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Consider $L^{p}(R^{d})$ with Lebesgue measure. Let

$f_{0}(x)=|x|^{- \alpha}$ , $|x|<1$ and zero other wise

$f_{1}(x)=|x|^{- \alpha}$ , $|x|\geq1$ and zero other wise

Show $f_{0}\in L^{p}$ iff $p\alpha<d$ and $f_{1}\in L^{p}$ iff $p\alpha>d$ ?

Is there any way to show that with out using polar coordinates?!

Thanks.

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Use the equivalence of the Euclidean norm $||x||_2=\sqrt{\sum_{i=1}^d x_i^2}$ with the $L^{\infty}$ norm $||x||_{\infty}=\max_{1\le i\le d}|x_i|$ and solve the same question with $||\cdot||_{\infty}$ instead. For the latter, one can do it by some explicit calculations only using Fubini's Theorem.

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  • $\begingroup$ We are in $L^{p}$ so we have to show that the integral of $|f|^{p}$ is finite... $\endgroup$
    – Math1
    Apr 29 '20 at 17:12
  • $\begingroup$ we are not in counting measure space! $\endgroup$
    – Math1
    Apr 29 '20 at 17:13
  • $\begingroup$ @Math1: I understand your question about showing the $L^p$ norm of some function $f$ on $\Omega_1=\mathbb{R}^d$ is finite. And I am telling you that you can do that by using the discrete $L^{\infty}$ norm of a vector $x$ seen as a function on $\Omega_2=\{1,2,.\ldots,d\}$. Also, the $L^p$ formulation of the question is a bad idea. It's just about some integral converging or not as a function of $\beta=p\alpha$. In other words, if you understand $p=1$ you understand the general case. $\endgroup$ Apr 29 '20 at 18:52

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