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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?

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    $\begingroup$ What do you mean by $X^3$? Your $\tilde{f}$ appears to be a function on $[0,1]^3$. That would suggest you're interested in $L^1([0,1]^3)$, which is not the same thing as the usual $X^3$ (the Cartesian product of three copies of $X$). $\endgroup$
    – Robert Israel
    Commented Oct 20, 2013 at 9:02
  • $\begingroup$ @RobertIsrael I meant $L^1([0,1]^3)$. Thank you for pointing it. $\endgroup$
    – user24451
    Commented Oct 20, 2013 at 12:25
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    $\begingroup$ Is $G$ a linear functional? Do you have any bounds on it? $\endgroup$
    – Robert Israel
    Commented Oct 20, 2013 at 17:25
  • $\begingroup$ @RobertIsrael Yes, it is linear. It even satisfies a Lipschitz condition. That's why I suspect a numerical scheme can work. $\endgroup$
    – user24451
    Commented Oct 20, 2013 at 17:28

3 Answers 3

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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.

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Given an upper bound $B$ on the norm of the linear functional $G$, all you need is to find $f$ with $G(f) > B$. But if you can find $f_0$ with $G(f_0) \ne 0$, $f = (B + 1) f_0/G(f_0)$ will do.

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If you choose any basis $f_n(x)$ for $L^1$ (the specific form of the basis and the measure space is of no consequence), then the the family $f_m(x_1)f_n(x_2)f_p(x_3)$ is a basis for $L^1(X^3)$ under a suitable ordering as a sequence. This follows from the fact that the latter is the triple (projective) tensor product of the original $L^1$-space and the theory of bases in tensor products of Banach spaces (for which see Gelbaum and de Lamadrid, Pac. Jour Math. 11 (1961) 1281-1286).

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