find $f$ such $f(f(x)+y)=2x+f(f(f(y))-x)$

Question

Find all function $f:R\to R$,such for any real number $x,y$ have $$f(f(x)+y)=2x+f(f(f(y))-x)$$

I have prove this one result:

$$f(f(0))=f(0)$$ proof:let $x=y=0$,then we have $$f(f(0))=f(f(f(0)))$$ and take $x=f(f(0)),y=0$,we have $$f(f(f(f(0))))=2f(f(0))+f(0)$$ so we have $$f(f(0))=-f(0)$$

use this links Ivan Loh methods to prove two reslut:

$f(x)$ is also injective

proof: For convenience, let $P(x, y)$ represent the equation $f(f(x)+y)=2x+f(f(f(y))-x)$.

$P(x, -f(x))$ gives $f(f(-f(f(x)))-x)=f(0)-2x$. Thus $f(x)$ is surjective.

Suppose $f(a)=f(b)$ for some $a, b$. Then $P(x, a)$ and $P(x, b)$ give $$f(f(x)+a)=2x+f(f(f(a))-x)=2x+f(f(f(b))-x)=f(f(x)+b)$$ Since $f(x)$ is surjective, $f(x+a)=f(x+b) \, \forall x \in \mathbb{R}$.

Now using the above equation and $P(a, y), P(b, y)$ give

\begin{align} 2a=f(f(a)+y)-f(f(f(y))-a) & =f(f(a)+y)-f(f(f(y))-(a+b)+b) \\ &=f(f(a)+y)-f(f(f(y))-(a+b)+a) \\ & =f(f(b)+y)-f(f(f(y))-b) \\ & =2b \end{align}

Thus $f(x)$ is also injective.

Finally, $P(0, y)$ gives $f(y+f(0))=f(f(f(y)))$, so since $f(x)$ is injective, $$f(f(y))=y+f(0) \, \forall y \in \mathbb{R}$$.

then I can't continue ,and I think this answer is $f(x)=x$.so How to prove it.and this problem is from my teacher to give me exercise .Thanks for you help! Now when post this post,This site automatically displays a prompt for a similar question to me(But it's different problem) :links

Answer 1

let $$p(x,y)=f(f(x)+y)-2x=f(f(f(y))-x)$$ we have from $p(x,-f(x))$ that $f$ is surgective. assume $f(a)=f(b)$ then from $$p(x,a)-p(x,b)$$ We have $$f(f(x)+a)=f(f(x)+b)$$ so we have if $a-b=t$ then $$f(x+t)=f(x)$$ because $f$ is surgective.now we have : $$p(x+t,y)-p(x,y)=0=2t$$ so $t=0$ now we have from $$p(0,y)=f(f(0)+y)=f(f(f(y)))$$ so we have $$T(y)=y+f(0)=f(f(y))$$ we have from $T(0)$ that $$f(0)=f(f(0))$$ so $f(0)=0$ and we rewrite $T(y)$ as $$y=f(f(y))$$ now rewrite $p(x,y)$ as $$p(x,y)=f(y+f(x))=2x+f(y-x)$$ let $$q(x,y)=p(x,y+x)=f(y+f(x)+x)=2x+f(y)$$ from $q(x,0)$ we have $$f(f(x)+x)=2x$$ so $f(x)+x$ is injective and surgective aswell. so we can just write $q(x,y)$ as $$q(x,y)=f(x+f(x)+y)=f(x+f(X))+f(y)$$ as $x+f(x)$ is injective and surgective. we can rewrite it as $$W(x,y)=f(x+y)=f(x)+f(y)$$ so we have that $f$ is additive. remember we had $f(f(x)+x)=2x$ so we have $$f(f(x))+f(x)=x+f(x)=2x$$ which gives $f(X)=x$ for all real $x$ and we are done

Answer 2

A partial answer for $f$ polynomial.

►Clearly $f(x)$ is not constant.

►Test for a linear polynomial $f(x)=ax+b$.

$$f(f(x)+y)=f(ax+b+y)=a(ax+b+y)+b=a^2x+ay+(ab+b)\\2x+f(fof(y)-x)=2x+f(a(ay+b)+b-x)=2x+a(a(ay+b)+b-x)+b\\2x+f(fof(y)-x)=(2-a)x+a^3y+(a^2+1)b$$ It follows $$a^2=2-a\Rightarrow a=1\text {and } a=-2 $$ and $$\begin{cases}a=a^3\\(a+1)b=(a^2+a+1)b\end{cases}$$ therefore $a=1$ and $2b=3b\Rightarrow b=0$.

Thus the only possible polynomial of degree $1$ is $\boxed{f(x)=x}$

►Test for $f(x)=a_nx^n+\cdots+a_1x+a_0$ where $n\gt1$ ( and $a_n\ne0$ of course)

Since the equality is valid for all $x,y$ we take $y=0$ and look for a contradiction. $$fof(x)=2x+f(fof(0)-x)\large?$$ We have $$ fof(x)=a_n(a_nx^n+\cdots+a_1x+a_0)^n+\cdots+a_1(a_nx^n+\cdots+a_1x+a_0)+a_0$$ Note that $fof(0)=f(a_0)=a_n(a_0)^n\cdots+a_1(a_0)+a_0$. $$2x+f(f(a_0)-x)=2x+a_n(f(a_0)-x)^n+\cdots+a_1(f(a_0)-x)+a_0$$ $LHS$ is of $n^2$ degree while $RHS$ is of $n$ degree which implies that all the coefficients from $x^{n^2}$ until $x^{n+1}$ should be zero. In particular we would have $a_na_n^{n}=0$. Contradiction.

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Is $f(x)=x$ the only solution $\large ?$

We have $$f(f(x)+y)=f(f(f(y))-x)+2x$$ for all $x,y$ so it is valid for $x=y=0$; therefore $$fof(0)=f(fof(0))$$ which means that $fof(0)$ is a fixed point of $f$. Making now $x=0$ we have $$f(f(0)+y)=f(fof(y))$$ so if $f(0)=0$ then $f(y)=fof(f(y))$ and $f(y)$ is a fixed point of $fof$ for all $y$ so $fof(x)=x$ for all $x$ in the codomain of $f$. It follows that if $f$ is surjective and $f(0)=0$ then the only function required is the identity $\boxed{f(x)=x}$.

Note that for all fixed $y$ the function $F_y(x)$ defined by $$F_y(x)=f(f(x)+y)-f(f(f(y))-x)=2x$$ is a bijection because $2x$ so is. I stop here. Whoever wants, if desired, can finish the solution following this way.