Question Number 227274 by Lara2440 last updated on 11/Jan/26
$$\mathrm{Does}\:\mathrm{condition} \\ $$$$\underset{{x}\in\mathbb{R}} {\mathrm{sup}}\:\mid\mid{f}^{\left(\mathrm{1}\right)} \left({x}\right)\mid\mid<\infty\:,\:\:\mathrm{Gaurantee}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{Bounded}\:\mathrm{in}\:\mathbb{R}? \\ $$
Answered by peace2 last updated on 12/Jan/26
$${non}\:{f}\left({x}\right)={x} \\ $$$${f}^{\mathrm{1}} \left({x}\right)=\mathrm{1} \\ $$$${supf}^{\mathrm{1}} \left({x}\right)=\mathrm{1}<\infty \\ $$$${f}\left({x}\right)={x}\:{noy}\:{bounded} \\ $$