Is my proof correct???












1












$begingroup$


The following question was asked in INMO 2005:



"Find all functions $f:Bbb{R} rightarrow Bbb{R} $ such that
$$f(x^2 + yf(z)) = xf(x) + zf(y)$$



for all $x,y,z in Bbb{R}$.



I am thinking of this as a solution:



"Put $x = y = 0 $ in the given equation, I get $$f(0) = zf(0)$$ for all $z$ in $R$ which means that $f(0) = 0$



Further take $y=0$ in the original equation to give me $$f(x^2)=xf(x)$$



Put $x = 0$ in the original equation to give $$f(yf(z)) = zf(y) *$$



In the above equation put $z=1$ to give $$f(yf(1)) = f(y)$$



Let $f(1) = c$ ,which gives me $f(cy) = f(y) **$
In the equation marked * put $y = 1$ to give $$f(f(z)) = zc$$



Taking $f$ on both sides gives $f(f(f(z))) = f(zc) = f(z)$ from **



Let $f(z) = x$ , say then I get $f(f(x)) = x$, but I know that $f(f(x)) = xc$ which means that $$c=1$$



Not put $z = 1$ and since $xf(x) = f(x^2)$ I get $$f(x^2 + y) = f(x^2) + f(y)$$ Now I replace x^2 by x to get $$f(x + y) = f(x) + f(y)$$ to give me the cauchy equation with solns $f(x) = 0$ for all x or $f(x) = x$(from f(1) = 1)



Now though I have given the solutions, there are two pints where I doubt my solution a little. They are when I replace$f(z)$ by some $x$. Is this step valid? And the other doubt is when I replace $x^2$ by $x$. Are these two steps valid and are there any errors in this solution?



Thanks in advance










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If you replace $f(z)$ by $x$ the equation you get will be valid for $x$ in the range of $f$. If you replace $x^{2}$ by $x$ your conclusion will only be valid for $x geq 0$. So these are indeed gaps in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Dec 21 '18 at 10:30










  • $begingroup$
    Moreover, unless additional properties (e.g. continuity) are assumed on $f$, $f(x) = cx$ is not the only solution of Cauchy's equation.
    $endgroup$
    – Song
    Dec 21 '18 at 10:35
















1












$begingroup$


The following question was asked in INMO 2005:



"Find all functions $f:Bbb{R} rightarrow Bbb{R} $ such that
$$f(x^2 + yf(z)) = xf(x) + zf(y)$$



for all $x,y,z in Bbb{R}$.



I am thinking of this as a solution:



"Put $x = y = 0 $ in the given equation, I get $$f(0) = zf(0)$$ for all $z$ in $R$ which means that $f(0) = 0$



Further take $y=0$ in the original equation to give me $$f(x^2)=xf(x)$$



Put $x = 0$ in the original equation to give $$f(yf(z)) = zf(y) *$$



In the above equation put $z=1$ to give $$f(yf(1)) = f(y)$$



Let $f(1) = c$ ,which gives me $f(cy) = f(y) **$
In the equation marked * put $y = 1$ to give $$f(f(z)) = zc$$



Taking $f$ on both sides gives $f(f(f(z))) = f(zc) = f(z)$ from **



Let $f(z) = x$ , say then I get $f(f(x)) = x$, but I know that $f(f(x)) = xc$ which means that $$c=1$$



Not put $z = 1$ and since $xf(x) = f(x^2)$ I get $$f(x^2 + y) = f(x^2) + f(y)$$ Now I replace x^2 by x to get $$f(x + y) = f(x) + f(y)$$ to give me the cauchy equation with solns $f(x) = 0$ for all x or $f(x) = x$(from f(1) = 1)



Now though I have given the solutions, there are two pints where I doubt my solution a little. They are when I replace$f(z)$ by some $x$. Is this step valid? And the other doubt is when I replace $x^2$ by $x$. Are these two steps valid and are there any errors in this solution?



Thanks in advance










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If you replace $f(z)$ by $x$ the equation you get will be valid for $x$ in the range of $f$. If you replace $x^{2}$ by $x$ your conclusion will only be valid for $x geq 0$. So these are indeed gaps in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Dec 21 '18 at 10:30










  • $begingroup$
    Moreover, unless additional properties (e.g. continuity) are assumed on $f$, $f(x) = cx$ is not the only solution of Cauchy's equation.
    $endgroup$
    – Song
    Dec 21 '18 at 10:35














1












1








1


1



$begingroup$


The following question was asked in INMO 2005:



"Find all functions $f:Bbb{R} rightarrow Bbb{R} $ such that
$$f(x^2 + yf(z)) = xf(x) + zf(y)$$



for all $x,y,z in Bbb{R}$.



I am thinking of this as a solution:



"Put $x = y = 0 $ in the given equation, I get $$f(0) = zf(0)$$ for all $z$ in $R$ which means that $f(0) = 0$



Further take $y=0$ in the original equation to give me $$f(x^2)=xf(x)$$



Put $x = 0$ in the original equation to give $$f(yf(z)) = zf(y) *$$



In the above equation put $z=1$ to give $$f(yf(1)) = f(y)$$



Let $f(1) = c$ ,which gives me $f(cy) = f(y) **$
In the equation marked * put $y = 1$ to give $$f(f(z)) = zc$$



Taking $f$ on both sides gives $f(f(f(z))) = f(zc) = f(z)$ from **



Let $f(z) = x$ , say then I get $f(f(x)) = x$, but I know that $f(f(x)) = xc$ which means that $$c=1$$



Not put $z = 1$ and since $xf(x) = f(x^2)$ I get $$f(x^2 + y) = f(x^2) + f(y)$$ Now I replace x^2 by x to get $$f(x + y) = f(x) + f(y)$$ to give me the cauchy equation with solns $f(x) = 0$ for all x or $f(x) = x$(from f(1) = 1)



Now though I have given the solutions, there are two pints where I doubt my solution a little. They are when I replace$f(z)$ by some $x$. Is this step valid? And the other doubt is when I replace $x^2$ by $x$. Are these two steps valid and are there any errors in this solution?



Thanks in advance










share|cite|improve this question











$endgroup$




The following question was asked in INMO 2005:



"Find all functions $f:Bbb{R} rightarrow Bbb{R} $ such that
$$f(x^2 + yf(z)) = xf(x) + zf(y)$$



for all $x,y,z in Bbb{R}$.



I am thinking of this as a solution:



"Put $x = y = 0 $ in the given equation, I get $$f(0) = zf(0)$$ for all $z$ in $R$ which means that $f(0) = 0$



Further take $y=0$ in the original equation to give me $$f(x^2)=xf(x)$$



Put $x = 0$ in the original equation to give $$f(yf(z)) = zf(y) *$$



In the above equation put $z=1$ to give $$f(yf(1)) = f(y)$$



Let $f(1) = c$ ,which gives me $f(cy) = f(y) **$
In the equation marked * put $y = 1$ to give $$f(f(z)) = zc$$



Taking $f$ on both sides gives $f(f(f(z))) = f(zc) = f(z)$ from **



Let $f(z) = x$ , say then I get $f(f(x)) = x$, but I know that $f(f(x)) = xc$ which means that $$c=1$$



Not put $z = 1$ and since $xf(x) = f(x^2)$ I get $$f(x^2 + y) = f(x^2) + f(y)$$ Now I replace x^2 by x to get $$f(x + y) = f(x) + f(y)$$ to give me the cauchy equation with solns $f(x) = 0$ for all x or $f(x) = x$(from f(1) = 1)



Now though I have given the solutions, there are two pints where I doubt my solution a little. They are when I replace$f(z)$ by some $x$. Is this step valid? And the other doubt is when I replace $x^2$ by $x$. Are these two steps valid and are there any errors in this solution?



Thanks in advance







proof-verification functional-equations






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share|cite|improve this question













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share|cite|improve this question








edited Dec 21 '18 at 11:07









dmtri

1,7612521




1,7612521










asked Dec 21 '18 at 10:24









saisanjeevsaisanjeev

1,081312




1,081312








  • 3




    $begingroup$
    If you replace $f(z)$ by $x$ the equation you get will be valid for $x$ in the range of $f$. If you replace $x^{2}$ by $x$ your conclusion will only be valid for $x geq 0$. So these are indeed gaps in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Dec 21 '18 at 10:30










  • $begingroup$
    Moreover, unless additional properties (e.g. continuity) are assumed on $f$, $f(x) = cx$ is not the only solution of Cauchy's equation.
    $endgroup$
    – Song
    Dec 21 '18 at 10:35














  • 3




    $begingroup$
    If you replace $f(z)$ by $x$ the equation you get will be valid for $x$ in the range of $f$. If you replace $x^{2}$ by $x$ your conclusion will only be valid for $x geq 0$. So these are indeed gaps in your argument.
    $endgroup$
    – Kavi Rama Murthy
    Dec 21 '18 at 10:30










  • $begingroup$
    Moreover, unless additional properties (e.g. continuity) are assumed on $f$, $f(x) = cx$ is not the only solution of Cauchy's equation.
    $endgroup$
    – Song
    Dec 21 '18 at 10:35








3




3




$begingroup$
If you replace $f(z)$ by $x$ the equation you get will be valid for $x$ in the range of $f$. If you replace $x^{2}$ by $x$ your conclusion will only be valid for $x geq 0$. So these are indeed gaps in your argument.
$endgroup$
– Kavi Rama Murthy
Dec 21 '18 at 10:30




$begingroup$
If you replace $f(z)$ by $x$ the equation you get will be valid for $x$ in the range of $f$. If you replace $x^{2}$ by $x$ your conclusion will only be valid for $x geq 0$. So these are indeed gaps in your argument.
$endgroup$
– Kavi Rama Murthy
Dec 21 '18 at 10:30












$begingroup$
Moreover, unless additional properties (e.g. continuity) are assumed on $f$, $f(x) = cx$ is not the only solution of Cauchy's equation.
$endgroup$
– Song
Dec 21 '18 at 10:35




$begingroup$
Moreover, unless additional properties (e.g. continuity) are assumed on $f$, $f(x) = cx$ is not the only solution of Cauchy's equation.
$endgroup$
– Song
Dec 21 '18 at 10:35










1 Answer
1






active

oldest

votes


















2












$begingroup$

As @Kavi Rama Murthy pointed out, substitution $f(z) mapsto x$ is only for $x$ in the range of $f$. However, if we assume that $f$ is not the trivial solution $0$, there is $x'neq 0$ such that
$$f(f(x')) = x'.$$ Since $f(f(x)) = cx$ for all real $x$, we have $c=1$ and it follows
$$
f(f(x)) = x,quad forall xinmathbb{R}.
$$
Secondly, $x^2mapsto x$ is only for non-negative $x$. Hence we get
$$
f(x+y) =f(x) +f(y) ,quad xgeq 0, yin mathbb{R}.tag{*}
$$
If one of $x,y$ is non-negative, $f(x+y) = f(x)+f(y)$ holds. If both are negative, we may use
$$
f(x^2) = xf(x) = f((-x)^2) = -xf(-x),quadforall x.
$$
Since $f$ is odd, $(*)$ holds for every real.



Finally, we obtained Cauchy's equation $(*)$. However, we can find non-continuous solution of $(*)$ using Hamel $mathbb{Q}$-basis of $mathbb{R}$. That is, $(*)$ does not necessarily lead to $f(x) = x$ (non-trivial solution.) To prove $f(x) = x$, let $F$ denote the set of all reals $x$ such that $x=f(x)$. We can see that $(i)$ $F$ is closed under addition, subtraction and multiplication, $(ii)$ $x^2 in F$ implies $pm xin F$. Proof of $(i)$ is quite straightforward if we notice that $f(zf(y))=yf(z)$. Proof of $(ii)$ uses the identity $f(x^2) = xf(x)$. Now, we are ready to prove $F=mathbb{R}.$ Note that
$$
f(x+f(x)) = f(x) +f(f(x)) = x+f(x),
$$
and
$$
f(x-f(x)) = f(x) +f(-f(x)) = f(x) -f(f(x)) = f(x) - x.
$$
Let $k= x-f(x)$. By $(ii)$, we have $kin F$ if $k^2 = x^2 + f(x)^2 -2xf(x)in F$. Note that $$ x+f(x)in F Rightarrow (x+f(x))^2 in F,$$
and that
$$ f(xf(x)) =xf(x) Rightarrow xf(x) in F.$$
These two facts show that $k^2 = (x+f(x))^2-4xf(x)in F$ and hence we get $kin F$.
Finally, from
$
f(k) = -k = k,
$
we get $x-f(x) = 0$ for all reals $x$ as desired.






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    2












    $begingroup$

    As @Kavi Rama Murthy pointed out, substitution $f(z) mapsto x$ is only for $x$ in the range of $f$. However, if we assume that $f$ is not the trivial solution $0$, there is $x'neq 0$ such that
    $$f(f(x')) = x'.$$ Since $f(f(x)) = cx$ for all real $x$, we have $c=1$ and it follows
    $$
    f(f(x)) = x,quad forall xinmathbb{R}.
    $$
    Secondly, $x^2mapsto x$ is only for non-negative $x$. Hence we get
    $$
    f(x+y) =f(x) +f(y) ,quad xgeq 0, yin mathbb{R}.tag{*}
    $$
    If one of $x,y$ is non-negative, $f(x+y) = f(x)+f(y)$ holds. If both are negative, we may use
    $$
    f(x^2) = xf(x) = f((-x)^2) = -xf(-x),quadforall x.
    $$
    Since $f$ is odd, $(*)$ holds for every real.



    Finally, we obtained Cauchy's equation $(*)$. However, we can find non-continuous solution of $(*)$ using Hamel $mathbb{Q}$-basis of $mathbb{R}$. That is, $(*)$ does not necessarily lead to $f(x) = x$ (non-trivial solution.) To prove $f(x) = x$, let $F$ denote the set of all reals $x$ such that $x=f(x)$. We can see that $(i)$ $F$ is closed under addition, subtraction and multiplication, $(ii)$ $x^2 in F$ implies $pm xin F$. Proof of $(i)$ is quite straightforward if we notice that $f(zf(y))=yf(z)$. Proof of $(ii)$ uses the identity $f(x^2) = xf(x)$. Now, we are ready to prove $F=mathbb{R}.$ Note that
    $$
    f(x+f(x)) = f(x) +f(f(x)) = x+f(x),
    $$
    and
    $$
    f(x-f(x)) = f(x) +f(-f(x)) = f(x) -f(f(x)) = f(x) - x.
    $$
    Let $k= x-f(x)$. By $(ii)$, we have $kin F$ if $k^2 = x^2 + f(x)^2 -2xf(x)in F$. Note that $$ x+f(x)in F Rightarrow (x+f(x))^2 in F,$$
    and that
    $$ f(xf(x)) =xf(x) Rightarrow xf(x) in F.$$
    These two facts show that $k^2 = (x+f(x))^2-4xf(x)in F$ and hence we get $kin F$.
    Finally, from
    $
    f(k) = -k = k,
    $
    we get $x-f(x) = 0$ for all reals $x$ as desired.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      As @Kavi Rama Murthy pointed out, substitution $f(z) mapsto x$ is only for $x$ in the range of $f$. However, if we assume that $f$ is not the trivial solution $0$, there is $x'neq 0$ such that
      $$f(f(x')) = x'.$$ Since $f(f(x)) = cx$ for all real $x$, we have $c=1$ and it follows
      $$
      f(f(x)) = x,quad forall xinmathbb{R}.
      $$
      Secondly, $x^2mapsto x$ is only for non-negative $x$. Hence we get
      $$
      f(x+y) =f(x) +f(y) ,quad xgeq 0, yin mathbb{R}.tag{*}
      $$
      If one of $x,y$ is non-negative, $f(x+y) = f(x)+f(y)$ holds. If both are negative, we may use
      $$
      f(x^2) = xf(x) = f((-x)^2) = -xf(-x),quadforall x.
      $$
      Since $f$ is odd, $(*)$ holds for every real.



      Finally, we obtained Cauchy's equation $(*)$. However, we can find non-continuous solution of $(*)$ using Hamel $mathbb{Q}$-basis of $mathbb{R}$. That is, $(*)$ does not necessarily lead to $f(x) = x$ (non-trivial solution.) To prove $f(x) = x$, let $F$ denote the set of all reals $x$ such that $x=f(x)$. We can see that $(i)$ $F$ is closed under addition, subtraction and multiplication, $(ii)$ $x^2 in F$ implies $pm xin F$. Proof of $(i)$ is quite straightforward if we notice that $f(zf(y))=yf(z)$. Proof of $(ii)$ uses the identity $f(x^2) = xf(x)$. Now, we are ready to prove $F=mathbb{R}.$ Note that
      $$
      f(x+f(x)) = f(x) +f(f(x)) = x+f(x),
      $$
      and
      $$
      f(x-f(x)) = f(x) +f(-f(x)) = f(x) -f(f(x)) = f(x) - x.
      $$
      Let $k= x-f(x)$. By $(ii)$, we have $kin F$ if $k^2 = x^2 + f(x)^2 -2xf(x)in F$. Note that $$ x+f(x)in F Rightarrow (x+f(x))^2 in F,$$
      and that
      $$ f(xf(x)) =xf(x) Rightarrow xf(x) in F.$$
      These two facts show that $k^2 = (x+f(x))^2-4xf(x)in F$ and hence we get $kin F$.
      Finally, from
      $
      f(k) = -k = k,
      $
      we get $x-f(x) = 0$ for all reals $x$ as desired.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        As @Kavi Rama Murthy pointed out, substitution $f(z) mapsto x$ is only for $x$ in the range of $f$. However, if we assume that $f$ is not the trivial solution $0$, there is $x'neq 0$ such that
        $$f(f(x')) = x'.$$ Since $f(f(x)) = cx$ for all real $x$, we have $c=1$ and it follows
        $$
        f(f(x)) = x,quad forall xinmathbb{R}.
        $$
        Secondly, $x^2mapsto x$ is only for non-negative $x$. Hence we get
        $$
        f(x+y) =f(x) +f(y) ,quad xgeq 0, yin mathbb{R}.tag{*}
        $$
        If one of $x,y$ is non-negative, $f(x+y) = f(x)+f(y)$ holds. If both are negative, we may use
        $$
        f(x^2) = xf(x) = f((-x)^2) = -xf(-x),quadforall x.
        $$
        Since $f$ is odd, $(*)$ holds for every real.



        Finally, we obtained Cauchy's equation $(*)$. However, we can find non-continuous solution of $(*)$ using Hamel $mathbb{Q}$-basis of $mathbb{R}$. That is, $(*)$ does not necessarily lead to $f(x) = x$ (non-trivial solution.) To prove $f(x) = x$, let $F$ denote the set of all reals $x$ such that $x=f(x)$. We can see that $(i)$ $F$ is closed under addition, subtraction and multiplication, $(ii)$ $x^2 in F$ implies $pm xin F$. Proof of $(i)$ is quite straightforward if we notice that $f(zf(y))=yf(z)$. Proof of $(ii)$ uses the identity $f(x^2) = xf(x)$. Now, we are ready to prove $F=mathbb{R}.$ Note that
        $$
        f(x+f(x)) = f(x) +f(f(x)) = x+f(x),
        $$
        and
        $$
        f(x-f(x)) = f(x) +f(-f(x)) = f(x) -f(f(x)) = f(x) - x.
        $$
        Let $k= x-f(x)$. By $(ii)$, we have $kin F$ if $k^2 = x^2 + f(x)^2 -2xf(x)in F$. Note that $$ x+f(x)in F Rightarrow (x+f(x))^2 in F,$$
        and that
        $$ f(xf(x)) =xf(x) Rightarrow xf(x) in F.$$
        These two facts show that $k^2 = (x+f(x))^2-4xf(x)in F$ and hence we get $kin F$.
        Finally, from
        $
        f(k) = -k = k,
        $
        we get $x-f(x) = 0$ for all reals $x$ as desired.






        share|cite|improve this answer











        $endgroup$



        As @Kavi Rama Murthy pointed out, substitution $f(z) mapsto x$ is only for $x$ in the range of $f$. However, if we assume that $f$ is not the trivial solution $0$, there is $x'neq 0$ such that
        $$f(f(x')) = x'.$$ Since $f(f(x)) = cx$ for all real $x$, we have $c=1$ and it follows
        $$
        f(f(x)) = x,quad forall xinmathbb{R}.
        $$
        Secondly, $x^2mapsto x$ is only for non-negative $x$. Hence we get
        $$
        f(x+y) =f(x) +f(y) ,quad xgeq 0, yin mathbb{R}.tag{*}
        $$
        If one of $x,y$ is non-negative, $f(x+y) = f(x)+f(y)$ holds. If both are negative, we may use
        $$
        f(x^2) = xf(x) = f((-x)^2) = -xf(-x),quadforall x.
        $$
        Since $f$ is odd, $(*)$ holds for every real.



        Finally, we obtained Cauchy's equation $(*)$. However, we can find non-continuous solution of $(*)$ using Hamel $mathbb{Q}$-basis of $mathbb{R}$. That is, $(*)$ does not necessarily lead to $f(x) = x$ (non-trivial solution.) To prove $f(x) = x$, let $F$ denote the set of all reals $x$ such that $x=f(x)$. We can see that $(i)$ $F$ is closed under addition, subtraction and multiplication, $(ii)$ $x^2 in F$ implies $pm xin F$. Proof of $(i)$ is quite straightforward if we notice that $f(zf(y))=yf(z)$. Proof of $(ii)$ uses the identity $f(x^2) = xf(x)$. Now, we are ready to prove $F=mathbb{R}.$ Note that
        $$
        f(x+f(x)) = f(x) +f(f(x)) = x+f(x),
        $$
        and
        $$
        f(x-f(x)) = f(x) +f(-f(x)) = f(x) -f(f(x)) = f(x) - x.
        $$
        Let $k= x-f(x)$. By $(ii)$, we have $kin F$ if $k^2 = x^2 + f(x)^2 -2xf(x)in F$. Note that $$ x+f(x)in F Rightarrow (x+f(x))^2 in F,$$
        and that
        $$ f(xf(x)) =xf(x) Rightarrow xf(x) in F.$$
        These two facts show that $k^2 = (x+f(x))^2-4xf(x)in F$ and hence we get $kin F$.
        Finally, from
        $
        f(k) = -k = k,
        $
        we get $x-f(x) = 0$ for all reals $x$ as desired.







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        edited Dec 21 '18 at 12:27

























        answered Dec 21 '18 at 11:03









        SongSong

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        18.6k21651






























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