# Fourier transform, domains, ranges, L^p-spaces

The Schwartz space on $\mathbb{R}^d$ is defined to be

$$S(\mathbb{R}^d) := \{f\in C^{\infty}(\mathbb{R}^d,\mathbb{C})\;|\; \|f\|_{S,N}<\infty\;\forall N\in\{0,1,2,3,\ldots\}\}$$

where

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

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

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

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.