# $\int_{s = - \infty}^{\infty} Rf(\varphi,s)h(s) ds = \int_{x \in \mathbb{R}^2} f(x) h(\langle \theta, x \rangle) dx$

I want to show the following: Let $$f \in L^1(\mathbb{R^2})$$ and let $$\varphi \in [0,2\pi]$$. Let $$h \in L^{\infty}(\mathbb R)$$. Then $$\int_{s = - \infty}^{\infty} Rf(\varphi,s)h(s) ds = \int_{x \in \mathbb{R}^2} f(x) h(\langle \theta, x \rangle) dx$$, where $$R$$ is the Radon transform. The Radon transform is defined as $$Rf(\varphi,s) = \int_{t = - \infty}^{\infty} f(s\theta + t\theta^{\bot}) dt$$ and $$\theta$$ is defined as $$\theta = \theta(\varphi) = (\cos\varphi, \sin\varphi), \theta^{\bot} = (-\sin\varphi, \cos\varphi)$$.

• Did you define theta ? – Thomas Mar 14 at 9:28
• Use the orthogonal change of coordinates $s = \left<x,\theta\right>$ for the right hand side and use Fubini's theorem. – r9m Mar 15 at 15:41
• @r9m Can you write it out? I am afraid I don't see the solution – Pazu Mar 15 at 15:42

Making the change of variables $$s = \langle x,\theta \rangle$$ and $$t = \langle x,\theta^\perp \rangle$$ so that we have $$x = s\theta + t\theta^\perp$$. Also note that the Jacobian of the transformation is $$1$$, i.e., $$\,dx = \,ds\,dt$$ since it is an orthogonal change of coordinate. Then we may write \begin{align*} \int_{\mathbb{R}^2} f(x) h(\langle \theta, x \rangle) \,dx &= \int_{\mathbb{R}^2} f(s\theta + t\theta^\perp) h(s)\,ds\,dt \\&= \int_{\mathbb{R}} \left(\int_{\mathbb{R}} f(s\theta + t\theta^\perp) \,dt\right) h(s) \,ds \\&= \int_{\mathbb{R}} Rf(\varphi,s)h(s)\,ds\end{align*} where, we used Fubini's theorem in the second line.