Find all functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that $$f(xyz)=f(xy+yz+xz)(f(x)+f(y)+f(z))$$ for all non-zero reals $x, y, z$ such that $xy+yz+xz\neq 0$.

I think that only two solutions are: $f(x)=\frac{1}{3}$ and $f(x)=\frac{1}{x}$.

I would appreciate any suggestions.

  • $\begingroup$ $f(x) = \frac 13$ is not an injective function. But, I haven't thought of any others, either. $\endgroup$ – Doug M Jun 23 '16 at 2:28

Plugging in $(x,y,z)=(1,1,-1)$ and $(x,y,z)=(1,-1,-1)$, we get $$\begin{eqnarray*} f(-1)&=&f(-1)(2f(1)+f(-1))\\ f(1)&=&f(-1)(2f(-1)+f(1)) \end{eqnarray*}$$ Solving, $f(1)=1$ or $\frac13$.

Case 1: $f(1)=1$

For any $a\neq0$, plugging in $(x,y,z)=(a,-a,1)$ gives $$ f(-a^2)=f(-a^2)(f(a)+f(-a)+1). $$ Hence $f(-a)=-f(a)$. For any $a,b>0$, putting $(x,y,z)=(\sqrt{a},-\sqrt{a},b)$ gives $$ f(-ab)=f(-a)f(b) $$ Hence $f(ab)=f(a)f(b)$.

Let $g(x)=\frac1{f(x)}$, so $g(ab)=g(a)g(b)$ for $a,b>0$. In particular if $a>0$ then $g(a)=g(\sqrt{a})^2>0$. The original equation becomes $$ g(xy+yz+zx)=g(xyz)\left(\frac1{g(x)}+\frac1{g(y)}+\frac1{g(z)}\right) =g(xy)+g(yz)+g(zx). $$ For any $a,b,c>0$, setting $(x,y,z)=(\sqrt{ac/b},\sqrt{ab/c},\sqrt{bc/a})$ gives $$ g(a+b+c)=g(a)+g(b)+g(c). $$ In particular if $a>b>0$ then $g(a)=g(b)+2g(\frac{a-b}2)>g(b)$. Note that $g(3)=3g(1)=3$, so $g(3^k)=3^k$ for all integers $k$. Thus for any $a,b>0$, $$ g(a+b)\leq g\left(a+b+3^k\right)=g(a)+g(b)+3^k. $$ Sending $k\rightarrow-\infty$, $g(a+b)\leq g(a)+g(b)$. On the other hand, pick $k$ such that $3^k<b$. Then $$ g(a)+g(b)\leq g(a)+g(b-3^k)+g(3^k)=g(a+b). $$ Hence $g(a+b)=g(a)+g(b)$. Since $g(1)=1$, $g(x)=x$ on positive rationals. Also $g$ is increasing, so $g(x)=x$ on positive reals. Since $f$ is odd, $f(x)=\frac1x$ for all $x\in\mathbb R^*$.

Case 2: $f(1)=\frac13$

For any $a\neq0$, plugging in $(x,y,z)=(a,-a,1)$ gives $$ f(-a^2)=f(-a^2)(f(a)+f(-a)+\frac13). $$ Hence $f(a)+f(-a)=\frac23$. Putting $(x,y,z)=(a,-a,b^2)$ gives $$ f(-a^2b^2)=f(-a^2)(\frac23+f(b^2))=f(-a^2)(\frac43-f(-b^2)). $$ Thus $$ f(-a^2b^2)+f(-a^2)f(-b^2)=\frac43f(-a^2). $$ Swapping $a$ and $b$, $f(-a^2)=f(-b^2)$. In particular, putting $b=1$, $f(-a^2)=\frac13$. Hence $f(a)=\frac13$ for $a<0$, and for $a>0$ we have $f(a)=\frac23-f(-a)=\frac13$ also.

  • $\begingroup$ Nice solution! Thank you! $\endgroup$ – MoM-ADMIN Jun 23 '16 at 10:13
  • $\begingroup$ How can you get conclusion "g(x) = x on positive real" from "g(x) = x on positive rationals"? $\endgroup$ – Zack Ni Jun 23 '16 at 12:09
  • $\begingroup$ @ Zack Ni, Because $g$ is increasing and $\mathbb{Q}$ is dense in $\mathbb{R}.$ $\endgroup$ – MoM-ADMIN Jun 23 '16 at 14:01

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.