I have the following problem:
Let $u,u_i:\Omega\rightarrow\mathbb{R}$ defined by: $$u(x):=\begin{cases}|x|^{-\alpha},\quad \text{with}\nobreakspace x\neq 0, \ \\ 0,\qquad \text{with} \nobreakspace x=0,\end{cases}$$ and $$u_i(x):=\begin{cases}\partial_i|x|^{-\alpha},\quad \text{with}\nobreakspace x\neq 0,\\0, \qquad\text{with}\nobreakspace x=0, \end{cases}$$ with $i=1,...,n$. Now for each case
$1) \Omega=B_1(0)\backslash\{0\}$,
$2) \Omega= \mathbb{R}^n\backslash\overline{B_1(0)}$,
$3) \Omega=\mathbb{R}^n\backslash\{0\}$
determine for which $\alpha \in \mathbb{R}\ u(x), u_i(x)$ are in $L^p(\Omega)$.
For the case $u(x)$:
In $1)$: For $u(x)$ to be in $L^p(\Omega)$ $p-\alpha$ must be greater than $-1$
In $2)$: For $u(x)$ to be in $L^p(\Omega)$ $p-\alpha$ must be less than $-1$
In $3)$: For $u(x)$ to be in $L^p(\Omega)$ there shouldn't be such an $\alpha$
For the case $u_i(x)$:
In $1)$: For $u_i(x)$ to be in $L^p(\Omega)$ $p-\alpha$ must be less than $1$
In $2)$: For $u_i(x)$ to be in $L^p(\Omega)$ $p-\alpha$ must be greater than $1$
In $3)$: For $u_i(x)$ to be in $L^p(\Omega)$ there shouldn't be such an $\alpha$
I got these results by only looking at $\mathbb{R}$ instead of $\mathbb{R}^n$. Are these statements true? How can I show this for $\mathbb{R}^n$? Can someone help me?