# Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.

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}

• No other answers used the mean value theorem, so I added one that does. Oct 30, 2014 at 1:30
• What about using Heine Cantor on each compact and this result ? math.stackexchange.com/questions/705961/… Sep 9, 2020 at 20:16

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 $\delta=\epsilon$.

• How did you get the 2 in the denominator? $$|x - a| (\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)})\le |x - a| (\frac{1}{2(1 + a^2)} + \frac{1}{2(1 + x^2)})\le |x-a|$$ Dec 13, 2014 at 8:15
• $\frac{|x|}{1+x^2} < 1$ 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 $\frac{|x|}{2(1+x^2)} + \frac{|a|}{2(1+a^2)} < 1$ May 29, 2018 at 16:06

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.

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

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

• This is actually true for all $x$, not just $x>0$. Mar 25, 2015 at 22:10

$$x^2 \geq 0 \implies 1+x^2 > 1 \implies \frac{1}{1+x^2} < 1$$

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 $$\delta = Inf\{1, \frac{\epsilon}{2|a|+1}\}$$

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.