text:
\begin{document}
\begin{abstract}
Let $X$ be a nonempty set and let $i,j \in \{1,2,3,4\}$. We say that
a binary operation $F:X^2\to X$ is $(i,j)$-selective if
$$
F(F(x_1,x_2),F(x_3,x_4))~=~F(x_i,x_j),
$$
for all $x_1,x_2,x_3,x_4\in X$. In this paper we provide
characterizations of the class of $(i,j)$-selective operations. We
also investigate some subclasses by adding algebraic properties such
as associativity or bisymmetry.
\end{abstract}

\keywords{$(i,j)$-selectiveness, transitivity, axiomatization,
associativity, bisymmetry.}

\subjclass[2010]{Primary 39B52}

\maketitle

\section{Introduction}

Let $X$ be a nonempty set and let $i,j\in\{1,2,3,4\}$. We say that
an operation $F\colon X^2 \to X$ is \emph{$(i,j)$-selective} if
$$
F(F(x_1,x_2),F(x_3,x_4))~=~F(x_i,x_j),
$$
for all $x_1,x_2,x_3,x_4\in X$. Also, we say that an operation
$F\colon X^2 \to X$ is \emph{biselective} if there exist
$i,j\in\{1,2,3,4\}$ such that $F$ is $(i,j)$-selective. Among these
operations, those which are $(1,3)$-selective are of particular
interest as they are \emph{transitive}, that is, satisfy the
functional equation
$$
F(F(x,z),F(y,z)) ~=~ F(x,y),
$$
for all $x,y,z \in X$ (see, e.g., \cite{Acz2006,Kan2009} and the
references therein). Also, we easily see that $(1,4)$-selective
operations are \emph{bisymmetric}, that is, satisfy the functional
equation
$$
F(F(x,y),F(u,v)) ~=~ F(F(x,u),F(y,v)),
$$
for all $x,y,u,v \in X$ (see, e.g., \cite{Acz2006}).

In this paper we investigate the class of $(i,j)$-selective
operations for every $i,j\in \{1,2,3,4\}$. In particular, we
characterize these operations with and without additional properties
such as associativity or bisymmetry.

The paper is organized as follows. After presenting the main
definitions, we show some basic results about $(i,j)$-selective
operations in Section 2. In particular, we prove that
$(i,j)$-selective operations with $j<i$ are constant (see
Proposition \ref{prop:ji}) as well as (2,3)-selective operations
(see Proposotion \ref{prop:sym2}). We also show that characterizing
the $(i,j)$-selective operations is equivalent to characterizing the
$(5-j,5-i)$-selective operations (see Lemma \ref{lem:leri}). In
Section 3 we characterize the $(1,3)$-selective operations (see
Theorem \ref{thm:utr1}). In Section 4 we characterize the
$(1,4)$-selective operations (see Theorem \ref{thm:chdb}) and in
Section 5 we describe the $(1,2)$-selective operations in
conjunction with additional properties such as associativity.
Finally, in Section 6 we summarize the main results and present some
open questions and directions of further investigations.

\section{Preliminaries}\label{s2}

In this section we introduce some basic definitions and present some
preliminary results.

\begin{definition}\label{de:def1}
An operation $F\colon X^2\to X$ is said to be
\begin{itemize}
\item \emph{idempotent} if $F(x,x)=x$ for all $x\in X$,
\item \emph{quasitrivial} (or \emph{selective}) if $F(x,y)\in\{x,y\}$ for all $x,y \in X$,
\item \emph{commutative} if $F(x,y) = F(y,x)$ for all $x,y \in X$,
\item \emph{anticommutative} if $\forall x,y\in X$: $F(x,y)=F(y,x) \Rightarrow x=y$,

\item \emph{associative} if
$$
F(x,F(y,z)) ~=~ F(F(x,y),z),
$$
for all $x,y,z \in X$,

\end{itemize}
\end{definition}

\begin{definition}
Let $F\colon X^2\to X$ be an operation.
\begin{itemize}
\item An element $e\in X$ is said to be a \emph{neutral element} of $F$ if
$F(e,x) = F(x,e) = x$ for all $x\in X$. It can be easily shown that
such a neutral element is unique.
\item An element $z\in X$ is said to be an \emph{annihilator} of $F$ if
$F(x,z) = F(z,x) = z$ for all $x\in X$. It can be easily shown that
such an annihilator is unique.
\item We denote the range of $F$ by $\ran(F)$. Clearly, $\ran(F)$ is nonempty since $X$ is nonempty.
\item An element $x\in X$ is said to be \emph{idempotent for $F$} if $F(x,x)=x$. We denote the set of all idempotent elements of $F$ by
$\id(F)$. Clearly, $\id(F)\subseteq \ran(F)$.
\end{itemize}
\end{definition}

Recall that a binary relation $R$ on $X$ is said to be
\begin{itemize}
\item \emph{reflexive} if $\forall x\in X$: $xRx$,
\item \emph{symmetric} if $\forall x,y\in X$: $xRy$ implies $yRx$,
\item \emph{transitive} if $\forall x,y,z\in X$: $xRy$ and $yRz$ implies $xRz$.
\end{itemize}

Recall also that an \emph{equivalence relation on $X$} is a binary
relation $\sim$ on $X$ that is reflexive, symmetric, and transitive.
For all $u\in X$, we use the notation $[u]_{\sim}$ to denote the
equivalence class of $u$, that is, $[u]_{\sim}=\{x\in X: x\sim u\}$.

Given $F \colon X^2 \to X$ we define the equivalence relation
$\sim_{F}$ on $X$ by
$$
x \sim_{F} y ~\Leftrightarrow~ F(x,x)~=~F(y,y) \qquad x,y\in X.
$$

\begin{fact}\label{fact:id}
If $F\colon X^2 \to X$ is an $(i,j)$-selective operation, then
$\id(F)\cap [x]_{\sim_F} = \{F(x,x)\}$ for all $x\in X$.
\end{fact}

\begin{proposition}\label{prop:ji}
An operation $F\colon X^2\to X$ is an $(i,j)$-selective operation
with $j<i$ if and only if $F$ is constant.
\end{proposition}

\begin{proof}
(Necessity) First, suppose that $F$ is $(4,1)$-selective (for
$(3,1)$-, $(4,2)$- and $(4,3)$-selective operations the proof is
similar).

By Fact \ref{fact:id}, we have $F(x,x)\in \id(F)$ for all $x\in X$.
If $x,y\in \id(F)$, then $$F(x,y)=F(F(x,x),F(y,y))=F(y,x),$$ by
$(4,1)$-selectiveness. Applying this, we get
$$x=F(x,x)=F(F(x,y),F(y,x))=F(F(y,x),F(x,y))=F(y,y)=y.$$
Thus, $|\id(F)|=1$ and we can assume that $\id(F) = \{u\}$. Hence, by Fact \ref{fact:id}, $F(x,x)=u$ for all $x\in X$.
Using $(4,1)$-selectiveness, we get
$$F(x,y)=F(F(y,y),F(x,x))=F(u,u)=u, \quad x,y\in X.$$

\

Now we suppose that $F$ is $(2,1)$-selective (the case where $F$ is
$(4,3)$-selective can be dealt with similarly). By Fact
\ref{fact:id}, we have $F(x,x)\in \id(F)$ for all $x\in X$. If
$x,y\in \id(F)$, then
$$x=F(x,x)=F(F(x,x),F(y,y))=F(x,y)$$ by $(2,1)$-selectiveness.
Applying that $x=F(x,y)$ and $y=F(y,x)$ for all $x,y\in \id(F)$, we
obtain $$x=F(x,x)=F(F(x,y),F(x,y))=F(y,x)=y.$$ Thus, $|\id(F)|=1$
and we can assume that $\id(F) = \{u\}$. Hence, by Fact
\ref{fact:id}, $F(y,y)=u$ for all $y\in X$. Using also
$(2,1)$-selectiveness of $F$ we get
$$F(y,z)=F(F(z,y),F(z,y))=u, \quad y,z\in X.$$

(Sufficiency) Obvious.
\end{proof}

 In the following two propositions we deal with the case where $F$ is an $(i,i)$-selective operation with $i\in \{1,2,3,4\}$.

\begin{proposition}
An operation $F\colon X^2\to X$ is $(2,2)$-selective (resp. $(3,3)$-selective) if and only if $F|_{\R^2(F)}$ is constant and $F(x,x)\in \id(F)$ for all $x\in X$.
\end{proposition}
\begin{proof}

(Necessity) Suppose that $F$ is $(2,2)$-selective (the case where
$F$ is $(3,3)$-selective can be dealt with similarly). By Fact
\ref{fact:id}, $F(z,z)\in \id(F)$ for all $z\in X$.  If $x,y\in
\id(F)$, then
$$F(x,y)=F(F(x,x),F(y,y))=F(x,x)=x.$$
Using this and $(2,2)$-selectiveness we obtain
$$x=F(x,x)=F(F(x,y),F(x,y))=F(y,y)=y.$$
Thus, we can assume that $\id(F) = \{x\}$ and by Fact \ref{fact:id},
$F(y,y)=x$ for all $y\in X$. Now let us assume that $u,v\in
\ran(F)$. Then there exist $a,b,c,d\in X$ such that $u=F(a,b)$ and
$v=F(c,d)$. Using $(2,2)$-selectiveness we obtain
$$F(u,v)=F(F(a,b),F(c,d))=F(b,b)=x,$$ which proves the statement.

(Sufficiency) Obvious.

\end{proof}

\begin{proposition}\label{prop:11-selective}
An operation $F\colon X^2 \to X$ is $(1,1)$-selective (resp.
$(4,4)$-selective) if and only if the following conditions hold.
\begin{enumerate}
\item[(a)] $F(x,y) = F(x,x)$ (resp. $F(x,y)=F(y,y)$) for all $x,y\in \ran(F)$.
\item[(b)] $F(x,y) \in [x]_{\sim_{F}}$   (resp. $F(x,y)\in [y]_{\sim_{F}}$) for all $x,y\in X$.
\end{enumerate}
\end{proposition}

\begin{proof} Suppose that $F$ is $(1,1)$-selective (the case where $F$ is $(4,4)$-selective can be dealt with similarly).

(Necessity) If $x,y\in \ran(F)$, then there exists $a,b,c,d\in X$
such that $F(x,y)=F(F(a,b),F(c,d))=F(a,a)$ by $(1,1)$-selectiveness.
Also, by $(1,1)$-selectiveness,
$$F(x,x)=F(F(x,y), F(x,y))=F(F(a,a),F(a,a))=F(a,a),$$
which gives that $F(x,y)=F(x,x)$.

Also, by $(1,1)$-selectiveness, we have $F(F(x,y),F(x,y))=F(x,x)$,
which shows that $F(x,y)\in [x]_{\sim_F}$ for all $x,y\in X$.

(Sufficiency) Let $x,y,u,v\in X$. By condition $\textrm{(a)}$, we obtain
$$F(F(x,y),F(u,v)) = F(F(x,y),F(x,y)).$$
Also, by condition $\textrm{(b)}$, we obtain $F(F(x,y),F(x,y))=F(x,x)$.
\end{proof}

\begin{remark}
In Figure \ref{fig:1}, we illustrate the partitioning of $X$ by
$\sim_F$, where $F\colon X^2\to X$ is $(1,1)$-selective.
\end{remark}

\begin{figure}[!ht]
\begin{center}
\begin{tikzpicture}

\draw[fill=black] (0.5,0) circle (0.05); \draw[fill=black] (3.5,0)
circle (0.05);

\node at (0.5,3) {$\vdots$}; \draw[fill=black] (0.5,1) circle
(0.05); \draw[fill=black] (3.5,1) circle (0.05);

\draw[fill=black] (0.5,2) circle (0.05); \draw[fill=black] (3.5,2)
circle (0.05);

\node at (3.5,3) {$\vdots$};

 \draw[blue] (0.5,1.5) ellipse (0.7 and 2);
 \draw[red] (0.5,1) ellipse (0.6 and 1.5);
 \draw[blue] (3.5,1.5) ellipse (0.7 and 2);
 \draw[red] (3.5,0.5) ellipse (0.5 and 1);

 \node at (5,0) {$\dots$};

 \draw[black, thick] (-0.5,-0.5)--(-0.5, 0.5)--(4.5,0.5)--(4.5,-0.5)--cycle;

 \node at (-1,0) {$\id(F)$};
 \node at (0.5,0.25) {$F(x,x)$};

 \node at (3.5,0.25) {$F(y,y)$};

 \node[blue] at (0.7,4) {$[x]_{\sim_F}$};
 \node[red] at (-1,2) {$\ran(F)$};

 \node[blue] at (3.7,4) {$[y]_{\sim_F}$};

 \node[blue] at (5,4) {$\dots$};

 \node[blue] at (5,2) {$\dots$};
 \node[blue] at (5,1) {$\dots$};

\end{tikzpicture}
\caption{}\label{fig:1}
\end{center}
\end{figure}

\begin{definition}\label{def1}

We say that an operation $F\colon X^2\to X$ is \emph{dual
transitive} if
$$
F(F(x,y),F(x,z)) ~=~ F(y,z),
$$
for all $x,y,z \in X$.
\end{definition}

The following lemma shows a strong connection between $(i,j)$-selective and \\ $(5-j,5-i)$-selective operations. The proof is omitted as it is straightforward.

\begin{lemma}\label{lem:leri} An operation $F:X^2\to X$ is $(i,j)$-selective (resp.\ transitive) if and only if the
operation $G:X^2\to X$ defined by $G(x,y)=F(y,x)$ for all $x,y\in X$
is $(5-j,5-i)$-selective (resp.\ dual transitive).
\end{lemma}

By Lemma \ref{lem:leri}, all the results for $(5-j,5-i)$-selective
operations can be deduced from those for $(i,j)$-selective
operations. Therefore we can focus only on $(1,2)$-, $(1,3)$-,
$(1,4)$-, and $(2,3)$-selective operations. Now we prove some useful
lemmas concerning these operations.

Recall that the {\it projection operations} are the binary
operations $\pi_1:X^2\to X$ and $\pi_2:X^2\to X$ defined by
$\pi_1(x,y)=x$ and $\pi_2(x,y)=y$ for all $x,y\in X$.

The following result provides a characterization of the
$(1,2)$-selective operations. Its proof is omitted as it is
straightforward.

\begin{lemma}\label{lem:s2}
An operation $F\colon X^2\to X$ is $(1,2)$-selective
if and only if $F|_{\ran(F)^2}=\pi_1|_{\ran(F)^2}$.
\end{lemma}

\begin{lemma}\label{lem:s1}
Let $F\colon X^2\to X$ be an operation that is $(1,4)$-selective
(resp.\ $(1,3)$-selective, $(2,3)$-selective). For all $x,y\in X$ we
have $F(x,y)=F(y,x)$ if and only if $F(x,x)=F(y,y)$. Moreover, if
there exist $x,y\in X$ such that any of the previous equalities
hold, then $F(x,y)=F(y,x)=F(x,x)=F(y,y)$.
\end{lemma}
\begin{proof}
We consider the case when $F$ is $(1,4)$-selective (the other cases
can be dealt with similarly).

(Necessity) Suppose that $F(x,y)=F(y,x)$, then
$$F(x,y)~=~F(F(x,y), F(x,y))~=~F(F(x,y), F(y,x))~=~F(x,x).$$
Similarly, we have that $F(x,y)=F(y,y)$.

(Sufficiency) Now assume that $F(x,x)=F(y,y)$, then
$$F(x,x)~=~F(F(x,x), F(x,x))~=~F(F(x,x), F(y,y))~=~F(x,y).$$
Similarly, we have that $F(x,x)=F(y,x)$.

The last statement of the lemma is now immediate.
\end{proof}

\begin{theorem}\label{prop:san}
Let $F\colon X^2\to X$ be an $(i,j)$-selective operation. Then the
following assertions are equivalent.
\begin{enumerate}
    \item[(i)] $F$ is commutative.
    \item[(ii)] $|\ran(F)|=1$ .
    \item[(iii)] $F$ has an annihilator.
    \item[(iv)] $|\id(F)|=1$.
\end{enumerate}
Moreover, we have that $F$ has a neutral element if and only if
$|X|=1$.
\end{theorem}

\begin{proof} By Lemma \ref{lem:leri} we only need to prove the result for $(i,j)$-selective operations where $(i,j)\in \{(1,2),(1,3),(1,4),(2,3)\}$. First, suppose that $F$ is $(1,4)$-selective (the cases where $F$ is $(1,3)$-selective or $(2,3)$-selective are similar).

$\textrm{(i)} \Rightarrow \textrm{(ii)}.$ If $F$ is commutative,
then by Lemma \ref{lem:s1}, $F(x,y)=F(x,x)$ for all $x,y\in X$. This
implies that $|\ran(F)|=1$.

$\textrm{(ii)} \Rightarrow \textrm{(iii)}.$ If $|\ran(F)|=1$, then
$F$ has clearly an annihilator.

$\textrm{(iii)} \Rightarrow \textrm{(iv)}.$ Let $a$ be the
annihilator of $F$. We clearly have that $a\in \id(F)$. Also, if
$x\in \id(F)$, then using $(1,4)$-selectiveness and the definition
of an annihilator, we get $ x = F(x,x) = F(F(x,a),F(a,x)) = F(a,a) =
a $ which shows that $|\id(F)|=1$.

$\textrm{(iv)} \Rightarrow \textrm{(i)}.$ If $|\id(F)|=1$, then
using Lemma \ref{lem:s1} we get that $F$ is commutative.

Let us now prove the last part of the statement. If $|X|=1$, then
$F$ has clearly a neutral element. Conversely, if $F$ has a neutral
element $e\in X$, then by Lemma \ref{lem:s1}, $x=F(x,e)=F(e,e)=e$
for all $x\in X$ which clearly implies that $|X|=1$.

\medskip

Now, suppose that $F$ is $(1,2)$-selective.

$\textrm{(i)} \Rightarrow \textrm{(ii)}.$ If $a,b\in R(F)$, then
using commutativity of $F$ and Lemma \ref{lem:s2} we get
$a=F(a,b)=F(b,a)=b$, which shows that $|\ran(F)|=1$.

$\textrm{(ii)} \Rightarrow \textrm{(iii)}.$ If $|\ran(F)|=1$, then
$F$ has clearly an annihilator.

$\textrm{(iii)} \Rightarrow \textrm{(iv)}.$ Let $a$ be the
annihilator of $F$. We clearly have that $a\in \id(F)$. Also, if
$x\in \id(F)$, then using $(1,2)$-selectiveness and the definition
of an annihilator, we get $ x = F(x,x) = F(F(x,x),F(a,a)) = F(x,a) =
a, $ which shows that $|\id(F)|=1$.

$\textrm{(iv)} \Rightarrow \textrm{(i)}.$ If $\id(F)=\{c\}$, then
using $(1,2)$-selectiveness we get
$$
F(x,y) ~=~ F(F(x,y),F(x,y)) ~=~ c ~=~ F(F(y,x),F(y,x)) ~=~ F(y,x),
$$
for all $x,y\in X$.

Let us now prove the last part of the statement. If $|X|=1$, then
$F$ has clearly a neutral element. Conversely, if $F$ has a neutral
element $e\in X$, then for all $x\in X$ we have $F(x,e)=x\in
\ran(F)$. On the other hand, by Lemma \ref{lem:s2}, $F(e,x)=e$ for
all $x\in \ran(F)$ which implies that $|X|=1$.
\end{proof}

As an important consequence of Theorem \ref{prop:san}, we provide
the characterization of $(2,3)$-selective operations.

\begin{proposition}\label{prop:sym2}
An operation $F\colon X^2 \to X$ is $(2,3)$-selective if and only if
$|\ran(F)|=1$.
\end{proposition}

\begin{proof}
(Necessity) We first show that $F(x,y)=F(y,x)$ for all $x,y\in X$.
Using four times $(2,3)$-selectiveness we obtain
\begin{eqnarray*}
  F(x,y) &=& F\left(F(x,x),F(y,y)\right) \\
  &=& F\left(F\left(F(y,x),F(x,y)\right),F\left(F(x,y),F(y,x)\right)\right) \\
  &=& F\left(F(x,y),F(x,y)\right) ~=~ F(y,x),
\end{eqnarray*}
which proves the commutativity of $F$. Now, it follows from Theorem
\ref{prop:san} that $|\ran(F)|=1$.

(Sufficiency) Obvious.
\end{proof}

\section{$(1,3)$-Selectiveness}

In the following result we provide a characterization of
$(1,3)$-selective operations. In the following $\sim_{F}$ denotes
the same equivalence relation as in Section \ref{s2}.

\begin{theorem}\label{thm:utr1}
Let $F:X^2\to X$ be an operation. Then, the following assertions are
equivalent.
\begin{enumerate}
\item[(i)] $F$ is $(1,3)$-selective.
\item[(ii)] $F(x,y)=F(u,v)\in [x]_{\sim_{F}}$ for all $x,y\in X$, $u\in [x]_{\sim_{F}}$, and $v\in [y]_{\sim_{F}}$.
\item[(iii)] $F(F(x,y),z) = F(x,z)$ and $F(x,F(y,z)) = F(x,y)$ for all $x,y,z\in X$.
\end{enumerate}
\end{theorem}

\begin{proof}
$\textrm{(i)} \Rightarrow \textrm{(ii)}.$ Let $x,y\in X$, $u\in
[x]_{\sim_{F}}$ and $v\in [y]_{\sim_{F}}$. Using
$(1,3)$-selectiveness we get
$$F(x,y)~=~F(F(x,x),F(y,y))~=~F(F(u,u),F(v,v))~=~F(u,v).$$
Also, using $(1,3)$-selectiveness, we get $F(F(x,y),F(x,y))=F(x,x),$
which shows that $F(x,y)=F(u,v)\in [x]_{\sim_{F}}$.

$\textrm{(ii)} \Rightarrow \textrm{(iii)}.$ Let $x,y,z\in X$.
Clearly, $x\in [x]_{\sim_{F}}$, $z\in [z]_{\sim_{F}}$, and by (ii)
we get $F(x,y)\in [x]_{\sim_{F}}$ and $F(y,z)\in [y]_{\sim_{F}}$.
Thus, by (ii), we obtain $F(F(x,y),z)=F(x,z)$ and
$F(x,F(y,z))=F(x,y)$.

$\textrm{(iii)} \Rightarrow \textrm{(i)}.$ Let $x,y,u,v\in X$. By
(iii), we obtain
$$F(F(x,y),F(u,v)) ~=~ F(F(x,y),u) ~=~ F(x,u),$$
which concludes the proof.
\end{proof}

\begin{remark}
In Figure \ref{fig:2}, we illustrate the partitioning of $X$ by
$\sim_F$, where $F\colon X^2\to X$ is $(1,3)$-selective.
\end{remark}

\begin{figure}[!ht]
\begin{center}
\begin{tikzpicture}

\draw[fill=black] (0.5,0) circle (0.05); \draw[fill=black] (2,0)
circle (0.05); \draw[fill=black] (3.5,0) circle (0.05);

\node at (0.5,3) {$\vdots$}; \draw[fill=black] (0.5,1) circle
(0.05); \draw[fill=black] (2,1) circle (0.05); \draw[fill=black]
(3.5,1) circle (0.05);

\node at (2,3) {$\vdots$}; \draw[fill=black] (0.5,2) circle (0.05);
\draw[fill=black] (2,2) circle (0.05); \draw[fill=black] (3.5,2)
circle (0.05);

\node at (3.5,3) {$\vdots$};

 \draw[blue] (0.5,1.5) ellipse (0.7 and 2);
 \draw[blue] (2,1.5) ellipse (0.7 and 2);
 \draw[blue] (3.5,1.5) ellipse (0.7 and 2);

 \node at (5,0) {$\dots$};

 \draw[black, thick] (-0.5,-0.5)--(-0.5, 0.5)--(4.5,0.5)--(4.5,-0.5)--cycle;

 \node at (-1,0) {$\id(F)$};
 \node at (0.5,0.25) {$F(x,x)$};
 \node at (2,0.25) {$F(y,y)$};
 \node at (3.5,0.25) {$F(z,z)$};

 \node[blue] at (0.7,4) {$[x]_{\sim_F}$};
 \node[blue] at (2.2,4) {$[y]_{\sim_F}$};
 \node[blue] at (3.7,4) {$[z]_{\sim_F}$};

 \node[blue] at (5,4) {$\dots$};

 \node[blue] at (5,2) {$\dots$};
 \node[blue] at (5,1) {$\dots$};

\end{tikzpicture}
\caption{}\label{fig:2}
\end{center}
\end{figure}

\begin{corollary}\label{cor:utr1}
If $F\colon X^2\to X$ is a $(1,3)$-selective operation, then
$F(x,x)=F(y,z)$ for all $x\in X$ and $y,z\in [x]_{\sim_{F}}$.
\end{corollary}

Now if we assume that $F$ is surjective, then we easily derive the
following characterization from Theorem \ref{thm:utr1}.

\begin{corollary}\label{thm:utr2}
An operation $F\colon X^2 \to X$ is $(1,3)$-selective and surjective
if and only if the following conditions hold.
\begin{enumerate}

    \item[(a)] $F(x,y)=F(u,v)\in [x]_{\sim_{F}},$ for all $x,y\in X$, $u\in [x]_{\sim_{F}}$, and $v\in [y]_{\sim_{F}}$.
    \item[(b)] For every $x\in X$ there exists $a,b\in X$ such that $F(F(a,a),F(b,b))=F(a,b)=x\in [a]_{\sim_{F}}$.
\end{enumerate}
\end{corollary}

\begin{remark}
Using condition \textrm{(a)} of Corollary \ref{thm:utr2}, we can reformulate
condition \textrm{(b)} of Corollary \ref{thm:utr2} in the following way. If
a $(1,3)$-selective operation $F$ is surjective, then for every
$x\in X$ there exists $y\in X$ such that $F(x,y)=x$.
\end{remark}

The following result shows that associativity and bisymmetry are
equivalent under $(1,3)$-selectiveness.
\begin{proposition}\label{prop:asbi}
Let $F:X^2 \to X$ be an $(1,3)$-selective operation. Then the
following assertions are equivalent.
\begin{enumerate}
    \item[(i)] $F$ is associative,
    \item[(ii)] $F(x,y)=F(x,z)$ for all $x,y,z\in X$,
    \item[(iii)] $F$ is bisymmetric,
    \item[(iv)] $F|_{\ran(F)\times X}=\pi_1|_{\ran(F)\times X}$.
\end{enumerate}
 \end{proposition}
\begin{proof}
$\textrm{(i)}\Rightarrow \textrm{(ii)}$. This follows from Theorem
\ref{thm:utr1}.

$\textrm{(ii)}\Rightarrow \textrm{(iii)}$. Let $x,y,u,v\in X$. Using
$(1,3)$-selectiveness and condition $(ii)$, we obtain
$$F(F(x,y),F(u,v))~=~F(x,u)~=~F(x,y)~=~F(F(x,u),F(y,v)),$$
which shows that $F$ is bisymmetric.

$\textrm{(iii)}\Rightarrow \textrm{(iv)}$. Let $x,y,z\in X$. By
$(1,3)$-selectiveness, $[z]_{\sim_{F}}=[F(z,y)]_{\sim_{F}}$ for all
$y\in X$. Using Theorem \ref{thm:utr1}, bisymmetry and
(1,3)-selectiveness, we obtain
$$F(F(x,y),z)~=~F(F(x,y),F(z,y))~=~F(F(x,z),F(y,y))=F(x,y).$$

$\textrm{(iv)}\Rightarrow \textrm{(i)}$. Let $x,y,z\in X$. By
Theorem \ref{thm:utr1} we have $F(x,y), F(x,z)\in [x]_{\sim_{F}}$.
Thus, using Theorem \ref{thm:utr1} and condition $(iv)$, we obtain
\begin{eqnarray*}
F(F(x,y),z) ~=~ F(x,z) ~=~ F(F(x,z),F(y,z)) ~=~ F(x,F(y,z)).
\end{eqnarray*}
\end{proof}

The following result provides characterizations of idempotent
$(1,3)$-selective operations.

\begin{proposition}
Let $F\colon X^2 \to X$ be an $(1,3)$-selective operation. Then the
following assertions are equivalent:
\begin{enumerate}
    \item[(i)] $F$ is quasitrivial.
    \item[(ii)] $F$ is idempotent.
    \item[(iii)] $|[x]_{\sim_{F}}|=1$ for all $x\in X$.
     \item[(iv)] $F=\pi_1$.
    \item[(v)] $F$ is anticommutative.
\end{enumerate}
\end{proposition}
\begin{proof}

$\textrm{(i)}\Rightarrow \textrm{(ii)}$. Obvious.

$\textrm{(ii)}\Rightarrow \textrm{(iii)}$. Obvious.

$\textrm{(iii)}\Rightarrow \textrm{(iv)}$. Let $x,y\in X$ with
$x\neq y$. Since $|[x]_{\sim_{F}}|=|[y]_{\sim_{F}}|=1$, we clearly
have that $x$ (resp. $y$) is the unique element of $[x]_{\sim_{F}}$
(resp. $[y]_{\sim_{F}}$). Also, by Theorem \ref{thm:utr1} we have
$F(x,y)\in [x]_{\sim_{F}}$ and $F(y,x)\in [y]_{\sim_{F}}$ and hence
$F(x,y)=x$ and $F(y,x)=y$.

$\textrm{(iv)}\Rightarrow \textrm{(v)}$. Obvious.

$\textrm{(v)}\Rightarrow \textrm{(i)}$. We proceed by contradiction.
Suppose that there exist $x,y\in X$ such that $F(x,y)\notin
\{x,y\}$. We clearly have $x\in [x]_{\sim_{F}}$ and by Theorem
\ref{thm:utr1} we have $F(x,y)\in [x]_{\sim_{F}}$. Thus, using
Corollary \ref{cor:utr1}, we obtain
$F(F(x,y),x)=F(x,x)=F(x,F(x,y))$, a contradiction.
\end{proof}

As an application of the structural description (see Theorem
\ref{thm:utr1} and Figure \ref{fig:2}) we can get the following
results.
\begin{proposition}\label{prop:fi1}
Let $F$ be a $(1,3)$-selective operation on a finite $X$. Then
$$|\ran(F)|\le |\id(F)|^2.$$
\end{proposition}

\begin{proof}
By Fact \ref{fact:id}, since $F$ is $(1,3)$-selective,  $F(y,y),
F(z,z)\in \id(F)$ for all $y,z\in X$. This clearly implies that the
number of ordered pairs of $\id(F)$ cannot be smaller than
$|\ran(F)|$.
\end{proof}

The set $\id(F)$ can be listed as $$\id(F)=\{x_1, \dots, x_k\}$$ for
some $k\ge 1$. We denote the cardinality of $[x_i]_{\sim_{F}}$
by $l_i$ for all $i \in \{1, \dots, k\}$.

\begin{corollary}\label{cornb}
Let $F$ be a surjective $(1,3)$-selective operation on a finite $X$
of cardinality $n\ge 1$ and $k=|\id(F)|$. Then $n\le k^2$ and $l_i\le k$
for all $1\le i\le k$.
\end{corollary}

\begin{proof}
This follows from Proposition \ref{prop:fi1} and Corollary
\ref{thm:utr2}.
\end{proof}

Let $s_k(n)$ denote the number of $(1,3)$-selective and surjective
operations on an $n$-element set $X$ with $|\id(F)|=k$. By Corollary \ref{cornb}, we have
$s_k(n)=0$ if $n> k^2$. Finding a closed-form expression for the number of
$(1,3)$-selective or surjective $(1,3)$-selective operations seems
hopeless. As an illustration of the characterization given in
Theorem \ref{thm:utr1}, we calculate $s_k(k^2)$.

\begin{proposition}
For all integer $k\ge 1$, we have
$$s_k(k^2)=\frac{k^2!}{((k-1)!)^{k-1}}.$$
\end{proposition}
\begin{proof}
We have $n=k^2$ and $k=|\id(F)|$. Hence by Corollary
\ref{cornb}, $l_i=k$ for all $1\le i\le k$ (see Figure \ref{fig:4}).

\begin{figure}[!ht]
\begin{center}
\begin{tikzpicture}

\draw[fill=black] (0,0) circle (0.05); \draw[fill=black] (1,0)
circle (0.05);

\draw[fill=black] (3,0) circle (0.05);

\node at (0,2) {$\vdots$}; \draw[fill=black] (0,1) circle (0.05);
\draw[fill=black] (1,1) circle (0.05);

\draw[fill=black] (3,1) circle (0.05);

\node at (1,2) {$\vdots$};

\draw[fill=black] (0,3) circle (0.05); \draw[fill=black] (1,3)
circle (0.05);

\draw[fill=black] (3,3) circle (0.05);

\node at (3,2) {$\vdots$};

 \draw[blue] (0,1.5) ellipse (0.3 and 2);
 \draw[blue] (1,1.5) ellipse (0.3 and 2);

 \draw[blue] (3,1.5) ellipse (0.3 and 2);
 \node at (2,0) {$\dots$};

 \draw[black, thick] (-0.5,-0.5)--(-0.5, 0.5)--(3.5,0.5)--(3.5,-0.5)--cycle;

 \node at (-1,0) {$\id(F)$};
 \node at (0,0.25) {$x_1$};
 \node at (1,0.25) {$x_2$};
 \node at (3,0.25) {$x_k$};

 \node[blue] at (0.2,4) {$[x_1]_{\sim_F}$};
 \node[blue] at (1.2,4) {$[x_2]_{\sim_F}$};
 \node[blue] at (3.2,4) {$[x_k]_{\sim_F}$};

 \node[blue] at (2,1) {$\dots$};
 \node[blue] at (2,2) {$\dots$};
 \node[blue] at (2,3) {$\dots$};
 \node[blue] at (2,4) {$\dots$};

 \node at (4,0) {$1$};
 \node at (4,1) {$2$};
 \node at (4,2) {$\vdots$};
 \node at (4,3) {$k$};
\end{tikzpicture}

\caption{} \label{fig:4}
\end{center}
\end{figure}

Let $X$ be a set with $k^2$ elements. First we choose the
equivalence classes for $\sim_F$. All of them have $k$
elements, thus this can be made by the multinomial coefficient
$\frac{k^2!}{(k!)^k}$. Now we choose one element from each class
that is also in $\id(F)$. This can be done in $k^k$ different ways.

By Theorem \ref{thm:utr1} and surjectivity of $F$, for all $x_i\in
\id(F)$ the elements of $[x_i]_{\sim_F}$ are of the form in $F(x_i,
x_j)$ for some $x_j\in \id(F)$. Since $l_i=|\id(F)|$, this implies
that $F(x_i, \cdot )$ is a bijection between $\id(F)$ and
$[x_i]_{\sim_F}$ with a fixed point $x_i\in \id(F)$. Thus for each
$i\in \{1, \dots, k\}$ there are $(k-1)!$ different permutations.
Consequently,
$$s_k(k^2)=\frac{k^2!}{(k!)^k}\cdot k^k\cdot k(k-1)!=
\frac{k^2!\cdot k}{((k-1)!)^{k-1}}.$$
\end{proof}
\begin{remark}
We observe that the number of isomorphism type of $(1,3)$-selective
and surjective operations on a $k^2$-element set $X$ with
$|\id(F)|=k$ is $1$.
\end{remark}

\section{$(1,4)$-Selectiveness}

In this section we characterize the class of operations $F\colon X^2
\to X$ that are $(1,4)$-selective (see Theorem \ref{thm:chdb}).

Clearly, any operation $F\colon X^2 \to X$ that is diagonal
bisymmetric is bisymmetric. The following result provides a
characterization and partial description of the class of
$(1,4)$-selective operations.

\begin{proposition}\label{lem:ass}
Let $F\colon X^2\to X$ be an operation. Then, the following
assertions are equivalent.
\begin{enumerate}
\item[(i)] $F$ is $(1,4)$-selective.
\item[(ii)] $F|_{R(F)^2}$ is $(1,4)$-selective and $F(x,y)=F(F(x,x),F(y,y))$ for all $x,y\in X$.
\item[(iii)] $F(F(x,y),z) ~=~ F(x,F(y,z)) ~=~ F(x,z)$ for all $x,y,z\in X$.
\end{enumerate}
\end{proposition}

\begin{proof}
$\textrm{(i)} \Rightarrow \textrm{(ii)}$. Obvious.

$\textrm{(ii)} \Rightarrow \textrm{(iii)}$. Let $x,y,z \in X$ and
let us prove that $F(F(x,y),z)=F(x,z)$ (the other case can be dealt
with similarly). Using our assumptions, we get
\begin{eqnarray*}
F(F(x,y),z) &=& F(F(F(x,y),F(x,y)),F(z,z)) \\
&=& F(F(F(x,y),F(x,y)),F(F(z,z),F(z,z))) \\
&=& F(F(x,y),F(z,z)) \\
&=& F(F(F(x,x),F(y,y)),F(F(z,z),F(z,z))) \\
&=& F(F(x,x),F(z,z)) ~=~ F(x,z),
\end{eqnarray*}

which concludes the proof.

$\textrm{(iii)} \Rightarrow \textrm{(i)}$. Let $x,y,u,v \in X$. By
assumption, we have $F(F(x,y),F(u,v)) = F(x,F(u,v)) = F(x,v),$ which
shows that $F$ is $(1,4)$-selective.
\end{proof}

\begin{corollary}\label{cor:ass}
If $F\colon X^2\to X$ is $(1,4)$-selective, then it is associative.
\end{corollary}

As another consequence we can prove the following.
\begin{proposition}\label{thm:main}
An operation $F\colon X^2 \to X$ is $(1,4)$-selective and
quasitrivial if and only if $F = \pi_1$ or $F = \pi_2$.
\end{proposition}

\begin{proof}
(Necessity) Let $x,y \in X, x \neq y$, then we have $F(x,y) \in
\{x,y\}$. We can suppose without loss of generality that $F(x,y) =
x$ (the other case can be dealt with similarly). Then, by
Proposition \ref{lem:ass} we have $F(F(x,z),y)=F(x,y)=x$ for all
$z\in X$. Thus, by quasitriviality we have $F(x,z)=x$ for all $z\in
X$.

(Sufficiency) Obvious.
\end{proof}

\begin{remark}\label{rem}
We observe that quasitriviality cannot be relaxed into idempotency
in Theorem \ref{thm:main}. Indeed, let us consider $X = \{a,b,c,d\}$
and the operation $F \colon X^2 \to X$ defined by $F(a,u)=F(c,u)=a$
and $F(b,u)=F(d,u)=d$ for all $u\in\{a,d\}$ and by $F(a,v)=F(c,v)=c$
and $F(b,v)=F(d,v)=b$ for all $v \in \{b,c\}$. It is not difficult
to see that $F$ is idempotent and $(1,4)$-selective, however it is
neither
$\pi_1$ nor $\pi_2$. It is also not hard to show that any idempotent and $(1,4)$-selective operation on $X$, that is not quasitrivial, can be given as the previous operation after a suitable permutation of the elements $a,b,c,d$.
\end{remark}

Now we provide a characterization of $(1,4)$-selective operations
using equivalence relations.

Given $F \colon X^2 \to X$ we define two binary relations
$\sim_{F,1}$ and $\sim_{F,2}$ on $X$ by
$$
x \sim_{F,1} y ~\Leftrightarrow~ F(x,y)=x \qquad x,y\in X,
$$
and
$$
x \sim_{F,2} y ~\Leftrightarrow~ F(x,y)=y \qquad x,y\in X.
$$

Thus, given $F\colon X^2 \to X$ we have that
$F|_{\{x,y\}^2}=\pi_1|_{\{x,y\}^2}$ (resp.\
$F|_{\{x,y\}^2}=\pi_2|_{\{x,y\}^2}$) for all $x,y\in X$ with $x
\sim_{F,1} y$ (resp.\ $x \sim_{F,2} y$).

\begin{remark}
We observe that for all $F\colon X^2 \to X$ the binary relation
$\sim_{F,2}$ on $X$ was already introduced in \cite[Definition
2.4]{Can2012} and was called the \emph{trace} of $F$.
\end{remark}

\begin{lemma}\label{lem:sim}
Let $F\colon X^2\to X$ be a $(1,4)$-selective operation. Then, the
following assertions hold.
\begin{enumerate}
\item[(a)] $\ran(F)=\id(F)$.
\item[(b)] $\sim_{F,1}$ and $\sim_{F,2}$ are transitive binary relations on
$X$.
\end{enumerate}
\end{lemma}

\begin{proof}
$\textrm{(a)}$. If $x\in \id(F)$, then we clearly have that $x\in
\ran(F)$. Conversely, if $x\in \ran(F)$, then there exist $a,b\in X$
such that $x=F(a,b)$. Thus, using $(1,4)$-selectiveness, we get
$F(x,x) = F(F(a,b),F(a,b)) = F(a,b) = x,$ which concludes the proof.

$\textrm{(b)}$. We only prove that $\sim_{F,1}$ is transitive (the
other case can be dealt with similarly). Let $x,y,z\in X$ such that
$x\sim_{F,1}y$ and $y\sim_{F,1}z$, that is, $F(x,y)=x$ and
$F(y,z)=y$. Using diagonal bisymmetry, we get
$$F(x,z)~=~F(F(x,y),F(y,z))~=~F(x,y)~=~x,$$
that is, $x\sim_{F,1}z$.
\end{proof}

\begin{remark}
We observe that Lemma \ref{lem:sim}(ii) is still valid if the
operation $F$ is only associative.
\end{remark}

\begin{proposition}\label{prop:id}
Let $F\colon X^2\to X$ be $(1,4)$-selective. Then, the following
assertions are equivalent.
\begin{enumerate}
\item[(i)] $F$ is anticommutative.
\item[(ii)] $F$ is surjective.
\item[(iii)] $F$ is idempotent.
\item[(iv)] $\sim_{F,1}$ is an equivalence relation on X.
\item[(v)] $\sim_{F,2}$ is an equivalence relation on X.
\end{enumerate}
\end{proposition}

\begin{proof}
$\textrm{(i)} \Rightarrow \textrm{(ii)}.$ Suppose to the contrary
that $F$ is not surjective and hence not idempotent. Thus, there
exists $x\in X$ such that $F(x,x)\neq x$. Using Proposition
\ref{lem:ass}, we obtain $F(F(x,x),x) = F(x,x) = F(x,F(x,x)),$ a
contradiction.

$\textrm{(ii)} \Rightarrow \textrm{(iii)}.$ This follows from Lemma
\ref{lem:sim}\textrm{(a)}.

$\textrm{(iii)} \Rightarrow \textrm{(iv)}.$ $\sim_{F,1}$ is clearly
reflexive since $F$ is idempotent. Also, by Lemma \ref{lem:sim}\textrm{(b)},
we have that $\sim_{F,1}$ is transitive. Now, let us show that
$\sim_{F,1}$ is symmetric. Let $x,y \in X$ such that $x \sim_{F,1}
y$, that is, $F(x,y) = x$. Then, using idempotency and
$(1,4)$-selectiveness, we get
$$F(y,x) ~=~ F(F(y,y),F(x,y)) ~=~ F(y,y) ~=~ y,$$
that is, $y \sim_{F,1} x$.

$\textrm{(iv)} \Rightarrow \textrm{(v)}.$ $\sim_{F,2}$ is clearly
reflexive since $\sim_{F,1}$ is reflexive. Also, by Lemma
\ref{lem:sim}\textrm{(b)} we have that $\sim_{F,2}$ is transitive. Now, let
us show that $\sim_{F,2}$ is symmetric. Let $x,y\in X$ such that
$x\sim_{F,2}y$, that is, $F(x,y)=y$. By Proposition \ref{lem:ass},
we have that $y\sim_{F,2}F(y,x)$ and by transitivity of $\sim_{F,2}$
we get $x\sim_{F,2}F(y,x)$, that is, $F(x,F(y,x))=F(y,x)$. By
Proposition \ref{lem:ass} and symmetry of $\sim_{F,1}$, we also have
that $x\sim_{F,1}F(y,x)$, that is, $F(x,F(y,x))=x$. Hence
$F(y,x)=x$, that is, $y\sim_{F,2}x$.

$\textrm{(v)} \Rightarrow \textrm{(i)}.$ Let $x,y\in X$ such that
$F(x,y) = F(y,x)$. Since $\sim_{F,2}$ is an equivalence relation on
$X$, it follows that $F$ is idempotent. Thus, using
$(1,4)$-selectiveness, we obtain
$$
x~=~F(x,x)~=~F(F(x,y),F(y,x))~=~F(F(y,x),F(x,y))~=~F(y,y)~=~y,
$$
which concludes the proof.
\end{proof}

The following result can be easily derived from Lemma
\ref{lem:sim}\textrm{(a)} and Proposition \ref{prop:id}.

\begin{corollary}\label{cor:main}
If $F\colon X^2 \to X$ is $(1,4)$-selective, then $F|_{\ran(F)^2}$
satisfies any of the conditions $(i)-(v)$ of Proposition
\ref{prop:id}.
\end{corollary}

The following lemma is a consequence of \cite[Lemma 1]{McL54} and
\cite[Lemma 2]{Kim58}.

\begin{lemma}\label{lem:kim}
Let $F\colon X^2 \to X$ be associative and idempotent. Then, the
following assertions are equivalent.
\begin{enumerate}
\item[(i)] $F(F(x,y),z) = F(x,z)$ for all $x,y,z\in X$.
\item[(ii)] $F(F(x,y),x) = x$ for all $x,y\in X$.
\item[(iii)] $F$ is anticommutative.
\end{enumerate}
\end{lemma}

\begin{proposition}
An operation $F\colon X^2 \to X$ is $(1,4)$-selective and satisfies
any of the conditions $(i)-(v)$ of Proposition \ref{prop:id} if and
only if it is associative, idempotent, and satisfies any of the
conditions $(i)-(iii)$ of Lemma \ref{lem:kim}.
\end{proposition}

\begin{proof}
(Necessity) This follows from Corollary \ref{cor:ass} and
Proposition \ref{prop:id}.

(Sufficiency) This follows from Lemma \ref{lem:kim} and Propositions
\ref{lem:ass} and \ref{prop:id}.

\end{proof}

\begin{proposition}\label{prop:cdb}
Let $F\colon X^2 \to X$ be an operation. The following assertions are equivalent.
\begin{enumerate}
\item[(i)] $F$ is $(1,4)$-selective and satisfies
any of the conditions $(i)-(v)$ of Proposition \ref{prop:id}.
\item[(ii)] $\sim_{F,1}$ and $\sim_{F,2}$ are equivalence relations on
$X$ such that $[x]_{\sim_{F,2}}\cap [y]_{\sim_{F,1}} = \{F(x,y)\}$
for all $x,y\in X$.
\item[(iii)] The following conditions hold.
\begin{enumerate}
\item[(a)] $\sim_{F,1}$ and $\sim_{F,2}$ are equivalence relations on $X$.
\item[(b)] For all $x,y,z\in X$ such that $x\in [y]_{\sim_{F,1}}$ there exists a unique $u\in [z]_{\sim_{F,1}}$ such that $x\sim_{F,2} u$.
\item[(c)] $F(x,y) = F(x,z)$ for all $x,y,z\in X$ such that $y\sim_{F,1} z$.
\end{enumerate}
\end{enumerate}
\end{proposition}

\begin{proof}
$\textrm{(i)}\Rightarrow \textrm{(ii)}$: By Proposition \ref{prop:id} we have that $\sim_{F,1}$
and $\sim_{F,2}$ are equivalence relations on $X$. Let $x,y\in X$
and let us show that $[x]_{\sim_{F,2}}\cap [y]_{\sim_{F,1}} =
\{F(x,y)\}$. By Proposition \ref{lem:ass} we have $F(x,y)\in
[x]_{\sim_{F,2}}\cap [y]_{\sim_{F,1}}$. Also, if $z \in
[x]_{\sim_{F,2}}\cap [y]_{\sim_{F,1}}$, then $z \sim_{F,1}\cap
\sim_{F,2} F(x,y)$ which by definition implies that $z=F(x,y)$.

$\textrm{(ii)}\Rightarrow \textrm{(iii)}$: Condition $\textrm{(a)}$ is clearly satisfied.

Let $x,y,z\in X$ such that $x\in [y]_{\sim_{F,1}}$. By $\textrm{(ii)}$, we have $[x]_{\sim_{F,2}}\cap [z]_{\sim_{F,1}} = \{F(x,z)\}$, which proves condition $\textrm{(b)}$.

Let $x,y,z\in X$ such that $y\sim_{F,1} z$, that is, $[y]_{\sim_{F,1}} = [z]_{\sim_{F,1}}$. By $\textrm{(ii)}$ and the previous assumption, we get
$$
\{F(x,y)\} = [x]_{\sim_{F,2}}\cap [y]_{\sim_{F,1}} = [x]_{\sim_{F,2}}\cap [z]_{\sim_{F,1}} = \{F(x,z)\},
$$
which proves condition $\textrm{(c)}$.

$\textrm{(iii)}\Rightarrow \textrm{(i)}$: Since $\sim_{F,1}$ and
$\sim_{F,2}$ are equivalence relations on $X$, it follows that $F$
is idempotent. Let $x,y,u,v\in X$ and let us show that
$F(F(x,y),F(u,v))=F(x,v)$. We clearly have that $t\in
[t]_{\sim_{F,1}}$ for all $t\in X$. By conditions $\textrm{(b)}$ and
$\textrm{(c)}$, we have that there exists a unique $s\in
[y]_{\sim_{F,1}}$ such that $F(x,y)=F(x,s)=s$, that is, $x\sim_{F,2}
s$. Also, by conditions $\textrm{(b)}$ and $\textrm{(c)}$, we have
that there exists a unique $t\in [v]_{\sim_{F,1}}$ such that
$F(u,v)=F(u,t)=t$, that is, $u\sim_{F,2} t$. Moreover, by conditions
$\textrm{(b)}$ and $\textrm{(c)}$, we have that there exists a
unique $z\in [v]_{\sim_{F,1}}$ such that $F(s,t)=F(s,z)=z$, that is,
$s\sim_{F,2} z$. By transitivity of $\sim_{F,2}$ we have that
$x\sim_{F,2} z$ and by condition $\textrm{(b)}$ we have that $z$ is
unique. Thus, we obtain $F(F(x,y),F(u,v))=F(x,v)=F(x,z)=z$ which
concludes the proof.
\end{proof}

\begin{remark}
In Proposition \ref{prop:cdb}\textrm{(iii)}, conditions $\textrm{(b)}$ and
$\textrm{(c)}$ can be replaced by the following two conditions.
\begin{enumerate}
\item[(b')] For all $x,y,z\in X$ such that $x\in [y]_{\sim_{F,2}}$ there exists a unique $u\in [z]_{\sim_{F,2}}$ such that $x\sim_{F,1} u$.
\item[(c')] $F(y,x) = F(z,x)$ for all $x,y,z\in X$ such that $y\sim_{F,2} z$.
\end{enumerate}
\end{remark}

The following corollary is an equivalent form of Proposition
\ref{prop:cdb}.

\begin{corollary}\label{cor:chiddb}
An operation $F\colon X^2 \to X$ is $(1,4)$-selective and satisfies
any of the conditions $(i)-(v)$ of Proposition \ref{prop:id} if and
only if the following conditions hold.
\begin{enumerate}
\item[(i)] $\sim_{F,1}$ and $\sim_{F,2}$ are equivalence relations on $X$ and for all $x,y\in X$ there exists a bijection $f\colon [x]_{\sim_{F,1}} \to [y]_{\sim_{F,1}}$ defined by
    $$f(u)~=~v ~\Leftrightarrow~ u~\sim_{F,2}~v, \qquad u\in [x]_{\sim_{F,1}}, v\in [y]_{\sim_{F,1}}.$$
\item[(ii)] $F(x,y) = F(x,z)$ for all $x,y,z\in X$ such that $y\sim_{F,1} z$.
\end{enumerate}
\end{corollary}

According to Corollary \ref{cor:chiddb}, any $(1,4)$-selective and
idempotent operation $F\colon X^2 \to X$ can be represented in a
grid form as follows. Two elements $x,y\in X$ belong to the same
column (resp.\ row) if and only if $x\sim_{F,1} y$ (resp.\
$x\sim_{F,2} y$). Conversely, any operation $F\colon X^2 \to X$ such
that $\sim_{F,1}$ and $\sim_{F,2}$ are equivalence relations and
that can be represented in such a grid form with the convention that
$F(x,y) = F(x,z)$ for all $x,y,z\in X$ such that $y\sim_{F,1} z$, is
$(1,4)$-selective and idempotent (see Figure \ref{fig:3}).

\begin{figure}[!ht]
\begin{center}
\begin{tikzpicture}

\draw[fill=black] (0,0) circle (0.05); \draw[fill=black] (1,0)
circle (0.05); \draw[fill=black] (2,0) circle (0.05);
\draw[fill=black] (3,0) circle (0.05);

\node at (0,4) {$\vdots$}; \draw[fill=black] (0,1) circle (0.05);
\draw[fill=black] (1,1) circle (0.05); \draw[fill=black] (2,1)
circle (0.05); \draw[fill=black] (3,1) circle (0.05);

\node at (1,4) {$\vdots$}; \draw[fill=black] (0,2) circle (0.05);
\draw[fill=black] (1,2) circle (0.05); \draw[fill=black] (2,2)
circle (0.05); \draw[fill=black] (3,2) circle (0.05);

\node at (2,4) {$\vdots$}; \draw[fill=black] (0,3) circle (0.05);
\draw[fill=black] (1,3) circle (0.05); \draw[fill=black] (2,3)
circle (0.05); \draw[fill=black] (3,3) circle (0.05);

\node at (3,4) {$\vdots$};

 \draw[black] (0,2) ellipse (0.3 and 2.5);
 \draw[black] (1,2) ellipse (0.3 and 2.5);
 \draw (2,2) ellipse (0.3 and 2.5);
 \draw (3,2) ellipse (0.3 and 2.5);
 \node at (4,0) {$\dots$};
 \node at (4,1) {$\dots$};
 \node at (4,2) {$\dots$};
 \node at (4,3) {$\dots$};
 \node at (4,4) {$\udots$};
 \draw[black] (2,0) ellipse (2.5 and 0.3);
 \draw[black] (2,1) ellipse (2.5 and 0.3);
 \draw[black] (2,2) ellipse (2.5 and 0.3);
 \draw[black] (2,3) ellipse (2.5 and 0.3);

 \node at (5.3,0) {$[y]_{\sim_{F,2}}$};
 \node at (5.3,3) {$[x]_{\sim_{F,2}}$};

\draw[->] (-0.9,-0.9)--(-0.05,-0.05); \node at (-1.05,-1.05)
{$F(y,x)$}; \draw[->] (-0.85,3)--(-0.07,3);
 \node at (-1.05,3) {$x$};
 \draw[->] (2,-0.85)--(2,-0.08);
 \node at (2,-1.05) {$y$};
 \draw[->] (-0.85,3)--(-0.08,3);
 \node at (-1.05,3) {$x$};
 \draw[->] (3.7,4.7)--(2.1,3.1);

\node at (4,5) {$F(x,y)$};

 \node at (0.2,5) {$[x]_{\sim_{F,1
 }}$};

 \node at (2.2,5) {$[y]_{\sim_{F,1}}$};

\end{tikzpicture}
\caption{}\label{fig:3}
\end{center}
\end{figure}

The following result, which is an immediate consequence of
Propositions \ref{lem:ass}, \ref{prop:id}, \ref{prop:cdb} and
Corollaries \ref{cor:main} and \ref{cor:chiddb}, provides a characterization of
$(1,4)$-selective operations.

\newpage

\begin{theorem}\label{thm:chdb}
Let $F\colon X^2 \to X$ be an operation and let $F'=F|_{\ran(F)^2}$.
Then, the following assertions are equivalent.
\begin{enumerate}
\item[(i)] $F$ is $(1,4)$-selective

\item[(ii)] $F'$ satisfies the conditions $(i)-(iii)$ of Proposition \ref{prop:cdb} and $F(u,v)=F(F(u,u),F(v,v))$ for all $u,v\in X$.
\item[(iii)] $F'$ satisfies the conditions $(i)$ and $(ii)$ of Corollary \ref{cor:chiddb} and $F(u,v)=F(F(u,u),F(v,v))$ for all $u,v\in X$.
\end{enumerate}
\end{theorem}

For all integer $n\geq 1$, let $X_n=\{1,\ldots,n\}$ and let
$\alpha(n)$ (resp.\ $\beta(n)$) denote the number of
$(1,4)$-selective operations on $X_n$ that are idempotent (resp.\
the number of isomorphism types of $(1,4)$-selective operations on $X_n$ that are
idempotent). In the following propositions we show
that $\alpha(n)=A121860(n)$ and $\beta(n)=d(n)=A000005(n)$ (see
\cite{Oeis}), where $d(n)$ denotes the number of positive integer
divisors of $n\in \mathbb{N}$.

\begin{proposition}
For all integer $n\geq 1$, we have
$$\alpha(n)~=~\sum_{d|n}\frac{n!}{d!\left(\frac{n}{d}\right)!}$$
\end{proposition}

\begin{proof}
By Corollary \ref{cor:chiddb}, counting the number of
$(1,4)$-selective operations on $X_n$ that are idempotent is
equivalent to counting the number of ways to partition $X_n$ into
$k$ equivalence classes of sizes $l,\ldots,l$ and the number of
bijections between two consecutive equivalence classes. Thus, we
have
$$
\alpha(n)~=~ \sum_{k,l \atop kl = n }\frac{{n\choose
l,\ldots,l}}{k!}(l!)^{k-1} ~=~ \sum_{k,l \atop kl = n
}\frac{n!}{k!l!},
$$
where the multinomial coefficient ${n\choose l,\ldots,l}$ provides
the number of ways to put the elements $1,\ldots,n$ into $k$ classes
of sizes $l,\ldots,l$ and $l!$ is the number of bijections between
two such classes.
\end{proof}

\begin{proposition}
For all integer $n\geq 1$, we have $\beta(n)=d(n)$.
\end{proposition}

\begin{proof}
By Corollary \ref{cor:chiddb}, counting the number of isomorphism types of
$(1,4)$-selective operations on $X_n$ that are idempotent is equivalent to counting the number of ways to arrange
the elements of $X_n$ in an unlabeled grid form, where two elements
$x,y\in X_n$ belong to the same column (resp.\ row) if and only if
$x\sim_{F,1} y$ (resp.\ $x\sim_{F,2} y$). Thus, $\beta(n)$ provides
the number of ways to write $n$ into a product of two elements
$k,l\in \{1,\ldots,n\}$. This is in turn the number of divisors of
$n$.
\end{proof}

\begin{corollary}
$\alpha(n)=2$ (resp.\ $\beta(n)=2$) if and only if $n$ is prime.
\end{corollary}

\begin{corollary}\label{cor:pi}
Let $F\colon X_n^2 \to X_n$ be $(1,4)$-selective and idempotent. If
$n$ is prime, then $F=\pi_1$ or $F=\pi_2$.
\end{corollary}

\begin{remark}
From Corollary \ref{cor:pi}, it follows that the example of
$(1,4)$-selective and idempotent operation described in Remark
\ref{rem} is the smallest example that is neither $\pi_1$ nor
$\pi_2$.
\end{remark}

\section{$(1,2)$-Selectiveness}

In Lemma \ref{lem:s2} we already gave a characterization of
$(1,2)$-selective operations. As a corollary we get the following
result if $F$ is surjective.
\begin{corollary}\label{thmpr1}
Let $F:X^2\to X$ be an operation that is $(1,2)$-selective. Then the
following assertions are equivalent.
\begin{enumerate}
\item[(i)] $F=\pi_1$,
\item[(ii)] $F$ is quasitrivial,
\item[(iii)] $F$ is idempotent,
\item[(iv)] $F$ is surjective.
\end{enumerate}
\end{corollary}

\begin{proof}
$\textrm{(i)}\Rightarrow \textrm{(ii)}\Rightarrow
\textrm{(iii)}\Rightarrow \textrm{(iv)}$: Obvious.

$\textrm{(iv)}\Rightarrow \textrm{(i)}$: This follows from Lemma \ref{lem:s2}.

\end{proof}

Now we characterize those operations that are bisymmetric.
\begin{proposition}\label{thmpr3}
A bisymmetric operation $F:X^2\to X$ is $(1,2)$-selective if and
only if the following two conditions hold.
\begin{enumerate}
\item[(i)] $F(x,y)=F(x,z)$ for all $x,y,z\in X$,
\item[(ii)] $F|_{\ran(F)\times X}=\pi_1|_{\ran(F)\times X}$.
\end{enumerate}
\end{proposition}

\begin{proof}
(Necessity) Let $x,y,z\in X$. Using bisymmetry and
$(1,2)$-selectiveness, we get
$$F(x,y)~=~F(F(x,y),F(z,z))~=~F(F(x,z),F(y,z))~=~F(x,z),$$
which proves (i).

Let $x\in \ran(F)$ and $y\in X$. Since $x\in \ran(F)$, there exist
$x_1, x_2\in X$ such that $F(x_1,x_2)=x$. Hence, using (i) and
$(1,2)$-selectiveness, we get
$$F(x,y)~=~F(x,x)~=~F(F(x_1, x_2), F(x_1, x_2))~=~F(x_1,x_2)~=~x,$$
which proves (ii).

(Sufficiency) Condition (ii) clearly implies that $F$ is
$(1,2)$-selective.

Let $x,y,u,v\in X$. Using condition (i) and $(1,2)$-selectiveness,
we get
$$
F(F(x,y),F(u,v)) ~=~ F(x,y) ~=~ F(x,u) ~=~ F(F(x,u),F(y,v)),
$$
which shows that $F$ is bisymmetric.
\end{proof}

In Figure \ref{fig:5} we illustrate a $(1,2)$-selective and
bisymmetric operation on $X$. The vertical lines express that
$F(x,\cdot)=x$ for all $x\in \ran(F)$. The dotted lines express that
the function $F$ is constant along those lines.

\begin{figure}[!ht]
    \centering

    \begin{tikzpicture}
    \draw (0,0) rectangle (4,4);
    \draw (0,0) rectangle (3,3);
    \draw (3,3)-- (3,4);
    \draw[dotted] (3.33,0)-- (3.33,4);
    \draw[dotted] (3.66,0)-- (3.66,4);

    \draw (0.33,0)-- (0.33,4);
    \draw (0.66,0)-- (0.66,4);
    \draw (1,0)-- (1,4);
    \draw (1.33,0)-- (1.33,1.7);
    \draw (1.66,0)-- (1.66,1.7);
    \draw (1.33,2.3)-- (1.33,4);
    \draw (1.66,2.3)-- (1.66,4);

    \draw (2,0)-- (2,4);
    \draw (2.33,0)-- (2.33,4);
    \draw (2.66,0)-- (2.66,4);
    \node at (3.7,-0.3) {$X\backslash\ran(F)$};
    \node at (-0.85,3.5) {$X\backslash\ran(F)$};
    \node at (1.5,2) {$\pi_1$};
    \node at (1.5,-0.3) {$\ran(F)$};
    \node at (-0.6,1.5) {$\ran(F)$};
    \end{tikzpicture}
     \caption{}
    \label{fig:5}
\end{figure}

In the following statement we provide a characterization of those
operations that are associative.
\begin{proposition}\label{prpr1}
An associative operation $F:X^2\to X$ is $(1,2)$-selective if and
only if the following two conditions hold.
\begin{enumerate}
\item[(i)] $F(x,y)=F(x,F(y,z))$ for all $x,y,z\in X$,
\item[(ii)] $F|_{\ran(F)\times X}=\pi_1|_{\ran(F)\times X}$.
\end{enumerate}
\end{proposition}

\begin{proof}
(Necessity) Let $x\in \ran(F)$ and $y\in X$. By Lemma \ref{lem:s2}
we have $F(x,x)=x$. Thus, using associativity and
$(1,2)$-selectiveness, we get
\begin{eqnarray*}
F(x,y) &=& F(F(x,x),y) \\
&=& F(F(F(x,x), x),y)~=~F(F(x,x), F(x,y))~=~F(x,x)~=~x,
\end{eqnarray*}
which proves (ii).

Let $x,y,z\in X$. Since $F(x,y)\in \ran(F)$, using (ii) and the
associativity of $F$, we get $F(x,y)=F(F(x,y),z)=F(x, F(y,z))$,
which proves (i).

(Sufficiency) Condition (ii) clearly implies that $F$ is
$(1,2)$-selective.

Let $x,y,z\in X$. Applying (ii) for $F(x,y)\in \ran(F)$ and (i) we
get $F(F(x,y),z)=F(x,y)=F(x,F(y,z))$, which shows that $F$ is
associative.
\end{proof}

As a consequence of Propositions \ref{thmpr3} and \ref{prpr1} we get
the following.

\begin{corollary}
Any $(1,2)$-selective and bisymmetric operation is associative.
\end{corollary}

Finally we show an example of $(1,2)$-selective and associative
operation that is not bisymmetric on $X=\{a,b,c,d\}$ (see Figure
\ref{fig:6}). The value $F(x,y)$ is represented above the
corresponding point $(x,y)$ in Figure \ref{fig:6} for all $x,y\in
X$. That is $F|_{\R^2_{F}}=\pi_1|_{\R^2_{F}}$ and
$F(d,a)=F(d,b)=F(d,d)=a$ and $F(d,c)=b$. By Lemma \ref{lem:s2}, $F$
is $(1,2)$-selective. It is also clear that $F$ is not bisymmetric
by Proposition \ref{thmpr3}. Using Proposition \ref{thmpr3} it can
be easily shown that $F$ is associative.

\begin{figure}[!ht]
    \centering

    \begin{tikzpicture}
    \draw[fill=black] (0,0) circle (0.05);
    \draw[fill=black] (1,0) circle (0.05);
    \draw[fill=black] (2,0) circle (0.05);
    \draw[fill=black] (3,0) circle (0.05);
    \draw[fill=black] (0,1) circle (0.05);
    \draw[fill=black] (1,1) circle (0.05);
    \draw[fill=black] (2,1) circle (0.05);
    \draw[fill=black] (3,1) circle (0.05);
    \draw[fill=black] (0,2) circle (0.05);
    \draw[fill=black] (1,2) circle (0.05);
    \draw[fill=black] (2,2) circle (0.05);
    \draw[fill=black] (3,2) circle (0.05);
      \draw[fill=black] (0,3) circle (0.05);
    \draw[fill=black] (1,3) circle (0.05);
    \draw[fill=black] (2,3) circle (0.05);
    \draw[fill=black] (3,3) circle (0.05);
       \draw[->] (-1,-0.5) -- (4,-0.5);
       \draw[->] (-0.5,-1) -- (-0.5,4);
    \draw (0,-0.6)--(0, -0.4);
    \draw (1,-0.6)--(1, -0.4);
    \draw (2,-0.6)--(2, -0.4);
    \draw (3,-0.6)--(3, -0.4);
    \draw (-0.6,0)--(-0.4,0);
    \draw (-0.6,1)--(-0.4,1);
    \draw (-0.6,2)--(-0.4,2);
    \draw (-0.6,3)--(-0.4,3);

    \node at (0,-0.9) {$a$};
    \node at (1,-0.9) {$b$};
    \node at (2,-0.9) {$c$};
       \node at (3,-0.9) {$d$};
    \node at (-0.9,0) {$a$};
    \node at (-0.9,1) {$b$};
    \node at (-0.9,2) {$c$};
    \node at (-0.9,3) {$d$};

    \node at (0,0.3 ) {$a$};
    \node at (0,1.3 ) {$a$};
    \node at (0,2.3 ) {$a$};
    \node at (0,3.3 ) {$a$};
     \node at (1,0.3 ) {$b$};
    \node at (1,1.3 ) {$b$};
    \node at (1,2.3 ) {$b$};
    \node at (1,3.3 ) {$b$};
     \node at (2,0.3 ) {$c$};
    \node at (2,1.3 ) {$c$};
    \node at (2,2.3 ) {$c$};
    \node at (2,3.3 ) {$c$};
     \node at (3,0.3 ) {$a$};
    \node at (3,1.3 ) {$a$};
    \node at (3,2.3 ) {$b$};
    \node at (3,3.3 ) {$a$};
    \end{tikzpicture}
     \caption{}
    \label{fig:6}
\end{figure}

\section{Conclusion and further directions}
In this article we introduced and investigated the $(i,j)$-selective
operations. First we showed some basic properties of these
operations. As a consequence we proved that any

$(i,j)$-selective operation with $j<i$ and any $(2,3)$-selective
operation is constant. We also characterized $(i,i)$-selective
operations. We described $(1,3)$-selective operations using the
equivalence relation $\sim_{F}$ and it turned out that it is enough
to understand $\id(F)$ (the set of idempotent elements). We also
proved that $(1,4)$-selective operations are bisymmetric and
associative. We characterized $(1,4)$-selective operations using
equivalence relations $\sim_{F,1}$ and $\sim_{F,2}$. Finally we
described $(1,2)$-selective operations. We studied the relation of
these operations with associativity, bisymmetry and other basic
properties.

In view of these results some questions arise. Now, we list them
below.

\begin{itemize}

    \item
    Let $n\geq 3$ be an integer and let $i_1,\ldots,i_n\in \{1,\ldots,n^2\}$.
    We say that an operation $F \colon X^n\to X$ is {\it $(i_1,\ldots,i_n)$-selective}, if
    $$F(F(x_{1},\ldots, x_{n}), \ldots, F(x_{n(n-1)+1}, \ldots,x_{n^2}))~=~F(x_{i_1}, \ldots, x_{i_n}),$$
    for all $x_{1},\ldots,x_{n^2}\in X$.\\
    Find characterizations of the class of $(i_1,\ldots,i_n)$-selective operations.
    \item Let $i,j \in \{1,2,3,4\}$. We say that the operations $F,G,H,K\colon X^2 \to X$ are {\it generalized $(i,j)$-selective} if
    $$F(G(x_1,x_2), H(x_3,x_4))~=~K(x_i, x_j), \qquad x_1,x_2,x_3,x_4\in X.$$

    Find characterizations of the class of generalized $(i,j)$-selective operations.
    \item Recall that an operation $F\colon X^2 \to X$ is said to be \emph{permutable} \cite{Acz2006,Bus2014} if it satisfies the following functional equation
    $$
    F(F(x,y),z) ~=~ F(F(x,z),y), \qquad x,y,z \in X.
    $$
    We observe that any $(1,2)$-selective operation that is bisymmetric is permutable. Find the conditions under which an $(i,j)$-selective operation is permutable.
\end{itemize}

\section*{Acknowledgements}

The authors would like to thank Jean-Luc Marichal for fruitful
discussions and valuable remarks. This research is supported by the
internal research project R-AGR-0500 of the University of Luxembourg
and by the Luxembourg National Research Fund R-AGR-3080.


text:
\begin{document}

\begin{center}

{\bf FOURIER TRANSFORMS IN EXPONENTIAL  REARRANGEMENT INVARIANT SPACES} \\

\vspace{3mm}

$ {\bf E.Ostrovsky^a, \ \ L.Sirota^b } $ \\

\vspace{2mm}

$ ^a $ Corresponding Author. {\it Department of Mathematics, Ben Gurion University of Negev,
84105, Beer - Sheva, Ben Gurion street, 2, Israel.}\\
\end{center}
E - mail: \ galaostr@cs.bgu.ac.il\\
\begin{center}
$ ^b $ {\it Department of Mathematics. Bar Ilan University, 59200, Ramat Gan, Israel.}\\
\end{center}
E - mail: \ sirota@zahav.net.il\\

\vspace{2mm}

{\bf Abstract}.  In this article we investigate the Fourier series and transforms for
the functions defined on the $ [0, 2 \pi]^ d $ or $ R^d $ and  belonging to  the exponential
Orlicz and some other rearrangement invariant (r.i.) spaces.\\

{\it Key words: }
 Orlicz and rearrangement invariant (r.i.) spaces, Fourier Integrals and Series,
Operators, Moment Inequalities, fundamental Function, equivalent Norms, slowly varying
Functions, Wavelets, Haar's Series.

{\it  Mathematics Subject Classification 2000.} Primary 42Bxx, 4202; Secondary 28A78, 42B08. \\
\section{Introduction. Notations. Problem Statement.}
 For the (complex) measurable function $ f = f(x) $ defined on the $ X = \{x\} = T^d = [0, \ 2 \pi]^d,
\ d = 1,2, \ldots $ or $ X = R^d $ we denote correspondently the Fourier coefficients and
transform
$$
c(n)=\int_{T^d} \exp( i(n,x) ) \ f(x) \ dx, \ \ F[f](t) = \int_{R^d} \exp(i(t,x)) \ f(x) \ dx,
$$
 where as usually
$$
F[f](t) \stackrel{def}{=}  \lim_{M \to \infty}\int_{|x| \le M} \exp(i(t,x)) \ f(x) \ dx,
$$
$$
x = (x_1,x_2,\ldots,x_d), \  n = (n_1,n_2, \ldots, n_d),  \ t = (t_1,t_2, \ldots, t_d), \
dx = \prod_{j=1}^d dx_j
$$
if $ x \in R^d, $ and $ dx = (2 \pi)^{-d} \ \prod_{j=1}^d dx_j $ in the case $ X = T^d; $
$$
n_i = 0, \pm 1, \pm 2, \ldots, \ (t,x) = \sum_{j=1}^d t_j x_j, \ |n| = \max_{j} |n_j|, \
|t| = \max_j |t_j|,
$$
$$
s_M[f](x) = (2 \pi)^{-d}  \sum_{n: |n| \le M} c(n) \exp(-i (n,x)), \ x \in T^d;
$$
$$
 S_M[f](x) = (2 \pi)^{-d} \int_{t: |t| \le M} \exp(-i(t,x)) \ F[f](t) \ dt, \ x \in R^d.
$$
 We will assume in this paper that the function $ f(\cdot) $ belongs to some Orlicz space
$ L(N) = L(N;X) $   with so - called {\it exponential
$ N \ - $ function } $ N = N(u) $ or some rearrangement invariant  space $ E $ with fundamental
function $ \phi_E(\delta), \ \delta > 0, $ and will investigate the properties of Fourier
transform of $ f, $ for example, the boundness of operators $ S_M[\cdot], \ s_M[\cdot] $ and
the convergence and divergence in {\it some} Orlicz norms
 $ s_M[f](\cdot) \to f, \ S_M[f](\cdot) \to f  $ as $ M \to \infty. $  \par
  Note than the case if the function $ N(\cdot) $ satisfies the $ \Delta_2 $ condition is
known; see, for example, [13], [11, p 193 - 197]. Our results are also some
generalization of [4]. The case of Hilbert transform and generalized functions
$ \ \{ f \} \ $ see in [2.] \par
 Now we will describe the using Orlicz and other  spaces. Let $ (X,A,\mu) $ be some
measurable space with non - trivial measure $ \mu. $ For the measurable complex function
$ f(x) = f_1(x) + i f_2(x), x \in X, f_{1,2}: X \to R^1 $ the symbol
$ |f|_p = |f|_p(X,\mu) $ will denote the usually $ L_p $ norm:
$$
|f|_p = ||f||L_p(X, \mu) = \left[ \int_X |f(x)|^p \ \mu(dx) \right]^{1/p}, \ p \ge 1.
$$
In the case $ X = R^d $ we introduce a new measure $ \nu(\cdot) $ (non - finite, in
general case): for all Borel set $ A \subset R^d $
$$
\nu(A) = \int_A \prod_{i=1}^d x_i^{-2} \ dx = \int_A \prod_{i=1}^d x_i^{-2} \ \cdot
\prod_{i=1}^d dx_i,
$$
and will denote  $ |f|_p(\nu) = $
$$
\left| \left| \left(\prod_{i=1}^d x_i \right) \cdot f \right| \right|L_p(X,\nu) =
\left[ \int_X \left|\prod_{j=1}^d x_j \right|^p \ \cdot |f(x)|^p \ \nu(dx)  \right]^{1/p} =
$$
$$
\left[\int_X \left| \prod_{j=1}^d x_j \right|^{p-2} \cdot |f(x)|^p  \ dx    \right]^{1/p}.
$$
 For arbitrary multiply sequence (complex, in general case)
$ c(n) = c(n_1,n_2,\ldots, n_d), \\ n_i = 0, \pm 1, \pm 2, \ldots,  n \in Z^d $ we denote
as usually
$$
|c|_p  =  \left[ \sum_{n} |c(n) |^p \right]^{1/p}, \ p \ge 1;
$$
and introduce the discrete analog of $ |f|_p(\nu) $ norm:
$$
|c|_p(\nu) = \left[ \sum_n \left|c(n) \right|^p \ \cdot  \left(
\left| \prod_{j=1}^d n_j \right|^{p-2} +1 \right) \right]^{1/p}, \ p \ge 2.
$$
 Let $ N = N(u) $ be some
$ N \ - $ Orlicz's function, i.e. downward convex, even, continuous
differentiable for all sufficiently greatest  values $ u, \ u \ge u_0, $
strongly  increasing in the right - side axis,
and such that $ N(u) = 0 \ \Leftrightarrow  u = 0; \ u \to \infty \ \ \Rightarrow
dN(u)/du \ \to \ \infty. $ We say that  $ N(\cdot) $ is an Exponential Orlicz Function,
briefly: $ N(\cdot) \in \ EOF, $ if $ N(u) $ has a view: for some continuous
differentiable strongly increasing downward convex in the domain $ [2,\infty] $
function $ W = W(u) $ such that $ u \to \infty \ \Rightarrow W^/(u) \to \infty $

$$
N(u) = N(W,u) = \exp(W(\log |u|)), \ |u| \ge e^2.
$$
 For the values $ u \in [-e^2,e^2] $ we define $ N(W,u) $ arbitrary  but so that the function
$ N(W,u) $ is even continuous convex strictly increasing in the right side axis
and such that $ N(u) = 0 \
\Leftrightarrow u = 0. $ The correspondent Orlicz space on $ T^d, \ R^d $ with usually
Lebesque measure  with $ N \ - $ function $ N(W,u) $
we will denote $ L(N) = EOS(W); \ EOS = \cup_{W} \{EOS(W)\} $ (Exponential Orlicz's Space).\par
 For example, let $ m = const > 0, \ r = const \in R^1, $
$$
N_{m,r}(u) = \exp \left[|u|^m \ \left(\log^{-mr}(C_1(r) + |u| ) \right) \right] -1,
$$
$ C_1(r) = e, \ r \le 0; \ C_1(r) = \exp(r), \ r > 0. $ Then $ N_{m,r}(\cdot) \in EOS. $
In the case $ r = 0 $ we will write $ N_m = N_{m,0}. $ \par
 Recall here that the Orlicz's norm on the arbitrary measurable space $ (X,A,\mu) $
$ ||f||L(N) = L(N,X,\mu) $ may be calculated by the formula (see, for example,
[11], p. 73; [6], p. 66)
$$
||f||L(N) = \inf_{v > 0}  \left \{ v^{-1} \left(1 + \int_X N( v|f(x)|) \ \mu(dx) \right) \right \}.
$$
 Recall also that the notation $ N_1(\cdot) << N_2(\cdot) $ for two Orlicz functions
$ N_1, N_2 $ denotes:
$$
\forall \lambda > 0 \ \Rightarrow \lim_{u \to \infty} N_1(\lambda u)/N_2(u) = 0.
$$
 We will denote for arbitrary Orlicz $ L(N) $ (and other r.i.) spaces by $ L^0(N) $ the
closure of all bounded functions with bounded support. \par
 Let $ \alpha $ be arbitrary number, $ \alpha \ge 1, $ and $ N(\cdot) \in EOS(W) $ for
some $ W = W(\cdot). $ We denote for such a function $ N = N(W,u) $
by $ N^{(\alpha)}(u) $ a new $ N \ - $ Orlicz's function such that
$$
N^{(\alpha)}(u) = C_1 \ |u|^{\alpha}, \ \ |u| \in [0, C_2];
$$
$$
N^{(\alpha)}(u) = C_3 + C_4 |u|, \ \ |u| \in (C_2, C_5];
$$
$$
N^{(\alpha) }(u) = N(u), \ \ |u| > C_5, \ \ 0 < C_2 < C_5 < \infty,\eqno(1.1)
$$
$$
C_{1,2,3,4,5} = C_{1,2,3,4,5}(\alpha,N(\cdot)).
$$
 In the case $ \alpha = m(j+1), \ m > 0, \ j = 0,1,2, \ldots $ the function $ N^{(\alpha)}_m(u) $
is equivalent to the following Trudinger's function:
$$
N_m^{(\alpha)}(u) \sim N_{[m]}^{(\alpha)}(u) = \exp \left(|u|^m \right) -\sum_{l=0}^{j}
u^{ml}/l!.
$$
 This method is described in [15], p. 42 - 47. Those Orlicz spaces are applied to the theory
of non - linear partial differential equations [15].\par
 We can define formally the spaces $ L(N^{(\alpha)}_m) $ at $ m = + \infty $ as a projective
limit at $ m \to \infty $ the spaces $ L(N^{(\alpha)}_m), $ but it is evident that
$$
L \left(N^{(\alpha)}_{\infty} \right) \sim  L_{\alpha} + L_{\infty},
$$
where the space $ L_{\infty} $ consists on all the a.e. bounded functions with norm
$$
|f|_{\infty} = vraisup_{x \in X} |f(x)|.
$$
 Of course, in the case $ X = T^d $
$$
L_{\alpha} + L_{\infty} \sim L_{\infty}.
$$
 Here and further we will denote by $  C_k = C_k(\cdot), k = 1,2,\ldots $ some positive finite
essentially and by $ C, C_0 $ non - essentially "constructive" constants.
 It is very simple to prove the existence of constants $ C_{1,2,3,4,5} = C_{1,2,3,4,5}(\alpha,
N(\cdot)) $ such that $ N^{(\alpha)} $ is some new exponential $ N \ $
Orlicz's function.  By the symbols $ K_j $ we will denote the "classical" absolute constants.\par
  Now we will introduce some {\it new } Banach spaces. Let $ \alpha = const \ge 1 $ and
$ \psi = \psi(p), \ p \ge \alpha $
be some continuous positive: $ \psi(\alpha) > 0 $  finite strictly increasing function such
that the function $ p \to p \log \psi(p) $ is downward convex and
$$
\lim_{p \to \infty} \psi(p) = \infty.
$$
The set of all those functions we will denote $ \Psi; \ \Psi = \{\psi\}. $ A particular case:

$$
\psi(p) = \psi(W;p) = \exp(W^*(p)/p),
$$
where
$$
W^*(p) = \sup_{z \ge \alpha} (pz - W(z))
$$
is so - called Young - Fenchel, or Legendre transform of $ W(\cdot). $ It follows from
theorem of Fenchel - Moraux that in this case
$$
W(p) = \left[ p \ \log \psi(W;p) \right]^*, \ \ p \ge p_0 = const \ge 2,
$$
and consequently  for all $ \psi(\cdot) \in \Psi $ we introduce the correspondent $ N \ - $
function by equality:
$$
N([\psi]) = N([\psi],u) =
\exp \left \{\left[p \log \psi(p) \right]^*(\log u) \right \}, \ u \ge e^2.
$$
Since $ \ \forall \ \psi(\cdot) \in \Psi, \ d = 0,1,\ldots \ \Rightarrow p^d \cdot \psi(p) \in \Psi, $
we can denote
$$
\psi_d(p) = p^d \cdot \psi(p), \ \  N_d([\psi]) = N_d([\psi],u) = N([\psi_d], u).
$$
For instance, if $ N(u)= \exp(|u|^m), \ u \ge 2, $ where $ \ m = const > 0, $ then
$$
N_d([\psi], u) \sim \exp \left(|u|^{m/(dm+1)} \right), \ u \ge 2.
$$
{\bf Definition 1. } We introduce for arbitrary such a function $ \psi(\cdot) \in \Psi $
the so - called $ G(\alpha;\psi) \ $ and $ G(\alpha; \psi,\nu) $
 norms  and correspondent Banach spaces $ G(\alpha; \psi), \ G(\alpha, \psi, \nu) $ as a
set of all measurable complex functions  with finite norms:

$$
||f||G(\alpha; \psi) = \sup_{p \ge \alpha} (|f|_p/\psi(p)),
$$
and  analogously,

$$
||f||G(\alpha; \psi,\nu) = \sup_{p \ge \alpha} (|f|_p(\nu)/\psi(p)).
$$
 For instance $ \psi(p) $ may be $ \psi(p) = \psi_m(p) = p^{1/m}, \ m = const > 0; $ in
this case we will write $ G(\alpha, \psi_m) = G(\alpha,m) $ and
$$
||f||G(\alpha,m) = \sup_{p \ge \alpha}  \left(|f|_p \ p^{-1/m} \right).
$$
 Also formally  we define
$$
||f||G(\alpha, m) = |f|_{\alpha} + |f|_{\infty}.
$$
{\it Remark 1.} It follows from Iensen inequality that in the case $ X = T^d $ all the spaces
$ G(\alpha_1; \psi), \ G(\alpha_2,\psi), 1 \le \alpha_1 < \alpha_2 < \infty $ are isomorphic:
$$
||f||G(\alpha; \psi) \le ||f||G(1; \psi) \le \max(1, \psi(\alpha)) \ ||f||G(\alpha; \psi)
$$
 It is false in the case $ X = R^d.$ \par
{\it Remark 2. } $ G(\alpha; \psi) $ is a rearrangement invariant (r.i.) space. $ G(\alpha,m) $
has a  fundamental function $ \phi(\delta; G(\alpha,m)), \ \delta > 0, $ where for any
rearrangement invariant space $ G $
$$
\phi(\delta;G) \stackrel{def}{=} ||I_A(\cdot)||G(\cdot), \ \  mes(A) = \delta \in (0,\infty),
$$
$ mes(A) $ denotes usually Lebesque measure of Borel set $ A. $  We have:
$$
 \phi(\delta; G(\alpha,m)) = (e \ m \ |\log \delta|)^{-1/m}, \ \ \delta > \exp(-\alpha/m),
$$
$$
\phi(\delta; G(\alpha,m)) = \alpha^{-1/m} \ \delta^{1/\alpha}, \ \delta \in (0, \exp(-\alpha/m)).
$$
Let us consider also another space
$ G(a,b,\alpha, \beta), \ 1 \le a < b < \infty; \ \alpha,\beta \ge 0. $ Here
$ X = R^d $ and we denote $ h = \min((a+b)/2; 2a). $  We introduce the function
$ \zeta: (a,b) \to R^1_+: $
$$
 \zeta(p)= \zeta(a,b,\alpha,\beta; p) = (p-a)^{\alpha}, \ p \in (a, h);
$$

$$
 \zeta(p) = (b-p)^{\beta}, \ p \in [h, b).
$$
 {\bf Definition 2.} The space $ G(a,b,\alpha,\beta) $ consists
on all the complex measurable functions  with finite norm:
$$
||f||G(a,b,\alpha,\beta) = \sup_{p \in (a,b)} \left[|f|_p \ \cdot \zeta(a,b,\alpha, \beta; p)
\right].
$$
 The space $ G(a,b,\alpha,\beta) $ is also a rearrangement invariant space. After some
calculations we receive the correspondence fundamental function
$ \phi(\delta; G(a,b,\alpha,\beta)). $ Namely, let us denote
 $ \delta_1 =  \exp(\alpha h^2/(h-a) ) $ and define for $ \delta \ge \delta_1 $
$$
p_1 = p_1(\delta) = \frac{\log \delta}{ 2 \alpha} - \left[\frac{\log^2 \delta}{4 \alpha^2} -
\frac{a \log \delta}{\alpha} \right]^{1/2},
$$

$$
\phi_1(\delta) = \delta^{1/p_1} \ (p_1 - a)^{\alpha};
$$
and for $ \delta \in (0, \delta_1) \ \Rightarrow $
$$
\phi_1(\delta) = \delta^{1/h} (h-a)^{\alpha}.
$$
 Further, set $ \delta_2 = \exp(-h^2 \beta/(b-h)), $ and for $ \delta \in (0, \delta_2) $
$ p_2 = p_2(\delta) = $
$$
- \frac{|\log \delta|}{2 \beta} +
\left[ \frac{\log^2 \delta}{4 \beta^2} + b \frac{|\log \delta|}{\beta} \right]^{1/2},
$$

$$
\phi_2(\delta) = \delta^{1/p_2(\delta)} (b-p_2(\delta))^{\beta},
$$
and for $ \delta \ge \delta_2  $ we define
$$
\phi_2(\delta) = \delta^{1/h} (b-h)^{\beta}.
$$
We can write:
$$
\phi(\delta; G(a,b,\alpha,\beta)) = \max \left[ \phi_1(\delta), \phi_2(\delta) \right].
$$
 Note than at $ \delta \to 0+ $
$$
\phi(\delta; G(a,b,\alpha,\beta)) \sim \max \left \{ \delta^{1/h}(h-a)^{\alpha},
(e \beta b^2)^{\beta} \ \delta^{1/\beta}\ |\log \delta|^{-\beta} \right\}
$$
and at $ \delta \to \infty  $
$$
\phi(\delta; G(a,b,\alpha,\beta)) \sim \max \left\{\delta^{1/h} (b-h)^{\beta},
(2a^2 \alpha e^{-2})^{\alpha} \delta^{1/a} \ (\log \delta) ^{-\alpha} \right\}.
$$
 For example, let us consider the function $ f(x) = f(a,b; x), \ x \in R^1 \to R: $
$ f(x) = 0, \ x \le 0; $
$$
f(x) = x^{-1/b}, \ x \in (0,1); \ \ f(x) = x^{-1/a}, \ x \in [1,\infty);
$$
then $ f(a,b,\cdot) \in G(a,b,1,1) $ and
$$
\forall \ \Delta \in (0, 1/2] \ \Rightarrow f \notin G(a,b,1-\Delta,1) \cup G(a,b, 1, 1-\Delta).
$$

 Analogously may be defined the "discrete" $ g(a,b,\alpha,\beta) $ spaces. Namely, let
$ c = c(n) = c(n_1,n_2,\ldots,n_d) $ be arbitrary multiply (complex) sequence. We say that
$ c \in g(a,b,\alpha,\beta) $  if

$$
||c||g(a,b, \alpha, \beta) \stackrel{def}{=} \sup_{p \in (a,b)}
\left[ |c|_p \ (p-a)^{\alpha} \ (b-p)^{\beta} \right].
$$
 It is evident that the non - trivial case of those spaces is only if $ \beta = 0; $
in this case we will write $ g(a,b, \alpha, 0) = g(a, \alpha) $ and
$$
||c||g(\alpha) = \sup_{p > a} |c|_p \ (p-a)^{\alpha}.
$$
 We denote also for $ \psi(\cdot) \in \Psi: \ ||c||g(\psi, \nu) = $
$$
 \sup_{p \ge 2} \left[|c|_p(\nu)/\psi(p) \right], \ \ ||c||_m(\nu) =
 \sup_{p \ge 2} \left[|c|_p(\nu) \ \cdot p^{-1/m} \right], \ \ m = const > 0.
$$

{\bf Our goal is investigating the boundness of Fourier operators and convergence (divergence)
 Fourier series and integrals in exponential Orlicz and $ G(a,b,\alpha, \beta),
\ \ g(\psi, \nu)  $ etc. spaces. } \\

 Note than our  Orlicz $ N \ - $ functions $ N \in EOS $ does not satisfy the so - called
$ \Delta_2 $ condition. \\
\section{ Formulations of main results.}
{\bf Theorem 1.} {\it Let $ X = [0, 2 \pi]^d  $ and  $ \psi \in \Psi. $ Then the Fourier
operators $ s_M[\cdot] $ are uniformly bounded in the space $ L(N[\psi]) $ into
the other Orlicz's  space} $ L(N_d[\psi]): $

$$
\sup_{M \ge 1} ||s_M[f]||L(N_d[\psi]) \le C_6(d,\psi) \ ||f||L(N[\psi]). \eqno(2.1)
$$

{\bf Theorem 2.} {\it Let now $ X = R^d, \ \psi \in \Psi $ and $ \alpha = const > 1. $
The Fourier operators $ S_M[\cdot] $ are uniformly bounded in the space $ L(N^{(\alpha)}[\psi]) $
into the space } $ L(N^{(\alpha)}_d[\psi]): $
$$
\sup_{M \ge 1} ||S_M[f]||L(N^{(\alpha)}_d [\psi]) \le C_7(\alpha,d, \psi) \
||f||L(N^{(\alpha)}[\psi]). \eqno(2.2)
$$
 Since the function $ N[\psi] $ does not satisfies the $ \Delta_2 $ condition, the assertions
(2.1) and (2.2) does not mean that in general case when $ f \in L(N_d^{\alpha}[\psi]) $
$$
\lim_{M \to \infty} ||s_M[f] - f||L(N_d[\psi]) = 0, \eqno(2.3)
$$
$$
\lim_{M \to \infty} ||S_M[f] - f||L(N_d^{(\alpha)}[\psi]) = 0; \eqno(2.4)
$$
see examples further. But it is evident that propositions (2.3) and (2.4) are true  if
correspondently
$$
f \in L^0(N_d[\psi]), \ \ f \in L^0(N_d^{(\alpha)} [\psi]).
$$
Also it is obvious that if $ f \in L(N_d[\psi]), X = [0, 2 \pi]^d $ or, in the case $ X = R^d,
\ f \in L(N_d^{(\alpha)}[\psi]), $ then for all EOF $ \Phi(\cdot) $  such that $ \Phi <<
N_d[\psi] $ or $ \Phi << N_d^{(\alpha)}([\psi]) $  the following impications hold:
 $$
 \forall f \in L(N_d[\psi]) \ \Rightarrow \lim_{M \to \infty} ||s_M[f] - f||L(\Phi) = 0, \ X =
[0,2\pi]^d; \eqno(2.5)
 $$

$$
\forall f \in L(N_d[\psi]) \ \Rightarrow \lim_{M \to \infty} ||S_M[f]-f||L(\Phi) = 0, \ X = R^d.
\eqno(2.6)
$$
{\bf Theorem 3.} {\it Let $ \Phi(\cdot) $ be an EOF and let
$ N(\cdot) =  L^{-1}(u) \in EOF, $ where $ L(y), \  y \ge \exp(2) $ is a positive slowly
varying at $ u \to \infty $ strongly increasing continuous differentiable
in the domain $ [\exp(2), \infty) $ function such that the function
$$
W(x) = W_L(x) = \log L^{-1}(\exp x), \ \ x \in [2, \infty)
$$
is again strong increasing to infty together with the derivative $ dW/dx. $
 In order to the impication (2.5) or,
correspondently, (2.6) holds, it is necessary and sufficient that $ \Phi << L(N_d[\psi]), $
or, correspondently } $ \Phi << L(N_d^{(\alpha)}[\psi]). $ \par
 For instance, the conditions of theorem 3 are satisfied for the functions $ N = N_{m,r}(u). $ \par
{\bf Theorem 4. } {\it Let $ f(\cdot) \in G(1,b, \alpha, 0), \alpha > 0. $
Then $ F[f] \in L(N^{(2)}_{1/\alpha}) $ and }
$$
\sup_{M \ge 1} ||S_M[f]||L \left(N^{(2)}_{1/\alpha} \right) \le C_8(\alpha, N) \
||f||G(1,b, \alpha, 0).  \eqno(2.7)
$$
{\bf Theorem 5. } {\it Let $ \{\phi_k(x), k = 1,2, \ldots \} $ be an orthonormal uniform
boumded: $ \sup_{k,x}|\phi_k(x)| < \infty $ sequence of functions
on some non - trivial measurable space $ (X, A, \mu) $ and  (in the $ L_2(X,\mu) $ sense) }
$$
f(x) = \sum_{k=1}^{\infty} c(k) \ \varphi_k(x).
$$
{\bf A).} {\it \ If $ c \in g(\psi, \nu), $ then }
$$
||f||L \left(N_1[\psi], X, \mu \right) \le C_9 \cdot( \max(1, \sup_{k,x} |\phi_k(x)|))
\cdot  ||c||g(\psi,\nu), \eqno(2.8)
$$
{\bf B.)} {\it Let $ c \in g(\alpha)  $ for some $ \alpha \in (0,1]. $ We assert that }
$$
||f|| L \left(N_{1/\alpha}, X, \mu \right) \le C_{10}(\alpha) \cdot (\max(1, \sup_{k,x} |\phi_k(x)|)
\cdot  \ ||c(\cdot)||g(\alpha).
$$
{\bf Theorem 6.} {\it If $ f \in G(\alpha; \psi,\nu), \ $ where $ \alpha \ge 2, $  then }
$$
  \sup_{M \ge 1} ||S_M[f]||L \left(N^{(\alpha)}_d [\psi] \right) \le C_{11}(\alpha,\psi, N, \nu) \
||f||G(\alpha; \psi,\nu). \eqno(2.9)
$$

\section{ Auxiliary results}

{\bf Theorem 7.} {\it Let  $ N(u) = N(W,u) = \exp(W(\log u)), \ u > e^2,  \
\psi(p) = \exp(W^*(p)/p), \ p \ge 2, $ and $ X = T^d. $
We propose that the Orlicz's norm $ ||\cdot|| L(N) $ and the  norm
$ ||\cdot||G(\psi) $ are equivalent.  Moreover, in this case  $ f \ne 0, \ f \in G(\psi) $
(or $ f \in L(N(W(\cdot),u) \ ) $ if and only if}  $ \ \exists C_{12},C_{13}, C_{14} \in
(0,\infty) \ \Rightarrow \forall \ u > C_{14} $
$$
T(|f|,u) \le C_{12} \exp \left(-W \left(\log \left(u/C_{13} \right) \right) \right), \eqno(3.1)
$$
{\it where for each measurable function } $ f: X \to R $
$$
T(|f|,u) = mes \{x: |f(x)| > u \}.
$$

{\bf Proof } of theorem 7. \ A). Assume at first  that $ f \in L(N), \ f \ne 0.$
Without loss of generality we suppose that $ ||f||L(N) = 1/2. $ Then
$$
\int_X N(W, |f(x)|) \ dx \le 1 < \infty.
$$
 The proposition (3.1) follows from Chebyshev's inequality such that in (3.1)
$ C_{12} = 1, \ C_{13} = C_{14} = 1/||f||L(N), \ f \ne 0. $ \par
B). Inversely, assume that $ f, \ f \ne 0 $ is a measurable function, $ f: X \to R^1 $ such
that
$$
T(|f|,u) \le \exp(-W(\log u)), \ u \ge e^2.
$$
 We have by virtue of properties of the function $ W: $
$$
\int_X N(|f(x)|/e^2) \ dx = \int_{ \{x: |f(x) \le e^2\} } + \int_{ \{x: |f(x)| > e^2} =
I_1 + I_2;
$$
$$
I_1 \le \int_X N(1) \ dx = N(1),
$$
$$
I_2 \le \sum_{k=2}^{\infty} \int_{e^k < |f| \le e^{k+1} }
\exp(W(|f(x)|/e^2)) \ dx \le
$$
$$
 \sum_{k=2}^{\infty} \exp((W(k-1)) \ T(|f|,k) \le
  \sum_{k=2}^{\infty} \exp(W(k-1) - W(k)) < \infty.
$$
 Thus, $ f \in L(N(W)) $ and
$$
\int_X N(|f(x)|/e^2) \ dx \le N(1) + \sum_{k=2}^{\infty} \exp(W(k-1) - W(k)) < \infty.
$$
C). Let now $ f \in G(\psi); $ without loss of generality we can assume that $ ||f||G(\psi)
=1. $ We deduce for $ p \ge 2:$
$$
\int_X |f(x)|^p \ dx \le \psi^p(p).
$$
We obtain using again the Chebyshev's  inequality:
$$
T(|f|,u) \le u^{-p} \psi^p(p) = \exp \left[-p \log u + p \log \psi(p) \right],
$$
and  after the minimization over $  p: \ u \ge \exp(2) \ \Rightarrow  $
$$
T(|f|,u) \le \exp \left( - \sup_{p \ge 2} (p \log u - p \log \psi(p)) \right) =
$$
$$
\exp \left( (p \log \psi(p))^*(\log u)  \right) = \exp(-W(\log u)).
$$
D). Suppose now that $ T(|f|,x) \le \exp(-W(\log x)), \ x \ge \exp(2). $ We conclude:
$$
\int_X |f(x)|^p \ dx = p \int_o^{\infty} x^{p-1} T(|f|,x) dx = p \int_0^{\exp(2)}
x^{p-1} T(|f|,x) dx +
$$
$$
p \int_{\exp(2)}^{\infty} x^{p-1} T(|f|,x) dx \le p \int_0^{\exp(2)} x^{p-1} dx + p
\int_{\exp(2)}^{\infty} x^{p-1} \ T(|f|,x) dx \le
$$

$$
 e^{2p} + \int_{\exp(2)}^{\infty} p x^{p-1} \exp(-W(\log x)) \ dx =
$$
$$
e^{2 p} + p \int_2^{\infty} \exp(py - W(y)) \ dy, \ p \ge 2.
$$
We obtain using Laplace's method and  theorem of Fenchel - Moraux ([8], p. 23 - 25):
$$
\int_X |f(x)|^p dx \le e^{2p} + C^p \exp \left(\sup_{y \ge 2} (py - W(y)) \right) = e^{2p} +
$$

$$
C^p \exp(W^*(p)) = e^{2p} + C^p \exp(p \log \psi(p)) \le C^p \ \psi^p(p).
$$
 Finally, $ ||f||G(\psi) < \infty. $ \par
For example, if $ m > 0, \ r \in R, $ then
$$
 f \in L \left(N_{m,r} \right) \Leftrightarrow  \sup_{p \ge 2} \left[|f|_p \ p^{-1/m} \
\log^{-r} p \right] < \infty \ \Leftrightarrow
$$
$$
T(|f|,u)  \le C_0(m,r) \exp \left( - C(m,r) u^m \left( \log^{-mr} u \right) \right), \ u \ge 2.
$$

{\it Remark 3.} If conversely
$$
T(|f|,x) \ge \exp(-W(\log x)), \ x \ge e^2,
$$
then for sufficiently large values of $ p; \ p \ge p_0 = p_0(W) \ge 2  $
$$
|f|_p \ge C_0(W) \ \psi(p), \ \  C_0(W) \in (0, \infty).\eqno(3.2)
$$

{\it Remark 4.} In this proof we used only the condition $ 0 < mes(X) < \infty. $ Therefore,
our  conclusions in theorem 7 are true in this more general case.\par

{\bf Theorem 8.} {\it Let $ \psi \in \Psi.$ We assert that $ f \in L^0(N[\psi]), $ or,
equally, $ f \in G^0(\psi) $ if and only if }

$$
\lim_{p \to \infty} |f|_p /\psi(p) = 0. \eqno(3.3)
$$
{\bf Proof.} It is sufficient by virtue of theorem 7 to consider only the case of
$ G(\psi) $ spaces.\par
 1. Denote $ G^{00}(\psi) = \{ f: \ \lim_{p \to \infty} |f|/\psi(p) = 0 \}. \ $ Let
$ f \in G^0(\psi), \  f \ne 0. $ Then for arbitrary $ \delta = const > 0 $ there exists a
constant $ B = B(\delta,f(\cdot)) \in (0, \infty) $ such that

$$
||f-f I(|f| \le B) \ ||G(\psi) \le \delta/2.
$$
Since  $ |f| I(|f| \le B)| \le B, $ we deduce
$$
|f I(|f| \le B)|_p/\psi(p) \le B/\psi(p).
$$
We obtain using triangular inequality  for sufficiently large values $ p: \ p \ge p_0(\delta)=
p_0(\delta,B) \ \Rightarrow $
$$
|f|_p /\psi(p) \le \delta/2 + B/\psi(p) \le \delta,
$$
as long as $ \psi(p) \to \infty $ at $ p \to \infty. $ Therefore $ G^0(\psi)
\subset G^{00}(\psi). $\par
(The set $ G^{00}(\psi) $ is a closed subspace of $ G(\psi) $ with respect to the $ G(\psi) $
norm and contains all bounded functions.) \par
2. Inversely, assume that $ f \in G^{00}(\psi). $ We deduce denoting $ f_B = f_B(x) =
f(x) I(|f|> B) $ for some $ B = const \in (0, \infty):$

$$
\forall Q \ge 2 \ \Rightarrow \lim_{B \to \infty} |f_B|_Q = 0.
$$
 Further,
$$
||f_B||G(\psi) = \sup_{p \ge 2} |f_B|_p/\psi(p) \le \max_{p \le Q} |f_B|_p/\psi(p) +
$$

$$
\sup_{p > Q} |f_B|_p/\psi(p) \stackrel{def}{=} \sigma_1 + \sigma_2;
$$

$$
\sigma_2 = \sup_{p > Q} |f_B|_P/\psi(p) \le \sup_{p \ge Q} (|f|_p/\psi(p)) \le \delta/2
$$
for sufficiently large $ Q $ as long as $ f \in G^{00}(\psi). $  Let us now estimate the
value $ \sigma_1: $
$$
\sigma_1 \le \max_{p \le Q} |f_B|_p/\psi(2) \le \delta/2
$$
for sufficiently large $ B = B(Q). $ Therefore,

$$
\lim_{B \to \infty} || f_B||G(\psi) = 0, \ \  f \in G^0(\psi).
$$
{\bf Theorem 9}. {\it Let $ \psi(\cdot) = \psi_N(\cdot), \theta(\cdot) = \theta_{\Phi}(\cdot) \ $
be a two functions on the classes $ \Psi $ with correspondent $ N \ - $ Orlicz's functions
$ N(\cdot), \Phi(\cdot): $

$$
N(u) = \exp \left \{\left[p \log \psi(p) \right]^*(\log u) \right\},
$$

$$
\Phi(u) = \exp \left \{ \left[ p \log \theta(p) \right]^*(\log u \right \}, \ \ u \ge \exp(2).
$$
We assert that  $ \lim_{p \to \infty} \psi(p)/\theta(p) = 0 $ if and only if $ N(\cdot) >>
\Phi(\cdot).$ } \\
{\bf Proof } of theorem 9. A). Assume at first that $ \lim_{p \to \infty} \psi(p)/\theta(p) = 0. $
Denote $ \epsilon(p) = \psi(p)/\theta(p), $ then $ \epsilon(p) \to 0, \ p \to \infty. $ \par
 Let $ \{ f_{\zeta}, \ \zeta \in Z \} $ be arbitrary bounded in the $ G(\psi) $ sense set of
functions:
$$
\sup_{\zeta \in Z} ||f_{\zeta}||G(\psi) = \sup_{\zeta \in Z} \sup_{p \ge 2} |f_{\zeta}|_p/\psi(p)
= C < \infty,
$$
then
$$
\sup_{\zeta \in Z} |f_{\zeta}|_p/\theta(p) \le C \epsilon(p) \to 0, \ p \to \infty.
$$
 It follows from previous theorem that $ \forall \zeta \in Z \ f_{\zeta} \in G^0(\theta) $ and
that the family $ \{f_{\zeta}, \ \zeta \in Z \} $ has uniform absolute continuous norm. Our
assertion follows from lemma 13.3 in the book [6].\par
B). Inverse, let $ \Phi(\cdot) << N(\cdot). $ Let us introduce the measurable function
$ f: X \to R $ such that $ \ \forall x \ge \exp(2) $

$$
\exp \left(-2 \left[p \log \psi(p) \right]^*(\log x) \right) \le T(|f|,x) \le
$$

$$
\exp \left( - \left[p \log \psi(p) \right]^*(\log x) \right).
$$
Then (see theorem 7)
$$
f(\cdot) \in G(\psi), \ \ \ C_{15}(\psi) \ \psi(p) \le |f|_p \le C_{14}(\psi) \ \psi(p), \ \  p \ge 2.
$$
 Since $ f \in G(\psi), \  \Phi << N, $ we deduce that $ f \in G^0(\theta), $
and, following,
$$
\lim_{p \to \infty} |f|_p/\theta(p) = 0.
$$
Therefore, $ \lim_{p \to \infty} \psi(p)/\theta(p) = 0. $ \par
{\bf Theorem 10}. {\it Let now $ X = R^d $ and $ \psi \in \Psi. $
We assert that the norms \\
$ ||\cdot|| L(N^{(\alpha)}, [\psi]) $ and $||\cdot|| G(\alpha, \psi), \ \alpha \ge 1 $
are equivalent.} \par
{\bf Proof.}
1. Let $ \forall p \ge \alpha \ \Rightarrow |f|_p \le \psi(p), \ f \ne 0. $  From
Chebyshev inequality follows that
$$
\lim_{v \to \infty} T(|f|,v) = 0.
$$
We can choose the values $ v, C_{17}, C_{18}, $ such that $ 0 < C_{17} < C_{18} < \infty,
\ v \in (0,\infty) $
and
$$
 C_{17} \le T(|f|,v) \le C_{18} < \infty.
$$
Let us consider for some sufficiently small value $ \epsilon \in (0, \epsilon_0), \
\epsilon_0 \in (0,1) $ the following integral:
$$
I_{\alpha, N}(f) = \int_X N^{(\alpha)}(\epsilon |f(x)| \ dx = I_1 + I_2,
$$
where
$$
I_1 = \int_{\{x: |f(x)| \le v \} } N^{(\alpha)}(\epsilon |f(x)|) \ dx,
\ I_2 = \int_{\{x: |f(x)|> v \} } N^{(\alpha)}(\epsilon |f(x)|) \ dx.
$$
Since for $ z \ge v $
$$
N^{(\alpha)}(z) \le C_{19}(\alpha, N(\cdot)) \ \cdot N( z),
$$
we have for the set $ X(v) = \{x, |f(x)| > v \}, $ using the result of theorem 7 for the
space with finite measure:
$$
I_2 = \int_{X(v)} N^{(\alpha)}(\epsilon |f(x)| ) \ dx \le C_{20}(\alpha,N,\epsilon) \
||f||L(N^{(\alpha)}, X(v)) \le
$$

$$
C_{21} (\alpha, \epsilon, \psi) \sup_{p \ge \alpha}\left[ ||f||L_p(X(v))/\psi(p) \right] \le
C_{21} \sup_{p \ge \alpha} |f|_p/\psi(p) < \infty.
$$
Further, since for $ z \in (0,v) \ \Rightarrow $
$$
N^{(\alpha)}(\epsilon z) \le C_{22}(v,\alpha, \epsilon) \ |z|^{\alpha},
$$
we have:
$$
I_1 \le C_{22}(\cdot) \int_X |f(x)|^{\alpha} \ dx < \infty.
$$
Thus, $ f \in L(N^{(\alpha)}[\psi]), \ ||f||L(N^{(\alpha)} [\psi]) < \infty.$ \par
2). We prove now the inverse inclusion. Let $ f \in L \left(N^{(\alpha)}[\psi] \right) $ and
$$
||f||L \left(N^{(\alpha)}[\psi] \right) = 1.
$$
Hence for some $ \epsilon > 0 $
$$
\int_X N^{(\alpha)}(\epsilon |f(x)|) dx < \infty.
$$
 It follows from the proof of
 theorem 7 and the consideration of two cases: $ |z| \le v; \ |z| > v $ the
following elementary inequality: at $ p \ge \alpha $ and for all $ z > 0 \ \Rightarrow $
$$
|z|^p \le C_{23} (\alpha, \epsilon,N) \ N^{(\alpha)} (\epsilon |z|) \cdot \psi^p(p).
$$
 We obtain for all values $ p, \ p \ge \alpha: $
$$
\int_{R^d} |f(x)|^p \ dx \le C_{24}^p(\alpha, \epsilon,\psi) \ \psi^p(p), \ \ \ ||f||G(\alpha;\psi) <
\infty.
$$
 Note that theorems 7 - 10 are some generalizations of [1, p. 201 - 233],
[15], p.29  and [8], section 1. The case of Hilbert transform of generalized functions $ \{ f \} $
is consider in [2].\\

\section{ Proofs of main results. }

 At first we consider the case Orlicz spaces, i.e. if the function $ f $ belongs to some
exponential Orlicz space.\par
{\bf Proof of theorems 1,2.} Let $ X = [0, \ 2\pi]^d $ and $ f \in L(N[\psi]) $ for some
$ \psi \in \Psi.$ Without los of generality we can assume that $ ||f||L(N[\psi]) = 1. $
From theorem 7 follows that
$$
\forall p \ge 2 \ \Rightarrow \ |f|_p \le C_{25}(\psi) \ \psi(p).
$$
 From the classical theorem of M. Riesz (see [3], p.100 - 103; [9]) follows the inequality:
$$
\sup_{M \ge 1} |s_M[f]|_p \le K_1^d \ p^d \ \psi(p), \ \ K_1 = 2 \pi.
$$
It follows again from theorem 7  that
$$
\sup_{M \ge 1} ||s_M[f] - f||L(N_d[\psi]) \le K_1^d + 1 < \infty.
$$
 For example, if $ N(u) = N_m(u) = \exp(|u|^m)-1 $ for some $ m = const \ge 1, $ then
$$
\sup_{M \ge 1} ||s_M[f] -f ||L(N_{m/(dm+1)}) \le C_{26}(d,m) \ ||f||L(N_m).  \eqno(4.1)
$$
 The "continual" analog of M.Riesz's inequality, namely, the case $ X = R, \ L(N) =
L_p(R), \ p \ge 2: $
$$
\sup_{M \ge 2} |S_M[f]|_p \le K_2^d \ p^d \ |f|_p, \ \ K_2 = 1
$$
 is proved, for example, in [10], p.187 - 188. [9].
This fact permit  us to prove also theorem 2. \par
{\bf Lemma 1.} {\it We assert that the "constant" $ m/(dm+1) $ in the estimation (3.5)
is exact. In detail, for all $ m \ge 1 $ there exists $ g = g_m(\cdot) \in L(N_m) $ such that
$ \forall  \Delta \in (0,m) $ }
$$
\sup_{M \ge 1} ||s_M[g]||L \left(N_{(m-\Delta)/(dm+1)} \right) = \infty.
$$
{\bf Proof} of lemma 1. It is enought to prove that
$$
\exists g \in L(N_m), \ \ ||H[g]||L(N_{(m-\Delta)/(dm+1)}) = \infty,
$$
where $ H[g] $ denotes the Hilbert transform on the $ [0, \ 2 \pi]^d, $
see [12], p. 193 - 197; [13]. Also it is enought to consider the case $ d = 1. $\par
 Let us introduce the function
$$
g(x) = g_m(x) = | \log(x/(2 \pi))|^{1/m}.
$$
Since for $ u > 0 $
$$
mes \{x: g_m(x) > u \} = \exp \left(- u^m \right),
$$
we conclude $ g_m(\cdot) \in L(N_m) \setminus L^0(N_m) $ (theorem 7).
 Further, it is very simple to verify using the formula for  Hilbert transform ([3], p.192)  that
$$
C_{28}(m) \left(| \log (x/ (2 \pi))|^{(m+1)/m} + 1 \right) \le |H[g_m](x)| \le
$$
$$
C_{29}(m) \left( |\log (x/(2 \pi))|^{(m+1)/m} + 1 \right).
$$
 Hence  $ \ \forall u \ge 2 $
$$
\exp \left( -C_{29}(m) u^{m/(m+1)} \right) \le mes \{ x, |H[g_m](x) > u \} \le
$$
$$
\exp \left(-C_{30}(m) \ u^{m/(m+1)} \right).
$$
 It follows again from theorem 7 that
$$
H[g_m] \in L(N_{m/(m+1)}) \setminus L(N_{m/(m+1)}).
$$
Hence $ \forall \Delta \in (0,m) \ \Rightarrow \ H[g_m] \notin L(N_{(m-\Delta)/(m+1)}). $ \par
{\bf Proof} of theorem 3. Let us consider the following function:
$$
z(x) = z_L(x) = \sum_{n=8}^{\infty} n^{-1} \ L(n) \ \sin(nx). \eqno(4.2)
$$
 It is known from the properties of slowly varying functions ([14], p. 98 - 101) that
the series (4.2) converge a.e. and at $ x \in (0, \ 2 \pi] $
$$
C_0 L(1/x) \le z(x) \le C L(1/x).
$$
Therefore, at $ u \in [\exp (2), \infty) $
$$
  L^{-1}(C u)  \le  T(|z|,u) \le   L^{-1}(C_0 u).
$$
 It follows from theorem 7 and (3.2)  that
$$
z(\cdot) \in L(N) \setminus L^0(N), \ N(u) = L^{-1}(u), \ u \ge \exp(2).
$$
 From theorem 8 follows that
$$
0 < C_0 \le |z|_p/\psi(p) \le C < \infty, \ p \ge 2, \ \psi(p) = \exp(W^*(p)/p). \eqno(4.3)
$$
 Note as a consequence that the series (4.2) {\it does not converge in the $ L(N) $ norm,}
as long as the system of functions $ \{\sin(nx) \} $ is bounded and hence in the case
when the series (4.2) {\it converge}  in the $ L(N) $ norm $ \Rightarrow z(\cdot) \in L^0(N). $   \par
 Let us suppose now that for some EOF $ \Phi(\cdot) $ with correspondence function $ \theta(p) $
the series (4.2) convergence in the $ L(\Phi) $ norm. Assume converse to the
assertion of theorem 3, or equally that
$$
\overline{\lim}_{p \to \infty} \theta(p)/\psi(p) > 0. \eqno(4.4)
$$
 Since the system of functions $ \{ \sin(nx) \} $ is bounded, $ z(\cdot) \in L^0(\Phi). $
By virtue of theorem 8 we conclude that
$$
\lim_{p \to \infty} |z|_p/\theta(p) = 0.
$$
 Thus, we obtain from (4.3)
$$
\lim_{p \to \infty} \psi(p)/\theta(p) = 0,
$$
in contradiction with (4.4). The cases $ X = [0, 2 \pi]^d, X = R^d $ are considered as well as
the case $ X = [0, 2 \pi]. $ \par
 Now we consider the case when $ f \in G(a,b,\alpha, \beta).$ \par
{\bf Proof } of theorem 4. Let $ f \in G(1,b,\alpha,0), \ ||f||G(1,b,\alpha,0) = 1. $
Then $ \forall q \in (1,2] $
$$
|f|_q \le (q-1)^{-\alpha}.
$$
We denote $ p = q/(q-1), $ then $ p \in [2, \infty). $ We will use the classical result of
Hardy - Littlewood, Hausdorff - Young ([3], p.193; [16], p. 93;  [17], p. 26):
$$
|F[f]|_p \le (2 \pi)^{1/2} \ |f|_q.
$$
We have in our case:
$$
|F[f]|_p \le (2 \pi)^{1/2} \ (q-1)^{-a} = (2 \pi)^{1/2} (p/(p-1) - 1)^{-a} < (2 \pi)^{1/2} \ p^a.
$$
Our proposition it follows  from theorem 10.\\
{\bf Proof } of theorem 5. {\bf A.} We will use the classical result of  Paley and
F.Riesz ([19], p.120).
Let $ \{ \phi_k(x), \ k = 1,2,\ldots \} $ be some orthonormal bounded sequence of functions.
 Then $ p \ge 2 \ \Rightarrow $
$$
|f|_p \le K_3 \cdot p \ \cdot \left( 1 + \sup_{k,x} |\phi_k(x)| \right) \cdot
 \left( \sum_k |c(k)|^p \left( |k|^{p-2} + 1 \right) \right)^{1/p}, \eqno(4.5)
$$
where $ K_3 $ is an absolute constant, $ f(x) = \sum_k c(k) \ \phi_k(x). $ \par
Let $ c(\cdot) \in g(\psi, \nu) $ and $ ||c||g(\psi,\nu) = 1. $
By definition of the $ g(\psi,\nu) $ norm
$$
  \sum_{k = 1}^{\infty} |c(k)|^p \ \left(k^{p-2} + 1 \right)  \le \ ||c||^p g(\psi, \nu) \cdot
\psi^p(p).
$$
Therefore
$$
|f|_p \le  K_3 \ C_{31} \ p \cdot \psi(p),
$$
and by virtue of theorem 7 $ \ f(\cdot) \in L(N_1[\psi]). $ \par
{\bf B.} Here we use the "discrete" inequality of Hausdorff - Young,  Hardy - Littlewood
(see [16], p.101; \ [17], p.26:)
$$
|f|_p \le K_4 \ |c|_q, \ p \ge 2, \ q = p/(p-1), \ \ K_4 = 2 \pi.
$$
If $ ||c||g(\alpha) =1, $ then
$$
|c|_q \le (q -1)^{\alpha},    |f|_p \le K_4 \ p^{\alpha}, \ p \ge 2.
$$
Again from theorem 7 follows that $ f \in L \left(N_{1/\alpha} \right). $ \par
{\bf Proof } of theorem 6. The analog of inequality (4.5) in the case
$$
F[f](t) = \int_R \exp(i t x) f(x) dx, \ d = 1,
$$
namely:
$$
|F[f]|_p \le K_5 \ p \ |f|_p(\nu),
$$
when $ f(\cdot) \in L_p(\nu) \subset G(\nu, \alpha, \psi), $ see, for example, in
([16], p. 108).  Hence for all $ \ p \ge \alpha $
$$
||F[f]||L \left(N_d^{(\alpha)}[\psi] \right) \le K_5 \ ||f||G(\alpha; \psi, \nu).
$$
(The generalization on the case $ d \ge 2 $ is evident). \par
 Note that the moment estimations for the wavelet transforms and Haar series are described
for example in the books [7], p. 21; [10], p.297  etc. It
is easy to generalize our results on the cases Haar's or wavelet series and transforms.\par
 In detail, it is true in this cases the moment estimation for the partial sums (wavelet's or
Haar's)
$$
|P_M[f]|_p \le K_6 \ |f|_p, \ X = [0,1], \ p \ge 1,
$$
where $ K_6 = 13 $ for Haar series on the interval $ X = [0,1] $  and $ K_6 = 1 $ for the classical
wavelet series, in both the cases $ X = [0,1] $ and $ \ X = R.$
 Hence $ \forall \psi \in \Psi, \ f(\cdot) \in L(N[\psi]) $
$$
\sup_{M \ge 1} ||P_M[f] - f||L(N[\psi]) \le (K_6+1) \ ||f||L(N[\psi]). \eqno(4.6)
$$
 But (4.6) does not mean in general case the convergence
$$
\lim_{M \to \infty} ||P_M[f] - f||L(N[\psi]) = 0, \eqno(4.7)
$$
as long as if (4.7) is true, then $ f(\cdot) \in L^0(N[\psi]) $ and conversely if
$ f \in L^0(N[\psi]), $ then (4.7) holds.\par
 For the different generalizations of wavelet series the estimation (4.6) with constants
$ K_6 $  not depending on $ p, \ p \ge 2 $ see, for example, in the books ([5], p. 221;
[18], p. 69). \\
{\bf Aknowledgement.} I am very grateful for support and attention to prof. V.Fonf (Beer - Sheva,
Israel).This investigation was partially supported by ISF (Israel Science Foundation) grant
$ N^o $ 139/03. \\

\newpage

{\bf References} \\

1. Buldygin V.V., Mushtary D.I., Ostrovsky E.I, Pushalsky M.I. {\it New Trends in Probability
Theory and Statistics.} Mokslas, 1992, Amsterdam, New York, Tokyo. \\
2. Chung Jaeyoung. {\it Hilbert Transform of Generalized Function of $L_p - $ Growth. }
Integral Transforms and Special Functions. 2001, $ N^o $ 2, p. 149 - 160.\\
3. Edwards R.E. {\it Fourier Series. A modern Introduction.} 1982, v.2; Springer Verlag, 1982.
Berlin, Heidelberg, Hong Kong, New York. \\
4. Gord Sinnamon. {\it The Fourier Transform in Weighted Lorentz Spaces.} Publ. Math., 2003,
v. 47, p. 3 - 29. \\
5. Hernandes E., Weiss G. {\it A First Course on Wavelets.} 1996, CRC Press, Boca Raton, New
York.\\
6. Krasnoselsky M.A., Routisky Ya. B.{\it Convex Functions and Orlicz Spaces.} P. Noordhoff
Ltd, 1961, Groningen. \\
7. Novikov I., Semenov E. {\it Haar Series and Linear Operators.} 1997, Kluwner Academic
Publishers.  Boston, Dorderecht, London. \\
8. Ostrovsky E.I. {\it Exponential estimates for random fields and their
 applications } (in Russian). 1999, OINPE, Obninsk.\\
9. Pichorides S.K. {\it On the best values of the constant in the theorem of
M.Riesz, Zygmund and Kolmogorov.} Studia Math., 44, 1972, 165 - 179.\\
10. Pinsky M.A. {\it Introduction to Fourier Analysis and Wavelets.} Books Coll., 2002,
Australia, Canada.\\
11. Rao M.M., Ren Z.D. {\it Theory of Orlicz Spaces.} Marcel Dekker Inc., 1991. New York,
Basel, Hong Kong.\\
12. Rao M.M., Ren Z.D. {\it Applications of Orlicz Spaces.} Marcel Dekker Inc.,
2002. New York, Basel, Hong Kong.\\
13. Ryan R. {\it Conjugate Functions in Orlicz Spaces.} Pacific Journal Math., 1963, v. 13,
p. 1371 - 1377.  \\
14. Seneta E. {\it Regularly Varying Functions.} Springer Verlag, 1985.
Russian edition, Moscow,  Science, 1985.\\
15. Taylor M.E. {\it Partial Differential Equations. } v.3, Nonlinear Equations.
Springer Verlag, 1991. Berlin, Heidelberg, New York.\\
16. Titchmarsh E.C. {\it Introduction to the Theory of Fourier Integrals. }
Claredon Press, Second Edition. 1948. Oxford.\\
17. Wolff T.H. {\it Lectures on Harmonic Analysis.}
AMS, 2003, Providence, Rhore Island.\\
18. Wong M.W. {\it Wavelet Transforms and Localization Operators.} Birkhauser Verlag, 2002.
Berlin, Basel, Boston.\\
19. Zygmund A. {\it Trigonometrical Series.} Cambridge, University Press, 1968, V.2.\\
\end{document}

config: arxiv


TITLE: Proof that $\Bbb R$ is uncountable.
QUESTION [1 upvotes]: Ok so, I was given given a proof of this by my lecturer, but I'm not understanding it fully,, and I was wondering if this suffices as a proof.
Proof:  It is enough to show that  $(0,1)$ is uncountable.  Let $x \in (0,1)$ and write $x$ in it's decimal expansion. i.e. $x= 0.a_1a_2a_3...$, for $a_i \in \{0,1,2....9\}$ for $i \in \Bbb N$  Next assume that $(0,1)$ is countable, that is $(0,1) =\{x_1, x_2, x_3,... \}$. We write  $x_1 = 0.a_{11}a_{12}a_{13}...$  $x_2 = 0.a_{21}a_{22}a_{23}...$
This is where I differ from my lecturer. He says:   let $y= b_1b_2b_3...$, where  $b_i = 2$ if $a_{ii} \neq 2$ and  $b_i = 3$ if $a_{ii} = 2$ for $i \in \Bbb N$  Then $y \in (0,1)$, but $y \neq x_i$, for all $i \in \Bbb N$  This contradiction proves that $(0,1)$ is uncountable.
What I'm saying is:
Let $b_i = a_{ii} + 1$.  Then $y$ will differ from $x_1$ in the first decimal position, from $x_2$ in the second decimal position,... and from $x_n$ in the the $n^{th}$ decimal position. Hence, the list $\{ x_1, x_2,... \}$ can never be an exhaustive list and so $(0,1)$ is uncountable.
Just want to know what you think. Thanks.

REPLY [3 votes]: Your proof doesn't work in case $a_{ii} = 9$. Perhaps you could include $b_i = 0$ if $a_{ii} = 9$.
The reason the instructor used $2$ and $3$ is to avoid the problem of having an infinite string of $9$'s.
To guarantee that each $x$ has a unique representation, you can require that each sequence $\{a_{ij}\}$ does not end in an infinite string of $9$'s. If it did, you could represent the number $x$ with a finite decimal representation. The problem is that the number you construct using the $b_i$ could end in an infinite string of $9$'s which will have a  different representation from every other number in the list, but could in fact equal one of those numbers. For instance, you might construct $0.4999999\ldots$ which already appeared in the list as $0.5000000\ldots$. Avoiding $9$'s altogether circumvents this problem.


TITLE: Finding the derivative $A'$ of $f((x_1,x_2,x_3)) = (x_1^2, x_3^2)$ by $\lim_{h\to 0} \frac{|A(x+h) -Ax-A'h|}{|h|} = 0$
QUESTION [1 upvotes]: I'm studying multivariable calculus and we defined a derivative of a function $f(x)$ from $\mathbb{R}^m$ to $\mathbb{R}^n$ to be a linear operator $A'$ such that
$$\lim_{h\to 0} \frac{|f(x+h)-f(x)-A'h|}{|h|} = 0$$
The simplest example may be the the linear transformations $A$, because by their linearity, $A(x+h) = Ax + Ah \implies A(x+h) -Ax = Ah$, therefore, in the limit definition, if we choose $A' = A$, we have
$$\lim_{h\to 0} \frac{|A(x+h) -Ax-A'h|}{|h|} =\lim_{h\to 0} \frac{0}{|h|} = 0$$
therefore, $A' = A$ because the derivative is unique and we just found one that worked.
Now the hard part comes when we need to find a derivative of a function that is not linear. For example:
$$f:\mathbb{R}^3\to \mathbb{R}^2\\f((x_1,x_2,x_3)) = (x_1^2, x_3^2)$$
I need to find $f':\mathbb{R}^3\to \mathbb{R}^2$ that satisfies
$$\lim_{h\to 0} \frac{|f(x+h)-f(x)-A'h|}{|h|} = \lim_{h\to 0} \frac{|((x_1+h_1)^2, (x_3+h_3)^2) - (x_1^2, x_3^2) - (x_1', x_2')|}{|(h_1,h_2,h_3)|} = 0$$
Where $A'h = (x_1', x_2')$
Let's expand the limit:
$$ \lim_{h\to 0} \frac{|(x_1^2 + 2x_1h_1 + h_1^2, x_3^2 + 2x_3h_3 + h_3^2) - (x_1^2, x_3^2) - (x_1', x_2')|}{\sqrt{h_1^2 + h_2^2 + h_3^2}} = \\ \lim_{h\to 0} \frac{|(2x_1h_1 + h_1^2, 2x_3h_3 + h_3^2) - (x_1', x_2')|}{\sqrt{h_1^2 + h_2^2 + h_3^2}}$$
The hard part is now to take $x_1', x_2'$, that is, the rule of $A'$ that will make it work. Below is something I tried but then I realized I can't even do that (write $x_1', x_2'$ in terms of $x_1, x_2, x_3$)

If I pick: $x_1'= 2x_1h_1$ and $x_2'= 2x_3h_3$ we end up with
$$\lim_{h\to 0} \frac{|(h_1^2, h_3^2)|}{\sqrt{h_1^2 + h_2^2 + h_3^2}}
 = \lim_{h\to 0}\frac{\sqrt{h_1^4 + h_3^4}}{\sqrt{h_1^2 + h_2^2 + h_3^2}}$$

How should I find $A'$?

REPLY [0 votes]: In this case $A'$ is the jacobian matrix and $h=(h_1,h_2,h_3)$, notably we have
$$A'=\begin{bmatrix}2x_1&0&0\\0&2x_2&0\end{bmatrix}$$
Indeed we have that
$$f(x_1+h_1,x_2+h_2,x_3+h_3)=((x_1+h_1)^2,(x_2+h_2)^2)=$$
$$=(x_1^2+2x_1h_1+h_1^2,x_2^2+2x_2h_2+h_2^2)=(x_1^2,x_2^2)+(2x_1h_1,2x_2h_2)+ (h_1^2,h_2^2) =$$
$$=f(x_1,x_2,x_3)+A'(x_1,x_2,x_3)\cdot (h_1,h_2,h_3)+(h_1^2,h_2^2)$$
with
$$A'(x_1,x_2,x_3)\cdot (h_1,h_2,h_3)=\begin{bmatrix}2x_1&0&0\\0&2x_2&0\end{bmatrix}\begin{bmatrix}h_1\\h_2\\h_3\end{bmatrix}=(2x_1h_1,2x_2h_2)$$
and
$$\lim_{h\to 0}\frac{(h_1^2,h_2^2)}{\sqrt{h_1^2+h_2^2+h_3^2}}\to(0,0)$$


TITLE: Discrete math functions proof
QUESTION [1 upvotes]: Let $\mathbb N_{\text{even}}$ be the set of all natural even numbers, and  $\mathbb N_{\text{odd}}$ be the set of all natural odd numbers,
the function $f:\mathcal P(\mathbb N)\to \mathcal P(\mathbb N_{\text{even}}) \times\mathcal P(\mathbb N_{\text{odd}})$ is defined like this: $$f(A) = (A \cap \mathbb N_{\text{even}}, A\cap \mathbb N_{\text{odd}})$$ Show the inverse function of $f$.
Obviously the inverse function would have to be something like this:
$$g:\mathcal P(\mathbb N_{\text{even}}) \times \mathcal P(\mathbb N_{\text{odd}}) \to \mathcal P(\mathbb N)$$
But I just can't seem to solve this problem, can someone guide me what is the best approach? Thanks.

REPLY [0 votes]: $$g(A\times B)=A\cup B.$$
Now prove this is indeed the inverse.


TITLE: Rudin 2.36 assumption?
QUESTION [1 upvotes]: I know this has been asked before but my problem is not the proof it self but the conclusion that we draw.
Theorem. If $ \{K_\alpha\} $ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\} $ is nonempty, then $\cap K_\alpha$ is nonempty.
Proof: Fix a member $K_1$ of $\{K_\alpha\}$ and put $G_\alpha=K_\alpha^c$. Assume that no point of $K_1$ belongs to every $K_\alpha$. Then the sets $G_\alpha$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices  $\alpha_1,…,\alpha_n$ such that $K_1\subset \cup_{i=1...n} G_{\alpha_i}$. But this means that $K_1\cap K_{\alpha_1}\cap\dots \cap K_{\alpha_n}$ is empty, in contradiction to our hypothesis.
How does this contradict our assumption? I can follow the proof but I fail to see what conclusion can be drawn. What is the hypothesis that this contradicts?
Thank you in advance

REPLY [5 votes]: Theorem. If $\{K_α\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_α\}$ is nonempty, then $\cap K_α$ is nonempty.

"What is the hypothesis that this contradicts?"

That the intersection of every finite subcollection of $\{K_a\}$ is nonempty.
That would mean  $K_1 \cap K_2 \cap ...... \cap K_{\alpha_n}$ is not empty.
But the proof showed it was.  That's a contradiction.