# K-transform

In mathematics, the K transform (also called the Single-Pixel X-ray Transform) is an integral transform introduced by R. Scott Kemp and Ruaridh Macdonald in 2016. The transform allows the structure of a N-dimensional inhomogeneous object to be reconstructed from scalar point measurements taken in the volume external to the object.
Gunther Uhlmann proved that the K transform exhibits global uniqueness on $\mathbb {R} ^{n}$, meaning that different objects will always have a different K transform. This uniqueness arises by the use of a monotone, nonlinear transform of the X-ray transform. By selecting the exponential function for the monotone nonlinear function, the behavior of the K transform coincides with attenuation of particles in matter as described by the Beer–Lambert law, and the K transform can therefore be used to perform tomography of objects using a low-resolution single-pixel detector.
An inversion formula based on a linearization was offered by Lai et al., who also showed that the inversion is stable under certain assumptions. A numerical inversion using the BFGS optimization algorithm was explored by Fichtlscherer.

## Definition

Let an object $f$ be a function of compact support that maps into the positive real numbers $f:\Omega \rightarrow \mathbb {R} _{0}^{+}.$ The K-transform of the object $f$ is defined as ${\mathcal {K}}:L^{1}(\Omega ,\mathbb {P} _{0}^{+})\rightarrow [0,1],$ ${\mathcal {K}}f(r)\equiv \int _{L_{D}(r)}e^{{\mathcal {P}}f(l)}\,dl,$ where $L_{D}(r)\equiv L(r)\cap L(D)$ is the set of all lines originating at a point $r$ and terminating on the single-pixel detector $D$, and ${\mathcal {P}}$ is the X-ray transform.

## Proof of global uniqueness

Let ${\mathcal {P}}f$ be the X-ray transform transform on $\mathbb {R} ^{n}$ and let ${\mathcal {K}}$ be the non-linear operator defined above. Let $L^{1}$ be the space of all Lebesgue integrable functions on $\mathbb {R} ^{n}$ , and $L^{\infty }$ be the essentially bounded measurable functions of the dual space. The following result says that $-{\mathcal {K}}$ is a monotone operator.
For $f,g\in L^{1}$ such that ${\mathcal {K}}f,{\mathcal {K}}g\in L^{\infty }$ then $\langle {\mathcal {K}}f-{\mathcal {K}}g,f-g\rangle \leq 0$ and the inequality is strict when $f\neq g$.
Proof. Note that ${\mathcal {P}}f(r,\theta )$ is constant on lines in direction $\theta$, so ${\mathcal {P}}f(r,\theta )={\mathcal {P}}f(E_{\theta }r,\theta )$, where $E_{\theta }$ denotes orthogonal projection on $\theta ^{\bot }$. Therefore:
$\langle {\mathcal {K}}f-{\mathcal {K}}g,f-g\rangle =\int _{\mathbb {R} ^{n}}\int _{\mathbb {S} ^{n-1}}\left(e^{-{\mathcal {P}}f(r,\theta )}-e^{-{\mathcal {P}}g(r,\theta )}\right)(f-g)(r)\,d\theta \,dr$
$=\int _{\mathbb {S} ^{n-1}}\int _{\mathbb {R} ^{n}}\left(e^{-{\mathcal {P}}f(r,\theta )}-e^{-{\mathcal {P}}g(r,\theta )}\right)(f-g)(r)\,dr\,d\theta$
$=\int _{\mathbb {S} ^{n-1}}\int _{\theta ^{\bot }}\left(e^{-{\mathcal {P}}f(E_{\theta }r,\theta )}-e^{-{\mathcal {P}}g(E_{\theta }r,\theta )}\right)\int _{\mathbb {R} }(f-g)(E_{\theta }r+s\theta )\,ds\,dr_{\!H}\,d\theta$
$=\int _{\mathbb {S} ^{n-1}}\int _{\theta ^{\bot }}\left(e^{-{\mathcal {P}}f(E_{\theta }r,\theta )}-e^{-{\mathcal {P}}g(E_{\theta }r,\theta )}\right)\left({\mathcal {P}}f(E_{\theta }r,\theta )-{\mathcal {P}}g(E_{\theta }r,\theta )\right)dr_{\!H}\,d\theta$
where $dr_{\!H}$ is the Lebesgue measure on the hyperplane $\theta ^{\bot }$. The integrand has the form $(e^{-s}-e^{-t})(s-t)$, which is negative except when $s=t$ and so $\langle {\mathcal {K}}f-{\mathcal {K}}g,f-g\rangle <0$ unless ${\mathcal {P}}f={\mathcal {P}}g$ almost everywhere. Then uniqueness for the X-Ray transform implies that $g=f$ almost everywhere. $\blacksquare$
Lai et al. generalized this proof to Riemannian manifolds.

## Applications

The K transform was originally developed as a means of performing a physical one-time pad encryption of a physical object. The nonlinearity of the transform ensures the there is no one-to-one correspondence between the density $f$ and the true mass $\int _{\mathbb {S} ^{n-1}}\int _{\mathbb {R} }f(x+s\theta )\,ds\,d\theta$, and therefore $f$ cannot be estimated from a single projection.