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Question Number 3878 by 123456 last updated on 23/Dec/15

$$f:[0,+\mathrm{\infty })\to \mathbb{R}$$$$f\left(ux\right)=x^{u}f\left(x\right)$$$$f\left(x\right)=?$$

Commented by Rasheed Soomro last updated on 23/Dec/15

$$f\left(ux\right)=x^{u}f\left(x\right)$$$$Letx=0$$$$f\left(0\right)=0^{u}f\left(0\right)=0$$$$x=1$$$$f\left(ux\right)=x^{u}f\left(x\right)\Rightarrow f\left(u\right)=f\left(1\right)$$$$\Rightarrow f\left(1\right)=f\left(u\right)\Rightarrow u=1..............................\left(i\right)$$$$So,f\left(ux\right)=x^{u}f\left(x\right)\Rightarrow x^u=1\Rightarrow x=1$$$$\mathcal{F}orx=1andu=1f\left(x\right)mayhave$$$$anydefinition.$$$$Forexamplef\left(x\right)=x^2$$$$f\left(ux\right)=x^{u}f\left(x\right)\Rightarrow {\left(ux\right)}^2=x^u\left(x^2\right)\Rightarrow 1=1forx=1,u=1$$

Commented by prakash jain last updated on 23/Dec/15

$$f\left(x\right)=0$$$$\mathrm{I}\mathrm{think}f\left(x\right)=0\mathrm{is}\mathrm{only}\mathrm{solution}.$$

Answered by RasheedSindhi last updated on 24/Dec/15

$$$$$$f\left(ux\right)=x^{u}f\left(x\right)..............\left(i\right)$$$$x=1\Rightarrow f\left(u\right)=f\left(1\right)\Rightarrow u=1$$$$u=1\Rightarrow \left(i\right)willbecome$$$$f\left(x\right)=xf\left(x\right)$$$$xf\left(x\right)-f\left(x\right)=0$$$$f\left(x\right)(x-1)=0$$$$f\left(x\right)=0\mid x=1$$

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