设F(x)=[e^(-x)]*f(x)
则F'(x)=[e^(-x)]'*f(x)+ [e^(-x)]*f'(x)
=-[e^(-x)]*f(x) + [e^(-x)]*f'(x)
=e^(-x)*[f'(x)-f(x)]
由已知：f'(x)>f(x) <=>f'(x)-f(x)>0
且e^(-x)>0
∴F'(x)>0
∴F(x)为定义在R上的单调增函数
∵a>0
∴F(a)>F(0)
而F(a)=[e^(-a)]*f(a)
F(0)=[e^(0)]*f(0)=f(0)
<=>[e^(-a)]*f(a)>f(0)
<=>f(a)>[e^(a)]*f(0)