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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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    What does the notation $x_1\cdot\left[x\in T\right]$ mean? What is $x_1$? What does the bracket mean? Are you taking a dot product?2012-03-04
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    It means $x_1$ when $x\in T$ and $0$ otherwise.2012-03-04
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    @J.D.: I suspect it's the [Iverson bracket](http://en.wikipedia.org/wiki/Iverson_bracket).2012-03-04
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    A basis function for what? Sure it could be a basis function, but one that might have bad properties in some circumstances -- it depends on how you want to use it.2012-03-04
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    Wouldn't http://scicomp.stackexchange.com/ be a more suitable place for this question?2012-03-04
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    Here is a nice pdf for FEM triangle basis functions. It might be helpful: http://people.sc.fsu.edu/~jburkardt/presentations/cg_lab_fem_basis_triangle.pdf2012-03-04
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    I flagged the question. I think moderators should move it scicomp.SE2012-03-05
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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).