The No. of Continuous function satisfying the condition $xf(y)+yf(x)=(x+y)f(x)f(y)$












5












$begingroup$


Could anyone tell me?



The No. of Continuous function satisfying the condition $$xf(y)+yf(x)=(x+y)f(x)f(y)$$ is



$1,2,3,$ or none of them?



or give me hints please.






continuity functional-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I can think of 2: $f(x)=0$ and $f(x)=1$. They are not interesting though.
    $endgroup$
    – Daryl
    Aug 29 '13 at 8:07






  • 1




    $begingroup$
    The trick usually in solving these equations is to plug in some interesting values. Otherwise you can't do much.
    $endgroup$
    – Patrick Da Silva
    Aug 29 '13 at 8:23










  • $begingroup$
    Related: Find all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$.
    $endgroup$
    – Martin Sleziak
    Dec 4 '18 at 18:39
















5












$begingroup$


Could anyone tell me?



The No. of Continuous function satisfying the condition $$xf(y)+yf(x)=(x+y)f(x)f(y)$$ is



$1,2,3,$ or none of them?



or give me hints please.






continuity functional-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I can think of 2: $f(x)=0$ and $f(x)=1$. They are not interesting though.
    $endgroup$
    – Daryl
    Aug 29 '13 at 8:07






  • 1




    $begingroup$
    The trick usually in solving these equations is to plug in some interesting values. Otherwise you can't do much.
    $endgroup$
    – Patrick Da Silva
    Aug 29 '13 at 8:23










  • $begingroup$
    Related: Find all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$.
    $endgroup$
    – Martin Sleziak
    Dec 4 '18 at 18:39














5












5








5


1



$begingroup$


Could anyone tell me?



The No. of Continuous function satisfying the condition $$xf(y)+yf(x)=(x+y)f(x)f(y)$$ is



$1,2,3,$ or none of them?



or give me hints please.






continuity functional-equations







share|cite|improve this question











$endgroup$




Could anyone tell me?



The No. of Continuous function satisfying the condition $$xf(y)+yf(x)=(x+y)f(x)f(y)$$ is



$1,2,3,$ or none of them?



or give me hints please.






continuity functional-equations




continuity functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 18:36









Martin Sleziak

44.7k9117272




44.7k9117272










asked Aug 29 '13 at 8:02









MarkovMarkov

17.2k957178




17.2k957178








  • 3




    $begingroup$
    I can think of 2: $f(x)=0$ and $f(x)=1$. They are not interesting though.
    $endgroup$
    – Daryl
    Aug 29 '13 at 8:07






  • 1




    $begingroup$
    The trick usually in solving these equations is to plug in some interesting values. Otherwise you can't do much.
    $endgroup$
    – Patrick Da Silva
    Aug 29 '13 at 8:23










  • $begingroup$
    Related: Find all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$.
    $endgroup$
    – Martin Sleziak
    Dec 4 '18 at 18:39














  • 3




    $begingroup$
    I can think of 2: $f(x)=0$ and $f(x)=1$. They are not interesting though.
    $endgroup$
    – Daryl
    Aug 29 '13 at 8:07






  • 1




    $begingroup$
    The trick usually in solving these equations is to plug in some interesting values. Otherwise you can't do much.
    $endgroup$
    – Patrick Da Silva
    Aug 29 '13 at 8:23










  • $begingroup$
    Related: Find all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$.
    $endgroup$
    – Martin Sleziak
    Dec 4 '18 at 18:39








3




3




$begingroup$
I can think of 2: $f(x)=0$ and $f(x)=1$. They are not interesting though.
$endgroup$
– Daryl
Aug 29 '13 at 8:07




$begingroup$
I can think of 2: $f(x)=0$ and $f(x)=1$. They are not interesting though.
$endgroup$
– Daryl
Aug 29 '13 at 8:07




1




1




$begingroup$
The trick usually in solving these equations is to plug in some interesting values. Otherwise you can't do much.
$endgroup$
– Patrick Da Silva
Aug 29 '13 at 8:23




$begingroup$
The trick usually in solving these equations is to plug in some interesting values. Otherwise you can't do much.
$endgroup$
– Patrick Da Silva
Aug 29 '13 at 8:23












$begingroup$
Related: Find all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$.
$endgroup$
– Martin Sleziak
Dec 4 '18 at 18:39




$begingroup$
Related: Find all $f:mathbb Rtomathbb R$ such that $forall x,yinmathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$.
$endgroup$
– Martin Sleziak
Dec 4 '18 at 18:39










1 Answer
1






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oldest

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9












$begingroup$

Assuming your functions are from $mathbb{R} to mathbb{R}$
$$
f(1)+f(1) = (1+1)f(1)f(1) Rightarrow f(1) = 0 text{ or } 1
$$
If $f(1) = 1$, then for any $xneq 0$
$$
x + f(x) = (x+1)f(x) Rightarrow f(x) = 1
$$
By continuity, $fequiv 1$
Similarly, if $f(1) = 0$, then $fequiv 0$.



Hence, the answer is 2.






share|cite|improve this answer









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    9












    $begingroup$

    Assuming your functions are from $mathbb{R} to mathbb{R}$
    $$
    f(1)+f(1) = (1+1)f(1)f(1) Rightarrow f(1) = 0 text{ or } 1
    $$
    If $f(1) = 1$, then for any $xneq 0$
    $$
    x + f(x) = (x+1)f(x) Rightarrow f(x) = 1
    $$
    By continuity, $fequiv 1$
    Similarly, if $f(1) = 0$, then $fequiv 0$.



    Hence, the answer is 2.






    share|cite|improve this answer









    $endgroup$


















      9












      $begingroup$

      Assuming your functions are from $mathbb{R} to mathbb{R}$
      $$
      f(1)+f(1) = (1+1)f(1)f(1) Rightarrow f(1) = 0 text{ or } 1
      $$
      If $f(1) = 1$, then for any $xneq 0$
      $$
      x + f(x) = (x+1)f(x) Rightarrow f(x) = 1
      $$
      By continuity, $fequiv 1$
      Similarly, if $f(1) = 0$, then $fequiv 0$.



      Hence, the answer is 2.






      share|cite|improve this answer









      $endgroup$
















        9












        9








        9





        $begingroup$

        Assuming your functions are from $mathbb{R} to mathbb{R}$
        $$
        f(1)+f(1) = (1+1)f(1)f(1) Rightarrow f(1) = 0 text{ or } 1
        $$
        If $f(1) = 1$, then for any $xneq 0$
        $$
        x + f(x) = (x+1)f(x) Rightarrow f(x) = 1
        $$
        By continuity, $fequiv 1$
        Similarly, if $f(1) = 0$, then $fequiv 0$.



        Hence, the answer is 2.






        share|cite|improve this answer









        $endgroup$



        Assuming your functions are from $mathbb{R} to mathbb{R}$
        $$
        f(1)+f(1) = (1+1)f(1)f(1) Rightarrow f(1) = 0 text{ or } 1
        $$
        If $f(1) = 1$, then for any $xneq 0$
        $$
        x + f(x) = (x+1)f(x) Rightarrow f(x) = 1
        $$
        By continuity, $fequiv 1$
        Similarly, if $f(1) = 0$, then $fequiv 0$.



        Hence, the answer is 2.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 29 '13 at 8:17









        Prahlad VaidyanathanPrahlad Vaidyanathan

        26.2k12152




        26.2k12152






























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