[proofplan] We write the Radon kernel as an oscillatory integral with phase $\phi(\omega,s,x,\lambda)=\lambda(x\cdot\omega-s)$. The critical equation in the frequency variable $\lambda$ gives precisely the incidence relation $x\cdot\omega=s$. The canonical relation is then obtained by differentiating the phase in the output variables $(\omega,s)$ and in the input variable $x$, with the input covector carrying the standard kernel sign $-d_x\phi$. Finally, the zero frequency is removed because a Fourier integral canonical relation is conic over nonzero phase frequencies. [/proofplan]

[step:Represent the Radon kernel by the incidence phase] Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$, and let $\mathbb{R}_0=\mathbb{R}\setminus\{0\}$. Let $\phi: S^1\times \mathbb{R}\times \mathbb{R}^2\times \mathbb{R}\to \mathbb{R}$ be the phase function defined by $\phi(\omega,s,x,\lambda)=\lambda(x\cdot\omega-s)$. With the stated normalization convention suppressed to the harmless factor $(2\pi)^{-1}$, the Schwartz kernel of $R$ is represented by \begin{align*} K_R(\omega,s,x)=\frac{1}{2\pi}\int_{\mathbb{R}} e^{i\lambda(x\cdot\omega-s)}\,d\mathcal{L}^1(\lambda). \end{align*} Let $\delta$ denote the Dirac delta distribution on $\mathbb{R}$. Equivalently, this is the distributional identity for $\delta$: \begin{align*} \delta(t)=\frac{1}{2\pi}\int_{\mathbb{R}} e^{i\lambda t}\,d\mathcal{L}^1(\lambda). \end{align*} with $t=x\cdot\omega-s$. Thus the relevant output variable is $y=(\omega,s)\in S^1\times\mathbb{R}$, the input variable is $x\in\mathbb{R}^2$, and the phase parameter is $\lambda\in\mathbb{R}$. The incidence function $F:S^1\times\mathbb{R}\times\mathbb{R}^2\to\mathbb{R}$ defined by $F(\omega,s,x)=x\cdot\omega-s$ is a submersion because $d_sF=-1$, so the phase is a clean homogeneous phase for the incidence hypersurface. The multiplicative constant and density normalization do not affect the canonical relation, since the canonical relation is determined by the phase derivatives on the nonzero-frequency critical set. [/step]

[step:Impose the critical equation in the phase variable]The phase derivative in the parameter variable is $\partial_\lambda \phi(\omega,s,x,\lambda)=x\cdot\omega-s$. Therefore the critical set in $S^1\times\mathbb{R}\times\mathbb{R}^2\times\mathbb{R}_0$ is $\Sigma_\phi=\{(\omega,s,x,\lambda): x\cdot\omega=s,\ \lambda\in\mathbb{R}_0\}$. The restriction $\lambda\ne 0$ is the usual removal of the zero section in the homogeneous phase variable.[/step]

[guided]The phase variable is $\lambda$, so the first task is to find the points where the oscillation is stationary with respect to $\lambda$. We compute the derivative of $\phi(\omega,s,x,\lambda)=\lambda(x\cdot\omega-s)$ with respect to $\lambda$: $\partial_\lambda\phi(\omega,s,x,\lambda)=x\cdot\omega-s$. The critical condition $\partial_\lambda\phi=0$ is therefore exactly $x\cdot\omega=s$. This is the incidence relation saying that the point $x\in\mathbb{R}^2$ lies on the oriented line with normal direction $\omega\in S^1$ and signed distance parameter $s\in\mathbb{R}$. Hence the phase-critical set, after excluding the zero frequency, is $\Sigma_\phi=\{(\omega,s,x,\lambda)\in S^1\times\mathbb{R}\times\mathbb{R}^2\times\mathbb{R}_0 : x\cdot\omega=s\}$. The exclusion of $\lambda=0$ is essential because the canonical relation of a Fourier integral operator lives in cotangent bundles with their zero sections removed; at $\lambda=0$ all covectors produced by differentiating this phase would vanish.[/guided]

[step:Differentiate in the output variables to obtain $\eta_\omega$ and $\eta_s$] Fix $(\omega,s,x,\lambda)\in\Sigma_\phi$. The derivative in the real variable $s$ is $\partial_s\phi(\omega,s,x,\lambda)=-\lambda$, so the $ds$-component of the output covector is $\eta_s=-\lambda$. For the $S^1$ component, let $v\in T_\omega S^1$. The differential of $\phi$ in the $\omega$ direction is $d_\omega\phi(\omega,s,x,\lambda)(v)=\lambda\,x\cdot v$. Thus $\eta_\omega=\lambda\,x|_{T_\omega S^1}\in T_\omega^*S^1$, where $x|_{T_\omega S^1}$ denotes the covector $v\mapsto x\cdot v$ on $T_\omega S^1$. [/step]

[step:Differentiate in the input variable and apply the kernel sign convention] The derivative in the input variable $x\in\mathbb{R}^2$ is the covector $d_x\phi(\omega,s,x,\lambda)(h)=\lambda\,\omega\cdot h$ for every $h\in T_x\mathbb{R}^2\cong\mathbb{R}^2$. Under the stated Fourier integral operator convention for kernels, the input covector is $-d_x\phi$. Therefore $\xi=-\lambda\omega\in T_x^*\mathbb{R}^2\cong\mathbb{R}^2$. [/step]

[step:Assemble the homogeneous canonical relation] Combining the critical equation and the covector computations gives $C_R=\{((\omega,s;\eta_\omega,\eta_s),(x;\xi)) : x\cdot\omega=s,\ \eta_\omega=\lambda\,x|_{T_\omega S^1},\ \eta_s=-\lambda,\ \xi=-\lambda\omega,\ \lambda\in\mathbb{R}_0\}$. This is exactly the canonical relation in $T^*(S^1\times\mathbb{R})\times T^*\mathbb{R}^2$. The parameter $\lambda\in\mathbb{R}_0$ makes the relation homogeneous and excludes the zero covectors, completing the proof. [/step]