$ and $f$ is periodic with period 1, if you want a $\delta$ for some $\epsilon$ and you need to consider $x \notin [0,1]$ you can use the periodicity to identify $x$ with some $y\in[0,1]$. – Gyu Eun Lee Jan 14 '13 at 22:47
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on $[0,1]$fix $\epsilon>0$, then we will get a $\delta>0$ such that for any $x_1,x_2$ with $|x_1-x_2|<\delta$ we have $f(x_1)-f(x_2)|<\epsilon$, now for this $\epsilon$ I can take $x=x_2-1\in [0,1]$ and $y=x_2\in [0,1]^c$ so that $|x-y|<\delta$ and $f(x)-f(y)=0$? – Myshkin Jan 14 '13 at 23:02