Orthonormal basis in $L^2(\Omega)$ for bounded $\Omega$ In the one dimension case, where $\Omega\subseteq{\bf R}$ is a bounded domain, for example $\Omega=[0,2\pi]$, one can find a orthonormal basis  $\{e_n\}_{n\in {\bf Z}}$ for $L^2(\Omega)$ 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?
 A: 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.
A: 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.
A: 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 $e_{m_1,\ldots,m_n}(x_1,\ldots,x_n)=(2\pi)^{-n/2}e^{i(m_1x_1+\cdots+m_nx_n)}$.
