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Question Number 3886 by 123456 last updated on 23/Dec/15

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

Commented by Yozzii last updated on 24/Dec/15

$$Letu=1.$$$$\therefore f\left(x\right)+g\left(x\right)=f\left(f\left(x\right)\right)+g(1+f\left(x\right)).$$$$f\left(x\right)=f\left(f\left(x\right)\right)$$$$Letf\left(x\right)=x\Rightarrow f\left(f\left(x\right)\right)=f\left(x\right)=x$$$$andf:[0,\mathrm{\infty })\to \mathbb{R}.$$$$\therefore g\left(x\right)=g(1+x)$$$$Periodicfunctionwithperiod1.$$$$g\left(x\right)=Asin2\pi x(A\in \mathbb{R})andg:[0,\mathrm{\infty })\to \mathbb{R}.$$$$\Rightarrow g(x+1)=Asin(2\pi +2\pi x)=Asin2\pi x=g\left(x\right)$$$$\therefore f\left(x\right)+g\left(x\right)=x+Asin2\pi xforexample$$$$$$$$g\left(x\right)=Asin2m\pi x+Bcos2n\pi x(n,m\in \mathbb{Z},A,B\in \mathbb{R})$$$$\because g(1+x)=Asin(2m\pi +2m\pi x)+Bcos(2n\pi +2\pi nx)$$$$g(1+x)=Asin2m\pi x+Bcos2n\pi x=g\left(x\right)$$$$$$$$f\left(x\right)=0orf\left(x\right)=x.$$$$$$

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