Parseval's identity

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Theorem (Parseval's Identity)

Let $H$ be a Hilbert space and let ${e_{n}}$ be a countable orthonormal basis of $H$ . Then for all $u \in H$ ,

\sum_{n} \lvert \langle u, e_{n} \rangle \rvert {#2} = \lvert \lvert u \rvert \rvert {#2}

Proof

We know that

u = \sum_{n} ⟨ u, e_{n} ⟩ e_{n}

from all elements of hilbert spaces with orthonormal bases can be written as sums of the basis elements. If the sum over $n$ is a finite sum, then we just expand the inner product $| | u | |^{2} = ⟨ u, u ⟩$ . Otherwise, by continuity of inner product, we have

\begin{align} \lvert \lvert u \rvert \rvert {#2} &= \lim_{ m \to \infty } \left\langle \sum_{n=1}^m \langle u, e_{n} \rangle e_{n}, \sum_{\ell=1}^m \langle u, e_{\ell} \rangle e_{\ell} \right\rangle \\ &= \lim_{ m \to \infty } \sum_{n, \ell = 1}^m\langle u, e_{n} \rangle \overline{\langle u, e_{\ell} \rangle } \langle e_{n, e_{\ell}} \rangle \\ (e_{n} \perp e_{\ell}, n \neq \ell) \implies&= \lim_{ m \to \infty } \sum_{n=1} \langle u, e_{n} \rangle \overline{\langle u, e_{n} \rangle } \\ &= \lim_{ m \to \infty } \sum_{n=1}^m \lvert \langle u, e_{n} \rangle \rvert {#2}

File	Last Modified
Functional Analysis Lecture 15	2025-07-15
separable Hilbert spaces are bijective with ell-2	2025-07-15

File

Last Modified

Functional Analysis Lecture 15

2025-07-15

separable Hilbert spaces are bijective with ell-2

2025-07-15

Parseval's identity

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