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Let $f \in L^1(\mathbb R^n)$ (or in case it helps, actually a probability density on $\mathbb R^n$). Define the Radon transform $R[f]:S_{n-1} \times \mathbb R \to \mathbb R$ of $f$ by

$$ R[f](w,b) := \int_{\mathbb R^n}\delta(x^\top w - b)f(x)\,dx = \int_{H_{w,b}}f(x)\,d\sigma(x), $$

where $S_{n-1}$ is the unit-sphere in $\mathbb R^n$ and $d\sigma$ is the areal measure on the hyperplane $H_{w,b}:=\{x \in \mathbb R^n \mid x^\top w = b\}$.

Question. Under what minimal additional conditions on $f$ is mapping $w \mapsto R[f](w,b)$ continuous continuous on $S_{n-1}$, for each $b \in \mathbb R$ ?

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  • $\begingroup$ Forget about continuity for a moment and focus on integrability: are you sure that $f \in L^1 (\mathbb R^n)$ is enough to deduce that $f | _{H_{w,b}} \in L^1 (H_{w,b})$ for all $w$ and $b$? Here is a counterexample: for some $w$ and $b$, let $g : H_{w,b} \to \mathbb R$ be measurable but not integrable. Extend it by $0$ to the whole $\mathbb R^n$ and call this extension $f$. Since $H_{w,b}$ is of codimension $1$, it is a null set, therefore $f=0 \in L^1 (\mathbb R^n)$. Nevertheless, $f |_{H_{w,b}} = g \notin L^1 (H_{w,b})$. So you need assumptions for integrability first. Continuity comes next. $\endgroup$
    – Alex M.
    May 16 at 5:51
  • $\begingroup$ Having $f$ continuous with compact support would fix integrability, but this is a very strong assumption that you may not like. $\endgroup$
    – Alex M.
    May 16 at 5:52
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    $\begingroup$ Indeed, the literature on Radon transform is really nasty. After some combing through the internet, it appears the "correct" domain of the Radon transform is not $L^1(\mathbb R^n)$ but $\mathcal S(\mathbb R^n)$, the Schwarz space on $\mathbb R^n$. These are functions whose derivatives of all orders decrease faster than any polynomial. In particular, this includes $\mathcal C_c^\infty(\mathbb R^n)$, i.e your proposal. $\endgroup$
    – dohmatob
    May 16 at 7:04
  • $\begingroup$ @AlexM. While your $g$ is non-measurable with respect to the induced measure on $H_{w,b}$ it is measurable on $\mathbb{R}^n$ (it is zero a.e.). In think that integrability of $f$ restricted to $H_{w,b}$ follows from Fubini, no? $\endgroup$
    – Dirk
    May 16 at 11:47
  • $\begingroup$ In 2D one can write the Radon transform down as a 1D integral depending on the angle and by the standard theorem on continuity of integrals w.r.t. parameters, you get continuity of the Radon transform $Rf$ w.r.t. the angle if $f$ is continuous. I think a similar reasoning should work in higher dimensions. $\endgroup$
    – Dirk
    May 16 at 12:29

2 Answers 2

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This is only a comment (but I amn't entitled) and applies also to your related question. Your formula involves three operations--composition, multiplication and integration. You finally ask when Euler type results hold--when do smoothness properties of the integrand carry over to the function defined by the integral.

In order to address your question, it is necessary to clarify the precise definitions in use. The presence of the $\delta$ function clearly implies that one has to work in the context of distribution theory--fortunately the requisite (elementary) theory has been around since the 60's. First composition: you are composing the Dirac function with a very regular function so no problem there. The product does, however, require more than integrability from $f$, say continuity. You then have to interpret the integral as a parametrised one in the sense of distributions but this theory is on record (J. Sebastião e Silva). Now comes the good news: any sensible version of Euler-type theorems (differentiating under the integral sign) holds without further restrictions.

Added after comments. Avoiding distributions just begs the question. Or do you really think you can integrate an $L^1$ function along a hyperspace? And in what sense are you carrying out the final integral? Given a simple and flexible notion of the parametrised integral of distributions which supplies existence without further restrictions and the Göttergeschenk of Euler theorems for free, why would you spurn it?

I will leave (and given the tone of the comments, leave means leave) with my wee conjecture: if $f$ is a continuous integrable function, then your expression is infinitely differentiable with respect to $w$ and the derivatives can be calculated by differentiating under the integral sign.

I am formulating this as a conjecture since I haven't had an opportunity to sit down and write out the details.

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    $\begingroup$ No distribution theory is needed here. What the OP means is to integrate $f$ on the hyperplane given by $\langle x, w \rangle = b$. $\endgroup$
    – Alex M.
    May 15 at 11:39
  • $\begingroup$ @terceira Thanks for the input. I second the comment of Alex M (ive also added explicit reformulation without Dirac delta distribution). Of course i'd be happy to hear that some powerful result from distribution theory (or anywhere else for that matter) allows one to deduce continuity... $\endgroup$
    – dohmatob
    May 15 at 12:00
  • $\begingroup$ @terceira Indeed, $L^1(\mathbb R^d)$ appears to not be the appropriate domain for the Radon transform; the Schwarz space $\mathcal S(\mathbb R^d)$ is. Also, $\mathcal S(\mathbb R^d)$ appears to be the dual space of the space of tempered distributions on $\mathbb R^d$. In this sense, you are right that the correct framework for studying my question is the theory of distributions. Please ignore my previous comment. $\endgroup$
    – dohmatob
    May 16 at 9:02
  • $\begingroup$ @terceira The comments where not meant to be offensive. I hope no offense was taken. Thanks again for the kind input. BTW, welcome to MO! $\endgroup$
    – dohmatob
    May 16 at 9:09
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    $\begingroup$ Integrating over a hyperplane can be done by Fubini (rotate and shift the space such that the hyperplane is a coordinate plane). $\endgroup$
    – Dirk
    May 16 at 11:49
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In the following I shall assume that $f$ is continuous with compact support.

It is known that if $\varphi : \mathbb R^n \to \mathbb R$ has $\nabla \varphi \ne 0$ at all the points of $H = \varphi^{-1} (0)$, then $$\int _{\mathbb R^n} f(v) \, \delta (\varphi(v)) \, \mathrm d v = \int _H \frac{f(s)} {\|\nabla \varphi(s)\|} \, \mathrm d s \ ,$$ the latter integral being with respect to the natural measure induced on $H$.

In your case $\varphi(x) = \langle w, x \rangle - b$, so $\| \nabla \varphi \| = \| w \| = 1$.

Since $w \ne 0$, let us assume that $w_n \ne 0$. Consider the (global) parametrization $h : \mathbb R^{n-1} \to H_{w,b}$ given by $$h(u_1, \dots, u_{n-1}) = \left( u_1, \dots, u_{n-1}, \frac 1 {w_n} (b - u_1 \, w_1 - \dots - u_{n-1} \, w_{n-1}) \right) .$$

In this paramerization, the coefficients of the Riemannian metric on $H_{w,b}$ are $g_{ii} = 1 + \frac {w_i ^2} {w_n ^2}$ and $g_{ij} = \frac {w_i w_j} {w_n ^2}$ for $i < j$, so your integral becomes $$\sqrt {\det (g_{ij})} \int _{\mathbb R ^{n-1}} f \left( u_1, \dots, u_{n-1}, \frac 1 {w_n} (b - u_1 \, w_1 - \dots - u_{n-1} \, w_{n-1}) \right) \ \mathrm d u_1 \dots \mathrm d u_{n-1} \ .$$

Since $f$ is continuous with compact support, the integral written above exists and is finite. Using Lebesgue's dominated convergence theorem, if the sequence $w^{(k)}$ converges to $w$ then you also get the convergence of the corresponding integrals.

The determinant, too, is continuous on some neighbourhood of some given $w$: if the coordinate $w_n$ is non-zero, then it is so on some neighbourhood $U$ of $w$ in $S^{n-1}$ (by the continuity of the coordinate functions), so you may safely divide by it; since all the operations involved in that determinant al algebraic, the determinant will depend continuously on $w$ in the neighbourhood $U$. Since continuity is a local property, this is enough.

To conclude: $f$ being continuous with compact support is enough for the continuity of its Radon transform (with respect to $w$ and $b$ jointly); it is up to you to decide whether this is too restrictive or not. This condition (continuity and compact support) is sufficient to ensure that the restriction of $f$ to any hyperplane $H_{w,b}$ is integrable. Absent it, I do not know what to replace it with.

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  • $\begingroup$ Thanks (upvoted). It appears one can remove the compact support assumption at the price of demanding more regularity of $f$. Indeed, see Theorem 2.4 of klein.mit.edu/~helgason/Radonbook.pdf. So the Schwaz space on $\mathbb R^n$ appears to be the appropriate replacement for compactness of support. $\endgroup$
    – dohmatob
    May 16 at 8:25
  • $\begingroup$ You'll have to check whether one may still apply the dominated convergence theorem in that setting - more specifically, the "domination" part (I, myself, have not checked this, so this is currently a vulnerability of my proof). $\endgroup$
    – Alex M.
    May 16 at 8:41
  • $\begingroup$ Well, Theorem 2.4 of the referenced book actually shows that the Radon transform bijectively maps the Schwarz space $\mathcal S(\mathbb R^n)$ to a subspace of Schwarz space $\mathcal S(\mathbb P^d)$, where $\mathbb P^d$ is the set of all hyperplanes in $\mathbb R^d$. This would immediately imply continuity of $R[f]$ for any $f \in \mathcal S(\mathbb R^d)$. $\endgroup$
    – dohmatob
    May 16 at 8:53
  • $\begingroup$ I think people just use the Schwartz space because it is convenient. If you made on estimate that holds for Schwartz space you can always try to go to the limit and get the same inequality in a different space… $\endgroup$
    – Dirk
    May 16 at 13:02

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