Here $W^{1, p}_{\text{loc}}(\mathbb{R}^2)$ is the usual locally integrable Sobolev space. I got as a result that $p>2$ and I would like to know if it is correct.
Clearly $u$ is itself locally integrable because $u$ is bounded. The derivatives $\partial_i u$ are zero outside the cone $|x_2|\leq |x_1|$. Also we have that $f\in L^p_{\text{loc}}(\mathbb{R}^2) \Leftrightarrow \int_{B(0, R)}|f|^p<\infty$ for all $R>0$. I use this characterization because I want to use spherical coordinates.
Now $\partial_1 u$ is of the form $\pm x_2/x_1^2$ and $\partial_2 u$ is of the form $\pm 1/x_1$ in the cone $|x_2|\leq |x_1|$. Because of symmetry it is enough to consider the sector $\theta\in [0, \pi/4]$. Here we get integrals
$$\int_0^{\pi/4} \frac{(\sin\theta)^p}{(\cos\theta)^{2p}}\text{d}\theta\int_0^R r^{1-p}\text{d}r $$
for $\partial_1 u$ and
$$\int_0^{\pi/4} \frac{1}{(\cos\theta)^{p}}\text{d}\theta\int_0^R r^{1-p}\text{d}r $$
for $\partial_2 u$. But the angular integrals are finite because $\theta$ is never close to $\pm \pi/2$ so the divergent part must come from the radial integral. And from that I conclude that $u\in W^{1, p}_{\text{loc}}(\mathbb{R}^2)\Leftrightarrow p>2$.
Is my reasoning correct here?
Everything is correct except the conclusion: $\int_0^Rr^{1-p}\,dr<\infty$ if and only if $1-p>-1$, that is, $p<2$.