Parseval's Identity for Inner Product Spaces

# Parseval's Identity for Inner Product Spaces

Recall from the Hilbert Bases (Orthonormal Bases) for Hilbert Spaces page that if $H$ is a Hilbert space then a sequence $(x_n)_{n=1}^{\infty}$ of orthonormal points in $H$ is said to be a Hilbert basis (or orthogonal basis) of $H$ if every $y \in H$ can be written as:

(1)
\begin{align} \quad y = \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n \end{align}

We will now state and prove Parseval's identity for Hilbert spaces. The result is very similar to Bessel's inequality but is stronger.

 Theorem 1 (Parseval's Identity for Hilbert Spaces): Let $H$ be a Hilbert space and let $(x_n)_{n=1}^{\infty}$ be a Hilbert basis of $H$. Then for every $y \in H$, $\displaystyle{\sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 = \| y \|^2}$.
• Proof: Let $y \in H$. From the theorem on the Bessel's Inequality for Inner Product Spaces page we have that the series $\displaystyle{\sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 \leq \| y \|^2}$ and so the series converges.
• By the theorem on the Convergence Criterion for Series in Hilbert Spaces page we also have that the series $\displaystyle{\sum_{n=1}^{\infty} \langle y, x_n \rangle x_n}$ converges to some $z \in H$, and since $(x_n)_{n=1}^{\infty}$ is a Hilbert basis we have that:
(2)
\begin{align} \quad y = \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n \end{align}
(3)
\begin{align} \quad \| y \|^2 = \biggr \| \lim_{N \to \infty} \sum_{n=1}^{N} \langle y, x_n \rangle x_n \biggr \|^2 = \lim_{N \to \infty} \biggr \| \sum_{n=1}^{N} \langle y, x_n \rangle x_n \biggr \|^2 = \lim_{N \to \infty} \sum_{n=1}^{N} |\langle y, x_n \rangle|^2 = \sum_{n=1}^{\infty} |\langle y, x_n \rangle|^2 \quad \blacksquare \end{align}