Restricted Radon data are square-integrable on fixed compact support

Choose $R>0$ with $\mathrm{supp}f\subseteq {\bar{B}}_R(0)$. If $\mathrm{\mid }t\mathrm{\mid }>R$, then $\mathrm{\mid }s\theta +t{\theta }^{\perp }\mathrm{\mid }\ge \mathrm{\mid }t\mathrm{\mid }>R$ and the integrand in [F1] is zero. On any compact set of $(\theta ,s)$, uniform continuity of $f$ on the compact image of $(s,\theta ,t)$ for $\mathrm{\mid }t\mathrm{\mid }\le R$ shows that the line integrals vary continuously: the difference of two such integrals is at most $2R$ times the uniform difference of their integrands. Hence the data function is continuous and measurable.

Put $D=\mathrm{diam}(\mathrm{\Omega })$ and $T_{\theta ,s}=\{t\in \mathbb{R}:s\theta +t{\theta }^{\perp }\in \mathrm{\Omega }\}$. Any two points of $T_{\theta ,s}$ differ by at most $D$, so this open set is contained in an interval of length at most $D$. Cauchy--Schwarz [F2] on $T_{\theta ,s}$, where the constant function $1$ has squared integral at most $D$, gives $\mathrm{\mid }{R}_{\mathcal{S}}f(\theta ,s){\mathrm{\mid }}^2\le D{\int }_{\mathbb{R}}\mathrm{\mid }f(s\theta +t{\theta }^{\perp }){\mathrm{\mid }}^{2}dt$.

The function $(\theta ,s,t)\mapsto \mathrm{\mid }f(s\theta +t{\theta }^{\perp }){\mathrm{\mid }}^2$ is bounded and vanishes unless $\mathrm{\mid }s\mathrm{\mid },\mathrm{\mid }t\mathrm{\mid }\le R$, so it is integrable on the finite-measure set $\mathrm{\Theta }\times [-R,R{]}^2$. Apply [F4] to exchange the iterated integrals, and for each $\theta$ apply [F3] to the orthogonal diffeomorphism $(s,t)\mapsto s\theta +t{\theta }^{\perp }$, whose absolute Jacobian is one. This gives ${\int }_{\mathrm{\Theta }}{\int }_{{\mathbb{R}}^2}\mathrm{\mid }f(s\theta +t{\theta }^{\perp }){\mathrm{\mid }}^{2}dsdtd\theta =\mathrm{\mid }\mathrm{\Theta }\mathrm{\mid }\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2({\mathbb{R}}^2)}^2$. Integrating step 1.2 yields the claimed bound and square-integrability.