The Fourier-slice identity for the Euclidean Radon transform

Fix $\theta$ and define $T(s,t)=s\theta +t{\theta }^{\perp }$ and $F(s,t)=f(T(s,t))$. The columns of $T$ are an orthonormal positively oriented basis, so $T$ is a linear diffeomorphism and $K:=T^{-1}(\mathrm{supp}f)$ is compact. Choose $M>0$ with $K\subset [-M,M{]}^2$. Then $F(s,t)=0$ for $\mathrm{\mid }t\mathrm{\mid }>M$, and $Rf(\theta ,s)={\int }_{-M}^{M}F(s,t)dt$; it vanishes unless $s$ belongs to the compact projection of $K$ onto the first coordinate.

Write $f=f_{\mathrm{R}}+i{f}_{\mathrm{I}}$ and $F_{\nu }=f_{\nu }\circ T$ for $\nu \in \{\mathrm{R},\mathrm{I}\}$. Each $f_{\nu }$ is real smooth; since $DT$ is constant, repeated applications of [F6] give ${\mathrm{\partial }}_s^k{F}_{\nu }(s,t)=(\theta \cdot \mathrm{\nabla }{)}^k{f}_{\nu }(T(s,t))$. Thus, for each integer $k\ge 0$, ${\mathrm{\partial }}_s^{k}F$ is continuous. On every compact $s$-interval, ${\mathrm{\partial }}_s^{k+1}F$ is uniformly continuous on the product with $[-M,M]$. For each real component, apply [F5] on $[0,1]$ to $u\mapsto {\mathrm{\partial }}_s^k{F}_{\nu }(s+uh,t)$. By [F6] its derivative is $h{\mathrm{\partial }}_s^{k+1}{F}_{\nu }(s+uh,t)$, continuous and hence integrable by [F2], with the same one-sided derivatives at the endpoints. Combining the two real equalities and dividing by $h$ gives, for sufficiently small nonzero $h$ of either sign, $\frac{{\mathrm{\partial }}_s^{k}F(s+h,t)-{\mathrm{\partial }}_s^{k}F(s,t)}{h}={\int }_0^1{\mathrm{\partial }}_s^{k+1}F(s+uh,t)du\mathrm{.}$ The right side converges uniformly for $t\in [-M,M]$ to ${\mathrm{\partial }}_s^{k+1}F(s,t)$ as $h\to 0$. Integrating in $t$ therefore proves inductively that $\frac{d^k}{d{s}^k}Rf(\theta ,s)={\int }_{-M}^M{\mathrm{\partial }}_s^{k}F(s,t)dt$ for every $k$, and uniform continuity also makes each derivative continuous in $s$. Thus $Rf(\theta ,\cdot )\in C_c^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})$, so its Fourier transform in [F1] is defined.

Put $H(s,t)=f(T(s,t))e^{-2\pi i\sigma s}$. This is continuous and supported in $K$. Choose a nondegenerate product rectangle containing $K$. By [F2], the real and imaginary parts of $H$ are Riemann integrable there, and all their one-variable sections are continuous and hence Riemann integrable.

By [F1], $\widehat{Rf}(\theta ,\sigma )$ is the iterated integral of $H$, first in $t$ and then in $s$. Apply the real-valued Riemann-Fubini theorem [F3] separately to $\mathrm{Re}H$ and $\mathrm{Im}H$ on the chosen rectangle, then combine the two equalities. Since $H$ vanishes outside that rectangle, this gives $\widehat{Rf}(\theta ,\sigma )={\int }_{{\mathbb{R}}^2}f(s\theta +t{\theta }^{\perp })e^{-2\pi i\sigma s}dsdt\mathrm{.}$

Apply [F4] to $T$ separately for the real and imaginary parts of the compactly supported Riemann integrand in step 3.1. Since $s=x\cdot \theta$ and $\mathrm{\mid }\det DT\mathrm{\mid }=1$, the integral becomes ${\int }_{{\mathbb{R}}^2}f(x)e^{-2\pi i\sigma x\cdot \theta }dx=\widehat{f}(\sigma \theta )$ by [F1]. This is the claimed identity.