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Two such functions would be $f(x) = x$ and $f(x) = -x$, but how would I know I've found all satisfactory functions?

E. Joseph
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Torn
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3 Answers3

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Those are the only possibilities.

Letting $x=1$ you get $$f(f(y))=y$$ for all real $y$. Applying $f$ to both sides of the original equation, the relation above gives $xf(y)=f(xy)$. Now let $y=1$; we get $xf(1)=f(x)$. So $f(x)=cx$ for some real number $c$. Now plug this back into the original equation, we get $c(xcy)=xy$, or $c^2xy=xy$. This gives $c=\pm 1$, so $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$.

JKRT
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  • Okay, that looks convincing, but if x or y are set to some specific number, why can we assume that the equation still holds when they aren't that number? – Torn Oct 28 '16 at 08:34
  • In the first step, when I set $x=1$, the other variable is left "free", so the equation $f(f(u))=u$ holds for arbitrary $u$. In particular, if you let $x$ and $y$ be "free variables" again, and write $u=xf(y)$, you get $f(f(xf(y))=xf(y)$ for all $x,y$. This, applied to the original equation, allows me to conclude $xf(y)=f(xy)$ for all $x,y$. The next step is similar...you let $y=1$ but the variable $x$ is left free, so the conclusion $f(x)=cx$ holds for arbitrary $x$. – JKRT Oct 28 '16 at 09:03
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Set $c=f(1)$. Then $$f(xc)=f(x f(1))=x \cdot 1 = x$$ so $c$ can't be zero. Thus, $f(x)=x/c$ and $c=f(1)=1/c$ yields $c=\pm 1$ both of which works.

H. H. Rugh
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First we check that $f(f(k))=k\, \forall k\in \mathbb{R}$, this fact is easy to prove making $x=1$ in the given functional equation.

Then $$f(f(xf(y)))=f(xy)$$ $$xf(y)=f(xy)$$ If $y=1$, We have $f(x)=xf(1)$.

Then back in the given equation. $$f(xf(y))=f(xyf(1))=xy(f(1))^2=xy$$

And from the last expression we have $f(1)=1$ or $f(1)=-1$.

Ricardo Largaespada
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