I'm trying to solve this exercice:
Find all possible values of $\alpha\in\mathbb{R}$ such that $f(x)=\min(|x|^\alpha,1)$ is in $W^{1,2}(\mathbb{R}^2)$ with $|x|=\sqrt{x_1^2+x_2^2}$.
First I control if $f\in L^2(\mathbb{R}^2)$. So I want to calculate or estimate ( I guess the answer is estimate)
$||f||_2^2 = \int_{\mathbb{R}^2}|f(x)| dx$
but I cannot do the integral because I don't know how to consider that $\min$.
Can you help me?
If $\alpha \ge 0$ then $|x|^\alpha \le 1$ if and only if $|x| \le 1$, and you can write $f$ as $$f(x) = \left\{ \begin{array}{ll} |x|^\alpha & |x| \le 1 \\ 1 & |x| > 1. \end{array} \right.$$ If $\alpha < 0$ then $|x|^\alpha \ge 1$ if and only if $|x| \le 1$, and you can write $f$ as $$f(x) = \left\{ \begin{array}{ll} 1 & |x| \le 1 \\ |x|^\alpha & |x| > 1. \end{array} \right.$$
In both cases you have $$\int_{\mathbb R^2} |f(x)|^2 \, dx = \int_{|x| \le 1} |f(x)|^2 \, dx + \int_{|x| > 1} |f(x)|^2 \, dx$$ and you can evaluate the integrals individually.