I was recently attempting a problem from Evans' PDE text and I solved it, but I'm not sure my answer is correct since it looks a bit fishy. I was hoping someone could verify it and correct me if I'm wrong.
The problem goes as follows:
Let U = $\{|x_1| < 1,|x_2| < 1\}$ denote the open unit square in $R^2$ centered at the origin. Define the function $u:U \mapsto R$ as follows:
$$ u(x_1,x_2) = \begin{cases} 1-x_1 \text{ if} & |x_1| \leq |x_2|, x_1\geq 0, \\ 1+x_1 \text{ if} & |x_1| \leq |x_2|, x_1\leq 0, \\ 1-x_2 \text{ if} & |x_2| \leq |x_1|, x_2\geq 0, \\ 1+x_2 \text{ if} & |x_2| \leq |x_1|, x_2\leq 0 \\ \end{cases} $$
For which $1 \leq p \leq \infty$ does u lie in $W^{1,p}(U)?$
Solution:
My answer is that u lies in the Sobolev space for all $1\leq p\leq\infty.$
Sketch of the argument:
First note that u is bounded on a set of finite measure so it lies in $L^{p}(U)$ for each $1 \leq p \leq \infty.$
Now it suffices to show that the weak first partials $u_{x_1}$ and $u_{x_2}$ also lie in $L^{p}(U)$ for each $1 \leq p \leq \infty.$ In particular first we show that these weak derivatives are actually functions and then we show that they are bounded.
To this extent we let $\phi$ be a test function on $U$ and consider the quantity:
$$\int_U{u \phi_{x_i}}$$
In order to simplify this quantity we divide the open unit square into 4 open regions $R_1,R_2,R_3, \text{ and } R_4$.
$$R_1 = \{|x_1| < |x_2|, x_1 > 0\}$$ $$R_2 = \{|x_1| < |x_2|, x_1 < 0\}$$ $$R_3 = \{|x_1| > |x_2|, x_2 > 0\}$$ $$R_4 = \{|x_1| > |x_2|, x_2 < 0\}$$
we orient each of these counter-clockwise and use Greens Theorem to simplify our integral as follows:
$$\int_{R_j} u \phi_{x_i} = \int_{\partial R_j} u \phi \nu_{i} dS - \int_{R_j} u_{x_i} \phi$$
Here {nu} denotes the outward pointing normal.
We note that $\phi$ vanishes on the outer boundary of the unit square being smooth and compactly supported in u (it's a test function).
If we picture each of the regions to be triangles then u agrees on the shared boundary of any two regions (the boundaries are precisely $x_1 = x_2$ and $x_1 = -x_2$).
However note that for two regions sharing a boundary their corresponding normals point in precisely opposite directions (due to being oriented in opposite directions).
So we have that:
$$\sum_{j=1}^4 \int_{\partial R_j} u \phi \nu_{i} dS = 0$$
We are left with the expression:
$$ \int_U{u \phi_{x_i}} = \sum_{j=1}^4 (\int_{\partial R_j} u \phi \nu_{i} dS - \int_{R_j} u_{x_i} \phi) = - \sum_{j=1}^4 \int_{R_j} u_{x_i} \phi) $$
However, we see that u is differentiable on each of the $R_j$'s and we may plug in $u_{x_i}$ for each of the regions for $i=1,2$ to find the weak partial derivatives of u i.e.,
$$u_{x_1} = 1_{R_1} - 1_{R_2}$$ $$u_{x_2} = 1_{R_3} - 1_{R_4}$$
However, both these (weak) partial derivates are bounded on sets of finite measure. So we see that they are in $L^{p}(U)$ for each $1 \leq p \leq \infty.$
We can finally conclude that u lies in $W^{1,p}(U)$ for each $1 \leq p \leq \infty. \: \square$
To reiterate, please let me know if this is correct or not, if not please let me know what the error is.