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In the book of Helgason which is called "Radon transform and integral geometry", he defines on page 2 the Radon transform on hyperplanes as:

$$ \hat{f}(\xi) = \int_{\xi} f(x) dm(x)$$

Where $dm(x)$ is a Euclidean measure on the hyperplane $\xi$.

Now, on page 3, he writes down that: $$(2) \hat{\left(\frac{\partial f}{\partial x_i}\right)}(\omega , p) = \omega_i \frac{\partial \hat{f}}{\partial p}(\omega , p)$$

where $\xi$, can be written as: $\xi \{ x \in \mathbb{R}^n : <x,\omega> = p \}$, where $<,>$ is the standard inner product of the Euclidean space.

I don't understand how did he arrive at equation $(2)$, anyone can help me on this?

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  • $\begingroup$ have you tried applying the chain rule? Remember that $p=<x,w>$. $\endgroup$ Oct 25, 2013 at 20:14
  • $\begingroup$ So we should have here (using Einstein's notation):$$\frac{\partial \hat{f}}{\partial <x,\omega>} = \frac{\partial \int_{\xi} f(x) dm(x)}{\partial <x,\omega>} = \frac{\partial \int_{\xi} f(x)dm(x)}{\partial x_i} \frac{\partial x_i}{\partial x_i\omega_i}+\frac{\partial \int_{\xi} f(x)dm(x)}{\partial p_i}\frac{\partial p_i}{\partial x_i\omega_i}$$ How to proceed from here? Thanks. $\endgroup$
    – Alan
    Oct 27, 2013 at 12:13
  • $\begingroup$ Ok, I see why this is true. Thanks. $\endgroup$
    – Alan
    Oct 27, 2013 at 12:44

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