Question

求助一道数学分析题目

Answer 1

令y=0,得2f(x)=2f(x)g(0),f(x)≠0，
∴g(0)=1
以x+y代x,得f(x+2y)+f(x)=2f(x+y)g(y),
同理，f(x)+f(x-2y)=2f(x-y)g(y),
相加得f(x+2y)+f(x-2y)+2f(x)=2[f(x+y)+f(x-y)]g(y)=4f(x)[g(y)]^2,
∴f(x+2y)+f(x-2y)=2f(x){2[g(y)]^2-1}.
以x+2y代x,得f(x+4y)+f(x)=2f(x+2y){2[g(y)^2-1},
同理，f(x)+f(x-4y)=2f(x-2y){2[g(y)]^2-1},
∴f(x+4y)+f(x-4y)+2f(x)=2[f(x+2y)+f(x-2y)]{2[g(y)]^2-1}
=4f(x){2[g(y)]^2-1}^2,
设h1(y)=2[g(y)]^2-1,h<n+1>(y)=h1[hn(y)],得
f(x+2y)+f(x-2y)=2f(x)h1(y),
f(x+4y)+f(-4y)=2f(x)h2(y),
依此类推，f(x+2^n*y)+f(x-2^n*y)=2f(x)hn(y).①
若存在y0≠0,使得p=|g(y0)|>1,则h1(y0)>[g(y0)]^2=p^2,
h2(y0)>(2p^2-1)^2>p^4,
……
hn(y0)>p^(2^n)→+∞，
由①，f(x+2^n*y0)或f(x-2^n*y0)无界，这与题设矛盾。
∴|g(y)|<=1.