# $W^{1,1}$ simplicial approximation.

Let $f$ be a continuous real-valued function defined on an $n$ dimesional simplex $\Sigma\subset \mathbb{R}^n$. The classical simplicial approximation scheme provides a sequence $f_k$ of piecewise affine functions converging uniformly to $f$: namely, on each simplex $S$ of the $k$-th baricentric subdivision of $\Sigma$, one takes ${f_k} _ {|S}$ to be the affine interpolation of the data of $f$ on the vertices of $S$. Precisely, if $\omega$ is a modulus of continuity for $f$, it follows easily that $$\big\| f - f_k \big\|_{\infty,\Sigma}\\ \le\\ \omega\bigg( \Big(\frac{n}{n+1}\Big )^k \operatorname{diam}\Sigma \bigg)\\ .$$

Notice that $\nabla f_k(x)$ is defined and locally constant at any point $x\in\Sigma$ not in the $(n-1)$-skeleton of the $k$-th barycentric subdivisions; in particular, a.e. in $\Sigma$. However, understanding the behaviour of this sequence is not as simple as before, even for smooth $f$. For instance, the norm of $\nabla f_k(x)$ is in general not bounded by the uniform norm of $\nabla f$.

Questions. Assuming that $f$ has bounded and continuous first order derivatives, can we conclude that $\nabla f_k$ converges to $\nabla f$ in $L^1(\Sigma)$? How to dominate pointwise $|\nabla f_k (x) |$ (independently of $k$), and how to bound $\|\nabla f_k -\nabla f\|_{1,\Sigma}$ ?

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It seems to me that this works using essentially the same proof you would use for proving that a piecewise linear approximation scheme for a $C^1$ function on an interval converges in $L^1$ (and maybe it even converges weakly in $L^\infty$?). The rough idea is that you are approximating an integral (of the gradient) by a Riemann sum (sum of the gradient times the area of each simplex). –  Deane Yang Jun 14 '12 at 12:27