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For fixed $\beta$ in the sphere $\mathbb S^2\subset \mathbb R^3$, we can define a "zonal derivative" via the distribution $\delta'(\langle \cdot, \beta \rangle)$ on $\mathbb S^2$ -- here $\delta'$ is the derivative of the Delta function, and $\langle \cdot, \cdot \rangle$ is the usual Euclidean scalar product on $\mathbb R^3$.

I'd like to combine two such distributions. To be more precise, given test function $u \in C^\infty(\mathbb S^2)$, I am interested in knowing more about the distribution that maps $u$ to

$ \tilde u(x) = \int_{\mathbb S^2} \int_{\mathbb S^2}\hspace{0.5em} u(\beta)\,\,\, \delta'(\langle \alpha, \beta \rangle) \,\,\, \delta'(\langle x, \beta \rangle) \,\,\,\,dS(\alpha) dS(\beta) .$

Here $dS$ is the (Euclidean) surface measure on the sphere.

My question: what can be said about $u \mapsto \tilde u$ as a differential operator? What order does this distribution have? This kind of object shows up very naturally in certain x-ray reconstruction problems, in particular in the more general form with weighting function $a \in C^\infty(\mathbb S^2 \times \mathbb S^2 )$:

$ \tilde u_a(x) = \int_{\mathbb S^2} \int_{\mathbb S^2}\,\,\,a(\alpha,x)\,\, u(\beta) \,\,\delta'(\langle \alpha, \beta \rangle) \delta'(\langle x, \beta \rangle)\,\,\, dS(\alpha) dS(\beta) .$

The tomographic reconstruction papers I've read using something like this usually go into coordinates quite quickly and are written from a very applied point of view. I'd be very interested in finding out more from the distribution-theoretic side.

Nathanael Schilling
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  • In what sense are you taking the derivative of $\delta$? This isn't a one variable problem, so the prime notation seems wrong. – A. Thomas Yerger Aug 28 '22 at 21:12
  • The delta function here acts only on one variable, $\langle \alpha, \beta \rangle$ denotes the scalar product, I don't think there is any ambiguity in the definition, you can always approximate with the usual mollifier construction but maybe I'm missing something... – Nathanael Schilling Aug 28 '22 at 22:11
  • That scalar product depends on both $\alpha$ and $\beta$ though. So with respect to what are you differentiating? – A. Thomas Yerger Aug 28 '22 at 22:16
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    If $t \mapsto h(t)$ is a function, then $h'(\langle \alpha, \beta \rangle)$ is well-defined. Now the delta function is not a function, but if you pick a sequence of functions $f_n \in C^\infty_c(\mathbb R)$ converging to $\delta$ in the space of distributions, their derivatives converge to $\delta'$. – Nathanael Schilling Aug 28 '22 at 22:30
  • OK, one more dumb question and then I will attempt a solution: In the one-variable case I'm familiar with, $\int u(x) \delta dx = u(0)$. But $0$ is not in the domain of integration here. What point is $\delta$ concentrated at? – A. Thomas Yerger Aug 29 '22 at 00:42
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    I'm not sure I understand your question. For fixed $\beta$, the distribution $\delta'(\langle \cdot, \beta \rangle)$ is concentrated on the great circle orthogonal to $\beta$. – Nathanael Schilling Aug 29 '22 at 07:09

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