Question Number 166805 by mathocean1 last updated on 28/Feb/22
$${f}\left({x}\right)=\frac{{x}}{\mathrm{1}+\mid{x}\mid}. \\ $$$${show}\:{that}\:\exists\:{K}\:\in\:\mathbb{R}_{+} \:{such}\:{that} \\ $$$$\forall\:{x},\:{y}\:\in\:\mathbb{R},\:\mid{f}\left({x}\right)−{f}\left({y}\right)\mid\leqslant{K}\mid{x}−{y}\mid \\ $$
Answered by TheSupreme last updated on 28/Feb/22
$${f}'\left({x}\right)=\frac{\mathrm{1}+\mid{x}\mid−\mid{x}\mid}{\left(\mathrm{1}+\mid{x}\mid\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\mathrm{1}+\mid{x}\mid\right)^{\mathrm{2}} }\leqslant\mathrm{1} \\ $$$${f}\left({x}\right)−{f}\left({x}+{h}\right)\leqslant{f}\left({x}\right)−\left({f}\left({x}+{h}\right)+{h}\right)\leqslant−{h} \\ $$$${f}\left({x}\right)−{f}\left({y}\right)\leqslant{x}−{y} \\ $$$$\frac{{f}\left({x}\right)−{f}\left({y}\right)}{{x}−{y}}\leqslant\mathrm{1} \\ $$$$ \\ $$