x>0时,00,n>0时,m+n > n,f(m+n) = f(m)*f(n) < f(n)
=> x>0时,f(x)单调递减.
f(0) = f(0)*f(0) => f(0) = 0 或 f(0)=1
当f(0) = 0 ,m>0 时,f(m+0) = f(m)*f(0) = 0 与题意矛盾
f(0) = 1
当m>0：
f(0) = f(m)*f(-m) = 1 => f(-m) = 1/f(m) => 当x1