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Question Number 3654 by prakash jain last updated on 17/Dec/15
f(xf(y)+x)=xy+f(x) f:R→R f(x)=?
$${f}\left({xf}\left({y}\right)+{x}\right)={xy}+{f}\left({x}\right) \\ $$$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)=? \\ $$
Commented by RasheedSindhi last updated on 18/Dec/15
f(xf(y)+x)=xy+f(x)....(i) Let y=0 f(xf(0)+x)=x(0)+f(x) f(x)=f( xf(0)+f(x) )........(ii) f(0)=f( 0.f(0)+f(0)) f(0)=f( f(0) ) x⇋y in (i) f(yf(x)+y)=xy+f(y).......(iii) y=x in (i) f( xf(x)+x )=x^2 +f(x) f(x)=f( xf(x)+x )−x^2 f(1)=f( f(1)+1 )−1 f(1)+1=f( f(1)+1 )
$${f}\left({xf}\left({y}\right)+{x}\right)={xy}+{f}\left({x}\right)….\left({i}\right) \\ $$$${Let}\:{y}=\mathrm{0} \\ $$$${f}\left({xf}\left(\mathrm{0}\right)+{x}\right)={x}\left(\mathrm{0}\right)+{f}\left({x}\right) \\ $$$${f}\left({x}\right)={f}\left(\:{xf}\left(\mathrm{0}\right)+{f}\left({x}\right)\:\right)……..\left({ii}\right) \\ $$$${f}\left(\mathrm{0}\right)={f}\left(\:\mathrm{0}.{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{0}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{0}\right)={f}\left(\:{f}\left(\mathrm{0}\right)\:\right) \\ $$$${x}\leftrightharpoons{y}\:{in}\:\left({i}\right) \\ $$$${f}\left({yf}\left({x}\right)+{y}\right)={xy}+{f}\left({y}\right)…….\left({iii}\right) \\ $$$${y}={x}\:{in}\:\left({i}\right) \\ $$$${f}\left(\:{xf}\left({x}\right)+{x}\:\right)={x}^{\mathrm{2}} +{f}\left({x}\right) \\ $$$${f}\left({x}\right)={f}\left(\:{xf}\left({x}\right)+{x}\:\right)−{x}^{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)={f}\left(\:{f}\left(\mathrm{1}\right)+\mathrm{1}\:\right)−\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)+\mathrm{1}={f}\left(\:{f}\left(\mathrm{1}\right)+\mathrm{1}\:\right) \\ $$
Commented by prakash jain last updated on 18/Dec/15
based on above f(x)=x is one solution, are there any other solutions?
$$\mathrm{based}\:\mathrm{on}\:\mathrm{above} \\ $$$${f}\left({x}\right)={x}\:\mathrm{is}\:\mathrm{one}\:\mathrm{solution},\:\mathrm{are}\:\mathrm{there}\:\mathrm{any}\:\mathrm{other} \\ $$$$\mathrm{solutions}? \\ $$
Commented by prakash jain last updated on 18/Dec/15
Let us say f(x)≠x and f(y)=0 for some y. f(x)=xf^(−1) (y)+f(x) f^(−1) (y)=0 so if f(x)≠x then either f(x)≠0 for all x∈R or f(0)=0.
$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:{f}\left({x}\right)\neq{x}\:\mathrm{and}\:{f}\left({y}\right)=\mathrm{0}\:{for}\:{some}\:{y}. \\ $$$${f}\left({x}\right)={xf}^{−\mathrm{1}} \left({y}\right)+{f}\left({x}\right) \\ $$$${f}^{−\mathrm{1}} \left({y}\right)=\mathrm{0} \\ $$$${so}\:{if}\:{f}\left({x}\right)\neq{x}\:{then}\:{either} \\ $$$${f}\left({x}\right)\neq\mathrm{0}\:\mathrm{for}\:\mathrm{all}\:{x}\in\mathbb{R}\:\mathrm{or}\:{f}\left(\mathrm{0}\right)=\mathrm{0}. \\ $$

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