Are piecewise linear functions dense in $W^{1,\infty}$ ?



Are piecewise linear functions dense in $W^{1,\infty}$?
Piecewise linear functions are ***not*** dense in $W^{1,\infty}(\Omega;\mathbb R^n)$ for any open set $\Omega\subset\mathbb R^m$.


If it were true for $\Omega$, it would also be true for any open subset, in particular any open ball.
Recall that the Sobolev space $W^{1,\infty}(\Omega;\mathbb R^n)$ is the space of Lipschitz functions $\Omega\to\mathbb R^n$ with the same norm if $\Omega$ is quasiconvex — and balls are quasiconvex.
(This and many other basic facts of Lipschitz functions can be found in [*Lectures on Lipschitz analysis*](http://www.math.jyu.fi/research/reports/rep100.pdf) by Heinonen.)
The Lipschitz condition behaves well when restricting to subspaces, which is not that obvious from the point of view of Sobolev spaces.


If piecewise linear functions were dense, for any $f\in W^{1,\infty}$ there would be a sequence of piecewise linear functions $f\_i$ tending to $f$.
Every Lipschitz function on a line can be extended to a Lipschitz function on the whole domain and the restriction of a piecewise linear function to a line is piecewise linear.
Therefore, if we restrict all functions to a line segment in $\Omega$ and evaluate only one component, we end up with the claim that every Lipschitz function $[a,b]\to\mathbb R$ can be approximated by a piecewise linear function in the Lipschitz norm.


Now, let $f:[a,b]\to\mathbb R$ be an arbitrary Lipschitz function and suppose there is a sequence of piecewise linear functions $f\_i:[a,b]\to\mathbb R$ converging to it.
Then $f\_i(a)\to f(a)$ and $f\_i'\to f'$ in $L^\infty$.
Since $f'$ can be any $L^\infty$ function, this means that piecewise constant functions are dense in $L^\infty$.
But this is false.
The function $g:[0,1]\to\mathbb R$, $g(x)=\sum\_{k=1}^\infty\chi\_{[2^{-2k},2^{1-2k}]}(x)$ is in $L^\infty$ but any piecewise constant function has at least distance $\frac12$ from it.


