Fourier-slice identity for Schwartz inputs

The Schwartz seminorms and the estimate proved in The Radon transform maps Schwartz functions to Schwartz offset data give $\mathrm{\mid }f(s\theta +t{\theta }^{\perp })\mathrm{\mid }\le C_N(1+\mathrm{\mid }s\mathrm{\mid }+\mathrm{\mid }t\mathrm{\mid }{)}^{-N}$ for every $N$, after changing the constant. For $N>2$ this is integrable on ${\mathbb{R}}^2$ in $(s,t)$. The offset integral is absolutely convergent, and [F6] applied to the absolutely integrable function $f(s\theta +t{\theta }^{\perp })e^{-2\pi i\sigma s}$ yields $\widehat{R_{\mathcal{S}}f}(\theta ,\sigma )={\int }_{{\mathbb{R}}^2}f(s\theta +t{\theta }^{\perp })e^{-2\pi i\sigma s}dsdt$. The Lebesgue Fourier-transform definitions in [F3] are applicable by [F8].

Choose a smooth radial cutoff ${\chi }_R$ equal to one on the radius-$R$ ball and supported in the radius-$2R$ ball, using [F4], and put $f_R={\chi }_{R}f$. Then $f_R\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$. The matrix with columns $\theta ,{\theta }^{\perp }$ is a positively oriented orthogonal matrix, so its determinant is one. Apply the Lebesgue change-of-variables formula [F7] to the integrable compactly supported function $x\mapsto f_R(x)e^{-2\pi i\sigma x\cdot \theta }$ and the map $(s,t)\mapsto s\theta +t{\theta }^{\perp }$; its pullback integral is ${\int }_{{\mathbb{R}}^2}{f}_R(s\theta +t{\theta }^{\perp })e^{-2\pi i\sigma s}dsdt=\widehat{f_R}(\sigma \theta )$.

As $R\to \mathrm{\infty }$, $f_R(x)\to f(x)$ and $\mathrm{\mid }{f}_R(x)\mathrm{\mid }\le \mathrm{\mid }f(x)\mathrm{\mid }$. The function $f$ is integrable by [F8], and the same Schwartz estimate in $(s,t)$ gives an integrable majorant for $f_R(s\theta +t{\theta }^{\perp })$. Dominated convergence [F5] passes both integrals in step 1.2 to their limits; step 1.1 identifies the data-side limit with $\widehat{R_{\mathcal{S}}f}(\theta ,\sigma )$, while the spatial-side limit is $\widehat{f}(\sigma \theta )$. This proves the formula.