Timeline for Let $f \in S(\mathbb R)$. Can we say $\widehat{|f|} \in L^{1}(\mathbb R)$?

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Nov 18, 2013 at 13:10	comment	added	Liviu Nicolaescu		Oops! I forgot that $f$ is a one-variable function. $(1+\|\xi\|^2)^{-1/2}$ is not in $L^2(\mathbb{R}^n)$ if $n\geq 2$.
Nov 18, 2013 at 5:27	vote	accept	Inquisitive
Nov 17, 2013 at 16:51	comment	added	Michael Renardy		It is more elementary than you think. If f is in S, it is Lipschitz continuous. It is then easy to prove that \|f\| is also Lipschitz continuous, and therefore in $H^1$.
Nov 17, 2013 at 8:20	comment	added	Inquisitive		Let $f\in S(\mathbb R)$ such that $\|f\|\not \in S(\mathbb R)$; (so $\widehat {\|f\|} \in L^{2}(\mathbb R)$ by Plancherel theorem). But why $(1+\|\xi\|^{2})^{\frac {1}{2}}\widehat {\|f\|} \in \mathbb L^{2}(\mathbb R) ?$
Nov 16, 2013 at 16:54	comment	added	abx		Because $(1+\xi ^2)^{-\frac{1}{2} }$ is in $L^2$, and the product of two functions of $L^2$ is in $L^1$.
Nov 16, 2013 at 16:28	comment	added	Liviu Nicolaescu		$g\in H^1\Longleftrightarrow (1+\|\xi\|^2)^{\frac{1}{2}}\widehat{g}(\xi)\in L^2$. Why does the last condition imply that $\widehat{g}\in L^1$?
Nov 16, 2013 at 11:48	history	answered	Michael Renardy	CC BY-SA 3.0