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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

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You can represent on every simplex $\sigma$ of your triangulation the derivative $D f_k$ by $$ D f_k = \frac{1}{\lvert \sigma \rvert} \sum_{i = 0}^n \int_{\sigma} \Bigl(\frac{1}{(1 - \beta_i (x))^n} - 1\Bigr)\, Df(x)[a_i - x]\,dx D\beta_i, $$ where $\beta_i$ are the maps that give barycentric coordinates $$ \beta_0 (x) + \dots + \beta_n (x) = 1 \quad \text{ and } \quad \beta_0 (x)a_0 + \dots + \beta_n (x)a_n = x. $$ For a proof, see J. Van Schaftingen, Approximation in Sobolev spaces by piecewise affine interpolation, [arXiv:1312.5986]).

The kernel appearing in this integral representation is absolutely integrable ($a_i - x$ controls $\beta_i (x)$), uniformly with respect to simplices that do not degenerate. This should imply in fact an $L^\infty$ approximation property under your assumptions. (If the function $f$ is merely in a Sobolev space, one can improve the convergence by translating the triangulation points.)

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