Restrictions of function decomposition in $R^3$ I'm interested in the properties of countable basis functions that span functions living in $\Bbb R^3$.  
Can I represent a $L^2$ normalizable function that has a point divergence, (for example, $\frac{1}{|\mathbf r|}e^{-m|\mathbf r|}$) with a countable basis whose members are everywhere finite?  Thanks.
 A: Short answer: Yes, such a "basis" exists.
The real function space $L^2(\mathbb{R}^d)$ is a Hilbert space with respect to the natural inner product $\langle f,g \rangle = \int_{\mathbb{R}^d} f g\; \mathrm d \omega$, where $\omega$ denotes the usual Lebesgue measure on $\mathbb{R}^d$.
Considered as a real vector space one can prove the existence of a basis, but this is perhaps not what the Question here is about.  Such a basis (a linearly independent spanning set) would be uncountable and non-constructively defined.
Instead what is likely wanted is a Schauder basis, a countable sequence $\mathscr{B} = \{b_1,b_2,\ldots\}$ of vectors of a complete normed linear space (a Banach space) $V$, with which any vector $v\in V$ can be uniquely expressed as a convergent series:
$$ v = \sum_{k=1}^\infty c_k b_k $$
where convergence is required in the norm of the Banach space:
$$ \lim_{n\to \infty} \left|\left| v - \sum_{k=1}^n c_k b_k \right|\right|_V = 0 $$
If each element $b_k \in \mathscr{B}$ is of unit norm $||b_k||_V = 1$, we say the Schauder basis is normalized.
A Hilbert space $H$ is a special kind of Banach space, since the norm is given by an inner product $\langle \cdot,\cdot \rangle_H$.  Thus one often asks for a Schauder basis of a Hilbert space that is orthogonal, i.e.
$$ i\neq j \implies \langle b_i,b_j \rangle_H = 0 $$
If such a Hilbert space basis is normalized as well as orthogonal, we say it is an orthonormal basis.
In 1910 Alfréd Haar constructed an orthonormal basis for $L^2([0,1])$ consisting of certain step functions.  In fact all his basis elements were dilations and translations of a single "mother" step function $\psi(x)$:
$$ \psi(x) = \begin{cases}
 1 & 0\le x \lt 1/2 \\ -1 & 1/2 \le x \le 1 \\ 0 & \; \text{  otherwise} 
\end{cases} $$
The Haar basis of $L^2([0,1])$ is then the nonzero functions on $[0,1]$ of the form:
$$ 2^{m/2} \psi(2^m x + n) \; \text{ where } m \in \mathbb{Z}_{\ge 0}, n \in \mathbb{Z} $$
By extending these to functions of the form $2^{m/2} \psi(2^m x + n)$ where $m,n \in \mathbb{Z}$, we get a Haar basis for $L^2(\mathbb{R})$.  By taking products of such univariate functions on the coordinate variables, one gets an orthonormal basis for $L^2(\mathbb{R}^d)$.
Each basis element is bounded (being piecewise constant with compact support). However the requirement that these be normalized means they are not bounded by a common limit; the narrower the "notch", the larger the constants.
The recent literature in this area often uses the term Haar wavelet to signify the simplest possible family of orthonormal basis functions with compact support.  Particular applications may benefit from adapting them and mitigating such drawbacks as their discontinuities.  See for example Krishtal et al (2007), "Some Simple Haar-Type Wavelets in Higher Dimensions".
