Lipschitz constant for $f(x) = 1/(1+x^2)$ from the definition of Lipschitz

Question

Lipschitz constant for $f(x) = 1/(1+x^2)$ from the definition of Lipschitz

Bit confused on how to do this from the definition instead of using any derivatives

So far I've reduced it to $$|f(x) - f(y)| \leqslant |(x+y)(x-y)|$$ by simplification and noting that $f(x) \le 1$.

Answer 1

\begin{align*} |f(x)-f(y)|&=\dfrac{|x-y||x+y|}{(1+|x|^{2})(1+|y|^{2})}\\ &\leq\dfrac{|x-y|(|x|+|y|)}{(1+|x|^{2})(1+|y|^{2})}, \end{align*} now \begin{align*} |x|+|y|=(|x|,1)\cdot(1,|y|)\leq(1^{2}+|x|^{2})^{1/2}(1^{2}+|y|^{2})^{1/2}, \end{align*} so \begin{align*} \dfrac{|x|+|y|}{(1+|x|^{2})(1+|y|^{2})}\leq\dfrac{1}{(1+|x|^{2})^{1/2}(1+|y|^{2})^{1/2}}<1. \end{align*}

Answer 2

Correct me if wrong:

$|f(x)-f(y)| =|\dfrac{y^2-x^2}{(1+x^2)(1+y^2)}|$

$= \dfrac{|y-x||x+y|}{(1+x^2)(1+y^2)}.$

Note: $|x+y| \le |x|+|y| \le$

$ \sqrt{x^2+1} + \sqrt{y^2+1}\le$

$(1+ x^2) +(1+y^2).$

Hence:

$|f(x)-f(y)| \lt$

$ |x-y|( \dfrac{1}{1+y^2} +\dfrac{1}{1+x^2}) \le$

$2 |x-y|$.

Note :$ |x| =\sqrt{x^2} \lt \sqrt{x^2+1} \le x^2+1.$