On the Core of a Low Dimensional Set-Valued Mapping

By Pavel Shvartsman

Department of Mathematics, Technion - Israel Institute of Technology,

32000 Haifa, Israel

e-mail<EMAIL_ADDRESS>

11footnotetext: Math Subject Classification: 46E35
Key Words and Phrases: Set-valued mapping, Lipschitz selection, Helly’s
theorem, the core of a set-valued mapping, Hausdorff distance, balanced
refinement.This research was supported by Grant No 2014055 from the United
States-Israel Binational Science Foundation (BSF).

###### Abstract

Let ${\mathfrak{M}}=({\mathcal{M}},\rho)$ be a metric space and let $X$ be a
Banach space. Let $F$ be a set-valued mapping from ${\mathcal{M}}$ into the
family ${\mathcal{K}}_{m}(X)$ of all compact convex subsets of $X$ of
dimension at most $m$. The main result in our recent joint paper with Charles
Fefferman (which is referred to as a “Finiteness Principle for Lipschitz
selections”) provides efficient conditions for the existence of a Lipschitz
selection of $F$, i.e., a Lipschitz mapping $f:{\mathcal{M}}\to X$ such that
$f(x)\in F(x)$ for every $x\in{\mathcal{M}}$.

We give new alternative proofs of this result in two special cases. When $m=2$
we prove it for $X=\mathbf{R}^{2}$, and when $m=1$ we prove it for all choices
of $X$. Both of these proofs make use of a simple reiteration formula for the
“core” of a set-valued mapping $F$, i.e., for a mapping
$G:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ which is Lipschitz with respect to
the Hausdorff distance, and such that $G(x)\subset F(x)$ for all
$x\in{\mathcal{M}}$.

##  1\. Introduction.

Let ${\mathfrak{M}}=({\mathcal{M}},\rho)$ be a pseudometric space, i.e.,
suppose that the “distance function”
$\rho:{\mathcal{M}}\times{\mathcal{M}}\to[0,+\infty]$ satisfies $\rho(x,x)=0$,
$\rho(x,y)=\rho(y,x)$, and $\rho(x,y)\leq\rho(x,z)+\rho(z,y)$ for all
$x,y,z\in{\mathcal{M}}$. Note that $\rho(x,y)=0$ may hold with $x\neq y$, and
$\rho(x,y)$ may be $+\infty$.

Let $(X,\|\cdot\|)$ be a real Banach space. Given a non-negative integer $m$
we let ${\mathcal{K}}_{m}(X)$ denote the family of all non-empty compact
convex subsets $K\subset X$ of dimension at most $m$. (We say that a convex
subset of $X$ has dimension at most $m$ if it is contained in an affine
subspace of $X$ of dimension at most $m$.) We let
${\mathcal{K}}(X)=\bigcup\\{{\mathcal{K}}_{m}(X):m=0,1,...\\}$ denote the
family of all non-empty compact convex finite-dimensional subsets of $X$.

By $\operatorname{Lip}({\mathcal{M}},X)$ we denote the space of all Lipschitz
mappings from ${\mathcal{M}}$ to $X$ equipped with the Lipschitz seminorm

$\|f\|_{\operatorname{Lip}({\mathcal{M}},X)}=\inf\,\\{\,\lambda>0:\|f(x)-f(y)\|\leq\lambda\,\rho(x,y)~{}~{}\text{for
all}~{}~{}x,y\in{\mathcal{M}}\,\\}.$

In this paper we study the following problem.

######  Problem 1.1.

Suppose that we are given a set-valued mapping $F$ which to each point
$x\in{\mathcal{M}}$ assigns a set $F(x)\in{\mathcal{K}}_{m}(X)$. A selection
of $F$ is a map $f:{\mathcal{M}}\to X$ such that $f(x)\in F(x)$ for all
$x\in{\mathcal{M}}$.

We want to know whether there exists a selection $f$ of $F$ in the space
$\operatorname{Lip}({\mathcal{M}},X)$. Such an $f$ is called a Lipschitz
selection of the set-valued mapping $F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$.

If a Lipschitz selection $f$ exists, then we ask how small we can take its
Lipschitz seminorm. See Fig. 1.

Fig. 1: $f:{\mathcal{M}}\to{\bf R}^{2}$ is a Lipschitz selection of the set-
valued mapping $F:{\mathcal{M}}\to{\mathcal{K}}({\bf R}^{2})$.

The following result provides efficient conditions for the existence of a
Lipschitz selection of an arbitrary set-valued mapping from a pseudometric
space into the family ${\mathcal{K}}_{m}(X)$. We refer to it as a “Finiteness
Principle for Lipschitz selections”, or simply as a “Finiteness Principle”.

######  Theorem 1.2.

(Fefferman,Shvartsman [16]) Fix $m\geq 1$. Let $({\mathcal{M}},\rho)$ be a
pseudometric space, and let $F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ for a
Banach space $X$. Let

(1.1) $\displaystyle
N(m,X)=2^{\ell(m,X)}~{}~{}~{}~{}~{}\text{where}~{}~{}~{}~{}~{}\ell(m,X)=\min\\{m+1,\dim
X\\}.$

Suppose that for every subset ${\mathcal{M}}^{\prime}\subset{\mathcal{M}}$
consisting of at most $N=N(m,X)$ points, the restriction
$F|_{{\mathcal{M}}^{\prime}}$ of $F$ to ${\mathcal{M}}^{\prime}$ has a
Lipschitz selection $f_{{\mathcal{M}}^{\prime}}$ with Lipschitz seminorm
$\|f_{{\mathcal{M}}^{\prime}}\|_{\operatorname{Lip}({\mathcal{M}}^{\prime},X)}\leq
1$.

Then $F$ has a Lipschitz selection $f$ with Lipschitz seminorm
$\|f\|_{\operatorname{Lip}({\mathcal{M}},X)}\leq\gamma$ where
$\gamma=\gamma(m)$ is a positive constant depending only $m$.

There is an extensive literature devoted to various versions of Finiteness
Principles for Lipschitz selections and related topics. We refer the reader to
the papers [1, 2, 4, 14, 15, 16, 23, 19, 20, 21, 24, 25, 26] and references
therein for numerous results in this direction.

We note that the “finiteness number” $N(m,X)$ in Theorem 1.2 is optimal; see
[24].

For the case of the trivial distance function $\rho\equiv 0$, Theorem 1.2
agrees with the classical Helly’s Theorem [9], except that the optimal
finiteness constant for $\rho\equiv 0$ is

$n(m,X)=\ell(m,X)+1=\min\\{m+2,\dim X+1\\}~{}~{}~{}~{}\text{in place
of}~{}~{}~{}~{}N(m,X)=2^{\ell(m,X)}.$

Thus, Theorem 1.2 may be regarded as a generalization of Helly’s Theorem.

Our interest in Helly-type criteria for the existence of Lipschitz selections
was initially motivated by some intriguing close connections of this problem
with the classical Whitney extension problem [27], namely, the problem of
characterizing those functions defined on a closed subset, say $E\subset{\bf
R}^{n}$, which are the restrictions to $E$ of $C^{m}$-smooth functions on
${\bf R}^{n}$. We refer the reader to the papers [5, 6, 7, 10, 11, 12, 13, 26]
and references therein for numerous results and techniques concerning this
topic.

One of the main ingredients of the proof of Theorem 1.2 is the construction of
a special set-valued mapping $G:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$
introduced in [16] which we call a “core” of the set-valued mapping $F$. In
fact each core is associated with a positive constant. Here are the relevant
definitions.

######  Definition 1.3.

Let $\gamma$ be a positive constant, and let
$F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ be a set-valued mapping. A set-valued
mapping $G:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ is said to be a $\gamma$-core
of $F$ if

(i). $G(x)\subset F(x)$ for all $x\in{\mathcal{M}}$;

(ii). $G$ is $\gamma$-Lipschitz with respect to Hausdorff distance, i.e.,

$\operatorname{d_{H}}(G(x),G(y))\leq\gamma\,\rho(x,y)~{}~{}~{}~{}~{}~{}\text{for
all}~{}~{}~{}x,y\in{\mathcal{M}}.$

We refer to a map $G$ as a core of $F$ if $G$ is a $\gamma$-core of $F$ for
some $\gamma>0$.

See Fig. 2, 3, 4.

Fig. 2: A set-valued mapping $F$ into a family of avocados and its Lipschitz
selection
with Lipschitz seminorm at most $\lambda$. Fig. 3: The core $G(x)$ is a convex
closed subset of $F(x)$. Fig. 4: The $\gamma$-core $G$ is $\gamma$-Lipschitz
with respect to the Hausdorff distance.

Recall that the Hausdorff distance $\operatorname{d_{H}}(A,B)$ between two
non-empty bounded subsets $A$ and $B$ of $X$ is defined by

(1.2)
$\displaystyle\operatorname{d_{H}}(A,B)=\inf\,\\{r>0:~{}A+B_{X}(0,r)\supset
B~{}~{}\text{and}~{}~{}B+B_{X}(0,r)\supset A\\}.$

Here and throughout this paper, for each $x\in X$ and $r>0$, we use the
standard notation $B_{X}(x,r)$ for the closed ball in $X$ with center $x$ and
radius $r$. We also let $B_{X}=B_{X}(0,1)$ denote the unit ball in $X$, and we
write $rB_{X}$ to denote the ball $B_{X}(0,r)$.

In Definition 1.3 $m$ can be any non-negative integer not exceeding the
dimension of the Banach space $X$. It can happen that a core
$G:\mathcal{M}\to\mathcal{K}_{m}(X)$ of a given set-valued mapping
$F:\mathcal{M}\to\mathcal{K}_{m}(X)$ in fact maps $\mathcal{M}$ into the
smaller collection $\mathcal{K}_{m^{\prime}}(X)$ for some integer
$m^{\prime}\in[0,m$). The next claim shows that the existence of some core
$G:\mathcal{M}\to\mathcal{K}_{m}(X)$ for $F$ implies the existence of a
(possibly different) core which maps $\mathcal{M}$ into $\mathcal{K}_{0}(X)$.
Since $\mathcal{K}_{0}(X)$ is identified with $X$, that core is simply a
Lipschitz selection of $F$.

######  Claim 1.4.

([16, Section 5]) Let $\gamma$ be a positive constant, let $m$ be a non-
negative integer, and let $G:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ be a
$\gamma$-core of a set-valued mapping $F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$
for some Banach space $X$. Then $F$ has a Lipschitz selection
$f:{\mathcal{M}}\to X$ with $\|f\|_{\operatorname{Lip}({\mathcal{M}},X)}\leq
C\,\gamma$ where $C=C(m)$ is a constant depending only on $m$.

In [16] we showed that this claim follows from Definition 1.3 and the
existence of the so-called “Steiner-type point” map
$\operatorname{St}:{\mathcal{K}}_{m}(X)\to X$ [25]. See formula (1.12).

In [16] given a set-valued mapping $F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$
satisfying the hypothesis of Theorem 1.2, we constructed a $\gamma$-core $G$
of $F$ with a positive constant $\gamma$ depending only on $m$. We produced
the core $G$ using a rather delicate and complicated procedure whose main
ingredients are families of Basic Convex Sets associated with $F$, metric
spaces with bounded Nagata dimension, ideas and methods of work [14] related
to the case ${\mathcal{M}}={\bf R}^{n}$, and Lipschitz selections on finite
metric trees. See [16] for more details.

In the present paper we suggest and discuss a different new geometrical method
for producing a core of a set-valued mapping. Its main ingredient is the so-
called balanced refinement of a set-valued mapping which we define as follows.

######  Definition 1.5.

Let $\lambda\geq 0$, let $({\mathcal{M}},\rho)$ be a pseudometric space, let
$X$ be a Banach space, and let $F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ be a
set-valued mapping for some non-negative integer $m$. For each
$x\in{\mathcal{M}}$ we consider the subset of $F(x)$ defined by

${\cal
BR\,}[{F}\\!:\\!{\lambda};{\rho}](x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+\lambda\,\rho(x,z)\,B_{X}\right].$

We refer to the set-valued mapping ${\cal
BR\,}[{F}\\!:\\!{\lambda};{\rho}]:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)\cup\\{\emptyset\\}$
as the $\lambda$-balanced refinement of the mapping $F$.

We note that any Lipschitz selection $f$ of a set-valued mapping
$F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ with
$\|f\|_{\operatorname{Lip}({\mathcal{M}},X)}\leq\lambda$ is also a Lipschitz
selection of the $\lambda$-balanced refinement of $F$, i.e.,

$f(x)\in{\cal BR\,}[{F}\\!:\\!{\lambda};{\rho}](x)~{}~{}~{}~{}\text{for
all}~{}~{}~{}~{}x\in{\mathcal{M}}.$

Various geometrical parameters of the set ${\cal
BR\,}[{F}\\!:\\!{\lambda};{\rho}](x)$ (such as diameter and width, etc.) may
turn out to be smaller than the same parameters for the set $F(x)$ which
contains it. When attempting to find Lipschitz selections of $F$ it may turn
out to be convenient for our purposes to search for them in the more
“concentrated” setting provided by the sets ${\cal
BR\,}[{F}\\!:\\!{\lambda};{\rho}](x)$. One can take this approach still
further by searching in even smaller sets which can be obtained from
consecutive iterations of balanced refinements of $F$, i.e., the sets which we
describe in the following definition.

######  Definition 1.6.

Let $\ell$ be a positive integer, and let
$\vec{\lambda}=\left\\{\lambda_{k}:1\leq k\leq\ell\right\\}$ be a finite
sequence of $\ell$ non-negative numbers $\lambda_{k}$. We set $F^{[0]}=F$,
and, for every $x\in{\mathcal{M}}$ and integer $k\in[0,\ell-1]$, we define

(1.3) $\displaystyle F^{[k+1]}(x)={\cal
BR\,}[{F^{[k]}}\\!:\\!{\lambda_{k+1}};{\rho}](x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F^{[k]}(z)+\lambda_{k+1}\,\rho(x,z)\,B_{X}\right].$

We refer to the set-valued mapping
$F^{[k]}:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)\,{{\cup}}\,\\{\emptyset\\}$,
$k\in[1,\ell]$, as the $k$-th order $(\vec{\lambda},\rho)$-balanced refinement
of $F$.

Clearly,

(1.4) $\displaystyle F^{[k+1]}(x)\subset
F^{[k]}(x)~{}~{}~{}~{}\text{on}~{}~{}~{}{\mathcal{M}}~{}~{}~{}\text{for
every}~{}~{}~{}k\in[0,\ell-1].$

(Put $z=x$ in the right hand side of (1.3).)

######  Remark 1.7.

Of course, for each integer $k\in[1,\ell]$ the set $F^{[k]}(x)$ also depends
on the sequence $\vec{\lambda}=\\{\lambda_{k}:1\leq k\leq\ell\\}$, on the
pseudometric space ${\mathfrak{M}}=({\mathcal{M}},\rho)$ and the Banach space
$X$. However, in all places where we use $F^{[k]}$’s, these objects, i.e.,
$\vec{\lambda}$, ${\mathfrak{M}}$ and $X$, are clear from the context.
Therefore, in these cases, we omit any mention of $\vec{\lambda}$,
${\mathfrak{M}}$ and $X$ in the notation of $F^{[k]}$’s. $\vartriangleleft$

We formulate the following

######  Conjecture 1.8.

Let $({\mathcal{M}},\rho)$ be a pseudometric space, and let $X$ be a Banach
space. Let $m$ be a fixed positive integer and (as in the formula (1.1) of
Theorem 1.2) let $N(m,X)$ denote the “finiteness number” $N(m,X)=2^{\ell}$
where $\ell=\ell(m,X)=\min\\{m+1,\dim X\\}$.

There exist a constant $\gamma\geq 1$ and a sequence
$\vec{\lambda}=\\{\lambda_{k}:1\leq k\leq\ell\\}$ of $\ell$ numbers
$\lambda_{k}$ all satisfying $\lambda_{k}\geq 1$ such that the following
holds:

Let $F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ be a set-valued mapping such
that, for every ${\mathcal{M}}^{\prime}\subset{\mathcal{M}}$ with
$\\#{\mathcal{M}}^{\prime}\leq N(m,X)$, the restriction
$F|_{{\mathcal{M}}^{\prime}}$ of $F$ to ${\mathcal{M}}^{\prime}$ has a
Lipschitz selection $f_{{\mathcal{M}}^{\prime}}:{\mathcal{M}}^{\prime}\to X$
with Lipschitz seminorm
$\|f_{{\mathcal{M}}^{\prime}}\|_{\operatorname{Lip}({\mathcal{M}}^{\prime},X)}\leq
1$.

Then the $\ell$-th order balanced refinement of the mapping $F$, namely the
set-valued mapping $F^{[\ell]}:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ is a
$\gamma$-core of $F$.

Here $F^{[\ell]}$ is defined as in Definition 1.6 using the particular
sequence $\vec{\lambda}$.

Our main results, Theorem 1.9 and Theorem 1.10 below, state that Conjecture
1.8 holds in two special cases, when either (i) $m=2$ and $\dim X=2$, or (ii)
$m=1$ and $X$ is an arbitrary Banach space. Note that in both of these cases
the above mentioned finiteness number $N(m,X)$ equals $4$.

######  Theorem 1.9.

Let ${\mathfrak{M}}=({\mathcal{M}},\rho)$ be a pseudometric space, and let $X$
be a two dimensional Banach space. Let $m=2$ so that the number $\ell(m,X)=2$.
In this case Conjecture 1.8 holds for every $\lambda_{1},\lambda_{2}$ and
$\gamma$ such that

(1.5) $\displaystyle\lambda_{1}\geq
e({\mathfrak{M}},X),~{}~{}~{}~{}~{}~{}\lambda_{2}\geq
3\lambda_{1},~{}~{}~{}~{}~{}~{}\gamma\geq\lambda_{2}\,(3\lambda_{2}+\lambda_{1})^{2}/(\lambda_{2}-\lambda_{1})^{2}.$

Here $e({\mathfrak{M}},X)$ denotes the Lipschitz extension constant of $X$
with respect to ${\mathfrak{M}}$. (See Definition 3.1.)

Thus, the following statement is true: Let
$F:{\mathcal{M}}\to{\mathcal{K}}(X)$ be a set-valued mapping from a
pseudometric space $({\mathcal{M}},\rho)$ into the family ${\mathcal{K}}(X)$
of all non-empty convex compact subsets of $X$. Given $x\in{\mathcal{M}}$ let

(1.6) $\displaystyle
F^{[1]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+\lambda_{1}\,\rho(x,z)B_{X}\right],~{}~{}~{}~{}F^{[2]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F^{[1]}(z)+\lambda_{2}\,\rho(x,z)B_{X}\right].$

Suppose that for every subset ${\mathcal{M}}^{\prime}\subset{\mathcal{M}}$
with $\\#{\mathcal{M}}^{\prime}\leq 4$, the restriction
$F|_{{\mathcal{M}}^{\prime}}$ of $F$ to ${\mathcal{M}}^{\prime}$ has a
Lipschitz selection with Lipschitz seminorm at most $1$.

Then for every $\lambda_{1},\lambda_{2}$ and $\gamma$ satisfying (1.5) the set

(1.7) $\displaystyle F^{[2]}(x)\neq\emptyset~{}~{}~{}~{}\text{for
every}~{}~{}~{}~{}x\in{\mathcal{M}}.$

Furthermore,

(1.8)
$\displaystyle\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\gamma\,\rho(x,y)~{}~{}~{}~{}\text{for
every}~{}~{}~{}~{}x,y\in{\mathcal{M}}.$

If $X$ is a Euclidean two dimensional space, (1.7) and (1.8) hold when (1.5)
is replaced by the weaker requirements that

(1.9) $\displaystyle\lambda_{1}\geq
e({\mathfrak{M}},X),~{}~{}~{}~{}~{}~{}\lambda_{2}\geq
3\lambda_{1},~{}~{}~{}~{}~{}~{}\gamma\geq\,\lambda_{2}\left\\{1+2\lambda_{2}\,{{/}}\left(\lambda_{2}^{2}-\lambda_{1}^{2}\right)^{\frac{1}{2}}\right\\}^{2}.$

In particular, in Section 3 we show that the mapping $F^{[2]}$ satisfies (1.7)
and (1.8) whenever $X$ is an arbitrary two dimensional Banach space and
$\lambda_{1}=4/3$, $\lambda_{2}=4$, $\gamma=100$. If $X$ is also Euclidean,
then one can set $\lambda_{1}=4/\pi$, $\lambda_{2}=12/\pi$ and $\gamma=38$.
Furthermore, we prove that if ${\mathcal{M}}$ is a subset of a Euclidean space
$E$, $\rho$ is the Euclidean metric in $E$, and $X$ is a two dimensional
Euclidean space, then properties (1.7) and (1.8) hold for $\lambda_{1}=1$,
$\lambda_{2}=3$, and $\gamma=25$.

In Section 5 we prove Theorem 5.5 which refines the result of Theorem 1.9 for
the space $X=\ell^{2}_{\infty}$, i.e., for ${\bf R}^{2}$ equipped with the
norm $\|x\|=\max\\{|x_{1}|,|x_{2}|\\}$, $x=(x_{1},x_{2})$. More specifically,
we show that in this case properties (1.7) and (1.8) hold whenever
$\lambda_{1}\geq 1$, $\lambda_{2}\geq 3\lambda_{1}$, and
$\gamma\geq\lambda_{2}\,(3\lambda_{2}+\lambda_{1})/(\lambda_{2}-\lambda_{1})$.
In particular, these properties hold for $\lambda_{1}=1$, $\lambda_{2}=3$ and
$\gamma=15$.

Let us now explicitly formulate the above mentioned second main result of the
paper. We prove it in Section 4. It deals with set-valued mappings from a
pseudometric space into the family ${\mathcal{K}}_{1}(X)$ of all bounded
closed line segments of an arbitrary Banach space $X$.

######  Theorem 1.10.

Let $({\mathcal{M}},\rho)$ be a pseudometric space. Let $m=1$ and let $X$ be a
Banach space with $\dim X>1$; thus, $\ell(m,X)=2$, see (1.1). In this case
Conjecture 1.8 holds for every $\lambda_{1},\lambda_{2}$ and $\gamma$ such
that

(1.10) $\displaystyle\lambda_{1}\geq 1,~{}~{}~{}~{}~{}~{}\lambda_{2}\geq
3\lambda_{1},~{}~{}~{}~{}~{}~{}\gamma\geq\lambda_{2}\,(3\lambda_{2}+\lambda_{1})/(\lambda_{2}-\lambda_{1}).$

Thus, the following statement is true: Let
$F:{\mathcal{M}}\to{\mathcal{K}}_{1}(X)$ be a set-valued mapping such that for
every subset ${\mathcal{M}}^{\prime}\subset{\mathcal{M}}$ with
$\\#{\mathcal{M}}^{\prime}\leq 4$, the restriction
$F|_{{\mathcal{M}}^{\prime}}$ of $F$ to ${\mathcal{M}}^{\prime}$ has a
Lipschitz selection with Lipschitz seminorm at most $1$.

Let $F^{[2]}$ be the mapping defined by (1.6). Then properties (1.7) and (1.8)
hold whenever $\lambda_{1}$, $\lambda_{2}$ and $\gamma$ satisfy (1.10). In
particular, one can set $\lambda_{1}=1$, $\lambda_{2}=3$ and $\gamma=15$.

If $X$ is a Euclidean space, the same statement is also true whenever, instead
of (1.10), $\lambda_{1},\lambda_{2}$ and $\gamma$ satisfy the weaker condition

(1.11) $\displaystyle\lambda_{1}\geq 1,~{}~{}~{}~{}~{}~{}\lambda_{2}\geq
3\lambda_{1},~{}~{}~{}~{}~{}~{}\gamma\geq\,\lambda_{2}+2\lambda_{2}^{2}\,{{/}}\left(\lambda_{2}^{2}-\lambda_{1}^{2}\right)^{\frac{1}{2}}.$

In particular, in this case, (1.7) and (1.8) hold whenever $\lambda_{1}=1$,
$\lambda_{2}=3$ and $\gamma=10$.

In Section 5.1 we note that Conjecture 1.8 also holds for a one dimensional
space $X$ and $m=1$. In this case the statement of the conjecture is true for
every $\lambda_{1}\geq 1$ and $\gamma\geq 1$. See Proposition 5.1.

Note that Theorem 1.9 tells us that for every set-valued mapping $F$
satisfying the hypothesis of this theorem, the mapping $F^{[2]}$ determined by
(1.6) with $\lambda_{1}=4/3$ and $\lambda_{2}=4$ provides a $\gamma$-core of
$F$ with $\gamma=100$. (See Definition 1.3.) In turn, Theorem 1.10 states that
the mapping $F^{[2]}$ corresponding to the parameters $\lambda_{1}=1$ and
$\lambda_{2}=3$ is a $15$-core of any $F$ satisfying the conditions of this
theorem.

We note that the proofs of Theorem 1.9 and Theorem 1.10 rely on Helly’s
Intersection Theorem and a series of auxiliary results about neighborhoods of
intersections of convex sets. See Section 2.

######  Remark 1.11.

Let us compare Conjecture 1.8 (and Theorems 1.9 and 1.10) with the Finiteness
Principle (FP) formulated in Theorem 1.2. First we note that FP is invariant
with respect to the transition to an equivalent norm on $X$, while the
statement of Conjecture 1.8 is not.

To express this more precisely, let $\|\cdot\|_{1}$ and $\|\cdot\|_{2}$ be two
equivalent norms on $X$, i.e., suppose that for some $\alpha\geq 1$ the
following inequality
$(1/\alpha)\,\|\cdot\|_{1}\leq\|\cdot\|_{2}\leq\alpha\,\|\cdot\|_{1}$ holds.
Clearly, if FP holds for $(X,\|\cdot\|_{1})$ then it immediately holds also
for $(X,\|\cdot\|_{2})$ (with the constant $\alpha^{2}\gamma$ instead of
$\gamma$). However the validity of Conjecture 1.8 for the norm $\|\cdot\|_{1}$
does not imply its validity for an equivalent norm $\|\cdot\|_{2}$ on $X$ (at
least we do not see any obvious way for obtaining such an implication). For
example, the validity of Conjecture 1.8 in $\ell^{n}_{\infty}$ (i.e., ${\bf
R}^{n}$ equipped with the uniform norm) does not automatically imply its
validity in the space $\ell^{n}_{2}$ (i.e., ${\bf R}^{n}$ with the Euclidean
norm).

We also note the following: in a certain sense, the result of Theorem 1.9 is
“stronger” than Theorem 1.2 (i.e., FP for the case of a two dimensional Banach
space $X$). Indeed, in this case, the hypotheses of FP and Theorem 1.9
coincide. Moreover, Theorem 1.9 ensures that the set-valued mapping $F^{[2]}$
is a core of $F$. This property of $F^{[2]}$ implies, via arguments in [16]
that the function

(1.12) $\displaystyle
f(x)=\operatorname{St}\,(F^{[2]})\,(x),~{}~{}~{}~{}~{}~{}x\in{\mathcal{M}},$

is a Lipschitz selection of $F$. Here
$\operatorname{St}:{\mathcal{K}}_{m}(X)\to X$ is the Steiner-type point map
[25].

Thus, FP (in the two dimensional case) follows immediately from Theorem 1.9.
However, it is absolutely unclear how the statement of Theorem 1.9 can be
deduced from FP. I would like to thank Charles Fefferman who kindly drew my
attention to this interesting fact. $\vartriangleleft$

Let us reformulate Conjecture 1.8 in a way which does not require the use of
the notion of a core of a set-valued mapping. We recall that the mapping
$F^{[\ell]}:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ which appears in Conjecture
1.8 is a $\gamma$-core of $F$ if

$\operatorname{d_{H}}(F^{[\ell]}(x),F^{[\ell]}(y))\leq\gamma\,\rho(x,y)~{}~{}~{}\text{for
all}~{}~{}~{}x,y\in{\mathcal{M}}.$

See part (ii) of Definition 1.3. Hence, given $x\in{\mathcal{M}}$,

(1.13) $\displaystyle F^{[\ell]}(x)\subset
F^{[\ell]}(y)+\gamma\,\rho(x,y)B_{X}~{}~{}~{}\text{for
every}~{}~{}~{}y\in{\mathcal{M}}.$

Let

$F^{[\ell+1]}(x)={\cal
BR\,}[{F^{[\ell]}}\\!:\\!{\gamma};{\rho}](x)=\bigcap_{y\in{\mathcal{M}}}\,\left[F^{[\ell]}(y)+\gamma\,\rho(x,y)\,B_{X}\right].$

Cf. (1.3). This and (1.13) imply the inclusion $F^{[\ell+1]}(x)\supset
F^{[\ell]}(x)$, $x\in{\mathcal{M}}$. On the other hand, (1.4) tells us that
$F^{[\ell+1]}(x)\subset F^{[\ell]}(x)$ proving that $F^{[\ell+1]}=F^{[\ell]}$
on ${\mathcal{M}}$.

These observations enable us to reformulate Conjecture 1.8 as follows.

######  Conjecture 1.12.

Let $({\mathcal{M}},\rho)$ be a pseudometric space, and let $X$ be a Banach
space. Let $m$ be a fixed positive integer and let $\ell=\ell(m,X)$, see(1.1).

There exists a sequence $\vec{\lambda}=\left\\{\lambda_{k}:1\leq
k\leq\ell+1\right\\}$ of $\ell+1$ numbers $\lambda_{k}$ all satisfying
$\lambda_{k}\geq 1$ such that, for every set-valued mapping
$F:{\mathcal{M}}\to{\mathcal{K}}_{m}(X)$ satisfying the hypothesis of the
Finiteness Principle (Theorem 1.2), the family $\\{F^{[k]}:k=1,...,\ell+1\\}$
of set-valued mappings constructed by formula (1.3) has the following
property:

(1.14) $\displaystyle
F^{[\ell]}(x)\neq\emptyset~{}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}~{}F^{[\ell+1]}(x)=F^{[\ell]}(x)~{}~{}~{}~{}\text{for
all}~{}~{}~{}x\in{\mathcal{M}}.$

We refer to (1.14) as a Stabilization Property of balanced refinements.

Thus, Theorem 1.9 and Theorem 1.10 tell us that a Stabilization Property of
balanced refinements holds whenever $\dim X=2$ or $m=1$ (and $X$ is an
arbitrary). More specifically, Theorem 1.9 shows that if $m=2$ and $\dim X=2$,
Conjecture 1.12 holds with $\ell=2$ and $\vec{\lambda}=\\{4/3,4,10^{2}\\}$.

In other words, in this case, $F^{[2]}(x)\neq\emptyset$ for each
$x\in{\mathcal{M}}$ and $F^{[3]}=F^{[2]}$ on ${\mathcal{M}}$. In turn, Theorem
1.10 states that the same property holds whenever $X$ is an arbitrary Banach
space, $m=1$, and $\vec{\lambda}=\\{1,3,15\\}$.

Acknowledgements. I am very thankful to Michael Cwikel for useful suggestions
and remarks. I am also very grateful to Charles Fefferman for stimulating
discussions and valuable advice.

The results of this paper were presented at the 12th Whitney Problems
Workshop, August 2019, the University of Texas at Austin, TX. I am very
thankful to all participants of that workshop for valuable conversations and
useful remarks. I am also grateful to the Conference Board of the Mathematical
Sciences and the University of Texas for supporting the Austin workshop.

##  2\. Neighborhoods of intersections of convex sets in a Banach space.

We first need to fix some notation.

Let $(X,\|\cdot\|)$ be a Banach space, and let $B_{X}$ be the unit ball in
$X$. Let $A$ and $B$ be non-empty subsets of $X$. We let $A+B=\\{a+b:a\in
A,b\in B\\}$ denote the Minkowski sum of these sets. By ${\mathfrak{C}}(X)$ we
denote the family of all convex closed non-empty subsets of $X$. We equip this
family with the Hausdorff distance.

Sometimes, for a given set ${\mathcal{M}}$, we will be looking simultaneously
at two distinct pseudometrics on ${\mathcal{M}}$, say $\rho$ and $\delta$. In
this case we will speak of a $\rho$-Lipschitz selection and $\rho$-Lipschitz
seminorm, or a $\delta$-Lipschitz selection and $\delta$-Lipschitz seminorm to
make clear which pseudometric we are using. Furthermore, given a mapping
$f:{\mathcal{M}}\to X$ we will write
$\|f\|_{\operatorname{Lip}(({\mathcal{M}};\rho),X)}$ or
$\|f\|_{\operatorname{Lip}(({\mathcal{M}};\delta),X)}$ to denote the Lipschitz
seminorm of $f$ with respect to the pseudometric $\rho$ or $\delta$
respectively.

For each finite set $S$, we let $\\#S$ denote the number of elements of $S$.

Given a Banach space $X$, Przesławski and Yost [21] introduced an important
geometrical characteristic of $X$, the so-called modulus of squareness of $X$.
Let us recall its definition.

We observe that for any $x,y\in X$ with $\|y\|<1<\|x\|$ there exists a unique
$z=z(x,y)$ with $\|z\|=1$ which belongs to the line segment $[x,y]$. We set

(2.1) $\displaystyle\omega(x,y)=\frac{\|x-z(x,y)\|}{\|x\|-1}$

and define a function $\xi:[0,1)\to[1,\infty)$ by

(2.2) $\displaystyle\xi_{X}(\beta)=\sup\,\\{\omega(x,y):x,y\in
X,\,\|y\|\leq\beta<1<\|x\|\,\\}.$

We also put

(2.3)
$\displaystyle\varphi(\beta)=(1+\beta)/(1-\beta)~{}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}~{}\psi(\beta)=(1-\beta^{2})^{-\frac{1}{2}},~{}~{}~{}~{}\beta\in[0,1).$

It is shown in [21], that for any Banach space $X$

(2.4) $\displaystyle\xi_{X}(\beta)\leq\varphi(\beta)~{}~{}~{}~{}\text{for
every}~{}~{}~{}\beta\in[0,1),$

and

(2.5) $\displaystyle\xi_{X}(\beta)=\psi(\beta)~{}~{}~{}~{}\text{for
every}~{}~{}~{}\beta\in[0,1),$

provided $X$ is a Euclidean space.

######  Theorem 2.1.

([21, Theorem 4]) Let $(S,\delta)$ be a metric space, let $X$ be a Banach
space, and let $f:S\to X$ and $g:S\to[0,\infty)$ be Lipschitz mappings. Let
$F:S\to{\mathfrak{C}}(X)$ be a Lipschitz (with respect to the Hausdorff
distance) set-valued mapping.

Suppose that there exists a constant $\gamma>1$ such that
$g(x)\geq\gamma\operatorname{dist}(f(x),F(x))$ for every $x\in S$. Then the
intersection mapping $G:S\to{\mathfrak{C}}(X)$ defined by

$G(x)=F(x)\cap B_{X}(f(x),g(x))$

is Lipschitz continuous on $S$ (with respect to $\operatorname{d_{H}}$) with
Lipschitz seminorm

$\|G\|_{\operatorname{Lip}(S,{\mathfrak{C}}(X))}\leq\|F\|_{\operatorname{Lip}(S,{\mathfrak{C}}(X))}+(\,\|F\|_{\operatorname{Lip}(S,{\mathfrak{C}}(X))}+\|f\|_{\operatorname{Lip}(S,X)}+\|g\|_{\operatorname{Lip}(S,{\bf
R})})\,\xi(1/\gamma).$

This theorem implies the following

######  Proposition 2.2.

Let $X$ be a Banach space, $a\in X$, $r\geq 0$, and let $C\subset X$ be a
convex set.

Suppose that $C\cap B_{X}(a,r)\neq\emptyset$. Then for every $s>0$ and every
$L>1$ the following inequality

$\operatorname{d_{H}}\left(C\cap B_{X}(a,Lr),(C+sB_{X})\cap
B_{X}(a,Lr+s)\right)\leq\left(1+2\,\xi_{X}\left(\tfrac{1}{L}\right)\right)\,s$

holds.

Proof. Let $S=\\{x,y\\}\subset{\bf R}$ where $x=0$ and $y=s$, and let
$\delta(x,y)=s$.

We define a mapping $f:S\to X$ and a function $g:S\to{\bf R}$ by letting
$f(x)=f(y)=a$ and $g(x)=Lr,g(y)=Lr+s$. Clearly,
$\|f\|_{\operatorname{Lip}(S,X)}=0$, and $\|g\|_{\operatorname{Lip}(S,{\bf
R})}=1$.

We put $\gamma=L$. We note that $C\cap B_{X}(a,r)\neq\emptyset$ so that

$\operatorname{dist}(f(x),F(x))=\operatorname{dist}(a,C)\leq r.$

Hence, $g(x)=Lr=\gamma\,r\geq\gamma\,\operatorname{dist}(f(x),F(x))$.

Then we define a mapping $F:S\to{\mathfrak{C}}(X)$ by setting $F(x)=C$ and
$F(y)=C+sB_{X}$. Clearly,

$\operatorname{d_{H}}(F(x),F(y))\leq s=\delta(x,y)~{}~{}~{}~{}\text{so
that}~{}~{}~{}~{}~{}\|F\|_{\operatorname{Lip}(S,{\mathfrak{C}}(X))}\leq 1.$

Thus, the conditions of Theorem 2.1 are satisfied for the metric space
$(S,\delta)$ and the mappings $f$, $g$ and $F$. This theorem tells us that the
mapping $G:S\to{\mathfrak{C}}(X)$ defined by

$G(u)=F(u)\cap B_{X}(f(u),g(u)),~{}~{}~{}~{}~{}~{}u\in S,$

is Lipschitz on $S$ with respect to the Hausdorff distance. Furthermore,

$\|G\|_{\operatorname{Lip}(S,{\mathfrak{C}}(X))}\leq\|F\|_{\operatorname{Lip}(S,{\mathfrak{C}}(X))}+(\,\|F\|_{\operatorname{Lip}(S,{\mathfrak{C}}(X))}+\|f\|_{\operatorname{Lip}(S,X)}+\|g\|_{\operatorname{Lip}(S,{\bf
R})})\,\xi(1/\gamma)\leq 1+2\,\xi_{X}\left(\tfrac{1}{L}\right).$

Hence,

$\displaystyle\operatorname{d_{H}}\left(C\cap B_{X}(a,Lr),(C+sB_{X})\cap
B_{X}(a,Lr+s)\right)$ $\displaystyle=$
$\displaystyle\operatorname{d_{H}}(G(x),G(y))$ $\displaystyle\leq$
$\displaystyle\|G\|_{\operatorname{Lip}(S,{\mathfrak{C}}(X))}\,\delta(x,y)\leq\left(1+2\,\xi_{X}\left(\tfrac{1}{L}\right)\right)\,s$

proving the proposition. $\blacksquare$

Proposition 2.2 implies the following important

######  Theorem 2.3.

Let $X$ be a Banach space, and let $C\subset X$ be a convex set. Let $a\in X$
and let $r\geq 0$. Suppose that

(2.6) $\displaystyle C\cap B_{X}(a,r)\neq\emptyset.$

Then for every $s>0$ and $L>1$

(2.7) $\displaystyle[C\cap
B_{X}(a,Lr)]+\theta(L)\,s\,B_{X}\supset(C+sB_{X})\cap B_{X}(a,Lr+s)$

where

(2.8) $\displaystyle\theta(L)=(3L+1)/(L-1).$

If $X$ is a Euclidean space then (2.7) holds with

(2.9) $\displaystyle\theta(L)=1+\frac{2L}{\sqrt{L^{2}-1}}\,.$

Proof. Let

$G=C\cap
B_{X}(a,Lr)~{}~{}~{}~{}\text{and}~{}~{}~{}~{}\widetilde{G}=(C+sB_{X})\cap
B_{X}(a,Lr+s).$

Definition (1.2) tells us that $\widetilde{G}\subset
G+\operatorname{d_{H}}(G,\widetilde{G})\,B_{X}$. In turn, Proposition 2.2
states that

$\operatorname{d_{H}}(G,\widetilde{G})\leq\left(1+2\,\xi_{X}\left(\tfrac{1}{L}\right)\right)\,s.$

Hence,

$\widetilde{G}\subset
G+\Theta(L)\,s\,B_{X}~{}~{}~{}~{}~{}\text{where}~{}~{}~{}~{}~{}\Theta(L)=1+2\,\xi_{X}\left(\tfrac{1}{L}\right).$

Now, let $X$ be an arbitrary Banach space. In this case, thanks to (2.3) and
(2.4), we have

$\Theta(L)=1+2\,\xi_{X}\left(\tfrac{1}{L}\right)\leq
1+2\frac{1+1/L}{1-1/L}=\frac{3L+1}{L-1}.$

This inequality and (2.8) imply inclusion (2.7) in the case under
consideration.

Finally, let $X$ be a Euclidean space. In this case, from (2.3), (2.5) and
(2.9), we have

$\Theta(L)=1+2\,\xi_{X}\left(\tfrac{1}{L}\right)=1+2\,(1-(1/L)^{2})^{-\frac{1}{2}}=1+\frac{2L}{\sqrt{L^{2}-1}}=\theta(L).$

The proof of the theorem is complete. $\blacksquare$

For the case of an arbitrary Banach space $X$, Theorem 2.3 was proved by
Przesławski and Rybinski [19, p. 279]. For the case of a Euclidean space $X$
see Przesławski, Yost [21, Theorem 4]. For similar results we refer the reader
to [1], [2, p. 369] and [4, p. 26].

For the sake of completeness, and for the reader’s convenience, below we give

A direct proof of Theorem 2.3. We follow the proof of Lemma 5.3 from [19, p.
279]. If $r=0$ then (2.7) holds trivially, so we assume that $r>0$. Without
loss of generality, we may also assume that $a=0$. Thus we should prove that

$[C\cap(LrB_{X})]+\theta s\,B_{X}\supset(C+sB_{X})\cap(LrB_{X}+sB_{X})$

provided $r>0$, $s>0$, $L>1$. Let

(2.10) $\displaystyle
z\in(C+sB_{X})\cap(LrB_{X}+sB_{X})=(C+sB_{X})\cap[(Lr+s)B_{X}].$

Prove that

(2.11) $\displaystyle z\in[C\cap(LrB_{X})]+\theta s\,B_{X}.$

Thanks to (2.10), $z\in(C+sB_{X})$ so that there exists an element $v\in C$
such that

(2.12) $\displaystyle\|v-z\|\leq s.$

If $\|v\|\leq Lr$, then $v\in C\cap(LrB_{X})$ proving (2.11).

Suppose that

(2.13) $\displaystyle\|v\|>Lr.$

Property (2.10) tells us that $\|z\|\leq Lr+s$ so that

(2.14) $\displaystyle\|v\|\leq\|z\|+s\leq Lr+2s.$

In turn, assumption (2.6) tells us that there exists an element $v^{\prime}\in
C$ such that

(2.15) $\displaystyle\|v^{\prime}\|\leq r<Lr.$

Choose $\lambda\in(0,1)$ such that the element

$\tilde{v}=\lambda v^{\prime}+(1-\lambda)v$

has the norm $\|\tilde{v}\|=Lr$. See Fig. 5.

Fig. 5: The points $z$, $v$, $\tilde{v}$ and $v^{\prime}$.

We know that $C$ is convex so that $[v^{\prime},v]\subset C$ proving that

(2.16) $\displaystyle\tilde{v}\in C\cap(LrB_{X}).$

Thanks to (2.15), (2.14) and the triangle inequality,

$Lr=\|\tilde{v}\|=\|\lambda v^{\prime}+(1-\lambda)v\|\leq\lambda
r+(1-\lambda)(Lr+2s)$

proving that

$\lambda\leq\frac{2s}{(L-1)r+2s}.$

Consequently, thanks to this inequality, (2.15) and (2.14)

$\|v-\tilde{v}\|=\lambda\|v-v^{\prime}\|\leq\lambda(\|v\|+\|v^{\prime}\|)\leq\frac{2s}{((L-1)r+2s)}\cdot(Lr+2s+r)\leq
2s(L+1)/(L-1).$

From this inequality and (2.12) we have

$\|z-\tilde{v}\|\leq\|z-v\|+\|v-\tilde{v}\|\leq s+2s(L+1)/(L-1)=\theta(L)\,s$

which together with (2.16) implies (2.11).

Let now $X$ be a Euclidean space. We modify the above proof after (2.16) as
follows.

We put $\beta=1/L$, and

(2.17) $\displaystyle
x={\tfrac{1}{Lr}}\,v,~{}~{}~{}~{}~{}~{}y={\tfrac{1}{Lr}}\,v^{\prime},~{}~{}~{}~{}~{}~{}w={\tfrac{1}{Lr}}\,\tilde{v}.$

Then, thanks to (2.13) and (2.15),

(2.18) $\displaystyle\|y\|\leq\beta<1<\|x\|.$

We note that for any $u,\tilde{u}\in X$ such that $\|\tilde{u}\|<1<\|u\|$,
there exists a unique point $w=w(u,\tilde{u})\in[u,\tilde{u}]$ with $\|w\|=1$.
Hence, thanks to (2.1),

(2.19)
$\displaystyle\omega(u,\tilde{u})=\frac{\|u-w(u,\tilde{u})\|}{\|u\|-1}.~{}~{}~{}~{}~{}~{}~{}~{}\text{See
Fig. 6.}$

Fig. 6: The points $u$, $w$ and $\tilde{u}$.

We also recall the definition of the function $\xi_{X}$, see (2.2):

(2.20)
$\displaystyle\xi_{X}(\beta)=\sup\,\\{\omega(u,\tilde{u}):u,\tilde{u}\in
X,\,\|\tilde{u}\|\leq\beta<1<\|u\|\,\\}.$

Prove that

(2.21) $\displaystyle\xi_{X}(\beta)=(1-\beta^{2})^{-\frac{1}{2}}.$

In fact, fix $u$ with $\|u\|>1$. One can easily see that
$\sup\,\\{\omega(u,\tilde{u}):\|\tilde{u}\|\leq\beta\\}$ is attained for some
$\tilde{u}$ with $\|\tilde{u}\|=\beta$, and the line segment $[\tilde{u},u]$
touches the sphere centered at the origin with radius $\beta$. Thus,
$u-\tilde{u}$ is perpendicular to $\tilde{u}$. See Fig. 7.

Fig. 7: The optimal position of the point $\tilde{u}$ for given $u$ with
$\|u\|=t>1$.

Hence, $\sup\\{\omega(u,\tilde{u}):\|\tilde{u}\|\leq\beta\\}=f(\|u\|)$ where

$f(t)=\frac{\sqrt{t^{2}-\beta^{2}}-\sqrt{1-\beta^{2}}}{t-1},~{}~{}~{}~{}t>1.$

The function $f$ is decreasing on $(1,+\infty)$ so that

$\sup_{t>1}f(t)=\lim_{t\to 1}f(t)=(1-\beta^{2})^{-\frac{1}{2}}$

proving (2.21).

We apply formula (2.21) to the points $x,y,w$ defined by (2.17), and to
$\beta=1/L$. We have

$\frac{\|v-\tilde{v}\|}{\|v\|-Lr}=\frac{\|x-w\|}{\|y\|-1}=\omega(x,y).$

See (2.19). Thanks to (2.18) and (2.20),

$\frac{\|v-\tilde{v}\|}{\|v\|-Lr}\leq\sup\,\\{\omega(u,\tilde{u}):u,\tilde{u}\in
X,\,\|\tilde{u}\|\leq\beta<1<\|u\|\,\\}=\xi_{X}(\beta)=\xi_{X}(1/L)$

so that, thanks to (2.21),

$\frac{\|v-\tilde{v}\|}{\|v\|-Lr}\leq\xi_{X}(1/L)=\frac{L}{\sqrt{L^{2}-1}}.$

In turn, thanks to (2.14), $\|v\|-Lr\leq 2s$, so that

$\|v-\tilde{v}\|\leq\frac{L}{\sqrt{L^{2}-1}}\,(\|v\|-Lr)\leq\frac{2sL}{\sqrt{L^{2}-1}}.$

This inequality and (2.12) imply the following:

$\|z-\tilde{v}\|\leq\|z-v\|+\|v-\tilde{v}\|\leq
s+\frac{2sL}{\sqrt{L^{2}-1}}=\left(1+2L/\sqrt{L^{2}-1}\,\right)\,s.$

This and (2.16) imply (2.11) with $\theta=\theta(L)$ defined by (2.9).

The proof of Theorem 2.3 is complete. $\blacksquare$

For the case of an arbitrary Banach space $X$, Theorem 2.3 was proved by
Przesławski and Rybinski [19, p. 279]. For the case of a Euclidean space $X$
see Przesławski, Yost [21, Theorem 4]. For similar results we refer the reader
to [1], [2, p. 369] and [4, p. 26].

######  Remark 2.4.

In general, assumption (2.6) in Theorem 2.3 cannot be omitted. For instance,
let $X=\ell^{2}_{\infty}$ and let $L=2$; then $\theta(L)=7$. Let $r\gg s>0$,
$a\in{\bf R}^{2}$, and let $C$ be a half-plane shown in Fig. 8. (In Fig. 8 –
11 we omit $X$ in the notation of $\ell^{2}_{\infty}$-balls.)

Fig. 8: The half-plane $C$ does not intersect the square $B(a,r)$.

The reader can easily see that in this case imbedding (2.7) does not hold
provided $r/s\to\infty$. Cf. Fig. 8 with Fig. 9 below.

Fig. 9: $\operatorname{diam}\\{[C+sB_{X}]\cap[B(a,2r)+sB]\\}\gg$
$\operatorname{diam}\\{[C\cap B(a,2r)]+7sB\\}$.

The obvious reason for this is the fact that if condition (2.6) is not met,
then the angle between the boundaries of $C$ and $B_{X}(a,2r)$ at the point of
their intersection is very small.

However, if $B_{X}(a,r)\cap C\neq\emptyset$, than, thanks to (2.7), the sets
$A_{1}=C\cap B_{X}(a,2r)+7sB_{X}$ and
$A_{2}=(C+sB_{X})\cap(B_{X}(a,Lr)+sB_{X})$ are “compatible”, i.e.,
$A_{1}\supset A_{2}$.

Note that in this case the angle between the boundaries of $C$ and
$B_{X}(a,2r)$ at the point of their intersection is not small. Cf. Fig. 10
with Fig. 11 below.

Fig. 10: The set $[C\cap B(a,2r)]+7sB$. Fig. 11: The set
$[C+sB_{X}]\cap[B(a,2r)+sB]$.

Let us recall the classical Helly Intersection Theorem for two dimensional
Banach spaces. It can be formulated as follows:

######  Theorem 2.5.

Let ${\mathcal{K}}$ be a collection of convex closed subsets of a two
dimensional Banach space $X$. Suppose that ${\mathcal{K}}$ is finite or at
least one member of the family ${\mathcal{K}}$ is bounded.

If every subfamily of ${\mathcal{K}}$ consisting of at most three elements has
a common point then there exists a point common to all of the family
${\mathcal{K}}$.

We conclude this section by stating and proving one more result which will be
one of the main tools for proving our main theorems.

######  Proposition 2.6.

Let $X$ be a two dimensional Banach space. Let $C,C_{1},C_{2}\subset X$ be
convex subsets, and let $r>0$. Suppose that

(2.22) $\displaystyle
C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rB_{X})\neq\emptyset.$

Then for every $L>1$ and every $\varepsilon>0$ the following inclusion

$\displaystyle[\\{(C_{1}\,{{\cap}}\,C_{2})+L\,rB_{X}\\}\,{{\cap}}\,C]+\theta(L)\,\varepsilon
B_{X}\supset$
$\displaystyle[(C_{1}\,{{\cap}}\,C_{2})+(L\,r+\varepsilon)B_{X}]\,{{\cap}}\,[\\{(C_{1}+rB_{X})\,{{\cap}}\,C\\}+\varepsilon
B_{X}]\,{{\cap}}\,[\\{(C_{2}+rB_{X})\,{{\cap}}\,C\\}+\varepsilon B_{X}]$

holds. The function $\theta$ in the above inclusion is as defined in Theorem
2.3. I.e., for an arbitrary Banach space $X$ that inclusion holds for
$\theta(L)=(3L+1)/(L-1)$, and if $X$ is Euclidean it also holds for
$\theta(L)=1+2L/\sqrt{L^{2}-1}$.

Proof. Let

$A=[(C_{1}\,{{\cap}}\,C_{2})+(L\,r+\varepsilon)B_{X}]\,{{\cap}}\,[\\{(C_{1}+rB_{X})\,{{\cap}}\,C\\}+\varepsilon
B_{X}]\,{{\cap}}\,[\\{(C_{2}+rB_{X})\,{{\cap}}\,C\\}+\varepsilon B_{X}]$

and let $a\in A$. Prove that

(2.23) $\displaystyle
a\in[\\{(C_{1}\,{{\cap}}\,C_{2})+L\,rB_{X}\\}\,{{\cap}}\,C]+\theta(L)\,\varepsilon
B_{X}.$

Let us show that if

(2.24) $\displaystyle
C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rB_{X})\,{{\cap}}\,B_{X}(a,Lr+\varepsilon)\neq\emptyset$

then (2.23) holds. Indeed, (2.24) provides the existence of a point $x\in X$
such that

(2.25) $\displaystyle x\in
C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rB_{X})\,{{\cap}}\,B_{X}(a,Lr+\varepsilon)\,.$

In particular, $x\in C+rB_{X}$ so that $B_{X}(x,r)\,{{\cap}}\,C\neq\emptyset$
proving that condition (2.6) of Theorem 2.3 holds. This theorem tells us that

(2.26) $\displaystyle[B_{X}(x,L\,r)\,{{\cap}}\,C]+\theta(L)\,\varepsilon
B_{X}\supset(C+\varepsilon B_{X})\,{{\cap}}\,B_{X}(x,L\,r+\varepsilon)\,.$

Since $a\in A$, we have $a\in C+\varepsilon B_{X}$. From (2.25) we learn that
and $a\in B_{X}(x,Lr+\varepsilon)$. Thus, the point $a$ belongs to the set
$(C+\varepsilon B_{X})\,{{\cap}}\,B_{X}(x,Lr+\varepsilon)$. Therefore, by
(2.26),

$a\in[B_{X}(x,Lr)\,{{\cap}}\,C]+\theta(L)\,\varepsilon
B_{X}=[(x+LrB_{X})\,{{\cap}}\,C]+\theta(L)\,\varepsilon B_{X}.$

But $x\in C_{1}\,{{\cap}}\,C_{2}$, see (2.25), and the required inclusion
(2.23) follows.

Thus, it remains to prove (2.24). Helly’s Theorem 2.5 tells us that this
property holds provided any three sets in the left hand side of (2.24) have a
common point. Note that, thanks to (2.22), this is true for $C_{1}$, $C_{2}$
and $C+rB_{X}$. Since the point $a\in A$, we have
$a\in[C_{1}\,{{\cap}}\,C_{2}]+(Lr+\varepsilon)B_{X}$, so that
$C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,B_{X}(a,Lr+\varepsilon)\neq\emptyset$.

Let us prove that

(2.27) $\displaystyle
C_{1}\,{{\cap}}\,(C+rB_{X})\,{{\cap}}\,B_{X}(a,Lr+\varepsilon)\neq\emptyset\,.$

The point $a\in A$ so that $a\in(C_{1}+rB_{X})\,{{\cap}}\,C+\varepsilon
B_{X}$. Let $b$ be a point nearest to $a$ on $(C_{1}+rB_{X})\,{{\cap}}\,C$,
and let $b_{1}\in C_{1}$ be a point nearest to $b$ on $C_{1}$. Prove that

(2.28) $\displaystyle b_{1}\in
C_{1}\,{{\cap}}\,(C+rB_{X})\,{{\cap}}\,B_{X}(a,Lr+\varepsilon)\,.$

Indeed, we have $\|a-b\|\leq\varepsilon$ and $\|b_{1}-b\|\leq r$. Thus,

(2.29) $\displaystyle b_{1}\in C_{1}~{}~{}~{}~{}\text{(by definition)~{}~{}
and}~{}~{}~{}~{}b_{1}\in C+rB_{X}~{}~{}~{}\text{(because $b\in C$)}.$

Furthermore,

$\|a-b_{1}\|\leq\|a-b\|+\|b-b_{1}\|\leq\varepsilon+r\leq\varepsilon+Lr$

proving that $b_{1}\in B_{X}(a,Lr+\varepsilon)$. Combining this property with
(2.29) we obtain (2.28) and (2.27). In a similar way we show that
$C_{2}\,{{\cap}}\,(C+rB_{X})\,{{\cap}}\,B_{X}(a,Lr+\varepsilon)\neq\emptyset$.

The proof of the proposition is complete. $\blacksquare$

##  3\. The main theorem for two dimensional Banach spaces.

In this section we prove Theorem 1.9

First, let us recall the notion of the Lipschitz extension constant
$e({\mathfrak{M}},X)$ which we use in the formulation of this theorem.

######  Definition 3.1.

Let ${\mathfrak{M}}=({\mathcal{M}},\rho)$ be a pseudometric space, and let $X$
be a Banach space. We define the Lipschitz extension constant
$e({\mathfrak{M}},X)$ of $X$ with respect to ${\mathfrak{M}}$ as the infimum
of the constants $\lambda>0$ such that for every subset
${\mathcal{M}}^{\prime}\subset{\mathcal{M}}$, and every Lipschitz mapping
$f:{\mathcal{M}}^{\prime}\to X$, there exists a Lipschitz extension
$\tilde{f}:{\mathcal{M}}\to X$ of $f$ to all of ${\mathcal{M}}$ such that
$\|\tilde{f}\|_{\operatorname{Lip}({\mathcal{M}},X)}\leq\lambda\|f\|_{\operatorname{Lip}({\mathcal{M}}^{\prime},X)}$.

######  Remark 3.2.

Recall several results about Lipschitz extension constants which we use in
this paper. In particular, thanks to the McShane-Whitney extension theorem,
$e({\mathfrak{M}},{\bf R})=1$ for every pseudometric space
${\mathfrak{M}}=({\mathcal{M}},\rho)$. Hence,
$e({\mathfrak{M}},\ell^{2}_{\infty})=1$ as well.

It follows from [22] and [8] that $e({\mathfrak{M}},X)\leq 4/3$ provided $X$
is an arbitrary two dimensional Banach space. See also [3]. Furthermore,
$e({\mathfrak{M}},X)\leq 4/\pi$ whenever $X$ is an arbitrary two dimensional
Euclidean space. See [22] and [17].

We also note that, thanks to Kirszbraun’s extension theorem [18],
$e({\mathfrak{M}},X)=1$ provided $X$ is a Euclidean space, ${\mathcal{M}}$ is
a subset of a Euclidean space $E$, and $\rho$ is the Euclidean metric in $E$.
$\vartriangleleft$

Proof of Theorem 1.9. Let ${\mathfrak{M}}=({\mathcal{M}},\rho)$ be a
pseudometric space, and let $X$ be a two dimensional Banach space. Let
$F:{\mathcal{M}}\to{\mathcal{K}}(X)$ be a set-valued mapping satisfying the
hypothesis of Theorem 1.9. Recall that, by this hypothesis, for every subset
$S\subset{\mathcal{M}}$ with $\\#S\leq 4$, the restriction $F|_{S}$ of $F$ to
$S$ has a $\rho$-Lipschitz selection $f_{S}:S\to X$ with
$\|f_{S}\|_{\operatorname{Lip}((S,\,\rho),X)}\leq 1$.

Let $\lambda_{1}$ and $\lambda_{2}$ be positive constants satisfying
inequalities (1.5), i.e., $\lambda_{1}\geq e({\mathfrak{M}},X)$ and
$\lambda_{2}\geq 3\lambda_{1}$. We set $L=\lambda_{2}/\lambda_{1}$. Thus, the
following inequalities

(3.1) $\displaystyle\lambda_{1}\geq
e({\mathfrak{M}},X),~{}~{}~{}~{}~{}~{}L\geq 3,$

hold. Then we introduce a new pseudometric on ${\mathcal{M}}$ defined by

(3.2)
$\displaystyle\operatorname{d}(x,y)=\lambda_{1}\,\rho(x,y),~{}~{}~{}~{}~{}x,y\in{\mathcal{M}}.$

This definition, Definition 3.1, the above hypothesis of Theorem 1.9 and the
inequality $\lambda_{1}\geq e({\mathfrak{M}},X)$ imply the following claim.

######  Claim 3.3.

Let $\widetilde{S}\subset{\mathcal{M}}$ be a finite set, and let
$S\subset\widetilde{S}$ be a set with $\\#S\leq 4$. Then there exists a
$\operatorname{d}$-Lipschitz mapping $\tilde{f}_{S}:\widetilde{S}\to X$ with
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((\widetilde{S},\operatorname{d}),X)}\leq
1$ such that $\tilde{f}_{S}(x)\in F(x)$ for every $x\in S$.

We introduce set-valued mappings

(3.3) $\displaystyle
F^{[1]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+\operatorname{d}(x,z)\,B_{X}\right],~{}~{}~{}~{}~{}x\in{\mathcal{M}},$

and

(3.4) $\displaystyle
F^{[2]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F^{[1]}(z)+L\operatorname{d}(x,z)\,B_{X}\right],~{}~{}~{}x\in{\mathcal{M}}.$

Thus, $F^{[1]}$ and $F^{[2]}$ are the first and the second order
$(\\{1,L\\},\operatorname{d})$-balanced refinements of $F$ respectively. See
Definition 1.6.

Our aim is to show that, if $L$ and $\lambda_{1}$ satisfy inequality (3.1),
then (i) $F^{[2]}(x)\neq\emptyset$ on ${\mathcal{M}}$, and (ii) the mapping
$F^{[2]}$ is $\operatorname{d}$-Lipschitz with respect to the Hausdorff
distance. We prove the statements (i) and (ii) in Proposition 3.8 and
Proposition 3.10 respectively.

We begin with the property (i). Its proof relies on a series of auxiliary
lemmas.

######  Lemma 3.4.

Let $X$ be a two dimensional Banach space, and let
${\mathcal{K}}\subset{\mathcal{K}}(X)$ be a collection of convex compact
subsets of $X$ with non-empty intersection.

Given $\tau>0$, let $B=\tau\,B_{X}$. Then

(3.5)
$\displaystyle\left(\,\bigcap_{K\in\,{\mathcal{K}}}K\right)+B=\bigcap_{K,K^{\prime}\in\,{\mathcal{K}}}\,\left\\{\,\left(\,K\,{{\cap}}\,K^{\prime}\right)+B\,\right\\}.$

Proof. Obviously, the right hand side of (3.5) contains its left hand side.
Let us prove the converse statement. Fix a point

(3.6) $\displaystyle
x\in\bigcap_{K,K^{\prime}\in\,{\mathcal{K}}}\,\left\\{\,\left(\,K\,{{\cap}}\,K^{\prime}\right)+B\,\right\\}$

and prove that $x\in\,{{\cap}}\,\\{K:K\in\,{\mathcal{K}}\\}+B$.

Clearly, it is true if
$B_{X}(x,\tau)\cap\,(\cap\\{K:K\in{\mathcal{K}}\\})\neq\emptyset$.

Let $S={\mathcal{K}}\,{{\cup}}\,\\{B_{X}(x,\tau)\\}$. Helly’s intersection
Theorem 2.5 tells us that this property holds provided
$\,{{\cap}}\,\\{K:K\in\,S^{\prime}\\}\neq\emptyset$ for every subfamily
$S^{\prime}\subset S$ consisting of at most three elements. Clearly, this is
true if $B_{X}(x,\tau)\notin S^{\prime}$ because there exists a point common
to all of the sets from ${\mathcal{K}}$.

Suppose that $B_{X}(x,\tau)\in S^{\prime}$. Then
$S^{\prime}=\\{B_{X}(x,\tau),K,K^{\prime}\\}$ for some
$K,K^{\prime}\in{\mathcal{K}}$. Thanks to (3.6),
$x\in\left(\,K{{\cap}}K^{\prime}\right)+B$ so that
$B_{X}(x,\tau)\,{{\cap}}\,K\,{{\cap}}\,K^{\prime}\neq\emptyset$.

Thus, $B_{X}(x,\tau)\cap\,(\cap\\{K:K\in{\mathcal{K}}\\})\neq\emptyset$, and
the proof of the lemma is complete. $\blacksquare$

######  Lemma 3.5.

For each $x\in{\mathcal{M}}$ the set $F^{[1]}(x)\in{\mathcal{K}}(X)$, i.e.,
$F^{[1]}(x)$ is a non-empty convex compact subset of $X$. Furthermore, for
every $x,z\in{\mathcal{M}}$ we have

$F^{[1]}(z)+L\operatorname{d}(x,z)B_{X}=\bigcap_{y^{\prime},y^{\prime\prime}\in{\mathcal{M}}}\left\\{\left[(F(y^{\prime})+\operatorname{d}(z,y^{\prime})B_{X})\,{{\cap}}\,(F(y^{\prime\prime})+\operatorname{d}(z,y^{\prime\prime})B_{X})\right]+L\operatorname{d}(x,z)B_{X}\right\\}.$

Proof. Prove that $F^{[1]}(x)\neq\emptyset$ for every $x\in{\mathcal{M}}$.
Indeed, formula (3.3) and Helly’s Theorem 2.5 tell us that
$F^{[1]}(x)\neq\emptyset$ provided

(3.7)
$\displaystyle[F(z_{1})+\operatorname{d}(x,z_{1})B_{X}]\,\,{{\cap}}\,\,[F(z_{2})+\operatorname{d}(x,z_{2})B_{X}]\,\,{{\cap}}\,\,[F(z_{3})+\operatorname{d}(x,z_{3})B_{X}]\neq\emptyset$

for every $z_{1},z_{2},z_{3}\in{\mathcal{M}}$.

To prove this property, we apply the hypothesis of Theorem 1.9 to the set
$S=\\{x,z_{1},z_{2},z_{3}\\}$. Thanks to this hypothesis, the restriction
$F|_{S}$ has a $\rho$-Lipschitz selection $f_{S}:S\to X$ with $\rho$-Lipschitz
seminorm at most $1$. Hence,

$\|f_{S}(x)-f_{S}(z_{i})\|\leq\rho(x,z_{i})\leq\lambda_{1}\rho(x,z_{i})=\operatorname{d}(x,z_{i})~{}~{}~{}\text{for
every}~{}~{}~{}i=1,2,3,$

proving that $f_{S}(x)$ belongs to the left hand side of (3.7). Thus, (3.7)
holds for arbitrary $z_{i}\in{\mathcal{M}}$, $i=1,2,3$, so that
$F^{[1]}(x)\neq\emptyset$.

Clearly, $F^{[1]}(x)$ is a convex bounded subset of $X$. See (3.3).

Finally, the second statement of the lemma immediately follows from this
property, Lemma 3.4 and formula (3.3). The proof of the lemma is complete.
$\blacksquare$

######  Lemma 3.6.

For every $x\in{\mathcal{M}}$ the set $F^{[2]}(x)$ admits the following
representation:

$F^{[2]}(x)=\bigcap_{u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}}\left\\{\left[(F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X})\,{{\cap}}\,(F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)B_{X})\right]+L\operatorname{d}(u,x)B_{X}\right\\}.$

Proof. The lemma is immediate from (3.4) and Lemma 3.5. $\blacksquare$

Given $x,u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}$ we set

(3.8) $\displaystyle
T_{x}(u,u^{\prime},u^{\prime\prime})=[(F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X})\,{{\cap}}\,(F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)B_{X})]+L\operatorname{d}(u,x)\,B_{X}\,.$

######  Remark 3.7.

Let us recall the standard definitions of a metric tree. Let $T=(V,E)$ be a
finite (graph theoretic) tree, where $V$ denotes the set of nodes of $T$, and
$E$ denotes the set of edges. Suppose we assign a positive number $\Delta(e)$
to each edge $e\in E$. Then for $x,y\in V$ we can define their distance
$\operatorname{d}_{T}(x,y)$ to be the sum of $\Delta(e)$ over all the edges
$e$ in the “minimal path” joining $x$ to $y$ as in Fig. 12.

Fig. 12: A minimal path joining nodes $x$ and $y$ in a tree. In this case,
$d(x,y)=\Delta(e_{1})+\Delta(e_{2})+...+\Delta(e_{5}).$

We call $\operatorname{d}_{T}$ a tree metric; $(V,\operatorname{d}_{T})$ is a
metric tree. We say that a mapping $g:V\to X$ agrees with the metric tree
${\mathfrak{M}}_{T}=(V,\operatorname{d}_{T})$ if the Lipschitz seminorm
$\|g\|_{\operatorname{Lip}((V,\operatorname{d}_{T}),X)}\leq 1$.

In these settings, given $x,u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}$,
the set $T_{x}(u,u^{\prime},u^{\prime\prime})$ defined by (3.8) admits the
following equivalent definition: Let $T$ be the metric tree with the family of
nodes $V=\\{x,u,u^{\prime},u^{\prime\prime}\\}$ and the set of edges
$E=\\{e_{1}=[u,x],e_{2}=[u^{\prime},u],e_{3}=[u^{\prime\prime},u]\\}$ and the
weights $w(e_{1})=L\,\operatorname{d}(u,x)$,
$w(e_{2})=\operatorname{d}(u^{\prime},u)$ and
$w(e_{3})=\operatorname{d}(u^{\prime\prime},u)$, as it shown in Fig. 13 below.

Fig. 13: $T_{x}(u,u^{\prime},u^{\prime\prime})$ is the orbit of $x$ with
respect to this diagram.

Then $T_{x}(u,u^{\prime},u^{\prime\prime})=\\{g(x)\\}$ where $g$ runs over all
Lipschitz mappings which agree with the metric tree
${\mathfrak{M}}_{T}=(V,\operatorname{d}_{T})$ such that $g(u^{\prime})\in
F(u^{\prime})$ and $g(u^{\prime\prime})\in F(u^{\prime\prime})$. We say that
the set $T_{x}(u,u^{\prime},u^{\prime\prime})$ is the orbit of $x$ with
respect to the diagram shown in Fig. 13. $\vartriangleleft$

In these settings, Lemma 3.6 reformulates as follows:

(3.9) $\displaystyle
F^{[2]}(x)=\bigcap_{u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}}T_{x}(u,u^{\prime},u^{\prime\prime}).$

######  Proposition 3.8.

For every $x\in{\mathcal{M}}$ the set $F^{[2]}(x)$ is non-empty.

Proof. Formula (3.9) and Helly’s Theorem 2.5 tell us that
$F^{[2]}(x)\neq\emptyset$ provided for every choice of elements
$u_{i},u^{\prime}_{i},u^{\prime\prime}_{i}\in{\mathcal{M}}$, $i=1,2,3$, we
have

(3.10) $\displaystyle
T_{x}(u_{1},u^{\prime}_{1},u^{\prime\prime}_{1})\,\,{{\cap}}\,\,T_{x}(u_{2},u^{\prime}_{2},u^{\prime\prime}_{2})\,\,{{\cap}}\,\,T_{x}(u_{3},u^{\prime}_{3},u^{\prime\prime}_{3})\neq\emptyset.$

We note that (3.10) holds provided there exists a mapping $g$ which agrees
with the diagram shown in Fig. 14 below.

Fig. 14: The mapping $g$ agrees with this diagram.

We set $r_{i}=\operatorname{d}(x,u_{i})$, $i=1,2,3$. Without loss of
generality, we may assume that $r_{1}\leq r_{2}\leq r_{3}$. For each
$i\in\\{1,2,3\\}$ we also set

(3.11) $\displaystyle
G(u^{\prime}_{i})=F(u^{\prime}_{i})+\operatorname{d}(u_{i}^{\prime},u_{i})B_{X}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}G(u^{\prime\prime}_{i})=F(u^{\prime\prime}_{i})+\operatorname{d}(u_{i}^{\prime\prime},u_{i})B_{X}.$

Note that the existence of a mapping $g$ on the set
$\\{x,u_{i},u^{\prime}_{i},u^{\prime\prime}_{i}:i=1,2,3\\}$ which agrees with
the diagram in Fig. 14 is equivalent to the existence of a mapping $g$ on
$\\{x,u_{1},u_{2},u_{3}\\}$ which agrees with the diagram shown in the
following Fig. 15.

Fig. 15: The mapping $g:\\{x,u_{1},u_{2},u_{3}\\}\to X$ agrees with this
diagram.

Let us prove that there exist points $y_{i}\in X$, $i=1,2,3$, such that

(3.12) $\displaystyle y_{i}\in
G(u^{\prime}_{i})\,\,{{\cap}}\,\,G(u^{\prime\prime}_{i})~{}~{}~{}~{}\text{for
every}~{}~{}~{}~{}i=1,2,3,$

and

(3.13) $\displaystyle\|y_{1}-y_{2}\|\leq
r_{1}+r_{2}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}\|y_{1}-y_{3}\|\leq
r_{1}+2r_{2}+r_{3}.~{}~{}~{}~{}~{}~{}~{}~{}~{}\text{See Fig. 16.}$

Fig. 16: The existence of points $\\{y_{i}:i=1,2,3\\}$ implies property
(3.10).

Let us see that the existence of the points $y_{i}$ with these properties
implies (3.10). Indeed, since $\|y_{1}-y_{3}\|\leq r_{1}+2r_{2}+r_{3}$, there
exists $z\in[y_{1},y_{3}]$ such that $\|z-y_{1}\|\leq r_{1}$ and
$\|z-y_{3}\|\leq 2r_{2}+r_{3}$. (For instance, one can set
$z=y_{1}+\tau(y_{3}-y_{1})$ with $\tau=r_{1}/(r_{1}+2r_{2}+r_{3})$.)

Hence,

$\|y_{2}-z\|\leq\|y_{2}-y_{1}\|+\|y_{1}-z\|\leq
r_{1}+r_{2}+r_{1}=2r_{1}+r_{2}.$

Recall that $r_{i}=\operatorname{d}(x,u_{i})$ and $r_{1}\leq r_{2}\leq r_{3}$.
From this and above inequalities, we have

(3.14) $\displaystyle\|z-y_{i}\|\leq
3r_{i}=3\operatorname{d}(x,u_{i}),~{}~{}~{}~{}i=1,2,3.$

Let us prove that $z\in T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})$ for
each $i\in\\{1,2,3\\}$. In fact, we know that $L\geq 3$, see (3.1).
Furthermore, by (3.12), $y_{i}\in
G(u^{\prime}_{i})\,{{\cap}}\,G(u^{\prime\prime}_{i})$, so that, thanks to
(3.11), (3.14) and definition (3.8),

$z\in[G(u^{\prime}_{i})\,\,{{\cap}}\,\,G(u^{\prime\prime}_{i})]+3\operatorname{d}(x,u_{i})B_{X}\subset[G(u^{\prime}_{i})\,\,{{\cap}}\,\,G(u^{\prime\prime}_{i})]+L\operatorname{d}(x,u_{i})B_{X}=T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})$

proving (3.10).

Thus, our aim is to prove the existence of points $y_{i}$ satisfying (3.12)
and (3.13). We will do this in three steps.

STEP 1. We introduce sets $W_{i}\subset X$, $i=1,...,4$, defined by

(3.15) $\displaystyle
W_{1}=G(u^{\prime}_{1}),~{}~{}~{}~{}~{}W_{2}=G(u^{\prime\prime}_{1}),~{}~{}~{}~{}W_{3}=[G(u^{\prime}_{2})\,{{\cap}}\,G(u^{\prime\prime}_{2})]+(r_{1}+r_{2})B_{X},$

and

(3.16) $\displaystyle
W_{4}=[G(u^{\prime}_{3})\,{{\cap}}\,G(u^{\prime\prime}_{3})]+(r_{1}+2r_{2}+r_{3})B_{X}.$

Obviously, there exist the points $y_{i}$ satisfying (3.12) and (3.13)
whenever

$W_{1}\,{{\cap}}\,W_{2}\,{{\cap}}\,W_{3}\,{{\cap}}\,W_{4}\neq\emptyset.$

See Fig. 17.

Fig. 17: The sets $W_{1}$, $W_{2}$, $W_{3}$ and $W_{4}$.

By Helly’s Theorem 2.5, this property holds provided any three members of the
family of sets $\\{W_{1},W_{2},W_{3},W_{4}\\}$ have a common point.

STEP 2. Prove that $W_{1}\,{{\cap}}\,W_{3}\,{{\cap}}\,W_{4}\neq\emptyset$. To
see this, we set

(3.17) $\displaystyle
V_{1}=G(u^{\prime}_{1})+(r_{1}+r_{2})B_{X},~{}~{}~{}~{}~{}V_{2}=G(u^{\prime}_{2}),~{}~{}~{}~{}V_{3}=G(u^{\prime\prime}_{2}),$

and

(3.18) $\displaystyle
V_{4}=[G(u^{\prime}_{3})\,{{\cap}}\,G(u^{\prime\prime}_{3})]+(r_{2}+r_{3})B_{X}.$

Let us show that if

(3.19) $\displaystyle
V_{1}\,{{\cap}}\,V_{2}\,{{\cap}}\,V_{3}\,{{\cap}}\,V_{4}\neq\emptyset,$

then $W_{1}\,{{\cap}}\,W_{3}\,{{\cap}}\,W_{4}$ is non-empty as well. Indeed,
this property, definitions (3.17) and (3.18) imply the existence of points
$z_{1}\in G(u^{\prime}_{1})$, $z_{2}\in
G(u^{\prime}_{2})\,{{\cap}}\,G(u^{\prime\prime}_{2})$, $z_{3}\in
G(u^{\prime}_{3})\,{{\cap}}\,G(u^{\prime\prime}_{3})$ such that
$\|z_{1}-z_{2}\|\leq r_{1}+r_{2}$ and $\|z_{2}-z_{3}\|\leq r_{2}+r_{3}$.
Hence,

$\|z_{1}-z_{3}\|\leq\|z_{1}-z_{2}\|+\|z_{2}-z_{3}\|\leq(r_{1}+r_{2})+(r_{2}+r_{3})=r_{1}+2r_{2}+r_{3}.$

Thus, thanks to (3.15) and (3.16), the point $z_{1}$ belongs to
$W_{1}\,{{\cap}}\,W_{3}\,{{\cap}}\,W_{4}$ proving that this set is non-empty.
smsk

Let us prove (3.19). Helly’s Theorem 2.5 tells us that (3.19) holds whenever
every three members of the family
${\mathcal{V}}=\\{V_{1},V_{2},V_{3},V_{4}\\}$ have a common point.

Let us prove this property. First, let us show that

(3.20) $\displaystyle V_{1}\,{{\cap}}\,V_{2}\,{{\cap}}\,V_{4}\neq\emptyset.$

Let
$S=\\{u^{\prime}_{1},u^{\prime}_{2},u^{\prime}_{3},u^{\prime\prime}_{3}\\}$
and let
$\widetilde{S}=\\{u_{1}^{\prime},u_{2},u_{2}^{\prime},u_{3},u_{3}^{\prime},u_{3}^{\prime\prime}\\}$.
Note that (3.20) holds if and only if there exists a mapping $\tilde{g}$ on
$\widetilde{S}$ which agrees with the diagram in Fig. 18 below.

Fig. 18: The mapping $\tilde{g}:\widetilde{S}\to X$ agrees with this diagram.

Note that, thanks to Claim 3.3, there exists a $\operatorname{d}$-Lipschitz
mapping $\tilde{f}_{S}:\widetilde{S}\to X$ with
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((\widetilde{S},\,\operatorname{d}),X)}\leq
1$ such that

$\tilde{f}_{S}(u^{\prime}_{1})\in
F(u^{\prime}_{1}),~{}~{}~{}~{}\tilde{f}_{S}(u^{\prime}_{2})\in
F(u^{\prime}_{2}),~{}~{}~{}~{}\tilde{f}_{S}(u^{\prime}_{3})\in
F(u^{\prime}_{3}),~{}~{}~{}~{}\text{and}~{}~{}~{}~{}\tilde{f}_{S}(u^{\prime\prime}_{3})\in
F(u^{\prime\prime}_{3}).$

Prove that

(3.21) $\displaystyle\tilde{f}_{S}(u_{2})\in
V_{1}\,{{\cap}}\,V_{2}\,{{\cap}}\,V_{4}.$

Indeed, $\tilde{f}_{S}(u^{\prime}_{2})\in F(u^{\prime}_{2})$ and
$\|\tilde{f}_{S}(u^{\prime}_{2})-\tilde{f}_{S}(u_{2})\|\leq\operatorname{d}(u^{\prime}_{2},u_{2})$,
so that

$\tilde{f}_{S}(u_{2})\in
F(u^{\prime}_{2})+\operatorname{d}(u^{\prime}_{2},u_{2})B_{X}=G(u^{\prime}_{2})=V_{2}.$

In the same way we prove that $\tilde{f}_{S}(u_{1})\in G(u^{\prime}_{1})$.

Note that
$\|\tilde{f}_{S}(u_{1})-\tilde{f}_{S}(u_{2})\|\leq\operatorname{d}(u_{1},u_{2})$
so that $\tilde{f}_{S}(u_{2})\in
G(u^{\prime}_{1})+\operatorname{d}(u_{1},u_{2})B_{X}$. By the triangle
inequality,

$\operatorname{d}(u_{1},u_{2})\leq\operatorname{d}(u_{1},x)+\operatorname{d}(x,u_{2})=r_{1}+r_{2}$

proving that

$\tilde{f}_{S}(u_{2})\in G(u^{\prime}_{1})+(r_{1}+r_{2})B_{X}=V_{1}.$

It remains to show that $\tilde{f}_{S}(u_{2})$ belongs to $V_{4}$. We know
that $\tilde{f}_{S}(u^{\prime}_{3})\in F(u^{\prime}_{3})$ and
$\tilde{f}_{S}(u^{\prime\prime}_{3})\in F(u^{\prime\prime}_{3})$. Furthermore,

$\|\tilde{f}_{S}(u_{3})-\tilde{f}_{S}(u^{\prime}_{3})\|\leq\operatorname{d}(u_{3},u^{\prime}_{3}),~{}~{}~{}~{}\|\tilde{f}_{S}(u_{3})-\tilde{f}_{S}(u^{\prime\prime}_{3})\|\leq\operatorname{d}(u_{3},u^{\prime\prime}_{3}).$

Hence,

$\tilde{f}_{S}(u_{3})\in[F(u^{\prime}_{3})+\operatorname{d}(u^{\prime}_{3},u_{3})B_{X}]\,{{\cap}}\,[F(u^{\prime\prime}_{3})+\operatorname{d}(u^{\prime\prime}_{3},u_{3})B_{X}]=G(u^{\prime}_{3})\,{{\cap}}\,G(u^{\prime\prime}_{3}).$

Furthermore,
$\|\tilde{f}_{S}(u_{2})-\tilde{f}_{S}(u_{3})\|\leq\operatorname{d}(u_{2},u_{3})$.
These properties of $\tilde{f}_{S}(u_{3})$ and the triangle inequality

$\operatorname{d}(u_{2},u_{3})\leq\operatorname{d}(u_{2},x)+\operatorname{d}(x,u_{3})=r_{2}+r_{3}$

imply the following:

$\tilde{f}_{S}(u_{2})\in[G(u^{\prime}_{3})\,{{\cap}}\,G(u^{\prime\prime}_{3})]+\operatorname{d}(u_{2},u_{3})B_{X}\subset[G(u^{\prime}_{3})\,{{\cap}}\,G(u^{\prime\prime}_{3})]+(r_{2}+r_{3})B_{X}=V_{4}.$

Thus, $\tilde{f}_{S}(u_{2})\in V_{1}\,{{\cap}}\,V_{2}\,{{\cap}}\,V_{4}$
proving (3.20).

In the same fashion we show that
$V_{1}\,{{\cap}}\,V_{3}\,{{\cap}}\,V_{4}\neq\emptyset$.

Next, let us prove that

(3.22) $\displaystyle
V_{2}\,{{\cap}}\,V_{3}\,{{\cap}}\,V_{4}=G(u^{\prime}_{2})\,{{\cap}}\,G(u^{\prime\prime}_{2})\,{{\cap}}\,\\{[G(u^{\prime}_{3})\,{{\cap}}\,G(u^{\prime\prime}_{3})]+(r_{2}+r_{3})B_{X}\\}\neq\emptyset.$

Following the scheme of the proof of (3.20), we put

$S=\\{u^{\prime}_{2},u^{\prime\prime}_{2},u^{\prime}_{3},u^{\prime\prime}_{3}\\}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}\widetilde{S}=\\{u_{2},u^{\prime}_{2},u^{\prime\prime}_{2},u_{3},u^{\prime}_{3},u^{\prime\prime}_{3}\\}.$

We note that (3.22) holds provided there exists a mapping
$\tilde{g}:\widetilde{S}\to X$ which agrees with the diagram on Fig. 19 below.

Fig. 19: The existence of $\tilde{g}:\widetilde{S}\to X$ which agrees with
this diagram implies (3.22).

Claim 3.3 tells us that there exists a $\operatorname{d}$-Lipschitz mapping
$\tilde{f}_{S}:\widetilde{S}\to X$ with Lipschitz seminorm
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((\widetilde{S},\,\operatorname{d}),X)}\leq
1$ such that $\tilde{f}_{S}(u^{\prime}_{i})\in F(u^{\prime}_{i})$ and
$\tilde{f}_{S}(u^{\prime\prime}_{i})\in F(u^{\prime\prime}_{i})$, $i=2,3$.

Then, following the scheme of the proof of (3.21), we show that
$\tilde{f}_{S}(u_{2})$ belongs to $V_{2}\,{{\cap}}\,V_{3}\,{{\cap}}\,V_{4}$
proving the required property (3.22).

Finally, following the same approach, we prove that

(3.23) $\displaystyle
V_{1}\,{{\cap}}\,V_{2}\,{{\cap}}\,V_{3}=[G(u^{\prime}_{1})+(r_{1}+r_{2})B_{X}]\,{{\cap}}\,G(u^{\prime}_{2})\,{{\cap}}\,G(u^{\prime\prime}_{2})\neq\emptyset.$

More specifically, we set
$S=\\{u^{\prime}_{1},u_{2},u^{\prime}_{2},u^{\prime\prime}_{2}\\}$. Then,
thanks to the hypothesis of Theorem 1.9, there exists a
$\operatorname{d}$-Lipschitz mapping $\tilde{f}_{S}:S\to X$ with
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((S,\,\operatorname{d}),X)}\leq 1$ such
that

$\tilde{f}_{S}(u^{\prime}_{1})\in
F(u^{\prime}_{1}),~{}~{}\tilde{f}_{S}(u^{\prime}_{2})\in
F(u^{\prime}_{2})~{}~{}~{}\text{and}~{}~{}\tilde{f}_{S}(u^{\prime\prime}_{2})\in
F(u^{\prime\prime}_{2}).$

Following the proof of (3.21), we show that $\tilde{f}_{S}(u_{2})\in
V_{1}\,{{\cap}}\,V_{2}\,{{\cap}}\,V_{3}$ completing the proof of (3.23).

Thus, (3.19) is proven, so that
$W_{1}\,{{\cap}}\,W_{3}\,{{\cap}}\,W_{4}\neq\emptyset$.

STEP 3. First, using a similar approach, we show that
$W_{2}\,{{\cap}}\,W_{3}\,{{\cap}}\,W_{4}\neq\emptyset$.

Next, we prove that

(3.24) $\displaystyle
W_{1}\,{{\cap}}\,W_{2}\,{{\cap}}\,W_{4}=G(u^{\prime}_{1})\,{{\cap}}\,G(u^{\prime\prime}_{1})\,{{\cap}}\,\\{[G(u^{\prime}_{3})\,{{\cap}}\,G(u^{\prime\prime}_{3})]+(r_{1}+2r_{2}+r_{3})B_{X}\\}\neq\emptyset.$

To see this, we set
$S=\\{u^{\prime}_{1},u^{\prime\prime}_{1},u^{\prime}_{3},u^{\prime\prime}_{3}\\}$
and
$\widetilde{S}=\\{u_{1},u^{\prime}_{1},u^{\prime\prime}_{1},u_{3},u^{\prime}_{3},u^{\prime\prime}_{3}\\}$.
Note that (3.24) holds whenever the distance between the set
$G(u_{1}^{\prime})\cap G(u_{1}^{\prime\prime})$ and the set
$G(u_{3}^{\prime})\cap G(u_{3}^{\prime\prime})$ is bounded by
$r_{1}+r_{2}+r_{3}$. See Fig. 20.

Fig. 20: The sets $W_{1}$, $W_{2}$ and $W_{4}$.

We note that Claim 3.3 tells us that there exists a
$\operatorname{d}$-Lipschitz mapping $\tilde{f}_{S}:\widetilde{S}\to X$ with
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((\widetilde{S},\,\operatorname{d}),X)}\leq
1$ such that $\tilde{f}_{S}(u^{\prime}_{i})\in F(u^{\prime}_{i})$ and
$\tilde{f}_{S}(u^{\prime\prime}_{i})\in F(u^{\prime\prime}_{i})$, $i=1,3$. The
reader can easily see that the point $\tilde{f}_{S}(u_{1})$ belongs to
$W_{1}\,{{\cap}}\,W_{2}\,{{\cap}}\,W_{4}$ so that (3.24) holds.

In a similar way we prove that

(3.25) $\displaystyle
W_{1}\,{{\cap}}\,W_{2}\,{{\cap}}\,W_{3}=G(u^{\prime}_{1})\,{{\cap}}\,G(u^{\prime\prime}_{1})\,{{\cap}}\,\\{[G(u^{\prime}_{2})\,{{\cap}}\,G(u^{\prime\prime}_{2})]+(r_{1}+r_{2})B_{X}\\}\neq\emptyset.$

More specifically, we set
$S=\\{u^{\prime}_{1},u^{\prime\prime}_{1},u^{\prime}_{2},u^{\prime\prime}_{2}\\}$,
$\widetilde{S}=\\{u_{1},u^{\prime}_{1},u^{\prime\prime}_{1},u_{2},u^{\prime}_{2},u^{\prime\prime}_{2}\\}$,
and apply Claim 3.3 to $S$ and $\widetilde{S}$. Thanks to this claim, there
exists a $\operatorname{d}$-Lipschitz mapping $\tilde{f}_{S}$ on
$\widetilde{S}$ with
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((\widetilde{S},\,\operatorname{d}),X)}\leq
1$ such that $\tilde{f}_{S}(u^{\prime}_{i})\in F(u^{\prime}_{i})$ and
$\tilde{f}_{S}(u^{\prime\prime}_{i})\in F(u^{\prime\prime}_{i})$, $i=1,2$. One
can readily see that the point $\tilde{f}_{S}(u_{1})\in
W_{1}\,{{\cap}}\,W_{2}\,{{\cap}}\,W_{3}$ proving the required property (3.25).

The proof of the proposition is complete. $\blacksquare$

We turn to the proof of inequality (1.8).

Note that, thanks to formula (3.9), for every $x,y\in{\mathcal{M}}$ we have

(3.26) $\displaystyle
F^{[2]}(x)=\bigcap_{u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}}T_{x}(u,u^{\prime},u^{\prime\prime})~{}~{}~{}\text{and}~{}~{}~{}F^{[2]}(y)=\bigcap_{u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}}T_{y}(u,u^{\prime},u^{\prime\prime}).$

######  Lemma 3.9.

For every $\tau>0$ and every $x\in{\mathcal{M}}$ the following representation

(3.27) $\displaystyle
F^{[2]}(x)+\tau\,B_{X}=\bigcap\,\,\left\\{\,\left[T_{x}(u,u^{\prime},u^{\prime\prime})\,{{\cap}}\,T_{x}(v,v^{\prime},v^{\prime\prime})\right]+\tau\,B_{X}\,\right\\}$

holds. Here the first intersection in the right hand side of (3.27) is taken
over all elements
$u,u^{\prime},u^{\prime\prime},v,v^{\prime},v^{\prime\prime}\in{\mathcal{M}}$.

Proof. The lemma is immediate from representation (3.26), Lemma 3.4 and
Proposition 3.8. $\blacksquare$

######  Proposition 3.10.

For every $x,y\in{\mathcal{M}}$ the following inequality

(3.28)
$\displaystyle\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\gamma_{0}(L)\operatorname{d}(x,y)$

holds. Here $\gamma_{0}(L)=L\,\theta(L)^{2}$ where $\theta(L)$ is the constant
from Theorem 2.3.

Proof. Let $x,y\in{\mathcal{M}}$ and let
$\tau=\gamma_{0}(L)\operatorname{d}(x,y)$. Prove that

(3.29) $\displaystyle
F^{[2]}(x)+\gamma_{0}(L)\operatorname{d}(x,y)B_{X}=F^{[2]}(x)+\tau
B_{X}\supset F^{[2]}(y)\,.$

Lemma 3.9 tells us that this inclusion holds provided

(3.30) $\displaystyle
A=[T_{x}(u,u^{\prime},u^{\prime\prime})\,{{\cap}}\,T_{x}(v,v^{\prime},v^{\prime\prime})]+\tau\,B_{X}\,\supset
F^{[2]}(y)$

for arbitrary
$u,u^{\prime},u^{\prime\prime},v,v^{\prime},v^{\prime\prime}\in{\mathcal{M}}$.
We note that the set $A$ is the orbit of $y$ with respect to the diagram shown
in Fig. 21. (I.e., the set $A$ consists of all points $a=g(y)$ where $g$ runs
over all mappings which agree with the diagram in Fig. 21.)

Fig. 21: The set $A$ is the orbit of $y$ with respect to this diagram.

In fact, we prove a stronger inclusion than (3.30), namely that

$\displaystyle A$ $\displaystyle\supset$ $\displaystyle
T_{y}(u,u^{\prime},u^{\prime\prime})\bigcap
T_{y}(v,v^{\prime},v^{\prime\prime})\bigcap
T_{y}(x,u^{\prime},v^{\prime})\bigcap$ $\displaystyle
T_{y}(x,u^{\prime},v^{\prime\prime})\bigcap
T_{y}(x,u^{\prime\prime},v^{\prime})\bigcap
T_{y}(x,u^{\prime\prime},v^{\prime\prime}).$

To prove the above inclusion, we introduce the following sets:

(3.31) $\displaystyle
C_{1}=F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X},~{}~{}~{}C_{2}=F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)B_{X},~{}~{}~{}C=T_{x}(v,v^{\prime},v^{\prime\prime})\,.$

Let

(3.32)
$\displaystyle\varepsilon=L\,\theta(L)\operatorname{d}(x,y)~{}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}~{}r=\operatorname{d}(x,u)\,.$

Then $\tau=\gamma_{0}(L)\operatorname{d}(x,y)=\theta(L)\,\varepsilon$, and

$A=[T_{x}(u,u^{\prime},u^{\prime\prime})\,{{\cap}}\,T_{x}(v,v^{\prime},v^{\prime\prime})]+\tau\,B_{X}=[(C_{1}\,{{\cap}}\,C_{2})+LrB_{X}]\,{{\cap}}\,C+\theta(L)\,\varepsilon\,B_{X}.$

See Fig. 22.

Fig. 22: The sets $C$, $C_{1}$ and $C_{2}$.

We want to apply Proposition 2.6 to the set $A$. To do this, we have to verify
condition (2.22) of this proposition, i.e., to show that

(3.33) $\displaystyle
C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rB_{X})\neq\emptyset.$

Let $S=\\{u^{\prime},u^{\prime\prime},v^{\prime},v^{\prime\prime}\\}$ and let
$\widetilde{S}=\\{x,u,u^{\prime},u^{\prime\prime},v,v^{\prime},v^{\prime\prime}\\}$.
Claim 3.3 tells us that there exists a $\operatorname{d}$-Lipschitz mapping
$\tilde{f}_{S}:\widetilde{S}\to X$ with $\operatorname{d}$-Lipschitz seminorm
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((\widetilde{S},\,\operatorname{d}),X)}\leq
1$ such that

$\tilde{f}_{S}(u^{\prime})\in
F(u^{\prime}),~{}~{}\tilde{f}_{S}(u^{\prime\prime})\in
F(u^{\prime\prime}),~{}~{}\tilde{f}_{S}(v^{\prime})\in
F(v^{\prime})~{}~{}~{}\text{and}~{}~{}~{}\tilde{f}_{S}(v^{\prime\prime})\in
F(v^{\prime\prime}).$

Prove that $\tilde{f}_{S}(u)$ belongs to the left hand side of (3.33). Indeed,
thanks to the inequality
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((\widetilde{S},\,\operatorname{d}),X)}\leq
1$, we have

$\|\tilde{f}_{S}(u^{\prime})-\tilde{f}_{S}(u)\|\leq\operatorname{d}(u^{\prime},u),~{}~{}~{}\|\tilde{f}_{S}(u^{\prime\prime})-\tilde{f}_{S}(u)\|\leq\operatorname{d}(u^{\prime\prime},u)~{}~{}~{}\|\tilde{f}_{S}(x)-\tilde{f}_{S}(u)\|\leq\operatorname{d}(x,u)=r.$

and $\|\tilde{f}_{S}(x)-\tilde{f}_{S}(u)\|\leq\operatorname{d}(x,u)=r$. Thanks
to these properties and (3.31), $\tilde{f}_{S}(u)\in C_{1}\,{{\cap}}\,C_{2}$.

In a similar way we show that $\tilde{f}_{S}(x)\in
T_{x}(v,v^{\prime},v^{\prime\prime})=C$, see (3.31) and (3.8). Hence, we have
$\tilde{f}_{S}(u)\in C+rB_{X}$. Thus,

$C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rB_{X})\ni\tilde{f}_{S}(u)$

proving (3.33).

We see that property (2.22) of Proposition 2.6 holds. This proposition tells
us that

$\displaystyle A$
$\displaystyle=[(C_{1}\,{{\cap}}\,C_{2})+LrB_{X}]\,{{\cap}}\,C+\theta(L)\,\varepsilon\,B_{X}$
$\displaystyle\supset[(C_{1}\,{{\cap}}\,C_{2})+(Lr+\varepsilon)B_{X}]\,{{\cap}}\,\\{[(C_{1}+rB_{X})\,{{\cap}}\,C]+\varepsilon
B_{X}\\}\,{{\cap}}\,\\{[(C_{2}+rB_{X})\,{{\cap}}\,C]+\varepsilon B_{X}\\}$
$\displaystyle=A_{1}\,{{\cap}}\,A_{2}\,{{\cap}}\,A_{3}.$

Prove that

(3.34) $\displaystyle A_{i}\supset F^{[2]}(y)~{}~{}~{}\text{for
every}~{}~{}~{}i=1,2,3.$

We begin with the set

$A_{1}=[C_{1}\,{{\cap}}\,C_{2}]+(Lr+\varepsilon)B_{X}.$

Thus,

$A_{1}=[(F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X})\,{{\cap}}\,(F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)B_{X})]+(L\operatorname{d}(u,x)+L\,\theta(L)\operatorname{d}(x,y))B_{X}\,.$

See (3.31). By the triangle inequality,

$\operatorname{d}(u,x)+\theta(L)\operatorname{d}(x,y)\geq\operatorname{d}(u,x)+\operatorname{d}(x,y)\geq\operatorname{d}(u,y).$

Hence,

$A_{1}\supset[(F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X})\,{{\cap}}\,(F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)B_{X})]+L\operatorname{d}(u,y)B_{X}=T_{y}(u,u^{\prime},u^{\prime\prime})\,.$

But $T_{y}(u,u^{\prime},u^{\prime\prime})\supset F^{[2]}(y)$, see (3.26), so
that $A_{1}\supset F^{[2]}(y)$.

We turn to the proof of the inclusion $A_{2}\supset F^{[2]}(y)$. Note that
$A_{2}$ is defined by

(3.35) $\displaystyle A_{2}=[(C_{1}+rB_{X})\,{{\cap}}\,C]+\varepsilon B_{X}.$

By the triangle inequality,

(3.36) $\displaystyle
C_{1}+rB_{X}=F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X}+\operatorname{d}(u,x)B_{X}\supset
F(u^{\prime})+\operatorname{d}(u^{\prime},x)B_{X}\,.$

Let

(3.37)
$\displaystyle\widetilde{C}=F(u^{\prime})+\operatorname{d}(u^{\prime},x)B_{X},~{}~{}~{}\widetilde{C}_{1}=F(v^{\prime})+\operatorname{d}(v^{\prime},v)B_{X},~{}~{}~{}\widetilde{C}_{2}=F(v^{\prime\prime})+\operatorname{d}(v^{\prime\prime},v)B_{X},$

and let

(3.38) $\displaystyle\tilde{r}=\operatorname{d}(v,x).$

In these settings

$C=T_{x}(v,v^{\prime},v^{\prime\prime})=[\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2}]+L\tilde{r}B_{X}.$

Let

(3.39)
$\displaystyle\widetilde{A}=\\{[(\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2})+L\tilde{r}B_{X}]\,{{\cap}}\,\widetilde{C}\\}+\varepsilon
B_{X}.$

Then, thanks to (3.35) and (3.36),

$A_{2}\supset[(F(u^{\prime})+\operatorname{d}(u^{\prime},x)B_{X})\,{{\cap}}\,C]+\varepsilon
B_{X}=\\{[(\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2})+L\tilde{r}B_{X}]\,{{\cap}}\,\widetilde{C}\\}+\varepsilon
B_{X}=\widetilde{A}\,.$

See Fig. 23.

Fig. 23: The sets $\widetilde{C}$, $\widetilde{C}_{1}$ and
$\widetilde{C}_{2}$.

Prove that $\widetilde{A}\supset F^{[2]}(y)$. We will do this by applying
Proposition 2.6 to the set $\widetilde{A}$. But first we must check the
hypothesis of this proposition, i.e., to show that

(3.40)
$\displaystyle\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2}\,{{\cap}}\,(\widetilde{C}+\tilde{r}B_{X})\neq\emptyset\,.$

To establish this property, we set
$S=\\{u^{\prime},v^{\prime},v^{\prime\prime}\\}$ and
$\widetilde{S}=\\{x,u^{\prime},v,v^{\prime},v^{\prime\prime}\\}$. Claim 3.3
tells us that there exists a $\operatorname{d}$-Lipschitz mapping
$\tilde{f}_{S}:\widetilde{S}\to X$ with $\operatorname{d}$-Lipschitz seminorm
$\|\tilde{f}_{S}\|_{\operatorname{Lip}((\widetilde{S},\,\operatorname{d}),X)}\leq
1$ such that

$\tilde{f}_{S}(u^{\prime})\in F(u^{\prime}),~{}~{}\tilde{f}_{S}(v^{\prime})\in
F(v^{\prime})~{}~{}~{}\text{and}~{}~{}~{}\tilde{f}_{S}(v^{\prime\prime})\in
F(v^{\prime\prime}).$

Combining these properties of $\tilde{f}_{S}$ with definitions (3.37) and
(3.38), we conclude that
$\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2}\,{{\cap}}\,(\widetilde{C}+\tilde{r}B_{X})\ni\tilde{f}_{S}(v)$
proving (3.40).

We recall that $\varepsilon=L\,\theta(L)\operatorname{d}(x,y)$, see (3.32), so
that

$\widetilde{A}=[\\{(\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2})+L\tilde{r}B_{X}\\}\,{{\cap}}\,\widetilde{C}]+L\,\theta(L)\operatorname{d}(x,y)B_{X},~{}~{}~{}~{}\text{(see
(\ref{TC-1})).}$

We apply Proposition 2.6 to $\widetilde{A}$ and obtain the following:

$\displaystyle\widetilde{A}$
$\displaystyle\supset\\{(\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2})+(L\tilde{r}+L\operatorname{d}(x,y))B_{X}\\}$
$\displaystyle\,{{\cap}}\,\\{[(\widetilde{C}_{1}+\tilde{r}B_{X})\,{{\cap}}\,\widetilde{C}]+L\operatorname{d}(x,y)B_{X}\\}\,{{\cap}}\,\\{[(\widetilde{C}_{2}+\tilde{r}B_{X})\,{{\cap}}\,\widetilde{C}]+L\operatorname{d}(x,y)B_{X}\\}$
$\displaystyle=\widetilde{A}_{1}\,{{\cap}}\,\widetilde{A}_{2}\,{{\cap}}\,\widetilde{A}_{3}.$

Prove that $\widetilde{A}_{i}\supset F^{[2]}(y)$ for every $i=1,2,3$. First,
let us show that

(3.41)
$\displaystyle\widetilde{A}_{1}=(\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2})+(L\tilde{r}+L\,\operatorname{d}(x,y))B_{X}\supset
F^{[2]}(y).$

By (3.38) and the triangle inequality,
$\tilde{r}+\operatorname{d}(x,y)=\operatorname{d}(v,x)+\operatorname{d}(x,y)\geq\operatorname{d}(v,y)$
so that

$\displaystyle\widetilde{A}_{1}$
$\displaystyle\supset(\widetilde{C}_{1}\,{{\cap}}\,\widetilde{C}_{2})+L\operatorname{d}(v,y)B_{X}$
$\displaystyle=[(F(v^{\prime})+\operatorname{d}(v^{\prime},v)B_{X})\,{{\cap}}\,(F(v^{\prime\prime})+\operatorname{d}(v^{\prime\prime},v)B_{X})]+L\operatorname{d}(v,y)B_{X}$
$\displaystyle=T_{y}(v,v^{\prime},v^{\prime\prime}).$

See (3.37) and (3.8). This inclusion and (3.26) imply (3.41).

Next, let us show that

(3.42)
$\displaystyle\widetilde{A}_{2}=[(\widetilde{C}_{1}+\tilde{r}B_{X})\,{{\cap}}\,\widetilde{C}]+L\operatorname{d}(x,y)B_{X}\supset
F^{[2]}(y)).$

Thanks to (3.37), (3.38) and the triangle inequality,

$\widetilde{C}_{1}+\tilde{r}B_{X}=F(v^{\prime})+\operatorname{d}(v^{\prime},v)B_{X}+\operatorname{d}(v,x)B_{X}\supset
F(v^{\prime})+\operatorname{d}(v^{\prime},x)B_{X}$

so that

$\widetilde{A}_{2}\supset[(F(v^{\prime})+\operatorname{d}(v^{\prime},x)B_{X})\,{{\cap}}\,(F(u^{\prime})+\operatorname{d}(u^{\prime},x)B_{X})]+L\,\operatorname{d}(x,y)B_{X}=T_{y}(x,u^{\prime},v^{\prime}).$

See (3.8). From this and (3.26), we have $\widetilde{A}_{2}\supset
T_{y}(x,u^{\prime},v^{\prime})\supset F^{[2]}(y)$, proving (3.42).

In the same way we show that

$\widetilde{A}_{3}=[(\widetilde{C}_{2}+\tilde{r}B_{X})\,{{\cap}}\,\widetilde{C}]+L\operatorname{d}(x,y)B_{X}\supset
T_{y}(x,u^{\prime},v^{\prime\prime})\supset F^{[2]}(y)\,.$

Combining this with (3.41) and (3.42), we obtain the required inclusion
$\widetilde{A}_{i}\supset F^{[2]}(y)$ for every $i=1,2,3$. In turn, this
proves that

$\widetilde{A}\supset
F^{[2]}(y)~{}~{}~{}~{}~{}\text{because}~{}~{}~{}~{}\widetilde{A}\supset\widetilde{A}_{1}\,{{\cap}}\,\widetilde{A}_{2}\,{{\cap}}\,\widetilde{A}_{3}\supset
F^{[2]}(y).$

We know that $A_{2}\supset\widetilde{A}$, so that $A_{2}\supset F^{[2]}(y)$.
In the same fashion we show that

$A_{3}=[(C_{2}+rB_{X})\,{{\cap}}\,C]+L\,\varepsilon\,B_{X}\supset F^{[2]}(y)$

proving (3.34). Hence, $A\supset
A_{1}\,{{\cap}}\,A_{2}\,{{\cap}}\,A_{3}\supset F^{[2]}(y)$ so that (3.30)
holds.

Thus, (3.29) is proved. By interchanging the roles of elements $x$ and $y$ in
this inclusion we obtain the inclusion
$F^{[2]}(y)+\gamma_{0}(L)\operatorname{d}(x,y)B_{X}\supset F^{[2]}(x)$. These
two inclusions imply inequality (3.28) proving the proposition. $\blacksquare$

We are in a position to complete the proof of Theorem 1.9.

Recall that $\lambda_{1},\lambda_{2}$ are parameters satisfying (1.5), and
$L=\lambda_{2}/\lambda_{1}$. Thus, $L$ and $\lambda_{1}$ satisfy (3.1). We
also recall that $\operatorname{d}=\lambda_{1}\,\rho$, see (3.2). Let $\gamma$
be a parameter satisfying (1.5).

In these settings, the mappings $F^{[1]}$ and $F^{[2]}$ defined by formulae
(3.3) and (3.4) are the first and the second order
$(\\{\lambda_{1},\lambda_{2}\\},\rho)$-balanced refinements of $F$
respectively. See Definition 1.6.

Proposition 3.8 tells us that, under these conditions,
$F^{[2]}(x)\neq\emptyset$ on ${\mathcal{M}}$. In turn, Proposition 3.10 states
that

$\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\gamma_{0}(L)\operatorname{d}(x,y)~{}~{}~{}~{}~{}\text{for
all}~{}~{}~{}~{}~{}x,y\in{\mathcal{M}}.$

Recall that $\gamma_{0}(L)=L\cdot\theta(L)^{2}$ where
$\theta=\theta(L)=(3L+1)/(L-1)$, see (2.8). Hence,
$\theta(L)=(3\lambda_{2}+\lambda_{1})/(\lambda_{2}-\lambda_{1})$ so that

$\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\gamma_{0}(L)\operatorname{d}(x,y)\leq
L\cdot\theta(L)^{2}\,\operatorname{d}(x,y))=\lambda_{2}\,\frac{(3\lambda_{2}+\lambda_{1})^{2}}{(\lambda_{2}-\lambda_{1})^{2}}\,\rho(x,y).$

Combining this inequality with the third inequality in (1.5), we obtain (1.8)
proving Theorem 1.9 for the parameters $\lambda_{1},\lambda_{2}$ and $\gamma$
satisfying (1.5).

In particular, one can set $\lambda_{1}=4/3$, $\lambda_{2}=4$, and
$\gamma=100$. Indeed, in this case $e({\mathfrak{M}},X)\leq\lambda_{1}=4/3$,
see Remark 3.2, so that $\lambda_{1},\lambda_{2}$ and $\gamma$ satisfy (1.5).

Next, let $X$ be a two dimensional Euclidean space, and let
$\lambda_{1},\lambda_{2}$ and $\gamma$ satisfy (1.9). In this case, we prove
(1.7) and (1.8) by replacing in the proof of Theorem 1.9 the function
$\theta=\theta(L)$ defined by (2.8) with the function
$\theta(L)=1+2\,L/\sqrt{L^{2}-1}$. We leave the details to the interested
reader.

In particular, we can set $\lambda_{1}=4/\pi$, $\lambda_{2}=12/\pi$,
$\gamma=38$. Indeed, in this case, $e({\mathfrak{M}},X)\leq 4/\pi$, see Remark
3.2, which implies (1.9) for these values of parameters $\lambda_{1}$,
$\lambda_{2}$ and $\gamma$.

Finally, suppose that $X$ is a Euclidean space, ${\mathcal{M}}$ is a subset of
a Euclidean space $E$, and $\rho$ is the Euclidean metric in $E$. In this case
$e({\mathfrak{M}},X)=1$, see Remark 3.2, so that one can set $\lambda_{1}=1$,
$\lambda_{2}=3$ and $\gamma=25$. Clearly, in this case inequalities (1.9)
hold.

The proof of Theorem 1.9 is complete. $\blacksquare$

##  4\. Balanced refinements of line segments in a Banach space.

In this section we prove Theorem 1.10.

Let $({\mathcal{M}},\rho)$ be a pseudometric space, and let $(X,\|\cdot\|)$ be
a Banach space. We assume that $\dim X>1$. Let us recall that
${\mathcal{K}}_{1}(X)$ is the family of all non-empty convex compacts in $X$
of dimension at most $1$ (i.e., the family of all points and all bounded
closed line segments in $X$).

We need the following version of one dimensional Helly’s Theorem.

######  Theorem 4.1.

Let ${\mathcal{K}}$ be a collection of closed convex subsets of $X$ containing
a set $K_{0}\in{\mathcal{K}}_{1}(X)$. If $K_{0}$ has a common point with any
two members of ${\mathcal{K}}$, then there exists a point common to all of the
collection ${\mathcal{K}}$.

Proof. We introduce a family
$\widetilde{{\mathcal{K}}}=\\{K\,{{\cap}}\,K_{0}:K\in{\mathcal{K}}\\}$, and
apply to $\widetilde{{\mathcal{K}}}$ one dimensional Helly’s Theorem. (See
part (a) of Lemma 5.2.) $\blacksquare$

We also need the following variant of Proposition 2.6 for the family
${\mathcal{K}}_{1}(X)$.

######  Proposition 4.2.

Let $X$ be a Banach space, and let $r\geq 0$. Let $C,C_{1},C_{2}\subset X$ be
convex closed subsets, and let $C_{1}\in{\mathcal{K}}_{1}(X)$. Suppose that

(4.1) $\displaystyle
C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rB_{X})\neq\emptyset.$

Then for every $L>1$ and every $\varepsilon>0$ the following inclusion

$[\\{(C_{1}\,{{\cap}}\,C_{2})+LrB_{X}\\}\,{{\cap}}\,C]+\theta(L)\,\varepsilon
B_{X}\supset[(C_{1}\,{{\cap}}\,C_{2})+(Lr+\varepsilon)B_{X}]\,{{\cap}}\,[\\{(C_{1}+rB_{X})\,{{\cap}}\,C\\}+\varepsilon
B_{X}]$

holds. Here $\theta(L)$ is the same as in Theorem 2.3, i.e.,
$\theta(L)=(3L+1)/(L-1)$ for an arbitrary $X$, and
$\theta(L)=1+2L/\sqrt{L^{2}-1}$ whenever $X$ is a Euclidean space.

Proof. Let

(4.2) $\displaystyle a\in[(C_{1}\cap
C_{2})+(Lr+\varepsilon)B_{X}]\cap\\{[(C_{1}+rB_{X})\cap C]+\varepsilon
B_{X}\\}.$

Prove that

(4.3) $\displaystyle a\in\\{[(C_{1}\cap C_{2})+LrB_{X}]\cap
C\\}+\theta(L)\,\varepsilon B_{X}\,.$

First, let us show that

(4.4) $\displaystyle C_{1}\cap C_{2}\cap(C+rB_{X})\cap
B_{X}(a,Lr+\varepsilon)\neq\emptyset.$

Recall that $C_{1}\in{\mathcal{K}}_{1}(X)$. Helly’s Theorem 4.1 tells us that
it is suffices to show that any two sets in the left hand size of (4.4) have a
common point with $C_{1}$.

First we note that $C_{1}$, $C_{2}$ and $C+rB_{X}$ have a common point. See
(4.1). We also know that

$a\in(C_{1}\cap C_{2})+(Lr+\varepsilon)B_{X},$

see (4.2), so that $C_{1}\cap C_{2}\cap B_{X}(a,Lr+\varepsilon)\neq\emptyset$.

Let us prove that

(4.5) $\displaystyle C_{1}\cap(C+rB_{X})\cap
B_{X}(a,Lr+\varepsilon)\neq\emptyset\,.$

Property (4.2) tells us that $a\in[(C_{1}+rB_{X})\cap C]+\varepsilon B_{X}$.
Let $b\in X$ be a point nearest to $a$ on $(C_{1}+rB_{X})\cap C$, and let
$b_{1}\in X$ be a point nearest to $b$ on $C_{1}$. Then $\|b_{1}-b\|\leq r$
and $\|a-b\|\leq\varepsilon$. Moreover, $b_{1}\in C_{1}\cap(C+rB_{X})$ and

$\|a-b_{1}\|\leq\|a-b\|+\|b-b_{1}\|\leq\varepsilon+r\leq\varepsilon+Lr,$

so that $b_{1}\in B_{X}(a,Lr+\varepsilon)$. Hence,

$b_{1}\in C_{1}\cap(C+rB_{X})\cap B_{X}(a,Lr+\varepsilon)$

proving (4.5).

Thus, (4.4) holds proving the existence of a point $x\in X$ such that

(4.6) $\displaystyle x\in C_{1}\cap C_{2}\cap(C+rB_{X})\cap
B_{X}(a,Lr+\varepsilon)\,.$

In particular, $x\in C+rB_{X}$ so that $B_{X}(x,r)\cap C\neq\emptyset$ proving
that condition (2.6) of Proposition 2.3 is satisfied. We apply this
proposition to $x$, $r$ and the set $C$ and get:

$[C\cap B_{X}(x,Lr)]+\theta(L)\,\varepsilon B_{X}\supset(C+\varepsilon
B_{X})\cap(B_{X}(x,Lr)+\varepsilon B_{X})=(C+\varepsilon B_{X})\cap
B_{X}(x,Lr+\varepsilon)\,.$

From (4.6) we learn that $a\in B_{X}(x,Lr+\varepsilon)$. In turn, (4.2) tells
us that

$a\in[(C_{1}+rB_{X})\cap C]+\varepsilon B_{X}\subset C+\varepsilon B_{X}.$

Hence, $(C+\varepsilon B_{X})\cap B_{X}(x,Lr+\varepsilon)\ni a$ proving that
$[C\cap B_{X}(x,Lr)]+\theta(L)\,\varepsilon B_{X}\ni a$.

Finally, (4.6) tells us that $x\in C_{1}\cap C_{2}$ proving the required
inclusion (4.3). $\blacksquare$

We recall that $\dim X>1$ so that the finiteness number
$N(1,X)=\min\\{2^{2},2^{\dim X}\\}=4$. Let
$F:{\mathcal{M}}\to{\mathcal{K}}_{1}(X)$ be a set-valued mapping. We suppose
that $F$ satisfies the hypothesis of Theorem 1.10, i.e., the following
statement is true.

######  Claim 4.3.

For every subset ${\mathcal{M}}^{\prime}\subset{\mathcal{M}}$ with
$\\#{\mathcal{M}}^{\prime}\leq 4$ the restriction
$F|_{{\mathcal{M}}^{\prime}}$ of $F$ to ${\mathcal{M}}^{\prime}$ has a
Lipschitz selection $f_{{\mathcal{M}}^{\prime}}:{\mathcal{M}}^{\prime}\to X$
with $\|f\|_{\operatorname{Lip}({\mathcal{M}}^{\prime},X)}\leq 1$.

Let $\vec{\lambda}=\\{\lambda_{1},\lambda_{2}\\}$, and let $F^{[1]}$ and
$F^{[2]}$ be the first and the second order $(\vec{\lambda},\rho)$-balanced
refinements of $F$. See Definition 1.6. Our aim is to show that if

(4.7) $\displaystyle\lambda_{1}\geq 1,~{}~{}~{}~{}~{}~{}\lambda_{2}\geq
3\lambda_{1},~{}~{}~{}~{}~{}~{}\gamma\geq\lambda_{2}\,(3\lambda_{2}+\lambda_{1})/(\lambda_{2}-\lambda_{1}),$

then the set-valued mapping $F^{[2]}$ is a $\gamma$-core of $F$ (with respect
to $\rho$), i.e.,

$F^{[2]}(x)\neq\emptyset~{}~{}\text{for
every}~{}~{}x\in{\mathcal{M}},~{}~{}~{}\text{and}~{}~{}~{}\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\gamma\rho(x,y)~{}~{}\text{for
all}~{}~{}x,y\in{\mathcal{M}}.$

To prove this, we set $L=\lambda_{2}/\lambda_{1}$ and introduce a new
pseudometric on ${\mathcal{M}}$ defined by

$\operatorname{d}(x,y)=\lambda_{1}\rho(x,y),~{}~{}~{}~{}~{}x,y\in{\mathcal{M}}.$

Note that, thanks to (4.7),

(4.8) $\displaystyle L\geq
3~{}~{}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}~{}~{}\rho\leq\operatorname{d}~{}~{}~{}~{}\text{on}~{}~{}~{}~{}{\mathcal{M}}.$

We also note that in these settings, for every $x\in{\mathcal{M}}$,

(4.9) $\displaystyle
F^{[1]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+\operatorname{d}(x,z)B_{X}\right]~{}~{}\text{and}~{}~{}F^{[2]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F^{[1]}(z)+L\operatorname{d}(x,z)B_{X}\right].$

Next, we need the following analog of Lemma 3.4.

######  Lemma 4.4.

Let ${\mathcal{K}}$ be a collection of convex closed subsets of $X$ containing
a set $K_{0}\in{\mathcal{K}}_{1}(X)$. Suppose that
$\,{{\cap}}\,\\{K:K\in\,{\mathcal{K}}\\}\neq\emptyset$. Then for every $r\geq
0$, we have

$\left(\,\bigcap_{K\in\,{\mathcal{K}}}K\right)+rB_{X}=\bigcap_{K\in\,{\mathcal{K}}}\,\left\\{\,\left[\,K\,{{\cap}}\,K_{0}\right]+rB_{X}\,\right\\}.$

Proof. Let
$\widetilde{{\mathcal{K}}}=\\{K\,{{\cap}}\,K_{0}:K\in{\mathcal{K}}\\}$.
Clearly, $\widetilde{{\mathcal{K}}}\subset{\mathcal{K}}_{1}(X)$. It is also
clear that the statement of the lemma is equivalent to the equality

$\left(\,\bigcap_{\widetilde{K}\in\,\widetilde{{\mathcal{K}}}}\widetilde{K}\right)+rB_{X}=\bigcap_{\widetilde{K}\in\,\widetilde{{\mathcal{K}}}}\,\left\\{\,\widetilde{K}+rB_{X}\,\right\\}$

provided
$\,{{\cap}}\,\\{\widetilde{K}:\widetilde{K}\in{\mathcal{K}}\\}\neq\emptyset$.
We prove this equality by a slight modification of the proof of Lemma 3.4.
More specifically, we obtain the result by using in that proof Helly’s Theorem
4.1 instead of Theorem 2.5. We leave the details to the interested reader.
$\blacksquare$

######  Lemma 4.5.

For every $x\in{\mathcal{M}}$ the set $F^{[1]}(x)$ belongs to the family
${\mathcal{K}}_{1}(X)$. Moreover, for every $x,z\in{\mathcal{M}}$, we have

(4.10) $\displaystyle
F^{[1]}(z)+L\operatorname{d}(x,z)B_{X}=\bigcap_{v\in{\mathcal{M}}}\left\\{\left[(F(v)+\operatorname{d}(z,v)B_{X})\,{{\cap}}\,F(z)\right]+L\operatorname{d}(x,z)B_{X}\right\\}.$

Proof. Let
${\mathcal{K}}=\\{F(z)+\operatorname{d}(z,x)B_{X}:z\in{\mathcal{M}}\\}.$
Clearly, ${\mathcal{K}}$ is a family of all bounded closed convex subsets of
$X$ containing the set $F(x)\in{\mathcal{K}}_{1}(X)$.

Theorem 4.1 tells us that the set
$F^{[1]}(x)=\,{{\cap}}\,\\{K:K\in{\mathcal{K}}\\}$ is non-empty whenever for
every $z^{\prime},z^{\prime\prime}\in{\mathcal{M}}$ the set

(4.11) $\displaystyle
E(x,z^{\prime},z^{\prime\prime})=F(x)\,{{\cap}}\,[F(z^{\prime})+\operatorname{d}(z^{\prime},x)B_{X}]\,{{\cap}}\,[F(z^{\prime\prime})+\operatorname{d}(z^{\prime\prime},x)B_{X}]\neq\emptyset.$

Fix $z^{\prime},z^{\prime\prime}\in{\mathcal{M}}$ and set
${\mathcal{M}}^{\prime}=\\{x,z^{\prime},z^{\prime\prime}\\}$. Thanks to Claim
4.3, there exists a function
$f_{{\mathcal{M}}^{\prime}}:{\mathcal{M}}^{\prime}\to X$ satisfying the
following conditions: $f_{{\mathcal{M}}^{\prime}}(x)\in F(x)$,
$f_{{\mathcal{M}}^{\prime}}(z^{\prime})\in F(z^{\prime})$,
$f_{{\mathcal{M}}^{\prime}}(z^{\prime\prime})\in F(z^{\prime\prime})$ and

$\|f_{{\mathcal{M}}^{\prime}}(z^{\prime})-f_{{\mathcal{M}}^{\prime}}(x)\|\leq\rho(z^{\prime},x)\leq\operatorname{d}(z^{\prime},x),~{}~{}~{}\|f_{{\mathcal{M}}^{\prime}}(z^{\prime\prime})-f_{{\mathcal{M}}^{\prime}}(x)\|\leq\rho(z^{\prime\prime},x)\leq\operatorname{d}(z^{\prime\prime},x).$

See (4.8). Then $f_{{\mathcal{M}}^{\prime}}(x)\in
E(x,z^{\prime},z^{\prime\prime})$ so that (4.11) holds. Hence,
$F^{[1]}(x)\neq\emptyset$ proving that $F^{[1]}(x)\in{\mathcal{K}}_{1}(X)$.

It remains to note that equality (4.10) is immediate from (4.9) and Lemma 4.4.

The proof of the lemma is complete. $\blacksquare$

######  Lemma 4.6.

For every $x\in{\mathcal{M}}$, the following equality

$F^{[2]}(x)=\bigcap_{u,u^{\prime}\in{\mathcal{M}}}\left\\{\left[(F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X})\,{{\cap}}\,F(u)\right]+L\operatorname{d}(u,x)B_{X}\right\\}$

holds.

Proof. The lemma is immediate from representation (4.9) and Lemma 4.5.
$\blacksquare$

Given $x,u,u^{\prime}\in{\mathcal{M}}$ we put

(4.12)
$\displaystyle\widetilde{T}_{x}(u,u^{\prime})=[(F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X})\,{{\cap}}\,F(u)]+L\operatorname{d}(u,x)\,B_{X}.$

Note that, in our terminology, the set $\widetilde{T}_{x}(u,u^{\prime})$ is
the orbit of $x$ with respect to the diagram shown in Fig. 24. Thus,
$\widetilde{T}_{x}(u,u^{\prime})$ is the family of all points $a=g(x)$ where
$g$ runs over all mappings which agree with the diagram in Fig. 24 below.

Fig. 24: $\widetilde{T}_{x}(u,u^{\prime})$ is the orbit of $x$ with respect to
this diagram.

Now, Lemma 4.6 reformulates as follows:

(4.13) $\displaystyle
F^{[2]}(x)=\bigcap_{u,u^{\prime}\in{\mathcal{M}}}\widetilde{T}_{x}(u,u^{\prime}).$

######  Proposition 4.7.

For every $x\in{\mathcal{M}}$ the set $F^{[2]}(x)$ is non-empty.

Proof. Recall that $F(x)\in{\mathcal{K}}_{1}(X)$. Furthermore, by (4.12),
$F(x)=\widetilde{T}_{x}(x,x)$. Therefore, by (4.13) and Helly’s Theorem 4.1,
the set $F^{[2]}(x)\neq\emptyset$ whenever for every
$u_{i},u^{\prime}_{i}\in{\mathcal{M}}$, $i=1,2$, we have

(4.14) $\displaystyle
F(x)\,{{\cap}}\,\widetilde{T}_{x}(u_{1},u^{\prime}_{1})\,{{\cap}}\,\widetilde{T}_{x}(u_{2},u^{\prime}_{2})\neq\emptyset.$

We note that (4.14) holds provided there exists a mapping $g$ which agrees
with the diagram shown in Fig. 25.

Fig. 25: The mapping $g$ agrees with this diagram.

Thanks to (4.12),

(4.15)
$\displaystyle\widetilde{T}_{x}(u_{i},u^{\prime}_{i})=[(F(u_{i}^{\prime})+\operatorname{d}(u_{i}^{\prime},u_{i})B_{X})\,{{\cap}}\,F(u_{i})]+L\operatorname{d}(u_{i},x)\,B_{X},~{}~{}~{}i=1,2.$

Fix elements $u_{1},u^{\prime}_{1},u_{2},u^{\prime}_{2}\in{\mathcal{M}}$ and
prove that property (4.14) holds.

First we note that, without loss of generality, one may assume that
$\rho(u_{1},x)\geq\rho(u_{2},x)$. Next, we introduce the following sets:

(4.16) $\displaystyle
G_{1}=F(u_{2}),~{}~{}~{}~{}G_{2}=F(u_{2}^{\prime})+\rho(u_{2},u_{2}^{\prime})B_{X},~{}~{}~{}~{}G_{3}=F(x)+\rho(u_{2},x)B_{X},$

and

(4.17) $\displaystyle
G_{4}=[(F(u_{1}^{\prime})+\rho(u_{1}^{\prime},u_{1})B_{X})\,{{\cap}}\,F(u_{1})]+\rho(u_{1},u_{2},)B_{X}\,.$

Prove that if

(4.18) $\displaystyle
G_{1}\,{{\cap}}\,G_{2}\,{{\cap}}\,G_{3}\,{{\cap}}\,G_{4}\neq\emptyset$

then (4.14) holds.

(Note that (4.18) holds whenever there exists a mapping $g$ which agrees with
the diagram in Fig. 26.)

Fig. 26: The existence of a mapping $g$ which agrees with this diagram implies
(4.18).

Indeed, let
$\widetilde{{\mathcal{M}}}=\\{u_{1}^{\prime},u_{1},x,u_{2},u_{2}^{\prime}\\}$.
This property and definitions (4.16), (4.17) imply the existence of a mapping
$g:\widetilde{{\mathcal{M}}}\to X$ with the following properties: $g(v)\in
F(v)$ on $\widetilde{{\mathcal{M}}}$,

(4.19)
$\displaystyle\|g(u_{1})-g(u_{1}^{\prime})\|\leq\rho(u_{1},u_{1}^{\prime}),~{}~{}~{}~{}\|g(u_{1})-g(u_{2})\|\leq\rho(u_{1},u_{2}),$

and

(4.20)
$\displaystyle\|g(u_{2})-g(u_{2}^{\prime})\|\leq\rho(u_{2},u_{2}^{\prime}),~{}~{}~{}~{}\|g(u_{2})-g(x)\|\leq\rho(u_{2},x).$

We establish (4.14) by showing that

(4.21) $\displaystyle g(x)\in
F(x)\,{{\cap}}\,\widetilde{T}_{x}(u_{1},u^{\prime}_{1})\,{{\cap}}\,\widetilde{T}_{x}(u_{2},u^{\prime}_{2}).$

In fact, from the above properties of $g$ it follows that $g(x)\in F(x)$. We
also know that $g(u_{2})\in F(u_{2})$, $g(u_{2}^{\prime})\in
F(u_{2}^{\prime})$. Thanks to (4.19), (4.20) and (4.8),

$\|g(u_{2})-g(u_{2}^{\prime})\|\leq\rho(u_{2},u_{2}^{\prime})\leq\operatorname{d}(u_{2},u_{2}^{\prime})~{}~{}~{}~{}\text{and}~{}~{}~{}~{}\|g(u_{2})-g(x)\|\leq\rho(u_{2},x)\leq
L\operatorname{d}(u_{2},x).$

From these properties of $g$ and definition (4.15), we have

$g(x)\in[(F(u_{2}^{\prime})+\operatorname{d}(u_{2}^{\prime},u_{2})B_{X})\,{{\cap}}\,F(u_{2})]+L\operatorname{d}(u_{2},x)\,B_{X}=\widetilde{T}_{x}(u_{2},u^{\prime}_{2}).$

It remains to show that $g(x)\in\widetilde{T}_{x}(u_{1},u^{\prime}_{1})$. As
we know,

(4.22) $\displaystyle g(x)\in F(x),~{}~{}~{}~{}~{}g(u_{1})\in
F(u_{1}),~{}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}~{}g(u_{1}^{\prime})\in
F(u_{1}^{\prime}).$

Furthermore, thanks to (4.19) and (4.8),

(4.23)
$\displaystyle\|g(u_{1})-g(u_{1}^{\prime})\|\leq\rho(u_{1},u_{1}^{\prime})\leq\operatorname{d}(u_{1},u_{1}^{\prime}).$

Let us estimate $\|g(u_{1})-g(x)\|$. From (4.19), (4.20) and the triangle
inequality, we have

$\displaystyle\|g(u_{1})-g(x)\|$
$\displaystyle\leq\|g(u_{1})-g(u_{2})\|+\|g(u_{2})-g(x)\|\leq\rho(u_{1},u_{2})+\rho(u_{2},x)$
$\displaystyle\leq\rho(u_{1},x)+\rho(x,u_{2})+\rho(u_{2},x)=\rho(u_{1},x)+2\rho(x,u_{2}).$

Recall that $\rho(u_{1},x)\geq\rho(u_{2},x)$. This and (4.8) yield

$\|g(u_{1})-g(x)\|\leq 3\rho(u_{1},x)\leq L\operatorname{d}(u_{1},x).$

This inequality, (4.22), (4.23) and (4.15) imply the required property
$g(x)\in\widetilde{T}_{x}(u_{1},u^{\prime}_{1})$ proving (4.21).

Thus, to complete the proof of the proposition, we have to prove that the sets
$G_{i}$, $i=1,...,4$, have a common point.

See Fig. 27.

Fig. 27: The sets $G_{1}$, $G_{2}$, $G_{3}$ and $G_{4}$.

Note that $G_{1}=F(u_{2})\in{\mathcal{K}}_{1}(X)$ so that, by Theorem 4.1,
this property holds provided

(4.24) $\displaystyle
G_{1}\,{{\cap}}\,G_{i}\,{{\cap}}\,G_{j}\neq\emptyset~{}~{}~{}~{}\text{for
every}~{}~{}~{}~{}2\leq i,j\leq 4,~{}i\neq j.$

First prove that $G_{1}\,{{\cap}}\,G_{2}\,{{\cap}}\,G_{3}\neq\emptyset.$

Let ${\mathcal{M}}_{1}=\\{u_{2}^{\prime},u_{2},x\\}$. Thanks to Claim 4.3,
there exists a mapping $f_{1}:{\mathcal{M}}_{1}\to X$ with the following
properties: $f_{1}(x)\in F(x)$, $f_{1}(u_{2})\in F(u_{2})$,
$f_{1}(u_{2}^{\prime})\in F(u_{2}^{\prime})$,

$\|f_{1}(u_{2})-f_{1}(x)\|\leq\rho(u_{2},x)~{}~{}~{}~{}\text{and}~{}~{}~{}~{}\|f_{1}(u_{2})-f_{1}(u_{2}^{\prime})\|\leq\rho(u_{2},u_{2}^{\prime}).$

These properties of $f_{1}$ and definition (4.16) tell us that
$f_{1}(u_{2})\in G_{1}\,{{\cap}}\,G_{2}\,{{\cap}}\,G_{3}$ proving that the
sets $G_{1}$, $G_{2}$ and $G_{3}$ have a common point.

Prove that $G_{1}\,{{\cap}}\,G_{2}\,{{\cap}}\,G_{4}\neq\emptyset.$

Let ${\mathcal{M}}_{2}=\\{u_{1}^{\prime},u_{1},u_{2}^{\prime},u_{2}\\}$. Using
Claim 4.3, we produce a mapping $f_{2}:{\mathcal{M}}_{2}\to X$ such that
$f_{2}(u_{i})\in F(u_{i})$, $f_{2}(u_{i}^{\prime})\in F(u_{i}^{\prime})$ for
every $i=1,2$,
$\|f_{2}(u_{1})-f_{2}(u_{1}^{\prime})\|\leq\rho(u_{1},u_{1}^{\prime})$,

$\|f_{2}(u_{1})-f_{2}(u_{2})\|\leq\rho(u_{1},u_{2})~{}~{}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}~{}~{}\|f_{2}(u_{2})-f_{2}(u_{2}^{\prime})\|\leq\rho(u_{2},u_{2}^{\prime}).$

These properties of $f_{2}$ and (4.16), (4.17) yield $f_{2}(u_{2})\in
G_{1}\,{{\cap}}\,G_{2}\,{{\cap}}\,G_{4}$ proving that the sets $G_{1}$,
$G_{2}$ and $G_{4}$ have a common point.

In the same way we show that
$G_{1}\,{{\cap}}\,G_{3}\,{{\cap}}\,G_{4}\neq\emptyset$. (We set
${\mathcal{M}}_{3}=\\{u_{1}^{\prime},u_{1},x,u_{2}\\}$, produce a
corresponding function $f_{3}:{\mathcal{M}}_{3}\to X$ and show that
$f_{3}(u_{2})\in G_{1}\,{{\cap}}\,G_{3}\,{{\cap}}\,G_{4}$.)

Thus, (4.24) holds, proving that the sets $G_{i}$ have a common point.

The proof of the proposition is complete. $\blacksquare$

In this section we set $\gamma_{0}=\gamma_{0}(L)=L\,\theta(L)$.

######  Proposition 4.8.

For every $x,y\in{\mathcal{M}}$ the following inequality

$\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\gamma_{0}(L)\operatorname{d}(x,y)$

holds.

Proof. Let $x,y\in{\mathcal{M}}$. Thanks to (4.13),

(4.25) $\displaystyle
F^{[2]}(x)=\bigcap_{u,u^{\prime}\in{\mathcal{M}}}\widetilde{T}_{x}(u,u^{\prime})~{}~{}~{}\text{and}~{}~{}~{}F^{[2]}(y)=\bigcap_{u,u^{\prime}\in{\mathcal{M}}}\widetilde{T}_{y}(u,u^{\prime}).$

Recall that

(4.26)
$\displaystyle\widetilde{T}_{x}(u,u^{\prime})=[(F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X})\,{{\cap}}\,F(u)]+L\operatorname{d}(u,x)\,B_{X}.$

Recall that the set $\widetilde{T}_{x}(u,u^{\prime})$ is the orbit of $x$ with
respect to the diagram of Fig. 24.

We know that $F^{[2]}(x)\neq\emptyset$, see Proposition 4.7, and
$\widetilde{T}_{x}(x,x)=F(x)\in{\mathcal{K}}_{1}(X)$. These properties, (4.25)
and Lemma 4.4 tell us that

(4.27) $\displaystyle
F^{[2]}(x)+\gamma_{0}(L)\operatorname{d}(x,y)\,B_{X}=\bigcap_{u,u^{\prime}\in{\mathcal{M}}}\,\,\left\\{\,\left[\widetilde{T}_{x}(u,u^{\prime})\,{{\cap}}\,F(x)\right]+\gamma_{0}(L)\operatorname{d}(x,y)\,B_{X}\,\right\\}.$

We fix $u,u^{\prime}\in{\mathcal{M}}$ and introduce a set

$\widetilde{A}=[\widetilde{T}_{x}(u,u^{\prime})\,{{\cap}}\,F(x)]+\gamma_{0}(L)\operatorname{d}(x,y)\,B_{X}.$

Clearly, this set is the orbit of $y$ with respect to the diagram in Fig. 28.

Fig. 28: $\widetilde{A}=\\{g(y)\\}$ where $g$ runs over all mappings which
agree with this diagram.

We also introduce sets

(4.28) $\displaystyle
C_{1}=F(u),~{}~{}~{}C_{2}=F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X},~{}~{}~{}~{}\text{and}~{}~{}~{}~{}C=F(x).$

Let

(4.29)
$\displaystyle\varepsilon=L\operatorname{d}(x,y)~{}~{}~{}\text{and}~{}~{}~{}r=\operatorname{d}(x,u)\,.$

In these settings, $\gamma_{0}(L)\operatorname{d}(x,y)=\theta(L)\,\varepsilon$
and

$\widetilde{A}=[\widetilde{T}_{x}(u,u^{\prime})\,{{\cap}}\,F(x)]+\gamma_{0}(L)\operatorname{d}(x,y)\,B_{X}=[\\{(C_{1}\,{{\cap}}\,C_{2})+LrB_{X}\\}\,{{\cap}}\,C]+\theta(L)\,\varepsilon\,B_{X}.$

Our aim is to prove that $\widetilde{A}\supset
F^{[2]}(y)=\,{{\cap}}\,\\{\widetilde{T}_{y}(z,z^{\prime}):z,z^{\prime}\in{\mathcal{M}}\\}$.
See Fig. 29.

Fig. 29: The sets $C_{1}$, $C_{2}$ and $C$.

In fact, we prove that

$\widetilde{A}\supset\widetilde{T}_{y}(u,u^{\prime})\cap\widetilde{T}_{y}(x,u^{\prime}).$

Let us apply Proposition 4.2 to the set $\widetilde{A}$. To do this, we have
to verify condition (4.1), i.e., to show that

(4.30) $\displaystyle
C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rB_{X})\neq\emptyset.$

Let $\widetilde{{\mathcal{M}}}=\\{x,u,u^{\prime}\\}$. Thanks to Claim 4.3,
there exists a $\rho$-Lipschitz selection $f_{\widetilde{{\mathcal{M}}}}$ of
the restriction $F|_{\widetilde{{\mathcal{M}}}}$ with
$\|f_{\widetilde{{\mathcal{M}}}}\|_{\operatorname{Lip}((\widetilde{{\mathcal{M}}};\rho),X)}\leq
1$. Thus, $f_{\widetilde{{\mathcal{M}}}}(u^{\prime})\in F(u^{\prime})$,
$f_{\widetilde{{\mathcal{M}}}}(u)\in F(u)$,
$f_{\widetilde{{\mathcal{M}}}}(x)\in F(x)$,

$\|f_{\widetilde{{\mathcal{M}}}}(u^{\prime})-f_{\widetilde{{\mathcal{M}}}}(u)\|\leq\rho(u^{\prime},u)~{}~{}~{}\text{and}~{}~{}~{}\|f_{\widetilde{{\mathcal{M}}}}(x)-f_{\widetilde{{\mathcal{M}}}}(u)\|\leq\rho(x,u).$

Let us prove that

(4.31) $\displaystyle f_{\widetilde{{\mathcal{M}}}}(u)\in
C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rB_{X}).$

Indeed, $f_{\widetilde{{\mathcal{M}}}}(u)\in F(u)=C_{1}$, see (4.28).
Furthermore, $f_{\widetilde{{\mathcal{M}}}}(u^{\prime})\in F(u^{\prime})$ and,
thanks to (4.8), $\rho\leq\operatorname{d}$ on ${\mathcal{M}}$. Hence,

$\|f_{\widetilde{{\mathcal{M}}}}(u^{\prime})-f_{\widetilde{{\mathcal{M}}}}(u)\|\leq\rho(u^{\prime},u)\leq\operatorname{d}(u^{\prime},u)$

proving that $f_{\widetilde{{\mathcal{M}}}}(u)\in C_{2}$, see (4.28). Finally,
by (4.28) and (4.29), $f_{\widetilde{{\mathcal{M}}}}(x)\in F(x)=C$ and

$\|f_{\widetilde{{\mathcal{M}}}}(x)-f_{\widetilde{{\mathcal{M}}}}(u)\|\leq\rho(x,u)\leq\operatorname{d}(x,u)=r,~{}~{}~{}~{}~{}\text{so
that}~{}~{}~{}~{}f_{\widetilde{{\mathcal{M}}}}(u)\in C+rB_{X}.$

Thus, (4.31) is true so that property (4.30) holds. Furthermore,
$C_{1}=F(u)\in{\mathcal{K}}_{1}(X)$, so that all conditions of the hypothesis
of Proposition 4.2 are satisfied. By this proposition,

$\displaystyle\widetilde{A}$
$\displaystyle=[\\{(C_{1}\,{{\cap}}\,C_{2})+LrB_{X}\\}\,{{\cap}}\,C]+\theta(L)\,\varepsilon\,B_{X}$
$\displaystyle\supset[(C_{1}\,{{\cap}}\,C_{2})+(Lr+\varepsilon)B_{X}]\,{{\cap}}\,[\\{(C_{1}+rB_{X})\,{{\cap}}\,C\\}+\varepsilon
B_{X}]$ $\displaystyle=\widetilde{A}_{1}\,{{\cap}}\,\widetilde{A}_{2}.$

Prove that $\widetilde{A}_{i}\supset F^{[2]}(y)$ for every $i=1,2$.

We begin with the set

$\widetilde{A}_{1}=C_{1}\,{{\cap}}\,C_{2}+(Lr+\varepsilon)B_{X}.$

Thanks to (4.28) and (4.29),

$\widetilde{A}_{1}=[\\{F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X}\\}\,{{\cap}}\,F(u)]+(L\operatorname{d}(u,x)+L\operatorname{d}(x,y))B_{X}\,.$

By the triangle inequality,

$\operatorname{d}(u,x)+\operatorname{d}(x,y)\geq\operatorname{d}(u,y)$

so that

$\widetilde{A}_{1}\supset[\\{F(u^{\prime})+\operatorname{d}(u^{\prime},u)B_{X}\\}\,{{\cap}}\,F(u)]+L\operatorname{d}(u,y)B_{X}=\widetilde{T}_{y}(u,u^{\prime}),~{}~{}~{}~{}~{}\text{see
(\ref{H-RM})}.$

But, by (4.25), $\widetilde{T}_{y}(u,u^{\prime})\supset F^{[2]}(y)$ which
implies the required inclusion $\widetilde{A}_{1}\supset F^{[2]}(y)$.

See Fig. 30.

Fig. 30: The set
$\widetilde{A}_{1}=C_{1}\,{{\cap}}\,C_{2}+(Lr+\varepsilon)B_{X}$ containing
the set $\widetilde{T}_{y}(u,u^{\prime})$.

We turn to the set
$\widetilde{A}_{2}=[(C_{1}+rB_{X})\,{{\cap}}\,C]+\varepsilon B_{X}$. By
(4.12), (4.28) and (4.29),

$\widetilde{A}_{2}=[(F(u)+\operatorname{d}(u,x)B_{X})\,{{\cap}}\,F(x)]+L\operatorname{d}(x,y)B_{X}=T_{y}(x,u).$

Thanks to (4.25), $\widetilde{T}_{y}(u,x)\supset F^{[2]}(y)$ proving that
$\widetilde{A}_{2}\supset F^{[2]}(y)$.

Thus,

$\widetilde{A}=[\widetilde{T}_{x}(u,u^{\prime})\,{{\cap}}\,F(x)]+\gamma_{0}(L)\operatorname{d}(x,y)\,B_{X}\supset\widetilde{A}_{1}\,{{\cap}}\,\widetilde{A}_{2}\supset
F^{[2]}(y)~{}~{}~{}~{}\text{for
every}~{}~{}~{}~{}u,u^{\prime}\in{\mathcal{M}}.$

From this and representation (4.27), we have
$F^{[2]}(x)+\gamma_{0}(L)\operatorname{d}(x,y)\,B_{X}\supset F^{[2]}(y)\,.$

By interchanging the roles of $x$ and $y$ we obtain also

$F^{[2]}(y)+\gamma_{0}(L)\operatorname{d}(x,y)\,B_{X}\supset F^{[2]}(x)\,.$

These two inclusions and (1.2) imply the statement of the proposition.
$\blacksquare$

We complete the proof of Theorem 1.10 as follows. We fix $\lambda_{1}$,
$\lambda_{2}$ and $\gamma$ satisfying inequalities (1.10). Then, by
Proposition 4.7, the set $F^{[2]}(x)\neq\emptyset$ for every
$x\in{\mathcal{M}}$.

In turn, Proposition 4.8 tells us that
$\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\gamma_{0}(L)\operatorname{d}(x,y)$
on ${\mathcal{M}}$. We recall that $L=\lambda_{2}/\lambda_{1}$,
$\operatorname{d}=\lambda_{1}\rho$, $\gamma_{0}(L)=L\,\theta(L)$ and
$\theta(L)=(3L+1)/(L-1)$. Hence,

$\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\gamma_{0}(L)\operatorname{d}(x,y)=L\left(\frac{3L+1}{L-1}\right)\operatorname{d}(x,y)=\frac{\lambda_{2}\,(3\lambda_{2}+\lambda_{1})}{(\lambda_{2}-\lambda_{1})}\,\rho(x,y).$

This inequality and (1.10) imply (1.8).

Thus, (1.7) and (1.8) hold provided $\lambda_{1}$, $\lambda_{2}$ and $\gamma$
satisfy inequalities (1.10). In particular, we can set $\lambda_{1}=1$,
$\lambda_{2}=3$ and
$\gamma=\lambda_{2}\,(3\lambda_{2}+\lambda_{1})/(\lambda_{2}-\lambda_{1})=15$.

Let now $X$ be a Euclidean space, and let $\lambda_{1}$, $\lambda_{2}$ and
$\gamma$ be the parameters satisfying inequalities (1.11).

In this case, we replace in the above calculations the constant
$\theta(L)=(3L+1)/(L-1)$ with $\theta(L)=1+2L/\sqrt{L^{2}-1}$. This leads us
to the required estimate

$\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\left\\{\lambda_{2}+2\lambda_{2}^{2}\,{{/}}\left(\lambda_{2}^{2}-\lambda_{1}^{2}\right)^{\frac{1}{2}}\right\\}\,\rho(x,y)\leq\gamma\,\rho(x,y)$

proving that (1.7) and (1.8) hold for $\lambda_{1}$, $\lambda_{2}$ and
$\gamma$ satisfying (1.11).

The proof of Theorem 1.10 is complete. $\blacksquare$

##  5\. The main theorem in $\ell^{2}_{\infty}$.

5.1 The case $X={\bf R}$.

######  Proposition 5.1.

Let $({\mathcal{M}},\rho)$ be a pseudometric space. Let $m=1$ and let $X={\bf
R}$; thus, $\ell=\ell(m,X)=1$, see (1.1). In this case Conjecture 1.8 holds
for every $\lambda_{1}\geq 1$ and $\gamma\geq 1$.

Thus, the following statement is true: Let $F$ be a set-valued mapping from
${\mathcal{M}}$ into the family ${\mathcal{K}}({\bf R})$ of all closed bounded
intervals in ${\bf R}$. Suppose that for every $x,y\in{\mathcal{M}}$ there
exist points $g(x)\in F(x)$ and $g(y)\in F(y)$ such that
$|g(x)-g(y)|\leq\rho(x,y)$.

Let $F^{[1]}(x)$, $x\in{\mathcal{M}}$, be the $\lambda_{1}$-balanced
refinement of the mapping $F$, i.e., the set

(5.1) $\displaystyle
F^{[1]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+\lambda_{1}\,\rho(x,z)\,I_{0}\right]~{}~{}~{}~{}\text{where}~{}~{}~{}~{}I_{0}=[-1,1].$

Then $F^{[1]}(x)\neq\emptyset$ for every $x\in{\mathcal{M}}$, and

$\operatorname{d_{H}}(F^{[1]}(x),F^{[1]}(y))\leq\gamma\,\rho(x,y)~{}~{}~{}~{}\text{for
all}~{}~{}~{}~{}x,y\in{\mathcal{M}}.$

Note that Proposition 5.1 easily follows from the one dimensional Helly
Theorem and a formula for a neighborhood of the intersection of intervals in
${\bf R}$. We formulate these statements in the following

######  Lemma 5.2.

Let ${\mathcal{K}}\subset{\mathcal{K}}({\bf R})$ be a collection of closed
bounded intervals in ${\bf R}$.

(a) (Helly’s Theorem in ${\bf R}$.) If the intersection of every two intervals
from ${\mathcal{K}}$ is non-empty, then there exists a point in ${\bf R}$
common to all of the family ${\mathcal{K}}$.

(b) Suppose that ${{\cap}}\\{K:K\in{\mathcal{K}}\\}\neq\emptyset$. Then for
every $r\geq 0$ we have

$\left(\,\bigcap_{K\in\,{\mathcal{K}}}K\right)+rI_{0}=\bigcap_{K\in\,{\mathcal{K}}}\,\left\\{\,K+rI_{0}\,\right\\}.$

Proof of part (b). In Lemma 3.4 we proved an analog of property (b) for ${\bf
R}^{2}$. We prove part (b) by replacing in that proof the Helly Theorem 2.5
with the one dimensional Helly Theorem formulated in part (a) of the present
lemma. We leave the details to the interested reader. $\blacksquare$

Proof of Proposition 5.1. We have to prove that the set $F^{[1]}(x)$ is non-
empty for each $x\in{\mathcal{M}}$, and for every $x,y\in{\mathcal{M}}$

(5.2)
$\displaystyle\operatorname{d_{H}}(F^{[1]}(x),F^{[1]}(y))\leq\gamma\,\rho(x,y).$

We know that the restriction $F|_{{\mathcal{M}}^{\prime}}$ of $F$ to every two
point subset ${\mathcal{M}}^{\prime}\subset{\mathcal{M}}$ has a Lipschitz
selection $f_{{\mathcal{M}}^{\prime}}:{\mathcal{M}}^{\prime}\to{\bf R}$ with
$\|f_{{\mathcal{M}}^{\prime}}\|_{\operatorname{Lip}({\mathcal{M}}^{\prime},{\bf
R})}\leq\gamma$. Thus, for every $z,z^{\prime}\in{\mathcal{M}}$ there exist
points

(5.3) $\displaystyle g(z)\in F(z),~{}g(z^{\prime})\in
F(z^{\prime})~{}~{}~{}\text{such
that}~{}~{}~{}|g(z)-g(z^{\prime})|\leq\gamma\,\rho(z,z^{\prime}).$

We recall that the set-valued mapping $F^{[1]}$ is defined by formula (5.1).

Prove that $F^{[1]}(x)\neq\emptyset$ for every $x\in{\mathcal{M}}$. Indeed,
thanks to (5.1) and Helly’s Theorem for intervals (part (i) of Lemma 5.2),
$F^{[1]}(x)\neq\emptyset$ provided

(5.4)
$\displaystyle(F(z)+\gamma\,\rho(x,z)\,I_{0})\cap(F(z^{\prime})+\gamma\,\rho(x,z^{\prime})\,I_{0})\neq\emptyset$

for every $z,z^{\prime}\in{\mathcal{M}}$.

We know that there exist points $g(z)$ and $g(z^{\prime})$ satisfying (5.3).
Let

$a=\min\\{g(z)+\gamma\,\rho(z,x),g(z^{\prime})+\gamma\,\rho(z^{\prime},x)\\}.$

Thanks to the inequality $|g(z)-g(z^{\prime})|\leq\gamma\,\rho(z,z^{\prime})$,
we have

$g(z)=\min\\{g(z),g(z^{\prime})+\gamma\,\rho(z^{\prime},z)\\}$

so that, by the triangle inequality,

$|a-g(z)|\leq\max\\{\gamma\,\rho(z,x),|\gamma\,\rho(z^{\prime},x)-\gamma\,\rho(z^{\prime},z)|\\}=\gamma\,\rho(z,x).$

We also know that $g(z)\in F(z)$, see (5.3), so that $a\in
F(z)+\gamma\,\rho(x,z)\,I_{0}$.

In the same way we show that $a\in
F(z^{\prime})+\gamma\,\rho(x,z^{\prime})\,I_{0}$ proving property (5.4).

Prove that

(5.5) $\displaystyle F^{[1]}(x)+\gamma\,\rho(x,y)\,I_{0}\supset
F^{[1]}(y)~{}~{}~{}~{}\text{for every}~{}~{}~{}~{}x,y\in{\mathcal{M}}.$

We know that $F^{[1]}(x)\neq\emptyset$ which enables us to apply part (b) of
Lemma 5.2 to the left hand side of (5.5). This lemma and definition (5.1) tell
us that

$\displaystyle F^{[1]}(x)+\gamma\,\rho(x,y)\,I_{0}$
$\displaystyle=\left\\{\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+\gamma\,\rho(x,z)\,I_{0}\right]\right\\}+\gamma\,\rho(x,y)\,I_{0}$
$\displaystyle=\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+(\gamma\,\rho(x,z)+\gamma\,\rho(x,y))\,I_{0}\right]$

so that, thanks to the triangle inequality,

$F^{[1]}(x)+\gamma\,\rho(x,y)\,I_{0}\supset\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+\gamma\,\rho(y,z)\,I_{0}\right]=F^{[1]}(y)$

proving (5.5). By interchanging the roles of $x$ and $y$ we obtain also

$F^{[1]}(y)+\gamma\,\rho(x,y)\,I_{0}\supset F^{[1]}(x).$

These two inclusions prove the required inequality (5.2).

The proof of Proposition 5.1 is complete. $\blacksquare$

5.2 Rectangular hulls of plane convex sets.

Let us fix some additional notation. We let ${\mathfrak{R}}({\bf R}^{2})$
denote the family of all bounded closed rectangles in ${\bf R}^{2}$ with sides
parallel to the coordinate axes $Ox_{1}$ and $Ox_{2}$.

Let $Q_{0}=B_{X}$ be the unit ball of the Banach space $X=\ell^{2}_{\infty}$,
i.e., the square $Q_{0}=[-1,1]^{2}$. Given $a\in{\bf R}^{2}$ and $r\geq 0$, we
set $rQ_{0}=[-r,r]^{2}$ and $Q(a,r)=rQ_{0}+a$.

######  Definition 5.3.

Let $S$ be a non-empty bounded convex closed subset in ${\bf R}^{2}$. We set

${\mathcal{H}}[S]=\,{{\cap}}\,\\{\Pi:\Pi\in{\mathfrak{R}}({\bf
R}^{2}),\Pi\supset S\\},$

and refer to ${\mathcal{H}}[S]$ as the “rectangular hull“ of the set $S$.

Note the following useful representation of the rectangular hull which easily
follows from Definition 5.3:

(5.6) $\displaystyle{\mathcal{H}}[S]=(S+Ox_{1})\,{{\cap}}\,(S+Ox_{2}).$

In the next section we need the following auxiliary

######  Lemma 5.4.

Let $K_{1},K_{2}\in{\mathcal{K}}({\bf R}^{2})$ be two convex compacts in ${\bf
R}^{2}$ with non-empty intersection. Let $\tau\geq 0$ and let
$Q=[-\tau,\tau]^{2}$. Then

(5.7)
$\displaystyle(K_{1}\,{{\cap}}\,K_{2})+Q=(K_{1}+Q)\,{{\cap}}\,(K_{2}+Q)\,{{\cap}}\,{\mathcal{H}}[(K_{1}\,{{\cap}}\,K_{2})+Q].$

Proof. Obviously, the right hand side of (5.7) contains its left hand side.

Let us prove the converse statement. Fix a point

(5.8) $\displaystyle
a\in(K_{1}+Q)\,{{\cap}}\,(K_{2}+Q)\,{{\cap}}\,{\mathcal{H}}[K_{1}\,{{\cap}}\,K_{2}+Q].$

Our aim is to prove that $a\in(K_{1}\,{{\cap}}\,K_{2})+Q$. Clearly, this
property holds if and only if
$Q(a,\tau)\,{{\cap}}\,K_{1}\,{{\cap}}\,K_{2}\neq\emptyset$. It is also clear
that $Q(a,\tau)=\Pi_{1}(a)\,{{\cap}}\,\Pi_{2}(a)$ where

$\Pi_{i}(a)=Q(a,\tau)+Ox_{i},~{}~{}~{}~{}~{}~{}~{}i=1,2.$

Thus, $a\in(K_{1}\,{{\cap}}\,K_{2})+Q$ provided

$K_{1}\,{{\cap}}\,K_{2}\,{{\cap}}\,\Pi_{1}(a)\,{{\cap}}\,\Pi_{2}(a)\neq\emptyset.$

See Fig. 31.

Fig. 31: TEXT

Thanks to Theorem 2.5, the family of sets
$\\{K_{1},K_{2},\Pi_{1}(a),\Pi_{2}(a)\\}$ has a common point provided any
three members of this family have a non-empty intersection.

Prove that it is true for $a$ satisfying (5.8). Indeed, $a\in K_{i}+Q$, so
that $K_{i}\,{{\cap}}\,Q(a,\tau)\neq\emptyset$, $i=1,2$. Hence,

$K_{i}\,{{\cap}}\,\Pi_{1}(a)\,{{\cap}}\,\Pi_{2}(a)=K_{i}\,{{\cap}}\,Q(a,\tau)\neq\emptyset,~{}~{}~{}~{}i=1,2.$

Next, thanks to (5.6) and (5.8), for every $i=1,2$,

$a\in{\mathcal{H}}[(K_{1}\,{{\cap}}\,K_{2})+Q]\subset(K_{1}\,{{\cap}}\,K_{2})+Q+Ox_{i}.$

Hence, $K_{1}\,{{\cap}}\,K_{2}\,{{\cap}}\,\Pi_{i}(x)\neq\emptyset$, $i=1,2$,
and the proof of the lemma is complete. $\blacksquare$

5.3 Balanced refinements of set-valued mappings in $\ell^{2}_{\infty}$.

In this section we refine the result of Theorem 1.9 for the space
$X=\ell^{2}_{\infty}$.

######  Theorem 5.5.

In the settings of Theorem 1.9 properties (1.7) and (1.8) hold provided
$X=\ell^{2}_{\infty}$ ,

(5.9) $\displaystyle\lambda_{1}\geq 1,~{}~{}~{}~{}~{}\lambda_{2}\geq
3\lambda_{1},~{}~{}~{}~{}\text{and}~{}~{}~{}~{}~{}\gamma\geq\lambda_{2}\,(3\lambda_{2}+\lambda_{1})/(\lambda_{2}-\lambda_{1}).$

In particular, (1.7) and (1.8) hold whenever $\lambda_{1}=1$, $\lambda_{2}=3$
and $\gamma=15$.

Proof. We mainly follow the scheme of the proof of Theorem 1.9 given in
Section 3.

Let $F:{\mathcal{M}}\to{\mathcal{K}}({\bf R}^{2})$ be a set-valued mapping
satisfying the hypothesis of Theorem 5.5. Thus, the next statement is true.

######  Claim 5.6.

For every ${\mathcal{M}}^{\prime}\subset{\mathcal{M}}$,
$\\#{\mathcal{M}}^{\prime}\leq 4$, the restriction
$F|_{{\mathcal{M}}^{\prime}}$ of $F$ to ${\mathcal{M}}^{\prime}$ has a
$\rho$-Lipschitz selection
$f_{{\mathcal{M}}^{\prime}}:{\mathcal{M}}^{\prime}\to\ell^{2}_{\infty}$ with
$\rho$-Lipschitz seminorm
$\|f_{{\mathcal{M}}^{\prime}}\|_{\operatorname{Lip}(({\mathcal{M}}^{\prime},\,\rho),\ell^{2}_{\infty})}\leq
1$.

Let $\lambda_{1}$ and $\lambda_{2}$ be positive constants satisfying
inequalities (5.9). We set $L=\lambda_{2}/\lambda_{1}$; thus, $L\geq 3$. Then
we introduce a pseudometric on ${\mathcal{M}}$ defined by
$\operatorname{d}(x,y)=\lambda_{1}\,\rho(x,y)$, $x,y\in{\mathcal{M}}$.

We let $F^{[1]}$ and $F^{[2]}$ denote the first and the second order
$(\\{1,L\\},\operatorname{d})$-balanced refinements of $F$ respectively. See
Definition 1.6. Thus,

$F^{[1]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F(z)+\operatorname{d}(x,z)\,Q_{0}\right]~{}~{}~{}\text{and}~{}~{}~{}F^{[2]}(x)=\bigcap_{z\in{\mathcal{M}}}\,\left[F^{[1]}(z)+L\operatorname{d}(x,z)\,Q_{0}\right],~{}~{}x\in{\mathcal{M}}.$

We also recall that $e({\mathfrak{M}},\ell^{2}_{\infty})=1$. In this case,
Lemma 3.5 and Proposition 3.8 tell us that $F^{[1]}(x)\neq\emptyset$ and
$F^{[2]}(x)\neq\emptyset$ for every $x\in{\mathcal{M}}$.

Let

$\tilde{\gamma}(L)=L\,\theta(L)~{}~{}~{}\text{where}~{}~{}~{}\theta(L)=(3L+1)/(L-1).$

Prove that

(5.10)
$\displaystyle\operatorname{d_{H}}(F^{[2]}(x),F^{[2]}(y))\leq\tilde{\gamma}(L)\,\operatorname{d}(x,y)~{}~{}~{}\text{for
every}~{}~{}~{}x,y\in{\mathcal{M}}.$

We recall that, thanks to formula (3.9),
$F^{[2]}(x)=\,{{\cap}}\,\\{T_{x}(u,u^{\prime},u^{\prime\prime}):u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}\\}$
where, given $u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}$,

(5.11) $\displaystyle
T_{x}(u,u^{\prime},u^{\prime\prime})=\\{[F(u^{\prime})+\operatorname{d}(u^{\prime},u)Q_{0}]\,{{\cap}}\,[F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)Q_{0}]\\}+L\operatorname{d}(u,x)\,Q_{0}.$

In particular, $T_{x}(u,u^{\prime},u^{\prime\prime})\neq\emptyset$ for all
$u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}$ (because
$F^{[2]}(x)\neq\emptyset$).

The next lemma is a refinement of the formula (3.27) for the special case of
$X=\ell^{2}_{\infty}$.

######  Lemma 5.7.

Let $\tau>0$ and let $Q=\tau Q_{0}=[-\tau,\tau]^{2}$. Then, for every
$x\in{\mathcal{M}}$, we have

(5.12) $\displaystyle
F^{[2]}(x)+Q=\bigcap_{v,u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}}\,\,\left\\{\,\left[T_{x}(u,u^{\prime},u^{\prime\prime})\,{{\cap}}\,\left(F(v)+\operatorname{d}(x,v)Q_{0}\right)\right]+Q\right\\}.$

Proof. Lemma 3.9 tells us that

(5.13) $\displaystyle
F^{[2]}(x)+Q=\bigcap\,\,\left\\{\,\left[T_{x}(u_{1},u^{\prime}_{1},u^{\prime\prime}_{1})\,{{\cap}}\,T_{x}(u_{2},u^{\prime}_{2},u^{\prime\prime}_{2})\right]+Q\,\right\\}$

where the intersection is taken over all
$u_{i},u^{\prime}_{i},u^{\prime\prime}_{i}\in{\mathcal{M}}$, $i=1,2$. Note
also that, by (5.11),

$F(v)+\operatorname{d}(x,v)Q_{0}=T_{x}(x,v,v).$

From this and (5.13) it follows that the right hand side of (5.12) contains
its left hand side.

Prove the converse statement. Fix a point

(5.14) $\displaystyle
a\in\bigcap_{v,u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}}\,\,\left\\{\,\left[T_{x}(u,u^{\prime},u^{\prime\prime})\,{{\cap}}\,\left(F(v)+\operatorname{d}(x,v)Q_{0}\right)\right]+Q\right\\}$

and show that $a\in F^{[2]}(x)+Q$. In view of formula (5.13), it suffice to
prove that for every
$u_{1},u^{\prime}_{1},u^{\prime\prime}_{1},u_{2},u^{\prime}_{2},u^{\prime\prime}_{2}\in{\mathcal{M}}$
the point $a$ belongs to the set $A$ defined by

(5.15) $\displaystyle
A=\left[T_{x}(u_{1},u^{\prime}_{1},u^{\prime\prime}_{1})\,{{\cap}}\,T_{x}(u_{2},u^{\prime}_{2},u^{\prime\prime}_{2})\right]+Q.$

To see this, given $i\in\\{1,2\\}$ we introduce the following sets:
$Q_{i}=L\operatorname{d}(u_{i},x)\,Q_{0}$,

(5.16) $\displaystyle
K^{\prime}_{i}=F(u^{\prime}_{i})+\operatorname{d}(u_{i},u^{\prime}_{i})Q_{0}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}K^{\prime\prime}_{i}=F(u^{\prime\prime}_{i})+\operatorname{d}(u_{i},u^{\prime\prime}_{i})Q_{0}.$

In these settings,
$T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})=K^{\prime}_{i}\,{{\cap}}\,K^{\prime\prime}_{i}+Q_{i}$,
$i=1,2$. See (5.11).

Note that $K^{\prime}_{i}\,{{\cap}}\,K^{\prime\prime}_{i}\neq\emptyset$
because $T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})\neq\emptyset$.
Therefore, thanks to Lemma 5.4,

(5.17) $\displaystyle
T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})=(K^{\prime}_{i}+Q_{i})\,{{\cap}}\,(K^{\prime\prime}_{i}+Q_{i})\,{{\cap}}\,{\mathcal{H}}[T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})],~{}~{}~{}~{}~{}i=1,2.$

Now, let us introduce the following families of sets:

${\mathcal{K}}^{+}=\\{K^{\prime}_{i}+Q_{i},K^{\prime\prime}_{i}+Q_{i}:i=1,2\\},~{}~{}{\mathcal{K}}^{++}=\\{{\mathcal{H}}[T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})]:i=1,2\\},~{}~{}{\mathcal{K}}={\mathcal{K}}^{+}{{\cup}}{\mathcal{K}}^{++}.$

Then, thanks to (5.15) and (5.17),
$A=[\,{{\cap}}\,\\{K:K\in{\mathcal{K}}\\}]+Q$.

We recall that, thanks to Proposition 3.8, the set $F^{[2]}(x)\neq\emptyset$
so that the left hand side of (5.13) is non-empty as well. From this and
(5.15) it follows that $A\neq\emptyset$ proving that
$\,{{\cap}}\,\\{K:K\in{\mathcal{K}}\\}\neq\emptyset$. Therefore, thanks to
Lemma 3.4,

$A=\,{{\cap}}\,\\{[K\,{{\cap}}\,K^{\prime}]+Q:K,K^{\prime}\in{\mathcal{K}}\\}.$

Thus, to prove that $a\in A$ it suffice to show that $a\in
K\,{{\cap}}\,K^{\prime}+Q$ for every $K,K^{\prime}\in{\mathcal{K}}$.

To do this, first let us note that, thanks to (5.16),

$K^{\prime}_{i}+Q_{i}=F(u^{\prime}_{i})+\operatorname{d}(u_{i},u^{\prime}_{i})Q_{0}+L\operatorname{d}(u_{i},x)\,Q_{0}\supset
F(u^{\prime}_{i})+(\operatorname{d}(u_{i},u^{\prime}_{i})+\operatorname{d}(u_{i},x))\,Q_{0}$

for every $i=1,2$. Therefore, thanks to the triangle inequality,

(5.18) $\displaystyle K^{\prime}_{i}+Q_{i}\supset
F(u^{\prime}_{i})+\operatorname{d}(u^{\prime}_{i},x)\,Q_{0}.$

In the same way we prove that

(5.19) $\displaystyle K^{\prime\prime}_{i}+Q_{i}\supset
F(u^{\prime\prime}_{i})+\operatorname{d}(u^{\prime\prime}_{i},x)\,Q_{0},~{}~{}~{}~{}i=1,2.$

Furthermore, we know that

(5.20)
$\displaystyle{\mathcal{H}}[T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})]\supset
T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i}),~{}~{}~{}~{}i=1,2.$

On the other hand, property (5.14) tells us that

(5.21) $\displaystyle a\in
T_{x}(u,u^{\prime},u^{\prime\prime})\,{{\cap}}\,[F(v)+\operatorname{d}(x,v)Q_{0})]+Q~{}~{}~{}~{}\text{for
every}~{}~{}~{}~{}u,u^{\prime},u^{\prime\prime},v\in{\mathcal{M}}.$

Combining this property with (5.18), (5.19) and (5.20), we conclude that

$a\in K\,{{\cap}}\,K^{\prime}+Q~{}~{}~{}\text{whenever
either}~{}~{}~{}K\in{\mathcal{K}}^{+},K^{\prime}\in{\mathcal{K}}^{++}~{}~{}~{}\text{or}~{}~{}~{}K,K^{\prime}\in{\mathcal{K}}^{+}.$

It remains to prove that

(5.22) $\displaystyle a\in
H_{1}\,{{\cap}}\,H_{2}+Q~{}~{}~{}\text{where}~{}~{}~{}H_{i}={\mathcal{H}}[T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})],~{}~{}~{}~{}~{}i=1,2.$

It is immediate from Lemma 5.2, part (b), that

$H_{1}\,{{\cap}}\,H_{2}+Q=(H_{1}+Q)\,{{\cap}}\,(H_{2}+Q)$

so that

$H_{1}\,{{\cap}}\,H_{2}+Q=\left\\{{\mathcal{H}}[T_{x}(u_{1},u^{\prime}_{1},u^{\prime\prime}_{1})]+Q\right\\}\,{{\cap}}\,\left\\{{\mathcal{H}}[T_{x}(u_{2},u^{\prime}_{2},u^{\prime\prime}_{2})]+Q\right\\}.$

From this and (5.20), we have

(5.23) $\displaystyle
H_{1}\,{{\cap}}\,H_{2}+Q\supset\left\\{T_{x}(u_{1},u^{\prime}_{1},u^{\prime\prime}_{1})+Q\right\\}\,{{\cap}}\,\left\\{T_{x}(u_{2},u^{\prime}_{2},u^{\prime\prime}_{2})+Q\right\\}.$

But, thanks to (5.21), $a\in
T_{x}(u_{i},u^{\prime}_{i},u^{\prime\prime}_{i})+Q$, $i=1,2$. Combining this
property with (5.23), we obtain the required property (5.22) completing the
proof of the lemma. $\blacksquare$

We are in a position to prove inequality (5.10). Our proof will follow the
scheme of the proof of Proposition 3.10.

Let $x,y\in{\mathcal{M}}$, and let
$\tau=\tilde{\gamma}(L)\operatorname{d}(x,y)$. (Recall that
$\tilde{\gamma}(L)=L\,\theta(L)$ and $\operatorname{d}=\lambda_{1}\,\rho$.)

Lemma 5.7 tells us that

(5.24) $\displaystyle F^{[2]}(x)+\tau
Q_{0}=\bigcap_{v,u,u^{\prime},u^{\prime\prime}\in{\mathcal{M}}}\,\,\left\\{\,\left[T_{x}(u,u^{\prime},u^{\prime\prime})\,{{\cap}}\,\left(F(v)+\operatorname{d}(x,v)Q_{0}\right)\right]+\tau
Q_{0}\right\\}.$

Let us fix elements $u,u^{\prime},u^{\prime\prime},v\in{\mathcal{M}}$ and a
set

(5.25)
$\displaystyle\widetilde{A}=\left[T_{x}(u,u^{\prime},u^{\prime\prime})\,{{\cap}}\,(F(v)+\operatorname{d}(x,v)Q_{0})\right]+\tau
Q_{0}.$

Our goal is to show that

$\widetilde{A}\supset
F^{[2]}(y)=\bigcap_{z,z^{\prime},z^{\prime\prime}\in{\mathcal{M}}}\,T_{y}(z,z^{\prime},z^{\prime\prime}).~{}~{}~{}~{}~{}~{}~{}~{}\text{See
(\ref{G-XP}).}$

In fact, we will prove a stronger imbedding:

$\widetilde{A}\,\supset\,\,T_{y}(u,u^{\prime},u^{\prime\prime})\,\bigcap\,T_{y}(x,u^{\prime},v)\,\bigcap\,T_{y}(x,u^{\prime\prime},v).$

We also note that the set $\widetilde{A}$ is the orbit of the element $y$ with
respect to the diagram shown in Fig. 32.

Fig. 32: $\widetilde{A}=\\{g(y)\\}$ where $g$ runs over all mappings which
agree with this diagram.

Let

(5.26) $\displaystyle
C_{1}=F(u^{\prime})+\operatorname{d}(u^{\prime},u)Q_{0},~{}~{}~{}C_{2}=F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)Q_{0},~{}~{}~{}C=F(v)+\operatorname{d}(x,v)Q_{0},$

and let $\varepsilon=L\,\operatorname{d}(x,y)$ and $r=\operatorname{d}(u,x)$.
Then

$\tau=\tilde{\gamma}(L)\operatorname{d}(x,y)=L\,\theta(L)\operatorname{d}(x,y)=\theta(L)\,\varepsilon.$

In these settings,
$T_{x}(u,u^{\prime},u^{\prime\prime})=(C_{1}\,{{\cap}}\,C_{2})+L\,r\,Q_{0}$,
see (5.11), and

$\widetilde{A}=[\\{(C_{1}\,{{\cap}}\,C_{2})+L\,r\,Q_{0}\\}\,{{\cap}}\,C]+\theta(L)\,\varepsilon\,Q_{0}.~{}~{}~{}~{}~{}~{}~{}~{}~{}\text{See
(\ref{A-LRP1}).}$

Let us apply Proposition 2.6 to the sets $C_{1},C_{2}$ and $C$ defined by
(5.26). To do this, we have to verify condition (2.22) of that proposition,
i.e., to prove that

(5.27) $\displaystyle
C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rQ_{0})\neq\emptyset.$

Let ${\mathcal{M}}^{\prime}=\\{u,u^{\prime},v\\}$. Then, thanks to Claim 5.6,
there exists a $\rho$-Lipschitz selection
$f_{{\mathcal{M}}^{\prime}}:{\mathcal{M}}^{\prime}\to\ell^{2}_{\infty}$ of the
restriction $F|_{{\mathcal{M}}^{\prime}}$ with
$\|f_{{\mathcal{M}}^{\prime}}\|_{\operatorname{Lip}(({\mathcal{M}}^{\prime},\,\rho),\ell^{2}_{\infty})}\leq
1$.

Because $e({\mathfrak{M}},\ell^{2}_{\infty})=1$ and
$\operatorname{d}=\lambda_{1}\rho\geq\rho$, the mapping
$f_{{\mathcal{M}}^{\prime}}:{\mathcal{M}}^{\prime}\to\ell^{2}_{\infty}$ can be
extended to a $\operatorname{d}$-Lipschitz mapping
$\tilde{f}:{\mathcal{M}}\to\ell^{2}_{\infty}$ defined on all of
${\mathcal{M}}$ with $\operatorname{d}$-Lipschitz seminorm

$\|\tilde{f}\|_{\operatorname{Lip}(({\mathcal{M}},\operatorname{d}),\ell^{2}_{\infty})}\leq\|f_{{\mathcal{M}}^{\prime}}\|_{\operatorname{Lip}(({\mathcal{M}}^{\prime},\,\rho),\ell^{2}_{\infty})}\leq
1.$

Thus, $\tilde{f}(u^{\prime})=f_{{\mathcal{M}}^{\prime}}(u^{\prime})\in
F(u^{\prime})$,
$\tilde{f}(u^{\prime\prime})=f_{{\mathcal{M}}^{\prime}}(u^{\prime\prime})\in
F(u^{\prime\prime})$, $\tilde{f}(v)=f_{{\mathcal{M}}^{\prime}}(v)\in F(v)$,

$\|\tilde{f}(u^{\prime})-\tilde{f}(u)\|\leq\operatorname{d}(u^{\prime},u),~{}~{}~{}\|\tilde{f}(u^{\prime\prime})-\tilde{f}(u)\|\leq\operatorname{d}(u^{\prime\prime},u)$

and

$\|\tilde{f}(x)-\tilde{f}(u)\|\leq\operatorname{d}(u,x)=r,~{}~{}~{}~{}~{}\|\tilde{f}(x)-\tilde{f}(v)\|\leq\operatorname{d}(v,x).$

Hence, $\tilde{f}(u)\in C_{1}\,{{\cap}}\,C_{2}$ and $\tilde{f}(x)\in C$, so
that $C_{1}\,{{\cap}}\,C_{2}\,{{\cap}}\,(C+rQ_{0})\ni\tilde{f}(u)$ proving
(5.27).

This enables us to apply Proposition 2.6 to the sets $C_{1}$, $C_{2}$ and $C$.
By this proposition,

$\displaystyle\widetilde{A}$
$\displaystyle=[\\{(C_{1}\,{{\cap}}\,C_{2})+LrQ_{0}\\}\,{{\cap}}\,C]+\theta(L)\,\varepsilon\,Q_{0}$
$\displaystyle\supset[(C_{1}\,{{\cap}}\,C_{2})+(Lr+\varepsilon)Q_{0}]\,{{\cap}}\,[\\{(C_{1}+rQ_{0})\,{{\cap}}\,C\\}+\varepsilon
Q_{0}]\,{{\cap}}\,[\\{(C_{2}+rQ_{0})\,{{\cap}}\,C\\}+\varepsilon Q_{0}]$
$\displaystyle=S_{1}\,{{\cap}}\,S_{2}\,{{\cap}}\,S_{3}.$

Prove that $S_{i}\supset F^{[2]}(y)$ for every $i=1,2,3$. We begin with the
set

$\displaystyle S_{1}$
$\displaystyle=(C_{1}\,{{\cap}}\,C_{2})+(Lr+\varepsilon)Q_{0}$
$\displaystyle=[\\{F(u^{\prime})+\operatorname{d}(u^{\prime},u)Q_{0}\\}\,{{\cap}}\,\\{F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)Q_{0}\\}]+(L\operatorname{d}(u,x)+L\operatorname{d}(x,y))Q_{0}\,.$

See (5.26). By the triangle inequality,
$\operatorname{d}(u,x)+\operatorname{d}(x,y)\geq\operatorname{d}(u,y)$ so that

$S_{1}\supset[\\{F(u^{\prime})+\operatorname{d}(u^{\prime},u)Q_{0}\\}\,{{\cap}}\,\\{F(u^{\prime\prime})+\operatorname{d}(u^{\prime\prime},u)Q_{0}\\}]+L\operatorname{d}(u,y)Q_{0}=T_{y}(u,u^{\prime},u^{\prime\prime})\,.$

See (5.11). But, thanks to (3.9), $T_{y}(u,u^{\prime},u^{\prime\prime})\supset