Under what hypothesis on the domain is the X-ray transform/John transform operator bounded?

Question

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Let $\mathcal{L}$=space of all lines $L$. The X-ray transform is defined here: https://en.wikipedia.org/wiki/X-ray_transform $Xf:\mathcal{L}\to \mathbb{R}$ is defined by: $Xf(L)=\int_{t \in L}f(t)|dt|$.

In this definition, the X-ray transform $Xf$ of a function $f$ is defined only for compactly supported continuous functions $f$ on $\mathbb{R}^d$.

Note that $X:\mathcal{C}_c(\mathbb{R}^d)\to \mathcal{S}$, where $\mathcal{S}$ is the space of continuous scalar valued functions on the set $\mathcal{L}$ of all lines L.

My questions are: 1) Under these conditions, is the transform operator $X$ bounded?

2) If we instead assume that $f$ is a rapidly decaying function, i.e., both $f, \nabla f \to 0$ at infinity, what can we say about the boundedness of $X$?

I'd very much appreciate a detailed explanation. Thank you!

Answer 1

Continuity of $X$ depends on the norms (or other topologies) you put on your function spaces. Let me give some examples. You can find a discussion of the Radon transform and the X-ray transform in various spaces in these lecture notes.

Example 1: $X:L^1(\mathbb R^d)\to L^1(\cal L)$ is continuous since $$ \int_{L\in\cal L}\left|\int_Lf\right| \leq \int_{L\in\cal L}\int_L|f| = \int_{\mathbb R^n}|f| $$ for a suitably scaled measure on $\cal L$.

Example 2: $X:C_c(\mathbb R^d)\to C_c(\cal L)$ is continuous. Let us define the map $h:\mathbb R^d\times S^{n-1}\to\cal L$ so that $h(x,v)$ is the line going through $x$ in direction $v$. This map is continuous and $\operatorname{spt}(Xf)\subset h(\operatorname{spt}(f)\times S^{n-1})$ so $X$ indeed maps compactly supported functions to compactly supported functions. If $f\in C_c(\mathbb R^d)$ and there is a compact convex set $K$ so that $Xf(L)=0$ whenever $L\cap K=\emptyset$, then $\operatorname{spt}(f)\subset K$ — this is Helgason's support theorem. Knowing this, it should not be too hard to show continuity.

Example 3: $X:C_c^k(\mathbb R^d)\to C_c^k(\cal L)$ is continuous for any $k\in\{1,2,\dots,\infty\}$.

About decay at infinity: It seems to me that if $f\to0$ and $\nabla f\to0$ at infinity, then $Xf$ should have the same behaviour. To make this rigorous, you need to specify what you want.

To play with the derivatives, observe that you can differentiate in $\cal L$ in essentially two kinds of directions: you can shift a line with constant direction or you can rotate a line around a point on the line. In both cases the derivative of $Xf$ at $L\in\cal L$ is the integral of a derivative of $f$ along $L$ with constant or linearly growing weight. Using this you can see, for example, that if $f$ and $\nabla f$ decay exponentially, so do $Xf$ and $\nabla Xf$.