f(x)定义域为R,存在x1≠x2,使得f(x1)≠f(x2),且f(x+y)=f(x)*f(y),证明x∈R时,f(x)>0
问题描述：
f(x)定义域为R,存在x1≠x2,使得f(x1)≠f(x2),且f(x+y)=f(x)*f(y),证明x∈R时,f(x)>0
答
f(x+y)=f(x)*f(y)
=> f(0)=f(0)*f(0)
=> f(0)=0,1
存在x1≠x2,使得f(x1)≠f(x2),且f(x)=f(x)*f(0)
=> f(0)=1 (若f(0)=0,则f(x)恒为0,与条件不符)
f(x)=f(x/2)*f(x/2)>=0,而由f(0)=f(x)*f(-x)=1,可知,f(x)不等于0
=> f(x)>0