I have the following Problem,
$\Omega \subset \mathbb{R}^n$ open and bounded with smooth boundary, $V\subset\mathbb{R}^n$ open with $V\subset\subset \Omega$ and $u:\Omega \rightarrow \mathbb{R}$.
There exists $C>0$, so that $||\nabla^h u||_{L^1(V)}\leq C$ for all $0<h<\frac{1}{2} \operatorname{dist}(V, \partial\Omega)$
I want to see now that for a any dimension $n$ it doesn't necessarily mean that $u\in W^{1,1}(V)$.
How do I find a counterexample for that? Anyone knows such an example or proof? Can't really figure it out.
I understand that $\nabla^h u$ is a vector that consists of functions $h^{-1}(u(x+he_j)-u(x))$, $j=1, \dots, n$, which are finite-difference approximations to partial derivatives of $u$. The statement is that the uniform bound on $L^1$ norm of such approximations is not enough to get an $L^1$ function that is the weak derivative of $u$.
The usual issue with $L^1$ bounds of this kind is that measures satisfy them without being an $L^1$ function itself. For example, take the Dirac delta $\delta_0$ and smoothen it out by convolution with a bump function supported on $[0, h]$. The result is a function with $L^1$ norm at most $1$; however, letting $h\to 0$ does not yield a limit in $L^1$.
In the spirit of the preceding paragraph, let $$u(x) = \begin{cases} 1, \quad &x_1 > 0 \\ 0 \quad & \text{otherwise} \end{cases}$$ (for example, on the unit ball). Then $\nabla^h u$ has norm $1/h$ in the thin slab $-h < x_1 \le 0$, and is zero elsewhere. So, $\|\nabla^h u\|_{L^1}$ is uniformly bounded but $u$ is not in $W^{1,1}$. Indeed, if it had a weak derivative $\partial u/\partial x_1 = g$, then $g$ would have to be $0$ for $x_1\ne 0$, where $u$ is smooth. But then $g=0$ a.e., which can't be the weak derivative of a nonconstant function.