Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In the one dimension case, where $\Omega\subseteq{\bf R}$ is a bounded domain, for example $\Omega=[0,2\pi]$, one can find the orthonormal basis for $L^2(\Omega)$:

$$\{e_n\}_{n\in {\bf Z}}$$

where $e_n(x)=\frac{1}{\sqrt{2\pi}}e^{inx}$.

In the high dimension, say, $\Omega\subseteq {\bf R}^n$ and $\Omega$ being bounded, can one still "construct" the orthonormal basis?

share|improve this question
    
I wonder if you really describe an orthonormal basis of $L^2(\Omega)$... You can't express discontinuous functions as a finite sum of $e_n$. –  Fezvez Apr 22 '11 at 16:49
1  
@Fezvez: He's thinking of a basis in the sense of Hilbert spaces, in which it makes sense to form infinite sums. –  Raskolnikov Apr 22 '11 at 17:05
2  
@Fezvez: You're right if you interpret the term orthonormal basis in the sense of linear algebra. However, what is meant is an orthonormal basis in the Hilbert space sense: every element can be uniquely writen as an (a priori) infinite sum $f = \sum a_n e_n$, where $a_n = \langle f, e_n \rangle$ and $\sum |a_n|^2 = \|f\|^2$ by Parseval's identity. See here and here for further information. –  t.b. Apr 22 '11 at 17:06
add comment

3 Answers 3

up vote 4 down vote accepted

The measure space $[0,2\pi]^n\subset \mathbb{R}^n$ with Lebesgue measure is a product of $n$ copies of $[0,2\pi]\subset\mathbb{R}$ with Lebesgue measure. If $(e_n)_n$ is an ONB for $L^2(\Omega)$ and $(f_n)_n$ is an ONB for $L^2(\Lambda)$, then $(e_m(\omega)f_n(\lambda))_{m,n}$ is an ONB for $L^2(\Omega\times\Lambda)$. So for example, $(e_{m_1,\ldots,m_n})_{m_j\in\mathbb Z}$ is an ONB for $L^2([0,2\pi]^n)$, where $e_{m_1,\ldots,m_n}(x_1,\ldots,x_n)=(2\pi)^{-n/2}e^{i(m_1x_1+\cdots+m_nx_n)}$.

share|improve this answer
add comment

As Jonas said, products will do if $\Omega$ is a parallelepiped. For general shapes, it's not so simple. However, you could use eigenfunctions of the Laplacian with, say, Dirichlet boundary conditions ($\phi = 0$ on $\partial \Omega$). How explicitly you can write these down will vary.

share|improve this answer
add comment

Every Hilbert space has an orthonormal basis. In particular, $L^2(\Omega)$ has an othonormal basis. A different problem is to find an explicit orthonormal basis. Some possibilties have already been mentioned by Jonas and Robert. Here is another possibility for the case of bounded $\Omega\subset\mathbb{R}^n$. The polynomials in the variables $\{x_1, \dots,x_n\}$ are dense in the space of continuous functions in $\bar\Omega$, which in turn are dense in $L^2(\Omega)$. Starting from the family of monomials $$ \{1,x_1,\dots,x_n,x_1^2,x_1x_2,\dots,x_n^2,x_1^3,\dots\} $$ construct an orthonormal basis by applying for instance the Gram-Schmidt orthogonalisation process.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.