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Find all real functions $f:\mathbb {R} \to \mathbb {R}$ satisfying the equation

$f(x^2+y.f(x))=x.f(x+y)$

My attempt - Clearly $f(0)=0$

Putting $x^2=x,y.f(x)=1$, we have $f(x+1)=x.f(x+y)$.

Now putting $x=x-1$,we have $f(x)=(x-1)f(x-1+y)$

Putting $x=0$ ,we have $f(0)=-1.f(y-1)$ or $f(y-1)=0$(\since $f(0)=0$)

Finally putting $y=(x+1)$ gives us $f(x)=0$.

This is one of the required functions. But $f(x)=x$ also satisfies the equation.How to achieve this? One of my friends said that the answer $f(x)=x$ could be obtained by using Cauchy theorem but when I searched the internet, I could not find any theorem of Cauchy related to functions .Does any such theorem exist. If yes what is it and how can it be used to solve the functional equation. Is there a way similar to the method of getting the first solution to achieve the second one?

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You proved that $f(0)=0$, we can continue from here

Suppose that there exist $k \neq 0$ such that $f(k)=0$. Then plugging $x=k,y=y-k$ gives, $f(k^2) = f(k^2 + (y-k)f(k)) = kf(k+(y-k))=kf(y)$, which means that $f(x)$ is a constant function and so $f(x)=0$.

Now suppose that there exist no $k \neq 0$ such that $f(k)=0$. Then plugging $x=x,y=-x$ gives $ f(x^2-xf(x)) = xf(0)= 0 $, which by assumption means $x^2=xf(x)$ or $f(x)=x$.

So we conclude that, possible functions are $f(x)=x,0 \forall x \in \mathbb{R} $

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  • $\begingroup$ Do u have any idea about Cauchy theorem?If yes can you please explain it to me.😊 $\endgroup$ – Snehil Sinha Feb 5 '15 at 15:26
  • $\begingroup$ I think , you friend mean Cauchy functional equation. $f(x+y)=f(x)+f(y)$ . It can be easily proved that, $f(x)=cx$ for rational numbers. But you can't say $f(x)=cx$ for all x, until some conditions are given to you , like continuity, monotonicity, boundness. If you are not given any such conditions, you can't say f(x)=kx. There exist many other awkward solution to this equation. $\endgroup$ – Shivang jindal Feb 5 '15 at 15:33
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    $\begingroup$ You can study the equation here: mathsolympiad.org.nz/wp-content/uploads/2009/01/… You will be able to get the idea of how to use it in olympiad problems. To see those awkard solutions[for better understanding] you can refer: sunejakobsen.files.wordpress.com/2010/12/cauchy_eng4.pdf $\endgroup$ – Shivang jindal Feb 5 '15 at 15:37
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    $\begingroup$ I don't understand how it becomes continuous, can you elaborate? :) $\endgroup$ – Sawarnik Feb 6 '15 at 14:46
  • $\begingroup$ ^ Sorry, i meant constant function -_- $\endgroup$ – Shivang jindal Feb 14 '15 at 3:36
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As you said $f(0)=0$ $$f(x^2+y.f(x))=x.f(x+y)$$ Take $y=0\implies f(x^2)=xf(x)$. So for $p>0$(p for positive,also I took this partition because $x^{1/2n}\mid n\in\mathbb N$ is defined only for $x>0$): $$f(p^2)=p(\sqrt pf(\sqrt p))=\lim_{n\to\infty}p^{\displaystyle \left(\sum_{k=0}^{n}\frac1{2^k}\right)}f\left(p^{1/n}\right)=p^2f(1)\\f(p)=kp\tag{$p>0$}$$ Now to find the function for $n<0$(n for negative): $$f((-x)^2)=f(x^2)\implies -xf(-x)=xf(x)\implies f(x)+f(-x)=0$$ $$f(n)=-f(-n)=-(-kn)=kn\tag{$n<0$}$$ So $f(x)$ is odd and we can say that it is: $$f(x)=kx$$

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    $\begingroup$ By your solution, one can conclude that, all solutions to functional equation $f(x^2)=xf(x)$ is of form $f(x)=mx$. But unfortunately this is wrong. See here, there are many examples : math.stackexchange.com/questions/109458/… $\endgroup$ – Shivang jindal Feb 5 '15 at 5:14
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    $\begingroup$ You assumed that function is continuous, which is WRONG assumption. $\endgroup$ – Shivang jindal Feb 5 '15 at 5:15
  • $\begingroup$ @Shivangjindal when(actually which line, to be precise) did I did that? $\endgroup$ – RE60K Feb 5 '15 at 12:07
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    $\begingroup$ (You assumed that function is continuous at 1) See , the link i gave. (2nd comment)[The solution is very similar to your solution] $\endgroup$ – Shivang jindal Feb 5 '15 at 12:15

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