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Question Number 26576 by gunawan last updated on 27/Dec/17

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

$$\mathrm{Find}\:\mathrm{all}\:{f}\::\:\mathrm{R}\rightarrow\mathrm{R}\:\mathrm{such}\:\mathrm{that} \\ $$$${f}\left({x}+{f}\left({x}\right)+{f}\left({y}\right)\right)={f}\left({y}+{f}\left({x}\right)\right)+{x}+{f}\left({y}\right)−{f}\left({f}\left({y}\right)\right)\: \\ $$$$\mathrm{for}\:\mathrm{all}\:{x},\:\mathrm{y}\:\in\:\mathrm{R} \\ $$

Commented by prakash jain last updated on 27/Dec/17

f(x+f(x)+f(y))=f(y+f(x)+x+f(y)−f(f(y)))? There is bracket missing on RHS. Is it at the end?

$${f}\left({x}+{f}\left({x}\right)+{f}\left({y}\right)\right)={f}\left({y}+{f}\left({x}\right)+{x}+{f}\left({y}\right)−{f}\left({f}\left({y}\right)\right)\right)?\: \\ $$$$ \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{bracket}\:\mathrm{missing}\:\mathrm{on}\:\mathrm{RHS}. \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}? \\ $$

Commented by gunawan last updated on 27/Dec/17

i′m sorry sir

$$\mathrm{i}'\mathrm{m}\:\mathrm{sorry}\:\mathrm{sir} \\ $$

Commented by prakash jain last updated on 27/Dec/17

f(x+f(x)+f(y))=f(y+f(x))+x+f(y)−f(f(y)) y=x=0 f(0+2f(0))=f(0+f(0))+0+f(0)−f(f(0)) f(2f(0))=f(0) assume f(0)=0 f(f(y))=f(y)+f(y)−f(f(y)) f(f(y))=f(y) f(x)=x (A) case f(0) ≠0 and f(a)=0 for some a x=y=a f(a+f(a)+f(a))=f(a+f(a))+a+f(a)−f(f(y>a)) 0=0+a+0−f(0) f(0)=a f(2f(0))=f(0)⇒f(2a)=a x=a,y=0 f(x+f(x)+f(y))=f(y+f(x))+x+f(y)−f(f(y)) f(a+f(a)+f(0))=f(0+f(a))+a+f(0)−f(f(0)) f(2a)=f(0)+a+f(0)−f(a) f(2a)=a+a+a−0=3a if f(a) =0 for some a ⇒a=0 case f(x)≠0 ∀x∈R continue

$${f}\left({x}+{f}\left({x}\right)+{f}\left({y}\right)\right)={f}\left({y}+{f}\left({x}\right)\right)+{x}+{f}\left({y}\right)−{f}\left({f}\left({y}\right)\right)\: \\ $$$${y}={x}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}+\mathrm{2}{f}\left(\mathrm{0}\right)\right)={f}\left(\mathrm{0}+{f}\left(\mathrm{0}\right)\right)+\mathrm{0}+{f}\left(\mathrm{0}\right)−{f}\left({f}\left(\mathrm{0}\right)\right) \\ $$$${f}\left(\mathrm{2}{f}\left(\mathrm{0}\right)\right)={f}\left(\mathrm{0}\right) \\ $$$${assume}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({f}\left({y}\right)\right)={f}\left({y}\right)+{f}\left({y}\right)−{f}\left({f}\left({y}\right)\right) \\ $$$${f}\left({f}\left({y}\right)\right)={f}\left({y}\right) \\ $$$${f}\left({x}\right)={x}\:\:\left({A}\right) \\ $$$${case}\:{f}\left(\mathrm{0}\right)\:\neq\mathrm{0}\:\mathrm{and}\:{f}\left({a}\right)=\mathrm{0}\:\mathrm{for}\:\mathrm{some}\:{a} \\ $$$${x}={y}={a} \\ $$$${f}\left({a}+{f}\left({a}\right)+{f}\left({a}\right)\right)={f}\left({a}+{f}\left({a}\right)\right)+{a}+{f}\left({a}\right)−{f}\left({f}\left({y}>{a}\right)\right)\: \\ $$$$\mathrm{0}=\mathrm{0}+{a}+\mathrm{0}−{f}\left(\mathrm{0}\right) \\ $$$${f}\left(\mathrm{0}\right)={a} \\ $$$${f}\left(\mathrm{2}{f}\left(\mathrm{0}\right)\right)={f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{2}{a}\right)={a} \\ $$$${x}={a},{y}=\mathrm{0} \\ $$$${f}\left({x}+{f}\left({x}\right)+{f}\left({y}\right)\right)={f}\left({y}+{f}\left({x}\right)\right)+{x}+{f}\left({y}\right)−{f}\left({f}\left({y}\right)\right)\: \\ $$$${f}\left({a}+{f}\left({a}\right)+{f}\left(\mathrm{0}\right)\right)={f}\left(\mathrm{0}+{f}\left({a}\right)\right)+{a}+{f}\left(\mathrm{0}\right)−{f}\left({f}\left(\mathrm{0}\right)\right) \\ $$$${f}\left(\mathrm{2}{a}\right)={f}\left(\mathrm{0}\right)+{a}+{f}\left(\mathrm{0}\right)−{f}\left({a}\right) \\ $$$${f}\left(\mathrm{2}{a}\right)={a}+{a}+{a}−\mathrm{0}=\mathrm{3}{a} \\ $$$${if}\:{f}\left({a}\right)\:=\mathrm{0}\:\mathrm{for}\:\mathrm{some}\:{a}\:\Rightarrow{a}=\mathrm{0} \\ $$$${case}\:{f}\left({x}\right)\neq\mathrm{0}\:\forall{x}\in\mathbb{R} \\ $$$${continue} \\ $$

Commented by gunawan last updated on 28/Dec/17

thanks Sir Yes

$$\mathrm{thanks}\:\mathrm{Sir} \\ $$$$\mathrm{Yes} \\ $$

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