令x=0,则f(0)=[f(0)]^, f(0)[1-f(0)]=0, ∵ f(0)≠0, ∴f(0)=1.
令y=-x.则f(x-x)=f(x)f(-x), ∴ f(x)f(-x)=f(0)=1>0,
(1) x>0时, -x<0, f(-x)>1, 即1/f(x)>1,∵ f(x)>0, ∴ 0<f(x)<1.
(2) 当x1<x2<0时, f(x1)=f[(x1-x2)+x2]=f(x1-x2)f(x2), ∵ x1-x2<0, ∴ f(x1-x2)>1, f(x2)>0, ∴ f(x1-x2)f(x2)>f(x2), 即f(x1)>f(x2), ∴ f(x)在(-∞,0)上是减函数, 同理可证f(x)在(0,+∞)上是减函数,又 f(0)=1, ∴ f(x)在(-∞,+∞)上是减函数.