Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x,y$ we have: $f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$

I have solved this problem but the solution is "bruteforce", so I wanted to ask if there is a more elegant way of approaching this.

  • 3
    $\begingroup$ you have found all solutions by bruteforce? $\endgroup$ – Börge Oct 8 '15 at 11:18

I claim that the only possible solutions to the functional equation $$f\big(xf(y)\big)=(1-y)f(xy)+x^2y^2f(y)\tag0\label0$$ are $f(x)=0$ and $f(x)=x-x^2$.

To show that, first let $y=1$ in \eqref{0} and you'll get $f\big(xf(1)\big)=x^2f(1)$. Now assuming $f(1)\neq0$ we'll have $f(x)=\frac{x^2}{f(1)}$ for every real number $x$, whcih leads to a contradiction using \eqref{0}. So we have: $$f(1)=0\tag1\label1$$ Next, let $x=1$ in \eqref{0} and you'll get: $$f\big(f(y)\big)=\big(1-y+y^2\big)f(y)\tag2\label2$$ Now, substituting $f(y)$ for $y$ in \eqref{0} and using \eqref{2} we get: $$f\Big(xf\big(f(y)\big)\Big)=\big(1-f(y)\big)f\big(xf(y)\big)+x^2f(y)^2f\big(f(y)\big)\\ \therefore f\Big(x\big(1-y+y^2\big)f(y)\Big)=\big(1-f(y)\big)f\big(xf(y)\big)+x^2f(y)^3\big(1-y+y^2\big)$$ So, if $f(y)\neq0$, letting $x=\frac1{f(y)}$ in the last equation and using \eqref{1}, we get: $$f\big(1-y+y^2\big)=\big(1-y+y^2\big)f(y)\ne0$$ Thus by \eqref{2}, $f\big(f(y)\big)=f\big(1-y+y^2\big)\ne0$ and so $f\Big(f\big(f(y)\big)\Big)=f\Big(f\big(1-y+y^2\big)\Big)$. Hence by \eqref{2}: $$\big(1-f(y)+f(y)^2\big)f\big(f(y)\big)=\Big(1-\big(1-y+y^2\big)+\big(1-y+y^2\big)^2\Big)f\big(1-y+y^2\big)\\ \therefore1-f(y)+f(y)^2=1-\big(1-y+y^2\big)+\big(1-y+y^2\big)^2\\ \therefore\Big(f(y)-\big(1-y+y^2\big)\Big)\Big(f(y)+\big(1-y+y^2\big)-1\Big)=0$$ $$\therefore f(y)=1-y+y^2\quad or\quad f(y)=y-y^2\tag3\label3$$ Now, if $f(y)=1-y+y^2$ then $f\big(1-y+y^2\big)=\big(1-y+y^2\big)^2$, but because $1-y+y^2\neq0$ we must have $f\big(1-y+y^2\big)=1-\big(1-y+y^2\big)+\big(1-y+y^2\big)^2$ or $f\big(1-y+y^2\big)=\big(1-y+y^2\big)-\big(1-y+y^2\big)^2$ by \eqref{3}. It's easy to check that the latter case leads to a contradiction and the former case leads to $y=0$ or $y=1$. But letting $x=y=0$ in \eqref{0} we have $f(0)=0$, and by \eqref{1} we have $f(1)=0$. So this case leads to a contradiction, too, since we assumed $f(y)\neq0$. So $f(y)$ can not be equal to $1-y+y^2$ and hence by \eqref{3}, $f(y)=y-y^2$.

Finally, if there is a real number $y$ such that $y\neq0$, $y\neq1$ and $f(y)=0$, then by \eqref{0} we get $f(0)=(1-y)f(xy)+0$ for every real number $x$, which means that $f$ is the constant zero function. This yields our first claim. It's easy to check that indeed the two mentioned functions satisfy \eqref{0}.

| cite | improve this answer | |
  • $\begingroup$ I came to the same solution... ;) even thogh I used some different steps ;) $\endgroup$ – Börge Oct 8 '15 at 12:15
  • $\begingroup$ @Börge Post your answer. It may be useful. $\endgroup$ – Mohsen Shahriari Oct 8 '15 at 12:17
  • $\begingroup$ If I may suggest a little simplification: from $(3)$ it easily follows that $f(x) = f(y) \implies$ $y = x$ or $y = 1-x$. Thus, when you get $f(f(y)) = f(1-y+y^2)$, you automatically get $f(y) = 1 - y+y^2$ or $f(y) = 1-(1- y +y^2)$, and can easily dismiss the first case because $1-y+y^2$ has no real roots, while $f$ does. $\endgroup$ – Ennar Oct 8 '15 at 14:16
  • $\begingroup$ @Ennar I think that there's a problem with your argument. I think in that step, there's still possible that we have $f(y)=1-y+y^2$ for some $y$ and $f(y)=y-y^2$ for some other $y$. It may also be useful to note that, this part of the proof is about the real numbers for which we have $f(y)\neq0$ and so having roots is irrelevant. $\endgroup$ – Mohsen Shahriari Oct 8 '15 at 16:41
  • $\begingroup$ Oh, ok, I was too hasty. $\endgroup$ – Ennar Oct 8 '15 at 16:56

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.