I am given a function as in the screenshot,
- Denote by $U$ the open square $\{ x\in \mathbb R^2 : |x_1|<1, \ |x_2 < 1\}.$ Define $$ u(x) = \begin{cases} 1-x_1 & x_1 > 0, |x_2|< x_1;\\ 1+x_1 & x_1 < 0, |x_2|<-x_1;\\ 1-x_2 & x_2 > 0, |x_1|<x_2;\\ 1-x_1 & x_1 < 0, |x_2|<-x_1.\\ \end{cases}$$ For which $1\le q \le \infty$ does $u$ belong to $W^{1,q}(U)$?
The function is trivially in $L^p(U)$, but I think the weak derivative does not exist. So I suppose that there is a function $u \in W^{1,q}(U)$. Then I try to find a test function which breaks the weak derivative equation, but im stuck here... Any advice is appreciated.
The function $u$ can be written $u = u_1 + u_2$ where $$u_1(x) = 1 - \max(|x_1|,\ |x_2|) = 1 - \|x\|_{\ell^\infty_2}$$ is the function in Evans Ch. 5 Exercise 3, and $$u_2(x) = \begin{cases} 2x_2 & x_2<0, |x_1|<-x_2;\\ 0 & \text{otherwise}. \end{cases}$$
Since $u_1\in \operatorname{Lip}(U) = W^{1,\infty}(U)\subset W^{1,q}(U)$ for any $q\in[1,\infty]$, it suffices to check if $u_2\in W^{1,q}$ for some $q$. Since this function has a jump discontinuity across the curve $C$ given by $x_2\le 0, |x_1| = -x_2$, it shouldn't have a weak derivative, so the answer should be no $q$.
To be more precise, suppose it was weakly differentiable. You can multiply $u_2$ by a smooth cutoff $\phi$ so that $\phi u_2$ is compactly supported in $x_2 \in [\frac12, \frac23]$, $x_1 > 0$, and $| |x_1| - (-x_2) | < 0.01$. We also enforce that near the curve $C$, away from the edges of the support of $\phi$, that $\phi \equiv 1$, and $|\phi|<1$ elsewhere. Let's call this set $D:=\{\phi \equiv 1\}$.
On the support of $\phi$, the function $\frac1{x_2}$ is smooth. So we can also multiply by this, and the resulting function $$ v(x) = \frac{\phi(x) u_2(x)}{x_2} $$ is also weakly differentiable. But on the set $D$, $v(x)$ is a rotation and translation of $\mathbb 1_{x_1 > 0}$ (both of these operations preserve weak differentiability).
So the question now reduces to showing that $I:=\mathbb 1_{x_1 > 0} $ is not weakly differentiable on say, $[-c,c]^2$ for any $c>0$ (this completes the proof by contradiction). For simplicity we set $c=1$.
Intuitively, $I=\mathbb 1_{x_1 > 0}$ has only the distributional derivative $\delta(x_1)$, so it is not weakly differentiable. To verify this pick a test function of the type $\phi(x) = g(x_1)f(x_2)$ with $f,g$ compactly supported on $[-1,1]$. Then if $I$ is weakly differentiable with weak partial derivative $J\in L^1_{\text{loc}}$ wrt $x_1$, we must have
$$ - \int_{[-1,1]^2} J \phi = \int_{[-1,1]^2} I \partial_1\phi = \left(\int_{-1}^1 f(y) dy\right) \int_0^1 g'(x) dx = - \left(\int_{-1}^1 f(y) dy\right) g(0).$$
Now let $\int_{-1}^1 f = 1$ and replace $g$ with a sequence $g_n$ of test functions with $g_n(0) \equiv 1$ but at the same time $g_n(x) \to 0$ as $n\to\infty$, for all $x\neq 0$. Then $\phi_n = f g_n \in L^\infty$ and converges a.e. to $0$, so $$ 0 = \lim_{n\to\infty} - \int_{[-1,1]^2} J \phi_n = \lim_{n\to\infty} - \left(\int_{-1}^1 f(y) dy\right) g_n(0) = -1, $$ which is a contradiction. This finishes the proof.