Ytukyg

find all function $f:Rto R$,and such for any $x,yin R$,and such
$$f(x+f(y)+yf(x))=y+f(x)+xf(y)$$

I have prove $$f(f(x))=x$$

proof:Let $y=0$,we have
$$f(x+f(0))=f(x)+xf(0)$$
let $x=0,yto x$,we have
$$f(f(x)+xf(0))=x+f(0)$$
so we have
$$f(f(x+f(0))=x+f(0)$$
then $$f(f(x))=x$$

and $f(x)=x$ is one solution,and I can't this only soluton.
It doesn't seem easy.Thank for you help！

This is only a partial result (EDIT: full result is given below), but I'll post it anyway. Let us write $(x,y)$ if $y = f(x)$ holds. Since $f(f(x)) = x$, $(x,y)$ is equivalent to $(y,x)$ and $(x,y)wedge (x,y')$ implies $y= y'$. What we can directly observe is that $0= f(f(0)) = f(0)$ by letting $x=y=0$. What is less trivial is $f(1) = 1$. Let $gamma =f(1)$. Then by letting $y = 1$ and $y= gamma$, we get
$$
(x+ gamma + f(x),1+ f(x) +gamma x)wedge ;(x + 1+gamma f(x), gamma + f(x) +x).$$
Hence, $1+ f(x) +gamma x = x + 1+gamma f(x)$ for all $x$, and if $gamma neq 1$, then $f(x) = x$, leading to contradiction.

We next show that $$(-frac{f(x) - f(x')}{x-x'}, -frac{x-x'}{f(x) - f(x')}),quad forall xneq x';cdots(*).$$ Assume $(a,b)wedge (x,y)wedge (x',y').$ Then, by the FE,
$$
(a+x+by, b+y +ax)wedge (a+x'+by', b+y' +ax').
$$ We can equate $b+y +ax$ and $ b+y' +ax'$ by letting $a= -frac{y-y'}{x-x'}$. Then $b$ should satisfy $a+x+by = a+x'+by'$, that is, $ b= -frac{x-x'}{y-y'}.$ Because $y = f(x), y' = f(x')$, this proves the claim.

From the previous claim, we know that $(-1,-1)$, i.e. $f(-1) = -1$. By letting $x' = 0, - 1$, we also know that $(x,y)$ implies
$$(-frac{y}{x}, -frac{x}{y})wedge (-frac{y+ 1}{x + 1}, -frac{x+ 1}{y + 1}).
$$If we write $-frac{y-y'}{x-x'}= k$, then $(-frac{1/k+ 1}{k + 1}, -frac{k+ 1}{1/k + 1}) = (-frac{1}{k}, -k)$ also shows that $(frac{y-y'}{x-x'}, frac{x-x'}{y-y'})$ whenever $xneq x'$. Next, by putting $y=-1$ in the original functional equation, we also have $(x,y)$ implies
$$(x-y-1, y-x-1).$$ Iterating this $n$ times we also have
$$(2^n(x-y)-1, 2^n(y-x)-1) ;cdots (**).
$$

Our next claim is that if $(x,x) vee (x, x+2^{1-N})$, then $(x+ 2^{-N}, x+2^{-N})$ ($Ngeq 0$.) To show this, let us write $(x+ 2^{-N}, alpha)$ and derive an equation about $alpha$. If $(x,x)$, $(*)$ implies $( 2^N(x-alpha ), frac{1}{2^N(x-alpha )})$. By applying $(**)$ to $(x+ 2^{-N}, alpha)$ we also have $(2^N(x-alpha), 2^N(alpha-x)-2)$. Hence, $2^N(alpha-x)-2 = frac{1}{2^N(x-alpha )}$ and this implies $2^N(x-alpha )=-1$ as desired. In case $(x, x+2^{1-N})$ can be dealt with similarly by noting that $(-2^N(x-alpha )-2, - frac{1}{2^N(x-alpha )+2})$.

A very similar argument can also prove that
$$(x,x) vee (x, x-2^{1-N})Rightarrow (x-2^{-N}, x-2^{-N}).$$
So far, all the tedious arguments show that $f(x) = x$ holds for every dyadic rationals $x=frac{j}{2^n}$, that is, the set $F$ of fixed points of $f$ contains every dyadic rational. The following are some other facts about $F$:
(i) If $(x,y)$, then $x+y+xy$ and $x+y+1$ belongs to $F$.

(ii) If $x,y in F$, then $x+y+xy in F$.

(iii) If $x in F$, then $xpm frac{j}{2^n}in F$. (Above claim)

And for some $(x,f(x))$ such that $xnotin F$, this generates so many non-fixed pairs
$$(pm left[frac{f(x) - q}{x-q}right]^r,pm left[frac{x- q}{f(x)-q}right]^r), quad (x+q+qf(x),f(x)+q+qx)_{q neq 1},$$ for all dyadic rational $q$ and $rin mathbf{N}$.
I tried but failed to get further ideas about how $F$ and $mathbf{R}setminus F$ looks like. But I wish this helps somehow.

(Note: Actually in the above proposition, $(x, xpm 2^{1-N})$ cannot occur since their difference cannot be in $F$.)

Following yesterday's post, I've completed proof: $f = id$ is the only solution of the equation.

I'll briefly review some facts already proved.

(i) $F + mathbf{Q}_{dyad} = F.$

(ii) If $x,y in F$, then $xy in F$, or equivalently, $Fcdot F = F$. (Since $x-1, y-1 in F$ implies $xy -1in F$ and thus $xy in F$.)

(iii) If $(x,y)$, then $x+y, ;xy + x+ y in F$.

Our claim starts with: If $0neq x in F$, then $frac{1}{x} in F$. Proof is simple. Let $(frac{1}{x},gamma).$ Then $gamma +frac{1}{x}in F$. Since $xin F$, $gamma x + 1 in F$, and $gamma x in F$. Note that $(gamma x, frac{1}{gamma x}).$(slope of $(frac{1}{x},gamma),(0,0).$) This shows $gamma = pm frac{1}{x}.$ If it were that $(frac{1}{x},-frac{1}{x})$, then $frac{1}{x^2} in F$. This implies that $xcdotfrac{1}{x^2}=frac{1}{x} in F$, as desired.

Our next claim is that $(x,y)$ implies $xy in F$. We start from $x+y, ;x+y+xy in F$. If $x+y =0$, it's already done. Otherwise, $frac{x+y+xy}{x+y} = 1+ frac{xy}{x+y} in F$, hence $frac{xy}{x+y}in F$. Then this implies $$(x+y)cdot frac{xy}{x+y} = xy in F.$$

Our almost final claim is that $0< y in F$ implies $sqrt{y} in F$. Suppose $(sqrt[4]{y}, gamma)$. Then $gamma^2 sqrt{y} in F$. Since $frac{1}{y} in F$, we have $frac{gamma^2}{sqrt{y}} in F.$ Notice that $(frac{gamma}{sqrt[4]{y}}, frac{sqrt[4]{y}}{gamma})$ and hence $(frac{gamma^2}{sqrt{y}}, frac{sqrt{y}}{gamma^2})$ Thus we have $gamma^4 = y$, $gamma = pm sqrt[4]{y}$. Suppose $gamma = sqrt[4]{y}$. Then, $sqrt[4]{y} in F$ implies $sqrt{y} in F$. Otherwise, $(sqrt[4]{y}, -sqrt[4]{y})$ implies $-sqrt{y} in F$. Thus $sqrt{y} in F.$

Finally we are ready to prove that $(x,y)$ implies $ x-y=d =0$. Assume to the contrary that $d>0$. Then, as I showed in yesterday's post, $(d-1, -d-1).$ But this implies $(-1+d)cdot(-1-d) = 1-d^2 in F$, and thus $d^2 in F.$ So $d$ must be in $F$ and $d-1$ also. This leads to $d-1 = -d-1$, contradicting $d>0$. So, the only solution of the functional equation is $f(x) = x$!

There is a proof given by Professor Wu Wei-chao in 2001, you may download the paper here.

Here is the overview of the proof:

• $f$ is one-to-one, and $$f(0)=0,tag{2}$$




• $forall yinBbb Rsetminus {0}$, $$fleft(-frac y{f(y)}right)=-frac{f(y)}y,qquad fleft(-frac{f(y)}yright)=-frac y{f(y)},tag{8,9}$$




• $$f(pm1)=pm1,tag{11,12}$$




• $forall xinBbb R$,
  $$f(x+f(x)+1)=f(x)+x+1,qquad f(x-f(x)-1)=-x+f(x)-1tag{13,14}$$




• $$f(pm2)=pm2,tag{23,29}$$




• $$(f(pmsqrt2))^2=2,qquad(f(pmfrac 1{sqrt2}))^2=frac12,tag{42,46}$$




• $forall xinBbb R$,
  $$f(3x-4)=-3f(-x)-4.tag{47}$$



• 
  

  Introduce 5 lemmas,

  
  
  

  
  

  Lemma 4. $forall ain Bbb R$, if $$tag{58}f(-x-a)=-f(x)-a,qquad (forall xinBbb R)$$ and $$tag{59}f(x-f(x)-dfrac {a^2}4)=f(x)-x-dfrac {a^2}4,qquad (forall xinBbb R)$$
  then
  $$f(x)=xqquad (forall xinBbb R).$$

  
  

  
  



• Saparate into 3 cases and finish the proof.

Case 1: $f(pmsqrt 2)=pmsqrt 2implies f(x)=x$ ($forall xinBbb R$).

Case 2: $f(pmsqrt 2)=mpsqrt 2, f(pmfrac 1{sqrt2})=pmfrac 1{sqrt2}implies$ contradition to injectivitiy of $f$.

Case 3: $f(pmsqrt 2)=mpsqrt 2, f(pmfrac 1{sqrt2})=mpfrac 1{sqrt2}implies$ contradition to $f(sqrt 2)=-sqrt 2$.

Wu also generalized the result.

Generalization: Suppose $F$ is a field or a ring, $Bbb Zsubseteq Fsubseteq Bbb C$, and $f:Fto F$ satisifies
$$f(x+f(y)+yf(x))=y+f(x)+xf(y),$$

• if $F=Bbb R$, then $f(x)=x$ ($forall xinBbb R$);




• if $F=Bbb C$, then $f(x)=x$ ($forall xinBbb C$) or $f(x)=overline x$ ($forall xinBbb C$);




• if $F=Bbb Q$, then $f(x)=x$ ($forall xinBbb Q$);




• if $F=Bbb Z$, then $f(x)=x$ ($forall xinBbb Z$);




• if $F={a+brmid forall a,binBbb Q}$, where $r$ is fixed irrational and $r^2inBbb Q$, then $f(x)=x$ ($forall xin F$) or $f(x)=overline x$ ($forall xin F$), here we define $overline{a+br}=a-br$;




• if $F={a+brmid forall a,binBbb Z}$, where $r$ is fixed irrational and $r^2inBbb Z$, then $f(x)=x$ ($forall xin F$) or $f(x)=overline x$ ($forall xin F$), here we define $overline{a+br}=a-br$.