TITLE: Determine the distribution function of two random variables
QUESTION [1 upvotes]: Suppose that (X,Y) has density function
$$f(x,y)=\frac{2}{(1+x+y)^3}\quad x,y\gt0 \quad$$
Determine the distribution of
(a) X + Y
(b) X − Y
My attempt at solving (a):
$$U=X+Y,V=Y$$
$$f_{U,V}(u,v)=f_{X,Y}(u-v,v)|J|=\frac{2}{(1+u-v+v)^3}*1=\frac{2}{(1+u)^3}$$
$$f_{U}(u)=\int f_{U,V}(u,v)dv=\frac{2v}{(1+u)^3}$$
Answer in the book is $\frac{2u}{(1+u)^3}$
My attempt at solving (b):
$$U=X-Y,V=Y$$
$$f_{U,V}(u,v)=f_{X,Y}(u+v,v)|J|=\frac{2}{(1+u+v+v)^3}*1=\frac{2}{(1+u+2v)^3}$$
$$f_{U}(u)=\int f_{U,V}(u,v)dv=\frac{-1}{2(2v+u+1)^2}$$
Answer in the book is $\frac{1}{2(1+|u|)^2}$
What did i do wrong? any feedback is appreciated!

REPLY [1 votes]: Look to the bounds of the integrals.   They are not indefinite.   Where do your $U,V$ "live"?
The support for $(X,Y)$ is $\{(x,y): 0<x, 0<y\}$
So setting $U=X+Y$, then the support for $(U,Y)$ is $\{(u,y): 0<y<u\}$ .   Because is $X$ is strictly positive then surely: $Y<X+Y$ .   Thus you must have:
$$f_U(u)=\int_{0}^u \frac 2{(1+(u-y)+y)^3}\mathsf d y=\dfrac{2u}{(1+u)^3}$$
Likewise setting $V=X-Y$, then the support for $(V,Y)$ is $\{(v,y):0< y, -y<v\}$.
$$f_V(v) = \int_{\max\{0,-v\}}^\infty \dfrac{2}{(1+(v+y)+y))^3}\mathsf d y$$
So...


TITLE: Bisection method absolute error
QUESTION [1 upvotes]: I know that $\varepsilon \le 2^{-n-1}(b_0 - a_0)$, how to conclude from this that I need $n = \lfloor log_2{\frac{b_0 - a_0}{2\varepsilon}}\rfloor+ 1$?
Using logs I get $ n \le log_2{\frac{b_0-a_0}{\varepsilon}} +1$ and not $n = \lfloor log_2{\frac{b_0 - a_0}{2\varepsilon}}\rfloor+ 1$ :(

REPLY [1 votes]: Hint: Try doing the algebra again (it's incorrect). Then, try bounding the expression you get on the right by adding 1, and think about what n means (number of iteration required to guarantee convergence to within epsilon of the actual zero).


text:
\begin{document}

\title{Variation of the light-like particle energy and its critical curve equations}
\author{ W. B. Belayev \\
 \normalsize Center for Relativity and Astrophysics,\\ \normalsize
185 Box, 194358, Sanct-Petersburg, Russia \\e-mail: wbelayev@yandex.ru }

\begin{abstract}
We consider variation of energy of the light-like particle in Riemann
space-time, find lagrangian, canonical momenta and forces. Equations of the
critical curve are obtained by the nonzero energy integral variation in
accordance with principles of the calculus of variations in mechanics. This
method is shown to not lead to violation of conformity of varied curve to the
null path in contradistinction of the interval variation. Though found
equations are differ from standard form of geodesics equations, for the
Schwarzschild space-time their solutions coincide with each other to within
parameter of differentiation.
\end{abstract}

\maketitle

Keywords: variations; null geodesics; energy integral; extremal
curve; Schwarzschild space-time \\

\section{Introduction}

One of postulates of general relativity is claim that in gravity
field in the absence of other forces the word lines of the
material particles and light rays are geodesics.
In differential geometry a geodesic line is defined as a curve, whose tangent vector is parallel propagated along itself \cite{a2}. Differential
equations of geodesic, which is a path of extremal length, can be
found also by the variation method with the aid of the virtual
displacements of coordinates $\mathbf{x}^{i}$ on a small quantity
$\mathbf{\omega}^{i}$. When we add variation to material particle coordinate, the meaning of the time-like interval slow changes, though that leaves it
time-like.

Finding of differential equations of the null geodesic, corresponding to the
light ray motion, by calculus of variations, described in \cite{a1}. In space-time with metrical coefficients $\mathbf{g}_{ij}$ it is considered variation of their first integral
\begin{eqnarray}\label{f1}
h=\mathbf{g}_{ij}\frac{\mathrm{d}\mathbf{x}^{i}}{\mathrm{d}\mu}\frac{\mathrm{d}\mathbf{x}^{j}}{\mathrm{d}\mu}\,,
\end{eqnarray}
where $\mu$ is affine parameter.
Deriving variation
for extremum determination we must admit arbitrary small
displacements of coordinates. The variation of integral of $h$
expanded in multiple Taylor series is written as
\begin{eqnarray}
 \mathrm{\delta}I=\int\limits_{\mu_0}^{\mu_1}\left[\sum\limits_{n=1}^{\infty}
 \frac{\mathrm{\partial}^{n}\mathbf{g}_{ij}}{\mathrm{\partial}^{\beta_{1}(n)}\mathbf{x}^{1}
 \ldots\mathrm{\partial}^{\beta_{4}(n)}\mathbf{x}^{4}}
 \frac{\mathrm{d}\mathbf{x}^{i}}{\mathrm{d}\mu}\frac{\mathrm{d}\mathbf{x}^{j}}{\mathrm{d}\mu}(\mathbf{\omega}^{1})^{\beta_1(n)}
 \ldots(\mathbf{\omega}^{4})^{\beta_4(n)}+\right.
\nonumber \\ \left.
+\mathbf{g}_{ij}\left(2\frac{\mathrm{d}\mathbf{x}^{i}}{\mathrm{d}
\mu}\frac{\mathrm{d}\mathbf{\omega}^{j}}{\mathrm{d}\mu}
 +\frac{\mathrm{d}\mathbf{\omega}^{i}}{\mathrm{d}\mu}\frac{\mathrm{d}\mathbf{\omega}^{j}}{\mathrm{d}\mu}\right)\right]\mathrm{d}\mu,
 \label{f2}
\end{eqnarray}
where $\mu_0, \mu_1$ are meanings of affine parameter in points, which
are linked by found geodesic, and it is fulfilled $\beta_{1}(n)+\ldots+\beta_{4}(n)=n$. With
 finding geodesic equations the
sum of terms containing variations of coordinates and their derivatives by $\mu$
in first power is equated to null, that gives the geodesic equations in form
\begin{eqnarray}
\frac{\mathrm{d}^2 \mathbf{x}^\lambda}{\mathrm{d}\mu ^2}+\Gamma^\lambda_{ij}
\frac{\mathrm{d}\mathbf{x}^i}{\mathrm{d}\mu}\frac{\mathrm{d}\mathbf{x}^j}{\mathrm{d}\mu}=0,
\label{fg}
\end{eqnarray}
where $\Gamma^\lambda_{ij}$ are Christoffel symbols:
\begin{eqnarray}
\Gamma^\lambda_{ij}=\frac{1}{2}\mathbf{g}^{\lambda\gamma}( \mathbf{g}_{i\gamma,j} +\mathbf{g}_{j\gamma,i} - \mathbf{g}_{ij,\gamma}).
\end{eqnarray}
Here a comma denotes partial differentiation.

Other terms of series in (\ref{f2}), containing small quantities
$\mathbf{\omega}^{i}$,
$\mathrm{d}\mathbf{\omega}^{j}/\mathrm{d}\mu$ in more high powers
or their products and being able to have nonzero values, don't
take into account. Thus such method admits violation of condition
$h=0$, which means that with certain coordinates variations the
interval \emph{a prior} becomes time-like or space-like. Since this interval accords with the light ray motion, that leads to the Lorentz-invariance violation in locality, namely, anisotropies.

The possibility of Lorentz symmetry break for the photon in vacuum by effects from the Plank scale is studied in \cite{cpt, anis}. At the contrary, it is shown in \cite{dhs} that a fundamental space-time discreteness need not contradict Lorentz invariance, and causal set's discreteness is in fact locally Lorentz invariant. Experiments \cite{eli} show exceptionally high precision of constancy of light speed confirmed a Lorentz symmetry in locality, and astrophysical tests don't detect isotropic Lorentz violation \cite{anis}.

In the method of calculus of variations in the large \cite{amm} ones are considered as possible paths along the manifold disregarding kind of interval, not as the trajectories of physical particles. This approach exceeds the limits of classical variational principle in mechanics, according as which virtual motions of the system are compared with cinematically possible motions.

Approximating time-like interval conforming in
general relativity to the material particle motion between fixed
points to null leads in physical sense to unlimited increase of
its mass, and the space-like interval doesn't conform to move of
any object. In this connection it should pay attention on speculation that discreteness at the Planck scale reveals maximum value of momentum for fundamental particles \cite{cam}.

Geodesic line must be extremal \cite{a2}, and the test particle moves along it only in the absence of non-gravity forces. Should photon have some rest mass variations of its path don't give different kinds of intervals, but this assumption doesn't confirm by experiments \cite{a3}. We examine choosing of energy so in order that application of variational principle to its integral for deriving of the isotropic critical curves equations would not lead to considering non-null paths.

\section{Definition of energy and its variation}

The interval in Riemann space-time is written in form
\begin{eqnarray}\label{f21}
\mathrm{d}s^2=\rho^2 \mathbf{g}_{11}\mathrm{d}\mathbf{x}^{12}+
2\rho
\mathbf{g}_{1p}\mathrm{d}\mathbf{x}^{1}\mathrm{d}\mathbf{x}^{p}+
\mathbf{g}_{pq}\mathrm{d}\mathbf{x}^{p}\mathrm{d}\mathbf{x}^{q},
\end{eqnarray}
where $\rho$ is some quantity, whose meaning is assumed to be equal 1. Putting down $\mathbf{x}^1$ as time coordinate, $\mathbf{x}^p \, (p=2,3,4)$ as space coordinates and considering $\rho$ as energy of light-like particle with $\mathrm{d}s=0$ we present it as
\begin{eqnarray}\label{f3}
\rho =\left\lbrace
-\mathbf{g}_{1p}\frac{\mathrm{d}\mathbf{x}^{p}}{\mathrm{d}\mu}+\sigma\left[\left(\mathbf{g}_{1p}\frac{\mathrm{d}\mathbf{x}^{p}}{\mathrm{d}\mu}\right)^{2}-
\mathbf{g}_{11}\mathbf{g}_{pq}\frac{\mathrm{d}\mathbf{x}^{p}}{\mathrm{d}\mu}\frac{\mathrm{d}\mathbf{x}^{q}}{\mathrm{d}\mu}\right]^{1/2}\right\rbrace
\left(\mathbf{g}_{11}\frac{\mathrm{d}\mathbf{x}^{1}}{\mathrm{d}\mu}\right)^{-1}\,,
\end{eqnarray}
where $\sigma$ takes meaning $\pm 1$.

With denotation of the velocity four-vector components as
$\mathbf{v}^{i}=\mathrm{d}\mathbf{x}^{i}/\mathrm{d}\mu$ energy
variation will be
\begin{eqnarray}\label{f31}
\delta \rho= \frac{\partial \rho}{\partial \mathbf{x}^{\lambda}}\delta
\mathbf{x}^{\lambda}+\frac{\partial \rho}{\partial \mathbf{v}^{\lambda}}\delta
\mathbf{v}^{\lambda}
\end{eqnarray}
After substitution

\begin{eqnarray}
\sigma\left[\left(\mathbf{g}_{1p}\frac{\mathrm{d}\mathbf{x}^{p}}{\mathrm{d}\mu}\right)^{2}-
\mathbf{g}_{11}\mathbf{g}_{pq}\frac{\mathrm{d}\mathbf{x}^{p}}{\mathrm{d}\mu}\frac{\mathrm{d}\mathbf{x}^{q}}{\mathrm{d}\mu}\right]^{1/2}=
\mathbf{g}_{1i}\frac{\mathrm{d}\mathbf{x}^{i}}{\mathrm{d}\mu}
\end{eqnarray}
the partial derivatives with respect to coordinates are written as
\begin{eqnarray}\label{f9}
\frac{\mathrm{\partial} \rho }{\mathrm{\partial}
\mathbf{x}^{\lambda}}=\frac{1}{\mathbf{g}_{11}\mathbf{v}^{1}}\left[-\frac{\mathrm{\partial}
\mathbf{g}_{1p}}{\mathrm{\partial}
\mathbf{x}^{\lambda}}\mathbf{v}^{p}+\frac{1}{2\mathbf{v}_{1}}\left(2\frac{\mathrm{\partial}
\mathbf{g}_{1p} }{\mathrm{\partial}
\mathbf{x}^{\lambda}}\mathbf{g}_{1q}-\frac{\mathrm{\partial}
\mathbf{g}_{11} }{\mathrm{\partial}
\mathbf{x}^{\lambda}}\mathbf{g}_{pq}-\frac{\mathrm{\partial}
\mathbf{g}_{pq} }{\mathrm{\partial}
\mathbf{x}^{\lambda}}\mathbf{g}_{11}\right)\mathbf{v}^{p}\mathbf{v}^{q}
\right]- \nonumber \\
-\frac{1}{\mathbf{g}_{11}}\frac{\mathrm{\partial}
\mathbf{g}_{11}}{\mathrm{\partial} \mathbf{x}^{\lambda}}\,.
\end{eqnarray}
This expression is reduced to
\begin{eqnarray}\label{f10}
\frac{\mathrm{\partial} \rho }{\mathrm{\partial}
\mathbf{x}^{\lambda}}=-\frac{1}{2\mathbf{v}_{1}\mathbf{v}^{1}}\frac{\mathrm{\partial}
\mathbf{g}_{ij}}{\mathrm{\partial}
\mathbf{x}^{\lambda}}\mathbf{v}^{i}\mathbf{v}^{j}\,.
\end{eqnarray}
The partial derivatives with respect to components of the velocity four-vector are
\begin{eqnarray}\label{f13}
\frac{\mathrm{\partial} \rho }{\mathrm{\partial}
\mathbf{v}^{\lambda}}=-\frac{\mathbf{v}_{\lambda}}{\mathbf{v}_{1}\mathbf{v}^{1}}\,.
\end{eqnarray}

For the particle, moving in empty space, lagrangian is taken in form
\begin{eqnarray}\label{f130}
L=-\rho,
\end{eqnarray}
and conforms to relation \cite{a4}:
\begin{eqnarray}\label{f131}
\rho=\mathbf{v}^{\lambda}\frac{\mathrm{\partial} L}{\mathrm{\partial}
\mathbf{v}^{\lambda}}-L ,
\end{eqnarray}
which is integral of the motion.
Obtained derivatives give meanings of canonical momenta
\begin{eqnarray}\label{f132}
\mathbf{p}_{\lambda}=\frac{\mathrm{\partial} L}{\mathrm{\partial}
\mathbf{v}^{\lambda}}=\frac{\mathbf{v}_{\lambda}}{\mathbf{v}^{1}\mathbf{v}_{1}}\,
\end{eqnarray}
and forces
\begin{eqnarray}\label{f133}
\mathbf{F}_{\lambda}=\frac{\mathrm{\partial} L }{\mathrm{\partial}
\mathbf{x}^{\lambda}}=\frac{1}{2\mathbf{v}^{1}\mathbf{v}_{1}}\frac{\mathrm{\partial}
\mathbf{g}_{ij}}{\mathrm{\partial}
\mathbf{x}^{\lambda}}\mathbf{v}^{i}\mathbf{v}^{j}\,.
\end{eqnarray}

\section{Equations of isotropic critical curve}

Motion equations are found from
variation of energy integral
\begin{eqnarray}\label{f4}
S=\int\limits_{\mu_0}^{\mu_1}\rho \mathrm{d}\mu,
\end{eqnarray}
where
$\mu_{0}$ and $\mu_{1}$ are fixed points. Energy $\rho$ has non-zero value, its variations leave interval to be light-like, and application of standard variational procedure yields Euler-Lagrange equations

\begin{eqnarray}\label{f8}
\frac{\mathrm{d}}{\mathrm{d}\mu}\frac{\mathrm{\partial}
\rho }{\mathrm{\partial} \dot{\mathbf{x}}^{\lambda}}-\frac{\mathrm{\partial} \rho }{\mathrm{\partial}
\mathbf{x}^{\lambda}}=0\,.
\end{eqnarray}

Critical curve equations are obtained by substitution of partial derivatives (\ref{f10}) and (\ref{f13}) in these equations. For the time coordinate we have
\begin{eqnarray}\label{f12}
\frac{\mathrm{d}\mathbf{v}^{1}}{\mathrm{d}\mu}+\frac{\mathbf{v}^{1}}{2\mathbf{v}_{1}}\frac{\mathrm{\partial}
\mathbf{g}_{ij}}{\mathrm{\partial}
\mathbf{x}^{1}}\mathbf{v}^{i}\mathbf{v}^{j}=0\,.
\end{eqnarray}
For finding of other three equations of motion the second term of (\ref{f8}) is presented in form
\begin{eqnarray}
\frac{\mathrm{d}}{\mathrm{d}\mu}\frac{\mathrm{\partial}
c}{\mathrm{\partial}
\mathbf{v}^{\lambda}}=\frac{1}{\mathbf{v}^{1}(\mathbf{v}_{1})^{2}}\left[(\mathbf{g}_{1p}\mathbf{v}_{\lambda}-\mathbf{g}_{p\lambda}
\mathbf{v}_{1})
\frac{\mathrm{d}\mathbf{v}^{p}}{\mathrm{d}\mu}-\left(\frac{\mathrm{\partial}
\mathbf{g}_{i\lambda}}{\mathrm{\partial}
\mathbf{x}^{j}}\mathbf{v}_1-\frac{\mathrm{\partial}
\mathbf{g}_{1i}}{\mathrm{\partial}
\mathbf{x}^{j}}\mathbf{v}_{\lambda}\right)\mathbf{v}^i \mathbf{v}^j
\right]+
 \nonumber \\
 +\frac{\mathbf{g}_{11}\mathbf{v}^{1}\mathbf{v}_{\lambda}+\mathbf{g}_{p\lambda} \mathbf{v}^{p}\mathbf{v}_{1}}{(\mathbf{v}^{1})^{2}\mathbf{v}_{1}^{2}}\frac{\mathrm{d}\mathbf{v}^{1}}{\mathrm{d}\mu}\,.
\label{f14}
\end{eqnarray}
Replacement of derivative $\mathrm{d}\mathbf{v}^{1}/\mathrm{d}\mu$
here on its expression, obtained from (\ref{f12}), and
substitution found terms in Euler-Lagrange equations gives
\begin{eqnarray}
(\mathbf{g}_{p\lambda}
\mathbf{v}_{1}-\mathbf{g}_{1p}\mathbf{v}_{\lambda})\frac{d\mathbf{v}^{p}}{d\mu}+\left[\frac{1}{2\mathbf{v}_{1}}\frac{\mathrm{\partial}
\mathbf{g}_{ij}}{\mathrm{\partial}
\mathbf{x}^{1}}(\mathbf{g}_{11}\mathbf{v}^{1}\mathbf{v}_{\lambda}+\mathbf{g}_{p\lambda}\mathbf{v}^{p}\mathbf{v}_{1})-\frac{1}{2}\frac{\mathrm{\partial}
\mathbf{g}_{ij}}{\mathrm{\partial}
\mathbf{x}^{\lambda}}\mathbf{v}_{1} +\frac{\mathrm{\partial}
\mathbf{g}_{i\lambda}}{\mathrm{\partial}
\mathbf{x}^{j}}\mathbf{v}_1- \frac{\mathrm{\partial}
\mathbf{g}_{1i}}{\mathrm{\partial}
\mathbf{x}^{j}}\mathbf{v}_{\lambda}\right]\mathbf{v}^i \mathbf{v}^j=0\,.
\label{f15}
\end{eqnarray}
These equations contain accelerations corresponded to the space coordinates and coupled with (\ref{f12}) describe motion of the test light-like particle along critical curve. They don't coincide to usual form (\ref{fg}) of the null geodesics equations.

\section{Isotropic curves in Schwarzschild space-time}

Central gravity field of a point body is described by the Schwarzschild metric. This is solution of Einstein's field equations for space-time, which is empty except central point. We chose units so, that a light velocity constant $c$ is rendered to $1$. At spherical coordinates $\mathbf{x}^i=(t,r,\theta,\varphi)$ the line element is
\begin{eqnarray}\label{f16}
\mathrm{d}s^{2}=\left(1-\frac{\alpha}{r}\right)\mathrm{d}t^{2}-\left(1-\frac{\alpha}{r}\right)^{-1}\mathrm{d}r^{2}-
r^{2}(\mathrm{d}\theta^{2}+{\sin}^{2}\theta
\mathrm{d}\varphi^{2}),
\end{eqnarray}
where $\alpha$ is constant.
Motion equations (\ref{f12}) and (\ref{f15}) lead to
\begin{eqnarray}\label{f17}
\frac{\mathrm{d}^{2}t}{\mathrm{d}\mu^2}=0
\end{eqnarray}
and
\begin{eqnarray}\label{f18}
\frac{\mathrm{d}^{2}r}{\mathrm{d}\mu^2}+\frac{\alpha}{2r^2}\left(1-
\frac{\alpha}{r}\right)\left(\frac{\mathrm{d}t}{\mathrm{d}\mu}\right)-
\frac{3\alpha}{2r(r-\alpha)}\left(\frac{\mathrm{d}r}{\mathrm{d}\mu}\right)^2-
 \nonumber \\
-(r-\alpha)\left[\left(\frac{\mathrm{d}\theta}{\mathrm{d}\mu}\right)^2+
{\sin}^{2}\theta\left(\frac{\mathrm{d}\varphi}{\mathrm{d}\mu}\right)^2\right]=0\,,
\end{eqnarray}
\begin{eqnarray}\label{f19}
\frac{\mathrm{d}^{2}\theta}{\mathrm{d}\mu^2}+\frac{2r-3\alpha}{r(r-\alpha)}\frac{\mathrm{d}r}{\mathrm{d}\mu}\frac{\mathrm{d}\theta}{\mathrm{d}\mu}+
\frac{1}{2}{\sin}
2\theta\left(\frac{\mathrm{d}\varphi}{\mathrm{d}\mu}\right)^2=0\,,
\end{eqnarray}
\begin{eqnarray}\label{f20}
\frac{\mathrm{d}^{2}\varphi}{\mathrm{d}\mu^2}+\frac{2r-3\alpha}{r(r-\alpha)}\frac{\mathrm{d}r}{\mathrm{d}\mu}\frac{\mathrm{d}\varphi}{\mathrm{d}\mu}+
2\cot\theta\frac{\mathrm{d}\theta}{\mathrm{d}\mu}\frac{\mathrm{d}\varphi}{d\mu}=0\,.
\end{eqnarray}
Metric (\ref{f16}) for the isotropic curve yields
\begin{eqnarray}\label{f21}
\left(1-\frac{\alpha}{r}\right)\left(\frac{\mathrm{d}t}{\mathrm{d}\mu}\right)^{2}
-\left(1-\frac{\alpha}{r}\right)^{-1}\left(\frac{\mathrm{d}r}{\mathrm{d}\mu}\right)^{2}-
r^2\left[\left(\frac{\mathrm{d}\theta}{\mathrm{d}\mu}\right)^2+
{\sin}^{2}\theta\left(\frac{\mathrm{d}\varphi}{\mathrm{d}\mu}\right)^2\right]=0\,.
\end{eqnarray}

Equations (\ref{f17}) and (\ref{f20}) are solvable by quadrature:
\begin{eqnarray}\label{f22}
\frac{\mathrm{d}t}{\mathrm{d}\mu}=A\,,
\end{eqnarray}
\begin{eqnarray}\label{f23}
\frac{\mathrm{d}\varphi}{\mathrm{d}\mu}=\frac{h}{r^{2}{\sin}^{2}\theta}\left(1-\frac{\alpha}{r}\right)
\,,
\end{eqnarray}
where $A,\,h$ are constants. Substituting
obtained solutions in equation (\ref{f21}) with $\theta=\pi/2$ we find
\begin{eqnarray}\label{f24}
\frac{\mathrm{d}r}{\mathrm{d}\mu}=\pm\left[\left(1-\frac{\alpha}{r}\right)^2
A^2-\left(\frac{h}{r}\right)^2\left(1-\frac{\alpha}{r}\right)^3\right]^{1/2}\,.
\end{eqnarray}
Though found equations (\ref{f12}) and (\ref{f15}) for the
Schwarzschild space-time are differ from standard form of null
geodesic equations, their solutions coincide with each other to within parameter of differentiation
\begin{eqnarray}\label{f25}
\mathrm{d}\mu=\mathrm{d}\mu_s\left(1-\frac{\alpha}{r}\right)^{-1},
\end{eqnarray}
where $\mu$ corresponds with standard solution \cite{a1}.

If we choose $\mu=t$ it follows from Eq. (\ref{f22}) that $A=1$. Canonical momenta (\ref{f132}) and forces (\ref{f133}) under considering constraint on $\theta$ are

\begin{eqnarray}\label{f251}
\mathbf{p}_{1}=1,\,\mathbf{p}_{2}=\mp\frac{1}{\left(1-\frac{\alpha}{r}\right)}
\sqrt{1-\frac{h^{2}}{r^{2}}\left(1-\frac{\alpha}{r}\right)},\,\mathbf{p}_{3}=0,\,
\mathbf{p}_{4}=-h;
\end{eqnarray}

\begin{eqnarray}\label{f252}
\mathbf{F}_{1}=\mathbf{F}_{3}=\mathbf{F}_{4}=0,\,
\mathbf{F}_{2}=\frac{\alpha}{r^{2}\left(1-\frac{\alpha}{r}\right)}-\frac{h^{2}}{r^{3}}+
\frac{\alpha h^2}{2r^{4}}.
\end{eqnarray}
A nonzero component of the contrvariant vector of canonical force is
\begin{eqnarray}\label{f253}
\mathbf{F}^{2}=-\frac{\alpha}{r^{2}}+\frac{h^{2}}{r^{3}}\left(1-\frac{\alpha}{r}\right)\left(1-\frac{\alpha}{2r}\right).
\end{eqnarray}
In so far as Newtonian limit of gravity theory with gravitational constant $G$ and mass $M$ requires $\alpha=2GM$, the first term of $\mathbf{F}^{2}$ yields twice Newton gravity force. That conforms to light deflection in central gravity field \cite{a34}, which is twice meaning being given by Newton gravity theory.

\section{Conclusion}

Presented form of energy allows applying of method of lagrangian for analysis of light-like particle motion. Virtual displacements of coordinates retain conformity of the trajectory of the light-like particle to the null path in Riemann space-time, not leaded to Lorentz-invariance violation. Obtained extremal isotropic curve equations for the Schwarzschild metric give the same solution, which follows from standard null geodesics equations to within appropriate parameter. At that found equations as whole are different from one, which are produced by variational method being applied to the path.

config: arxiv


TITLE: Subgroup of elements of order at most $2^{m}$
QUESTION [8 upvotes]: The problem A5 in Putnam 2009 reads as follows:

Is there a finite abelian group $G$ such that the product of the
  orders of all its elements is $2^{2009}$?

The answer is No. I am reading the official solution here. The solution starts by observing that if such group existed, it would be a 2-group. By structure theorem for finitely generated abelian groups, we may write
$$G\cong\prod_{i=1}^{\infty} (\mathbb{Z}/2^{i}\mathbb{Z})^{e_i}$$
for some nonnegative integers $e_1, e_2, ...$ all but finitely many of which are $0$. I am having trouble understanding the step that immediately follows:

For any nonnegative integer $m$, the elements of $G$ of order at most
  $2^m$ form a subgroup isomorphic to  $$\prod_{i=1}^{\infty}
 (\mathbb{Z}/2^{\min(i, m)}\mathbb{Z})$$

I can see that the elements of order at most $2^{m}$ forms a subgroup. But I don't see why this subgroup must be isomorphic to the group $\prod_{i=1}^{\infty}
 (\mathbb{Z}/2^{\min(i, m)}\mathbb{Z})$. Can someone shed light on this matter? Perhaps even show an explicit isomorphism if possible?
I appreciate any input.

REPLY [3 votes]: Just look at each factor individually.
The subgroup of $\Bbb{Z}/2^i\Bbb{Z}$ if elements of order at most $2^m$ is isomorphic to $\Bbb{Z}/2^{\min(i, m)}\Bbb{Z}$, because, as a subgroup of a cyclic group, this must be cyclic, and if $ i < m$ then all $2^i$ elements have order at most $2^m$, otherwise only $2^m$ of them do because $\Bbb{Z}/2^i\Bbb{Z}$ is cyclic, so the subgroup is generated by a single element $g$, and, because $g$ has order $2^m$, the subgroup generated by $g$ must have $2^m$ elements (if $g$ had order less that $2^m$, then all of the elements of the subgroup would also have order less than $2^m$, but clearly, if $i\ge m$, $\Bbb{Z}/2^i\Bbb{Z}$ has an element of order exactly $2^m$).
Since the order of an element of a product of groups is the LCM of the orders of it's projections onto the factors, an element of order at most $2^m$ in $G$ is given by simply choosing an element of order at most $2^m$ from each factor, which is exactly the desired isomorphism.


text:
To prove joint convexity of $f_{\alpha, z}(H,\rho, \sigma)$ from equation (\ref{f1}), the next few results are required. Recall the equation from (\ref{functional})
\begin{equation}\label{functional1}
(A,B)\mapsto\mathrm{Tr}(B^{\frac{q}{2}}K^*A^pKB^{\frac{q}{2}})^s.
\end{equation}

\begin{theorem}\cite[Theorem 1.1]{CFLconjecture}\label{Reduced1}
Fix any invertible matrix $\cK$. Suppose that $p\geq q$ and $s>0$.
\begin{enumerate}
    \item If $0\leq q\leq p\leq 1$ and $0<s\leq\frac{1}{p+q}$, then the map from (\ref{functional1}) is jointly concave.
    \item If $-1\leq q\leq p\leq 0$ and $s>0$, then the map from (\ref{functional1}) is jointly convex.
    \item If $-1\leq q \leq 0$, $1\leq p\leq2$, $(p,q)\neq (1,-1)$ and $s\geq\frac{1}{p+q}$, then the map from (\ref{functional1}) is jointly convex.
\end{enumerate}
\end{theorem}

\begin{theorem}\cite[Theorem 3.3]{CFLconjecture}\label{zhangresult}
For $r_i>0$, $i\in\{0,1,2\}$ such that $\frac{1}{r_0}=\frac{1}{r_1}+\frac{1}{r_2}$, we have that for any invertible $X,Y\in\cB(\cH)$ that
\begin{equation}
    \mathrm{Tr}|XY|^{r_1} = \max\left\{ \frac{r_1}{r_0} \mathrm{Tr}|XZ|^{r_0}-\frac{r_1}{r_2}\mathrm{Tr}|Y^{-1}Z|^{r_2} : Z\in\cB(\cH) \text{ and invertible} \right\}.
\end{equation}

\end{theorem}

\begin{proposition}\cite[Proposition 5]{summaryPaper}\label{fixedB}
For a fixed operator $B$, the map on positive operators
$$A\mapsto \mathrm{Tr}\left[(B^*A^pB)^{\frac{1}{p}}\right] $$ \begin{enumerate}
    \item is concave for $0\leq p\leq 1$, with $p\neq0$.
    \item is convex for $1\leq p\leq 2$, with $p\neq0$.
\end{enumerate}
\end{proposition}

\begin{proposition}\label{sum}
If $f:\cD(\cH)\times\cD(\cH)\times\cP(\cH)\mapsto[0,\infty)$ is defined as
$$f(A,B,H) := g(A,H) + h(B,H), $$
where $g$ and $h$ are continuous, the functional $g$ is convex in $A$ and the functional $h$ is convex in $B$, then $f$ is jointly convex in $(A,B)$.
\end{proposition}

\begin{proof}
For all $i$ such that $0\leq\lambda_i\leq1$, with $\sum_i\lambda_i=1$, by convexity of $g$ and $h$,
$$g\left(\sum_i\lambda_iA_i, H\right) \leq \sum_i\lambda_ig(A_i, H) $$ and
$$h\left(\sum_i\lambda_iB_i, H\right) \leq \sum_i\lambda_ih(B_i, H). $$
Thus
\begin{align}
    f\left(\sum_i\lambda_iA_i,\sum_i\lambda_iB_i,H \right) & = g\left(\sum_i\lambda_iA_i, H\right) + h\left(\sum_i\lambda_iB_i, H\right) \\
    & \leq \sum_i\lambda_ig(A_i, H) + \sum_i\lambda_ih(B_i, H) \\
    & = \sum_i\lambda_i\left(g(A_i, H) + h(B_i, H)\right) \\
    & = f(A_i, B_i, H),
\end{align}
as desired.
\end{proof}

\begin{proposition}\label{maxconvex}
If $f(A,B,H)$ is jointly convex in $A$ and $B$, then so is $$\sup_{H>0}\{f(A,B,H)\}.$$
\end{proposition}

\subsection{Certain Trace Functional and its Convexity}

Here are the technical results that helped us prove the main result.   Recall that $$\Psi_{\alpha,z}(\rho||\sigma):=\mathrm{Tr}\left[\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^z\right]. $$

\begin{lemma}\label{function2}
Let $\rho$ and $\sigma\in\cD(\cH)$, and assume that $\alpha>1$ and $z>1$. For any positive operator $H$, define
\begin{equation}\label{f1}
     f_{\alpha,z}(H,\rho,\sigma) :=      z\mathrm{Tr}(\sigma^{\frac{z-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{z-\alpha}{2z}}H) - (z-1)\mathrm{Tr}\left[(\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}})^{\frac{z}{z-1}}\right].
\end{equation}
Then
\begin{equation}
    \Psi_{\alpha,z}(\rho||\sigma)=\sup_{H>0} f_{\alpha,z,H}(\rho,\sigma),
\end{equation}
where the supremum is achieved whenever $H = \sigma^{\frac{1-z}{2z}}(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}})^{z-1}\sigma^{\frac{1-z}{2z}}$.

\end{lemma}
\begin{proof}

For $X$ and $Y$ $\in\cP(\cH)$, and for a choice of  $1\leq p,q\leq\infty$, such that $1=\frac{1}{p}+\frac{1}{q}$, one has
\begin{align}
    \mathrm{Tr}(XY) =& \mathrm{Tr}(X^*Y) \label{xpos1}\\
    \leq& |\mathrm{Tr}(X^*Y)| \\
    \leq& \left(\mathrm{Tr}|X|^p\right)^{\frac{1}{P}}\left(\mathrm{Tr}|Y|^q\right)^{\frac{1}{q}} \label{baug1}\\
    =& \left(\mathrm{Tr}X^p\right)^{\frac{1}{P}}\left(\mathrm{Tr}Y^q\right)^{\frac{1}{q}} \label{sa} \label{xypos1}\\
    \leq& \frac{1}{p}\mathrm{Tr}X^p + \frac{1}{q} \mathrm{Tr}Y^q, \label{young1}
\end{align}
where (\ref{xpos1}) is from the assumption that $X$ is positive, (\ref{baug1}) is from Theorem 2 in \cite{baumgartner}, (\ref{sa}) is because $X$ and $Y$ are self adjoint, and (\ref{young1}) is from Young's inequality stated as Theorem \ref{Young}.
Thus,
\begin{equation}\label{inequality1}
    p\mathrm{Tr}(XY) - \frac{p}{q}\mathrm{Tr}(Y)^{q} \leq \mathrm{Tr}X^{p},
\end{equation}
 with equality if and only if
$ X^{p}=Y^{q}.$
Take  positive operators
$$ X=(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}) \text{ and } Y=(\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}}), $$
where $H$ is some positive operator in $\cB(\cH)$. Then (\ref{inequality1}) becomes
\begin{equation}\label{newIneq1}
    p\mathrm{Tr}(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}}) - \frac{p}{q}\mathrm{Tr}\left[(\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}})^{q}\right] \leq \mathrm{Tr}\left[(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}})^{p}\right].
\end{equation}
Equality happens
\begin{align}
    \text{if and only if } & X^{p}=Y^{q} \\
    \text{if and only if } & \left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^{p}=\left(\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}}\right)^{q} \\
    \text{if and only if } & \sigma^{\frac{1-z}{2z}}\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^{\frac{p}{q}}\sigma^{\frac{1-z}{2z}}=H.
\end{align}
By construction, $H$ is unique, hence the left hand side of (\ref{inequality1}) becomes
\begin{align}
     &\hspace*{-1cm} p\mathrm{Tr}\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\sigma^{\frac{z-1}{2z}}\sigma^{\frac{1-z}{2z}}\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^{\frac{p}{q}}\sigma^{\frac{1-z}{2z}}\sigma^{\frac{z-1}{2z}}\right) - \frac{p}{q}\mathrm{Tr}\left[\left(\sigma^{\frac{z-1}{2z}}\sigma^{\frac{1-z}{2z}}\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^{\frac{p}{q}}\sigma^{\frac{1-z}{2z}}\sigma^{\frac{z-1}{2z}}\right)^{q}\right] \\
     = & p\mathrm{Tr}\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^{\frac{p}{q}}\right) - \frac{p}{q}\mathrm{Tr}\left[\left(\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^{\frac{p}{q}}\right)^{q}\right] \\
     = & p\mathrm{Tr}\left(\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^{\frac{p}{q}+1}\right) - \frac{p}{q}\mathrm{Tr}\left[\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^p\right] \\
     = & \left(p-\frac{p}{q}\right)\mathrm{Tr}\left[\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^p\right]
     \\
     = & \mathrm{Tr}\left[\left(\sigma^{\frac{1-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{1-\alpha}{2z}}\right)^p\right].
\end{align}
where $1=\frac{1}{p}+\frac{1}{q}$ implies that $p+q=pq$ which implies $\frac{p}{q}+1 = p$ and $p-\frac{p}{q}=1$. For a choice of $p=z$ and $q=\frac{z}{z-1}$, where $z>1$, define the left hand side of (\ref{newIneq1}) as
\begin{equation}\label{function1}
    f_{\alpha,z}(H,\rho,\sigma) :=      z\mathrm{Tr}(\sigma^{\frac{z-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{z-\alpha}{2z}}H) - (z-1)\mathrm{Tr}\left[(\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}})^{\frac{z}{z-1}}\right].
\end{equation} Then the desired result is obtained.
\end{proof}

\begin{proposition}\label{jconvexity}
Whenever $1<z\leq \alpha\leq 2z$ and for a fixed $H>0$, one has that $f_{\alpha,z}(H,\rho,\sigma)$ from equation (\ref{f1}) is jointly convex in $\rho$ and $\sigma$.
\end{proposition}

\begin{proof}
Fix $H>0$. Let $p = \frac{\alpha}{z}$, $q = \frac{z-\alpha}{z}$, $X=\rho^{\frac{p}{2}}$, and $Y = \sigma^{\frac{q}{2}}H^{\frac{1}{2}}$. Then $X$ is invertible because $\rho$ is, and similarly, $Y$ is invertible because $H$ and $\sigma$ are. Note that \begin{align}
    \mathrm{Tr}\left(\sigma^{\frac{z-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{z-\alpha}{2z}}H\right) & = \mathrm{Tr}\left(H^{\frac{1}{2}}\sigma^{\frac{z-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{z-\alpha}{2z}}H^{\frac{1}{2}}\right) \\
    & = \mathrm{Tr}\left(H^{\frac{1}{2}}\sigma^{\frac{z-\alpha}{2z}}\rho^{\frac{\alpha}{2z}}\rho^{\frac{\alpha}{2z}}\sigma^{\frac{z-\alpha}{2z}}H^{\frac{1}{2}}\right) \\
    & = \mathrm{Tr}\left(H^{\frac{1}{2}}\sigma^{\frac{q}{2}}\rho^{\frac{p}{2}}\rho^{\frac{p}{2}}\sigma^{\frac{q}{2}}H^{\frac{1}{2}}\right)\\
    & = \mathrm{Tr}\left(Y^*X^*XY\right)\\
    & = \mathrm{Tr}|XY|^2.
\end{align}
Since $1<z\leq\alpha\leq 2z$, it follows that $1\leq p = \frac{\alpha}{z}\leq 2$ and $0\leq -q = -\frac{z-\alpha}{z}\leq 1$. Both $p$ and $-q$ are both positive numbers, so set $(r_0,r_1,r_2) = (\frac{2}{p},2,-\frac{2}{q})$.
Then by Theorem \ref{zhangresult},
\begin{equation}\label{partialsup}
\mathrm{Tr}\left(\sigma^{\frac{z-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{z-\alpha}{2z}}H\right) = \max\left\{ p\mathrm{Tr}|\rho^{\frac{p}{2}}Z|^{\frac{2}{p}}+q\mathrm{Tr}|H^{-\frac{1}{2}}\sigma^{-\frac{q}{2}}Z|^{-\frac{2}{q}} \right\},
\end{equation}
where $Z$ is invertible. Since $1\leq p\leq 2$, it follows from Proposition \ref{fixedB} part 2 that
$$\rho\mapsto \mathrm{Tr}|\rho^{\frac{p}{2}}Z|^{\frac{2}{p}} = \mathrm{Tr}\left[\left(Z^*\rho^p Z\right)^{\frac{1}{p}}\right] $$ is convex in $\rho$.
For the second term in (\ref{partialsup}), choose $\Phi_Z(X):= Z^*XZ$, which is linear, positive, and self adjoint. Let $\sigma = X$, $-q=p$, and $A=Z^{-1}H^{-1}(Z^{-1})^*$. Then $A$ is positive. Moreover, for any $y$, since
\begin{align}
    \langle Z^*(Z^{-1})^*y,y\rangle
    & = \langle (Z^{-1})^*y,Zy\rangle \\
    & = \langle y,Z^{-1}Zy\rangle \\
    & = \langle y,y\rangle,
\end{align}
it follows that $Z^*(Z^{-1})^* = \mathbbm{1}$, and similarly, $(Z^{-1})^*Z^* = \mathbbm{1}$ so then $(Z^*)^{-1} = (Z^{-1})^*$. Thus rewrite $A= Z^{-1}H^{-1}(Z^*)^{-1}$. Then by Proposition \ref{newPick},
\begin{align}
    \mathrm{Tr}|H^{-\frac{1}{2}}\sigma^{-\frac{q}{2}}Z|^{-\frac{2}{q}} & = \mathrm{Tr}\left(\left[(H^{-\frac{1}{2}}\sigma^{-\frac{q}{2}}Z)^*(H^{-\frac{1}{2}}\sigma^{-\frac{q}{2}}Z)\right]^{-\frac{1}{q}}\right) \\
    & = \mathrm{Tr}\left(\left[Z^*\sigma^{-\frac{q}{2}}H^{-1}\sigma^{-\frac{q}{2}}Z\right]^{-\frac{1}{q}}\right) \\
    & = \mathrm{Tr}\left(\left[Z^*\sigma^{-\frac{q}{2}}ZZ^{-1}H^{-1}(Z^*)^{-1}Z^*\sigma^{-\frac{q}{2}}Z\right]^{-\frac{1}{q}}\right) \\
    & = \mathrm{Tr}\left[\left\{\Phi_Z\left(X^{\frac{p}{2}}\right)A \,\Phi_Z\left(X^{\frac{p}{2}}\right)\right\}^{\frac{1}{p}}\right]
\end{align} is concave in $\sigma$. Since $q<0$, it follow that
$$\sigma\mapsto q\mathrm{Tr}|H^{-\frac{1}{2}}\sigma^{-\frac{q}{2}}Z|^{-\frac{2}{q}} $$ is convex in $\sigma$. By Propositions \ref{sum} and \ref{maxconvex}, one concludes that $\mathrm{Tr}\left(\sigma^{\frac{z-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{z-\alpha}{2z}}H\right)$ is the maximum of a sum of two convex functionals and thus is itself convex in $\rho$ and $\sigma$.
$ $

On the other hand, since $\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}}$ and $H^{\frac{1}{2}}\sigma^{\frac{z-1}{z}}H^{\frac{1}{2}}$ have the same nonzero eigenvalues,
$$\mathrm{Tr}\left(\left[\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}}\right]^{\frac{z}{z-1}}\right)=\mathrm{Tr}\left(\left[H^{\frac{1}{2}}\sigma^{\frac{z-1}{z}}H^{\frac{1}{2}}\right]^{\frac{z}{z-1}}\right).$$
So choosing $B=H^{\frac{1}{2}}$, $A=\sigma$, and $p=\frac{z-1}{z}$, by Proposition \ref{fixedB} part 1, $\mathrm{Tr}\left(\left[\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}}\right]^{\frac{z}{z-1}}\right)$ is concave in $\sigma$ whenever $0\leq\frac{z-1}{z}\leq 1$. Hence $$-(z-1)\mathrm{Tr}\left(\left[\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}}\right]^{\frac{z}{z-1}}\right)$$ is convex in $\sigma$.By Proposition \ref{sum},
$$ f_{\alpha,z}(H,\rho,\sigma)=      z\mathrm{Tr}\left(\sigma^{\frac{z-\alpha}{2z}}\rho^{\frac{\alpha}{z}}\sigma^{\frac{z-\alpha}{2z}}H\right) - (z-1)\mathrm{Tr}\left(\left[\sigma^{\frac{z-1}{2z}}H\sigma^{\frac{z-1}{2z}}\right]^{\frac{z}{z-1}}\right) $$ is a sum of two convex functionals in $\rho$ and $\sigma$ as desired.
\end{proof}


text:
\begin{document}

\title[] {Bounds on Convex Bodies in Pairwise Intersecting Minkowski Arrangement of Order $\mu$}

\author[Vikt\'oria F\"oldv\'ari]{Vikt\'oria F\"oldv\'ari}

\address{Vikt\'oria F\"oldv\'ari,
Institute of Mathematics, E\"otv\"os Lor\'and University, Budapest, Hungary,
}
\email{foldvari@cs.elte.hu}

\subjclass[2010]{52C15, 51N20, 52A37}
\keywords{Minkowski arrangement, Bezdek--Pach Conjecture, homothets, packing}

\begin{abstract}
  A generalization of pairwise intersecting Minkowski arrangement of centrally symmetric convex bodies is
  the pairwise intersecting Minkowski arrangement of order $\mu$. Here, the homothetic copies of a centrally
  symmetric convex body are so that none of their interiors intersect the $\mu$-kernel of any other.
  We give general upper and lower bounds on the cardinality of such arrangements, and study two special cases:
  For $d$-dimensional translates in classical pairwise intersecting Minkowski arrangement we prove that the sharp upper bound
  is $3^d$.
  For $\mu=1$ the general version yields to another known problem:
  The Bezdek--Pach Conjecture asserts that the maximum number of pairwise
  touching positive homothetic copies of a convex body in $\Re^d$ is $2^d$. We
  verify the conjecture on the plane, that is, when $d=2$. Indeed, we show that
  the number in question is four for any planar convex body.
\end{abstract}
\maketitle

\section{Introduction}

A \emph{positive homothetic copy} of a convex body (i.e. a compact convex set
with non-empty interior) $K$ in Euclidean $d$-space $\Red$ is a set of the form
$\lambda K+t$ where $\lambda >0$ and $t\in\Red$.
Two sets on the plane are said to \emph{touch} each other if they intersect but
their interiors are disjoint.

Pairwise intersecting homothets of a centrally symmetric convex body in the $d$-dimensional Euclidean
space are in \emph{pairwise intersecting Minkowski arrangement} if
none of them contains the center of any other in its interior.

Polyanskii \cite{P17} recently proved that such a family of convex bodies has at most $3^{d+1}$
members. This result was improved by Naszódi and Swanepoel \cite{NS17} showing an upper bound of $2\cdot 3^d$, which is still not sharp.
The conjectured maximum number of elements is $3^d$.

In Section \ref{sec:Minkowski} we prove the following upper bound on the cardinality of
a family containing translates of a centrally symmetric convex
body in pairwise intersecting Minkowski arrangement:

\begin{thm}\label{thm:Minkowski3d}
In $\mathbb{R}^d$ a pairwise intersecting Minkowski arrangement consisting of translates of a centrally symmetric convex body $K$ contains at most $3^d$ elements. This bound is sharp, equality holds if and only if $K$ is a $d$-dimensional parallelotope.
\end{thm}

We show a construction for arbitrary centrally symmetric convex body that gives a linear lower bound on the cardinality of maximal pairwise intersecting Minkowski arrangements of translates.

\begin{thm}\label{thm:lowerbound}
For a centrally symmetric convex body $K\subset \mathbb{R}^d$ ($d\geq 2$) a maximal pairwise intersecting Minkowski arrangement consisting of translates of $K$ has at least $2d+3$ elements.
\end{thm}

In Section \ref{sec:muMinkowski} we introduce some generalizations of the problem based on an idea of Böröczky and Szabó \cite{BSZ04}.
For $0\leq \mu \leq 1$ they defined the \emph{$\mu$-kernel} of a centrally symmetric convex body $K$ as $\mu K$ concentric to $K$.

Using this notion, for homothets of a centrally symmetric convey body we can consider a
\emph{pairwise intersecting Minkowski arrangement of order $\mu$}, where the homothets are pairwise intersecting but none of their interiors intersect
the center of any other.

We prove an upper bound on the cardinality of such an arrangement, then, for some special bodies we verify the existence of an exponential lower bound.

\begin{thm}\label{thm:muMink}
In $\mathbb{R}^d$ a pairwise intersecting Minkowski arrangement of order $\mu$ consisting of translates of a centrally symmetric convex body $K$ contains at most $ \left( 1 + \frac{2}{1+\mu}  \right)^d $ elements.
\end{thm}

\begin{thm}\label{thm:exponential}
For $\mu<\sqrt{2}-1$ a pairwise intersecting $\mu$-Minkowski arrangement of translates of a euclidean ball can have exponentially many elements.
\end{thm}

The generalized notion of Minkowski arrangement provides a connection between the original problem of pairwise intersecting
Minkowski arrangements, and the Bezdek--Pach Problem \cite{BC88}, detailed in Section \ref{sec:BezdekPach}.

In 1962, Danzer and Gr\"unbaum \cite{DG62} proved that the maximum cardinality
of a family
of pairwise touching translates of a convex body $K$ in $\Red$ is $2^d$, which
bound is attained if, and only if, $K$ is an affine image of a cube.
Petty \cite{P71} showed that every convex body in the plane (resp., in 3-space)
has three (resp., four) pairwise touching translates.
As an extension of this problem, Bezdek and Pach \cite{BC88} conjectured in
1988 that the maximum number of pairwise touching positive homothetic copies of
a convex body in $\Re^d$ is $2^d$. They showed that any such family of
homothetic copies has at most $3^d$ elements, and if $K$ is a $d$-dimensional
Euclidean ball, then the maximum is equal to $d+2$. Naszódi
\cite{N06} improved the first estimate of Bezdek and Pach by proving the upper
bound $2^{d+1}$. In \cite{LN09}, L\'angi and Naszódi proved
(using a result \cite{BB03} of Bezdek and Brass about one-sided Hadwiger
numbers) the upper bound $3\cdot2^{d-1}$ in the case when $K$ is centrally
symmetric.

In Section \ref{sec:BezdekPach} of the present paper we show that the conjecture holds on the plane, moreover,
every planar convex body has four pairwise touching homothets.
\begin{thm}\label{thm:BPplane}
For any convex body $K$ in $\Re^2$, the maximum number of pairwise touching
positive homothetic copies of $K$ is exactly four.
\end{thm}

For two points $a,b$ in $\Re^2$, we denote the closed (resp., open) line
segment connecting them by $[a,b]$ (resp., $(a,b)$). We use the standard
notations $\conv$, $\bd$ and $\intof$ to denote the convex hull, the boundary and the interior of a set
in $\Re^2$, respectively.

$\mathbf{Acknowledgement:}$ This research was supported by the ÚNKP-17-3 New National Excellence Program of the Ministry of Human Capacities.

\section{Bounds on pairwise intersecting Minkowski arrangements}\label{sec:Minkowski}

\begin{defn}
Homothets of a centrally symmetric convex body are said to be in \emph{pairwise intersecting Minkowski
arrangement} if they are pairwise intersecting but none of them contains the center of any other
in its interior.
\end{defn}

The conjecture is that in $\mathbb{R}^d$ a pairwise intersecting Minkowski arrangement consisting of homothets of a centrally symmetric convex body contains at most $3^d$ elements. Here we prove this upper bound for the case, when all the homothets in the arrangement are translates of the given body.

\subsection{Proof of Theorem \ref{thm:Minkowski3d}}
$\ $

Any centrally symmetric convex body can be considered as the unit ball of a normed space. Thus, it is enough to show that the statement holds for the unit ball of any normed space.
It is easy to see that having a Minkowski arrangement is equivalent to the following two conditions on the distances between centers: none of them can be farther than 2, nor closer than 1 to any other.

\begin{lem}\label{lem:terfogatlemma}
Consider a centrally symmetric convex body $K$ in $\mathbb{R}^d$ and $v_1,v_2,...,v_n \in R^d$, so that $1 \leq \norm{v_i-v_j}_K \leq \lambda $ for any $i\neq j$. Then $n\leq (\lambda+1)^d$.
\end{lem}

\begin{proof}
Because of the assumption, for $i\in \{1,...,n\}$ different indices the bodies $v_i+\frac{1}{2}K$ are disjoint.
Let $Q=\conv \left[  \bigcup\limits_{i=1}^n \left( v_i+\frac{1}{2}K\right)\right]$. Since $\mathrm{diam}_K (Q) \leq \lambda+1$, using the isodiametric inequality for Minkowski spaces we get that
\newpage
\begin{equation}\label{terfogatlemma}
\frac{n}{2^d}\vol{K} \leq \vol{Q} \leq \vol{\frac{\lambda+1}{2}K}.
\end{equation}
From this, $n \leq (\lambda+1)^d$ follows.
\end{proof}

Applying this lemma for $\lambda=2$, we get that the number of points with pairwise distances between $1$ and $2$ is at most $3^d$, what is equivalent to the statement of Theorem \ref{thm:Minkowski3d}.

To reach this bound, \eqref{terfogatlemma} has to hold with two equalities. From the following lemma of Groemer \cite{G61} we can see that this happens if and only if $K$ is a $d$-dimensional parallelotope.

\begin{lem}\label{Groemer}
Suppose that $K\subset \mathbb{R}^d$ is a convex body such that for some $1<t\in \mathbb{R}$ the body $tK$ can be decomposed into translates of $K$. Then $K$ is a $d$-dimensional parallelotope and $t$ is an integer. The partition is unique.
\end{lem}
\qed

\subsection{Proof of Theorem \ref{thm:lowerbound}}
$\ $
\\

First we show a construction of seven bodies in $\mathbb{R}^2$, then the lower bound $2d+3$ for the higher dimensional cases will follow recursively.
In $\mathbb{R}^2$, consider an affine-regular hexagon inscribed to $K$ that is symmetric to the center of $K$. There exist seven translates of this hexagon in Minkowski arrangement, shown in Figure \ref{fig:hatszog}, where if a center point is contained in the boundary of another hexagon, then it is a vertex. Translate $K$ in a way that the center points are the same as the centers of the above hexagons. Now a center of any translate is either not contained in another body, or lies on its boundary. Furthermore, these translates share a common point, so they are pairwise intersecting. This means, that the construction gives a Minkowski arrangement.

\begin{figure}[h]
\centering
\includegraphics[width=0.4\textwidth]{hatszog_3.png}
\caption{}
\label{fig:hatszog}
\end{figure}

For a centrally symmetric convex body $K\subset \mathbb{R}^d$ denote by $M_d(K)$ the maximal number of translates in Minkowski arrangement. It is easy to see, that for any such $K_1\in \mathbb{R}^1$, $M_1(K_1)=3$, and we showed that for $K_2\in \mathbb{R}^2$, $M_2(K_2)\geq 7$.

In dimension $d\geq 3$ consider the translates of $K$ in the cartesian coordinate system. Let $x_1,...x_d$ denote the coordinate axes and suppose that $K$ is $o$-symmetric. Using the above planar construction, we can take 7 translates of $K$ in Minkowski arrangement such that their centers lie in the $x_1x_2$ plane. Along each coordinate axis $x_3,...,x_d$ we can add two further translates of $K$ to the arrangement so that they contain $o$ on their boundary. This way we get that $M_{d+1}(K)\geq M_{d}(K)+2$.\\
\qed

\section{Minkowski arrangements of order $\mu$}\label{sec:muMinkowski}

The notion of pairwise intersecting Minkowski arrangement can be generalized in the following way:

\begin{defn}
For $K$ centrally symmetric convex body and $0\leq\mu\leq 1$ the translate of $\mu K$  concentric to $K$ is called the \emph{$\mu$-kernel of $K$}.
\end{defn}

\begin{defn}
Homothets of a centrally symmetric convex body are said to be in \emph{pairwise intersecting Minkowski arrangement of order $\mu$}
if they are pairwise intersecting but none of their interiors intersects the $\mu$-kernel of any other body.
\end{defn}

This generalization provides a connection between the original Minkowski arrangement
and the Bezdek--Pach Conjecture, where the homothets of a convex body are in pairwise touching position. In both problems we consider
pairwise intersecting Minkowski arranements of order $\mu$, but in the earlier version $\mu=0$, while in the latter case $\mu=1$.

The proof of Theorem \ref{thm:muMink} goes the same way as of Theorem \ref{thm:Minkowski3d}.
In this case, the distance between two centers is at least $1+\mu$ but at most $2$. After applying a homothety, this is equivalent to the problem when the distances are between 1 and $\frac{2}{1+\mu}$. Using Lemma \ref{lem:terfogatlemma} we get the result.
\qed

The bound in Theorem \ref{thm:muMink} gives the earlier result $3^d$ for $\mu=0$ and $2^d$ for the pairwise touching case, when $\mu=1$.

Notice that the value of $\mu$ plays an important role, that is, it is not enough to know whether $\mu=0$ or not. For example when $\mu=\frac{1}{2}$ the above theorem in the plane gives 5 as an upper bound, but there exists a sufficiently small positive $\mu$ for which we can find a planar $\mu$-Minkowski arrangement of 6 translates.

For some special convex bodies we can prove exponential lower bound using probabilistic methods.

\section{Proof of the upper bound in Theorem~\ref{thm:BPplane}}\label{sec:BezdekPach}

In this section, we show that the maximum number of pairwise touching
positive homothetic copies of a convex body $K\subset \Re^2$ is at most four.

Let $K_1,K_2,\ldots,K_n$ be a family of pairwise touching positive
homothetic copies of a planar convex body $K$.

If there is a point that belongs to four of the homothets, then we can
enlarge (or shrink) each of the four bodies from that point as a center, to
obtain four
touching translates of $K$. By the result of Danzer and Gr\"unbaum (see the Introduction),
 this implies that $K$ is a parallelogram. It is easy to see that in this case the
family does not have a fifth member. Thus, from this point on, we will assume
that no point belongs to four of the homothets.

If there is a point that belongs to three of the homothets and $K$ has at least
four members, then we will show that this point also belongs to a fourth body.

\begin{prop}\label{nincs_haromszoros}
Let $n\geq 4$ and $K_1,\  K_2,\ K_3,\ K_4$ pairwise touching positive homothets
of the convex body $K\subset \mathbb{R}^2$. If $K_1\cap K_2\cap K_3\neq \emptyset$,
 then $K_1\cap K_2\cap K_3\cap K_4\neq \emptyset$.
\end{prop}
\begin{proof}
Let  $p_{123}\in K_1\cap K_2\cap K_3$. For $i\in\{1,2,3\}$ and a line $e$ through
$p_{123}$ for which $e\cap K_i=\emptyset$ denote by $H^i_e$ the halfplane bounded by
$e$ that contains $K_i$.
Let $C_i=\bigcap\limits_{\substack{p_{123}\in e \\ e\cap K_i=\emptyset}} H^i_e$
the smallest cone with vertex $p_{123}$ containing $K_i$.

Now we show that $\intof C_i \cap \intof C_j=\emptyset$
for any $i\neq j$, $i,j\in\{1,2,3\}$.

Suppose that for a pair $i\neq j$ there exists $c\in(\intof C_i \cap \intof C_j)$.
Then the line $p_{123}c$ intersects the interior of both $K_i$ and $K_j$ because
$C_i$ and $C_j$ are the smallest cones containing $K_i$ and $K_j$ respectively.
 Hence due to the convexity of the bodies, $K_i$ overlaps $K_j$, which is a contradiction.

Suppose that $K_1\cap K_2\cap K_3\cap K_4= \emptyset$. Then $p_{123}\notin K_4$,
thus there exists a supporting line $L_4$ of $K_4$ that does not go through $p_{123}$
and separates $K_4$ from $p_{123}$. $K_4$ touches $K_1$, $K_2$ and $K_3$, hence each of
these three bodies has a point in both of the closed halfplanes bounded by $L_4$.
From this it follows that $L_4$ intersects the cones $C_1$, $C_2$ and $C_3$. For every
$i\in\{1,2,3\}$, $L_4\cap C_i$ is a connected subset of $L_4$, thus there is a middle
one of them. Without loss of generality we can assume that this one is $K_1$. Let $v_1$ be
the vector for which $p_{123}=x_1+v_1$. The image of $p_{123}$ by the homothety that maps
$K_1$ to $K_4$ is the point $x_4+\frac{\lambda_4}{\lambda_1}\cdot v_1 \in L_4$. The same
homothety maps $C_1$ to the cone
$C_1^{\prime}:=C_1+\left(x_4-x_1+v_1\cdot \left(\frac{\lambda_4}{\lambda_1}-1\right)\right)$.
(Figure \ref{fig:nincsharomszoros}) As $K_1\subset C_1$ and the bodies are positive
homothets, $K_4\subset C_1^{\prime}$ follows. At least one pair of the bounding
lines of $C_1$ and $C_1^{\prime}$ are different, thus due to the fact that
$\intof C_i \cap \intof C_j=\emptyset$ for any $i\neq j$, $i,j\in\{1,2,3\}$
 $C_1^{\prime}$ is disjoint to at least one of the cones $C_2$ and $C_3$. But in this case
$K_4$ cannot touch the body lying in this cone, which is a contradiction.

\end{proof}

Thus, it is enough to consider the case when no point belongs to three of the homothets.

\begin{prop}\label{prop:k5sikbarajzolhato}
Let $K_1$, $K_2$,...,$K_n\subset \Re^2$ be pairwise touching convex bodies,
such that no three share a common point. Then $n\leq 4$.
\end{prop}
\begin{proof}
For each $i\in \{1,...,n\}$ choose an interior point $p_i\in K_i$. The bodies are pairwise touching, so we can draw a curve between any two of the chosen points $p_i$, $p_j$ so that it goes in $K_i \cup K_j$. Since no three of the bodies share a common point, these curves intersect only in the interior of the bodies. It is easy to see that we can eliminate these intersections with a perturbation. This way we draw the complete graph of $n$ vertices on the plane, from which $n \leq 4$ follows immediately.
\end{proof}

We also show here a longer version of the proof:

Using the Jordan curve theorem, it is easy to show that the complement of
$K_1\cup K_2\cup K_3$ in $\Re^2$ has two connected components, one bounded
and one unbounded. We call the closure of the bounded component the
\emph{internal region} surrounded by $K_1,K_2, K_3$.

\begin{prop}\label{prop:elhelyezkedes}
Let $K_1$, $K_2$, $K_3$, $K_4\subset \Re^2$ be pairwise touching convex bodies,
such that no three share a common point. Then one of them lies in the internal
region surrounded by the other three.
\end{prop}

First observe that $n \leq 4$ follows from Proposition~\ref{prop:elhelyezkedes}.
Indeed, we may assume that $K_4$ is in the internal region $I$ surrounded by
$K_1,K_2,K_3$.
Suppose that $n\geq 5$. Since $K_5$ touches $K_4$, it must also lie in $I$.
On the other hand, $K_5$ touches $K_1,K_2,K_3$ at points that do not belong to
$K_4$. Now, $( \bd I)\setminus K_4$ is the union of three open arcs, and $K_4$
must have a point on at least two of these arcs to touch $K_1,K_2,K_3$.
However, then the interior of $K_5$ intersects the interior of at least one set
from $K_1,\ldots,K_4$, a contradiction.

\begin{proof}[Proof of Proposition~\ref{prop:elhelyezkedes}]
For each $i,j \in \{1,2,3,4\}$, $i\neq j$, choose a point $p_{ij} \in K_i \cap
K_j$. Due to our assumptions, these are pairwise distinct points. It follows that
\begin{equation}\label{eq:nocrossing}
 (p_{ik},p_{il})\cap (p_{jm},p_{jn})=\emptyset \mbox{ for any }
i,j,k,l\in\{1,2,3,4\} \mbox{ with } i\neq j.
\end{equation}

Note that no three of the six touching points are collinear, since no three
homothets of $K$ share a point in common.

We examine the convex hull $C=\conv(p_{12},\ldots,p_{45})$ of the six touching
points.

Assume that $C$ is a hexagon. Consider the points $p_{12}$, $p_{13}$, $p_{14}$.
We can select two of them, say $p_{12}$ and $p_{14}$ such that they are second
neighbors in the natural cyclic order of the vertices of $C$, thus
$p_{23},\ p_{24}$ or $p_{34}$ is a common neighbour of them
($p_{23}$ on Figure \ref{fig:hatszog}). Consider the segment connecting this
common neighbour and another point of $p_{23},\ p_{24}$ and $p_{34}$ ($p_{34}$ on Figure \ref{fig:hatszog}). Then $(p_{12},p_{14})$ and this segment intersect,
contradicting \eqref{eq:nocrossing}.

Next, assume that $C$ is a pentagon. Without loss of generality we may assume
that $p_{12}$ is the point which is not a vertex of $C$. Every vertex of $C$
has a 3 or a 4 in its indices. Thus, $p_{34}$ has a second neighbor in the
cyclic ordering of the verticies of $C$, which has $3$ or $4$ in its indices.
We consider the case when it is $p_{23}$ (see Figure \ref{fig:otszog}), the
other cases can be handled in the same manner. We connect $p_{12}$ with each of
the two neighbors of $p_{34}$. One of these two line segments intersects
$(p_{34},p_{23})$ contradicting \eqref{eq:nocrossing}.

Next, assume that $C$ is a quadrilateral. Without loss of generality we may
assume that $p_{12}$ is one of the two points that are not vertices of $C$. It
is sufficient to examine those two cases, when the other such point is $p_{13}$
or $p_{34}$, any other arrangement can be obtained from these by changing the
roles (Figure \ref{fig:negyszog}).\\
If $p_{13}$ is not a vertex of the convex hull: The cyclic order of the
vertices of $C$ is $p_{23}, p_{34},p_{14}, p_{24}$, otherwise the diagonals of
the quadrilateral would contradict \eqref{eq:nocrossing}. Then the segment
$(p_{24},p_{34})$ and the polygon $(p_{23},p_{12},p_{14})$ connect two opposite
vertices of the quadrilateral, therefore they intersect each other. Again,
contradicting \eqref{eq:nocrossing}.\\
If $p_{34}$ is not a vertex of the convex hull then the cyclic order of the
vertices of $C$ is $p_{23}, p_{13},p_{14}, p_{24}$, and a very similar argument
to the previous case yields the desired contradiction.

Finally, assume that $C$ is a triangle. The vertices of the triangle have
6 indices in total, so by the pigeon hole principle, some index occurs in at
least two of the vertices. Thus, we will assume that $p_{12}$ and $p_{13}$
are vertices of $C$. Examine which point of the remaining four can be the third
vertex of $C$. It cannot be $p_{14}$, as in this case the other three touching
points
would be in the interior of $K_1$, which is a contradiction. If the third vertex was
$p_{24}$ then one of the three line segments
$(p_{34},p_{12}),(p_{34},p_{13}),(p_{34},p_{24})$ would intersect
one of the three line segments
$(p_{14},p_{12}),(p_{14},p_{13}),(p_{14},p_{24})$ contradicting
\eqref{eq:nocrossing} (see Figure \ref{fig:haromszog}). The same reasoning
shows that the third vertex of $C$ cannot be $p_{34}$. Thus the third vertex is
$p_{23}$. This is indeed possible, and in this case $K_4$ is in the internal
region surrounded by $K_1,K_2,K_3$, finishing the proof of
Proposition~\ref{prop:elhelyezkedes}.

\begin{figure}[h]
\centering
\includegraphics[width=0.65\textwidth]{nincsharomszoros.pdf}
\caption{}
\label{fig:nincsharomszoros}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=0.65\textwidth]{korulzar.pdf}
\caption{}
\label{fig:korulzar}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=0.5\textwidth]{kb_6szog.pdf}
\caption{}
\label{fig:hatszog}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=0.5\textwidth]{kb_5szog.pdf}
\caption{}
\label{fig:otszog}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=\textwidth]{kb_4szog.pdf}
\caption{}
\label{fig:negyszog}
\end{figure}
\begin{figure}[h]
\centering
\includegraphics[width=\textwidth]{kb_3szog.pdf}
\caption{}
\label{fig:haromszog}
\end{figure}

\end{proof}

\begin{rem}
The following proposition is also true and can be proved the same way considering further cases:

If $K_1,\ K_2,\ K_3,\ K_4\subset \mathbb{R}^2$ are pairwise touching convex bodies and $K_1\cap K_2\cap K_3\cap K_4 = \emptyset$, then one of them lies in the internal region surrounded by the other three.
\end{rem}

\section{Proof of the lower bound in Theorem~\ref{thm:BPplane}}

In this section, we show that for any planar convex body $K$, there are four pairwise touching
homothets of $K$.

Consider two distinct parallel support lines of $K$ that each touch $K$ at one
point: $x_1$ and $x_2$. The existence of such pair of lines follows from
Theorem~2.2.11. of \cite{Sch14} (Theorem~2.2.9. of \cite{Sch93}), but may also
be proved as an exercise.

Let $K_1=K$ and $K_2=K+x_2-x_1$. Let $f$ be the line through the single point
of contact, $x_2$, of $K_1$ and $K_2$ parallel to $x_2-x_1$. On both
sides of $f$, there is a translate of $K$ that touches both $K_1$ and $K_2$.
Indeed, if we push $K$ around $K_1$ so that it always touches $K_1$ then, by
continuity, such two positions will be found.

If on both sides we can find such translates of $K$ that also contains $x_2$
then $x_2$ is a common point of four translates of $K$ and we are done. Thus we
assume that at least one of these translates does not contain $x_2$. We call
this translate $K_3$.
\newpage
Now, $K_1,K_2,K_3$ are pairwise touching translates of $K$ that do not share a
common point. It follows that they surround a bounded region $R$ with non-empty
interior. Consider the largest homothet $K_4$ of $K$ contained in $R$. To
finish the proof, we claim that $K_4$ touches $K_1,K_2$ and $K_3$. Indeed,
assume $K_4$ touches only two of them, say $K_1$ and $K_2$. Consider a line
that
separates $K_4$ and $K_1$, and another line that separates $K_4$ and $K_2$. Let
$u_1$ and $u_2$ be the unit normal vectors of these two lines respectively,
pointing away from $K_4$. Clearly, if the origin is not in $\conv(u_1,u_2)$
then
$K_4$ can be moved a little inside $R$ so that it does not touch either $K_1,
K_2$ or $K_3$. Then, we may enlarge $K_4$ slightly within $R$ contradicting the
maximality of $K_4$. Thus $o\in\conv(u_1,u_2)$, that is $u_1=-u_2$. However in
this case, $K_1$ and $K_2$ are strictly separated, which is a contradiction.

\section{A topological note}

In this section we formulate a more general version of
Proposition~\ref{prop:elhelyezkedes}. This generalization is not needed for the
proof of Theorem~\ref{thm:BPplane}, but may of some interest.

An \emph{arc} in the plane is the image of an injective continuous map
of the $[0,1]$ interval into the plane.
A \emph{Jordan curve} in the plane is the image of an injective continuous map
of the circle into the plane.
We will call the closed bounded region bounded by a Jordan curve a \emph{Jordan
region}.

Let $K_1,K_2,K_3$ be three pairwise touching Jordan regions whose pairwise
intersections are non-empty arcs (which may be degenerate, that is a single
point). Using the Jordan curve theorem, it is easy to show that the complement
of
$K_1\cup K_2\cup K_3$ in the plane has two connected components, one
bounded and one unbounded. We call the closure of the bounded component the
\emph{internal region} surrounded by $K_1,K_2, K_3$, and the closure of the
unbounded component the \emph{external region}.

\begin{prop}\label{prop:elhelyezkedesTop}
Let $K_1,K_2,K_3,K_4$ be four pairwise touching Jordan regions whose pairwise
intersections are non-empty arcs (which may be degenerate, that is a single
point). Then one of them lies in the internal region surrounded by the other
three.
\end{prop}

\begin{proof}[An outline of the proof of
Proposition~\ref{prop:elhelyezkedesTop}]

We will call the image of the non-negative reals under an injective mapping
into the plane an \emph{unbounded path} if it is an unbounded subset of the
plane. The image of $0$ is the \emph{starting point} of the unbounded path.

Assume that $K_1$ is not in the internal region
surrounded by $K_2,K_3,K_4$. Then there is a point $p_1$ on the boundary of
$K_1$ that does not belong to either of the other three sets, and from which
there is an unbounded path, $\gamma_1$ which is disjoint from the other three
sets.
Similarly, if $K_2$ is not in the internal region surrounded by the other
three, then there is a point $p_2$ on the boundary of
$K_2$ that does not belong to either of the other three sets, and from which
there is an unbounded path, $\gamma_2$ which is disjoint from the other three
sets. And the same holds for $K_3$ yielding $p_3$ and $\gamma_3$.

We may assume that $\gamma_1,\gamma_2$ and $\gamma_3$ are pairwise disjoint.
Now, $\gamma_1\cup \gamma_2 \cup \gamma_3$ partition the external
region of $K_1,K_2,K_3$ into three parts. And $K_4$ is in one of these three
parts. However, each part only intersects two of the sets $K_1,K_2,K_3$, which
 is a contradiction.
\end{proof}

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