$W^{1,\infty}$ function that cannot be approximated by Smooth Functions

Question

I am trying to find a function $u \in W^{1,\infty}$ that cannot be approximated by smooth functions. It seems like it should be an easy construction but I am blanking. Thanks.

Answer 1

For example take $u(x):=|x|$ on $\mathbb{R}$. I assume that you can prove that $u\in W^{1,\infty}(\mathbb{R})$ (more generally, it is not so difficult to see that $W^{1,\infty}(\mathbb{R})$ consists exactly of the Lipschitz functions).

The weak derivative of $u$ is $v(x)=\text{sgn}(x)$ (which equals $1$ on $\mathbb{R}^+$ and $-1$ on $\mathbb{R}^-$).
Now take any smooth $f\in C^{\infty}(\mathbb{R})\cap W^{1,\infty}(\mathbb{R})$: if $f'(0)\ge 0$ then $$\lim_{x\to 0^-}|f'(x)-v(x)|=|f'(0)-(-1)|\ge 1,$$ so $\|f'-v\|_\infty\ge 1$; if instead $f'(0)\le 0$ then $$\lim_{x\to 0^+}|f'(x)-v(x)|=|f'(0)-1|\ge 1,$$ so again $\|f'-v\|_\infty\ge 1$.

In any case you get $\|f-u\|_{W^{1,\infty}}\ge 1$.

But if you weaken your requirement the situation improves: there is always a sequence of smooth functions $f_n$ such that $f_n\to u$ in $L^\infty$ and $f_n'\rightharpoonup u'$ in the weak-$*$ topology of $L^\infty$.

Answer 2

Any Lipschitz function that is not $C^1$ will do, since convergence in the $W^{1,\infty}$ norm implies uniform convergence (for the function and its derivatives) in $C^1$.

Answer 3

$u(x)=|x|$ for $-1<x<1$? (Weak) $u'(x)=H(x)=1[x>0]+0[x<0]$ (where $[]$ is Iverson bracket), that's discontinuous, so no uniform ($L^\infty$) convergence of continuous (let alone smooth) function sequence.