# Condition on a function to be in $L^{p}(R^{d})$ space

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

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

Thanks.

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.
• We are in $L^{p}$ so we have to show that the integral of $|f|^{p}$ is finite... Apr 29 '20 at 17:12
• @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. Apr 29 '20 at 18:52