Csaba Vincze (2013)

University of Debrecen

If the Euclidean distance is substituted by the distance function arising from the 1-norm then we have another notion of generalized conics with applications in geometric tomography [58]. Geometrictomography focuses on problems of reconstructing homogeneous ( often convex) objects from tomographic data (X-rays, projections, sections etc.). A key theorem in this area states that any convex planar body can be determined by parallel X-rays in a set of four directions whose slopes have a transcendental cross-ratio [26].

Let K be a compact body (a compact set is called a body if it is the closure of its interior) in the coordinate plane,

${F}_{K}(x,y):=\frac{1}{A\left(K\right)}{f}_{K}(x,y),$ |
(10.15) |

where

${f}_{K}(x,y):={\int}_{K}{d}_{1}\left(\right(x,y),(\alpha ,\beta \left)\right)d\alpha d\beta ,$ |

A(K) is the area of K and

${d}_{1}(v,w)=|{v}^{1}-{w}^{1}|+|{v}^{2}-{w}^{2}|$ |

is the distance function induced by the taxicab norm (1-norm)

$\left|\right(x,y){|}_{1}=|x|+|y|.$ |

**Excercise 10.3.1 **
*Using the notatios *

$x<K:=\left\{\right(\alpha ,\beta )\in K|x<\alpha \},K<x:=\{(\alpha ,\beta )\in K|\alpha <x\},$ |

$y<K:=\left\{\right(\alpha ,\beta )\in K|y<\beta \},K<y:=\{(\alpha ,\beta )\in K|\beta <y\},$ |

$x=K:=\left\{\right(\alpha ,\beta )\in K|\alpha =x\},y=K:=\{(\alpha ,\beta )\in K|\beta =y\}$ |

prove that

${D}_{1}{F}_{K}(x,y)=\frac{A(K<x)}{A\left(K\right)}-\frac{A(x<K)}{A\left(K\right)},$ |

${D}_{2}{F}_{K}(x,y)=\frac{A(K<y)}{A\left(K\right)}-\frac{A(y<K)}{A\left(K\right)}.$ |

Hint. For the proof see [57].

**Corollary 10.3.2 **
*The global minimizer of the generalized conic function associated to K bisects the area in the sense that the vertical and horizontal lines through this point cut the body K into two parts with equal area. *

Using these formulas the partial derivatives can be expressed by the Cavalieri's principle as follows:

$A(K<x){=\int}_{-\mathrm{\infty}}^{x}Y\left(s\right)ds\text{}and\text{}A(Ky){=\int}_{-\mathrm{\infty}}^{y}X\left(t\right)dt,$ |

where the functions X and Y give the one-dimensional measure of sections with the coordinate lines, respectively. These are called X-ray functions into the coordinate directions. Lebesgue's differentiation theorem shows that

${D}_{1}{D}_{1}{F}_{K}(x,y)=\frac{2}{A\left(K\right)}Y\left(x\right)\text{}and\text{}{D}_{2}{D}_{2}{F}_{K}(x,y)=\frac{2}{A\left(K\right)}X\left(y\right)$ |

holds almost everywhere. Therefore the function 10.15 measuring the average "taxicab" distance can be considered as an accumulation of coordinate X-rays' information. The following figures show different polygons with the same coordinate X-rays to illustrate the difficulties of the reconstruction.

As we can see the inner singularities (together with the endpoints of the support intervals) of the coordinate X-ray functions determine a grid having the possible vertices. This means (among others) that we have only finitely many different polygons with the same coordinate X-rays. Another important consequence is that any convex polygon can be successively determined[14] by three X-rays because we can choose the third direction in such a way that it is not parallel to any line joining the points of the grid.

**Excercise 10.3.3 **
*Prove that the singularities of the coordinate X-rays correspond to the vertices of the convex polygon in the plane. *

Hint. Any convex polygon P in the plane can be written into the form

$P=\left\{\right(x,y\left)\right|f\left(x\right)\le y\le g\left(x\right)\},a\le x\le b$ |

where f is a convex, g is a concave function of the variable. Therefore the vertical X-ray function has the simple form Y(x)=g(x) - f(x). On the other hand the support of Y is an interval [a,b]. It can be easily seen that the difference of a concave and a convex function is concave and, consequently, the X-ray functions of convex compact planar bodies are continuous and differentiable almost anywhere. Moreover,if Y is differentiable at an inner point x then

${f\mathrm{\text{'}}}_{+}\left(x\right)-{f\mathrm{\text{'}}}_{-}\left(x\right)={g\mathrm{\text{'}}}_{+}\left(x\right)-{g}_{-\mathrm{\text{'}}}\left(x\right),$ |

where the signs + and - refer to the right and left hand side derivatives of the functions, respectively. Since f is convex,

${f\mathrm{\text{'}}}_{-}\left(x\right)\le {f\mathrm{\text{'}}}_{+}\left(x\right)$ |

and, consequently, for a concave function,

${g\mathrm{\text{'}}}_{-}\left(x\right)\ge {g}_{+\mathrm{\text{'}}}\left(x\right)$ |

which means that

${f\mathrm{\text{'}}}_{-}\left(x\right)={f\mathrm{\text{'}}}_{+}\left(x\right)\text{}and\text{}{g\mathrm{\text{'}}}_{-}\left(x\right)\le {g}_{+\mathrm{\text{'}}}\left(x\right),$ |

i.e. Y is differentiable at x if and only if P has no vertex along the vertical line at x.

**Definition **The box of a compact convex planar body means the circumscribed rectangle with parallel sides to the coordinate directions.

**Excercise 10.3.4 **
*Prove that the set of compact convex planar bodies having the same box is convex in the sense that if K and L have a common box then it isa box for any convex combination *

$M:=\lambda K+(1-\lambda )L$ |

too. Conclude that

${Y}_{M}\left(x\right)\ge \lambda {Y}_{K}\left(x\right)+(1-\lambda ){Y}_{L}\left(x\right),$ |
(10.16) |

${X}_{M}\left(y\right)\ge \lambda {X}_{K}\left(y\right)+(1-\lambda ){X}_{L}\left(y\right).$ |
(10.17) |

Express the coordinate X-rays of M in terms of the coordinate X-rays of K and L.

Hint. Use the infimal convolution of functions.

Inequalities 10.16 and 10.17 imply, by the Cavalieri's principle, that the volume is a concave function on the set of compact convex planar bodies having a common box; in general see the Brunn-Minkowski inequality in subsection 4.2.1. As another consequence we have

${f}_{\lambda K+(1-\lambda )L}\ge \lambda {f}_{K}+(1-\lambda ){f}_{L}.$ |

For generalizations and applications see [59]. The goal of this section is to present a positive reconstructing result for the class of generalized 1-conics

${f}_{K}(x,y)=const.$ |

The proof will be presented in a more general situation. Let K be a compact body and consider the distance function

${d}_{p}\left(\right(x,y),(\alpha ,\beta \left)\right)=\left(\right|x-\alpha {|}^{p}+|y-\beta {|}^{p}{)}^{1/p}$ |

induced by the p-norm, where p is greater or equal than one. Define the class of generalized p-conics as the boundary of the level sets

${f}_{K}^{p}(x,y)\le c$ |
(10.18) |

of generalized p-conic functions

${f}_{K}^{p}(x,y):={\int}_{K}{d}_{p}\left(\right(x,y),(\alpha ,\beta \left)\right)d\alpha d\beta ;especially{f}_{K}={f}_{K}^{1}.$ |

**Theorem 10.3.5 **
*Let C be a solid generalized p-conic and suppose that C* is a compact body with the same area as C. If the generalized p-conic functions associated to C and C* coincide then C is equal to C* except on a set of measure zero. *

**Proof **Let C be the level set 10.18 of the generalized p-conic function associated with K and suppose that C* is a compact body
with the same area as C such that the generalized p-conic functions associated with C and C* coincide. By the Fubini theorem

${\int}_{C}{f}_{K}^{p}={\int}_{K}{f}_{C}^{p}={\int}_{K}{f}_{{C}^{\mathrm{*}}}^{p}={\int}_{{C}^{\mathrm{*}}}{f}_{K}^{p}$ |
(10.19) |

and thus

${\int}_{C\backslash {C}^{\mathrm{*}}}{f}_{K}^{p}={\int}_{C}{f}_{K}^{p}-{\int}_{C\cap {C}^{\mathrm{*}}}{f}_{K}^{p}\stackrel{}{=}{\int}_{{C}^{\mathrm{*}}}{f}_{K}^{p}-{\int}_{C\cap {C}^{\mathrm{*}}}{f}_{K}^{p}={\int}_{{C}^{\mathrm{*}}\backslash C}{f}_{K}^{p}.$ |
(10.20) |

Since

${\int}_{C\backslash {C}^{\mathrm{*}}}{f}_{k}^{p}\le cA(C\backslash {C}^{\mathrm{*}})$ |
(10.21) |

and

${\int}_{{C}^{\mathrm{*}}\backslash C}{f}_{k}^{p}\ge cA({C}^{\mathrm{*}}\backslash C)$ |
(10.22) |

we have that

$A(C\backslash {C}^{\mathrm{*}})\ge A({C}^{\mathrm{*}}\backslash C).$ |
(10.23) |

But C and C* have the same area and, consequently, equality holds in formula 10.23. Therefore equalities hold in both 10.21 and 10.22. This means that C does not contain any subset of positive measure which is disjoint from C* and vice versa:

$A(C\backslash {C}^{\mathrm{*}})=A({C}^{\mathrm{*}}\backslash C)=0$ |
(10.24) |

showing that C is equal to C* except on a set of measure zero. ▮

**Corollary 10.3.6 **
*Let C and C* be generalized p-conics. If the generalized p-conic functions associated to C and C* coincide then C=C*. *

**Proof **Since both of the sets C and C* are generalized p-conics they have a symmetric role in 10.23 showing that they have the same area. To finish the proof we
use the previous theorem for the compact convex sets C and C*.

**Corollary 10.3.7 **
*Generalized 1-conics are determined by their X-rays in the coordinate directions among compact bodies. *

**Proof **In case of p=1 the condition for the generalized conic functions implies automatically that C and C* have the same area. To finish the proof we use the previous theorem for the sets C and C*.

**Remark **The problem of
determination of convex bodies by a finite set of X-rays was posed by P. C. Hammer at the A.M.S. Symposium on Convexity in 1961. X-rays can be considered as original but special examples for tomographic quantities. Another question is how to recognize a convex planar body from its angle function. The notion was introduced by J. Kincses [34]. Conditions for distinguishability are formulated in terms of the tangent homomorphism using thetheory of dynamical systems [33]. Like the coordinate X-ray pictures the author prove that distinguishability is typical in the sense of Baire category: a set is of *first category* if it is the countable union of nowhere dense sets (not typical cases) and of second category otherwise (typical cases).

**Example **Consider the square

$N:=conv\left\{\left(\mathrm{0,0}\right),\left(\mathrm{1,0}\right),\left(\mathrm{1,1}\right),\left(\mathrm{0,1}\right)\right\};$ |

for any point (x,y) in N

${f}_{N}(x,y)=(x-(1/2){)}^{2}+(y-(1/2){)}^{2}+(1/2)$ |

and circles can be interpreted as generalized 1-conics with N as the set foci. Therefore they are determined by their X-rays into the coordinate directions among compact bodies.

**Excercise 10.3.8 **
*Prove that for any point in N *

${f}_{N}(x,y)=(x-(1/2){)}^{2}+(y-(1/2){)}^{2}+(1/2).$ |

**Excercise 10.3.9 **
*Consider the triangle *

$T:=conv\left\{\left(\mathrm{0,0}\right),\left(\mathrm{1,0}\right),\left(\mathrm{1,1}\right)\right\}.$ |

Prove that for any point in the box of T

${f}_{T}(x,y)=\frac{{x}^{3}-{y}^{3}}{3}+{y}^{2}-\frac{x-y}{2}+\frac{1}{2}.$ |

**Excercise 10.3.10 **
*Let P be a convex polygon in the plane. Prove that generalized conics with P as the set of foci are the union of adjacent algebraic curves of degree at most three. *

Hint. For the partition of the level curves use the grid of the polygon determined by the vertices.

**Excercise 10.3.11 **
*Consider the unit disk *

$D=\left\{\right(x,y\left)\right|{x}^{2}+{y}^{2}\le 1\}.$ |

Prove that for any point in the box of D

${f}_{D}(x,y)=2{x}^{2}\sqrt{1-{x}^{2}}+2x\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\left(x\right)+\frac{4}{3}(1-{x}^{2}{)}^{3/2}+$ |

$2{y}^{2}\sqrt{1-{y}^{2}}+2y\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\left(y\right)+\frac{4}{3}(1-{y}^{2}{)}^{3/2}.$ |