Show $f(x)=\frac{x}{1+x^{2}}$ is lipschitz continuous.

Question

I have to show $\frac{x}{1+x^{2}}$ is lipschitz continuous. Hence I have to show $$|f(x)-f(y)|= \left| \frac{x}{1+x^{2}}- \frac{y}{1+y^{2}} \right| < M|x-y|,$$ for some $M \in \mathbb{R}$. I know I have to use some form of algebraic factorization but I don't see how. Obviously

$$\left| \frac{x}{1+x^{2}}- \frac{y}{1+y^{2}} \right|=\left| \frac{x+xy^{2}-y-yx^{2}}{(1+x^{2})(1+y^{2})} \right| $$ but how to proceed?

Any tips or hints would be much appreciated.

Answer 1

We have that $$\begin{align}\left| \frac{x}{1+x^{2}}- \frac{y}{1+y^{2}} \right|&=\left| \frac{(1-xy)(x-y)}{(1+x^{2})(1+y^{2})} \right|\\&\leq \frac{1+|xy|}{(1+x^{2})(1+y^{2})} \cdot |x-y|\leq |x-y| \end{align}$$ because $$1+|xy|\leq 1+\frac{x^2+y^2}{2}\leq 1+x^2+y^2+x^2y^2= (1+x^{2})(1+y^{2}).$$

Answer 2

We have $|f'(x)|=\frac{|1-x^2|}{(1+x^2)^2} \le \frac{1+x^2}{(1+x^2)^2} = \frac{1}{1+x^2} \le 1.$

Now invoke the mean value theorem.