Let f:R→R satisfy

1. f(a)≠0 for some a in R
2. f(xf(y))=yf(x) for all x,y in R

Prove that f(-x)=-f(x) for all x in R.

Obviously f1(x)=x and f2(x)=1/x (with f2(0):=0) are solutions. They are pretty much the only solutions.

We have
f(f(x)*f(y))=y*f(f(x))=x*y

Applying f on both sides gives (due to f(f(x))=x, see my previous post)

(1) f(x)*f(y)=f(x*y)

From (1) it is easy to show that f(x^r)=f(x)^r for all rational numbers r. Next compute f(x) for all rational powers of 2: If x=2^r then f(x)=f(2)^r=f(2)^(log(x)/log(r))=x^(log(f(2))/log(2)).

So f(x)=x^c for some c. From f(f(x)) follows x=f(f(x))=f(x^c)=f(x)^c. Therefore c^2=1, i.e. c=-1 or c=-1.

So we have shown that f, when restricted to all rational powers of 2 (i.e. to x=2^r) is the identity or the reciprocal function.

Because the rational powers of 2 and their complement are dense in the positive real numbers, that means there can be no other solutions, except those that are discontinuous everywhere. I suspect those weird examples could be constructed easily.

Posted by JLo on 2006-12-29 08:44:43