Bessel's Inequality, and Parseval's Identity

Bessel's Inequality

Theorem 1. Let ($V, \langle \cdot, \cdot \rangle$) be an inner product space, and let $S = \{v_1, ..., v_n\}$ be an orthonormal subset of $V$. Then $\forall x \in V$, $$||x||^2 \geq \sum_{i=1}^n |\langle x, v_i \rangle|^2.$$

Proof. Let $\langle S \rangle = W$. Then $! \exists y \in W, z \in W^{\perp}$ such that $x = y + z$ by Theorem 1. Thus we have $$||x||^2 = ||y||^2 + ||z||^2 = \sum_{i=1}^n |\langle y, v_i \rangle|^2 + ||z||^2 \geq \sum_{i=1}^n |\langle x, v_i \rangle|^2. \blacksquare$$

Parseval's Identity

Theorem 2. In the notaion of Theorem 1, if $V$ is finite-dimensional and $S$ is an orthonormal basis for $V$, then $\forall x, y \in V$, $$\langle x, y \rangle = \sum_{i=1}^n \langle x, v_i \rangle \overline{\langle y, v_i\rangle}.$$

Proof. Denote $x = \sum_{i=1}^n \langle x, v_i \rangle v_i, y = \sum_{j=1}^n \langle y, v_j \rangle v_j$. Then we have $$\langle x, y\rangle = \langle \sum_{i=1}^n \langle x, v_i \rangle v_i, \sum_{j=1}^n \langle y, v_j \rangle v_j \rangle \\ = \sum_{i=1}^n \langle x, v_i \rangle \sum_{j=1}^n \overline{\langle y, v_j \rangle} \langle v_i, v_j \rangle \\ = \sum_{i=1}^n \langle x, v_i \rangle \overline{\langle y, v_i \rangle}. \blacksquare$$

Corollary

Corollary. Let $\langle \cdot, \cdot \rangle '$ denote the standard inner product of $F^n$. Then $\forall x, y \in V$, $\langle x, y \rangle = \langle [x]_{\beta}, [y]_{\beta} \rangle '$.