In order to prove that $|x|^{\mu} \in H^{2}(B_1(0))=W^{2,2}(B_1(0))$, I tried to show that the Sobolev norm is bounded, i.e. $||~|x|^{\mu}~||_{2,2,B_1(0)}<\infty$, where $B_1(0)$ is the open unit ball centered around the origin.
I'm using the following definition of the Sobolev norm: $$||u||_{k,p,\Omega}= \sum_{|\beta|\leq k} \left(\int_{\Omega} |D^{\beta}u|^p\right)^{1/p},$$ where $|\beta|$ is a multi-index and where $D^{\beta}$ is the $\beta$-th weak partial derivative.
Thus, $$||~|x|^{\mu}~||_{2,2,B_1(0)}=\sum_{|\beta|\leq 2}\left(\int_{B_1(0)} |D^{\beta}|x|^{\mu}|^2\right)^{1/2}=\left(\int_{B_1(0)} |x|^{2\mu} \right)^{1/2}+\left(\int_{B_1(0)} (D|x|^{\mu})^2\right)^{1/2}+\left(\int_{B_1(0)} (D^2|x|^{\mu})^2\right)^{1/2}.$$
Computing the weak derivatives is the first obstacle for me as I have not performed many computations with weak derivatives. My attempt is the following:
$$D|x|^{\mu}=\sum_i \frac{\partial}{\partial x_i} |x|^{\mu} = \sum_i \frac{x_i \mu |x|^{\mu-1}}{|x|}= \sum_i x_i \mu |x|^{\mu-2}.$$
Similarly,
$$D^2|x|^{\mu}=\sum_{i,j} \frac{\partial^2}{\partial x_j \partial x_i} |x|^{\mu} = \sum_{i,j} x_i x_j \mu (\mu-2) |x|^{\mu-4}.$$
Should those expressions be correct, I wouldn't know how to show that they are square integrable, since we have expressions of the form $|x|^r$ where $r<0$. Only for the first term, i.e. $\left(\int_{B_1(0)} |x|^{2\mu} \right)^{1/2}$, I would be confident to conclude that it is bounded since $0 \leq |x|^{2\mu} \leq 1$ for any $0<\mu<1$ and for any $x\in B_1(0)$.
Any help is much appreciated. As I'm an undergraduate student not that experienced in this domain I would appreciate details and possibly elementary explanations.