So, I'm facing this problem, find all values of $\beta \in R$ : $u(x):= \max\{||x||^\beta -5,0\} \in W^{1,2}(R^2)$. For past exam contents I've seen that this kind of exercise, with this function defined by max or min comes often.
if the function $u(x) \in W^{1,2}(R^2)$ we must have $||u(x)||_{W^{1,2}(R^2)} < +\infty$.
By definition $||u(x)||_{W^{1,2}(R^2)} = ||u(x)||_{L^2(R^2)} + ||\nabla u(x)||_{L^2(R^2)}$.
1) the $|| u(x) ||$ is a module in this context? The notation confuse me.
2) I've tried to solve this, starting from the norm in $L^2(R^2)$, because all the Sobolev norm must be limited; I end stuck at this point:
${||u(x)||_{L^2(R^2)}}^2 =\int_{- \infty}^0 ((-x)^\beta -5)^2dx+=\int_{0}^{ +\infty} ((+x)^\beta -5)^2dx$
Am I supposed to extract the condition from here? Because there are some $x$ that do not have the $\beta$ and lead to divergence.
The norm for the $0$ is of course some constant $c \in R^+$
What am I missing?
I think that $x=(x_1,x_2)\in\mathbb{R}^2$ so the $L^2$ norm should be computed in $\mathbb{R}^2$ and not in $\mathbb{R}$ as you are doing. If $\beta>0$, then $||x||^\beta>5$ for all $||x||$ large. In this case you have that $$\lim_{||x||\to\infty}u(x)=\infty$$ and so your function cannot be in $L^2(\mathbb{R}^2)$.
If $\beta=0$, then $u=0$.
If $\beta<0$ then $u$ is zero outside of a ball of radius $5^{1/\beta}$. Hint: Use polar coordinates to study the integrability of $u$ and $\nabla u$ in that ball.