If $f \in C_{c}^{\infty}\left(\mathbb{R}^{2}\right)$, the Radon transform of $f$ is the function $$R f(s, \omega):=\int_{-\infty}^{\infty} f\left(s \omega+t \omega^{\perp}\right) d t, \quad s \in \mathbb{R}, \omega \in S^{1} .$$

From the definition, Radon transform captures integral of functional along the line.

$R^{*}$ is the backprojection operator defined as $$R^{*}: C^{\infty}\left(\mathbb{R} \times S^{1}\right) \rightarrow C^{\infty}\left(\mathbb{R}^{2}\right), \quad R^{*} h(y)=\int_{S^{1}} h(y \cdot \omega, \omega) d \omega$$

Above is an adjoint of the Radon transform. I think it captures the integral of a function over the circle of radius $y\cdot \omega$ passing through the point $y$. Is this geometrical interpretation is correct?

Any help or hint will be greatly appreciated.

  • 1
    $\begingroup$ If you consider the space of (co)oriented lines on the plane as the cylinder $x^2 + y^2 = 1$ in Euclidean $3$-space, this is an integral over an ellipse obtained by intersecting the cylinder with a plane through the origin. See page 11 in this paper (published in a book called MASS Selecta) for some intuition going back and forth between the geometry of the space of oriented lines and Euclidean geometry in the plane. citeseerx.ist.psu.edu/viewdoc/… $\endgroup$ May 30, 2021 at 9:25


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