If one has a function $u:\mathbb{R}^{n}\to\mathbb{R}$ given by $u(x)=|x|^{\alpha}-1$ for $|x|<1$, and $0$ for $|x|\ge1$, how is it that we can determine the values of $\alpha$ and $p\in[1,\infty]$, for which $Du(x)$ belongs to $L^{p}(\mathbb{R}^n)$?
My first thought is to find $Du(x)$ and then I guess one has to determine the values of $\alpha$ and $p$ that satisfy the integral
$$\left(\int_{S} |Du(x)|^{p} d\mu\right)^{1/p}$$
Is this even on the right track or am I totally missing something? Honestly, I am not totally familiar with Lp-spaces and this is really throwing me off.
We calculate $Du = \alpha|x|^{\alpha-2}x$, so $|Du|^p = \alpha^p|x|^{p(\alpha-1)}$ on the unit ball and zero otherwise. In general the power function $|x|^\beta$ is integrable over the unit ball $B_1(0)$ in $\mathbb{R}^n$ if and only if $\beta > -n$; indeed, if we switch to polar coordinates we have $$ \int_{B_1(0)}|x|^\beta\ dx = \int_0^1\int_{\partial B_r(0)} |rz|^\beta \ dS(z)dr = |\partial B_0(1)|\int_0^1 r^{n-1 +\beta}\ dr $$ where the one dimensional integral on the right converges iff $n-1 + \beta > -1 \implies \beta > -n$. Therefore the integral of $|Du|^p = \alpha^p|x|^{p(\alpha-1)}$ is finite iff $p(\alpha - 1) > -n$. If we fix $p \in [1,\infty)$ then this means we can allow any $\alpha > 1 - \frac{n}{p}$. If $p = \infty$, then clearly all we need for $Du$ to be in $L^\infty$ is for $|x|^{\alpha-1}$ to be bounded on the unit ball, which means $\alpha - 1 \geq 0 \implies \alpha \geq 1$. We can think of this as the limiting case of the inequality $\alpha > 1 - \frac{n}{p}$ as $p\to\infty$.