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Among other things, the Fourier transform maps functions from $L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$, $L^1(\mathbb{R}^n) \to C_0(\mathbb{R}^n)$ (continuous functions vanishing at infinity), and $\mathcal{S}(\mathbb{R}^n)\to\mathcal{S}(\mathbb{R}^n)$ (Schwartz space, or the space of rapidly decreasing functions).

I'm interested in looking more closely at the codomain of the second mapping. Since $C_0 \subset L^\infty$, every $L^1$ function is mapped by the Fourier transform into an $L^\infty$ function. However, I was wondering if it is easy to find specific examples of functions that are only in $L^1$, but are mapped into $L^1$ or $L^2$ (or possibly any other $L^p$ for $1 \leq p < \infty$).

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    $\begingroup$ Please clarify what it means for a function to be "only in" a function space as opposed to being "in" that space. For example, the zero function is in L^1 and its Fourier transform is in any L^p. That is certainly not something that you have in mind as an example. $\endgroup$
    – KCd
    Feb 4, 2011 at 7:44
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    $\begingroup$ By "only," I meant it so that the function in question wouldn't obviously be mapped into the codomain in consideration. Since $L^2$ functions are mapped into $L^2$, a function that is in $L^1 \cap L^2$ would necessarily be mapped into $L^2$, but it wouldn't be a very illuminating example (the zero function that you demonstrated would fit this category). Sorry for the confusion. $\endgroup$
    – user1736
    Feb 4, 2011 at 20:52

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By only in $L^1$ I understand that the function $f$ is in $L^1$ but not in $L^p$ for any $p>1$. Such a function is for instance $f(x)=\min(|x|^{-a},|x|^{-b})$, where $0<a<1<b$ (and $f(0)=0$). Then $\hat f\in L^\infty$, but $\hat f\not\in L^p$ for $1\le p\le2$. This a consequence of the inversion formula and the Hausdorff Young inequality: if $g\in L^p$, $1\le p\le2$, then $\|\hat g\|_q\le C\|g\|_p$ for a constant $C$ independent of $g$ and $p^{-1}+q^{-1}=1$. However, it could still happen $\hat f\in L^p$ for some $p\in(2,\infty)$.

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    $\begingroup$ Haussforg is a rather interesting typo... :) $\endgroup$ Feb 4, 2011 at 13:58
  • $\begingroup$ @Willie: thanks for the comment. I have corrected the typo. $\endgroup$ Feb 4, 2011 at 15:42

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