设f(x)对任意的实数x1,x2有f(x1+x2)=f(x1)f(x2),且f'(0)=1,证明f'(x)=f(x)

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  • 证明:

    (i)设f(x)在定义域内恒不为零,

    由原式得:|f(x+y)|=|f(x)|*|f(y)|

    从而:ln|f(x+y)|=ln|f(x)|+ln|f(y)|

    等式两边同时对y求导得:(x+y)'f'(x+y)/f(x+y)=f'(y)/f(y)+0

    移项整理:f'(x+y)=f(x+y)f'(y)/f(y)=f'(y)f(x)

    取y=0得:f'(x+0)=f'(x)=f'(0)f(x)=f(x)

    (ii)当f(x)在某点处(不妨设在点a处)为零时,即f(a)=0时,

    可知对任意的x,成立:f(x+a)=f(x)f(a)=f(x)*0=0

    由x的任意性,知x+a能取遍整个实数空间,则f在实数域内恒为0,亦满足f'(x)=f(x)(=0)

    综合(i)和(ii)可知,原命题成立

    证毕.