# polynomial interpolation problem

Consider a polygon $S$ in the plane. Then, we perform a triangulation of this polygon. Consider the space $Q$ the space of functions such that the restriction of any function in the space $Q$ to any triangle in the polygon $S$ is a linear polynomial and continuous at the center of the triangle edges. Can you provide a counter-example of a function in $Q$ which is not globally continuous on $S$?

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What makes you think such a counter-example exists? – nbubis Dec 8 '12 at 14:30
Any boundary conditions? (At the outer edges.) – WimC Dec 8 '12 at 14:32

It's not hard to construct a counterexample if you consider a specific pair of triangles (one counterexample will do). Let, say, $T_1$ with the vertices $(0,0),(1,0),(1,1)$ and $T_2$ with the vertices $(1,1),(0,1),(0,0)$. The midpoint of the common edge $e$ is then $(1/2,1/2)$. If you prescribe the value $0$ for the interpolant $v$ there, you can pick the remaining values such that $v|_{T_1}$ and $v|_{T_2}$ have opposite sign on $e$. One example is $v|_{T_1} = 1-x-y$ and $v|_{T_2} = x+y-1$, for which $$v|_{T_1}(1/2,1/2) = 0 = v|_{T_2}(1/2,1/2),$$ but $$v|_{T_1}(0,0) = 1 \quad\text{and}\quad v|_{T_2}(0,0) = -1.$$ Hence, $v$ is not continuous on $e$.