Let $X=L^1\left([0,1]^3\right)$,
for numerical purpose, what are the possible basis function for $X$? 


In finite element method, the basis functions are tooth functions, or polynomial functions.  

Is there a generalization of this idea to product space?


I ask because I would like to numerically find some ball $B(f,1)$, with center $f$ and radius 1, such that
$$
G(f')>0 \text{ is true for all } f' \in B(f,1), \tag{1}
$$ 
where $G(\cdot)$ is some functional which can only be valued numerically. 


My numerical search algorithm is a brute force approach. That is, calculate $G(\tilde f)$ for a great number of the basis function $\tilde f$, until a ball $B(\tilde f, 1)$ is identified.


My first idea will be to use piecewise linear function as basis functions for $X$
$$
\tilde f\left(x\_1, x\_2, x\_3\right) = \sum\_{i=0}^n \chi\_{\left(a\_i,b\_i\right)}\left(x\_1\right) \cdot \chi\_{\left(c\_i,d\_i\right)}\left(x\_2\right) \cdot \chi\_{\left(e\_i,f\_i\right)}\left(x\_3\right),
$$  

where $x\_1,x\_2, x\_3 \in [0,1]$.  

But then, how to choose the pieces $\left\{a\_i,b\_i, c\_i, \ldots,f\_i\right\}$?  

How many pieces to use, i.e., how to choose $n$?  

Also, I'm not sure if piecewise functions can reliably identifies the ball in (1)?


Apart from that, my idea does not seems to be a generalization of finite element basis functions.


What will be good basis functions for $X$ so as to numerically determine (1)?  

If, instead of $L^1$, I use $L^2$, is the situation better?



What are the basis functions for a product space?
**Edit:** Note that I understood the question for $L^1([0,1]^3)$, c.f. Robert Israels comment.


First, the term "basis" for general Banach spaces, especially ugly spaces such as $L^1$, can be complicated. As far as I remember $L^1$ is one of the spaces which do not possess an "unconditional" basis. On $L^2$ the situation is much nicer as you probably know. For the product $L^2([0,1]^3)$ you could use your favorite orthonormal basis $(\psi\_n)$ of $L^2([0,1])$ (e.g. Fourier, Haar,...) and take the tensor-product basis; basically as you proposed: $\Psi\_{n,m,l}(x\_1,x\_2,x\_2) = \psi\_n(x\_1)\psi\_m(x\_2)\psi\_l(x\_3)$. Without more details about your problem I can't suggest anything more specific.


