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How to show $f(x)$=$\frac{1}{1+x^2}$ is uniform continuous on $\Bbb R$.

Although, of course for any interval $[a,b]$, this function is continuous and bounded, therefore also uniformly continuous. Following Continuous Extension Theorem it is uniformly continuous on any $(a,b)$. Therefore proceeding this way, we can show it is uniformly continuous on $ \Bbb R$.

I wish to prove the same analytically. I assumed there exists $x,u \in \Bbb R$, such that $ |x-u|< \delta$.

Now,

$|f(x)-f(u)|$=$\frac {|x^2-u^2|}{|(1+x^2)(1+u^2)|}$ $\le$ $\frac{|x-u||x+u|}{x^2u^2}$ $\le$ $\delta$$\frac{|x+u|}{x^2u^2}$.

Here I stuck. I wish to find an $\epsilon$ so that the $|f(x)-f(y)|\lt \epsilon$, where $\delta$ depends only on $\epsilon$, not on $x$. But unable to do that. Tried to apply A.M-G.M inequality but could not find a fruitful result. What to do?

Peter
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L--
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    Note that uniform continuity on each interval $(a, b)$ does not imply uniform continuity in $\Bbb R$. – Martin R Jan 03 '18 at 13:12
  • Also: https://math.stackexchange.com/q/2101969, https://math.stackexchange.com/q/485133, https://math.stackexchange.com/q/2212000, https://math.stackexchange.com/q/593313 – all found quickly with Approach0 – Martin R Jan 03 '18 at 13:20
  • Will you please explain why uniform continuity on each interval $(a,b)$ does not imply uniform continuity in $\Bbb R$. I am not getting it actually... @Martin R – L-- Jan 03 '18 at 13:48
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    Example: $f(x) = x^2$. – Martin R Jan 03 '18 at 13:50
  • Take $f (x) = x^2$. It is uniformly continuous on every closed interval but not on $\mathbb {R} $ –  Jan 03 '18 at 13:50
  • Yes, I got it. Thank you so much....@ Martin R, @Math_QED – L-- Jan 03 '18 at 13:52

5 Answers5

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Note that the function is Lipschitz, as for any $x,y$ we have $$ |f(x)-f(y)|\leq \sup_{x\in \mathbb{R}}|f'(x)||x-y|=\frac{3\sqrt{13}}{16}|x-y| $$ Allowing you to choose $$ \delta=\frac{16\epsilon}{3\sqrt{13}} $$

operatorerror
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If $f'$ is bounded in $\mathbb{R}$, then $f$ is uniform continuous. Indeed you know that $$ \left|f\left(x\right)-f\left(u\right)\right| \leq \underset{c \in \mathbb{R}}{\text{sup}} \ f'\left(c\right) \left|x-u\right| $$ Hence the result. ( Lipschitz continuous imples uniformly continous )

Atmos
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You can try to prove the following general result which is interesting by itself. A continuous real function defined on $\mathbb R$ which has finite limits on $\pm \infty$ is uniform continuous.

Then you just have to apply this result to your function $f$.

mathcounterexamples.net
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To complement the other answers; if you do wish to continue on your path:

$|f(x)-f(u)|$=$\color{blue}{\frac {|x^2-u^2|}{|(1+x^2)(1+u^2)|}}$ $\le$ $\frac{|x-u||x+u|}{x^2u^2}$ $\le$ $\delta$$\frac{|x+u|}{x^2u^2}$.

Picking up at the blue step:

$$\begin{align} \frac {|x^2-u^2|}{|(1+x^2)(1+u^2)|} & \le |x-u|\frac{|x|+|u|}{(1+x^2)(1+u^2)} \\[6pt] & \le|x-u|\left(\frac{|x|}{1+x^2}+\frac{|u|}{1+u^2}\right) \\[6pt] & \le 2|x-u| \end{align}$$

StackTD
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$f(x)=\frac{1}{1+x^2}\implies f'(x)=-\frac{2x}{(1+x^2)^2} $

For $|x|\le1$ we have $$\frac{|x|}{(1+x^2)^2} \le\frac{1}{(1+x^2)^2}\le 1$$

For $|x|\ge1$ we have $$|x|\le x^2 \le (1+x^2)^2\implies \frac{|x|}{(1+x^2)^2} \le 1$$

Hence,

$$|f'(x)|=\frac{2|x|}{(1+x^2)^2} \le 2\implies |f(x)-f(y)|\le 2|x-y|$$

This shows that $f$ is Lipschitz therefore, uniformly continuous.