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I'm trying to prove that $\vert x \vert^{-1}$ can be decomposed into an $L^p+L^q$ functions, where $p<3$ and $q>3$, the integration is over $\mathbb{R}^3$, but I'm still unable to proof/ understand this. The I'm going on about it is, fix $r>0$, so $\vert x \vert^{-p}$ over the ball $B(0,r)$, is finite, however, I'm not sure about the behavior as $x$ is close to zero. I'd appreciate any help with this.

pitariver
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SamKC71
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1 Answers1

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You can use the decomposition $|x|^{-1}=\chi_{B(0,1)}|x|^{-1} + (1-\chi_{B(0,1)})|x|^{-1}$. Using polar coordinates, we compute \begin{align*} \int_{\mathbb{R}^3} \big( \chi_{B(0,1)}|x|^{-1}\big)^p\, dx = \int_0^1 \int_{\partial B(0,r)} r^{-p}\,dSdr = |\partial B(0,1)|\int_0^1 r^{-p+2}dr < \infty\quad \text{since }p<3, \end{align*} and \begin{align*} \int_{\mathbb{R}^3} \big( (1-\chi_{B(0,1)})|x|^{-1}\big)^q\, dx = \int_1^\infty \int_{\partial B(0,r)} r^{-q}\,dSdr = |\partial B(0,1)|\int_1^\infty r^{-q+2}dr < \infty\quad \text{since }q>3. \end{align*}

StarBug
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