Question Number 3878 by 123456 last updated on 23/Dec/15
$${f}:\left[\mathrm{0},+\infty\right)\rightarrow\mathbb{R} \\ $$$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right) \\ $$$${f}\left({x}\right)=? \\ $$
Commented by Rasheed Soomro last updated on 23/Dec/15
$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right) \\ $$$${Let}\:{x}=\mathrm{0}\:\: \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}^{{u}} {f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1} \\ $$$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right)\Rightarrow{f}\left({u}\right)={f}\left(\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)={f}\left({u}\right)\Rightarrow{u}=\mathrm{1}…………………………\left({i}\right) \\ $$$${So},\:\:\:\:{f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right)\Rightarrow{x}^{{u}} =\mathrm{1}\Rightarrow{x}=\mathrm{1} \\ $$$$\mathcal{F}{or}\:{x}=\mathrm{1}\:{and}\:{u}=\mathrm{1}\:{f}\left({x}\right)\:{may}\:{have} \\ $$$${any}\:\:{definition}. \\ $$$${For}\:{example}\:{f}\left({x}\right)={x}^{\mathrm{2}} \\ $$$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right)\Rightarrow\left({ux}\right)^{\mathrm{2}} ={x}^{{u}} \left({x}^{\mathrm{2}} \right)\Rightarrow\mathrm{1}=\mathrm{1}\:{for}\:{x}=\mathrm{1},{u}=\mathrm{1} \\ $$
Commented by prakash jain last updated on 23/Dec/15
$${f}\left({x}\right)=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{think}\:{f}\left({x}\right)=\mathrm{0}\:\mathrm{is}\:\mathrm{only}\:\mathrm{solution}. \\ $$
Answered by RasheedSindhi last updated on 24/Dec/15
$$ \\ $$$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right)…………..\left({i}\right) \\ $$$${x}=\mathrm{1}\Rightarrow{f}\left({u}\right)={f}\left(\mathrm{1}\right)\Rightarrow{u}=\mathrm{1} \\ $$$${u}=\mathrm{1}\Rightarrow\:\left({i}\right)\:{will}\:{become} \\ $$$${f}\left({x}\right)={xf}\left({x}\right) \\ $$$${xf}\left({x}\right)−{f}\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{0}\:\mid\:\:{x}=\mathrm{1} \\ $$