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Prove that the function $$f(x) = \frac1{1+x^2}$$ is uniformly continuous on the interval $[0, \infty)$.

One part of my proof is $$\begin{align*} \left|\frac1{1+x^2}-\frac1{1+y^2}\right| &= \left|\frac{y^2-x^2}{(1+x^2)(1+y^2)}\right| \\ & = \frac{|y-x||y+x|}{|1+x^2||1+y^2|}\end{align*}$$

My question is how to make that to equal to $\epsilon$?

AlexR
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afsdf dfsaf
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1 Answers1

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There's no need of the hypothesis $x,y\in[0,\infty)$, they can be arbitrary real numbers.

Notice that $$ \frac{|x|}{1+x^2}\le1 $$ so you can work with $$ \frac{|x-y|\,|x+y|}{(1+x^2)(1+y^2)}\le |x-y|\left(\frac{|x|}{(1+x^2)(1+y^2)}+\frac{|y|}{(1+x^2)(1+y^2)}\right) $$

On the other hand you have $$ f'(x)=\frac{-2x}{(1+x^2)^2} $$ and so $$ |f'(x)|=2\frac{|x|}{1+x^2}\frac{1}{1+x^2}\le 2 $$ so the function has bounded derivative, which implies it's Lipschitz, hence uniformly continuous.

egreg
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  • $< |x-y|\left( \frac{1}{1+y^2} + \frac{1}{1+x^2} \right)< \epsilon, \space \therefore |x-y| < \frac{\epsilon}{\left( \frac{1}{1+y^2} + \frac{1}{1+x^2} \right)}?$ Is this correct? Or am I totally off? – Zhoe Dec 05 '13 at 00:43
  • @Zhoe We have proved that $|f(x)-f(y)|\le 2|x-y|$; so if we set $\delta=\varepsilon/2$, we have that if $|x-y|<\delta$, then $|f(x)-f(y)|<\varepsilon$. – egreg Dec 05 '13 at 00:48
  • I understand that...So, $\left( \frac{|x|}{(1+y^2)(1+x^2)} + \frac{|y|}{1+x^2)(1+y^2)} \right) < 2?$ – Zhoe Dec 05 '13 at 00:53
  • @egreg: you notice that $$ \frac{|x|}{1+x^2}\le1 $$. it is because we need to make $$ \frac{|x-y|,|x+y|}{(1+x^2)(1+y^2)}\le |x-y|\left(\frac{|x|}{(1+x^2)(1+y^2)}+\frac{|y|}{(1+x^2)(1+y^2)}\right) $$ < |x-y|2? – afsdf dfsaf Dec 05 '13 at 01:45
  • @afsdfdfsaf Yes, that's the idea, which is confirmed by the derivative. – egreg Dec 05 '13 at 10:46
  • @egreg: we take a step further to say $$ \frac{|x|}{1+x^2}<1 $$ right? – afsdf dfsaf Dec 05 '13 at 14:28