Radon transform on Schwartz functions

For $f\in \mathcal{S}({\mathbb{R}}^2)$, $\theta =({\theta }_1,{\theta }_2)\in S^1$, $s\in \mathbb{R}$, and ${\theta }^{\perp }=(-{\theta }_2,{\theta }_1)$, define $R_{\mathcal{S}}f(\theta ,s)={\int }_{\mathbb{R}}f(s\theta +t{\theta }^{\perp })dt$, where the absolutely convergent integral is with respect to one-dimensional Lebesgue measure. Here $\mathcal{S}({\mathbb{R}}^2)$ is the space of smooth complex functions for which every seminorm ${\sup }_{x\in {\mathbb{R}}^2}\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\beta }f(x)\mathrm{\mid }$, for multi-indices $\alpha ,\beta$, is finite. The parameter $\theta$ is the unit normal to the line, so $(\theta ,s)$ and $(-\theta ,-s)$ represent the same unoriented line. Under the Axiom of Countable Choice, the Lebesgue integral in this definition agrees on compactly supported smooth inputs with the compact-support Riemann line integral, so the normalization is the same.

Compactly supported smooth domain, line normal, and normalization; current body inspected

Schwartz space and its seminorms

Countable Choice for the Lebesgue/Riemann integral agreement

Agreement of bounded Riemann and Lebesgue integrals under Countable Choice

Mikko Salo, On geometric inverse problems and microlocal analysis, §2.1

Venkataswaran P. Krishnan and Eric Todd Quinto, Microlocal Analysis in Tomography, §1.3.2