Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $f(xf(y))+f(yf(x)=2xy$.

By putting $x=y=0$ we get $f(0)=0$ and by putting $x=y=1$ we get $f(f(1))=1$. Let $y=f(1)\Rightarrow f(x)+f(f(x)f(1))=2x$, which tells us that $f$ is an injective function. The only solutions I came up with so far are $f(x)=x$ and $f(x)=-x$.

  • 3
    $\begingroup$ $f(xf(x))=x^2=f(-xf(-x))$ and therefore $xf(x)=-xf(-x)$. Hence $f(x)=-f(-x)$. $\endgroup$ – Pp.. Jan 6 '15 at 19:47
  • $\begingroup$ @Pp : That is definition of all $ odd functions $ $\endgroup$ – Narasimham Jan 6 '15 at 20:14
  • 1
    $\begingroup$ @Narasimham Yes, it is a property that $f$ must have. $\endgroup$ – Pp.. Jan 6 '15 at 20:15

Putting $x=1$ in $f(x)+f(f(x)f(1))=2xf(1)$ we get that $f(1)=\pm 1$

Let $f(1)=1$.

$x\rightarrow xf(x), y\rightarrow \frac{1}{x} \Rightarrow f(xf(x)f(\frac{1}{x}))+f(x)=2f(x) \rightarrow f(x)f(\frac{1}{x})=1$.

Putting now $x\rightarrow \frac{1}{x}, y \rightarrow 1$ we get $\frac{1}{f(x)}+ \frac{1}{f(f(x))}=\frac{2}{x}$. Using $f(f(x))=2x-f(x)$ we get $(f(x)-x)^2=0$, and there we get one of our two solutions, $f(x)=x$. We get $f(x)=-x$ as the other solution by putting $f(1)=-1$.


This is all I have found so far:

Just keep feeding the snake with its own tail. As you noted $f(f(1))=1$. So with $x=y=f(1)$ we then see that $1=f(f(1))=(f(1))^2$ so that $f(1)=\pm 1$. As you almost correctly noted, we have for $y=f(1)$ that $$ f(x)+f(f(x)f(1))=2xf(1) $$ which indeed shows that $f$ is injective. With that observation, the comment by Pp.. works to conclude that $f$ is odd since $f(xf(x))=x^2=f(-xf(-x))$ then implies $f(x)=-f(-x)$.

Now, if we knew that $f$ was a polynomial it would be easy to conlude that $f(xf(x))=x^2$ implies $\deg f(\deg f+1)=2$ so $f$ has to have degree 1 in that case. Since $f(0)=0$ and $f(1)=\pm 1$ that would then imply $f(x)=\pm x$.

But if $f$ is not a polynomial, all I can tell this far is that $f$ is an injective, odd function with $f(0)=0$ and $f(1)=\pm 1$ and a bucnh of functional equations attached to it. Maybe someone else can elaborate further ...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.