Is the function $f(x,y)=y \cdot \text{sign}(x) \in W^{1,p}\big((-1,1) \times (-1,1)\big)$ for some $p \ge 1$?
I think not but I am not sure if my reasoning below is correct. I would like to get some feedback about it, or even better, a suggestion for an easier approach.
My proof:
If $f$ were Sobolev, then its weak partial derivatives would have to be $\frac{\partial{f}}{\partial{x}}=0,\frac{\partial{f}}{\partial{y}}=\text{sign}(x)$.
Thus, for any smooth test function $\phi \in C_C^{\infty}((-1,1)^2)$,
$$0= \int_{(-1,1)^2} \frac{\partial{\phi}}{\partial{x}} f=\int_{(-1,1)^2 \cap \{x>0 \}} \frac{\partial{\phi}}{\partial{x}} y-\int_{(-1,1)^2 \cap \{x<0 \}} \frac{\partial{\phi}}{\partial{x}} y=$$
$$ \int_{(0,1) \times (-1,1)} \frac{\partial{\phi}}{\partial{x}} y-\int_{(-1,0) \times (-1,1)} \frac{\partial{\phi}}{\partial{x}} y=$$
$$ \int_{ \partial {\big((0,1) \times (-1,1)}\big)} \phi y\nu_1-\int_{ \partial {\big((-1,0) \times (-1,1)}\big)} \phi y \tilde \nu_1,$$
where $\nu=(\nu_1,\nu_2),\tilde \nu=(\tilde \nu_1,\tilde \nu_2)$ are the outward unit normal vector fields on the domains $U=(0,1) \times (-1,1),\tilde U=(-1,0) \times (-1,1)$ respectively.
Looking at the domains, we see that on the common boundary $\{0\} \times (-1,1)$, we have $\nu=(-1,0),\tilde \nu=(1,0)$. Since the on the rest of the boundaries, $\phi=0$, we obtain
$$ 0=-\int_{ \{0\} \times (-1,1)} \phi y-\int_{ \{0\} \times (-1,1)} \phi y =-2 \int_{ \{0\} \times (-1,1)} \phi(0,y) y dy$$
which doesn't seem to be zero for an arbitrary compactly supported smooth function $\phi$.
Motivation: I am trying to find a Sobolev map $\mathbb{R}^2 \to \mathbb{R}^2$ whose differential zig-zags between a fixed invertible matrix and its negative. If the above $f$ were Sobolev, then $f(x,y)=(|x|,y \text{sign}(x))$ would be an example.
Here is an alternative argument:
Since the restriction of a variable in a Sobolev function is a.e Sobolev, if $f(x,y)=y \cdot \text{sign}(x)$ were Sobolev, then so was the function $x \to \text{sign}(x)$, which is not the case.
One way to see that $\text{sign}(x)$ is not Sobolev is that if it were, then its weak derivative should be identically zero, hence it must be constant. (Alternatively, it is not continuous, and any Sobolev function in one dimension is absolutely continuous).