Find the function equation $f(x+f(y)+yf(x))=y+f(x)+xf(y)$











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




I have proven that $$f(f(x))=x$$




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 but I can't argue this to be the only solution.
It doesn't seem easy.Thank for you help!










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  • 1




    Are we at least assuming continuity/differentiability?
    – MisterRiemann
    Nov 21 at 12:13










  • no,This condtion,if have I can to solve
    – function sug
    Nov 21 at 12:24










  • Specializing the equation to the case $y = x$ gives $f(hat x) = hat x$, where $hat x = x + f(x) + x f(x)$, so we can immediately conclude that $fvert_A(x) = x$ on the image of $x mapsto x + f(x) + x f(x)$.
    – Travis
    Nov 21 at 12:34










  • The problem was given at IMO 1995, see for example here artofproblemsolving.com/community/c6h506p1611 but with a much stronger constraint on its domain and monotonicity. Note that in that case the nontrivial $f(x) = -frac{x}{x+1}$ is also a solution(which obviously doesn't work here because $x=-1$ is in the domain. All in all, I expect a hard solution for this problem.
    – Sorin Tirc
    Nov 21 at 12:47












  • Note that $f(f(x))=x$ means that f is an involution. Other common involutions include $f:xrightarrow-x$ and $f:xrightarrow frac{1}{x}$ and $f:xrightarrow frac{-1}{x}$ even if these don't work here. or you can try this involution
    – TheD0ubleT
    Nov 22 at 9:02

















up vote
7
down vote

favorite
6












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




I have proven that $$f(f(x))=x$$




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 but I can't argue this to be the only solution.
It doesn't seem easy.Thank for you help!










share|cite|improve this question




















  • 1




    Are we at least assuming continuity/differentiability?
    – MisterRiemann
    Nov 21 at 12:13










  • no,This condtion,if have I can to solve
    – function sug
    Nov 21 at 12:24










  • Specializing the equation to the case $y = x$ gives $f(hat x) = hat x$, where $hat x = x + f(x) + x f(x)$, so we can immediately conclude that $fvert_A(x) = x$ on the image of $x mapsto x + f(x) + x f(x)$.
    – Travis
    Nov 21 at 12:34










  • The problem was given at IMO 1995, see for example here artofproblemsolving.com/community/c6h506p1611 but with a much stronger constraint on its domain and monotonicity. Note that in that case the nontrivial $f(x) = -frac{x}{x+1}$ is also a solution(which obviously doesn't work here because $x=-1$ is in the domain. All in all, I expect a hard solution for this problem.
    – Sorin Tirc
    Nov 21 at 12:47












  • Note that $f(f(x))=x$ means that f is an involution. Other common involutions include $f:xrightarrow-x$ and $f:xrightarrow frac{1}{x}$ and $f:xrightarrow frac{-1}{x}$ even if these don't work here. or you can try this involution
    – TheD0ubleT
    Nov 22 at 9:02















up vote
7
down vote

favorite
6









up vote
7
down vote

favorite
6






6





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




I have proven that $$f(f(x))=x$$




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 but I can't argue this to be the only solution.
It doesn't seem easy.Thank for you help!










share|cite|improve this question















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




I have proven that $$f(f(x))=x$$




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 but I can't argue this to be the only solution.
It doesn't seem easy.Thank for you help!







functions






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edited Dec 2 at 17:11









Mostafa Ayaz

13.4k3836




13.4k3836










asked Nov 21 at 12:08









function sug

3101438




3101438








  • 1




    Are we at least assuming continuity/differentiability?
    – MisterRiemann
    Nov 21 at 12:13










  • no,This condtion,if have I can to solve
    – function sug
    Nov 21 at 12:24










  • Specializing the equation to the case $y = x$ gives $f(hat x) = hat x$, where $hat x = x + f(x) + x f(x)$, so we can immediately conclude that $fvert_A(x) = x$ on the image of $x mapsto x + f(x) + x f(x)$.
    – Travis
    Nov 21 at 12:34










  • The problem was given at IMO 1995, see for example here artofproblemsolving.com/community/c6h506p1611 but with a much stronger constraint on its domain and monotonicity. Note that in that case the nontrivial $f(x) = -frac{x}{x+1}$ is also a solution(which obviously doesn't work here because $x=-1$ is in the domain. All in all, I expect a hard solution for this problem.
    – Sorin Tirc
    Nov 21 at 12:47












  • Note that $f(f(x))=x$ means that f is an involution. Other common involutions include $f:xrightarrow-x$ and $f:xrightarrow frac{1}{x}$ and $f:xrightarrow frac{-1}{x}$ even if these don't work here. or you can try this involution
    – TheD0ubleT
    Nov 22 at 9:02
















  • 1




    Are we at least assuming continuity/differentiability?
    – MisterRiemann
    Nov 21 at 12:13










  • no,This condtion,if have I can to solve
    – function sug
    Nov 21 at 12:24










  • Specializing the equation to the case $y = x$ gives $f(hat x) = hat x$, where $hat x = x + f(x) + x f(x)$, so we can immediately conclude that $fvert_A(x) = x$ on the image of $x mapsto x + f(x) + x f(x)$.
    – Travis
    Nov 21 at 12:34










  • The problem was given at IMO 1995, see for example here artofproblemsolving.com/community/c6h506p1611 but with a much stronger constraint on its domain and monotonicity. Note that in that case the nontrivial $f(x) = -frac{x}{x+1}$ is also a solution(which obviously doesn't work here because $x=-1$ is in the domain. All in all, I expect a hard solution for this problem.
    – Sorin Tirc
    Nov 21 at 12:47












  • Note that $f(f(x))=x$ means that f is an involution. Other common involutions include $f:xrightarrow-x$ and $f:xrightarrow frac{1}{x}$ and $f:xrightarrow frac{-1}{x}$ even if these don't work here. or you can try this involution
    – TheD0ubleT
    Nov 22 at 9:02










1




1




Are we at least assuming continuity/differentiability?
– MisterRiemann
Nov 21 at 12:13




Are we at least assuming continuity/differentiability?
– MisterRiemann
Nov 21 at 12:13












no,This condtion,if have I can to solve
– function sug
Nov 21 at 12:24




no,This condtion,if have I can to solve
– function sug
Nov 21 at 12:24












Specializing the equation to the case $y = x$ gives $f(hat x) = hat x$, where $hat x = x + f(x) + x f(x)$, so we can immediately conclude that $fvert_A(x) = x$ on the image of $x mapsto x + f(x) + x f(x)$.
– Travis
Nov 21 at 12:34




Specializing the equation to the case $y = x$ gives $f(hat x) = hat x$, where $hat x = x + f(x) + x f(x)$, so we can immediately conclude that $fvert_A(x) = x$ on the image of $x mapsto x + f(x) + x f(x)$.
– Travis
Nov 21 at 12:34












The problem was given at IMO 1995, see for example here artofproblemsolving.com/community/c6h506p1611 but with a much stronger constraint on its domain and monotonicity. Note that in that case the nontrivial $f(x) = -frac{x}{x+1}$ is also a solution(which obviously doesn't work here because $x=-1$ is in the domain. All in all, I expect a hard solution for this problem.
– Sorin Tirc
Nov 21 at 12:47






The problem was given at IMO 1995, see for example here artofproblemsolving.com/community/c6h506p1611 but with a much stronger constraint on its domain and monotonicity. Note that in that case the nontrivial $f(x) = -frac{x}{x+1}$ is also a solution(which obviously doesn't work here because $x=-1$ is in the domain. All in all, I expect a hard solution for this problem.
– Sorin Tirc
Nov 21 at 12:47














Note that $f(f(x))=x$ means that f is an involution. Other common involutions include $f:xrightarrow-x$ and $f:xrightarrow frac{1}{x}$ and $f:xrightarrow frac{-1}{x}$ even if these don't work here. or you can try this involution
– TheD0ubleT
Nov 22 at 9:02






Note that $f(f(x))=x$ means that f is an involution. Other common involutions include $f:xrightarrow-x$ and $f:xrightarrow frac{1}{x}$ and $f:xrightarrow frac{-1}{x}$ even if these don't work here. or you can try this involution
– TheD0ubleT
Nov 22 at 9:02












2 Answers
2






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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 $f(x) = x$!






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    up vote
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    +25










    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 $overline{a+br}=a-br$.








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      2 Answers
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      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 $f(x) = x$!






      share|cite|improve this answer



























        up vote
        4
        down vote













        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 $f(x) = x$!






        share|cite|improve this answer

























          up vote
          4
          down vote










          up vote
          4
          down vote









          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 $f(x) = x$!






          share|cite|improve this answer














          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 $f(x) = x$!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 at 2:17

























          answered Nov 24 at 7:04









          Song

          1,940112




          1,940112






















              up vote
              2
              down vote



              +25










              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 $overline{a+br}=a-br$.








              share|cite|improve this answer



























                up vote
                2
                down vote



                +25










                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 $overline{a+br}=a-br$.








                share|cite|improve this answer

























                  up vote
                  2
                  down vote



                  +25







                  up vote
                  2
                  down vote



                  +25




                  +25




                  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 $overline{a+br}=a-br$.








                  share|cite|improve this answer














                  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 $overline{a+br}=a-br$.









                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 27 at 16:35

























                  answered Nov 27 at 15:54









                  Tianlalu

                  2,9901936




                  2,9901936






























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