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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.

Mohsen Shahriari
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CryoDrakon
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1 Answers1

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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 $f(y)$ cannot 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}.

Mohsen Shahriari
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  • I came to the same solution... ;) even thogh I used some different steps ;) – Börge Oct 08 '15 at 12:15
  • @Börge Post your answer. It may be useful. – Mohsen Shahriari Oct 08 '15 at 12:17
  • 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. – Ennar Oct 08 '15 at 14:16
  • @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. – Mohsen Shahriari Oct 08 '15 at 16:41
  • Oh, ok, I was too hasty. – Ennar Oct 08 '15 at 16:56