PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (210)
Orphanage
Unclass'd (1)
Unproven (473)
Corrections (34)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: Very low Entry average rating: No information on entry rating
[parent] $L^2$-spaces are Hilbert spaces (Theorem)

Let $ (X, \mathfrak{B}, \mu)$ be a measure space. Let $ L^2(X)$ denote the $ L^2$-space associated with this measure space, i.e. $ L^2(X)$ consists of measurable functions $ f:X \longrightarrow \mathbb{C}$ such that

$\displaystyle \Vert f\Vert _2 := \left (\int_X \vert f\vert^2 d\mu \right)^{\frac{1}{2}} < \infty $
identified up to equivalence almost everywhere.

It is known that all $ L^p$-spaces, with $ 1\leq p \leq \infty$, are Banach spaces with respect to the $ L^p$-norm $ \;\Vert\cdot\Vert _p$. For $ L^2(X)$ we can say even more:

Theorem - $ L^2(X)$ is an Hilbert Space with respect to the inner product $ \langle \cdot, \cdot \rangle$ defined by

$\displaystyle \langle f, g \rangle = \int_X f\overline{g} \;d\mu $

Proof :

Sequilinearity follows from the linearity of the Lebesgue integral. The conjugate symmetry is evident.

Positive definiteness holds by construction: If $ \int_X \vert f\vert^2 d\mu = 0$, then $ \vert f\vert^2$ (and therefore $ f$) is zero almost everywhere, thus the equivalence class of $ f$ is the equivalence class of the zero function (which is the additive neutral element of the space).

Completeness is proved for the general case of $ L^p$-spaces in this article.$ \square$

Remarks

  • The spaces $ \mathbb{C}^n$ or $ \mathbb{R}^n$ with the usual inner product are particular examples of $ L^2(X)$, choosing $ X = \{1, \dots, n\}$ with the counting measure.
  • Choosing appropriate spaces $ X$ it can be shown that all Hilbert spaces are isometrically isomorphic to a $ L^2$-space.



Anyone with an account can edit this entry. Please help improve it!

"$L^2$-spaces are Hilbert spaces" is owned by asteroid. [ full author list (2) ]
(view preamble)

View style:

See Also: $L^p$-space

Other names:  square integrable functions form an Hilbert space

This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: isometrically isomorphic, counting measure, neutral element, additive, function, equivalence class, positive, symmetry, conjugate, inner product, Hilbert space, Banach spaces, almost everywhere, equivalence, measurable functions, measure space
There are 3 references to this entry.

This is version 4 of $L^2$-spaces are Hilbert spaces, born on 2007-09-14, modified 2007-10-15.
Object id is 9939, canonical name is L2SpacesAreHilbertSpaces.
Accessed 783 times total.

Classification:
AMS MSC: 46C05 (Functional analysis :: Inner product spaces and their generalizations, Hilbert spaces :: Hilbert and pre-Hilbert spaces: geometry and topology )
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

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