Let $u:B(0,1)\subseteq \mathbb{R}^N$, with $N\geq 2$ such that $$ u(x) = \begin{cases} 1 & \text{if } x_N \geq 0,\\\\ 0 & \text{if } x_N <0. \end{cases} $$ I want to show that $u\not\in W^{1,p}(B(0,1))$ for $1\leq p \leq +\infty$.
My attempt:
To avoid the case $x_N=0$, let's define following sets
$$ \begin{align} S^1_\varepsilon & = \left\{x\in B(0,1)\,:\, x_N <-\varepsilon\right\}, \\\\ S^2_\varepsilon & = \left\{x\in B(0,1)\,:\, x_N >\varepsilon\right\}, \\\\ S^3_\varepsilon & = \left\{x\in B(0,1)\,:\, -\varepsilon < x_N <\varepsilon\right\}. \end{align} $$
Since in $S^1_\varepsilon$ and $S^2_\varepsilon$ it's possible to use integration by parts and by dominated convergence when $\varepsilon \to 0$ $$ \int_{B(0,1)} u \frac{\partial \phi}{\partial x_i} \, dx = - \int_{B(0,1)} \frac{\partial u }{\partial x_i} \phi \, dx + 2 \int_{\left\{x\in B(0,1) \, :\, x_N=0\right\} } u\phi n_i d\, \sigma(x) \quad \forall \phi\in C_c^\infty (B(0,1)). $$ Hence $$ 2 \int_{\left\{x\in B(0,1) \, :\, x_N=0\right\} } u\phi n_i d\, \sigma(x) = 0 \quad \forall \phi\in C_c^\infty (B(0,1)). $$ with $n_i$ a component of normal vector.
It's correct? How can I conclude? My idea is that is false, but I'm not sure how to write.
