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It is necessary for me to find the shape functions on $L^2(\Omega)$(piecewise constant functions), I searched a lot but that, could not find anything, all of them are about higher degree polynomial. If we define the space as follow: $${\cal Dp0}=\{v \in L^2(\Omega):v|_T \in {\cal p0(T)}, \forall T \in {\cal T}\}$$ where ${\cal T}$ is triangulation of the domain. How can I find them?

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  • $\begingroup$ This is essentially what Finite Volume Method consists of. $\endgroup$ – Paul Mar 31 '16 at 1:47
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$p_0(T)$ is the space of all functions that are constant on cell $T$. The shape function that corresponds to this is the function that is 1 on cell $T$ and zero everywhere else. Its derivative is then obviously zero in the cell itself, as well as outside the cell, but it is undefined at the interface.

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    $\begingroup$ @Rosa, Wolfgang's point about cells is that you need to discretize $\Omega$ into a finite number of subregions, usually called elements or cells. Triangles, in two dimensions, are popular. $\endgroup$ – Bill Barth Mar 30 '16 at 19:58
  • $\begingroup$ Ah, yes, the question originally included this information. Yes, $T$ are the cells of a mesh. $\endgroup$ – Wolfgang Bangerth Mar 30 '16 at 22:26
  • $\begingroup$ sorry for removing it, I added some information on it again. $\endgroup$ – Rosa Mar 31 '16 at 7:57

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