$f(y+zf(x))=f(y)+xf(z) $ with $x,y,zin mathbb{R}$











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Find all $fcolonmathbb{R} rightarrow mathbb{R}$ such that



$f(y+zf(x))=f(y)+xf(z)$ with $x,y,zin mathbb{R}$



When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?










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  • 1




    Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
    – 5xum
    Nov 22 at 13:49










  • I'd narrow down the number of functions by noting that these would hold for $x=y=z$
    – Rhys Hughes
    Nov 22 at 14:00










  • For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
    – Paul
    Nov 22 at 14:29















up vote
1
down vote

favorite












Find all $fcolonmathbb{R} rightarrow mathbb{R}$ such that



$f(y+zf(x))=f(y)+xf(z)$ with $x,y,zin mathbb{R}$



When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?










share|cite|improve this question




















  • 1




    Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
    – 5xum
    Nov 22 at 13:49










  • I'd narrow down the number of functions by noting that these would hold for $x=y=z$
    – Rhys Hughes
    Nov 22 at 14:00










  • For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
    – Paul
    Nov 22 at 14:29













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Find all $fcolonmathbb{R} rightarrow mathbb{R}$ such that



$f(y+zf(x))=f(y)+xf(z)$ with $x,y,zin mathbb{R}$



When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?










share|cite|improve this question















Find all $fcolonmathbb{R} rightarrow mathbb{R}$ such that



$f(y+zf(x))=f(y)+xf(z)$ with $x,y,zin mathbb{R}$



When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?







functional-equations






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edited Nov 22 at 15:02









Ivan Neretin

8,75021535




8,75021535










asked Nov 22 at 13:47









Trong Tuan

766




766








  • 1




    Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
    – 5xum
    Nov 22 at 13:49










  • I'd narrow down the number of functions by noting that these would hold for $x=y=z$
    – Rhys Hughes
    Nov 22 at 14:00










  • For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
    – Paul
    Nov 22 at 14:29














  • 1




    Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
    – 5xum
    Nov 22 at 13:49










  • I'd narrow down the number of functions by noting that these would hold for $x=y=z$
    – Rhys Hughes
    Nov 22 at 14:00










  • For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
    – Paul
    Nov 22 at 14:29








1




1




Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
– 5xum
Nov 22 at 13:49




Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
– 5xum
Nov 22 at 13:49












I'd narrow down the number of functions by noting that these would hold for $x=y=z$
– Rhys Hughes
Nov 22 at 14:00




I'd narrow down the number of functions by noting that these would hold for $x=y=z$
– Rhys Hughes
Nov 22 at 14:00












For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
– Paul
Nov 22 at 14:29




For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
– Paul
Nov 22 at 14:29










1 Answer
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By applying $z = 0$, we have $f(0) = 0$.



$f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.



Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.



Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.



By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.



By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.



So $f$ satisfies $f(y + z) = f(y)+ f(z)$.



By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.



Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.



It can be demonstrated by the following steps :




  1. Show that $f(x) = x$ for all $xinmathbb{Q}$.

  2. Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)

  3. Deduce that $f$ is strictly increasing.

  4. For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.






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    By applying $z = 0$, we have $f(0) = 0$.



    $f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.



    Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.



    Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.



    By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.



    By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.



    So $f$ satisfies $f(y + z) = f(y)+ f(z)$.



    By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.



    Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.



    It can be demonstrated by the following steps :




    1. Show that $f(x) = x$ for all $xinmathbb{Q}$.

    2. Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)

    3. Deduce that $f$ is strictly increasing.

    4. For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.






    share|cite|improve this answer



























      up vote
      1
      down vote













      By applying $z = 0$, we have $f(0) = 0$.



      $f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.



      Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.



      Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.



      By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.



      By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.



      So $f$ satisfies $f(y + z) = f(y)+ f(z)$.



      By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.



      Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.



      It can be demonstrated by the following steps :




      1. Show that $f(x) = x$ for all $xinmathbb{Q}$.

      2. Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)

      3. Deduce that $f$ is strictly increasing.

      4. For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        By applying $z = 0$, we have $f(0) = 0$.



        $f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.



        Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.



        Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.



        By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.



        By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.



        So $f$ satisfies $f(y + z) = f(y)+ f(z)$.



        By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.



        Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.



        It can be demonstrated by the following steps :




        1. Show that $f(x) = x$ for all $xinmathbb{Q}$.

        2. Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)

        3. Deduce that $f$ is strictly increasing.

        4. For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.






        share|cite|improve this answer














        By applying $z = 0$, we have $f(0) = 0$.



        $f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.



        Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.



        Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.



        By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.



        By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.



        So $f$ satisfies $f(y + z) = f(y)+ f(z)$.



        By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.



        Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.



        It can be demonstrated by the following steps :




        1. Show that $f(x) = x$ for all $xinmathbb{Q}$.

        2. Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)

        3. Deduce that $f$ is strictly increasing.

        4. For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 16:06

























        answered Nov 22 at 15:49









        曾靖國

        3868




        3868






























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