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The problem

Consider a $C²$ function $f: X \to \mathbb{R}$, for some compact set $X \subset \mathbb{R}^d$ with $C^1$ boundary, say $\partial X$. I am only interested in $d\in \{2,3\}$.

Then, consider the quantity -- called the spherical mean,
$$ f_x(t) := \int_{S^{d-1}} f(x + t \omega) \mathbb{1}(x + t \omega\in X ) \mathrm{d}\sigma^{d-1}(\omega). $$ What is known about the regularity of of the map $X \to \mathbb{R}: x\mapsto f_x(t)$? In particular, if the function $f$ is bounded below on $X$, can the map be $C^{1,\alpha}$ (differentiable with $\alpha$-Hölder gradient) for almost all $t \in [0, \operatorname{diam}(X)]$. If yes, what is the maximal value of $\alpha$? Can we get a better $\alpha$ if $\partial X$ is $C^2$? Can one get a grasp of the dependency on $t$ of the Hölder constant, when defined?

Attempt I tried to find some references. I can find counterexamples where it does not work for certain $t$'s, hence the statement for almost all $t$. I also tried to do the computations by hand relying on the definitions. I end up dealing with differences of indicators on sets which are quite tedious and I am wondering if there is no theorem in (harmonic) analysis that already fulfils my needs.

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  • $\begingroup$ If the circle $\{x+t\omega: \omega\in S^1\}$ isn't contained in $X$, it is not clear how the integral is defined. If, on the other hand, some ball around $x$ with radius $r>t$ is contained in $X$, isn't this just differentiation of a parameter integral? $\endgroup$
    – Jochen Wengenroth
    Commented Jan 12, 2023 at 17:36
  • $\begingroup$ @JochenWengenroth: Sorry for the lack of clarity. The function is needs to be extended to zero outside of its support. It is not as simple as you point, the variation of the intersection of the boundary and the sphere makes it tricky. These differences of indicators (that appear if you aim at using the dominated convergence theorem) are the crucial point. $\endgroup$
    – Gilles Mordant
    Commented Jan 12, 2023 at 18:13
  • $\begingroup$ What exactly is your definition of $C^{1,\alpha}$-differentiability at a single point? $\endgroup$
    – fedja
    Commented Jan 12, 2023 at 23:21
  • $\begingroup$ It is not at a single point. It is the classical definition, i.e., $ \lVert \nabla f_x(t) - \nabla f_y(t) \rVert \le C_t \lVert x-y \rVert^\alpha$. Still, this is not possible for all $t$, which is the reason why I am wondering whether one can expect this for almost all $t$ or alternatively $\int_I \lVert \nabla f_x(t) - \nabla f_y(t) \rVert \mathrm{d}t \le C \lVert x-y \rVert^\alpha$, for $I:= [0, \operatorname{diam}(X)]$, for instance. $\endgroup$
    – Gilles Mordant
    Commented Jan 13, 2023 at 9:14

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