Find all injective functions $f : \mathbb{R} \rightarrow \mathbb{R}$, such that for any $x, y \in \mathbb{R}$, they satisfy the condition: $f(xy) = f(f(x)f(y))$.

Attempt: Because f is injective, $xy=f(x)f(y)$. What now?

  • 1
    $\begingroup$ well, what if $y=1$? $\endgroup$
    – lulu
    Commented May 20 at 20:20
  • $\begingroup$ Really, you gotta try basic stuff like $x,y$ being $0,1$. $\endgroup$
    – copper.hat
    Commented May 20 at 20:29

1 Answer 1


Let's start by checking the case when $x=0$, then we get $0=f(0)f(y)$. This means $f(0)=0$ because $f(y)=0$ would mean $f$ is not injective!

Now that we've settled the $x=0$ case, we can divide through to make,


This implies that $\frac{f(x)}{x}=c$ for some constant $c$, because otherwise we could hold $y$ constant and change $x$ to break the equality.

At this point it's down hill, you simply plug back in $f(x)=cx$ to $xy=f(x)f(y)$ to find $c^2=1$. So $f(x)=x$ and $f(x)=-x$ are the only possible solutions. And we can easily check that they are.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .