For $d\in \mathbb{N}$ and $\alpha>0$, consider the function $$f:\mathbb{R}^d\to \mathbb{R}, \ \ \ \ \ x\mapsto ||x||_2^{-\alpha}.$$ a) Compute the range of all $\alpha$ such that $f \in L_{\text{loc}}^1(\mathbb{R}^d)$.
b) For $p\in (1,\infty)$ determine all $\alpha$ such that $f\in W^{1,p}([-1,1]^d)$.
I have just started learning Sobolev spaces and am struggling with this innocent-looking question.
For a) I need to check for what values $\alpha$ the integral $\int_B |f| dx$ is finite for any ball $B$. I think the problem why $f$ would not be locally integrable is $0$ so that we split the integral and look at whether $\int_{B({0,\epsilon})}(x_1^2+\dots+x_d^2)^{-\alpha/2} dx_1\dots dx_d$ is finite as $\epsilon\to 0$. I struggle with integrating this as I have $n$ variables and don't see an immediate antiderivative. I could say $x_i^2<\epsilon^2$ on $B(0,\epsilon)$ but then the integral is dominated by $d^{-\alpha/2}\epsilon^{-\alpha/2}$ which does never go to zero…
I think b) is stronger than a). We need to find $\alpha$ so that there are weak derivatives $(v_i)_{1\leq i \leq d}$ which satisfy a) and $$\int_{[-1,1]^d} v_i\phi_{x_i} dx=-\int_{[-1,1]^d} v_i \phi dx$$ for all $\phi\in C_c^{\infty}(\mathbb{R^d})$. We also need that $v_i$ is in $L^p$. I am quite stuck how to find for what values of $\alpha$ these conditions hold. Can someone point me in the right direction?
The only thing to really do is apply polar integration. Let $\textrm{d}S$ denote the surface measure on the $d-1$ sphere, $S^{d-1}$. Then, since $||x||_2^{-\alpha}$ depends only on the norm, we get
$$ \int_{B(0,\varepsilon)} ||x||_2^{-\alpha}\textrm{d}x=\int_0^{\varepsilon} \int_{S^{d-1}} r^{d-1} r^{-\alpha}\textrm{d}S(\theta)\textrm{d}r=dS(S^{d-1})\int_0^{\varepsilon} r^{d-\alpha-1}\textrm{d}r, $$ which is finite, if and only if $d-\alpha-1>-1$, i.e., when $\alpha<d.$
Note now that your function is classically differentiable away from $0$ with $\partial_j ||x||_2^{-\alpha}=-\alpha x_j||x||_2^{-\alpha-2}$. Thus, this determines the derivative on $\mathbb{R}^d\setminus \{0\}.$ Since $\{0\}$ is a null-set, $C_c^{\infty}(\mathbb{R}^d\setminus\{0\})$ is dense in $L^p(\mathbb{R}^d)$, so this is our only possible candidate for a derivative (and, similarly, on $[-1,1]^d$). Note that, by the above, since $||x_j||x||_2^{-\alpha-2}||\leq||x||_2^{-\alpha-1}$ it is in $L^p([-1,1]^d)$ when $d>p(\alpha+1)$.I've skipped a bit on why this is strict, but you can make a similar bound from below on any part of the $\varepsilon$-ball which cuts away some cone-shaped segement around the line $x_j=0$.
Note, for any $\phi\in C_c^{\infty}([-1,1]^d),$ we have, by usual integration by parts and dominated convergence that, at least for $\alpha<d-1$,
\begin{align} \int_{[-1,1]^d} \partial_j\phi'(x)||x||_2^{-\alpha}\textrm{d}x&=\lim_{\varepsilon\to 0^+}\int_{[-1,1]^d\setminus B(0,\varepsilon)} \partial_j\phi(x)||x||_2^{-\alpha}\textrm{d}x\\ &=\lim_{\varepsilon\to 0^+} \left(-\int_{[-1,1]^d\setminus B(0,\varepsilon)} \phi(x) \partial_j||x||_2^{-\alpha}\textrm{d}x+\varepsilon^{d-1}\int_{S^{d-1}}\nu_j\phi(\varepsilon \theta)||\varepsilon \theta||_2^{-\alpha}\textrm{d}S(\theta)\right), \end{align}
where $\nu_j$ is the $j$'th component of the outer unit normal of $\varepsilon S^{d-1}$ (which naturally coincides with that of $S^{d-1}$). Now, simply continuing the calculation, we get, again by Dominated Convergence
\begin{align} \int_{[-1,1]^d} \partial_j\phi(x)||x||_2^{-\alpha}\textrm{d}x &= \int_{[-1,1]^d} \phi(x)\partial_j||x_2||^{-\alpha}\textrm{d}x+\lim_{\varepsilon\to 0^+}\varepsilon^{d-1} \int_{S^{d-1}} \nu_j \varphi(\varepsilon \theta) \varepsilon^{-\alpha} \textrm{d}S(\theta) \\ &=\int_{[-1,1]^d} \phi(x)\partial_j||x_2||^{-\alpha}\textrm{d}x+\lim_{\varepsilon\to 0^+}\varepsilon^{d-\alpha-1}\int_{S^{d-1}} \nu_j \varphi(\varepsilon \theta) \textrm{d}S(\theta) \\ &=\int_{[-1,1]^d} \phi(x)\partial_j||x_2||^{-\alpha}\textrm{d}x, \end{align} since, $|\int_{S^{d-1}} \nu_i \phi(\varepsilon(\theta))\textrm{d}\theta|$ is bounded by $\textrm{d}S(S^{d-1})||\phi||_{\infty}$.
Hence, whenever both $||x||_2^{-\alpha}$ and $\partial_j||x||_2^{-\alpha}$ are in $L^p([-1,1]^d),$ we can also justify that the latter is the weak $j$-derivative of the former. Hence, our analysis of, when the two are in $L^p_{loc}$ exactly matches, when $||x||_2^{-\alpha}$ lies in the Sobolev space.