- #1

- 2,112

- 18

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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\}\}

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where

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\|f\|_{S,N} := \underset{|\alpha|,|\beta|\leq N}{\textrm{max}}\;\underset{x\in\mathbb{R}^d}{\textrm{sup}}\; |x^{\alpha}\partial^{\beta}f(x)|.

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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 [itex]\mathcal{F}:S(\mathbb{R}^d)\to S(\mathbb{R}^d)[/itex]. Since the Schwartz space is dense in [itex]L^p(\mathbb{R}^d)[/itex], [itex]1\leq p < \infty[/itex], it is possible to obtain a continuous extension of the Fourier transform onto the [itex]L^p(\mathbb{R}^d)[/itex] 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 [itex]L^p[/itex] when [itex]p>1[/itex].

My question deals with the range of the Fourier transform. Am I correct to guess, that we have

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\mathcal{F}(L^p)=L^q

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with [itex]1/p+1/q=1[/itex]? It is a known result, that [itex]\mathcal{F}(L^2)=L^2[/itex]. It is also easy to show for example that [itex]\|\mathcal{F}f\|_{\infty}\leq \|f\|_1[/itex], but I'm not sure how to show equality. [itex]\mathcal{F}(L^p)=L^q[/itex] would seem plausible result, but I couldn't find it with a quick skim over the Rudin's Fourier transform chapter at least.