Inverse Radon Transform of a function in the Schwartz class

Question

This question comes from reading through Stein and Shakarchi's Fourier Analysis, page 206.

Consider the two Schwartz spaces $\mathcal{S}(\mathbb{R}^3)$ and $\mathcal{S}(\mathbb{R}\times S^2)$, where by the latter space we mean the space of all continuous functions $F(t,\gamma)$ that are indefinitely differentiable in $t$, with continuous derivatives on $\mathbb{R}\times S^2$, and that satisfy $$\sup_{t\in\mathbb{R},\,\gamma\in S^2} |t|^k\left|\frac{\partial^\ell F}{\partial t^\ell}(t,\gamma)\right| < \infty \quad\text{for all integers $k,\ell\geq 0$.}$$

The Radon transform $\mathcal{R}:\mathcal{S}(\mathbb{R}^3)\to\mathcal{S}(\mathbb{R}\times S^2)$ is defined by $$\mathcal{R}(f)(t,\gamma) = \int_{\mathcal{P}_{t,\gamma}} f$$ where $\mathcal{P}_{t,\gamma}=\{x\in\mathbb{R}^3:x\cdot\gamma=t\}$ is the plane in $\mathbb{R}^3$ passing through and perpendicular to $t\gamma$. The book shows that $\mathcal{R}(f)\in\mathcal{S}(\mathbb{R}\times S^2)$ when $f\in\mathcal{S}(\mathbb{R}^3)$.

Given $F\in\mathcal{S}(\mathbb{R}\times S^2)$, define its dual Radon transform by $$\mathcal{R}^*(F)(x) = \int_{S^2} F(x\cdot\gamma,\gamma)\,d\sigma(\gamma).$$ This definition basically integrates $F$ over all planes passing through $x$.

The book asserts that $\mathcal{R}^*$ maps $\mathcal{S}(\mathbb{R}\times S^2)$ into $\mathcal{S}(\mathbb{R}^3)$ and does not provide a proof. I have tried without success to check this. What I need to show is that $$\sup_{x\in\mathbb{R}^3} |x|^k\left|\left(\frac{\partial}{\partial x}\right)^\alpha\mathcal{R}^*(F)(x)\right| < \infty \quad\text{for all integers $k\geq 0$ and multi-indices $\alpha$.}$$ Now \begin{align*} |x|^k\left|\left(\frac{\partial}{\partial x}\right)^\alpha\mathcal{R}^*(F)(x)\right| &= |x|^k\left|\int_{S^2} \frac{\partial^{|\alpha|}F}{\partial t^{|\alpha|}}(x\cdot\gamma,\gamma)\gamma^\alpha\,d\sigma(\gamma)\right| \\ &\leq \int_{S^2} |x|^k\left|\frac{\partial^{|\alpha|}F}{\partial t^{|\alpha|}}(x\cdot\gamma,\gamma)\right|\,d\sigma(\gamma). \end{align*} Here I get stuck. It seems the basic problem I cannot get around is that no matter how large $x$ is, there are always planes passing through $x$ which go through or near the origin where $F$ and its derivatives can be large. So I think I need to split the integral up and estimate.

Fix $x$ (we can assume $|x|\geq 1$) and choose $\eta\in S^2$ so that $x=s\eta$, with $s>0$. We can split $S^2$ into a "band" where $|\gamma\cdot\eta|\leq\sin\alpha$ (thinking of $\alpha$ as small and to be determined later—integrating over the band integrates over the planes through $x$ which pass through or near the origin) and two "caps" where $|\gamma\cdot\eta|\geq\sin\alpha$. I then want to write $$\int_{\text{band}} |x|^k\left|\frac{\partial^{|\alpha|}F}{\partial t^{|\alpha|}}(x\cdot\gamma,\gamma)\right|\,d\sigma(\gamma) \leq C|x|^k\sin\alpha$$ (here I use the facts that the area of the band is $4\pi\sin\alpha$ and the derivatives of $F$ are bounded) and \begin{align*} \int_{\text{caps}} |x|^k\left|\frac{\partial^{|\alpha|}F}{\partial t^{|\alpha|}}(x\cdot\gamma,\gamma)\right|\,d\sigma(\gamma) &= \int_{\text{caps}} \frac{|x|^k}{|x\cdot\gamma|^k}|x\cdot\gamma|^k\left|\frac{\partial^{|\alpha|}F}{\partial t^{|\alpha|}}(x\cdot\gamma,\gamma)\right|\,d\sigma(\gamma) \\ &\leq \frac{D}{(\sin\alpha)^k} \end{align*} (here I use the fact that $F\in\mathcal{S}(\mathbb{R}\times S^2)).$

There is no way to choose $\alpha$ so that both terms are bounded in $x$. But I don't see any way to improve my estimations. Or maybe I am approaching this in an entirely wrong way?

Update, 08/10/15: I haven't made any more progress on this problem, can anyone help?

Answer 1

The reason you have trouble showing the claim is that it is actually false.

To see this, we let $$ f:\mathbb{R}\times S^{2}\to\mathbb{R},\left(x,\gamma\right)\mapsto e^{-x^{2}}. $$ Then $f$ is certainly a Schwartz function in your sense, but we will see that $\mathcal{R}^{\ast}f$ is not. To this end, note with $e_{3}=\left(0,0,1\right)$ that we have $$ \left(\mathcal{R}^{\ast}f\right)\left(te_{3}\right)=\int_{S^{2}}f\left(\left\langle te_{3},\gamma\right\rangle ,\gamma\right)\,{\rm d}\sigma\left(\gamma\right)=\int_{S^{2}}e^{-t^{2}\gamma_{3}^{2}}\,{\rm d}\sigma\left(\gamma\right). $$ We will now use the well-known Layer-Cake formula (see $L^p$-norm of a non-negative measurable function for $p=1$) to compute this integral. To this end, note for $\alpha>0$ and $t\neq0$ that \begin{eqnarray*} M_{\alpha} & := & \left\{ \gamma\in S^{2}\,\mid\, e^{-t^{2}\gamma_{3}^{2}}>\alpha\right\} \\ & = & \left\{ \gamma\in S^{2}\,\mid\,-t^{2}\gamma_{3}^{2}>\ln\alpha\right\} \\ & = & \left\{ \gamma\in S^{2}\,\mid\,\gamma_{3}^{2}<\frac{-\ln\alpha}{t^{2}}\right\} \\ & = & \begin{cases} \left\{ \gamma\in S^{2}\,\mid\,\left|\gamma_{3}\right|<\frac{\sqrt{-\ln\alpha}}{t}\right\} , & \text{if }\alpha\in\left(0,1\right),\\ \emptyset, & \text{if }\alpha\geq1. \end{cases} \end{eqnarray*} Thus, the Layer-Cake formula yields $$ \int_{S^{2}}e^{-t^{2}\gamma_{3}^{2}}\,{\rm d}\sigma\left(\gamma\right)=\int_{0}^{\infty}\sigma\left(M_{\alpha}\right)\,{\rm d}\alpha=\int_{0}^{1}\sigma\left(\left\{ \gamma\in S^{2}\,\mid\,\left|\gamma_{3}\right|<\frac{\sqrt{-\ln\alpha}}{t}\right\} \right)\,{\rm d}\alpha, $$ where $\sigma$ denotes the usual surface measure/Hausdorff measure on the unit sphere.

Our next step is thus to compute the surface measure of the spherical cap $$ C_{\lambda}:=\left\{ \gamma\in S^{2}\,\mid\,\left|\gamma_{3}\right|<\lambda\right\} $$ for $0<\lambda<1$. Using symmetry of the surface measure, it is not hard to see $\sigma\left(C_{\lambda}\right)=2\sigma\left(C_{\lambda}^{+}\right)$ with $$ C_{\lambda}^{+}=\left\{ \gamma\in S^{2}\,\mid\,0<\gamma_{3}<\lambda\right\} . $$

To compute the surface measure of this cap, we will use the parametrization $$ \Phi:B_{1}\left(0\right)\subset\mathbb{R}^{2}\to S^{2}\cap\left(\mathbb{R}^{2}\times\left(0,\infty\right)\right),x\mapsto\left(x,\:\sqrt{1-\left|x\right|^{2}}\right). $$ I leave it to you to verify that the Gramian determinant of $\Phi$ is given by $$ g\left(x\right)=\sqrt{\det\left(\left[D\Phi\left(x\right)\right]^{T}\cdot D\Phi\left(x\right)\right)}=\frac{1}{\sqrt{1-\left|x\right|^{2}}}. $$ Note that we have \begin{eqnarray*} \Phi^{-1}\left(C_{\lambda}^{+}\right) & = & \left\{ x\in B_{1}\left(0\right)\subset\mathbb{R}^{2}\,\mid\,\sqrt{1-\left|x\right|^{2}}<\lambda\right\} \\ & = & \left\{ x\in B_{1}\left(0\right)\subset\mathbb{R}^{2}\,\mid\,1-\left|x\right|^{2}<\lambda^{2}\right\} \\ & = & \left\{ x\in B_{1}\left(0\right)\subset\mathbb{R}^{2}\,\mid\,1-\lambda^{2}<\left|x\right|^{2}\right\} \\ & = & \left\{ x\in B_{1}\left(0\right)\subset\mathbb{R}^{2}\,\mid\,\sqrt{1-\lambda^{2}}<\left|x\right|\right\} . \end{eqnarray*} Thus, using polar coordinates (we have an integral in $\mathbb{R}^{2}$), we get \begin{eqnarray*} \sigma\left(C_{\lambda}^{+}\right) & = & \int_{\Phi^{-1}\left(C_{\lambda}^{+}\right)}g\left(x\right)\,{\rm d}x\\ & = & \int_{\sqrt{1-\lambda^{2}}<\left|x\right|<1}\frac{1}{\sqrt{1-\left|x\right|^{2}}}\,{\rm d}x\\ & = & 2\pi\cdot\int_{\sqrt{1-\lambda^{2}}}^{1}r\cdot\frac{1}{\sqrt{1-r^{2}}}\,{\rm d}r\\ & \overset{\omega=r^{2}}{=} & \pi\cdot\int_{1-\lambda^{2}}^{1}\frac{1}{\sqrt{1-\omega}}\,{\rm d}\omega\\ & \overset{\gamma=1-\omega}{=} & \pi\cdot\int_{0}^{\lambda^{2}}\gamma^{-1/2}\,{\rm d}\gamma\\ & = & \pi\cdot\frac{\gamma^{1/2}}{1/2}\bigg|_{0}^{\lambda^{2}}=2\pi\cdot\lambda \end{eqnarray*} and thus $$ \sigma\left(C_{\lambda}\right)=4\pi\cdot\lambda\text{ for }0<\lambda<1. $$ For $\lambda\geq1$, we have $$ C_{\lambda}=\left\{ \gamma\in S^{2}\,\mid\,\left|\gamma_{3}\right|<\lambda\right\} =S^{2} $$ and hence $$ \sigma\left(C_{\lambda}\right)=\begin{cases} 4\pi\cdot\lambda, & \text{if }0<\lambda<1,\\ 4\pi, & \text{if }\lambda\geq1. \end{cases} $$

Hence, \begin{eqnarray*} \sigma\left(M_{\alpha}\right) & = & \begin{cases} \sigma\left(C_{\frac{1}{t}\cdot\sqrt{-\ln\alpha}}\right), & \text{if }\alpha\in\left(0,1\right),\\ 0, & \text{if }\alpha\geq1, \end{cases}\\ & = & \begin{cases} 4\pi\cdot\frac{\sqrt{-\ln\alpha}}{t}, & \text{if }e^{-t^{2}}<\alpha<1,\\ 4\pi, & \text{if }0<\alpha\leq e^{-t^{2}}\\ 0, & \text{else}, \end{cases} \end{eqnarray*} which finally yields for $t>0$ that \begin{eqnarray*} \left(\mathcal{R}^{\ast}f\right)\left(te_{3}\right)=\int_{S^{2}}e^{-t^{2}\gamma_{3}^{2}}\,{\rm d}\sigma\left(\gamma\right) & = & \int_{0}^{e^{-t^{2}}}4\pi\,{\rm d}\alpha+\frac{4\pi}{t}\cdot\int_{e^{-t^{2}}}^{1}\sqrt{-\ln\alpha}\,{\rm d}\alpha\\ & = & \frac{4\pi}{e^{t^{2}}}+\frac{4\pi}{t}\cdot\int_{e^{-t^{2}}}^{1}\sqrt{-\ln\alpha}\,{\rm d}\alpha. \end{eqnarray*} Hence, \begin{eqnarray*} t\cdot\left(\mathcal{R}^{\ast}f\right)\left(te_{3}\right) & = & 4\pi\cdot\frac{t}{e^{t^{2}}}+4\pi\cdot\int_{e^{-t^{2}}}^{1}\sqrt{-\ln\alpha}\,{\rm d}\alpha\\ & \xrightarrow[t\to\infty]{} & 4\pi\cdot\int_{0}^{1}\sqrt{-\ln\alpha}\,{\rm d}\alpha\\ & \overset{\gamma=-\ln\alpha}{=} & 4\pi\cdot\int_{0}^{\infty}\sqrt{\gamma}\cdot e^{-\gamma}\,{\rm d}\gamma\\ & = & 4\pi\cdot\int_{0}^{\infty}\gamma^{\frac{3}{2}-1}\cdot e^{-\gamma}\,{\rm d}\gamma\\ & = & 4\pi\cdot\Gamma\left(\frac{3}{2}\right)\\ & = & 4\pi\cdot\frac{\sqrt{\pi}}{2}=2\cdot\pi^{3/2}, \end{eqnarray*} where $\Gamma$ denotes the usual Gamma function (see https://en.wikipedia.org/wiki/Gamma_function and https://en.wikipedia.org/wiki/Particular_values_of_the_Gamma_function).

But this shows $$ \left(\mathcal{R}^{\ast}f\right)\left(te_{3}\right)\sim\frac{2\cdot\pi^{3/2}}{t}\text{ as }t\to\infty, $$ so that $\mathcal{R}^{\ast}f$ is no Schwartz function.