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The definition of uniform continuity states that a function is uniformly continuous if, given any challenge $\epsilon > 0$ that there exists a response $\delta > 0$ for every value $x_1,x_2 \in R$ s.t. $|f(x_1)-f(x_2)| < \epsilon$ implies that $|x_1-x_2| < \delta$.

I found that for the function $f(x)$, in order to satisfy $|x1-x2| < \delta$,

$$\left|\frac{1}{1+x_1^2}-\frac{1}{1+x_2^2}\right| < \epsilon$$

I simplified that to $\left|\frac{x_2^2-x_1^2}{(1+x_1^2)(1+x_2^2)}\right| < \epsilon$, but I wasn't sure how to simplify or modify that expression to get $\left|x_1-x_2\right|$ in terms of $x_1, x_2$, and $\epsilon$. Any help would be appreciated!

Steven Stadnicki
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Louis
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  • This is not the definition of uniformly continuous. The implication goes the other way (i.e. $|x_1-x_2| < \delta$ implies that $|f(x_1) - f(x_2)| < \epsilon$). – Joshua Mundinger Nov 24 '14 at 03:18
  • Also you have to be careful and observe that $\delta$ can only depend on $\epsilon$ and nothing else. The definition for uniform convergence is much stricter than normal convergence. – user60887 Nov 24 '14 at 03:38
  • More generally, every continuous function in $\mathbb{R}$ with limits at infinity is uniformly continuous. – Aloizio Macedo Nov 24 '14 at 03:47
  • Roll pack your hiding of the question because if you dont, someone else will. – dustin Nov 24 '14 at 20:50
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    It is considered very bad form to destructively edit a question after receiving your answer. – Steven Stadnicki Nov 24 '14 at 20:50
  • @Louis I'm assuming you were trying to delete your question, but couldn't because it already had responses. I'm not sure why you wanted to delete it, but you should know if you plan to continue using the site that once someone answers your question, you can't delete the question. It remains on the site. So if these happen to be homework questions and you are trying to cover up your tracks, it doesn't work here. I'm not trying to accuse you of anything, but I'm just letting you know how the site runs. – layman Nov 24 '14 at 20:52
  • @StevenStadnicki It is my hypothesis that individuals who do that are hiding a question that is worth credit in a course. – dustin Nov 24 '14 at 20:52
  • @dustin I suspect that is often the case, but don't want to presume the worst (and didn't want to accuse here). – Steven Stadnicki Nov 24 '14 at 20:53

3 Answers3

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This is one of the times where instead of proving the result directly, it's often easier to prove that the function is differentiable with a bounded derivative. That in turn will imply that the function is lipshitz continuous, which implies uniformly continuous. Your derivative is $f'(x)=\frac {-2x} {(1+x^2)^2}$, note $|f'(x)|=2|\frac x {(1+x^2)^2}|$, and since $|x|\le |(1+x^2)^2|$, we have $|f'(x)|\le 2$, hence bounded derivative, and we're done

I highly recommend this technique because it's often easier than playing with epsilons and deltas :)

Alan
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$\left|\dfrac{1}{1+x_1^2} - \dfrac{1}{1+x_2^2}\right| \leq |x_1-x_2|\cdot \dfrac{|x_1|+|x_2|}{(1+x_1^2)(1+x_2^2)} \leq |x_1-x_2|\cdot \left(\dfrac{|x_1|}{1+x_1^2} + \dfrac{|x_2|}{1+x_2^2}\right) = |x_1-x_2|\cdot \left(\dfrac{1}{2} + \dfrac{1}{2}\right) = |x_1-x_2|$.

DeepSea
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  • First, it may not be a good idea to split a line that way. Imo, it's better to close the line with the dollar sign (or double) and begin a new one. Second, how did you get your one-before-the-last equality? – Timbuc Nov 24 '14 at 03:47
  • I know that: why did you write an equality, though? And is that comma after that $;x_1;$ at the right end of the first line a typo? – Timbuc Nov 24 '14 at 03:50
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From where you got:

$$\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|=|x-y|\,\left|\frac{x+y}{(1+x^2)(1+y^2)}\right|\le|x-y|$$

since

$$\frac{|x+y|}{(1+x^2)(1+y^2)}\le1\iff |x+y|\le1+x^2+y^2+x^2y^2\;\iff$$

$$\iff \pm(x+y)\le x^2y^2+x^2+y^2+1\iff x^2y^2+\left(x\pm\frac12\right)^2+\left(y\pm\frac12\right)^2+\frac12\ge0$$

Timbuc
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