Let $f :\mathbb R\to \mathbb R$ be a continuous function such that $f(x+1)=f(x) , \forall x\in \mathbb R$ i.e. $f$ is of period $1$ , then how to prove that $f$ is uniformly continuous on $\mathbb R$ ?
2 Answers
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Take any interval $[x,x+1]$. Then f is a continuous function defined on a compact interval, so it is uniformly-continuous there. Then use the pair $\delta -\epsilon$ that works in $[0,1]$ anywhere in the Real line by periodicity of $f$. Another way of looking at this is that this is a function defined on the compact set $[0,1]$/~ , where $0$~$1$, which is homeomorphic to $S^1$, a compact space. Then $f$ is continuous, defined on a compact space, so uniformly-continuous.
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$\begingroup$ Well, this doesn't work exactly. Say $x = 0$ (this choice is as good as any). Then pairs separated by delta could have representatives like $\delta/2$ and $1-\delta/2$. So you either need to use $2\epsilon$ or an interval $[0,2]$ or some other trick. $\endgroup$– ronnoMay 31, 2014 at 8:36
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$\begingroup$ @ronno: Please see my edit; I am defining $f$ on (the compact space) $S^1$. $\endgroup$ May 31, 2014 at 8:44
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$\begingroup$ And I think the only problems you may run into are at the corners $x, x+1$ , and not in the center. $\endgroup$ May 31, 2014 at 8:48
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$\begingroup$ The $S^1$ argument can be made to work, although I don't think it's complete as it stands. And I agree that the issue is at the endpoints. I just wanted to mention that the $\delta$ that works on $[0,1]$ for a given $\epsilon$ need not work on $\mathbb{R}$. $\endgroup$– ronnoMay 31, 2014 at 8:56
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$\begingroup$ I don't see anything wrong with it; a continuous function on a compact set is uniformly-continuous, and f periodic of period 1 is equivalent to a function defined on $S^1$. $\endgroup$ May 31, 2014 at 9:05
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First take $I = [0,2]$. Since $I$ is compact and $f$ is continuous on $I$ we know $f$ is uniformly continuous on $I$. Now let $x,y \in \mathbb{R}$. Let $\epsilon > 0$, since $f$ is uniformly continuous we know $\forall z,w \in I \; \exists \delta' > 0$ where $$ |z - w| < \delta' \implies |f(x) - f(y)| < \epsilon $$ Now put $\delta = \min \{ \delta', 1 \}$. For every $x,y \in \mathbb{R}$ so that $$ | x - y | < \delta $$ we know that $\exists n \in \mathbb{Z}$ so that $x + n,y+n \in I$. So, finally $$ \begin{eqnarray} | x - y | < \delta & \implies & | x + n - y - n | < \delta \\ & \implies & | x + n - y - n | < \delta' \text{ since } \delta \le \delta'\\ & \implies & | f(x+n) - f(y + n) | < \epsilon \\ & \implies & |f(x) - f(y)| < \epsilon \end{eqnarray} $$
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$\begingroup$ Any chance the downvoter can provide constructive criticism? I feel like my last line needs explanation but I'm unable to come up with something concrete. $\endgroup$– DanZimmMay 31, 2014 at 10:54
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$\begingroup$ @LuisValerin I don't follow why you can take $x'$ and $y'$ in the same period. Could you explain? $\endgroup$– DanZimmMay 31, 2014 at 11:14
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$\begingroup$ @LuisValerin I still don't follow what you're saying. We know for $x,y \in I$ that the implication holds. So seeing that $x',y' \in I$ we have the implication that says $|f(x')-f(y')| < \epsilon$ which in turns means $|f(x)-f(y)| < \epsilon$. Now if $x,y$ are within $\min\{\delta',1-\delta'\}$ then $x',y'$ are within $\delta'$ so it follows. $\endgroup$– DanZimmMay 31, 2014 at 11:22
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$\begingroup$ You right we cannot take in the same period. $\endgroup$– ValentMay 31, 2014 at 11:49
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1$\begingroup$ You should assume $\delta \le 1$ to ensure the existence of $n$. Anyway, changed to +1. $\endgroup$– ronnoJun 3, 2014 at 6:51