- #1
jostpuur
- 2,112
- 19
The Schwartz space on \mathbb{R}^d is defined to be
<br /> S(\mathbb{R}^d) := \{f\in C^{\infty}(\mathbb{R}^d,\mathbb{C})\;|\; \|f\|_{S,N}<\infty\;\forall N\in\{0,1,2,3,\ldots\}\}<br />
where
<br /> \|f\|_{S,N} := \underset{|\alpha|,|\beta|\leq N}{\textrm{max}}\;\underset{x\in\mathbb{R}^d}{\textrm{sup}}\; |x^{\alpha}\partial^{\beta}f(x)|.<br />
Alpha and beta are multi-indexes. It turns out, that when Fourier transform is defined on this space, with the integral formula, one obtains a continuous mapping \mathcal{F}:S(\mathbb{R}^d)\to S(\mathbb{R}^d). Since the Schwartz space is dense in L^p(\mathbb{R}^d), 1\leq p < \infty, it is possible to obtain a continuous extension of the Fourier transform onto the L^p(\mathbb{R}^d) too. I have not seen explicit counter examples yet, but I've heard that one cannot define the Fourier transform directly with the integral formula in L^p when p>1.
My question deals with the range of the Fourier transform. Am I correct to guess, that we have
<br /> \mathcal{F}(L^p)=L^q<br />
with 1/p+1/q=1? It is a known result, that \mathcal{F}(L^2)=L^2. It is also easy to show for example that \|\mathcal{F}f\|_{\infty}\leq \|f\|_1, but I'm not sure how to show equality. \mathcal{F}(L^p)=L^q would seem plausible result, but I couldn't find it with a quick skim over the Rudin's Fourier transform chapter at least.
<br /> S(\mathbb{R}^d) := \{f\in C^{\infty}(\mathbb{R}^d,\mathbb{C})\;|\; \|f\|_{S,N}<\infty\;\forall N\in\{0,1,2,3,\ldots\}\}<br />
where
<br /> \|f\|_{S,N} := \underset{|\alpha|,|\beta|\leq N}{\textrm{max}}\;\underset{x\in\mathbb{R}^d}{\textrm{sup}}\; |x^{\alpha}\partial^{\beta}f(x)|.<br />
Alpha and beta are multi-indexes. It turns out, that when Fourier transform is defined on this space, with the integral formula, one obtains a continuous mapping \mathcal{F}:S(\mathbb{R}^d)\to S(\mathbb{R}^d). Since the Schwartz space is dense in L^p(\mathbb{R}^d), 1\leq p < \infty, it is possible to obtain a continuous extension of the Fourier transform onto the L^p(\mathbb{R}^d) too. I have not seen explicit counter examples yet, but I've heard that one cannot define the Fourier transform directly with the integral formula in L^p when p>1.
My question deals with the range of the Fourier transform. Am I correct to guess, that we have
<br /> \mathcal{F}(L^p)=L^q<br />
with 1/p+1/q=1? It is a known result, that \mathcal{F}(L^2)=L^2. It is also easy to show for example that \|\mathcal{F}f\|_{\infty}\leq \|f\|_1, but I'm not sure how to show equality. \mathcal{F}(L^p)=L^q would seem plausible result, but I couldn't find it with a quick skim over the Rudin's Fourier transform chapter at least.
