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Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.

Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$

Then suppose $x, a$ are elements of $\Bbb R. $ Now \begin{align} |f(x) - f(a)| &= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| \\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right| \\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)} \\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)} \\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right] \end{align}

I don't know how to simplify more. Can someone please help me finish? Thank very much.

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You are nearly finishing the proof.

$$|x - a| (\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)})\le |x - a| (\frac{1}{2(1 + a^2)} + \frac{1}{2(1 + x^2)})\le |x-a|$$

Take $\delta=\epsilon$.

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    $\begingroup$ How did you get the 2 in the denominator? $$|x - a| (\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)})\le |x - a| (\frac{1}{2(1 + a^2)} + \frac{1}{2(1 + x^2)})\le |x-a|$$ $\endgroup$ – Su003 Dec 13 '14 at 8:15
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    $\begingroup$ $\frac{|x|}{1+x^2} < 1$ and $\frac{|a|}{1+a^2} < 1$, so we add the two inequalities to get - $\frac{|x|}{1+x^2} + \frac{|a|}{1+a^2} < 1+1$ so $\frac{|x|}{2(1+x^2)} + \frac{|a|}{2(1+a^2)} < 1$ $\endgroup$ – BAYMAX May 29 '18 at 16:06
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According to the mean value theorem, $$ \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| = f'(c)|x-a| $$ where $f(x)=\dfrac 1 {1+x^2}$ and $c$ is somewhere between $x$ and $a$. But $|f'(c)|\le\max |f'|$, the absolute maximum value of $|f'|$. In order for this to make sense, you need to show that $|f'|$ does have an absolute maximum value. But that is not hard. So you have $$ |f(x)-f(a)|\le M|x-a| \text{ for ALL values of $x$ and $a$}, $$ (where $M$ is the absolute maximum of $|f'|$). So $f$ is Lipschitz-continuous and therefore uniformly continuous.

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Hint: $$\frac{|x|}{(1+x^2)(1+a^2)} \leq \frac{|x|}{1+x^2} < 1$$ and $$\frac{|a|}{(1+x^2)(1+a^2)} \leq \frac{|a|}{1+a^2} < 1$$

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Hint: use the inequality $$x>0\implies x < 1 + x^2 $$ (if $x<1$ it is true; otherwise via multiplication by $x$, $x>1\implies x^2>x$)

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  • $\begingroup$ This is actually true for all $x$, not just $x>0$. $\endgroup$ – StevenClontz Mar 25 '15 at 22:10
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$$x^2 \geq 0 \implies 1+x^2 > 1 \implies \frac{1}{1+x^2} < 1$$

Using the above inequality,

$$ \begin{align} |f(x)-f(a)| &\leq |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}\\ &\leq |x - a||x+a|\\ &\leq |x - a|(|x|+|a|) \end{align} $$ Choose $$|x-a| \leq 1 \implies |x|\leq |a|+1$$ Now, $$ \begin{align} |f(x)-f(a)| &\leq |x - a|(|x|+|a|)\\ &\leq |x-a|(2|a|+1) \end{align} $$ Choose $$|x-a| < \frac{\epsilon}{2|a|+1} $$ Now, $$ \begin{align} |f(x)-f(a)| &\leq \epsilon \end{align} $$ where $$\delta = Inf\{1, \frac{\epsilon}{2|a|+1}\}$$

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

Then observes that; For $|x|\le1$ $$\frac{|x|}{(1+x^2)^2} \le\frac{1}{(1+x^2)^2}\le 1$$

and for $|x|\ge1$ $$|x|\le x^2 \le (1+x^2)^2\implies \frac{|x|}{(1+x^2)^2} \le 1$$

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 which implies, that $f$ is uniformly continuous.

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