Bessel不等式 设\(H\)是Hilbert空间，\((e_n)_{n\geq 1}\)是\(H\)中的规范正交序列，则对任意\(x\in H\)，有， \[ \sum_{n=1}^{\infty}|\langle x, e_n \rangle|^2\leq ||x||^2 \] Proof 对任意的\(n\)，有 \[ \begin{aligned} ||x-\sum_{k=1}^{n}\langle x, e_k \rangle e_k||^2&=\langle x-\sum_{k=1}^{n}\langle x, e_k \rangle e_k,x-\sum_{k=1}^{n}\langle x, e_k \rangle e_k \rangle\\ &=\langle x, x \rangle+\langle \sum_{k=1}^{n}\langle x, e_k \rangle e_k, \sum_{k=1}^{n}\langle x, e_k \rangle e_k \rangle-2Re\{\langle x, \sum_{k=1}^{n}\langle x, e_k \rangle e_k \rangle\}\\ &=||x||^2+\sum_{k=1}^{n}\langle x, e_k \rangle\langle e_k, \sum_{k=1}^{n}\langle x, e_k \rangle e_k \rangle-2Re\{\sum_{k=1}^{n}\overline{\langle x, e_k \rangle}\langle x, e_k \rangle\}\\ &=||x||^2+\sum_{k=1}^{n}\langle x, e_k \rangle\langle e_k, \langle x, e_k \rangle e_k \rangle-2Re\{\sum_{k=1}^{n}\langle x, e_k \rangle^2\}\\ &=||x||^2+\sum_{k=1}^{n}\langle x, e_k \rangle\overline{\langle x, e_k \rangle}\langle e_k, e_k \rangle-2\sum_{k=1}^{n}\langle x, e_k \rangle^2\\ &=||x||^2+\sum_{k=1}^{n}\langle x, e_k \rangle^2\cdot1-2\sum_{k=1}^{n}\langle x, e_k \rangle^2\\ &=||x||^2-\sum_{k=1}^{n}\langle x, e_k \rangle^2\geq0 \end{aligned} \] 于是不等式得证。

Parserval恒等式 设\(H\)是Hilbert空间，\((e_n)_{n\geq 1}\)是\(H\)中的规范正交基，则对任意\(x\in H\)，有， \[ \sum_{n=1}^{\infty}|\langle x, e_n \rangle|^2=||x||^2 \] 更一般地，\(x，y\in H\)，有， \[ \langle x, y \rangle=\sum_{n=1}^{\infty}\langle x, e_n \rangle\langle e_n, y \rangle \]

证明明天补上。