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Question Number 180759 by depressiveshrek last updated on 16/Nov/22

Find all functions f : R→R for all x, y ∈R such that  f(f(x−y)−yf(x))=xf(y)

$${Find}\:{all}\:{functions}\:{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:{for}\:{all}\:{x},\:{y}\:\in\mathbb{R}\:{such}\:{that} \\ $$$${f}\left({f}\left({x}−{y}\right)−{yf}\left({x}\right)\right)={xf}\left({y}\right) \\ $$

Answered by aleks041103 last updated on 17/Nov/22

x=0  ⇒f(f(−y)−yf(0))=0  y=0  ⇒f(f(x))=xf(0)  x=y  ⇒f(f(0)−xf(x))=xf(x)  ⇒f(f(0)−z)=z  v=f(0)−z⇒z=f(0)−v  ⇒f(v)=f(0)−v  ⇒f(f(0)+y−x−yf(0)+yx)=x(f(0)−y)  f(0)−f(0)−y+x+yf(0)−yx=xf(0)−yx  x−y+yf(0)=xf(0)  y=0⇒x=xf(0)⇒f(0)=1  ⇒x−y+y=x→true  ⇒f(x)=1−x

$${x}=\mathrm{0} \\ $$$$\Rightarrow{f}\left({f}\left(−{y}\right)−{yf}\left(\mathrm{0}\right)\right)=\mathrm{0} \\ $$$${y}=\mathrm{0} \\ $$$$\Rightarrow{f}\left({f}\left({x}\right)\right)={xf}\left(\mathrm{0}\right) \\ $$$${x}={y} \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{0}\right)−{xf}\left({x}\right)\right)={xf}\left({x}\right) \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{0}\right)−{z}\right)={z} \\ $$$${v}={f}\left(\mathrm{0}\right)−{z}\Rightarrow{z}={f}\left(\mathrm{0}\right)−{v} \\ $$$$\Rightarrow{f}\left({v}\right)={f}\left(\mathrm{0}\right)−{v} \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{0}\right)+{y}−{x}−{yf}\left(\mathrm{0}\right)+{yx}\right)={x}\left({f}\left(\mathrm{0}\right)−{y}\right) \\ $$$${f}\left(\mathrm{0}\right)−{f}\left(\mathrm{0}\right)−{y}+{x}+{yf}\left(\mathrm{0}\right)−{yx}={xf}\left(\mathrm{0}\right)−{yx} \\ $$$${x}−{y}+{yf}\left(\mathrm{0}\right)={xf}\left(\mathrm{0}\right) \\ $$$${y}=\mathrm{0}\Rightarrow{x}={xf}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\Rightarrow{x}−{y}+{y}={x}\rightarrow{true} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{1}−{x} \\ $$

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