Question

f(x)=f(2x)且f(x)在x=0处连续,证明f(x)是常值函数

Answer 1

f(x)=f(2x), 所以f(x)=f(2x)=f(4x)=...=f((2^n)x),

如果令y=(2^n)x,则有x=y/(2^n),

则有f(y)=f(y/(2^n))

因为f(x)在x=0处连续,所以limf(x)=f(0)(x→0)

对于任意的y0有f(y0)=f(y0/(2^n)),且n是任意的正整数

所以f(y0)=f(y0/(2^n))=limf(y0/(2^n))(n→+∞)=f(0)

即f(x)=f(0), 结论得证.