Hölder continuity of Radon transform of smooth function

Question

Given an integrable function (e.g a probability density function) $f:\mathbb R^n \to \mathbb R$, let $R[f]$ be its Radon transform defined by $$ R[f](w,b) := \int_{\mathbb R^n} \delta(x^\top w - b)f(x)\,dx, $$ for every $(w,b) \in \mathbb R^{n+1}$. Here, $\delta$ is the Dirac delta distribution.

Question. In terms of smoothness of $f$, what is a sufficient condition to ensure that (1) $\|R[f]\|_\infty < \infty$, and (2) there exist constants $\alpha,C \in (0,\infty)$ such that for every $w \in \mathbb R^d$, the function $b \mapsto R[f](w,b)$ is $(\alpha,C)$-Hölder continuous, i.e., $$ \big|R[f](w,b')-R[f](w,b)\big| \le C|b'-b|^\alpha, \tag{+} $$ for all $w \in \mathbb R^d$ and $b,b' \in \mathbb R$ ?

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Related: Smoothness of Radon transform

Answer 1

• It turns out that the "correct" domain of definition of the Radon transform is the Schwarz space $\mathcal S(\mathbb R^n)$ of infinitely-differential functions on $\mathbb R^n$ with derivatives which decrease faster than any polynomial.

• Also, it is well-known that for any integer $k$, $\mathcal S(\mathbb R^n)$ is dense in $H^k(\mathbb R^n)$, the fractional Sobolev space of order $k$. Taking $\alpha \in (0,1)$, we know thanks to Lemma 1 of Oberlin and Stein (1982) Mapping Properties of the Radon Transform that, there exists a constant $C_\alpha \in (0,\infty)$ such that for any appropriately integrable function $g:\mathbb R \to \mathbb R$, it holds that $$ \sup_{b \in \mathbb R}|\Delta_t g(b)| \le C_\alpha\|g\|_{H^k(\mathbb R)}\cdot |t|^\alpha, \tag{1} $$ where $k := \alpha+1/2$ and $\Delta_t g(b) := g(b+t)-g(b)$. Applying (1) with $g = R_w[f]:b \mapsto R[f](w,b)$, for fixed nonzero $w \in \mathbb R^n$, and recalling the Central Slice Theorem by which $\widehat{R_w[f]}(s) = \|w\|_2\cdot \widehat{f}(sw)$, we deduce that $$ \frac{|\Delta_t R_w[f](b)|^2}{C_\alpha^2|t|^{2\alpha}} \le \ldots \le \|w\|^{-2\alpha}\|f\|_{H^k(\mathbb R^n)}, $$ from which it follows that $R_w[f]$ is Hoelder-continuous of order $\alpha$.