6
$\begingroup$

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 $1 \leq p < \infty$).

$\endgroup$
  • 2
    $\begingroup$ 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 '11 at 7:44
  • 1
    $\begingroup$ 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 '11 at 20:52
4
$\begingroup$

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 $p\in(2,\infty)$.

$\endgroup$
  • 2
    $\begingroup$ Haussforg is a rather interesting typo... :) $\endgroup$ – Willie Wong Feb 4 '11 at 13:58
  • $\begingroup$ @Willie: thanks for the comment. I have corrected the typo. $\endgroup$ – Julián Aguirre Feb 4 '11 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.