Find all functions $f: \Bbb{R} \to \Bbb{R} $ such that for every $x, y \in \mathbb R, f(x\cdot f(y)) = x\cdot y$.

Now, I imagine $f(x)$ is some sort of a linear function, since $f(y)$ is basically just a constant. But I lack ideas on how to approach this problem. I might be able to find some functions that meet the criteria (for example $f(x) = x$), but how can I be sure I've found them all?


If we put $x=0$ we get $f(0)=0 \;\;\;(*)$.

If we put $x=1$ we got $f(f(y))=y$ so $f$ is bijective. Since $f$ is injective and we have $$f(xf(y))=xy=f(yf(x))$$ we have also $xf(y)=yf(x)$ so ${f(x)\over x} = {f(y)\over y} $ for all $xy\ne 0$. Since left side is independent of $y$ we conclude that ${f(x)\over x}=a$ for some real $a$. So $f(x)=ax$ for all $x\ne 0$. Since we have also $(*)$ we have $f(x)=ax$ for all $x$.

Now if we put this into starting formula we get $a=\pm 1$, so we have two solutions.


For each $a\in \Bbb R$, let $f_a$ be some solution with $f_a(1) = a$. Since the functional equation is valid for all $x, y$, it's specifically valid when $y = 1$. Assuming $a \neq 0$, we set $x = \frac ta$, and get $$ f_a\left(\frac ta\cdot a\right) = \frac ta\implies f_a(t) = \frac ta $$ Additionally, we assumed that $f_a(1) = a$, which means $\frac 1a = a$, and we get $a = \pm 1$. We see that both the solutions $f_{-1}$ and $f_1$ actually work.

On the other hand, if $a = 0$, which is to say $f_0(1) = 0$, we get for all $x$ that $$ f_0(x\cdot 0) = x $$ which implies that $f_0(0)$ takes on all real values simultaneously, which is not possible.

So we're left with only the two solutions $f_{-1}$ and $f_1$.

  • $\begingroup$ Comments about general approach: Assuming some arbitrary value for $f(1)$ or $f(0)$ (or other specific values that might be nicely related to the hape of the equation) is usually helpful, as is playing around with inserting relations and constants for the different variables on the original equation. Here I did both, and it worked out nicely. $\endgroup$ – Arthur Nov 10 '17 at 9:22

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