Showing that a function is Uniformly continuous

Question

Show that the function $f(x)= \frac{1}{1+x^2}$ for $x \in \mathbb{R}$ is uniformly continuous on $\mathbb{R}$.

Attempt: Let $\epsilon>0$ be given . Choose $\delta =1$. Then, for $|x-u|<1 \iff u-1<x<u+1$ we have that:

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

Then I tried to manipulate this expression further, tried to put bounds on the expression so that I have something of the sort $M|x-u|$ that I could say, is, less than $\epsilon$. However, I realised that whatever I did, I'd end up with a $\delta$ that is not only dependent on $\epsilon$ but on the value of $u$ as well.

Next, I tried if I could show that the function is lipschitz; however, I failed at doing that so either. (Perhaps, it is not lipschitz as uniform continuity does not necessary imply that the function is lipschitz, hence, I may be wrong there).

My question is:

Can anyone show me how to prove that the function is uniformly continuous on $\mathbb{R}$? I would appreciate if the answer would entail the thought process that went behind to arrive at the answer so I can possibly employ/use similar reasoning to prove uniform continuity of other functions.

Thank you.

Answer 1

For $\forall\epsilon>0$, $\exists \delta>0$ such that for $\forall x,y\in \mathbb{R}$: $$|x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon.$$ We have: $$|f(x)-f(y)|=\left|\frac{1}{1+x^2}-\frac{1}{1+y^2}\right|=\left|\frac{y^2-x^2}{(1+x^2)(1+y^2)}\right|=\left|\frac{(x-y)(x+y)}{(1+x^2)(1+y^2)}\right|=\\ \frac{|x+y|}{(1+x^2)(1+y^2)}\cdot |x-y|\le \frac{|x|+|y|}{(1+x^2)(1+y^2)}\cdot |x-y|=\left(\frac{|x|}{1+x^2}+\frac{|y|}{1+y^2}\right)\cdot |x-y|<\\ \left(\frac12+\frac12\right)\cdot |x-y|=|x-y|=\delta=\epsilon.$$ Note that the following inequalities are used: $$|x+y|\le |x|+|y|,\\ 1+x^2\ge 2|x|.$$

Answer 2

Note that $\bigl|f'(x)\bigr|=\left|-\frac{2x}{(1+x^2)^2}\right|\leqslant\frac{3\sqrt3}8$. Therefore, by the intermediate value theorem, if $x\neq y$,$$\bigl|f(x)-f(y)\bigr|\leqslant\frac{3\sqrt3}8|x-y|.$$You can also use the fact that\begin{align}\left|\frac{y^2-x^2}{(1+x^2)(1+y^2)}\right|&\leqslant\left(\frac{|x|}{1+x^2}\times\frac1{1+y^2}+\frac{|y|}{1+y^2}\times\frac1{1+x^2}\right)|y-x|\\&\leqslant|y-x|.\end{align}