---
abstract: |
    For each ${\small b\in\left(  0,\,\infty\right)  }$ we intend to generate a decreasing sequence of subsets $\left(  \mathcal{Y}_{b}^{\left(  n\right)
    }\right)  \subset Y_{\mathrm{conc}}$ depending on $b$ such that whenever $n\in\mathbb{N}$, then $\mathcal{A}\cap\mathcal{Y}_{b}^{\left(  n\right)  }%
    $ is dense in $\mathcal{Y}_{b}^{\left(  n\right)  }$ and the following four sets $\mathcal{Y}_{b}^{\left(  n\right)  }$, $\mathcal{Y}_{b}^{\left(
    n\right)  }\backslash\left(  \mathcal{A}\cap\mathcal{Y}_{b}^{\left(  n\right)
    }\right)  $, $\mathcal{A}\cap\mathcal{Y}_{b}^{\left(  n\right)  }$ and $\mathcal{Y}_{\mathrm{conc}}$ are pairwise equinumerous. Among others we also show that if $f$ is any measurable function on a measure space $\left(
    \Omega,\mathcal{F},\lambda\right)  $ and $p\in\left[  1,\infty\right)  $ is an arbitrary number then the quantities $\left\Vert f\right\Vert _{L^{p}}$ and $\sup_{\Phi\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}}\left(  \Phi\left(
    1\right)  \right)  ^{-1}\left\Vert \Phi\circ\left\vert f\right\vert
    \right\Vert _{L^{p}}$ are equivalent, in the sense that they are both either finite or infinite at the same time.
address: |
    Institute of Mathematics\
    University of Miskolc\
    H-3515 Miskolc–Egyetemváros\
    Hungary
author:
- 'N. K. Agbeko'
date: 'March 24th, 2006'
title: 'Bijections and metric spaces induced by some collective properties of concave Young-functions'
---

Introduction
============

We know that concave functions play major roles in many branches of mathematics for instance probability theory ([@BURK1973], [@GARS1973], [@MOGY1981], say), interpolation theory (cf. [@TRIEB1978], say), weighted norm inequalities (cf. [@GARFRAN1985], say), and functions spaces (cf. [@SINN2002], say), as well as in many other branches of sciences. In the line of [@BURK1973], [@GARS1973] and [@MOGY1981], the present author also obtained in martingale theory some results in connection with certain collective properties or behaviors of concave Young-functions (cf. [@AGB1986], [@AGB1989]). The study presented in [@AGB2005] was mainly motivated by the question why strictly concave functions possess so many properties, worth to be characterized using appropriate tools that await to be discovered.

We say that a function $\Phi:\left[  0,\,\infty\right)  \rightarrow\left[
0,\,\infty\right)  $ belongs to the set $\mathcal{Y}_{\mathrm{conc}}$ (and is referred to as a concave Young-function) if and only if it admits the integral representation$$\Phi\left(  x\right)  =\int\nolimits_{0}^{x}\varphi\left(  t\right)  dt,
\label{id1}%$$ (where $\varphi:\left(  0,\,\infty\right)  \rightarrow\left(  0,\,\infty
\right)  $ is a right-continuous and decreasing function such that it is integrable on every finite interval $\left(  0,\,x\right)  $) and $\Phi\left(
\infty\right)  =\infty$. It is worth to note that every function in $\mathcal{Y}_{\mathrm{conc}}$ is strictly concave.

We will remind some results obtained so far in [@AGB2005].

We shall say that a concave Young-function $\Phi$ satisfies the *density-level property* if $A_{\Phi}\left(  \infty\right)  <\infty$, where $A_{\Phi}\left(  \infty\right)  :=\int\nolimits_{1}^{\infty}%
\frac{\varphi\left(  t\right)  }{t}dt$. All the concave Young-functions possessing the density-level property will be grouped in a set $\mathcal{A}$.

In Theorems \[theo1\] and \[theo2\] (cf. [@AGB2005]), we showed that the composition of any two concave Young-functions satisfies the density-level property if and only if at least one of them satisfies it. These two theorems show that concave Young-functions with the density-level property behave like left and right ideal with respect to the composition operation.

We also proved ([@AGB2005], Lemma 5, page 12) that if $\Phi\in
\mathcal{Y}_{\mathrm{conc}}$, then there are constants $C_{\Phi}>0$ and $B_{\Phi}\geq0$ such that$$A_{\Phi}\left(  \infty\right)  -B_{\Phi}\leq%
%TCIMACRO{\dint _{0}^{\infty}}%
%BeginExpansion
{\displaystyle\int_{0}^{\infty}}
%EndExpansion
\frac{\Phi\left(  t\right)  }{\left(  t+1\right)  ^{2}}dt\leq C_{\Phi}%
+A_{\Phi}\left(  \infty\right)  .$$ This led us to the idea to search for a Lebesgue measure (described here below) with respect to which every concave Young-function turns out to be square integrable ([@AGB2005], Lemma 6, page 13), i.e. $\mathcal{Y}%
_{\mathrm{conc}}\subset L^{2}:=L^{2}\left(  \left[  0,~\infty\right)
,~\mathcal{M},~\mu\right)  $, where $\mathcal{M}$ is a $\sigma$-algebra (of $\left[  0,~\infty\right)  $) containing the Borel sets and $\mu
:\mathcal{M}\rightarrow\left[  0,~\infty\right)  $ is a Lebesgue measure defined by $\mu\left(  \left[  0,~x\right)  \right)  =\frac{1}{3}\left(
1-\frac{1}{\left(  x+1\right)  ^{3}}\right)  $ for all $x\in\left[
0,~\infty\right)  $. The mapping $d:L^{2}\times L^{2}\rightarrow\left[
0,~\infty\right)  $, defined by$$\operatorname*{d}\left(  f,g\right)  =\sqrt{%
%TCIMACRO{\dint _{\left[  0,~\infty\right)  }}%
%BeginExpansion
{\displaystyle\int_{\left[  0,~\infty\right)  }}
%EndExpansion
\left(  f-g\right)  ^{2}d\mu}=\sqrt{%
%TCIMACRO{\dint _{0}^{\infty}}%
%BeginExpansion
{\displaystyle\int_{0}^{\infty}}
%EndExpansion
\frac{\left(  f\left(  x\right)  -g\left(  x\right)  \right)  ^{2}}{\left(
x+1\right)  ^{4}}dx}, \label{dist}%$$ is known to be a semi-metric.

Further on, we proved in ([@AGB2005], Theorem 8, page 16) that $\mathcal{A}$ is a dense set in $\mathcal{Y}_{\mathrm{conc}}$.

Throughout this communication $\Phi_{\operatorname{id}}$ will denote the identity function defined on the half line $\left[  0,\text{ }\infty\right)  $ and we write $\left\Vert \Phi\right\Vert :=\sqrt{%
%TCIMACRO{\dint _{\left[  0,\text{ }\infty\right)  }}%
%BeginExpansion
{\displaystyle\int_{\left[  0,\text{ }\infty\right)  }}
%EndExpansion
\Phi^{2}d\mu}$ whenever $\Phi\in\mathcal{Y}_{\mathrm{conc}}$.

We intend to generate a decreasing sequence of subsets $\left(  \mathcal{Y}%
_{b}^{\left(  n\right)  }\right)  \subset\mathcal{Y}_{\mathrm{conc}}$ depending on $b$ such that whenever $n\in\mathbb{N}$, then $\mathcal{A}%
\cap\mathcal{Y}_{b}^{\left(  n\right)  }$ is dense in $\mathcal{Y}%
_{b}^{\left(  n\right)  }$ and the following four sets $\mathcal{Y}%
_{b}^{\left(  n\right)  }$, $\mathcal{Y}_{b}^{\left(  n\right)  }%
\backslash\left(  \mathcal{A}\cap\mathcal{Y}_{b}^{\left(  n\right)  }\right)
$, $\mathcal{A}\cap\mathcal{Y}_{b}^{\left(  n\right)  }$ and $\mathcal{Y}%
_{\mathrm{conc}}$ are pairwise equinumerous. We shall also prove that the two pairs $\left(  \mathcal{Z}^{\ast\left(  n\right)  },\operatorname*{dist}%
\right)  $ and $\left(  \mathcal{Z}^{\left(  n\right)  },\operatorname*{dist}%
\right)  $ are metric spaces, where $\mathcal{Z}^{\ast\left(  n\right)
}=\left\{  \mathcal{Y}_{b}^{\left(  n\right)  }:{\small b\in\left(
0,\,\infty\right)  }\right\}  $ and $\mathcal{Z}^{\left(  n\right)  }=\left\{
\mathcal{A}_{b}^{\left(  n\right)  }:{\small b\in\left(  0,\,\infty\right)
}\right\}  $ for each $n\in\mathbb{N}$ and the distance between any two sets $\mathcal{F}$ and $\mathcal{G}$ in $\mathcal{Y}_{\mathrm{conc}}$ being defined by $$\begin{aligned}
\operatorname*{dist}\left(  \mathcal{F},\mathcal{G}\right)   &  :=\sup\left\{
\inf\left\{  \operatorname{d}\left(  \Phi,\Psi\right)  :\Psi\in\mathcal{G}%
\right\}  :\Phi\in\mathcal{F}\right\} \\
&  =\sup\left\{  \inf\left\{  \operatorname{d}\left(  \Phi,\Psi\right)
:\Phi\in\mathcal{F}\right\}  :\Psi\in\mathcal{G}\right\}  .\end{aligned}$$ We show in the last section that if $f$ is any measurable function on a measure space $\left(  \Omega,\mathcal{F},\lambda\right)  $ and $p\in\left[
1,\infty\right)  $ is an arbitrary number then the quantities $\left\Vert
f\right\Vert _{L^{p}}$ and $\sup_{\Phi\in\widetilde{\mathcal{Y}_{\mathrm{conc}%
}}}\left(  \Phi\left(  1\right)  \right)  ^{-1}\left\Vert \Phi\circ\left\vert
f\right\vert \right\Vert _{L^{p}}$ are equivalent, in the sense that they are both either finite or infinite at the same time, where $\widetilde
{\mathcal{Y}_{\mathrm{conc}}}$ is a proper subset of $\mathcal{Y}%
_{\mathrm{conc}}$.We then use this subset to express the value of $\left\Vert
f\right\Vert _{L^{p}}$ whenever $\left\Vert f\right\Vert _{L^{p}}<\infty$.

Bijections between subsets of $\mathcal{Y}_{\mathrm{conc}}$
===========================================================

We first anticipate that there are as many elements in each of the sets $\mathcal{A}$ and $\mathcal{Y}_{\mathrm{conc}}\backslash\mathcal{A}$ as there exist in $\mathcal{Y}_{\mathrm{conc}}$, showing how broad the set of concave Young-functions possessing the density-level property and its complement really are.

\[theo1\]The sets $\mathcal{A}$, $\mathcal{Y}_{\mathrm{conc}}$ and $\mathcal{Y}_{\mathrm{conc}}\backslash\mathcal{A}$ are pairwise equinumerous.

We first show that there is a bijection between $\mathcal{A}$ and $\mathcal{Y}_{\mathrm{conc}}$. In fact, since $\mathcal{A}$ is a proper subset of $\mathcal{Y}_{\mathrm{conc}}$ there is an injection from $\mathcal{A}$ to $\mathcal{Y}_{\mathrm{conc}}$, as a matter of fact, the identity mapping from $\mathcal{A}$ into $\mathcal{Y}_{\mathrm{conc}}$ will do. Fix any number $\alpha\in\left(  0,\,1\right)  $ and define the mapping $S_{\alpha
}:\mathcal{Y}_{\mathrm{conc}}\rightarrow\mathcal{A}$ by $S_{\alpha}\left(
\Phi\right)  =\Phi^{\alpha}$. We point out that this mapping exists in virtue of Theorem 2 in [@AGB2005]. It is not hard to see that $S_{\alpha}$ is an injection. Then the Schröder-Bernstein theorem entails that there exists a bijection between $\mathcal{A}$ and $\mathcal{Y}_{\mathrm{conc}}$. To complete the proof it is enough to show that there is a bijection between $\mathcal{A}$ and $\mathcal{Y}_{\mathrm{conc}}\backslash\mathcal{A}$. In fact, fix arbitrarily some $\Phi\in\mathcal{Y}_{\mathrm{conc}}\backslash\mathcal{A}$ and define the function $h_{\Phi}:\mathcal{A}\rightarrow\mathcal{Y}_{\mathrm{conc}%
}\backslash\mathcal{A}$ by $h_{\Phi}\left(  \Delta\right)  =\Delta+\Phi$. Obviously, $h_{\Phi}$ is an injection. Now, fix any $\Delta\in\mathcal{A}$ and define the function $f_{\Delta}:\mathcal{Y}_{\mathrm{conc}}\backslash
\mathcal{A}\rightarrow\mathcal{A}$ by $f_{\Delta}\left(  \Phi\right)
=\Delta\circ\Phi$. We point out that this function always exists due to Theorem \[theo2\] in [@AGB2005]. It is not difficult to show that $f_{\Delta}$ is an injection if we take into account that $\Delta$ is an invertible function. Consequently, the Schröder-Bernstein theorem guarantees the existence of a bijection between $\mathcal{A}$ and $\mathcal{Y}_{\mathrm{conc}}\backslash\mathcal{A}$. Therefore, we can conclude on the validity of the argument.

Write $\mathcal{A}_{b}:=\left\{  \Phi\in\mathcal{A}:\Phi\left(  b\right)
=b\right\}  $ and $\mathcal{Y}_{b}:=\left\{  \Phi\in\mathcal{Y}_{\mathrm{conc}%
}:\Phi\left(  b\right)  =b\right\}  $ for every number $b\in\left(
0,\,\infty\right)  $.

Let us denote by $\mathcal{Z}:=\left\{  \mathcal{A}_{b}:b\in\left(
0,\,\infty\right)  \right\}  $ and $\mathcal{Z}^{\ast}:=\left\{
\mathcal{Y}_{b}:b\in\left(  0,\,\infty\right)  \right\}  $.

It is obvious that $\mathcal{A}_{b}\subset\mathcal{Y}_{b}$ for every number $b\in\left(  0,\,\infty\right)  $ and $\mathcal{Z}\cap\mathcal{Z}^{\ast
}=\varnothing$.

\[lem1\]For every number $b\in\left(  0,\,\infty\right)  $ the identities $\mathcal{A}_{b}=\left\{  \frac{b\Phi}{\Phi\left(  b\right)  }:\Phi
\in\mathcal{A}\right\}  $ and $\mathcal{Y}_{b}=\left\{  \frac{b\Phi}%
{\Phi\left(  b\right)  }:\Phi\in\mathcal{Y}_{\mathrm{conc}}\right\}  $ hold true.

Pick any function $\Psi\in\mathcal{A}_{b}$. Then $\Psi\in\mathcal{A}$ and $\Psi\left(  b\right)  =b$, so that $\Psi=\frac{b\Psi}{\Psi\left(  b\right)
}\in\left\{  \frac{b\Phi}{\Phi\left(  b\right)  }:\Phi\in\mathcal{A}\right\}
$, i.e. $\mathcal{A}_{b}\subset\left\{  \frac{b\Phi}{\Phi\left(  b\right)
}:\Phi\in\mathcal{A}\right\}  $. To show the reverse inclusion consider any function $\Psi\in\left\{  \frac{b\Phi}{\Phi\left(  b\right)  }:\Phi
\in\mathcal{A}\right\}  $. Then necessarily there must exist some $\Phi
\in\mathcal{A}$ such that $\Psi=\frac{b\Phi}{\Phi\left(  b\right)  }$. It is obvious that $\Psi\in\mathcal{A}$ and $\Psi\left(  b\right)  =b$, i.e. $\Psi\in\mathcal{A}_{b}$. Hence, $\left\{  \frac{b\Phi}{\Phi\left(  b\right)
}:\Phi\in\mathcal{A}\right\}  \subset\mathcal{A}_{b}$. These two inclusions yield that $\mathcal{A}_{b}=\left\{  \frac{b\Phi}{\Phi\left(  b\right)  }%
:\Phi\in\mathcal{A}\right\}  $. The proof of identity $\mathcal{Y}%
_{b}=\left\{  \frac{b\Phi}{\Phi\left(  b\right)  }:\Phi\in\mathcal{Y}%
_{\mathrm{conc}}\right\}  $ can be similarly carried out.

\[def1\]A proper subset $\mathcal{G}$ of $\mathcal{A}$ is said to be maximally bounded if each of the sets $\mathcal{G}$ and $\mathcal{A}%
\backslash\mathcal{G}$ is equinumerous with $\mathcal{A}$, i.e. there is a bijection between $\mathcal{A}$ and $\mathcal{G}$, and $\operatorname*{diam}%
(\mathcal{G})<\infty$, where $\operatorname*{diam}(\mathcal{G}):=\sup\left\{
\operatorname*{d}\left(  \Phi_{1},\Phi_{2}\right)  :\Phi_{1},~\Phi_{2}%
\in\mathcal{G}\right\}  $ is the diameter of $\mathcal{G}$.

We note that Definition \[def1\] makes sense for the two reasons here below.

On the one hand we assert that $\operatorname*{diam}(\mathcal{A})=\sup\left\{
\operatorname*{d}\left(  \Phi_{1},\Phi_{2}\right)  :\Phi_{1},~\Phi_{2}%
\in\mathcal{A}\right\}  =\infty$. In fact, fix some $\Phi\in\mathcal{A}$ and define a sequence $\left(  \Phi_{n}\right)  \subset\mathcal{Y}%
_{\mathrm{conc}}$ by $\Phi_{2n}=4n\Phi$ and $\Phi_{2n-1}=\left(  2n-1\right)
\Phi$, $n\in\mathbb{N}$. It is clear that $\left(  \Phi_{n}\right)
\subset\mathcal{A}$ and $\operatorname*{d}\left(  \Phi_{2n},\Phi
_{2n-1}\right)  =\left(  2n+1\right)  \left\Vert \Phi\right\Vert $, $n\in\mathbb{N}$. Hence, $\operatorname*{diam}(\mathcal{A})=\infty$.

On the other hand the set $\left\{  \left(  \Phi\left(  1\right)  \right)
^{-1}\Phi:\Phi\in\mathcal{Y}_{\mathrm{conc}}\right\}  $ is of finite diameter. In fact for any $\Phi$, $\Psi\in\mathcal{Y}_{\mathrm{conc}}$ we have, via Lemma \[lem3\] in [@AGB2005], that $$\operatorname*{d}\left(  \left(  \Phi\left(  1\right)  \right)  ^{-1}%
\Phi,\left(  \Psi\left(  1\right)  \right)  ^{-1}\Psi\right)  \leq\left\Vert
\left(  \Phi\left(  1\right)  \right)  ^{-1}\Phi\right\Vert +\left\Vert
\left(  \Psi\left(  1\right)  \right)  ^{-1}\Psi\right\Vert \leq2\left\Vert
S\right\Vert <\infty.$$

Let us define two relations $\mathrm{\bot}\subset\mathcal{A}\times\mathcal{A}%
$ and $\mathrm{\bot}^{\ast}\subset\mathcal{Y}_{\mathrm{conc}}\times
\mathcal{Y}_{\mathrm{conc}}$ as follows:

1.  We say that $\left(  \Phi,\Psi\right)  \in\mathrm{\bot}$, where $\left(
    \Phi,\Psi\right)  \in\mathcal{A}\times\mathcal{A}$, (and write $\Phi
    \mathrm{\bot}\Psi$) if and only if there is some constant $c\in\left(
    0,\,\infty\right)  $ such that $\Psi\left(  x\right)  =c\Phi\left(  x\right)
    $ for all $x\in\left(  0,\,\infty\right)  $.

2.  We say that $\left(  \Phi,\Psi\right)  \in\mathrm{\bot}^{\ast}$, where $\left(  \Phi,\Psi\right)  \in\mathcal{Y}_{\mathrm{conc}}\times\mathcal{Y}%
    _{\mathrm{conc}}$, (and write $\Phi\mathrm{\bot}^{\ast}\Psi$) if and only if there is some constant $c\in\left(  0,\,\infty\right)  $ such that $\Psi\left(  x\right)  =c\Phi\left(  x\right)  $ for all $x\in\left(
    0,\,\infty\right)  $.

It is not hard to see that $\mathrm{\bot}$ and $\mathrm{\bot}^{\ast}$ are equivalence relations on $\mathcal{A}$ and $\mathcal{Y}_{\mathrm{conc}}$ respectively, i.e. they are reflexive, symmetric and transitive. Their corresponding equivalence classes are respectively$$\begin{aligned}
p_{\mathrm{\bot}}\left(  \Psi\right)   &  :=\left\{  \Phi:\Phi\in
\mathcal{A}\text{ and }\Phi\mathrm{\bot}\Psi\right\}  ,\text{ }\Psi
\in\mathcal{A}\\
p_{\mathrm{\bot}^{\ast}}\left(  \Delta\right)   &  :=\left\{  \Phi:\Phi
\in\mathcal{Y}_{\mathrm{conc}}\text{ and }\Phi\mathrm{\bot}^{\ast}%
\Delta\right\}  ,\text{ }\Delta\in\mathcal{Y}_{\mathrm{conc}}%\end{aligned}$$ and their respective induced factor (or quotient) sets can be given by$$\begin{aligned}
\mathcal{A}/\mathrm{\bot}  &  :=\left\{  \mathcal{C}:\mathcal{C}%
\subset\mathcal{A}\text{ and }\mathcal{C}=p_{\mathrm{\bot}}\left(
\Psi\right)  \text{ for some }\Psi\in\mathcal{A}\right\}  ,\\
\mathcal{Y}_{\mathrm{conc}}/\mathrm{\bot}^{\ast}  &  :=\left\{  \mathcal{C}%
:\mathcal{C}\subset\mathcal{Y}_{\mathrm{conc}}\text{ and }\mathcal{C}%
=p_{\mathrm{\bot}^{\ast}}\left(  \Delta\right)  \text{ for some }\Delta
\in\mathcal{Y}_{\mathrm{conc}}\right\}\end{aligned}$$

One can easily verify that for all $\Psi\in\mathcal{A}$ and $\Delta
\in\mathcal{Y}_{\mathrm{conc}}$ the equivalence classes $p_{\mathrm{\bot}%
}\left(  \Psi\right)  $ and $p_{\mathrm{\bot}^{\ast}}\left(  \Delta\right)
$ are of continuum size or magnitude.

\[theo2\]Let $b\in\left(  0,\,\infty\right)  $ be any fixed number.**Part I.** Define the mapping $f:\mathcal{A}%
\rightarrow\mathcal{A}_{b}$ by $f\left(  \Phi\right)  =\frac{b}{\Phi\left(
b\right)  }\Phi$. Then there is a unique mapping $g:\mathcal{A}/\mathrm{\bot
}\rightarrow\mathcal{A}_{b}$ for which the diagram$$\xymatrix{\mathcal{A} \ar[r]^{p_{\mathrm{\bot}}} \ar[dr]_f & \mathcal{A}/\mathrm{\bot} \ar[d]^g\\ ~ & \mathbf{b} }
\label{com3}%$$ commutes *(*i.e. $f=g\circ p_{\mathrm{\bot}}$*)* and moreover, the mapping $g$ is a bijection.**Part II.** Define the mapping $f^{\ast}:\mathcal{Y}_{\mathrm{conc}}\rightarrow\mathcal{Y}_{b}$ by $f^{\ast
}\left(  \Delta\right)  =\frac{b}{\Delta\left(  b\right)  }\Delta$. Then there is a unique mapping $g^{\ast}:\mathcal{Y}_{\mathrm{conc}}/\mathrm{\bot}%
^{\ast}\rightarrow\mathcal{Y}_{b}$ for which the diagram$$\xymatrix{\mathcal{Y}_{\mathrm{conc}} \ar[r]^{p_{\mathrm{\bot}^{\ast}}} \ar[dr]_{f^{\ast}} & \mathcal{Y}_{\mathrm{conc}}/\mathrm{\bot}^{\ast} \ar[d]^{g^{\ast}}\\ ~ & \mathbf{b}^{\ast} }
\label{com4}%$$ commutes *(*i.e. $f^{\ast}=g^{\ast}\circ p_{\mathrm{\bot}^{\ast}}%
$*)* and moreover, the mapping $g^{\ast}$ is a bijection.

We point out that the proof of Theorem \[theo2\] is obvious.

\[prop1\]Let $b\in\left(  0,\,\infty\right)  $ be an arbitrarily fixed number.**Part I.** There is a bijection between $\mathcal{Y}_{b}$ and $\mathcal{Y}_{\mathrm{conc}}$.**Part II.** There is a bijection between $\mathcal{A}_{b}$ and $\mathcal{A}$.

We shall only show the first part because the other case can be similarly proved. To this end, write $\mathcal{Y}_{bb}:=\left\{  b\Phi:\Phi
\in\mathcal{Y}_{\mathrm{conc}}\right\}  $. We note that $\mathcal{Y}_{bb}$ and $\mathcal{Y}_{\mathrm{conc}}$ are equinumerous for the reasons that $\mathcal{Y}_{bb}\subset\mathcal{Y}_{\mathrm{conc}}$ and the function $F:\mathcal{Y}_{\mathrm{conc}}\rightarrow\mathcal{Y}_{bb}$, defined by $F\left(  \Phi\right)  =b\Phi$, can be easily shown to be an injection. Thus it will be enough to prove that $\mathcal{Y}_{bb}$ and $\mathcal{Y}_{b}$ are equinumerous. In fact, consider the function $Q:\mathcal{Y}_{b}\rightarrow
\mathcal{Y}_{bb}$ defined by $Q\left(  \frac{b}{\Phi\left(  b\right)  }%
\Phi\right)  =b\Phi$. We shall just point out that function $Q$ can be easily shown to be a bijection, which ends the proof.

\[cor1\]Let $b\in\left(  0,\,\infty\right)  $ be arbitrary. Then the following six sets $\mathcal{A}$, $\mathcal{A}_{b}$, $\mathcal{Y}_{b}$, $\mathcal{A}/\mathrm{\bot}$, $\mathcal{Y}_{\mathrm{conc}}/\mathrm{\bot}^{\ast
}$and $\mathcal{Y}_{\mathrm{conc}}$ are equinumerous.

We note that $\mathcal{A}$ and $\mathcal{Y}_{\mathrm{conc}}$ are equinumerous (by Theorem \[theo1\]) and, by Theorem 2, $\mathcal{A}/\mathrm{\bot}$ and $\mathcal{A}_{b}$ are equinumerous. On the other hand $\mathcal{A}$ and $\mathcal{A}_{b}$ are equinumerous as well as $\mathcal{Y}_{b}$ and $\mathcal{Y}_{\mathrm{conc}}$ are (by Proposition \[prop1\]). Thus $\mathcal{A}_{b}$ and $\mathcal{Y}_{\mathrm{conc}}$ are equinumerous. Therefore, as $\mathcal{Y}_{b}$ and $\mathcal{Y}_{\mathrm{conc}}/\mathrm{\bot
}^{\ast}$ are equinumerous (by Theorem 2), we can conclude on the validity of the argument.

\[rem1\]Let $b_{1}$ and $b_{2}\in\left(  0,\,\infty\right)  $ be two arbitrary distinct numbers. Then $\mathcal{A}_{b_{1}}\cap\mathcal{A}_{b_{2}}$ and $\mathcal{Y}_{b_{1}}\cap\mathcal{Y}_{b_{2}}$ are empty sets.

\[rem2\]Let $b_{1}$ and $b_{2}\in\left(  0,\,\infty\right)  $ be two arbitrary distinct numbers. Then $\mathcal{A}_{b_{1}}\cup\mathcal{A}_{b_{2}%
}\notin\mathcal{Z}$ and $\mathcal{Y}_{b_{1}}\cup\mathcal{Y}_{b_{2}}%
\notin\mathcal{Z}^{\ast}$.

\[rem3\]Fix arbitrarily a number $b\in\left(  0,\,\infty\right)  $. Then it is easily seen that the function $h_{b}:\left[  0,~\infty\right)
\rightarrow\left[  0,~\infty\right)  $, defined by $h_{b}\left(  x\right)
=x+b$, is square integrable with respect to measure $\mu$ and, moreover, $C_{b}:=%
%TCIMACRO{\dint _{0}^{\infty}}%
%BeginExpansion
{\displaystyle\int_{0}^{\infty}}
%EndExpansion
\frac{\left(  h_{b}\left(  x\right)  \right)  ^{2}}{\left(  x+1\right)  ^{4}%
}dx=\frac{1}{3}\left(  b^{2}+b+1\right)  <\infty$.

\[rem4\]If $\Phi\in\mathcal{Y}_{b}$, then $\Phi\left(  x\right)  \leq
h_{b}\left(  x\right)  $ for all $x\in\left[  0,\,\infty\right)  $.

Fix any $\Phi\in\mathcal{Y}_{b}$. As $\Phi$ is a concave function its graph must lie below the tangent of equation $y=\varphi\left(  b\right)  \left(
x-b\right)  +b$ at point $\left(  b,b\right)  $ since $\Phi\left(  b\right)
=b$. Consequently, for all $x\in\left[  0,\,\infty\right)  $ we have:$$\begin{aligned}
\Phi\left(  x\right)   &  \leq\varphi\left(  b\right)  \left(  x-b\right)
+b\leq\varphi\left(  b\right)  x+b=b\varphi\left(  b\right)  \frac{x}{b}+b\\
&  \leq\Phi\left(  b\right)  \frac{x}{b}+b=h_{b}\left(  x\right)  .\end{aligned}$$

\[prop2\]Let $b\in\left(  0,\,\infty\right)  $ be any number. Then $\mathcal{Y}_{b}$ is of finite diameter.

Let $b\in\left(  0,\,\infty\right)  $ be the source of $\mathcal{Y}_{b}%
\in\mathcal{Z}^{\ast}$. We need to prove that $\mathcal{Y}_{b}$ has a finite diameter. In fact, consider two arbitrary functions $\Phi_{1}$, $\Phi_{2}%
\in\mathcal{Y}_{b}$. Then$$\operatorname{d}\left(  \Phi_{1},\Phi_{2}\right)  =\left\Vert \Phi_{1}%
-\Phi_{2}\right\Vert \leq\left\Vert \Phi_{1}\right\Vert +\left\Vert \Phi
_{2}\right\Vert \leq\sqrt{2C_{b}}\text{,}%$$ via Remarks \[rem4\] and \[rem3\]. Therefore,$$\operatorname*{diam}\left(  \mathcal{Y}_{b}\right)  :=\sup\left\{
\operatorname{d}\left(  \Phi_{1},\Phi_{2}\right)  :\Phi_{1},~\Phi_{2}%
\in\mathcal{Y}_{b}\right\}  \leq\sqrt{2C_{b}}<\infty.$$

\[theo3\]Let $b\in\left(  0,\,\infty\right)  $ be any number. Then $\mathcal{Y}_{b}$ is maximally bounded.

We just point out that the proof follows from the conjunction of both Propositions \[prop2\] and \[prop1\].

In the sequel $H_{\left[  0,~1\right]  }$ will stand for the collection of all finite sequences $\left(  t_{1},~\ldots,~t_{k}\right)  \subset\left[
0,~1\right]  $ such that $t_{1}+\ldots+t_{k}=1$.

For any fixed $b\in\left(  0,\,\infty\right)  $ and every counting number $n\in\mathbb{N}$ write $\operatorname*{X}\limits_{i=1}^{n}\mathcal{A}_{b}$ (resp. $\operatorname*{X}\limits_{i=1}^{n}\mathcal{Y}_{b}$) for the $n$-fold Descartes product of $\mathcal{A}_{b}$ (resp. $\mathcal{Y}_{b}$).

For $n=1$ let us set $\mathcal{A}_{b}^{\left(  1\right)  }=\mathcal{A}_{b}$, $\mathcal{Y}_{b}^{\left(  1\right)  }=\mathcal{Y}_{b}$ and whenever $n\geq2$, write $\mathcal{Y}_{b}^{CO\left(  n\right)  }=\left\{  \Delta_{1}\circ
\Delta_{2}\circ\ldots\circ\Delta_{n}:\left(  \Delta_{1},\Delta_{2}%
,\ldots,\Delta_{n}\right)  \in\operatorname*{X}\limits_{i=1}^{n}%
\mathcal{Y}_{b}\right\}  $, $$\mathcal{A}_{b}^{CO\left(  n\right)  }=\left\{  \Phi_{1}\circ\ldots\circ
\Phi_{n}:\left(  \Phi_{1},\ldots,\Phi_{n}\right)  \in\operatorname*{X}%
\limits_{i=1}^{n}\mathcal{Y}_{b}\text{ and }\Phi_{j}\in\mathcal{A}_{b}\text{
for some index }j\right\}  ,$$ $\mathcal{Y}_{b}^{\left(  n\right)  }=\left\{  \sum_{i=1}^{k}t_{i}\Delta
_{i}:\Delta_{1},~\Delta_{2},~\ldots,~\Delta_{k}\in\mathcal{Y}_{b}^{CO\left(
n\right)  },~\left(  t_{1},~\ldots~t_{k}\right)  \in H_{\left[  0,~1\right]
}\right\}  $, $\mathcal{A}_{b}^{\left(  n\right)  }=\left\{  \sum_{i=1}%
^{k}t_{i}\Phi_{i}:\Phi_{1},~\Phi_{2},~\ldots,~\Phi_{k}\in\mathcal{A}%
_{b}^{CO\left(  n\right)  },~\left(  t_{1},~\ldots~t_{k}\right)  \in
H_{\left[  0,~1\right]  }\right\}  $.

Further, for $n=1$ write $\mathcal{Z}^{\left(  1\right)
}=\mathcal{Z}$, $\mathcal{Z}^{\ast\left(  1\right)  }=\mathcal{Z}^{\ast}$ and, for $n\in\mathbb{N}\backslash\left\{  1\right\}  $ write $\mathcal{Z}^{\left(
n\right)  }:=\left\{  \mathcal{A}_{b}^{\left(  n\right)  }:b\in\left(
0,\,\infty\right)  \right\}  $ and $\mathcal{Z}^{\ast\left(  n\right)
}:=\left\{  \mathcal{Y}_{b}^{\left(  n\right)  }:b\in\left(  0,\,\infty
\right)  \right\}  $.

\[rem5\]For any pair of numbers $n\in\mathbb{N}$ and $b\in\left(
0,\,\infty\right)  $ the set $\mathcal{A}_{b}^{\left(  n\right)  }$ is a proper subset of $\mathcal{Y}_{b}^{\left(  n\right)  }$.

\[rem6\]For any pair of numbers $n\in\mathbb{N}$ and $b\in\left(
0,\,\infty\right)  $ we have $\mathcal{A}_{b}^{\left(  n\right)  }%
\subset\mathcal{A}_{b}^{\left(  1\right)  }=\mathcal{A}_{b}$.

We point out that Remark \[rem6\] is a direct consequent of Theorem 2 in [@AGB2005], page 6.

\[rem7\]Let $b\in\left(  0,\,\infty\right)  $, $n\in\mathbb{N}$ and $k\geq
n$ be arbitrary numbers. Then *(1)* $\Phi_{1}\circ\Phi_{2}%
\circ\ldots\circ\Phi_{k}\in\mathcal{A}_{b}^{CO\left(  n\right)  }$ whenever $\Phi_{1},~\Phi_{2},~\ldots,~\Phi_{k}\in\mathcal{Y}_{b}^{\left(  1\right)  }$ and $\Phi_{j}\in\mathcal{A}_{b}^{\left(  1\right)  }$ for some index $j\in\left\{  1,~\ldots,~k\right\}  $*(2)* $\Delta_{1}\circ
\Delta_{2}\circ\ldots\circ\Delta_{k}\in\mathcal{Y}_{b}^{CO\left(  n\right)  }$ whenever $\Delta_{1},~\Delta_{2},~\ldots,~\Delta_{k}\in\mathcal{Y}%
_{b}^{\left(  1\right)  }$.

Note that $\Phi_{1}\circ\Phi_{2}\circ\ldots\circ\Phi_{k}=\Phi_{1}\circ\Phi
_{2}\circ\ldots\circ\Phi_{n-1}\circ\Psi_{1}$ and $\Delta_{1}\circ\Delta
_{2}\circ\ldots\circ\Delta_{n-1}\circ\Psi_{2}$, where $\Psi_{1}=\Phi_{n}%
\circ\Phi_{n+1}\circ\ldots\circ\Phi_{k}$ and $\Psi_{2}=\Delta_{n}\circ
\Delta_{n+1}\circ\ldots\circ\Delta_{k}$. From this simple observation the result easily follows.

From Remark \[rem7\] the following result can be easily derived, since it implies that $\mathcal{A}_{b}^{CO\left(  n+1\right)  }$ is a proper subset of $\mathcal{A}_{b}^{CO\left(  n\right)  }$ and, $\mathcal{Y}_{b}^{CO\left(
n+1\right)  }$ is also a proper subset of $\mathcal{Y}_{b}^{CO\left(
n\right)  }$.

\[lem2\]Let $b\in\left(  0,\,\infty\right)  $ and $n\in\mathbb{N}$ be arbitrary numbers. Then the following two assertions are valid.*(1)* The set $\mathcal{A}_{b}^{\left(  n+1\right)  }$ is a proper subset of $\mathcal{A}_{b}^{\left(  n\right)  }$.*(2)* The set $\mathcal{Y}_{b}^{\left(  n+1\right)  }$ is a proper subset of $\mathcal{Y}%
_{b}^{\left(  n\right)  }$.

\[theo4\]For any fixed pair of numbers $n\in\mathbb{N}$ and $b\in\left(
0,\,\infty\right)  $, the two sets $\mathcal{A}_{b}^{\left(  n\right)  }$ and $\mathcal{A}_{b}$ are equinumerous.

Throughout the proof we shall fix any counting number $n\in\mathbb{N}$. We first note that the identity function $I_{\operatorname{id}}:\mathcal{A}%
_{b}^{\left(  n\right)  }\rightarrow\mathcal{A}_{b}$ is an injection, since $\mathcal{A}_{b}^{\left(  n\right)  }\subset\mathcal{A}_{b}$. Next, pick any $\Delta\in\mathcal{A}_{b}$ and define the function $f_{\Delta}:\mathcal{A}%
_{b}\rightarrow\mathcal{A}_{b}^{\left(  n\right)  }$ by $f_{\Delta}\left(
\Phi\right)  =\underset{\left(  n-1\right)  \text{-fold}}{\underbrace
{\Delta\circ\ldots\circ\Delta}}\circ\Phi$. We show that $f_{\Delta}$ is an injection. In fact, let $\Phi_{1}$, $\Phi_{2}\in\mathcal{A}_{b}$ be arbitrary and assume that $f_{\Delta}\left(  \Phi_{1}\right)  =f_{\Delta}\left(
\Phi_{2}\right)  $. Then taking into account that $\Delta$ is an invertible function we can easily deduce that $\Phi_{1}=\Phi_{2}$, i.e. $f_{\Delta}$ is an injection. Therefore, the Schröder-Bernstein theorem entails that there is a bijection between $\mathcal{A}_{b}$ and $\mathcal{A}_{b}^{\left(
n\right)  }$. This was to be proved.

\[prop3\]For any pair of numbers $n\in\mathbb{N}$ and $b\in\left(
0,~\infty\right)  $ the sets $\mathcal{A}_{b}^{\left(  n\right)  }$ and $\mathcal{Y}_{b}^{\left(  n\right)  }\backslash\mathcal{A}_{b}^{\left(
n\right)  }$ are equinumerous.

Let $\Phi\in\mathcal{Y}_{b}^{\left(  n\right)  }\backslash\mathcal{A}%
_{b}^{\left(  n\right)  }$ and $\left(  \alpha,\beta\right)  \in H_{\left[
0,~1\right]  }$ be arbitrarily fixed. Define the function $h_{\Phi}^{\left(
\alpha,\beta\right)  }:\mathcal{A}_{b}^{\left(  n\right)  }\rightarrow
\mathcal{Y}_{b}^{\left(  n\right)  }\backslash\mathcal{A}_{b}^{\left(
n\right)  }$ by $h_{\Phi}^{\left(  \alpha,\beta\right)  }\left(
\Delta\right)  =\alpha\Delta+\beta\Phi$. It is clear that $h_{\Phi}^{\left(
\alpha,\beta\right)  }$ is actually an injection. Now, fix any $\Delta
\in\mathcal{A}_{b}^{\left(  n\right)  }$ and define the function $f_{\Delta
}:\mathcal{Y}_{b}^{\left(  n\right)  }\backslash\mathcal{A}_{b}^{\left(
n\right)  }\rightarrow\mathcal{A}_{b}^{\left(  n\right)  }$ by $f_{\Delta
}\left(  \Phi\right)  =\Delta\circ\Phi$. We note that this function always exists because of the inclusion $\mathcal{A}_{b}^{\left(  n\right)  }%
\subset\mathcal{A}$ and Theorem \[theo2\] in [@AGB2005]. Here too we can easily check that $f_{\Delta}$ is an injection. Therefore, The Schröder-Bernstein theorem yields the result to be proven.

For any pair of numbers $n\in\mathbb{N}$ and $b\in\left(  0,~\infty\right)  $ the following five sets $\mathcal{Y}_{b}^{\left(  n\right)  }$, $\mathcal{A}%
_{b}^{\left(  n\right)  }$, $\mathcal{Y}_{b}^{\left(  n\right)  }%
\backslash\mathcal{A}_{b}^{\left(  n\right)  }$, $\mathcal{A}$ and $\mathcal{Y}_{\mathrm{conc}}$ are pairwise equinumerous.

The metrization of sets $\mathcal{Z}^{\left(  n\right)  }$ and $\mathcal{Z}^{\ast\left(  n\right)  }$
=====================================================================================================

We shall only deal with the metrization of sets $\mathcal{Z}$ and $\mathcal{Z}^{\ast}$ since all the results in this section can be easily extended to the sets $\mathcal{Z}^{\left(  n\right)  }$ and $\mathcal{Z}%
^{\ast\left(  n\right)  }$.

Whenever $\Phi\in\mathcal{Y}_{\mathrm{conc}}$ write $G_{\Phi}:=\left\{
\left(  x,\Phi\left(  x\right)  \right)  :x\in\left(  0,~\infty\right)
\right\}  $ for the graph of $\Phi$ on $\left(  0,~\infty\right)  $ and $G_{\Phi}^{a||b}:=\left\{  \left(  x,\Phi\left(  x\right)  \right)
:x\in\left[  a,~b\right]  \right\}  $ for the graph of $\Phi$ on the interval $\left[  a,~b\right]  $ where $a<b$ are any non-negative numbers.

\[rem8\]Let $b_{1}$ and $b_{2}\in\left(  0,\,\infty\right)  $ be two arbitrary distinct numbers. If $b_{1}<b_{2}$, then the following two assertions hold true:*(1)* For all $\Phi_{1}\in\mathcal{A}%
_{b_{1}}$ and $\Phi_{2}\in\mathcal{A}_{b_{2}}$ the inequality $\Phi
_{1}\left(  b_{2}\right)  <\Phi_{2}\left(  b_{1}\right)  $ holds.$\emph{(2)}$ For all $\Phi_{1}\in\mathcal{Y}_{b_{1}}$ and $\Phi_{2}%
\in\mathcal{Y}_{b_{2}}$ the inequality $\Phi_{1}\left(  b_{2}\right)
<\Phi_{2}\left(  b_{1}\right)  $ holds.

Suppose that $b_{1}<b_{2}$ and fix arbitrarily two functions $\Phi_{1}%
\in\mathcal{Y}_{b_{1}}$ and $\Phi_{2}\in\mathcal{Y}_{b_{2}}$. Obviously, $\Phi_{1}$ must hit $\Phi_{\operatorname{id}}$ *prior to* $\Phi_{2}$. Hence, $G_{\Phi_{1}}^{b_{1}||b_{2}}$ lies below $G_{\Phi_{2}}^{b_{1}||b_{2}}$. But since $G_{\Phi_{1}}^{b_{1}||\infty}$ lies above the graph of the line of equation $y=b_{1}$ in the interval $\left(  b_{1},\,\infty\right)  $, we have as an aftermath that $\Phi_{1}\left(  b_{1}\right)  <\Phi_{1}\left(
b_{2}\right)  <\Phi_{2}\left(  b_{1}\right)  $. To end the proof we note that assertion (2) can be similarly shown.

The binary relations $\prec$ and $\preceq$ , defined on $\mathcal{Z}$ respectively by $\mathcal{A}_{b_{1}}\prec\mathcal{A}_{b_{2}}$ if and only if $\Phi_{1}\left(  b_{2}\right)  <\Phi_{2}\left(  b_{1}\right)  $ for all pairs $\left(  \Phi_{1},\Phi_{2}\right)  \in\mathcal{A}_{b_{1}}\times\mathcal{A}%
_{b_{2}}$, and by $\mathcal{A}_{b_{1}}\preceq\mathcal{A}_{b_{2}}$ if and only if $\mathcal{A}_{b_{1}}\prec\mathcal{A}_{b_{2}}$ or $\mathcal{A}_{b_{1}%
}=\mathcal{A}_{b_{2}}$. We point out that The binary relations $\prec$ and $\preceq$ can be similarly defined on $\mathcal{Z}^{\ast}$.

We point out that the law of trichotomy is valid on $\left(  \mathcal{Z}%
,\preceq\right)  $ and $\left(  \mathcal{Z}^{\ast},\preceq\right)  $, i.e. whenever $\left(  \mathcal{A}_{b_{1}},\mathcal{A}_{b_{2}}\right)
\in\mathcal{Z}\times\mathcal{Z}$ or $\left(  \mathcal{A}_{b_{1}}%
,\mathcal{A}_{b_{2}}\right)  \in\mathcal{Z}^{\ast}\times\mathcal{Z}^{\ast}$, then precisely one of the following holds: $\mathcal{A}_{b_{1}}=\mathcal{A}%
_{b_{2}}$, $\mathcal{A}_{b_{1}}\prec\mathcal{A}_{b_{2}}$, $\mathcal{A}_{b_{2}%
}\prec\mathcal{A}_{b_{1}}$. Hence, we can easily check that $\left(
\mathcal{Z},\preceq\right)  $ and $\left(  \mathcal{Z}^{\ast},\preceq\right)
$ are chains, i.e. they are totally ordered sets.

\[theo5\]The functions $f_{1}:\left(  0,\,\infty\right)  \rightarrow
\mathcal{Z}$ and $f_{2}:\left(  0,\,\infty\right)  \rightarrow\mathcal{Z}%
^{\ast}$, defined respectively by $f_{1}\left(  p\right)  =\mathcal{A}_{p}$ and $f_{2}\left(  p\right)  =\mathcal{Y}_{p}$, are order preserving bijections.

We show that the function $f_{1}:\left(  0,\,\infty\right)  \rightarrow
\mathcal{Z}$, $f_{1}\left(  p\right)  =\mathcal{A}_{p}$, is an order preserving bijection. In fact, it is not hard to see via Remark \[rem1\] that $f_{1}$ is an injection. Now pick any element $\mathcal{C}\in\mathcal{Z}%
$. Obviously, there must exist some number $p\in\left(  0,\,\infty\right)  $ such that $\mathcal{C}=\mathcal{A}_{p}=f_{1}\left(  p\right)  $, i.e. $f_{1}$ is a surjection. Consequently, $f_{1}$ is a bijection. To end the proof of this part we simply point out that the bijection $f_{1}$ is order preserving in virtue of Remark \[rem5111\]. Finally, we note that we can similarly prove that $f_{2}$ is also an order preserving bijection.

Since the sets $\left(  \mathcal{Z},\preceq\right)  $ and $\left(
\mathcal{Z}^{\ast},\preceq\right)  $ are chains it is natural to look for a metric on them. We shall do this in the following two results. But before that let us recall the definitions of some distances known in the literature (cf. [@KUR1966], say). If $\Phi\in\mathcal{Y}_{\mathrm{conc}}$ is any function and $\mathcal{F}$, $\mathcal{G}\subset\mathcal{Y}_{\mathrm{conc}}$ are arbitrary non-empty subsets, then we define the distance from the point $\Phi$ to the set $\mathcal{G}$ by $\rho\left(  \Phi,\mathcal{G}\right)
:=\inf\left\{  \operatorname{d}\left(  \Phi,\Psi\right)  :\Psi\in
\mathcal{G}\right\}  =\inf\left\{  \operatorname{d}\left(  \Psi,\Phi\right)
:\Psi\in\mathcal{G}\right\}  =\rho\left(  \mathcal{G},\Phi\right)  $ and the distance between the two sets $\mathcal{F}$ and $\mathcal{G}$ by $$\begin{aligned}
\operatorname*{dist}\left(  \mathcal{F},\mathcal{G}\right)   &  :=\sup\left\{
\inf\left\{  \operatorname{d}\left(  \Phi,\Psi\right)  :\Psi\in\mathcal{G}%
\right\}  :\Phi\in\mathcal{F}\right\} \\
&  =\sup\left\{  \inf\left\{  \operatorname{d}\left(  \Phi,\Psi\right)
:\Phi\in\mathcal{F}\right\}  :\Psi\in\mathcal{G}\right\}  .\end{aligned}$$

First we find sufficient conditions for which the distance from a point to a subset (both in $\mathcal{Y}_{\mathrm{conc}}$) should be positive, in order to guarantee that the distance between two sets in $\mathcal{Y}_{\mathrm{conc}}$ have sense.

\[lem3\]Let $b_{1}$ and $b_{2}\in\left(  0,\,\infty\right)  $ be two arbitrary distinct numbers. Then $\rho\left(  \mathcal{Y}_{b_{1}},\Phi
_{2}\right)  >0$ and $\rho\left(  \mathcal{A}_{b_{1}},\Phi_{2}\right)
>0$ whenever $\Phi_{2}\in\mathcal{Y}_{b_{2}}$.

It is enough to show that $\rho\left(  \mathcal{A}_{b_{1}},\Phi_{2}\right)
>0$ whenever $\Phi_{2}\in\mathcal{Y}_{b_{2}}$. In fact, suppose in the contrary that $\rho\left(  \mathcal{A}_{b_{1}},\Phi_{2}\right)  =0$ for some $\Phi_{2}\in\mathcal{Y}_{b_{2}}$. Then there can be extracted some sequence $\left(  \Delta_{n}\right)  \subset\mathcal{A}_{b_{1}}$ such that $\operatorname{d}\left(  \Delta_{n},\Phi_{2}\right)  >\operatorname{d}\left(
\Delta_{n+1},\Phi_{2}\right)  $, $n\in\mathbb{N}$, and $\lim_{n\rightarrow
\infty}\operatorname{d}\left(  \Delta_{n},\Phi_{2}\right)  =\rho\left(
\mathcal{A}_{b_{1}},\Phi_{2}\right)  =0$. We point out that this can be done because of the definition of the infimum. For each $n\in\mathbb{N}$ let us set $\Gamma_{n}:=\inf_{k\geq n}\left(  \Delta_{k}-\Phi_{2}\right)  ^{2}%
$. Clearly, $\left(  \Gamma_{n}\right)  $ is a non-decreasing sequence of measurable functions with its corresponding sequence of integrals $\left(
\int_{\left[  0,\text{ }\infty\right)  }\Gamma_{n}d\mu\right)  $ been bounded above by $C_{b_{1}}+C_{b_{2}}<\infty$, see Remark \[rem3\]. Then by the Beppo Levi’s Theorem we can derive that sequence $\left(  \Gamma_{n}\right)  $ converges almost everywhere to some integrable measurable function $\Gamma$ and $\int_{\left[  0,\text{ }\infty\right)  }\Gamma d\mu=\lim_{n\rightarrow
\infty}\int_{\left[  0,\text{ }\infty\right)  }\Gamma_{n}d\mu\leq
\lim_{n\rightarrow\infty}\operatorname{d}\left(  \Delta_{n},\Phi_{2}\right)
=0$, meaning that $\lim_{n\rightarrow\infty}\inf_{k\geq n}\Delta_{k}=\Phi_{2}$ almost everywhere. There are two cases to be clarified. First assume that $b_{1}<b_{2}$. Obviously, $\mu\left(  \left(  b_{1},~b_{2}\right)  \right)
>0$, so that there must be at least one point $x_{0}\in\left(  b_{1}%
,~b_{2}\right)  $ such that $\lim_{n\rightarrow\infty}\inf_{k\geq n}\Delta
_{k}\left(  x_{0}\right)  =\Phi_{2}\left(  x_{0}\right)  $. But since $b_{1}<b_{2}$ the concave property implies that the graph of $\Phi_{2}%
$ (resp. the graph of each function $\inf_{k\geq n}\Delta_{k}$) lies above (resp. below) the graph of the line of equation $y=x$ in the interval $\left(
b_{1},~b_{2}\right)  $. Consequently, $\lim_{n\rightarrow\infty}\inf_{k\geq
n}\Delta_{k}\left(  x_{0}\right)  \leq x_{0}<\Phi_{2}\left(  x_{0}\right)  $. This, however, is absurd since $\lim_{n\rightarrow\infty}\inf_{k\geq n}%
\Delta_{k}\left(  x_{0}\right)  =\Phi_{2}\left(  x_{0}\right)  $. Considering the second case when $b_{1}>b_{2}$ we can similarly get into a contradiction. Therefore, the statement is valid.

\[lem4\]Let $b$ and $c\in\left(  0,\,\infty\right)  $ be two arbitrary numbers. Then the following assertions are equivalent:*(1)* The equality $b=c$ holds.*(2)* The sets $\mathcal{Y}_{b}$ and $\mathcal{Y}_{c}$ are equal.*(3)* The equality $\operatorname*{dist}\left(  \mathcal{Y}_{b},\mathcal{Y}_{c}\right)  =0$ holds.

We first note that the chain of implications (1) $\rightarrow$ (2) $\rightarrow$ (3) is obviously true. Thus we need only show the conditional (3) $\rightarrow$ (1). In fact, assume that $\operatorname*{dist}\left(
\mathcal{Y}_{b},\mathcal{Y}_{c}\right)  =0$ but $b\neq c$. Then $\rho\left(
\mathcal{Y}_{b},\Delta\right)  =0$ for all $\Delta\in\mathcal{Y}_{c}$. Nevertheless, this contradicts Lemma \[lem3\], since $b\neq c$. Therefore, the argument is valid.

We can similarly prove that:

\[lem5\]Let $b$ and $c\in\left(  0,\,\infty\right)  $ be two arbitrary numbers. Then the following assertions are equivalent*:(1)* The equality $b=c$ holds.*(2)* The sets $\mathcal{A}%
_{b}$ and $\mathcal{A}_{c}$ are equal.*(3)* The equality $\operatorname*{dist}\left(  \mathcal{A}_{b},\mathcal{A}_{c}\right)  =0$ holds.

\[theo6\]Let $b$ and $c\in\left(  0,\,\infty\right)  $ be two arbitrary numbers. Then the quantities $\operatorname*{dist}\left(  \mathcal{A}%
_{b},\mathcal{A}_{c}\right)  $ and $\operatorname*{dist}\left(
\mathcal{Y}_{b},\mathcal{Y}_{c}\right)  $ define metrics on $\mathcal{Z}$ and $\mathcal{Z}^{\ast}$ respectively. Hence, the couples $\left(  \mathcal{Z}%
,\operatorname*{dist}\right)  $ and $\left(  \mathcal{Z}^{\ast}%
,\operatorname*{dist}\right)  $ are metric spaces.

We need only show that $\operatorname*{dist}\left(  \mathcal{Y}_{b}%
,\mathcal{Y}_{c}\right)  $ is a metric on the set $\mathcal{Z}^{\ast}$, because the other case can be similarly proved. In fact, we first point out that the condition $\operatorname*{dist}\left(  \mathcal{Y}_{b},\mathcal{Y}%
_{c}\right)  \geq0$ is obvious and, by Lemma \[lem4\] the equality holds if and only if $\mathcal{Y}_{b}=\mathcal{Y}_{c}$. We also note that the symmetry property trivially holds true. We are now left with the proof of the triangle inequality. In fact, let $\mathcal{Y}_{b_{j}}\in\mathcal{Z}^{\ast}$ and $\Phi_{j}\in\mathcal{Y}_{b_{j}}$ ($j\in\left\{  1,~2,~3\right\}  $) be arbitrary. Then by Proposition 5 (cf. [@AGB2005], page 15) we have that $\operatorname{d}\left(  \Phi_{1},\Phi_{3}\right)  \leq\operatorname{d}\left(
\Phi_{1},\Phi_{2}\right)  +\operatorname{d}\left(  \Phi_{2},\Phi_{3}\right)
$. Next, by taking the infimum over $\Phi_{3}\in\mathcal{Y}_{b_{3}}$ it follows that $$\rho\left(  \Phi_{1},\mathcal{Y}_{b_{3}}\right)  \leq\operatorname{d}\left(
\Phi_{1},\Phi_{2}\right)  +\rho\left(  \Phi_{2},\mathcal{Y}_{b_{3}}\right)
\leq\operatorname{d}\left(  \Phi_{1},\Phi_{2}\right)  +\operatorname*{dist}%
\left(  \mathcal{Y}_{b_{2}},\mathcal{Y}_{b_{3}}\right)  ,$$ i.e. $\rho\left(  \Phi_{1},\mathcal{Y}_{b_{3}}\right)  \leq\operatorname{d}%
\left(  \Phi_{1},\Phi_{2}\right)  +\operatorname*{dist}\left(  \mathcal{Y}%
_{b_{2}},\mathcal{Y}_{b_{3}}\right)  $. Finally, taking the infimum over $\Phi_{2}\in\mathcal{Y}_{b_{2}}$ yields $\rho\left(  \Phi_{1},\mathcal{Y}%
_{b_{3}}\right)  \leq\rho\left(  \Phi_{1},\mathcal{Y}_{b_{2}}\right)
+\operatorname*{dist}\left(  \mathcal{Y}_{b_{2}},\mathcal{Y}_{b_{3}}\right)
$, so that $$\operatorname*{dist}\left(  \mathcal{Y}_{b_{1}},\mathcal{Y}_{b_{3}}\right)
\leq\operatorname*{dist}\left(  \mathcal{Y}_{b_{1}},\mathcal{Y}_{b_{2}%
}\right)  +\operatorname*{dist}\left(  \mathcal{Y}_{b_{2}},\mathcal{Y}_{b_{3}%
}\right)  .$$ This was to be proven.

By the law of trichotomy it is not hard to see that $\left(  \mathcal{Z}%
,\preceq\right)  $ and $\left(  \mathcal{Z}^{\ast},\preceq\right)  $ are lattices. Here too, the supremum and infimum binary operations on the lattices $\left(  \mathcal{Z},\preceq\right)  $ and $\left(  \mathcal{Z}^{\ast}%
,\preceq\right)  $ will be denoted by the usual symbols $\vee$ and $\wedge$ respectively. We also point out that $\left(  \mathcal{Z},\preceq\right)  $ and $\left(  \mathcal{Z}^{\ast},\preceq\right)  $ are infinite graphs. Between two vertices $\mathcal{A}_{b_{1}}$, $\mathcal{A}_{b_{2}}\in\mathcal{Z}$ we can define the edge in two different ways: one by $e=\operatorname*{dist}\left(
\mathcal{A}_{b_{1}},\mathcal{A}_{b_{2}}\right)  \in\left(  0,~\infty\right)  $ and the other one by $\mathcal{A}_{e}\in\mathcal{Z}$ where $e=\operatorname*{dist}\left(  \mathcal{A}_{b_{1}},\mathcal{A}_{b_{2}}\right)
$. These two edges can apply for the vertices of $\mathcal{Z}^{\ast}$ as well.

Dense subsets in $\mathcal{Y}_{b}^{\left(  n\right)  }$
=======================================================

\[theo7\]Let $b\in\left(  0,\,\infty\right)  $ be an arbitrary number. Then $\mathcal{A}_{b}$ is a dense set in $\mathcal{Y}_{b}$.

Fix arbitrarily any function $\Psi\in\mathcal{Y}_{b}$. Then there is some $\Phi\in\mathcal{Y}_{\mathrm{conc}}$ such that $\Psi=\frac{b\Phi}{\Phi\left(
b\right)  }$ (by Lemma \[lem1\]). Define $\Psi_{n}\left(  x\right)
=\frac{b\left(  \Phi\left(  x\right)  \right)  ^{1-1/(n+1)}}{\left(
\Phi\left(  b\right)  \right)  ^{1-1/(n+1)}}$, for all $x\in\left[
0,\,\infty\right)  $ and $n\in\mathbb{N}$. As we know from Theorem \[theo2\] (cf. [@AGB2005], page 6) function $\Phi^{1-1/(n+1)}\in\mathcal{A}$ for all $\Phi\in\mathcal{Y}_{\mathrm{conc}}$, $n\in\mathbb{N}$. Then $\left(  \Psi
_{n}\right)  \subset\mathcal{A}$ (via Lemma \[lem1\], [@AGB2005], page 5). Hence, $\left(  \Psi_{n}\right)  \subset\mathcal{A}_{b}$, since $\Psi
_{n}\left(  b\right)  =b$ for all $n\in\mathbb{N}$. We can easily show that $\left(  \Psi_{n}\right)  $ converges pointwise to $\Psi$. By Remark \[rem4\] it ensues that $\Psi\left(  x\right)  \leq h_{b}\left(  x\right)  $ and $\Psi_{n}\left(  x\right)  \leq h_{b}\left(  x\right)  $ for all $x\in\left[  0,\,\infty\right)  $ and $n\in\mathbb{N}$, where $h_{b}\left(
x\right)  =x+b$, $x\in\left[  0,\,\infty\right)  $. We know via Remark \[rem3\] that function $h_{b}$ is square integrable. Then by applying twice the Dominated Convergence Theorem one can verify that$$\lim_{n\rightarrow\infty}%
%TCIMACRO{\dint _{\left[  0,\,\infty\right)  }}%
%BeginExpansion
{\displaystyle\int_{\left[  0,\,\infty\right)  }}
%EndExpansion
\Psi_{n}^{2}d\mu=%
%TCIMACRO{\dint _{\left[  0,\,\infty\right)  }}%
%BeginExpansion
{\displaystyle\int_{\left[  0,\,\infty\right)  }}
%EndExpansion
\Psi^{2}d\mu\text{ \ and }\lim_{n\rightarrow\infty}%
%TCIMACRO{\dint _{\left[  0,\,\infty\right)  }}%
%BeginExpansion
{\displaystyle\int_{\left[  0,\,\infty\right)  }}
%EndExpansion
\Psi_{n}\Psi d\mu=%
%TCIMACRO{\dint _{\left[  0,\,\infty\right)  }}%
%BeginExpansion
{\displaystyle\int_{\left[  0,\,\infty\right)  }}
%EndExpansion
\Psi^{2}d\mu,$$ so that $\lim_{n\rightarrow\infty}\operatorname{d}\left(  \Psi,\Psi
_{n}\right)  =0$, because $\Psi\left(  x\right)  \Psi_{n}\left(  x\right)
\leq\left(  h_{b}\left(  x\right)  \right)  ^{2}$ for all $x\in\left[
0,\,\infty\right)  $ and $n\in\mathbb{N}$ (by Remark \[rem4\]). This was to be proven.

\[theo8\]Fix any pair of numbers $n\in\mathbb{N}\backslash\left\{
1\right\}  $ and $b\in\left(  0,~\infty\right)  $. Then $\mathcal{A}%
_{b}^{\left(  n\right)  }$ is dense in $\mathcal{Y}_{b}^{\left(  n\right)  }$.

Pick arbitrarily some $\Delta\in\mathcal{Y}_{b}^{\left(  n\right)  }$. Since obviously $\mathcal{Y}_{b}^{CO\left(  n\right)  }$ is a proper subset of $\mathcal{Y}_{b}^{\left(  n\right)  }$, we will have two cases to take into consideration. First assume that $\Delta\in\mathcal{Y}_{b}^{CO\left(
n\right)  }$. This means that there can be found a counting number $k\geq n$ and a finite sequence $\Phi_{1},~\ldots,~\Phi_{k}\in\mathcal{Y}_{b}^{\left(
1\right)  }=\mathcal{Y}_{b}$ such that $\Delta=\Phi_{1}\circ\ldots\circ
\Phi_{k}$. Fix any integer $j\in\mathbb{N}$ and write $\Delta_{j}=\Psi
_{j}\circ\Delta$, where $\Psi_{j}\left(  x\right)  =\left(  b^{1/j}x\right)
^{j/(j+1)}$, $x\in\left[  0,~\infty\right)  $. Clearly, $\Psi_{j}%
\in\mathcal{A}_{b}^{\left(  1\right)  }$ for all $j\in\mathbb{N}$. Then applying Theorem \[theo2\] in [@AGB2005] and via the structure of set $\mathcal{A}_{b}^{CO\left(  n\right)  }$, we can deduce that $\Delta_{j}%
\in\mathcal{A}_{b}^{CO\left(  n\right)  }$ for all $j\in\mathbb{N}$. It is not difficult to see that sequence $\left(  \Delta_{j}\right)  $ converge pointwise to $\Delta$. By Remark \[rem4\] we observe that $\Delta\leq h_{b}%
$, $\Delta_{j}\leq h_{b}$ and hence, $\Delta\Delta_{j}\leq\left(
h_{b}\right)  ^{2}$ on $\left[  0,~\infty\right)  $. Then recalling twice the Dominated Convergence Theorem we can easily verify that $$\lim_{j\rightarrow\infty}\int_{\left[  0,~\infty\right)  }\left(  \Delta
_{j}\right)  ^{2}d\mu=\int_{\left[  0,~\infty\right)  }\Delta^{2}d\mu
=\lim_{j\rightarrow\infty}\int_{\left[  0,~\infty\right)  }\Delta\Delta
_{j}d\mu.$$ Consequently, $\lim_{j\rightarrow\infty}\operatorname{d}\left(  \Delta
,\Delta_{j}\right)  =0$. In the second case we can suppose that $\Delta
\in\mathcal{Y}_{b}^{\left(  n\right)  }\backslash\mathcal{Y}_{b}^{CO\left(
n\right)  }$. Then without loss of generality we may choose $\Phi_{1}%
,~\ldots,~\Phi_{k}\in\mathcal{Y}_{b}^{CO\left(  n\right)  }$, whose graphs are pairwise distinct, and some finite sequence $\left(  t_{1},~\ldots
~t_{k}\right)  \in H_{\left[  0,~1\right]  }$ with $\left(  t_{1}%
,~\ldots~t_{k}\right)  \subset\left(  0,~1\right)  $ such that $\Delta
=\sum_{i=1}^{k}t_{i}\Phi_{i}$. Consider $\Delta_{j}=\sum_{i=1}^{k}t_{i}\left(
\Psi_{j}\circ\Phi_{i}\right)  $, where $\Psi_{j}\left(  x\right)  =\left(
b^{1/j}x\right)  ^{j/(j+1)}$, $x\in\left[  0,~\infty\right)  $, $j\in
\mathbb{N}$. Clearly, on the one hand we have that $\left(  \Delta_{j}\right)
\subset\mathcal{A}_{b}^{\left(  n\right)  }$ because $\left(  \Psi_{j}%
\circ\Phi_{i}\right)  \subset\mathcal{A}_{b}^{CO\left(  n\right)  }$ for every fixed index $i\in\left\{  1,~\ldots~k\right\}  $ and on the other hand $\lim_{j\rightarrow\infty}\operatorname{d}\left(  \Phi_{i},\Psi_{j}\circ
\Phi_{i}\right)  =0$, $i\in\left\{  1,~\ldots~k\right\}  $, because of the first part of this proof. Consequently, by the Minkowski inequality we can observe that $\lim_{j\rightarrow\infty}\operatorname{d}\left(  \Delta
,\Delta_{j}\right)  \leq\sum_{i=1}^{k}t_{i}\lim_{j\rightarrow\infty
}\operatorname{d}\left(  \Phi_{i},\Psi_{j}\circ\Phi_{i}\right)  =0$. This completes the proof.

Some criterium on the $L^{p}$-norm
==================================

The result here below is worth being mentioned, which is an answer to the second open problem in [@AGB2005].

\[theo9\]Let $\Phi\in\mathcal{Y}_{\mathrm{conc}}$ be arbitrary. Then the following assertions are equivalent.*(1)* $\lim_{t\rightarrow
\infty}\frac{\Phi\left(  t\right)  }{t}=\lim_{t\rightarrow\infty}%
\varphi\left(  t\right)  \in\left(  0,\,\infty\right)  $.*(2)* There is some constant $c\in\left[  1,\,\infty\right)  $ such that $c\Phi
>\Phi_{\operatorname{id}}$ on $\left(  0,\,\infty\right)  $.*(3)* There is some constant $c\in\left[  1,\,\infty\right)  $ and some strictly concave function $\Delta:\left[  0,\,\infty\right)  \rightarrow\left[
0,\,\infty\right)  $, differentiable on $\left(  0,\,\infty\right)  $ and vanishing at the origin such that $c\Phi=\Phi_{\operatorname{id}}+\Delta$ on $\left[  0,\,\infty\right)  $.

We first prove the conditional (1)$\rightarrow$(2). In fact, assume that $\lim_{t\rightarrow\infty}\frac{\Phi\left(  t\right)  }{t}\in\left(
0,\,\infty\right)  $ but in the contrary for every counting number $k\in\mathbb{N}$ there is some $x_{k}\in\left(  0,\,\infty\right)  $ for which $k\Phi\left(  x_{k}\right)  \leq x_{k}$. Obviously, $\limsup
\limits_{k\rightarrow\infty}\frac{\Phi\left(  x_{k}\right)  }{x_{k}}\leq
\lim_{k\rightarrow\infty}k^{-1}=0$ which is absurd since $\limsup
\limits_{k\rightarrow\infty}\frac{\Phi\left(  x_{k}\right)  }{x_{k}}\in\left(
0,\,\infty\right)  $ by the assumption. Next we show the implication (2)$\rightarrow$(3). In fact, assume that there is some constant $c\in\left[
1,\,\infty\right)  $ such that $c\Phi>\Phi_{\operatorname{id}}$ on $\left(
0,\,\infty\right)  $ and write $\Delta:=c\Phi-\Phi_{\operatorname{id}}$. Clearly, $\Delta:\left[  0,\,\infty\right)  \rightarrow\left[  0,\,\infty
\right)  $ is a function such that $\Delta\left(  0\right)  =0$ and $\Delta$ is positive on $\left(  0,\,\infty\right)  $. We also note that $\Delta$ is differentiable on $\left(  0,\,\infty\right)  $. Writing $\delta$ for the derivative of $\Delta$, we can observe that $\delta=c\varphi-1$ on $\left(
0,\,\infty\right)  $. To show that $\Delta$ is strictly concave it is enough if we prove that $$\left(  y-x\right)  \delta\left(  y-0\right)  <\Delta\left(  y\right)
-\Delta\left(  x\right)  <\left(  y-x\right)  \delta\left(  x+0\right)
=\left(  y-x\right)  \delta\left(  x\right)$$ for all $x$, $y\in\left(  0,\,\infty\right)  $ with $x<y$ (where, $\delta\left(  t-0\right)  $ respectively is the left derivative and $\delta\left(  t+0\right)  $ the right derivative of $\Delta$ at point $t$). In fact, fix arbitrarily two numbers $x$, $y\in\left(  0,\,\infty\right)  $ such that $x<y$. But since $\Phi$ is strictly concave we have that$$\left(  y-x\right)  \varphi\left(  y-0\right)  <\Phi\left(  y\right)
-\Phi\left(  x\right)  <\left(  y-x\right)  \varphi\left(  x+0\right)
=\left(  y-x\right)  \varphi\left(  x\right)$$ which easily leads to$$c\varphi\left(  y-0\right)  <\frac{c\Phi\left(  y\right)  -c\Phi\left(
x\right)  }{y-x}<c\varphi\left(  x+0\right)  =c\varphi\left(  x\right)  .$$ Hence,$$c\varphi\left(  y-0\right)  -1<\frac{c\Phi\left(  y\right)  -c\Phi\left(
x\right)  }{y-x}-1<c\varphi\left(  x\right)  -1,$$ i.e. $$\left(  y-x\right)  \delta\left(  y-0\right)  <\Delta\left(  y\right)
-\Delta\left(  x\right)  <\left(  y-x\right)  \delta\left(  x\right)  .$$ This ends the proof of the implication (2)$\rightarrow$(3). In the last step, we just point out that the conditional (3)$\rightarrow$(1) is obvious. Therefore, we can conclude on the validity of the argument.

Denote $\widetilde{\mathcal{Y}_{\mathrm{conc}}}:=\left\{  \Phi\in
\mathcal{Y}_{\mathrm{conc}}:\lim_{t\rightarrow\infty}\frac{\Phi\left(
t\right)  }{t}>0\right\}  $. It is not difficult to check that $\widetilde
{\mathcal{Y}_{\mathrm{conc}}}=\left\{  \Delta\in\mathcal{Y}_{\mathrm{conc}%
}:c\Delta>\Phi_{\operatorname{id}}\text{ on }\left(  0,\,\infty\right)  \text{
for some }c\in\left[  1,\,\infty\right)  \right\}  $. Write $T_{\Delta
}=\left\{  c\in\left[  1,\,\infty\right)  :c\Delta>\Phi_{\operatorname{id}%
}\text{ on }\left(  0,\,\infty\right)  \right\}  $, $\Delta\in\widetilde
{\mathcal{Y}_{\mathrm{conc}}}$.

Some few words about set $\widetilde{\mathcal{Y}_{\mathrm{conc}}}$.

\[rem9\]Let $\alpha\in\left(  0,\,\infty\right)  $ be arbitrary. Then $\alpha\Delta\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}$ provided that $\Delta\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}$.

Whenever $\Delta\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}$ we can choose a corresponding $c\in T_{\Delta}$ such that $c\Delta>\Phi_{\operatorname{id}}$ on $\left(  0,\,\infty\right)  $. Now choose a constant $t_{0}\in\left(
1,\,\infty\right)  $ such that $\alpha t_{0}\geq c$. Hence, $t_{0}\left(
\alpha\Delta\right)  \geq c\Delta>\Phi_{\operatorname{id}}$ on $\left(
0,\,\infty\right)  $, i.e. $\alpha\Delta\in\widetilde{\mathcal{Y}%
_{\mathrm{conc}}}$.

\[rem10\]Every function $\Delta\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}$ can be written as the sum of a finite number of elements of $\widetilde
{\mathcal{Y}_{\mathrm{conc}}}$. Conversely, the sum of a finite number of elements of $\widetilde{\mathcal{Y}_{\mathrm{conc}}}$ also belongs to $\widetilde{\mathcal{Y}_{\mathrm{conc}}}$.

Next, we show that the quantities $\left\Vert f\right\Vert _{L^{p}}$ and $\sup_{\Phi\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}}\left(  \Phi\left(
1\right)  \right)  ^{-1}\left\Vert \Phi\circ\left\vert f\right\vert
\right\Vert _{L^{p}}$ are equivalent, in the sense that they are both either finite or infinite at the same time. This provides a kind of criterium for a measurable function to belong to $L^{p}$.

\[theo10\]Let $f$ be any measurable function on an arbitrarily fixed measure space $\left(  \Omega,\mathcal{F},\lambda\right)  $ and $p\in\left[
1,\infty\right)  $ be any number. Then$$\left\Vert f\right\Vert _{L^{p}}\leq\sup_{\Phi\in\widetilde{\mathcal{Y}%
_{\mathrm{conc}}}}\left(  \Phi\left(  1\right)  \right)  ^{-1}\left\Vert
\Phi\circ\left\vert f\right\vert \right\Vert _{L^{p}}\leq\left\Vert
f\right\Vert _{L^{p}}+\lambda\left(  \Omega\right)  .$$

Pick any function $\Phi\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}$. Then$$\int_{\Omega}\left(  \left(  \Phi\left(  1\right)  \right)  ^{-1}\Phi
\circ\left\vert f\right\vert \right)  ^{p}d\lambda\leq\int_{\Omega}\left(
\left\vert f\right\vert +1\right)  ^{p}d\lambda$$ because $\Delta\leq\left(  \Phi_{\operatorname{id}}+1\right)  \Delta\left(
1\right)  $ for all $\Delta\in\mathcal{Y}_{\mathrm{conc}}$. Consequently, via the Minkowski inequality, it follows that $\left(  \Phi\left(  1\right)
\right)  ^{-1}\left\Vert \Phi\circ\left\vert f\right\vert \right\Vert _{L^{p}%
}\leq\left\Vert f\right\Vert _{L^{p}}+\lambda\left(  \Omega\right)  $, which proves the inequality on the right hand-side of the above chain. To show the left side inequality fix any $\Delta\in\mathcal{Y}_{\mathrm{conc}}$ and write $\Delta_{n}=n^{-1}\Delta$, $n\in\mathbb{N}$. Clearly, $\left(  \Delta
_{n}\right)  \subset\mathcal{Y}_{\mathrm{conc}}$. It is also evident that $\Phi_{\operatorname{id}}+\Delta_{n}\in\widetilde{\mathcal{Y}_{\mathrm{conc}}%
}$, $n\in\mathbb{N}$. Then $$\begin{aligned}
\sup_{\Phi\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}}\left(  \Phi\left(
1\right)  \right)  ^{-1}\left\Vert \Phi\circ\left\vert f\right\vert
\right\Vert _{L^{p}}  &  \geq\left(  1+n^{-1}\right)  ^{-1}\left\Vert \left(
\Phi_{\operatorname{id}}+\Delta_{n}\right)  \circ\left\vert f\right\vert
\right\Vert _{L^{p}}\\
&  =\left(  1+n^{-1}\right)  ^{-1}\left\Vert \left\vert f\right\vert
+n^{-1}\left\vert f\right\vert \right\Vert _{L^{p}}\geq\left(  1+n^{-1}%
\right)  ^{-1}\left\Vert f\right\Vert _{L^{p}}.\end{aligned}$$ Passing to the limit yields $\sup_{\Phi\in\widetilde{\mathcal{Y}%
_{\mathrm{conc}}}}\left(  \Phi\left(  1\right)  \right)  ^{-1}\left\Vert
\Phi\circ\left\vert f\right\vert \right\Vert _{L^{p}}\geq\left\Vert
f\right\Vert _{L^{p}}$. Therefore, we have obtained a valid argument.

\[theo11\]Let $\left(  \Omega,\mathcal{F},\lambda\right)  $ be any measure space and on it let $f$ be any measurable function. Then$$\lambda\left(  \left\vert f\right\vert \geq\varepsilon\right)  =\inf\left\{
\inf\left\{  \lambda\left(  \Delta\circ\left\vert f\right\vert \geq\varepsilon
c^{-1}\right)  :c\in T_{\Delta}\right\}  :\Delta\in\widetilde{\mathcal{Y}%
_{\mathrm{conc}}}\right\}$$ for every number $\varepsilon\in\left[  0,\,\infty\right)  $.

Throughout the proof $\varepsilon\in\left[  0,\,\infty\right)  $ will be any fixed number. We first note that the assertion is trivial when $\left( \left\vert f\right\vert =\infty\right)
\neq\varnothing$. We shall then prove it when $\left(  \left\vert
f\right\vert <\infty\right)  \neq\varnothing$. Pick some $\Delta\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}$ and $c\in
T_{\Delta}$ such that $c\Delta>\Phi_{\operatorname{id}}$ on $\left( 0,\,\infty\right)  $. It is not hard to see that $\left(
\left\vert f\right\vert \geq\varepsilon\right)  =\left(
\Delta\circ\left\vert f\right\vert \geq\Delta\left(
\varepsilon\right)  \right)  \subset\left(
\Delta\circ\left\vert f\right\vert \geq\varepsilon c^{-1}\right)  $ and thus$$\lambda\left(  \left\vert f\right\vert \geq\varepsilon\right)  =\lambda\left(
\Delta\circ\left\vert f\right\vert \geq\Delta\left(  \varepsilon\right)
\right)  \leq\lambda\left(  \Delta\circ\left\vert f\right\vert \geq\varepsilon
c^{-1}\right)  .$$ Consequently, $$\lambda\left(  \left\vert f\right\vert \geq\varepsilon\right)  \leq
\inf\left\{  \inf\left\{  \lambda\left(  \Delta\circ\left\vert f\right\vert
\geq\varepsilon c^{-1}\right)  :c\in T_{\Delta}\right\}  :\Delta\in
\widetilde{\mathcal{Y}_{\mathrm{conc}}}\right\}  .$$ To prove the converse statement, we need show that $$\lambda\left(  \left\vert f\right\vert \geq\varepsilon\right)  \geq
\inf\left\{  \inf\left\{  \lambda\left(  \Delta\circ\left\vert f\right\vert
\geq\varepsilon c^{-1}\right)  :c\in T_{\Delta}\right\}  :\Delta\in
\widetilde{\mathcal{Y}_{\mathrm{conc}}}\right\}  .$$ In fact, for any $n\in\mathbb{N}$ set $\Delta_{n}=\Phi_{\operatorname{id}%
}+n^{-1}\left(  1-e^{-\Phi_{\operatorname{id}}}\right)  $. It is not difficult to see that $\Delta_{n}\in\mathcal{Y}_{\mathrm{conc}}$ and $\Delta_{n}%
>\Phi_{\operatorname{id}}$ on $\left(  0,\,\infty\right)  $, $n\in\mathbb{N}$. This means that $\left(  \Delta_{n}\right)  \subset\widetilde{\mathcal{Y}%
_{\mathrm{conc}}}$ and moreover, $1\in T_{\Delta_{n}}$, $n\in\mathbb{N}$. Consequently, $$\begin{aligned}
\inf\left\{  \inf\left\{  \lambda\left(  \Delta\circ\left\vert f\right\vert
\geq\varepsilon c^{-1}\right)  :c\in T_{\Delta}\right\}  :\Delta\in
\widetilde{\mathcal{Y}_{\mathrm{conc}}}\right\}   &  \leq\lambda\left(
\Delta_{n}\circ\left\vert f\right\vert \geq\varepsilon\right)  \\
&  =\lambda\left(  \left\vert f\right\vert +n^{-1}\left(  1-e^{-\left\vert
f\right\vert }\right)  \geq\varepsilon\right)  \text{.}%\end{aligned}$$ However, as $\left(  \Delta_{n}\right)  $ is a decreasing sequence it is obvious that $\left(  \Delta_{n+1}\circ\left\vert f\right\vert \geq
\varepsilon\right)  \subset\left(  \Delta_{n}\circ\left\vert f\right\vert
\geq\varepsilon\right)  $, $n\in\mathbb{N}$. Thus having passed to the limit we can observe that $$\inf\left\{  \inf\left\{  \lambda\left(  \Delta\circ\left\vert f\right\vert
\geq\varepsilon c^{-1}\right)  :c\in T_{\Delta}\right\}  :\Delta\in
\widetilde{\mathcal{Y}_{\mathrm{conc}}}\right\}  \leq\lambda\left(  \left\vert
f\right\vert \geq\varepsilon\right)  .$$ Therefore, the proof is a valid argument.

\[theo12\]Let $f\in L^{p}\left(  \Omega,\mathcal{F},\lambda\right)  $, $p\geq1$, where $\left(  \Omega,\mathcal{F},\lambda\right)  $ is any given measure space. Then $$\left\Vert f\right\Vert _{L^{p}}=\inf\left\{  \inf\left\{  c\left\Vert
\Delta\circ\left\vert f\right\vert \right\Vert _{L^{p}}:c\in T_{\Delta
}\right\}  :\Delta\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}\right\}  .$$

Pick arbitrarily some $\Delta\in\widetilde{\mathcal{Y}_{\mathrm{conc}}}$ and $c\in T_{\Delta}$ such that $c\Delta>\Phi_{\operatorname{id}}$ on $\left(
0,\,\infty\right)  $. Clearly, $c\left\Vert \Delta\circ\left\vert f\right\vert
\right\Vert _{L^{p}}\geq\left\Vert f\right\Vert _{L^{p}}$. We can then easily observe that $$\inf\left\{  \inf\left\{  c\left\Vert \Delta\circ\left\vert f\right\vert
\right\Vert _{L^{p}}:c\in T_{\Delta}\right\}  :\Delta\in\widetilde
{\mathcal{Y}_{\mathrm{conc}}}\right\}  \geq\left\Vert f\right\Vert _{L^{p}}.$$ To prove the converse of this inequality consider the sequence $\left(
\Delta_{n}\right)  \subset\widetilde{\mathcal{Y}_{\mathrm{conc}}}$, where $\Delta_{n}=\Phi_{\operatorname{id}}+n^{-1}\left(  1-e^{-\Phi
_{\operatorname{id}}}\right)  >\Phi_{\operatorname{id}}$ on $\left(
0,\,\infty\right)  $, $n\in\mathbb{N}$. Then as $1\in T_{\Delta_{n}}$, $n\in\mathbb{N}$, we have$$\inf\left\{  \inf\left\{  c\left\Vert \Delta\circ\left\vert f\right\vert
\right\Vert _{L^{p}}:c\in T_{\Delta}\right\}  :\Delta\in\widetilde
{\mathcal{Y}_{\mathrm{conc}}}\right\}  \leq\left\Vert \Delta_{n}%
\circ\left\vert f\right\vert \right\Vert _{L^{p}}.$$ Since $\left(  \Delta_{n}\right)  $ is a decreasing sequence it ensues that $\left(  \Delta_{n}\circ\left\vert f\right\vert \right)  $ is also a decreasing sequence which tends to $\left\vert f\right\vert $. As every member of sequence $\left(  \Delta_{n}\circ\left\vert f\right\vert \right)  $ is dominated by $\Delta_{1}\circ\left\vert f\right\vert \in L^{p}$, then by applying the Dominated Convergence Theorem it will entail that $$\inf\left\{  \inf\left\{  c\left\Vert \Delta\circ\left\vert f\right\vert
\right\Vert _{L^{p}}:c\in T_{\Delta}\right\}  :\Delta\in\widetilde
{\mathcal{Y}_{\mathrm{conc}}}\right\}  \leq\left\Vert f\right\Vert _{L^{p}}.$$ This completes the proof.

\[cor3\]Suppose that $h:\mathbb{R\rightarrow R}$ is a continuous function. Then $\left\vert h\right\vert =\frac{1}{\sqrt[p]{\lambda\left(  \Omega\right)
}}\inf\left\{  \inf\left\{  \left(  \Delta\circ\left\vert h\right\vert
\right)  c:c\in T_{\Delta}\right\}  :\Delta\in\widetilde{\mathcal{Y}%
_{\mathrm{conc}}}\right\}  $.

Fix any number $x\in\mathbb{R}$ and let $f\in L^{p}\left(  \Omega
,\mathcal{F},\lambda\right)  $ be the constant function defined by $f\equiv
h\left(  x\right)  $ on $\Omega$. Then by applying Theorem \[theo10\] we can easily deduce the result.

Given any number $k\in\mathbb{N}$ characterize all pairs of functions $\Phi$ and $\Delta\in\mathcal{Y}_{\mathrm{conc}}$ such that $\left\vert \left\{
x\in\left(  0,~\infty\right)  :\Phi\left(  x\right)  =\Delta\left(  x\right)
\right\}  \right\vert =k$.

Characterize all pairs of functions $\Phi$ and $\Delta\in\mathcal{Y}%
_{\mathrm{conc}}$ such that the sets $\left(  0,~\infty\right)  $ and $\left\{  x\in\left(  0,~\infty\right)  :\Phi\left(  x\right)  =\Delta\left(
x\right)  \right\}  $ should be equinumerous.

[99]{}

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