Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f \in L^{4/3}(\mathbb{R}^2)$ and denote its Fourier transform by $\mathscr{F}(f)$. Is it true that the function $g:\mathbb{R}^2 \rightarrow \mathbb{C}$ defined by $$g(x)=|x|^{-1}\mathscr{F}(f)(x)$$ is in $L^{4/3}(\mathbb{R}^2)$ also?

Simply appealing to Hausdorff-Young and Hölder's inequality doesn't suffice.

share|improve this question
1  
I'm pretty sure one can do it with the Marcinkiewicz interpolation theorem. (Or is this overkill?) –  Hendrik Vogt Jun 27 '11 at 11:44
1  
@Hendrik: I am not quite seeing it. What are you proposing to use as the end points? –  Willie Wong Jun 27 '11 at 13:37
2  
@Willie: $4/3-\varepsilon$ and $4/3+\varepsilon$, for a suitable $\varepsilon>0$. The point is that $x\mapsto|x|^{-1}$ is in the weak $L_2$ space. –  Hendrik Vogt Jun 27 '11 at 13:52
1  
(Egads, my interpolation theory is rusty.) Of course: $f\mapsto g$ is $L^1\to L^{2,w}$ and $L^2 \to L^{1,w}$. I was looking at the scale the wrong way and forgot to dualize. –  Willie Wong Jun 27 '11 at 14:39
1  
@Willie: I took the freedom to use your end points in my answer :-) –  Hendrik Vogt Jun 27 '11 at 15:18
add comment

1 Answer

As I wrote in the comments, this can be proved with the Marcinkiewicz interpolation theorem. Here's the main steps:

It's well-known that $\mathscr{F}\colon L^1(\mathbb R^2)\to L^\infty(\mathbb R^2)$ and $\mathscr{F}\colon L^2(\mathbb R^2)\to L^2(\mathbb R^2)$ is bounded. Moreover, the function $x\mapsto|x|^{-1}$ lies in the weak $L^2$ space $L^{2,\rm w}(\mathbb R^2)$. For the operator $T$ defined by $$ Tf(x) := |x|^{-1}\mathscr{F}(f)(x) $$ we thus obtain that $T: L^1(\mathbb R^2)\to L^{2,\rm w}(\mathbb R^2)$ and $T: L^2(\mathbb R^2)\to L^{1,\rm w}(\mathbb R^2)$. In other words, $T\,$ is of weak type $(1,2)$ and of weak type $(2,1)$. Now the Marcinkiewicz interpolation theorem yields: $\,T\,$ is of strong type $(p,q)$ (i.e., $T: L^p(\mathbb R^2)\to L^q(\mathbb R^2)$ is bounded) if $p$ and $q$ are such that $p\le q$ and $$ \frac1p = \frac\theta1 + \frac{1-\theta}2 = \frac{1+\theta}2, \quad \frac1q = \frac\theta2 + \frac{1-\theta}1 = 1-\frac\theta2, $$ where $\theta\in(0,1)$. For $\theta=\frac12$ we obtain $p=q=\frac43$. Thus, in particular, $g=Tf \in L^{4/3}(\mathbb{R}^2)$ for all $f \in L^{4/3}(\mathbb{R}^2)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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