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.

x=y=0 gives

(1) f(0)=0


Define c:=f(1). First let's determine c. We have

(2) f(c)=f(1*f(1))=1*f(1)=c

Also

(3) f(c^2)=f(c*f(c))=c*f(c)=c^2

We also have

(4) f(c*x)=f(x*f(1))=1*f(x)=f(x)

Setting x=c gives

f(c^2)=f(c)=c

Together with (3) we get c^2=c, i.e. c=0 or c=1. Because of (4) and (1), c=0 is impossible, otherwise f(x)=0 for all x, which is excluded as a valid solution. Hence

(5) c=f(1)=1

Now we have

(6) f(f(y))=f(1*f(y))=y*f(1)=y

This means f is a self-inverse function, i.e. its graph is symmetrical w.r.t. the axis y=x in the coordinate system. The only self-inverse function with (1) and (5) is f(x)=x. Well, allright, that would only be true if f were continuous, which isn't specified, so there is a tiny bit missing...


OK, at least we know f is bijective. (6) gives us

Now f(f(-1)*f(-1))=-1*f(f(-1))=1. Applyinf f on both sides and using (6) again gives

f(-1)*f(-1)=f(1)=1

This means f(-1)=-1 (f(-1)=1 is excluded because f(1)=1 and f bijective)

Finally f(-x)=f(-1*f(f(x)))=f(x)*f(-1)=-f(x).

Edited on December 28, 2006, 7:53 pm

Posted by JLo on 2006-12-28 19:26:48