> [!theorem] > > Let $\ch$ be an [[Inner Product|inner product space]] and $\seqi{x}$ be an [[Orthonormal Set|orthonormal]] set, then > $ > \sum_{i \in I}\abs{\angles{x, x_i}}^2 \le \norm{x}^2 > $ > where $\bracs{i \in : \angles{x, x_i} \ne 0}$ is countable. > > *Proof*. Let $F \subset I$ be a finite subset, then > $ > \begin{align*} > 0 &\le \norm{x - \sum_{i \in F}\angles{x, x_i}x_i} \\ > &= \norm{x}^2 - 2\re{\angles{x, \sum_{i \in F}\angles{x, x_i}x_i}} + \norm{\sum_{i \in F}\angles{x, x_i}x_i}^2 \\ > &= \norm{x}^2 - 2\sum_{i \in I}\ol{\angles{x, x_i}} \cdot \angles{x, x_i} + \sum_{i \in F}\abs{\angles{x, x_i}}^2 \\ > &= \norm{x}^2 - 2 \sum_{i \in I}\abs{\angles{x, x_i}}^2 + \sum_{i \in F}\abs{\angles{x, x_i}}^2 \\ > &= \norm{x}^2 - \sum_{i \in I}\abs{\angles{x, x_i}}^2 > \end{align*} > $ > by the [[Pythagorean Theorem|Pythagorean theorem]].