Decide whether or not $f(x)=\frac{1}{x^2+1}$ is uniformly continuous on $\mathbb{R}$.
The definition for a function to be uniformly continuous is $\forall \epsilon >0, \exists \delta>0$ such that if $x,y \in D$ and $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$
I need help seeing if it is uniformly continuous
Best Answer
You have $$\begin{aligned} \left\vert f(x)-f(y) \right\vert &= \left\vert \frac{1}{x^2+1}-\frac{1}{y^2+1}\right\vert\\ &=\frac{\left\vert x^2-y^2 \right\vert}{(x^2+1)(y^2+1)}\\ &\le \left\vert x-y\right\vert \frac{\vert x \vert +\vert y \vert}{(x^2+1)(y^2+1)}\\ &\le \vert x-y \vert \end{aligned}$$
as $2\vert x \vert \le x^ 2+1 \le (x^2+1)(y^2+1)$ and similar inequality for $\vert y \vert$.
This enables to prove the claim that $\delta=\epsilon$ works in the definition of uniform continuity.