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Let $\mathbb{B}_1(0)$ be the unit ball centered at origin. Note that $n$ is the dimension of the space. For what relation of $k,p,n,a $ is $$f(x) = |x|^a$$ in $W^{k,p} (\mathbb{B}_1(0))$ and $W^{k,p} (\mathbb{R}^n \setminus \mathbb{B}_1(0))$?

My work: So far, I have considered the case where $a$ is negative and $k=1$. Assuming these two conditions, I have found that $f\in W^{1,p}(B_1(0))$ if and only if $a > \frac{p-n}{p}$. Additionally, I have found that $f\notin W^{1,p}(B_0(1))$ if $p\geq n$.

The questions that I currently do not know how to handle are

  1. How does $k>1$ change the problem?
  2. How does the space $\mathbb{R}^n\setminus B_1(0)$ affect the weak derivative?

For question 1, my problem mostly comes from not fully understanding the notion of $D^{\alpha}f(x)$ and how to find such a weak derivative. For question 2, since $f$ has a continuous derivative away from $0$, can I take the weak derivative to be the normal derivative in $\mathbb{R}^n\setminus B_1(0)$? If so, determining for which $p,n,a$ (if any) is $f \in W^{1,p}(\mathbb{R}^n\setminus B_1(0))$ may not be too difficult, but, once I consider $k>1$, I'm back to question 1.

chris7347
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  • $k>1$: instead of applying the argument to $f$, apply it to derivatives of $f$. (2) changing the domain changes the requirements on the exponents: in the case $B_1(0)$ you want exponents not too negative to get integrability, for $\mathbb R^d \setminus B_1(0)$ you want exponents negative to get decay and integrability at infinity.
    • – daw Dec 08 '22 at 10:16
  • @daw For $\mathbb{R}^d \setminus B_1(0)$ with $a>0, k=1$, I know that $|Df| = |a||x|^{a-1}$. Is it sufficient to say that $|Df| \in L^1$ when $a-1 < 0$ or is there more to be done to prove this. – chris7347 Dec 08 '22 at 17:07