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Conditions under which a function is in L^2.

  1. May 18, 2013 #1
    Let [itex]f[/itex] be the function [itex]f(x) = \|x\|^\alpha[/itex] and let [itex]\phi \in S(\mathbb{R})[/itex] s.t. [itex]\phi(0) \neq 0[/itex]. Show that [itex]f\phi \in L^2(\mathbb{R}) \Leftrightarrow \alpha > -1/2[/itex]

    The way to go is to verify that the integral of the square is finite. The solutions manual first integrates over [itex]\|x\| > 1[/itex] (okay there), then [itex]\|x\| \leq 1[/itex], where the confusion sets in. The explanation is that since [itex]\|f(x)\phi(x)\|^2 \rightarrow \|x\|^{2\alpha}\phi^2(0) + o(1)[/itex] as [itex]\|x\| \rightarrow 0[/itex], we have [itex]\int {f\phi}^2 < \infty \Leftrightarrow \int \|x\|^{2\alpha}\phi^2(0) < \infty[/itex].

    My question: I understand that the value at zero is a potential hangup, but why are we allowed to make this IFF statement? Where does the limit at zero come in? And why o(1)?
  2. jcsd
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