Let f:R→R satisfy

f(a)≠0 for some a in R

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 selfinverse function, i.e. its graph is symmetrical w.r.t. the axis y=x in the coordinate system. The only selfinverse 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 20061228 19:26:48 