# Conditions under which a function is in L^2.

1. May 18, 2013

### QIsReluctant

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

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

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)?