How to show a piecewise quadratic interpolant is $H^1$

Question

I am preparing for a final exam and came across this question:

Suppose that $\Omega\subset\mathbb{R}^2$ is an open bounded domain with triangulation $\mathscr{T}$. Suppose that $v_h$ is a piecewise quadratic interpolant of some continuous function with interpolation points on each element as indicated in Figure 1. [Figure one is a triangle with a node at each vertex and each midpoint.] Show that $v_h\in H^1(\Omega)$ and that $\|v_h\|^2_{H^1(\Omega)}=\sum_{K\in\mathscr{T}}\|v_h\|^2_{H^1(K)}$.

[HINT: For $w\in C^\infty(\overline{\Omega})$, consider $\int_\Omega \nabla v\cdot\nabla w$ and use the divergence theorem to move derivatives from $v$ to $w$.]

My questions are:

1. Why isn't this obvious since $v_h$ is a piecewise polynomial on a bounded domain? The integral of $v_h$ is obviously bounded. Is the only problem possibly where two elements in $\mathscr{T}$ meet?

2. How is this hint supposed to help?

Thank you in advance!

Answer 1

1. Yes, you can have problems between two elements. However, it is easy to show that $v_h$ is continuous.
2. Using the continuity of $v_h$ and the fact that $v_h$ is piecewise smooth, you can calculate its weak derivative. Contrary to the hint, I would start with $\int_\Omega v_h \, \partial_i w \, \mathrm{d}x$, to obtain the $i$th weak derivative of $v_h$.