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In FEM, with a triangular mesh over $R^2$, could $\phi\left(x\right)=x_1\cdot\left[x\in T\right]$ be a basis function for the triangle $T$ with vertices in $\left(0,0\right), \left(0,1\right), \left(1,0\right)$? My doubts come from the fact it is not contiuous, $\phi\left(\left(1,0\right)\right)=1$ but $\phi\left(\left(1+\epsilon,0\right)\right)=0\ for\ \epsilon>0$.

Edit: basis of the trial space.

Edit: I forgot to add I want piecewise linear trial space, so the the question pretty much is about a convenient basis for it.

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    @J.D. I am not entirely convinced that scicomp is better for this question. It seems to be asking about the theory behind the FEM, which would be on topic here.2012-03-05

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Well l I guess not, $\phi$ may be like that on triangle T, but (with the mesh consisting of triangles with horizontal, vertical and slope=-1 edges) in the other five mesh triangles which have $\left(0,0\right)$ as a vertex, $\phi$ (I'm talking piecewise linear here) should be 1 at $\left(0,0\right)$ and disappear at the edge that is opposite to (0,0).