Question

设f(x)是定义在R上的函数，对任意x,y∈R，恒有f(x+y)=f(x)·f(y)，当x>0时，有0<f(x)<1.

⑴求证：f(0)=1,且当x<0时，f(x)>1
⑵证明：f(x)在R上单调递减

Answer 1

证明：令x＞0，y=0，得 f(x+0)=f(x)·f(0) 0<f(x)<1 则 f(0)=1
令x+y=0，x＞0，得 f(0)=f(x)·f(-x) 0<f(x)<1 则 f(-x)＞1
故 f(0)=1,且当x<0时，f(x)>1
证明：设x1＜x2，则 x2-x1＞0 由(1)，得 f(x)＞0 f(x2-x1)＜1
f(x2)=f(x1+x2-x1)=f(x1)·f(x2-x1)＜f(x1) 即 f(x)在R上单调递减