uniform continuity of $f(x) = \frac{x}{1 +x^2}$ on $\mathbb{R}$

Question

$f(x) = \frac{x}{1 +x^2}$ on $\mathbb{R}$ to comment about its uniform continuity, I tried to figure out the definition " A function $f(x)$ is said to be uniformly continuous on a set $S$, if for given $\epsilon > 0$, there exists $\delta > 0$ such that $x$, $y \in S$, $|x − y| < \delta ⇒ |f(x) − f(y)| < \epsilon$" but couldn't get where to start.

then i thought about breaking down the interval into some closed and open intervals and as $f(x)$ is continuous in closed bounded interval so it would be uniformly continuous as well. also i observed that limit $x$ tending to infinity and minus infinity exist and is equal to zero but don't know how to use it.

any hint would be highly appreciated

Answer 1

It is enough to prove that $f$ is Lipschitz continuous. By simple calculation, $$ \left| {f(x) - f(y)} \right| = \frac{{\left| {xy - 1} \right|}}{{(x^2 + 1)(y^2 + 1)}}\left| {x - y} \right|. $$ Observe that $$ \left| {xy - 1} \right| \le |xy|+1=\sqrt {x^2 y^2 } + 1 \le \frac{{x^2 + y^2 }}{2} + 1 = \frac{{(x^2 + 1) + (y^2 + 1)}}{2}. $$ Whence $$ \frac{{\left| {xy - 1} \right|}}{{(x^2 + 1)(y^2 + 1)}} \le \frac{{(x^2 + 1) + (y^2 + 1)}}{{2(x^2 + 1)(y^2 + 1)}} = \frac{1}{{2(y^2 + 1)}} + \frac{1}{{2(x^2 + 1)}} \le \frac{1}{2} + \frac{1}{2} = 1. $$ Therefore, $|f(x)-f(y)|\le |x-y|$ for all $x,y\in \mathbb R$.

Answer 2

Hint

By MVT,

$f(x)-f(y)=f^\prime(c)(x-y)$

with $|f^\prime(c)|=\bigg{|}\dfrac{1-c^2}{(1+c^2)^2} \bigg{|}\leq \ \dfrac{1+c^2}{(1+c^2)^2}\ \leq \dfrac{1}{(1+c^2)} \leq 1$

follows $|f(x)-f(y)|\leq |x-y|$ for all $x,y \in \mathbb{R}$.