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

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?

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

## 1 Answer

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,

$$\frac{f(x)}{x}\frac{f(y)}{y}=1$$

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.