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Consider the function $w:\mathbb{R}^4\setminus\{0\}\to \mathbb{R}, w(x)=|x|^{-\frac{4}{5}p}(1+|x|^3)$. Firstly, I am asked to find the values of $p\ge 1$ such that $w\in L^1(B_1(0))$ and then I am asked to find the values of $p\ge 1$ such that $w\in L^1(\mathbb{R}^4\setminus\overline{B_1(0)})$, where in both cases $B_1(0)$ stand for the open unit ball in $\mathbb{R}^4$.

According to what I learned in the course where I found this problem, I have to use the coarea formula. To be clear, I have to use the fact that $$\int_{B_{R}(0)}f dx=\int_0^R\left(\int_{\partial B_r(0)}fd\sigma \right)dr$$ for an integrable function $f$ (I think that it also works if $f$ is not integrable, but only positive).

So, off we go. According to the coarea formula, $$\int_{B_1(0)}|w(x)|dx=\int_0^1\left(\int_{\partial B_r(0)}|x|^{-\frac{4}{5}p}(1+|x|^3)d\sigma(x) \right)dr=\int_0^1r^{-\frac{4}{5}p}(1+r^3)|\partial B_r(0)|dr,$$ where I used that $w$ is a radial function. So, if we denote by $\omega_4$ the area of the unit sphere in $\mathbb{R}^4$, we have that $|\partial B_r(0)|=\omega_4\cdot r^3$, so we want $$\int_0^1 r^{-\frac{4}{5}p+3}(1+r^3)dr<\infty\iff \int_0^1 (r^{4-\frac{4}{5}p}+r^{6-\frac{4}{5}p})dr<\infty \iff p<\frac{25}{4}.$$ Thus, $1\le p <\frac{25}{4}$ is the answer to the first part of the problem.

Now, I am a bit unsure about the second part. I don't know if by the coarea formula I am allowed to write $$\int_{\mathbb{R}^4\setminus B_1(0)}|w(x)|dx=\int_1^\infty\left(\int_{\partial B_r(0)}|w|d\sigma \right)dr$$ and then of course proceed as above. I sure know that I can write $$\int_{\mathbb{R}^4}|w(x)|dx=\int_0^\infty \left(\int_{\partial B_r(0)}|w|d\sigma \right)dr,$$ but I don't know about the other one.

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For the first question, your answer is wrong because you added a superfluous factor of $r$ (the exponent went from 3 to 4). The correct result is that $1\le p< 5$. For the second question, just use the formula you know applied to $$w(x)\,\cdot1_{|x|\ge 1}$$

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  • $\begingroup$ thank you! Where did the exponent go from $3$ to $4$? Not that it matters, I was kind of sure that I messed up the computations. And for the second question: that's a really nice idea! Just to be sure, applying my formula to $w(x)\cdot 1_{|x|\ge 1}$ gives me precisely what I wanted, i.e. that $\int_{\mathbb{R}^4\setminus B_1(0)}|w(x)|dx=\int_1^\infty\left(\int_{\partial B_r(0)}|w|d\sigma \right)dr $, right? $\endgroup$ Jan 27, 2023 at 22:14
  • $\begingroup$ I saw where I messed up the exponent, thanks a bunch! $\endgroup$ Jan 27, 2023 at 22:17

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