Theorem 1. If \(\mathcal{X}=\{x_{1}, \ldots, x_{n}\}\) is any finite orthonormal set in an inner product space, if \(x\) is any vector, and if \(\alpha_{i}=(x, x_{i})\) , then (Bessel’s inequality) \[\sum_{i}|\alpha_{i}|^{2} \leq\|x\|^{2}.\] The vector \(x^{\prime}=x-\sum_{i} \alpha_{i} x_{i}\) is orthogonal to each \(x_{j}\) and, consequently, to the subspace spanned by \(\mathcal{X}\) .
Proof. For the first assertion: \begin{align} 0 &\leq \|x^{\prime}\|^{2}\\ &= (x^{\prime}, x^{\prime})\\ &= \Big(x-\sum_{i} \alpha_{i} x_{i}, x-\sum_{j} \alpha_{j} x_{j}\Big) \\ &= (x, x)-\sum_{i} \alpha_{i}(x_{i}, x)-\sum_{j} \bar{\alpha}_{j}(x, x_{j})+\sum_{i} \sum_{j} \alpha_{i} \bar{\alpha}_{j}(x_{i}, x_{j}) \\ &= \|x\|^{2}-\sum_{i}|\alpha_{i}|^{2}-\sum_{i}|\alpha_{i}|^{2}+\sum_{i}|\alpha_{i}|^{2} \\ &= \|x\|^{2}-\sum_{i}|\alpha_{i}|^{2}; \end{align} for the second assertion: \[(x^{\prime}, x_{j})=(x, x_{j})-\sum_{i} \alpha_{i}(x_{i}, x_{j})=\alpha_{j}-\alpha_{j}=0.\] ◻
Theorem 2. If \(\mathcal{X}\) is any finite orthonormal set in an inner product space \(\mathcal{V}\) , the following six conditions on \(\mathcal{X}\) are equivalent to each other.
- The orthonormal set \(\mathcal{X}\) is complete.
- If \((x, x_{i})=0\) for \(i=1, \ldots, n\) , then \(x=0\) .
- The subspace spanned by \(\mathcal{X}\) is the whole space \(\mathcal{V}\) .
- If \(x\) is in \(\mathcal{V}\) , then \(x=\sum_{i}(x, x_{i}) x_{i}\) .
- If \(x\) and \(y\) are in \(\mathcal{V}\) , then (Parseval’s identity) \[(x, y)=\sum_{i}(x, x_{i})(x_{i}, y).\]
- If \(x\) is in \(\mathcal{V}\) , then \[\|x\|^{2}=\sum_{i}|(x, x_{i})|^{2}.\]
Proof. We shall establish the implications
(1) \(\implies\) (2) \(\implies\) (3) \(\implies\) (4) \(\implies\) (5) \(\implies\) (6) \(\implies\) (1).
Thus we first assume (1) and prove (2), then assume (2) to prove (3), and so on till we finally prove (1) assuming (6).
(1) \(\implies\) (2). If \((x, x_{i})=0\) for all \(i\) and \(x \neq 0\) , then we may adjoin \(x /\|x\|\) to \(\mathcal{X}\) and thus obtain an orthonormal set larger than \(\mathcal{X}\) .
(2) \(\implies\) (3). If there is an \(x\) that is not a linear combination of the \(x_{i}\) , then, by the second part of Theorem 1, \(x^{\prime}=x-\sum_{i}(x, x_{i}) x_{i}\) is different from \(0\) and is orthogonal to each \(x_{i}\) .
(3) \(\implies\) (4). If every \(x\) has the form \(x=\sum_{j} \alpha_{j} x_{j}\) , then \[(x, x_{i})=\sum_{j} \alpha_{j}(x_{j}, x_{i})=\alpha_{i}.\]
(4) \(\implies\) (5). If \(x=\sum_{i} \alpha_{i} x_{i}\) and \(y=\sum_{j} \beta_{j} x_{j}\) , with \(\alpha_{i}=(x, x_{i})\) and \(\beta_{j}=(y, x_{j})\) , then \begin{align} (x, y) &= \Big(\sum_{i} \alpha_{i} x_{i}, \sum_{j} \beta_{j} x_{j}\Big)\\ &= \sum_{i} \alpha_{i} \bar{\beta}_{j}(x_{i}, x_{j})\\ &= \sum_{i} \alpha_{i} \bar{\beta}_{i}. \end{align}
(5) \(\implies\) (6). Set \(x=y\) .
(6) \(\implies\) (1). If \(\mathcal{X}\) were contained in a larger orthogonal set, say if \(x_{0}\) is orthogonal to each \(x_{i}\) , then \[\|x_{0}\|^{2}=\sum_{i}|(x_{0}, x_{i})|^{2}=0,\] so that \(x_{0}=0\) . ◻