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?

linearfunctional? Do you have any bounds on it? $\endgroup$