PlanetMath (more info)
 Math for the people, by the people. Sponsor PlanetMath
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Owner confidence rating: Medium Entry average rating: No information on entry rating
Bessel inequality (Theorem)

Let $\Hilb$ be a Hilbert space, and suppose $e_1, e_2, \ldots \in \Hilb$ is an orthonormal sequence. Then for any $x\in\Hilb$ $$ \sum_{k=1}^{\infty}\size{\scalar{x}{e_k}}^2 \le \norm{x}^2. $$

Bessel's inequality immediately lets us define the sum $$ x' = \sum_{k=1}^{\infty}\scalar{x}{e_k}e_k. $$ The inequality means that the series converges.

For a complete orthonormal series, we have Parseval's theorem, which replaces inequality with equality (and consequently $x'$ with $x$ .




"Bessel inequality" is owned by ariels.
(view preamble | get metadata)

View style:


Attachments:
proof of Bessel inequality (Proof) by ariels
Log in to rate this entry.
(view current ratings)

Cross-references: equality, Parseval's theorem, complete, converges, series, inequality, sum, sequence, orthonormal, Hilbert space
There are 3 references to this entry.

This is version 2 of Bessel inequality, born on 2002-06-10, modified 2002-06-11.
Object id is 3089, canonical name is BesselInequality.
Accessed 10205 times total.

Classification:
AMS MSC46C05 (Functional analysis :: Inner product spaces and their generalizations, Hilbert spaces :: Hilbert and pre-Hilbert spaces: geometry and topology )

Pending Errata and Addenda
None.
[ View all 1 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | prove | add result | add corollary | add example | add (any)