Is $f(x)=\frac{1}{x^2+1}$ uniformly continuous on $\mathbb{R}$?

Question

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

Answer 1

Hint: There exists some $R$ such that $|f(x)| < \frac{\epsilon}{2}$ if $|x| >R$.

Hint: $f(x)$ is continuous on the closed interval $[-R-1, R+1]$ therefore uniformly continuous.

Answer 2

We shall use the elementary fact that $\frac {2|ab|} {a^2 + b^2} \le 1$ (because $(|a| - |b|)^2 \ge 0$).

Notice that $f'(x) = \frac {-2x} {(x^2+1)^2}$ which means $|f'(x)| = \frac {2 |1 \cdot x|} {x^2 + 1} \frac 1 {x^2 + 1} \le 1 \cdot 1 = 1$ for all $x \in \Bbb R$.

Using Lagrange's mean value theorem, for arbitrary $x, y \in \Bbb R$ there exist $z$ between $x$ and $y$ such that $|f(x) - f(y)| = |f'(z)| \ |x-y|$. Using the previous paragraph, $|f'(z)| \le 1$, which implies that for arbitrary $x,y \in \Bbb R$ we have $|f(x) - f(y)| \le |x-y|$, showing that $f$ is a Lipschitz function. It is then easy to prove that it is also uniformly continuous (just take $\delta = \varepsilon$).

This (showing Lipschitzianity using Lagrange's mean value theorem) is a standard approach to showing uniform continuity.

Answer 3

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.