证明两个函数傅里叶级数相等的充要条件
1个回答
首先命题等价于:在[-π,π]可积的2π周期函数f(x),Fourier系数全为0的充要条件是∫{-π,π} |f(x)|dx = 0.
充分性很容易:0 ≤ |∫{-π,π} f(x)dx| ≤ ∫{-π,π} |f(x)|dx = 0.
0 ≤ |∫{-π,π} f(x)sin(nx)dx| ≤ ∫{-π,π} |f(x)|·|sin(nx)|dx ≤ ∫{-π,π} |f(x)|dx = 0.
0 ≤ |∫{-π,π} f(x)cos(nx)dx| ≤ ∫{-π,π} |f(x)|·|cos(nx)|dx ≤ ∫{-π,π} |f(x)|dx = 0.
故所有Fourier系数全为0.
必要性用Parseval恒等式:由f(x)在[-π,π]可积,|f(x)|²也在[-π,π]可积.
并成立Parseval恒等式:∫{-π,π} |f(x)|²dx = |a0|²+∑{1 ≤ n} (|an|²+|bn|²).
由f(x)的Fourier系数全为0,可知∫{-π,π} |f(x)|²dx = 0.
再由Cauchy不等式:(∫{-π,π} |f(x)|dx)² ≤ (∫{-π,π} |f(x)|²dx)·(∫{-π,π} 1²dx) = 0.
即得∫{-π,π} |f(x)|dx = 0.
