Interpretation of the integral "with respect to a plane wave" in terms of Radon transform

Question

This question might have a formulation in higher dimensions, but for now let's deal with the 2 dimensional Radon transform: $\newcommand{\R}{\mathbb{R}}$ $$ Rf(\varphi,s)=\int_{-\infty}^\infty f(s\theta+t\theta^\perp)ds, $$ where $\theta=[\cos(\varphi),\sin(\varphi)]^\intercal\in S^1$ and $s\in\Bbb{R}$. As $t\in\R$ varies, $s\theta+t\theta^\perp$ describes a line in $\R^2$; thus $Rf$ gives the integral over lines, which justifies its use in tomography.

Now, one can show that if $f\in L^1(\R^2)$ and $h\in L^\infty(\R)$, we have the following Projection Slice Theorem:

$$ \int_{-\infty}^\infty Rf(\varphi,s)h(s)ds=\int_{x\in\Bbb{R}^2}f(x)h(\theta\cdot x)dx\tag{1} $$

In particular, if we take $h(s)=e^{-i\xi s}$, we have the extremely useful Fourier Slice Theorem:

$$ \mathcal{F_s}Rf(\varphi,\xi)=\sqrt{2\pi}\hat{f}(\xi\theta) $$ Here $\mathcal{F}_s$ is the Fourier transform with respect to only the second variable, and $\hat{f}$ is the two-dimensional Fourier transform (recall $\theta\in S^1$).

Now, in the book "The Radon transform, inverse problems, and tomography", Quinto describes the general projection slice theorem (1) as follows:

"Integrating $Rf$ with respect to $h(s)$ is the same as integrating $f$ with respect to the plane wave in the direction $\theta, h(\theta\cdot x)$."

This seems to make good sense when $h(s)=e^{-i\xi s}$, so that $h(\theta\cdot x)$ describes a harmonic plane wave and thus integrating with respect to $h$ gives frequency data. But what about this general statement? What does it mean, intuitively, to integrate "with respect to a plane wave", and why might this be useful? Can we interpret the integral of $f(x)h(\cdot)$ as a kind of inner product?

Answer 1

Sure you can't - but somehow you can. Obviously, $x\mapsto h(\theta\cdot x)$ is not an integrable function (if not $\equiv 0$) since it is constant along lines perpendicular to $\theta$. However, if $f$ is not $L^2$ but $L^1$, then you can view the integral as duality pairing: If $h$ is bounded, i.e. $h\in L^\infty$, then it makes sense to write $$ \int f(x) h(\theta\cdot x)dx = \langle f,h(\theta\cdot\ )\rangle_{L^1\times L^\infty}. $$ The situation is somehow similar to the situation for the Fourier transform: Of course, the integration against a harmonic plane wave is not an inner product in $L^2$ but makes sense in an $L^1$-way. But one can intuitively think of it like an inner product against an "uncountable orthonormal basis" to keep the analogy with Fourier series or discrete Fourier transforms.