text:
\begin{document}

\newcommand{\eps}{\epsilon}
\newcommand{\f}{{\rm 1st}}
\newcommand{\s}{{\rm 2nd}}
\newcommand{\bof}{{\bf 1st}}
\newcommand{\bos}{{\bf 2nd}}
\newcommand{\boH}{{\bf H}}
\newcommand{\boV}{{\bf V}}
\newcommand{\Neg}{\!-\!}

\title{Who Wins Domineering on Rectangular Boards?}

\author{Michael Lachmann \inst{1}, Cristopher Moore \inst{1,2}, and
Ivan Rapaport \inst{3}}

\institute{
Santa Fe Institute, 1399 Hyde Park Road, Santa Fe NM 87501
{\tt dirk@santafe.edu}
\and
Computer Science Department and Department of Physics and Astronomy,
University of New Mexico,
Albuquerque NM 87131
{\tt moore@cs.unm.edu}
\and
Departamento de Ingenieria Matematica, Facultad de Ciencias Fisicas y
Matematicas, Universidad de Chile, Casilla 170/3, Correo 3, Santiago,
Chile {\tt irapapor@dim.uchile.cl}}

\maketitle

\begin{abstract}
Using mostly elementary considerations, we find out who wins the game
of Domineering on all rectangular boards of width 2, 3, 5, and 7.  We
obtain bounds on other boards as well, and prove the existence of
polynomial-time strategies for playing on all boards of width 2, 3, 4,
5, 7, 9, and 11.  We also comment briefly on toroidal and cylindrical
boards.
\end{abstract}

\section{Introduction}

Domineering or Crosscram is a game invented by G\"oran Andersson and
introduced to the public in \cite{gardner}.  Two players, say Vera and
Hepzibah, have vertical and horizontal dominoes respectively.  They
start with a board consisting of some subset of the square lattice and
take turns placing dominoes until one of them can no longer move.  For
instance, the $2 \times 2$ board is a win for the first player, since
whoever places a domino there makes another space for herself while
blocking the other player's moves.

A beautiful theory of combinatorial games of this kind, where both
players have perfect information, is expounded in \cite{conway,bcg}.
Much of its power comes from dividing a game into smaller subgames,
where a player has to choose which subgame to make a move in.  Such a
combination is called a {\em disjunctive sum}.  In Domineering this
happens by dividing the remaining space into several components, so
that each player must choose in which component to place a domino.

Each game is either a win for Vera, regardless of who goes first, or
Hepzibah regardless of who goes first, or the first player regardless
of who it is, or the second regardless of who it is.  These correspond
to a value $G$ which is positive, negative, fuzzy, or zero, i.e.\ $G >
0$, $G < 0$, $G \,\|\, 0$, or $G = 0$.  (By convention Vera and
Hepzibah are the left and right players, and wins for them are
positive and negative respectively.)  However, we will often
abbreviate these values as $G=V$, $H$, $\f$, or $\s$.  We hope this
will not confuse the reader too much.

In this paper, we find who wins Domineering on all rectangles,
cylinders, and tori of width 2, 3, 5, and 7.  We also obtain bounds on
boards of width 4, 7, 9, and 11, and partial results on many others.
We also comment briefly on toroidal and cylindrical boards.

Note that this is a much coarser question than calculating the actual
game-theoretic values of these boards, which determine how they act
when disjunctively summed with other games.  Berlekamp \cite{berl} has
found exact values for $2 \times n$ rectangles with $n$ odd, and
approximate values to within an infinitesimal or `ish' (which
unfortunately can change who wins in unusual situations) for other
positions of width 2 and 3.  In terms of who wins, the $8 \times 8$
board and many other small boards were recently solved by Breuker,
Uiterwijk and van den Herik using a computer search with a good system
of transposition tables in \cite{buh,uiter,recent}.  We make use of
these results below.

\section{$2 \times n$ boards}

On boards of width 2, it is natural to consider dividing it into two
smaller boards of width 2.  At first glance, Vera (the vertical
player) has greater power, since she can choose where to do this.
However, she can only take full advantage of this if she goes first.
Hepzibah (the horizontal player) has a greater power, since whether
she goes first or second, she can divide a game into two simply by
{\em not} placing a domino across their boundary.  We will see that,
for sufficiently large $n$, this gives Hepzibah the upper hand.

We will abbreivate the value of the $2 \times n$ game as $[n]$.

Let's look at what happens when Vera goes first, and divides a board
of length $m+n+1$ into one of length $m$ and one of length $n$.  If
she can win on both these games, i.e.\ if $[m]=[n]=V$, clearly she
wins.  If $[m]=V$ and $[n]=\s$, Hepzibah will
eventually lose in $[m]$ and be forced to play in $[n]$, whereupon Vera replies there and wins.  Finally, if $[m]=[n]=\s$, Vera replies to Hepzibah in both and wins.  Since
Vera can win if she goes first, $[m+n+1]$ must be a win
either for the first player or for V.  This gives us the following
table for combining boards of lengths $m$ and $n$ into boards of
length $m+n+1$:
\begin{equation} \begin{array}{r|cc}
[m+n+1] & \s & V \\ \hline
     \s & \f \mbox{ or } V & \f \mbox{ or } V \\
      V & \f \mbox{ or } V & \f \mbox{ or } V
\end{array} \label{tab1} \end{equation}
This table can be summarized by the equation
\begin{equation}
\mbox{If $[m] \ge 0$ and $[n] \ge 0$, then $[m+n+1] \,\,\|\!\!> 0$.}
\label{eqn1}
\end{equation}

Hepzibah has a similar set of tools at her disposal.  By declining to
ever place a domino across their boundaries, she can effectively play
$[m+n]$ as a sum of $[m]$ and $[n]$ for whichever $m$ and $n$ are the
most convenient.  If Hepzibah goes first, she can win whenever
$[m]=\f$ and either $[n]=\s$ or $[n]=H$, by playing first in $[m]$ and
replying to Vera in $[n]$.  If $[m]=[n]=H$ she wins whether she goes
first or second, and if $[m]=\s$ and $[n]=H$, the same is true since
she plays in $[n]$ and replies to Vera in $[m]$.  Finally, if
$[m]=[n]=\s$, she can win if she goes second by replying to Vera in
both games.  This gives the table
\begin{equation} \begin{array}{r|ccc}
[m+n] & \f & \s & H \\ \hline
   \f & ?                & \f \mbox{ or } H & \f \mbox{ or } H \\
   \s & \f \mbox{ or } H & \s \mbox{ or } H & H \\
    H & \f \mbox{ or } H & H & H
\end{array} \label{tab2} \end{equation}
which can be summarized by the equation
\begin{equation}
[m+n] \le [m] + [n] \label{eqn2}.
\end{equation}
This simply states that refusing to play across a vertical boundary
can only make things harder for Hepzibah.

These two tables alone, in conjunction with some search by hand and by
computer, allow us to determine the following values.  Values derived
from smaller games using Tables \ref{tab1} and \ref{tab2} are shown in
plain, while those found in other ways, such as David Wolfe's
Gamesman's Toolkit \cite{wolfe}, our own search program, or
Berlekamp's solution for odd lengths \cite{berl} are shown in bold.
\[ \begin{array}{lcclcclcclc}
0 & \bos & \hspace{5mm}
           & 10 & \f   & \hspace{5mm}
                         & 20 & H    & \hspace{5mm}
                                       & 30 & H \\
1 & \boV & & 11 & \f   & & 21 & H    & & 31 & \boH \\
2 & \bof & & 12 & H    & & 22 & \boH & & 32 & H \\
3 & \bof & & 13 & \bos & & 23 & \f   & & 33 & H \\
4 & \boH & & 14 & \f   & & 24 & H    & & 34 & H \\
5 & \boV & & 15 & \f   & & 25 & H    & & 35 & H \\
6 & \f   & & 16 & H    & & 26 & H    & & 36 & H \\
7 & \f   & & 17 & H    & & 27 & \f   & & 37 & H \\
8 & H    & & 18 & \bof & & 28 & H    & & 38 & H \\
9 & \boV & & 19 & \f   & & 29 & H    & & 39 & H
\end{array} \]
In fact, $[n]$ is a win for Hepzibah for all $n \ge 28$.

Some discussion is in order.  Once we know that $[4]=H$, we have
$[4k]=H$ for all $k \ge 1$ by Table \ref{tab2}.  Combining Tables
\ref{tab1} and \ref{tab2} gives $[6]=[7]=\f$, since these are both
$\f$ or $V$ and $\f$ or $H$.  A similar argument gives $[10]=[11]=\f$
and $[14]=[15]=\f$, once we learn through search that $[13]=\s$ (which
is rather surprising, and breaks an apparent periodicity of order 4).

Combining $[13]$ with multiples of 4 and with itself gives
$[13+4k]=[26+4k]=[39+4k]=H$ for $k \ge 1$.  Since $26=24+2=13+13$, we
have $[26]=H$ since it is both $\f$ or $H$ and $\s$ or $H$, giving
$[39]=H$ since 39=26+13.

Since $19=9+9+1=17+2$, we have $[19]=\f$ since it is both $\f$ or $V$
and $\f$ or $H$.  Similarly $23=9+13+1=21+2$ and $27=13+13+1=25+2$ so
$[23]=[27]=\f$.  A computer search gives $[22]=H$, and since
$35=22+13$ we have $[35]=H$.

So far, we have gotten away without using the real power of game
theory.  However, for $[31]$ we have found no elementary proof, and it
is too large for our search program.  Therefore, we turn to
Berlekamp's beautiful solution for $2 \times n$ Domineering when $n$
is odd (Ref. \cite{berl}), evaluate it with the Gamesman's Toolkit
\cite{wolfe}, and find the following (see Refs. \cite{conway,bcg,berl}
for notation):
\begin{equation}
[31] \,=\, \frac12 - 15 \cdot \left( \frac14 + \int^{3/4} * \right)
           + \int^{3/4} \int_{1/2}^{1/2*} 3\frac78 \,=\,
\left\{ 2 \,|\, 0 \,\|\, {-\frac12} \,|\, {-2} \right\} |\, {-\frac52}
 \,<\, 0
\end{equation}
Thus $[31]$ is negative and a win for Hepzibah.  This closes the last
loophole, telling us who wins the $2 \times n$ game for all $n$.

\section{Boards of width 3, 4, 5, 7, 9, 11 and others}

The situation for rectangles of width 3 is much simpler.  While
Equation~\ref{eqn1} no longer holds since Vera cannot divide the board
in two with her first move, Equation~\ref{eqn2} holds for all widths,
since Hepzibah can choose not to cross a vertical boundary between two
games.  A quick search shows that $[3 \times n]=H$ for $n=4,5,6$ and
$7$, so we have for width 3
\[ \mbox{$[1]=V$, $[2]=[3]=\f$, and $[n]=H$ for all $n \ge 4$.} \]
For width 5, we obtain
\[ \mbox{$[1]=[3]=V$, $[2]=[4]=H$, $[5]=\s$, and $[n]=H$ for
all $n \ge 6$.} \]
For width 7, Breuker, Uiterwijk and van den Herik found by computer
search (Refs. \cite{buh,uiter,recent}) that $[4]=[6]=[9]=[11]=H$.
Then $[8]=[10]=H$, and combining this with searches on small boards we
have
\[ \mbox{$[1]=[3]=[5]=V$, $[2]=[7]=\f$, $[4]=[6]=H$, and $[n]=H$
for all $n \ge 8$.} \]

In all these cases, we were lucky enough that $[n]=\s$ or $H$ for
enough small $n$ to generate all larger $n$ by addition.  This becomes
progressively rarer for larger widths.  However, we have some partial
results on other widths.  For width 4, Uiterwijk and van den Herik
(Refs. \cite{uiter,recent}) found by computer search that
$[8]=[10]=[12]=[14]=H$, so $[n]=H$ for all even $n \ge 8$.  They also
found that $[15]=[17]=H$, so
\[ [4 \times n]=H \mbox{ for all } n \ge 22. \]
This leaves $[4 \times 19]$ and $[4 \times 21]$ as the only unsolved
boards of width 4.

As a general method, whenever we can find a length for which Hepzibah
wins by some positive number of moves (rather than by an
infinitesimal), then she wins on any board long enough to contain a
sufficient number of copies of this one to overcome whatever advantage
Vera might have on smaller boards.  Game-theoretically, if $[n] < -r$,
then $[m] < 0$ whenever $m \ge (1/r) \max_{\,l < n} \,[l]$.

For width 9, for instance, we have $[1]=4$, $[2]=\frac32 \,|\, 0
\,\|\, {-\frac12} \,|\, {-\frac52}$, $[3]=5 \,|\, 3 \,\|\,
\frac{11}{4} \,|\, \frac14$, and $[4] \le [2]+[2] = 1 \,|\, {-\frac12}
\,\|\, {-1} \,|\, {-\frac52} < {-\frac12}$.  By summing these, it is
easy to show that
\[ [9 \times n]=H \mbox{ for all } n \ge 22. \]
Similarly, for width 11 we have $[1]=5$ and $[2]=1 \,|\, \left\{
\frac12 \,|\, {-1} \,\|\, {-\frac32} \,|\, {-\frac72} \right\}$.  Then
$[8] \le [2]+[2]+[2]+[2] = 1 \,|\, {-\frac12} \,\|\, {-1} \,|\,
{-\frac52} < {-\frac12}$ and $[16] \le [8]+[8] \le {-\frac32}$, so
\[ [11 \times n]=H \mbox{ for all } n \ge 56. \]
Unfortunately, for all other widths greater than 7, either [2] or
$[2]+[2]$ is positive and $[3]$ is as well, so without some way to
calculate values for $[4]$ or more we can't establish this kind of
bound.  Nor do we know of any length for which Hepzibah wins on width
8, or a proof that she wins any board of width 6 by a positive amount.

To get results on boards of other widths, we can use a variety of
tricks.  First of all, just as Hepzibah can choose to cross a vertical
boundary between games, Vera can choose not to cross a horizontal one.
Thus Equation~\ref{eqn2} is one of a dual pair,
\begin{eqnarray}
[m \times (n_1+n_2)] & \le & [m \times n_1] + [m \times n_2] \\
{}[(m_1+m_2) \times n] & \ge & [m_1 \times n] + [m_2 \times n]
\end{eqnarray}
Another useful rule is that $[n \times n]=\f$ or $\s$, since neither
player can have an advantage on a square board.  In fact, in
game-theoretic terms $[n \times n]+[n \times n]=0$, so if Hepzibah
goes second she can win by mimicking Vera's move, rotated $90^\circ$,
in the other board.  More generally we have
\[ \begin{array}{l}
     \mbox{If $[n \times n]=\f$, then } [n \times kn]=
       \left\{ \begin{array}{ll}
         \s \mbox{ or } H & \mbox{ for even } k > 1 \\
         \f \mbox{ or } H & \mbox{ for odd } k > 1
       \end{array} \right. \\
     \mbox{If $[n \times n]=\s$, then $[n \times kn]=\s$ or $H$
       for all $k>1$.}
\end{array} \]
For instance, this tells us that $[6 \times 12]=\s$ or $H$, and since
$[6 \times 4]=\f$ and $[6 \times 8]=H$ (Ref. \cite{buh}) we also have
$[6 \times 12]=\f$ or $H$.  Therefore $[6 \times 12]=H$ and
\[ [6 \times (4+4k)]=H \mbox{ for all } k \ge 1. \]
We can also use our addition rules backward; since no two games can
sum to a square in a way that gives an advantage to either player,
\begin{equation}
\mbox{If $m < n$ and $[m \times n]=\s$ or $V$, then $[(n-m) \times n]
\ne V$}
\end{equation}
and similarly for the dual version.

Using the results of Refs. \cite{berl} and \cite{buh}, some computer
searches of our own, and a program that propagates these rules as much
as possible gives the table shown in Figure~\ref{bigtab}.  It would be
very nice to deduce who wins on some large squares; the $9 \times 9$
square is the largest known so far (Ref. \cite{recent}).  We note that
if $[9 \times 13]=\f$ then $[13 \times 13]=\f$ since $[4 \times
13]=V$.

\begin{figure}
\begin{center}
\psfig{figure=domtable.eps,width=5in}
\end{center}
\caption{What we know so far about who wins Domineering on rectangular
boards.  $1$, $2$, $H$ and $V$ mean a win for the first player, second
player, Hepzibah and Vera respectively.  Things like ``1h'' mean either
$\f$ or $H$ (i.e.\ all we know is that Hepzibah wins if she goes first)
and ``-v'' means that it is not a win for Vera all the time.  Values
outlined in black are those provided by search or other methods; all
others are derived from these using our rules or by symmetry.}
\label{bigtab}
\end{figure}

\section{Playing on cylinders and tori}

On a torus, Hepzibah can choose not to play across a vertical boundary
and Vera can choose not to play across a horizontal one.  Thus cutting
a torus, or pasting a rectangle along one pair of edges, to make a
horizontal or vertical cylinder gives the inequalities shown in
Figure~\ref{torusfig}.  Note that there is no obvious relation between
the value of a rectangle and that of a torus of the same size.

\begin{figure}
\centerline{\psfig{figure=torus.eps,width=3in}}
\caption{Inequalities between rectangular, cylinder, and toroidal
boards of the same size.  Cutting vertically can only hurt Hepzibah,
while cutting horizontally can only hurt Vera.}
\label{torusfig}
\end{figure}

While it is easy to find who wins on tori and cylinders of various
small widths, we do the analysis here only for tori of width 2.  Since
Vera's move takes both squares in the same column, these boards are
equivalent to horizontal cylinders like those shown on the left of
Fig.~\ref{torusfig} (or, for that matter, M\"obius strips or Klein
bottles).  Therefore, if Hepzibah can win on the rectangle of length
$n$, she can win here as well.

The second player has slightly more power here than she did in on the
rectangle, since the first player has no control over the effect of
her move.  If Vera goes first, she simply converts a torus of length
$n$ into a rectangle of length $n-1$, and if Hepzibah goes first, Vera
can choose where to put the rectangle's vertical boundary, in essence
choosing Hepzibah's first move for her.  On the other hand, in the
latter case Hepzibah gets to play again, and can treat the remainder
of the game as the sum of two rectangles and a horizontal space.

These observations give the following table for tori of width 2:
\[ \begin{array}{cc|c}
[n]_{\rm rect} & [n-1]_{\rm rect} & [n]_{\rm torus} \\ \hline
H & & H \\
\f & \f \mbox{ or } H & H \\
\f & \s \mbox{ or } V & \f \\
\s \mbox{ or } V & \f \mbox{ or } H & \s \mbox{ or } H
\end{array} \]
These and our table for $2 \times n$ rectangles determine $[n]_{\rm
torus}$ for all $n$ except 5, 9, and 13.  Vera loses all of these if
she plays first, since she reduces the board to a rectangle which is a
win for Hepzibah.  For 5 and 9, Vera wins if Hepzibah plays first by
playing in such a way that Hepzibah's domino is in the center of the
resulting rectangle, creating a position which has zero value.  Thus
these boards are wins for the $\s$ player.  For 13, Hepzibah wins
since (as the Toolkit tells us) all of Vera's replies to Hepzibah
leave us in a negative position.

Our computer searches show that $n \times n$ tori are wins for the
$\s$ player when $n=1$, $3$, or $5$, and for the $\f$ player when
$n=2$, $4$, or $6$.  We conjecture that this alternation continues,
and that square tori of odd and even size are wins for the $\s$ and
$\f$ players respectively.  We note that a similar argument can be
used to show that 9 is prime.

\section{Polynomial-time strategies}

While correctly playing the sum of many games is PSPACE-complete in
general \cite{morris}, the kinds of sums we have considered here are
especially easy to play.  For instance, if $[m]$ and $[n]$ are both
wins for Hepzibah, she can win on $[m+n]$ by playing wherever she
likes if she goes first, and replying to Vera in whichever game Vera
chooses thereafter.  Thus if we have strategies for both these games,
we have a strategy for their sum.  All our additive rules are of this
kind.

Above, we showed for a number of widths that boards of any length can
be reduced to sums of a finite number of lengths.  Since each of these
can be won with some finite strategy, and since sums of them can be
played in a simple way, we have proved the following theorem:

\begin{theorem}
For boards of width 2, 3, 4, 5, 7, 9, and 11, there exist
polynomial-time strategies for playing on boards of any length.
\end{theorem}

Note that we are not asking that the strategy produce optimum play, in
which Hepzibah (or on small boards, $\f$, $\s$ or Vera) wins by as
many moves as possible, but only that it tells her how to win.

In fact, we conjecture that this theorem is true for boards of any
width.  This would follow if for any $m$ there exists an $n$ such that
Hepzibah wins by some positive number of moves, which in turn implies
that there is some $n'$ such that she wins on all boards longer than
$n'$.  A similar conjecture is made in \cite{uiter}.  Note, however,
that this is not the same as saying that there is a single
polynomial-time strategy for playing on boards of any size.  The size
or running time of the strategy could grow exponentially in $m$, even
if it grows polynomially when $m$ is held constant.

{\bf Acknowledgements.}  We thank Elwyn Berlekamp, Aviezri Fraenkel,
and David Wolfe for helpful communications, and Jos Uiterwijk for
sharing his group's recent results.  I.R.\ also thanks the Santa Fe
Institute for hosting his visit, and FONDECYT 1990616 for their
support.  Finally, C.M. thanks Molly Rose and Spootie the Cat.


text:
\begin{document}
\title{How Arzel\`a and Ascoli would have proved Pego theorem for $L^1(G)$ (if they lived in the $21^{\text{st}}$ century)?}
\author{Mateusz Krukowski}
\affil{Institute of Mathematics, \L\'od\'z University of Technology, \\ W\'ol\-cza\'n\-ska 215, \
90-924 \ \L\'od\'z, \ Poland \\ \vspace{0.3cm} e-mail: mateusz.krukowski@p.lodz.pl}
\maketitle

\begin{abstract}
In the paper we make an effort to answer the question ``What if Arzel\`a and Ascoli lived long enough to see Pego theorem?''. Giulio Ascoli and Cesare Arzel\`a died in 1896 and 1912, respectively, so they could not appreciate the characterization of compact families in $L^2(\reals^N)$ provided by Robert L. Pego in 1985. Unlike the Italian mathematicians, Pego employed various tools from harmonic analysis in his work (for instance the Fourier transform or the Hausdorff-Young inequality). Our article is meant to serve as a bridge between Arzel\`a-Ascoli theorem and Pego theorem (for $L^1(G)$ rather than $L^2(G)$, $G$ being a locally compact abelian group). In a sense, the former is the ``raison d'\^etre'' of the latter, as we shall painstakingly demonstrate.
\end{abstract}

\smallskip
\noindent
\textbf{Keywords : } Banach algebras, $C^*-$algebra of a locally compact abelian group, Arzel\`a-Ascoli theorem, Pego theorem, equicontinuity and equivanishing, compactness in function spaces\\
\vspace{0.2cm}
\\
\textbf{AMS Mathematics Subject Classification (2010): } 43A15, 43A20, 43A25, 46J10, 46L05

\section{Introduction}

In 1985 Robert L. Pego came up with the following characterization of compact families in $L^2(\reals^N):$

\begin{thm}(classical Pego theorem, comp. \cite{Pego})\\
A bounded family $\Ffamily\subset L^2(\reals^N)$ is relatively compact if and only if
\begin{itemize}
	\item $\widehat{\Ffamily}$ is equicontinuous, i.e. for every $\eps>0$ there exists an open neighbourhood $U_0\in\reals^N$ of the zero vector such that
	$$\forall_{\substack{y\in U_0\\ f\in\Ffamily}}\ \int_{\reals^N}\ |\widehat{f}(x+y) - \widehat{f}(x)|^2 < \eps,$$
	\item $\widehat{\Ffamily}$ is equivanishing, i.e. for every $\eps>0$ there exists a compact set $K$ (we denote it by $K\Subset \reals^N$) such that
	$$\forall_{f\in\Ffamily}\ \int_{\reals^N\backslash K}\ |\widehat{f}(x)|^2 < \eps.$$
\end{itemize}
\label{classicalPego}
\end{thm}

The heart of Pego's reasoning is that equicontinuity in the ``time domain'' is equivalent to equivanishing in the ``frequency domain'' (i.e. the domain of the Fourier transform). The same is true if we look at equivanishing in the ``time domain'' and equicontinuity in the ``frequency domain'' $-$ they are also equivalent. If we couple these observations with the Kolmogorov-Tamarkin-Riesz theorem (comp. \cite{HancheOlsenHolden}), we obtain an elegant characterization of compact families in $L^2(\reals^N)$ via the Fourier transform.

Nearly 30 years after Pego's discovery, Przemys\l{}aw G\'orka published a paper (comp. \cite{Gorka}) in which he generalized Theorem \ref{classicalPego} by replacing $\reals^N$ with a locally compact abelian group $G$. Originally, G\'orka's result contained a rather technical restriction, which the author (collaborating with Tomasz Kostrzewa) proved to be obsolete 2 year later (comp. \cite{GorkaKostrzewa}) . G\'orka's approach bears a striking resemblance to Pego's $-$ it employs Plancherel theorem to demonstrate the ``equicontinuity$-$equivanishing swap'' and invokes Weil's theorem (comp. Chapter 12 in \cite{Weil}, p. 52), which generalizes Kolmogorov-Tamarkin-Riesz theorem.

All three mathematicians (Pego, G\'orka and Kostrzewa) view the Fourier transform as a linear and bounded operator on ``$L^2-$space'' (be it $L^2(\reals^N)$ or $L^2(G)$). However, the Fourier transform is originally defined on ``$L^1-$space'' and only then it is extended to ``$L^2-$space''. This motivates the question: ``\textit{Is there a counterpart of the Pego theorem for $L^1(G)?$}'' Our paper is devoted to answering this question in the affirmative.

In Chapter \ref{Chapter:Banachalgebras} we briefly recall the notions related to Banach ($*-$)algebras and $C^*-$algebras. The chapter contains a reflextion on the question ``\textit{What if $L^1(G)$ were a $C^*-$algebra?}'' We argue that if it were the case, we could apply the celebrated Gelfand-Naimark theorem directly to $L^1(G),$ which would save us a considerable amount of effort. Unfortunately, $L^1(G)$ is not a $C^*-$algebra and overcoming this hurdle is the primary goal of the consecutive chapter.

Chapter \ref{Chapter:Cstaralgebra} commences with the introduction of the convolution operator $\Psi.$ Our main source of reference at this point is Deitmar's and Echterhoff's ``Principles of Harmonic analysis'', although we also suggest an alternative approach to the construction of $\Psi.$ The purpose of the convolution operator is to ``copy'' $L^1(G)$ into the $C^*-$algebra $\BB(L^2(G))$ as ``faithfully'' as possible. Theorem \ref{normofconvolutionoperator} and Corollary \ref{Psinotisometry} that it may happen that $\|\Psi(F)\| < \|f\|_1,$ so $C^*(G)$ (the ``copy'' of $L^1(G)$) is not as faithful as we would wish it to be. Nevertheless, $\Psi$ is still an injective algebra $*-$homomorphism $-$ this is not ideal, but definitely very far from tragic.

Chapter \ref{Chapter:ArzelaAscoliPego} is the final part of our investigations. The beginning of the chapter focuses on Arzel\`a-Ascoli theorem, whose classic version is well-established in the literature. Unfortunately, as far as the author is aware, the same remark does not apply for the generalizations of this result. In the paper, we characterize relatively compact families in $C_0(X)$ (where $X$ is a locally compact group) via the ``collapse-of-the-topologies'' technique. This method can be traced as far back as the classic monographs by Kelley and Munkres (comp. \cite{Kelley} and \cite{Munkres}). Once we have the Arzel\`a-Ascoli theorem at our disposal, we apply this result to characterize relatively compact families in $L^1(G)$ via the Fourier transform. This counterpart of the classic Pego theorem is the climax of the paper.

\section{Beyond Banach algebras}
\label{Chapter:Banachalgebras}

As the ancient proverb goes, ``\textit{A journey of a thousand miles begins with a single step}'' (comp. \cite{LaoTse}, \mbox{Chapter 64}). In accordance with the Chinese wisdom, our journey has to begin somewhere and we feel that the realm of Banach algebras is a good place to start. Following Eberhard Kaniuth's ``A Course in Commutative Banach Algebras'' (comp. \cite{Kaniuth}, p. 1) we say that a normed linear space $(\algebra,\|\cdot\|)$ over the field of complex numbers $\complex$ is a \textit{normed algebra} if it is an algebra and the norm is \textit{submultiplicative}, i.e.
$$\forall_{f,g \in \algebra}\ \|f\cdot g\| \leqslant \|f\|\cdot \|g\|.$$

\noindent
A normed algebra $\algebra$ is said to be a \textit{Banach algebra} if it is a Banach space (a plethora of examples of Banach algebras is provided in \cite{Bobrowski}, Chapter 6). A map $\Psi:\algebra_1\longrightarrow \algebra_2$ between two Banach algebras $\algebra_1$ and $\algebra_2$ is a \textit{Banach algebra homomorphism} if it is continuous, $\complex-$linear and \textit{multiplicative}, i.e.
$$\forall_{f,g\in\algebra_1}\ \Psi(f\cdot g) = \Psi(f)\cdot \Psi(g).$$

With every commutative Banach algebra $\algebra$, one can associate the so-called \textit{structure space} $\Delta_{\algebra}$ (comp. \cite{DeitmarEchterhoff}, p. 43 or \cite{Folland}, p. 5 or \cite{Kaniuth}, p. 46). It is the set of all non-zero Banach algebra homomorphisms $m:\algebra \longrightarrow \complex.$ The elements of $\Delta_{\algebra}$ are typically referred to as the \textit{multiplicative linear functionals}, which is why we use the letter $m$ to denote them.

Suppose that $\algebra$ is a Banach algebra and that $*:\algebra\longrightarrow \algebra$ is a map such that for every $f,g\in\algebra,\ \lambda\in\complex$ we have
\begin{itemize}
	\item $(f+g)^* = f^* + g^*,$
	\item $(\lambda f)^* = \overline{\lambda}f^*,$
	\item $(f\cdot g)^* = g^*\cdot f^*,$
	\item $(f^*)^* = f,$
	\item $\|f^*\| = \|f\|.$
\end{itemize}

\noindent
The pair $(\algebra,*)$ is called a \textit{Banach $*-$algebra} and the map $*:\algebra\longrightarrow\algebra$ is called an \textit{involution} (comp. \cite{DeitmarEchterhoff}, p. 49). A map $\Psi:\algebra_1\longrightarrow \algebra_2$ between two Banach $*-$algebras $\algebra_1$ and $\algebra_2$ is a \textit{Banach $*-$algebra homomorphism} if, in addition to being a Banach algebra homomorphism, it satisfies
$$\forall_{f\in\algebra_1}\ \Psi\left(f^*\right) = \Psi(f)^*.$$

Arguably the most common example of a Banach $*-$algebra is $\BB(\Hilbert),$ the space of bounded and linear operators on a given Hilbert space $\Hilbert$. The involution $*:\BB(\Hilbert)\longrightarrow\BB(\Hilbert)$ maps every operator $T$ to its \textit{adjoint operator} (comp. Theorem 4.10 in \cite{Rudin} or Proposition 5.4 in \cite{SteinShakarchi}, p. 183), i.e. a unique bounded and linear operator  $T^*$ satisfying
$$\forall_{u,v\in\Hilbert}\ \langle Tu|v \rangle = \langle u|T^*v \rangle,$$

\noindent
where $\langle\cdot|\cdot\rangle$ denotes the inner product in $\Hilbert.$ The relevance of $\BB(\Hilbert)$ in our discussion will manifest itself in Chapter \ref{Chapter:Cstaralgebra}, where it sets the stage for the $C^*-$algebra of a locally compact abelian group $G$.

The second example of a Banach $*-$algebra, which permeates the mathematical literature (Lemma 2.6.2 in \cite{DeitmarEchterhoff}, p. 50 or \cite{Dixmier}, p. 1 or \cite{Kaniuth}, p. 18), is the space $L^1(G),$ where $G$ is (as always in this paper) a locally compact abelian group. The \textit{convolution} defined by
$$\forall_{x\in G}\ f\star g(x) := \int_G\ f(x-y)\cdot g(y)\ dy$$

\noindent
is the algebra multiplication in $L^1(G)$ and the involution is given by
$$\forall_{x\in G}\ f^*(x) := \overline{f(-x)}.$$

\noindent
The importance of this example is self-evident: after all, the main goal of the paper is to provide a Pego-like characterization of relatively compact families in $L^1(G).$

Before closing this chapter we introduce one more concept for future progress: a Banach $*-$algebra $\algebra$, which satisfies the \textit{$C^*-$property}:
\begin{gather}
\forall_{f\in \algebra}\ \|f\cdot f^*\| = \|f\|^2,
\label{Cstarproperty}
\end{gather}

\noindent
is called a $C^*-$algebra (comp. \cite{BelfiDoran}, p. 5 or \cite{Davidson}, p. 1 or \cite{DeitmarEchterhoff}, p. 49 or \cite{Dixmier}, p. 8 or \cite{Kaniuth}, p. 66). Property (\ref{Cstarproperty}) is easily memorized as the functional analytic version of $z\cdot \overline{z} = |z|^2$ for $z\in\complex$ in complex analysis. $\BB(\Hilbert),$ which we mentioned earlier, is an example of a (noncommutative) $C^*-$algebra (comp. \cite{Dixmier}, p. 1 or \cite{DeitmarEchterhoff}, p. 49 or \cite{Kaniuth}, p. 66), but what about $L^1(G)$? If $L^1(G)$ were a (commutative) $C^*-$algebra, then by the celebrated Gelfand-Naimark theorem (comp. Theorem 7.1 in \cite{BelfiDoran}, p. 22 or Theorem I.3.1 in \cite{Davidson}, p. 7 or Theorem 2.6.7 in \cite{DeitmarEchterhoff}, p. 53 or Theorem 2.4.5 in \cite{Kaniuth}, p. 69), it would be isometrically $*-$isomorphic to $C_0(X),$ the space of continuous functions, which vanish at infinity ($X$ would be a certain locally compact space). This, in turn, would greatly simplify our work, as characterizing relatively compact families in $L^1(G)$ would then amount to characterizing relatively compact families in $C_0(X)$. Unfortunately, $L^1(G)$ is \textit{not} a $C^*-$algebra, unless $G$ is trivial (comp. Theorem 2.6.2 in \cite{DeitmarEchterhoff}, p. 50). This entails that if we want to write down Pego theorem for $L^1(G)$ then we need to overcome the obstacle of it not being a $C^*-$algebra. Resolving this issue lies at the heart of our next chapter.

\section{Convolution operator and the \texorpdfstring{$C^*-$}{Lg}algebra of a locally compact abelian group}
\label{Chapter:Cstaralgebra}

As foreshadowed in Chapter \ref{Chapter:Banachalgebras}, the current section will be devoted to the construction of a $C^*-$algebra of a locally compact abelian group $G$. This object will serve as the ``enhanced version'' of $L^1(G),$ which is not a $C^*-$algebra.

To begin with, for fixed $f\in L^1(G)$ and $\varphi\in L^2(G)$ we consider the map
\begin{gather}
\phi \mapsto \int_G\ f(y)\cdot \langle \phi|T_y\varphi \rangle_2\ dy,
\label{linearfunctional}
\end{gather}

\noindent
where $T_y:L^2(G)\longrightarrow L^2(G)$ is the \textit{translation operator} $T_y\varphi(x) := \varphi(x-y)$ and $\langle\cdot|\cdot\rangle_2$ stands for the inner product in $L^2(G)$. On page 70 in \cite{DeitmarEchterhoff}, Deitmar and Echterhoff prove that the map (\ref{linearfunctional}) is a linear and bound functional on $L^2(G)$. By Riesz representation theorem (Theorem 4.11 in \cite{Brezis}, p. 97 or Theorem 6.52 in \cite{RenardyRogers}, p. 196) there exists a function $\Psi_{f,\varphi}\in L^2(G)$ such that
\begin{gather}
\forall_{\phi \in L^2(G)}\ \langle \phi|\Psi_{f,\varphi}\rangle_2 = \int_G\ f(y)\cdot \langle \phi|T_y\varphi \rangle_2\ dy.
\label{Rieszcorollary}
\end{gather}

It is easy to observe that for every $\alpha_1,\alpha_2\in\complex$ and $\varphi_1,\varphi_2\in L^2(G)$ we have
$$\Psi_{f,\alpha_1\varphi_1+\alpha_2\varphi_2} = \alpha_1\Psi_{f,\varphi_1} + \alpha_2\Psi_{f,\varphi_2}.$$

\noindent
Furthermore, (\ref{Rieszcorollary}) leads to the following estimate (for details see \cite{DeitmarEchterhoff}, p. 70):
$$\|\Psi_{f,\varphi}\|_2 \leqslant \|f\|_1\cdot \|\varphi\|_2.$$

\noindent
Consequently, we may define a linear and bounded map $\Psi_f: L^2(G)\longrightarrow L^2(G)$ by $\Psi_f(\varphi) := \Psi_{f,\varphi}.$

Deitmar and Echterhoff go on to prove (Lemma 3.3.1 in \cite{DeitmarEchterhoff}, p. 70) that
\begin{gather}
\forall_{\varphi\in L^1(G)\cap L^2(G)}\ \Psi_f(\varphi) = f\star \varphi = \varphi \star f.
\label{Psiandconvolution}
\end{gather}

\noindent
Going off on a tangent for a brief moment, let us remark that we could define $\Psi_f$ in a slightly different manner than Deitmar and Echterhoff do. In this alternative approach (comp. \cite{Kaniuth}, p. 95), we would begin with \textit{defining} $\Psi_f(\varphi) := f\star \varphi$ for every $\varphi\in L^1(G)\cap L^2(G) -$ this definition is obviously inspired by the property (\ref{Psiandconvolution}) in Deitmar-Echterhoff's point of view. Since $L^1(G)\cap L^2(G)$ is dense in $L^2(G),$ we would subsequently extend the operator $\Psi_f:L^1(G)\cap L^2(G)\longrightarrow L^2(G)$ (by Theorem 1.6 in \cite{ReedSimon}, p. 9) to all of $L^2(G).$ This extension is unique, so our operator $\Psi_f$ would have no choice but to coincide with the one defined by Deitmar and Echterhoff.

Going back to the main line of reasoning, we define the map $\Psi : L^1(G)\longrightarrow \BB(L^2(G))$ by the formula $\Psi(f) := \Psi_f.$ Lemma 3.3.2 in \cite{DeitmarEchterhoff}, p. 70 (or Theorem 2.7.7 in \cite{Kaniuth}, p. 95) proves that $\Psi$ is an injective algebra $*-$homomorphism between Banach $*-$algebras $L^1(G)$ and $\BB(L^2(G)).$ What more can we say about $\Psi$?

\begin{lemma}(known in the mathematical folklore)\\
If $f\in L^1(G)$ and $f\geqslant 0$ then $\|\Psi(f)\| = \|f\|_1.$
\label{normofPsiandfinL1}
\end{lemma}
\begin{proof}
First, let us remark that it is clear to see that $\Psi$ is a homeomorphism due to the general version of the open mapping theorem (comp. Theorem 2.11 in \cite{Rudin}, p. 48). However, we claim that even more is true, so a more detailed analysis is required.

We fix $f\in L^1(G)$ and observe that
$$\forall_{\phi\in L^1(G)\cap L^2(G)}\ \|\Psi_f(\phi)\|_2 \stackrel{(\ref{Psiandconvolution})}{=} \|f\star \phi\|_2 \leqslant \|f\|_1\cdot \|\phi\|_2,$$

\noindent
where the last inequality follows from Young's convolution inequality (comp. Corollary 20.14 in \cite{HewittRoss}, p. 293). We instantly conclude that $\|\Psi_f\|\leqslant \|f\|_1,$ which is the effortless part of the proof. Henceforth, we focus our attention on demonstrating that $\|\Psi_f\|\geqslant \|f\|_1.$

Since $f\in L^1(G)$ then $\supp(f)$ is contained in a $\sigma-$compact set (comp. Corolarry 1.3.5 in \cite{DeitmarEchterhoff}, p. 10). Hence
$$\supp(f)\subset \bigcup_{n=1}^{\infty}\ K_n,$$

\noindent
where $K_n \Subset G$ (meaning $K_n$ is compact) for every $n\in\naturals$. It is easy to see that without loss of generality, we may assume that
\begin{itemize}
	\item $0\in K_n$ for every $n\in\naturals$ (adjoining one point does not change compactness of $K_n$),
	\item $K_n\subset K_{n+1}$ for every $n\in\naturals$ (this is a standard trick of substituting $D_n:= \bigcup_{m=1}^n\ K_m$ for $K_n$ if necessary),
	\item $K_n$ is symmetric for every $n\in\naturals$ (we may substitute $K_n\cup (-K_n)$ for $K_n$ if necessary).
\end{itemize}

\noindent
Consequently, we have that
\begin{gather}
\forall_{n\geqslant 2}\ K_n - K_n \supset 0 - K_n = -K_n = K_n \supset K_{n-1}.
\label{inequalitythatweneedattheend}
\end{gather}

Furthermore, let us define a family of functions $\phi_n := \mu(K_n)^{-\frac{1}{2}}\cdot \mathds{1}_{K_n}$ for every $n\in\naturals,$ where $\mathds{1}_{K_n}$ is the characteristic function of $K_n.$ An easy calculation shows that $\|\phi_n\|_2 = 1$ for every $n\in\naturals.$  In order to conclude that $\|\Psi(f)\|\geqslant \|f\|_1,$ it is enough to demonstrate that
$$\|\Psi_f(\phi_n)\| = \|f\star \phi_n\|_2 \geqslant \|f\|_1 - \eps$$

\noindent
for a fixed $\eps>0$ and $n$ large enough. To this end we fix $\eps>0$ and choose $N\in\naturals$ such that
\begin{gather}
\int_{K_N}\ f(x)\ dx \geqslant \|f\|_1 - \eps.
\label{choiceofNforepsilon}
\end{gather}

\noindent
Let us remark that the assumption ``$f\geqslant 0$'' plays a pivotal role in the choice of $N$: if $f$ could have been negative then it would be possible for the integral $\int_{K_N}\ f(x)\ dx$ to be negative and $\|f\|_1 - \eps$ to be positive. This would lead to contradiction. In other words, the choice of $N$ is possible due to the assumption that $f$ is non-negative.

Finally, we have
\begin{gather*}
\forall_{n> N}\ \|f\star \phi_n\|_2^2 = \int_G\left|\int_G\ f(x-y)\cdot \phi_n(y)\ dy\right|^2\ dx = \frac{1}{\mu(K_n)}\int_G\ \left(\int_{K_n}\ f(x-y)\ dy\right)^2\ dx \\
\stackrel{y\mapsto x-y}{=} \frac{1}{\mu(K_n)}\int_G \left(\int_{x-K_n}\ f(y)\ dy\right)^2\ dx \geqslant \frac{1}{\mu(K_n)}\int_{K_n} \left(\int_{x-K_n}\ f(y)\ dy\right)^2\ dx \\
\stackrel{(\ref{inequalitythatweneedattheend})}{\geqslant} \frac{1}{\mu(K_n)}\int_{K_n} \left(\int_{K_N}\ f(y)\ dy\right)^2\ dx \stackrel{(\ref{choiceofNforepsilon})}{\geqslant} (\|f\|_1 - \eps)^2,
\end{gather*}

\noindent
which concludes the proof.
\end{proof}

The above lemma is elegant and charming, but the assumption ``$f\geqslant 0$'' limits its applicability to a certain degree. If the assumption of nonnegativity is removed, then the equality $\|\Psi(f)\| = \|f\|_1$ need not hold. In order to see this, let us recall that for every function $f\in L^1(G)$ we define its Fourier transform (comp. Chapter 1.7 in \cite{DeitmarEchterhoff}, p. 29 or Chapter 4.2 in \cite{Folland}, p. 93 or Chapter 2.8.4 in \cite{HavinNikolski}, p. 260 or Chapter 8 in \cite{HewittRoss2}, p. 209) with the formula
\begin{gather}
\forall_{\chi\in\widehat{G}}\ \widehat{f}(\chi) := \int_G\ f(x)\cdot \overline{\chi(x)}\ dx,
\label{Fouriertransform}
\end{gather}

\noindent
where $\widehat{G}$ is the dual group (comp. Chapter 7 in \cite{Deitmar}, p. 101 or Chapter 3.1 in \cite{DeitmarEchterhoff}, p. 63 or Chapter 4.1 in \cite{Folland}, p. 87 or Chapter 2.8.1 in \cite{HavinNikolski}, p. 255 or Chapter 6.23 in \cite{HewittRoss}, p. 355). We will be particularly interested in the dual group $\widehat{\integers}$, whose elements (characters) are the maps $\chi_{\alpha}:\integers\longrightarrow S^1$ given by
$$\chi_{\alpha}(k) := e^{2\pi i k\alpha},$$

\noindent
where $\alpha\in [0,1)$ (comp. Proposition 7.1.1 in \cite{Deitmar}, p. 101). If $[0,1)$ is treated as topological group with addition $\bmod 1$ then it can be shown (comp. Proposition 7.1.6 in \cite{Deitmar}, p. 106) that the map $\hh:\widehat{\integers}\longrightarrow ([0,1),+_{\bmod 1})$ given by
\begin{gather}
\hh(\chi_{\alpha}) := \alpha
\label{hisomorphism}
\end{gather}

\noindent
is a homeomorphic isomorphism. This fact allows for a convenient description of the Haar measure on $\widehat{\integers},$ due to the following result:

\begin{lemma}(known in the mathematical folklore)\\
Let $G_1$ and $G_2$ be locally compact (not necessarily abelian) groups and let $h:G_1\longrightarrow G_2$ be a homeomorphic isomorphism. If $\mu_1,\mu_2$ are (left) Haar measures on $G_1$ and $G_2$ respectively, then $\mu_1 = c\cdot\mu_2\circ h$ for some constant $c>0.$
\label{Haarmeasuresandhomeomorphiciso}
\end{lemma}
\begin{proof}
Since a Haar measure is unique up to a positive constant (comp. Theorem 2.20 in \cite{Folland}, p. 39 or Theorem 3.3.2 in \cite{ReiterStegman}, p. 83) then it suffices to check that $\mu_2\circ h$ is a Haar measure on $G_1$. First of all, it is a nontrivial map since $\mu_2\circ h(G_1) = \mu_2(G_2) > 0.$ Obviously, we have $\mu_2\circ h(\emptyset) = \mu_2(\emptyset) = 0.$ Next, if $(A_n)_{n\in\naturals}\subset G_1$ is a sequence of pairwise disjoint Borel-measurable sets then the sequence $(h(A_n))_{n\in\naturals}\subset G_2$ is also pairwise disjoint and we have
$$\mu_2\circ h\left(\bigcup_{n\in\naturals}\ A_n\right) = \mu_2\left(\bigcup_{n\in\naturals}\ h(A_n)\right) = \sum_{n\in\naturals}\ \mu_2\circ h(A_n).$$

\noindent
This demonstrates that $\mu_2\circ h$ is a nontrivial Borel measure.

If $K$ is a compact subset of $G_1$ then $h(K)$ is compact in $G_2$ and $\mu_2\circ h(K) < \infty.$ Furthermore, to see that $\mu_2\circ h$ is inner regular we take an open set $U\subset G_1$ (then $h(U)$ is open in $G_2$) and observe that
\begin{equation*}
\begin{split}
\mu_2\circ h(U) &= \sup \left\{\mu_2(D)\ :\ D\subset h(U), D\Subset G_2 \right\} \\
&= \sup \left\{\mu_2\circ h(h^{-1}(D))\ :\ h\circ h^{-1}(D)\subset h(U), D\Subset G_2\right\}\\
&= \sup \left\{\mu_2\circ h(h^{-1}(D))\ :\ h^{-1}(D)\subset U, D\Subset G_2\right\} \\
&= \sup \left\{\mu_2\circ h(K)\ :\ K\subset U, K\Subset G_1\right\}.
\end{split}
\end{equation*}

\noindent
Analogously, if $A\subset G_1$ is a Borel-measurable set then $h(A)$ is Borel-measurable in $G_2$ and
\begin{equation*}
\begin{split}
\mu_2\circ h(A) &= \inf \left\{\mu_2(U)\ :\ h(A)\subset U, U\ \text{is open in}\ G_2\right\} \\
&= \inf \left\{\mu_2\circ h(h^{-1}(U))\ :\ h(A)\subset h(h^{-1}(U)), U\ \text{is open in}\ G_2\right\}\\
&= \inf \left\{\mu_2\circ h(h^{-1}(U))\ :\ A\subset h^{-1}(U), U\ \text{is open in}\ G_2\right\}\\
&= \inf \left\{\mu_2\circ h(V)\ :\ A\subset V, V\ \text{is open in}\ G_2\right\},
\end{split}
\end{equation*}

\noindent
which establishes that $\mu_2\circ h$ is outer regular. Last but not least, if $x\in G_1$ and $A\subset G_1$ is a Borel-measurable set then
$$\mu_2\circ h(x\cdot A) = \mu_2\left(h(x)\cdot h(A)\right) = \mu_2\circ h(A),$$

\noindent
due to left-invariance of $\mu_2$ (an the fact that $h$ is an isomorphism). This concludes the proof.
\end{proof}

We now apply Lemma \ref{Haarmeasuresandhomeomorphiciso} to describe the Haar measure and Haar integral on $\widehat{\integers}.$ In the corollary below, we use the term ``\textit{normalized} Haar measure'', which means that the measure of the whole (compact) group is equal to $1$.

\begin{cor}(known in the mathematical folklore)\\
The normalized Haar measure $\mu_{\widehat{\integers}}$ on $\widehat{\integers}$ satisfies
\begin{gather}
\mu_{\widehat{\integers}} = \lambda|_{[0,1)}\circ \hh,
\label{Haarmeasureonwidehatintegers}
\end{gather}

\noindent
where $\lambda$ is the classical Lebesgue measure on $\reals$. Consequently, if $F\in L^1(\widehat{\integers})$ then
\begin{gather}
\int_{\widehat{\integers}}\ F(\chi)\ d\chi = \int_0^1\ F\circ\hh^{-1}(\alpha)\ d\alpha.
\label{howtocalculateintegraloverwidehatintegers}
\end{gather}
\end{cor}
\begin{proof}
First of all, we remark that $([0,1), +_{\bmod 1})$ is a compact group, since it is homeomorphic to the circle group $S^1$ (or the quotient group $\reals/\integers$ for that matter). Furthermore, $\lambda|_{[0,1)}$ is the normalized Haar measure on $([0,1), +_{\bmod 1}).$ By Lemma \ref{Haarmeasuresandhomeomorphiciso} we have that $\lambda|_{[0,1)}\circ \hh$ is the normalized Haar measure on $\widehat{\integers}.$ This demonstrates the equality (\ref{Haarmeasureonwidehatintegers}).

Regarding the second part of the theorem, we begin by observing that if $F = \mathds{1}_A$ for some measurable $A\subset \widehat{\integers},$ then
\begin{gather*}
\int_{\widehat{\integers}}\ F(\chi)\ d\chi = \int_{\widehat{\integers}}\ \mathds{1}_A(\chi)\ d\chi = \mu_{\widehat{\integers}}(A) \stackrel{(\ref{Haarmeasureonwidehatintegers})}{=} \lambda|_{[0,1)}\circ \hh(A) = \int_0^1\ \mathds{1}_{\hh(A)}(\alpha)\ d\alpha\\
=\int_0^1\ \mathds{1}_A(\hh^{-1}(\alpha))\ d\alpha = \int_0^1\ F\circ\hh^{-1}(\alpha)\ d\alpha,
\end{gather*}

\noindent
which is congruent with (\ref{howtocalculateintegraloverwidehatintegers}). The rest of the argument follows a classical technique. We take the liberty of not transcribing the whole reasoning in detail, but provide just a simple sketch: since (\ref{howtocalculateintegraloverwidehatintegers}) is true for the characteristic functions (as we have just demonstrated) then it is also true for simple functions (with non-negative coefficients). Moreover, given a non-negative function $F\in L^1(\widehat{\integers})$ we approximate it by a nondecreasing sequence of non-negative simple functions (comp. Theorem 2.10 in \cite{FollandRealAnalysis}, p. 47) and (\ref{howtocalculateintegraloverwidehatintegers}) for such $F$ follows from the monotone convergence theorem (comp. Theorem 2.14 in \cite{FollandRealAnalysis}, p. 50). Next, any real-valued integrable function can be represented as a difference of two non-negative integrable functions, so (\ref{howtocalculateintegraloverwidehatintegers}) holds for any real-valued $F\in L^1(\widehat{\integers}).$ Last but not least, every complex-valued $F\in L^1(\widehat{\integers})$ can be decomposed as $F_1+iF_2,$ where $F_1,F_2\in L^1(\widehat{\integers})$ are real-valued. This implies that (\ref{howtocalculateintegraloverwidehatintegers}) holds true for every function in $L^1(\widehat{\integers}).$
\end{proof}

In the next theorem we calculate the norm $\|\Psi_f\|$ for $f\in L^1(\integers).$ This computation will prove to be invaluable in Corollary \ref{Psinotisometry}, in which we demonstrate the existence of a function $g\in L^1(\integers)$ such that $\|\Psi_g\| < \|g\|_1.$

The theorem below bears a dim resemblance to Corollary 1.2 in \cite{GohbergGoldbergKaashoek}, p. 217. The result is also mentioned (without any reference or proof) in \cite{Kovrizhkin}. However, as we were unable to find a reference with proof for this result, we have taken the liberty of proving it ourselves.

\begin{thm}
If $G=\integers$ then
$$\forall_{f\in L^1(\integers)}\ \|\Psi_f\| = \|\widehat{f}\|_{\infty},$$

\noindent
where $\|\cdot\|_{\infty}$ denotes the supremum norm.
\label{normofconvolutionoperator}
\end{thm}
\begin{proof}
We fix $f\in L^1(\integers)$ and observe that
\begin{gather}
\|\Psi_f\| = \sup_{\substack{\phi\in C_c(\integers)\\ \|\phi\|_2 = 1}}\ \|\Psi_f(\phi)\|_2 = \sup_{\substack{\phi\in C_c(\integers)\\ \|\phi\|_2 = 1}}\ \|f\star \phi\|_2
= \sup_{\substack{\phi\in C_c(\integers)\\ \|\phi\|_2 = 1}}\ \|\widehat{f\star\phi}\|_2 = \sup_{\substack{\phi\in C_c(\integers)\\ \|\phi\|_2 = 1}}\ \|\widehat{f}\cdot\widehat{\phi}\|_2,
\label{calculatethenormpsif}
\end{gather}

\noindent
where we have employed both the Plancherel theorem (comp. Theorem 3.4.8 in \cite{DeitmarEchterhoff}, p. 77 or Theorem 4.25 in \cite{Folland}, p. 99 or Theorem 4.4.1 in \cite{Kaniuth}, p. 219 ) and the property $\widehat{f\star\phi} = \widehat{f}\cdot\widehat{\phi}$ (comp. Theorem 8.3.1 in \cite{Deitmar}, p. 120 or Lemma 1.7.2 in \cite{DeitmarEchterhoff}, p. 30). Furthermore, we have
\begin{gather*}
\forall_{\substack{\phi\in C_c(\integers)\\ \|\phi\|_2 = 1}}\|\widehat{f}\cdot\widehat{\phi}\|_2^2 = \int_{\widehat{\integers}}\ |\widehat{f}(\chi)\cdot\widehat{\phi}(\chi)|^2\ d\chi\leqslant \|\widehat{f}\|_{\infty}^2 \cdot \int_{\widehat{\integers}}\ |\widehat{\phi}(\chi)|^2\ d\chi = \|\widehat{f}\|_{\infty}^2\cdot \|\widehat{\phi}\|_2^2 = \|\widehat{f}\|_{\infty}^2\cdot \|\phi\|_2^2 = \|\widehat{f}\|_{\infty}^2,
\end{gather*}

\noindent
which (coupled with (\ref{calculatethenormpsif})) establishes that $\|\Psi_f\|\leqslant \|\widehat{f}\|_{\infty}.$

For the converse, we fix $\alpha_*\in[0,1)$ and define a sequence of functions $\Phi_n:\widehat{\integers}\longrightarrow\complex$ by
\begin{gather*}
\Phi_n(\chi_{\alpha}) := \left\{\begin{array}{cl}
\sqrt{n} & \text{ if}\ |\alpha-\alpha_*| < \frac{1}{2n},\\
0 & \text{ otherwise.}
\end{array}\right.
\end{gather*}

\noindent
Naturally $\Phi_n\in L^2(\widehat{\integers})$ and we have
\begin{gather}
\|\widehat{f}\cdot \Phi_n\|_2^2 = \int_{\widehat{\integers}}\ |\widehat{f}(\chi)\cdot \Phi_n(\chi)|^2\ d\chi \stackrel{(\ref{howtocalculateintegraloverwidehatintegers})}{=} \int_0^1\ |\widehat{f}\circ \hh^{-1}(\alpha)|^2\cdot |\Phi_n\circ \hh^{-1}(\alpha)|^2\ d\alpha \geqslant n\int_{\alpha_*-\frac{1}{2n}}^{\alpha_*+\frac{1}{2n}}\ |\widehat{f}\circ \hh^{-1}(\alpha)|\ d\alpha.
\label{prepareforHospital}
\end{gather}

\noindent
By a standard application of de l'H\^ospital's rule we conclude that
\begin{gather}
\limsup_{n\rightarrow\infty}\ \|\widehat{f}\cdot \Phi_n\|_2 \stackrel{(\ref{prepareforHospital})}{\geqslant} |\widehat{f}\circ \hh^{-1}(\alpha_*)| = |\widehat{f}(\chi_{\alpha_*})|.
\label{applicationofHospital}
\end{gather}

\noindent
By density of $C_c(\integers)$ in $L^2(\integers)$ (comp. Proposition 7.9 in \cite{FollandRealAnalysis}, p. 217) we pick a sequence $(\phi_n)_{n\in\naturals}\subset C_c(\integers)$ such that
\begin{gather}
\|\phi_n - \widecheck{\Phi_n}\|_2 \stackrel{n\rightarrow\infty}{\longrightarrow} 0,
\label{choiceofphin}
\end{gather}

\noindent
where $\widecheck{\Phi_n}\in L^2(\integers)$ denotes the inverse Fourier transform of $\Phi_n.$ In particular, (\ref{choiceofphin}) means that $\|\phi_n\|\stackrel{n\rightarrow\infty}{\longrightarrow} 1$, since (again by Plancherel theorem) $\|\widecheck{\Phi_n}\|_2 = \|\Phi_n\|_2 = 1$ for every $n\in\naturals.$

Finally, we have
\begin{equation*}
\begin{split}
\|\Psi_f\| &\geqslant \limsup_{n\rightarrow \infty}\ \frac{\|\Psi_f(\phi_n)\|_2}{\|\phi_n\|_2} = \limsup_{n\rightarrow\infty}\ \|f\star\phi_n\|_2 = \limsup_{n\rightarrow\infty}\ \|\widehat{f\star\phi_n}\|_2= \limsup_{n\rightarrow\infty}\ \|\widehat{f}\cdot\widehat{\phi_n}\|_2 \\
&\geqslant \limsup_{n\rightarrow\infty} \bigg( \|\widehat{f}\cdot\Phi_n\|_2 - \|\widehat{f}\cdot (\widehat{\phi_n} - \Phi_n)\|_2\bigg) \geqslant \limsup_{n\rightarrow\infty} \bigg( \|\widehat{f}\cdot\Phi_n\|_2 - \|\widehat{f}\|_{\infty} \cdot \|\widehat{\phi_n} - \Phi_n\|_2\bigg) \\
&= \limsup_{n\rightarrow\infty} \bigg( \|\widehat{f}\cdot\Phi_n\|_2 - \|\widehat{f}\|_{\infty} \cdot \|\phi_n - \widecheck{\Phi_n}\|_2\bigg) \stackrel{(\ref{choiceofphin})}{=} \limsup_{n\rightarrow\infty}\  \|\widehat{f}\cdot\Phi_n\|_2 \stackrel{(\ref{applicationofHospital})}{\geqslant} |\widehat{f}(\chi_{\alpha_*})|.
\end{split}
\end{equation*}

\noindent
As the choice of $\alpha_*\in[0,1)$ (equivalently $\chi_{\alpha_*}\in\widehat{\integers}$) was arbitrary, we are done.
\end{proof}

\begin{cor}
There exists a function $g\in L^1(\integers)$ such that $\|\widehat{g}\|_{\infty} < \|g\|_1,$ and consequently $\|\Psi(g)\| < \|g\|_1.$
\label{Psinotisometry}
\end{cor}
\begin{proof}
We put
\begin{gather*}
g(n) := \left\{\begin{array}{cl}
1 & \text{ for }n=0,1 \\
-1 & \text{ for }n=2 \\
0 & \text{ otherwise}. \\
\end{array}\right.
\end{gather*}

\noindent
Obviously $g\in L^1(\integers)$ with $\|g\|_1 = 3.$ Subsequently, we perform the following calculation:
\begin{gather*}
\forall_{y\in[0,1]}\ |\widehat{g}(y)|^2 = \left|\sum_{n\in\integers}\ g(n)\cdot \exp(-2\pi iny)\right|^2 =  \left|1 + \exp(-2\pi iy) - \exp(-4\pi iy)\right|^2\\
= \left|1+\cos(2\pi y)-\cos(4\pi y) - i\big(\sin(2\pi y) - \sin(4\pi y)\big)\right|^2 \\
= \big(1+\cos(2\pi y)-\cos(4\pi y)\big)^2 + \big(\sin(2\pi y) - \sin(4\pi y)\big)^2\\
= 3-2\cos(4\pi y) \leqslant 5,
\end{gather*}

\noindent
where we have ommitted a number of trivial applications of trigonometric identities. The above estimate demonstrates that
$$\|\widehat{g}\|_{\infty} \leqslant \sqrt{5} < 3 = \|g\|_1,$$

\noindent
which concludes the proof.
\end{proof}

A rather sad upshot of Corollary \ref{Psinotisometry} is that on some groups (like $\integers$ as we demonstrated), the operator $\Psi$ is not an isometry. Nevertheless, we use this map to define the \textit{group $C^*-$algebra} as the norm$-$closure of $\Psi(L^1(G))$ inside the $C^*-$algebra $\BB(L^2(G)).$ Since $\Psi$ is an injective algebra $*-$homomorphism between $L^1(G)$ and $\BB(L^2(G))$ (comp. Lemma 3.3.2 in \cite{DeitmarEchterhoff}, p. 70 or Theorem 2.7.7 in \cite{Kaniuth}, p. 95) then $\Psi(L^1(G))$ is a commutative, normed $*-$subalgebra of $\BB(L^2(G)),$ which satisfies the $C^*-$property. Taking the closure simply ``completes'' $\Psi(L^1(G))$ in $\BB(L^2(G))$, turning it into a commutative $C^*-$algebra, which we denote by $C^*(G).$

\section{Arzel\`a, Ascoli and Pego}
\label{Chapter:ArzelaAscoliPego}

At last, we have arrived at the main chapter of the paper. We already have the proper ``canvas'' to work with, namely the $C^*-$algebra $C^*(G),$ where $G$ is (as always) a locally compact abelian group. As we demonstrate in the first theorem of the current chapter, $C^*(G)$ is \textit{de facto} the space of continuous functions (on the dual group $\widehat{G}$), which vanish at infinity. Therefore, employing a correct version of the Arzel\`a-Ascoli theorem we should be able to characterize relatively compact families in $C^*(G)$. This approach will bear fruit in the form of Pego theorem for $L^1(G),$ i.e. a characterization of relatively compact families in $L^1(G)$ via the Fourier transform.

\begin{thm}
There exists an isometric $*-$isomorphism $\Omega$ between $C_0(\widehat{G})$ and $C^*(G).$
\label{C0widehatGCstarG}
\end{thm}
\begin{proof}
By Theorem 3.3.3 in \cite{DeitmarEchterhoff}, p. 71 we know that the map
\begin{gather}
\Psi_{\Delta} : \Delta_{C^*(G)}\longrightarrow \Delta_{L^1(G)}\hspace{0.4cm} \text{given by}\hspace{0.4cm} \Psi_{\Delta}(m) := m\circ \Psi
\label{definitionPsiDelta}
\end{gather}

\noindent
is a homeomorphism. We easily derive an explicit formula for the inverse of $\Psi_{\Delta}:$
\begin{gather}
\forall_{m\in \Delta_{L^1(G)}}\ \Psi_{\Delta}^{-1}(m) = m\circ \Psi^{-1}.
\label{inversePsi}
\end{gather}

\noindent
By Theorem 3.2.1 in \cite{DeitmarEchterhoff}, p. 67 the map $M:\widehat{G}\longrightarrow \Delta_{L^1(G)}$ given by
\begin{equation}
M(\chi) := m_{\chi},\hspace{0.4cm} \text{where}\hspace{0.4cm} m_{\chi}(f) := \widehat{f}(\chi) \stackrel{(\ref{Fouriertransform})}{=} \int_G\ f(x)\cdot \overline{\chi(x)}\ dx,
\label{definitionM}
\end{equation}

\noindent
is a homeomorphism. Consequently, the map $M^{-1}\circ \Psi_{\Delta}$ is a homeomorphism between $\Delta_{C^*(G)}$ and $\widehat{G}.$ As in Corollary 2.2.13 in \cite{Kaniuth}, p. 57 we establish that the map
\begin{gather}
\Theta: C_0(\widehat{G})\longrightarrow C_0\left(\Delta_{C^*(G)}\right)\hspace{0.4cm} \text{given by} \hspace{0.4cm} \Theta(F) := F \circ M^{-1}\circ \Psi_{\Delta}
\label{definitionTheta}
\end{gather}

\noindent
is an isometric algebra isomorphism. Furthermore, we easily note that
$$\Theta(F^*) = F^*\circ M^{-1}\circ \Psi_{\Delta} = \overline{F\circ M^{-1}\circ \Psi_{\Delta}} = \Theta(F)^*,$$

\noindent
so $\Theta$ is in fact an isometric $*-$isomorphism. Again, we easily derive an explicit formula for the inverse of $\Theta:$
\begin{gather}
\forall_{f\in C_0\left(\Delta_{C^*(G)}\right)}\ \Theta^{-1}(f) = f\circ \Psi_{\Delta}^{-1}\circ M.
\label{inverseTheta}
\end{gather}

By the celebrated Gelfand-Naimark theorem (comp. Theorem 7.1 in \cite{BelfiDoran}, p. 22 or Theorem I.3.1 in \cite{Davidson}, p. 7 or Theorem 2.6.7 in \cite{DeitmarEchterhoff}, p. 53 or Theorem 2.4.5 in \cite{Kaniuth}, p. 69), the Gelfand transform $\Gamma: C^*(G)\longrightarrow C_0(\Delta_{C^*(G)})$ given by
$$\Gamma(f) := \widehat{f}, \hspace{0.4cm}\text{where}\hspace{0.4cm} \widehat{f}(m) := m(f),$$

\noindent
is an isometric $*-$isomorphism. In conclusion, the map $\Omega : C_0(\widehat{G})\longrightarrow C^*(G)$ given by
\begin{gather}
\Omega:= \Gamma^{-1}\circ \Theta
\label{definitionofOmega}
\end{gather}

\noindent
is an isometric $*$-isomorphism.
\end{proof}

The relation between the newly-introduced map $\Omega$ and the convolution operator $\Psi$ (the injective $*-$homomorphism constructed in Chapter \ref{Chapter:Cstaralgebra}) is pictured on the diagram below:
$$L^1(G) \stackrel{\Psi}{\longrightarrow} \Psi(L^1(G))\subset C^*(G) \stackrel{\Omega^{-1}}{\longrightarrow} C_0(\widehat{G}),$$

\noindent
Our approach to characterize relatively compact families in $L^1(G)$ is rather straightforward $-$ the key idea is that
\begin{center}
\textit{relative compactness of $\Ffamily\subset L^1(G)$ is equivalent (because $\Psi$ and $\Omega^{-1}$ are homeomorphisms) to relative compactness of $\Omega^{-1}\circ \Psi(\Ffamily)\subset C_0(\widehat{G})$.}
\end{center}

\noindent
Hence, characterizing relatively compact families in $L^1(G)$ should not be more difficult than finding the Arzel\`a-Ascoli theorem for $C_0(\widehat{G})$.

The classic version of Arzel\`a-Ascoli theorem can be found in countless sources: Theorem 4.25 in \cite{Brezis}, p. 111 or Theorem 23.2 in \cite{Choquet}, p. 99 or Theorem 6.3.1 in \cite{Dixmiertopology}, p. 69 or Theorem 4.43 in \cite{FollandRealAnalysis}, p. 137 or Corollary 10.49 in \cite{Knapp}, p. 479 or Theorem 11.28 in \cite{Rudinrealandcomplex}, p. 245 or Theorem A5 in \cite{Rudin}, p. 394 etc.

\begin{thm}(classic version of Arzel\`a-Ascoli theorem)\\
Let $X$ be a compact (Hausdorff) space. The family $\Ffamily\subset C(X)$ is relatively compact if and only if
\begin{itemize}
	\item $\Ffamily$ is pointwise bounded, i.e. for every $x\in X$ there exists $M_x>0$ such that
	$$\forall_{f\in\Ffamily}\ |f(x)| < M_x,$$
	\item $\Ffamily$ is equicontinuous at every point, i.e. for every $x\in X$ and $\eps>0$ there exists an open neighbourhood $U_x$ of $x$ such that
	$$\forall_{\substack{y\in U_x,\\ f\in\Ffamily}}\ |f(y)- f(x)| < \eps.$$
\end{itemize}
\label{classicalAA}
\end{thm}

However, in our case the domain $X = \widehat{G}$ need not be compact. In fact, if $\widehat{G}$ is compact then $G$ must be finite (comp. Proposition 4.35 in \cite{Folland}, p. 103), which we do not assume. Despite best efforts, we were unable to find a good reference for a theorem characterizing relatively compact families in $C_0(X),$ where $X$ is a locally compact group. Even worse, we came across a version of Arzel\`a-Ascoli theorem that is palpably wrong!

\begin{thm}(comp. Theorem A.1.4 in \cite{Kaniuth}, p. 320)\\
Let $X$ be a locally compact Hausdorff space and $\Ffamily\subset C_0(X).$ Suppose that $\Ffamily$ satisfies the following two conditions:
\begin{itemize}
	\item $\Ffamily$ is pointwise bounded,
	\item $\Ffamily$ is equicontinuous at every point.
\end{itemize}

\noindent
Then $\Ffamily$ is relatively compact in $(C_0(X),\|\cdot\|_{\infty})$.
\label{wrongtheorem}
\end{thm}

If this theorem were true then it would work in particular for $X = \integers$. We observe that for such a choice of $X$, the equicontinuity condition becomes obsolete $-$ every family $\Ffamily\subset C_0(\integers)$ is equicontinuous since we may always pick $U_x = \{x\}.$ Thus it suffices to consider a sequence of characteristic functions $\Ffamily = \left(\mathds{1}_{\{n\}}\right)_{n\in\naturals},$ which is obviously pointwise bounded. However, this sequence does not contain any convergent subsequence, so $\Ffamily$ cannot be relatively compact (contrary to what Theorem \ref{wrongtheorem} implies)! This demonstrates that Theorem \ref{wrongtheorem} cannot be true.

In the pursuit of the proper characterization of relatively compact families in $C_0(X)$ we came across Exercise 17 on page 182 in John B. Conway's ``A Course in Functional Analysis'' (comp. \cite{Conway}), which is likely the closest to what we need. Unfortunately, Conway does not provide the proof, leaving this task to the reader. Discouraged by the fruitless search for a proper reference, we have decided to provide a full proof of the Arzel\`a-Ascoli theorem for $C_0(X)$ ourselves. Our approach is inspired by Chapter 7 in Kelley's ``General Topology'' (comp. \cite{Kelley}) as well as Chapters 46 and 47 in Munkres' ``Topology'' (comp. \cite{Munkres}). In their monographs, Kelley and Munkres employ the technique of ``\textit{collapsing topologies}'', which we briefly summarize. There are three main topologies on $C_0(X):$
\begin{itemize}
	\item \textit{Topology of pointwise convergence} $\tau_{pc}.$ The family
	$$\{f\in C_0(X)\ :\ |f(x) - f_*(x)| < \eps\}_{x\in X, f_*\in C_0(X), \eps>0}$$

	\noindent
	forms a subbase for this topology (comp. \cite{Munkres}, p. 281).
	\item \textit{Topology of uniform convergence on compact sets} $\tau_{ucc}.$ The family
	$$\{f\in C_0(X)\ :\ \forall_{x\in K}\ |f(x)-f_*(x)| < \eps\}_{K\Subset X, f_*\in C_0(X), \eps>0}$$

	\noindent
	forms a base for this topology (comp. \cite{Munkres}, p. 283).

	\item \textit{Topology of uniform convergence} $\tau_{uc}.$ The family
	$$\{f\in C_0(X)\ :\ \forall_{x\in X}\ |f(x)-f_*(x)| < \eps\}_{f_*\in C_0(X), \eps>0}$$

	\noindent
	forms a base for this topology.
\end{itemize}

If we consider $C_0(X)$ with the topology of pointwise convergence $\tau_{pc},$ then the Arzel\`a-Ascoli theorem is equivalent to Tychonoff's theorem (comp. Theorem 1 in \cite{Kelley}, p. 218). The crucial observation is that if $\Ffamily\subset C_0(X)$ is equicontinuous, then $\tau_{pc}$ and $\tau_{ucc}$ coincide on $\Ffamily$ (comp. Theorem 15 in \cite{Kelley}, p. 232). We say, rather informally, that the topologies ``collapse onto each other'' on $\Ffamily$. This line of reasoning lies at the heart of Arzel\`a-Ascoli theorem for $C_0(X)$ with the topology $\tau_{ucc}.$

At this point it is at least plausible to believe that there exists some condition on the family $\Ffamily\subset C_0(X)$ under which the topologies $\tau_{ucc}$ and $\tau_{uc}$ coincide. We simply need to work out what this condition might be.

\begin{lemma}
Let $X$ be a locally compact space. If the family $\Ffamily\subset C_0(X)$ is equivanishing, i.e.
\begin{gather}
\forall_{\eps>0}\ \exists_{K\Subset X}\ \forall_{f\in\Ffamily}\ \sup_{x\in X\backslash K}\ |f(x)| < \eps,
\label{assumptionequivanishing}
\end{gather}

\noindent
then the topologies $\tau_{ucc}$ and $\tau_{uc}$ coincide on $\Ffamily,$ i.e. $\tau_{co}|_{\Ffamily} = \tau_{uc}|_{\Ffamily}.$
\label{equivanishingandtopologycollapse}
\end{lemma}
\begin{proof}
By Theorem 46.7 in \cite{Munkres}, p. 285 we know that $\tau_{ucc}\subset \tau_{uc}$ so we only need to prove the reverse inclusion holds on $\Ffamily$. We fix $f_*\in C_0(X),\ \eps>0$ and define a $\tau_{uc}-$open set
$$U_{f_*,\eps} := \big\{f\in\Ffamily\ :\ \forall_{x\in X}\ |f(x) - f_*(x)| < \eps\big\}.$$

\noindent
Our task is as follows: for every $g\in U_{f_*,\eps}$ we need to define $\tau_{ucc}-$open set
$$V_{K, g, \delta} := \big\{f\in\Ffamily\ :\ \forall_{x\in K}\ |f(x)-g(x)| < \delta\big\}$$

\noindent
such that $V_{K, g, \delta}\subset U_{f_*,\eps}.$

We fix $g\in U_{f_*,\eps}$ and define
\begin{gather}
d:= \|g-f_*\|_{\infty}.
\label{wedefined}
\end{gather}

\noindent
We pick $\delta>0$ such that
\begin{gather}
3\delta + d < \eps.
\label{choiceofdelta}
\end{gather}

\noindent
By (\ref{assumptionequivanishing}) there exists $K\Subset X$ such that
\begin{gather}
\forall_{f\in\Ffamily}\ \sup_{x\in X\backslash K}\ |f(x)| < \delta.
\label{wechooseK}
\end{gather}

\noindent
Finally, if $f\in V_{K,g,\delta}$ then
\begin{gather*}
\|f-f_*\|_{\infty} \leqslant \|f-g\|_{\infty} + \|g-f_*\|_{\infty} \stackrel{(\ref{wedefined})}{\leqslant} \sup_{x\in K}\ |f(x)-g(x)| + \sup_{x\in X\backslash K}\ |f(x)- g(x)|+ d \\
\leqslant \delta + \sup_{x\in X\backslash K}\ |f(x)| + \sup_{x\in X\backslash K}\ |g(x)| + d \stackrel{(\ref{wechooseK})}{\leqslant} 3\delta + d \stackrel{(\ref{choiceofdelta})}{<} \eps,
\end{gather*}

\noindent
which concludes the proof.
\end{proof}

With Lemma \ref{equivanishingandtopologycollapse} at our disposal we are ready to characterize relatively compact families in $C_0(X).$

\begin{thm}(Arzel\`a-Ascoli theorem for $C_0(X)$)\\
Let $X$ be a locally compact space. A family $\Ffamily\subset C_0(X)$ is relatively compact with respect to $\|\cdot\|_{\infty}-$norm (equivalently $\tau_{uc}-$topology) if and only if
\begin{description}
	\item[\hspace{0.4cm} (AA1)] $\Ffamily$ is pointwise bounded,
	\item[\hspace{0.4cm} (AA2)] $\Ffamily$ is equicontinuous at every point,
	\item[\hspace{0.4cm} (AA3)] $\Ffamily$ is equivanishing.
\end{description}
\label{AAforC0X}
\end{thm}
\begin{proof}
For the ``\textit{if}'' part, we observe that if $\Ffamily\subset C_0(X)$ is both pointwise bounded and equicontinuous at every point then it is relatively $\tau_{ucc}-$compact (comp. Theorem 17 in \cite{Kelley}, p. 233 or Theorem 47.1 in \cite{Munkres}, p. 290). Moreover, since $\tau_{ucc}$ and $\tau_{uc}$ coincide on $\Ffamily$ (due to condition (\textbf{AA3}) and Lemma \ref{equivanishingandtopologycollapse}) then $\Ffamily$ is also relatively $\tau_{uc}-$compact.

As far as the ``\textit{only if}'' part is concerned, Theorem 17 in \cite{Kelley}, p. 233 (or Theorem 47.1 in \cite{Munkres}, p. 290) demonstrates that a relatively $\tau_{uc}-$compact family $\Ffamily\subset C_0(X)$ is both pointwise bounded and equicontinuous at every point. Consequently, it remains to prove that $\Ffamily$ is equivanishing. We fix $\eps>0$ and (due to relative $\tau_{uc}-$compactness) we pick an $\eps-$net $(f_n)_{n=1}^N,$ i.e. a finite sequence of functions in $\Ffamily$ such that
\begin{gather}
\forall_{f\in\Ffamily}\ \exists_{n=1,\ldots,N}\ \|f-f_n\|_{\infty} < \eps.
\label{choiceofepsnet}
\end{gather}

\noindent
Next, we let $K\Subset G$ be such that
\begin{gather}
\forall_{n=1,\ldots,N}\ \sup_{x\in X\backslash K}\ |f_n(x)| < \eps.
\label{choiceofKcompact}
\end{gather}

\noindent
Finally, we have
\begin{gather*}
\forall_{\substack{n=1,\ldots,N\\ f\in\Ffamily}}\ \sup_{x\in X\backslash K}\ |f(x)| \leqslant \sup_{x\in X\backslash K}\ |f(x) - f_n(x)| + \sup_{x\in X\backslash K}\ |f_n(x)| \stackrel{(\ref{choiceofKcompact})}{<} \|f-f_n\|_{\infty} + \eps.
\end{gather*}

\noindent
Due to (\ref{choiceofepsnet}) and the arbitrary choice of $\eps>0$ we conclude that $\Ffamily$ is equivanishing, which ends the proof.
\end{proof}

Before we conclude the paper with Pego theorem for $L^1(G)$ let us introduce one final improvement to Theorem \ref{AAforC0X}. The idea for this innovation dates back to the work of Vladimir N. Sudakov $-$ a modern exposition of his techniques can be found in \cite{GorkaRafeiro} or \cite{HancheOlsenHoldenMalinnikova}.

We need the following technical lemma:

\begin{lemma}
Let $X$ be a locally compact (not necessarily abelian) group and let $K\Subset X.$ If a family $\Ffamily \subset C_0(X)$ is equicontinuous at every point, then for every $\eps>0$ there exists an open neighbourhood $V_e$ of the neutral element $e\in X$ such that
$$\forall_{\substack{y\in V_e\\ f\in\Ffamily}}\ \sup_{x\in K}\ |f(yx) - f(x)| < \eps.$$
\label{technicallemmasudakov}
\end{lemma}
\begin{proof}
Fix $\eps>0.$ For every $x\in K$ we choose an open neighbourhood $U_x$ of the neutral element (note that $U_x$ need not be a neighbourhood of $x$) such that
\begin{gather}
\forall_{\substack{y\in U_xx\\ f\in\Ffamily}}\ |f(y)- f(x)| < \eps,
\label{equicontSudakov}
\end{gather}

\noindent
according to the definition of equicontinuity in Theorem \ref{classicalAA}. Furthermore, for every $x\in K$ we let $V_x$ be an open neighbourhood of the identity element such that $V_x^2\subset U_x.$ The family $(V_xx)_{x\in K}$ is an open cover of $K$ so (by the compactness of $K$) we choose a finite subcover $(V_{x_n}x_n)_{n=1}^N.$ Let $V_e$ be defined by the formula
\begin{gather}
V_e := \bigcap_{n=1}^N\ V_{x_n}.
\label{definitionofVeSudakov}
\end{gather}

For every $x\in K$ let $ind(x)\in \{1,\ldots,N\}$ denote an index such that $x\in V_{x_{ind(x)}}x_{ind(x)}.$ Consequently, the fact that
$$\forall_{x\in K}\ x\in V_{x_{ind(x)}}x_{ind(x)} \subset U_{x_{ind(x)}}x_{ind(x)}$$

\noindent
coupled with (\ref{equicontSudakov}) implies that
\begin{gather}
\forall_{\substack{x\in K\\ f\in\Ffamily}}\ |f(x) - f(x_{ind(x)})| < \eps.
\label{ineq1Sudakov}
\end{gather}

\noindent
Furthermore, since
$$\forall_{\substack{y\in V_e\\ x\in K}}\ yx \in V_eV_{x_{ind(x)}}x_{ind(x)} \stackrel{(\ref{definitionofVeSudakov})}{\subset} V_{x_{ind(x)}}^2x_{ind(x)}\subset U_{x_{ind(x)}}x_{ind(x)},$$

\noindent
then (again using (\ref{equicontSudakov})) we have
\begin{gather}
\forall_{\substack{y\in V_e\\ x\in K\\ f\in\Ffamily}}\ |f(yx) - f(x_{ind(x)})| < \eps.
\label{ineq2Sudakov}
\end{gather}

\noindent
Finally, we have
\begin{gather*}
\forall_{\substack{y\in V_e\\ f\in\Ffamily}}\ \sup_{x\in K}\ |f(yx) - f(x)| \leqslant \sup_{x\in K}\ |f(yx) - f(x_{ind(x)})| + \sup_{x\in K}\ |f(x_{ind(x)}) - f(x)| \stackrel{(\ref{ineq1Sudakov}),\ (\ref{ineq2Sudakov})}{\leqslant} 2\eps,
\end{gather*}

\noindent
which concludes the proof.
\end{proof}

Our next theorem demonstrates that under certain circumstances, the assumption (\textbf{AA1}) in Theorem \ref{AAforC0X} becomes redundant.

\begin{thm}
Let $X$ be a locally compact (not necessarily abelian) group such that for any open neighbourhood $U_e$ of the neutral element $e\in X$ there exists an element $x_*\in U_e$ such that the sequence $(x_*^n)_{n\in\naturals}$ is not contained in any compact set. If a family $\Ffamily\subset C_0(X)$ is equicontinuous at every point and equivanishing then it is pointwise bounded, i.e. (\textbf{AA2}) and (\textbf{AA3}) together imply (\textbf{AA1}).
\label{AASudakov}
\end{thm}
\begin{proof}
Let $\Ffamily\subset C_0(X)$ be a family, which is both equicontinuous at every point and equivanishing. By Lemma \ref{technicallemmasudakov} there exists an open neighbourhood $V_e$ of the neutral element such that
\begin{gather}
\forall_{\substack{y\in V_e\\ f\in\Ffamily}}\ \sup_{x\in K}\ |f(yx) - f(x)| < 1.
\label{choiceofUe}
\end{gather}

\noindent
Furthermore there exists $K\Subset X$ such that
\begin{gather}
\sup_{\substack{x\in X\backslash K\\ f\in\Ffamily}}\ |f(x)| < 1.
\label{equivanish1}
\end{gather}

For the set $V_e$ we pick $x_*\in V_e$ such that the sequence $(x_*^n)_{n\in\naturals}$ is not contained in any compact set. Since we have
\begin{gather*}
\forall_{f\in\Ffamily}\ \sup_{x\in K}\ |f(x)| \leqslant \sup_{x\in K}\ |f(x_*x) - f(x)| + \sup_{x\in K}\ |f(x_*x)| \stackrel{(\ref{choiceofUe})}{<} 1 + \sup_{x\in K}\ |f(x_*x)| = 1 + \sup_{x\in x_*K}\ |f(x)|,
\end{gather*}

\noindent
then by an inductive reasoning we obtain that
\begin{gather}
\forall_{\substack{f\in\Ffamily\\ n\in\naturals}}\ \sup_{x\in K}\ |f(x)| \leqslant n + \sup_{x\in x_*^nK}\ |f(x)|.
\label{byinductivereasoning}
\end{gather}

\noindent
It remains to note that there exists $N\in\naturals$ such that $x_*^NK\cap K = \emptyset$ since otherwise this would lead to $(x_n)\subset KK^{-1},$ which contradicts our assumption. Finally, we have
$$\forall_{f\in\Ffamily}\ \|f\|_{\infty} \leqslant \sup_{x\in K}\ |f(x)| + \sup_{x\in X\backslash K}\ |f(x)|\stackrel{(\ref{byinductivereasoning}), (\ref{equivanish1})}{<} N + \sup_{x\in x_*^NK}\ |f(x)| + 1 \leqslant N+2,$$

\noindent
which concludes the proof.
\end{proof}

\begin{cor}
A family $\Ffamily\subset C_0(\reals^N)$ is relatively compact (with respect to $\|\cdot\|_{\infty}-$norm) if and only if
\begin{itemize}
	\item $\Ffamily$ is equicontinuous at every point,
	\item $\Ffamily$ is equivanishing.
\end{itemize}
\end{cor}

We have finally reached the climax of the paper $-$ the characterization of relatively compact families in $L^1(G)$ via the Fourier transform. The theorem below serves as $L^1(G)-$counterpart of the classic Pego theorem demonstrated in \cite{Pego} and generalized in \cite{Gorka} and \cite{GorkaKostrzewa}.

\begin{thm}(Pego theorem for $L^1(G)$)\\
A family $\Ffamily\subset L^1(G)$ is relatively compact if and only if
\begin{description}
	\item[\hspace{0.4cm} (P1)] $\widehat{\Ffamily}$ is bounded in $C_0(\widehat{G}),$
	\item[\hspace{0.4cm} (P2)] $\widehat{\Ffamily}$ is equicontinuous at every point (character), i.e. for every $\chi\in\widehat{G}$ and $\eps>0$ there exists an open neighbourhood $U_{\chi}\in\widehat{G}$ of $\chi$ such that
	$$\forall_{\substack{\eta\in U_{\chi}\\ f\in\Ffamily}}\ |\widehat{f}(\eta) - \widehat{f}(\chi)| < \eps,$$
	\item[\hspace{0.4cm} (P3)] $\widehat{\Ffamily}$ is equivanishing, i.e. for every $\eps>0$ there exists $K\Subset \widehat{G}$ such that
	$$\sup_{\chi\in\widehat{G}\backslash K}\ |\widehat{f}(\chi)| < \eps.$$
\end{description}

Furthermore, if $\widehat{G}$ is such that for any open neighbourhood $U_{\mathds{1}}$of the identity $\mathds{1}$ there exists an element $\chi_*\in U_{\mathds{1}}$ such that the sequence $(\chi_*^n)_{n\in\naturals}$ is not contained in any compact set, then the condition (\textbf{P1}) is redundant.
\label{Pegotheorem}
\end{thm}
\begin{proof}
Let $f\in\Ffamily$. Then
\begin{gather*}
\Omega^{-1}\circ\Psi(f) \stackrel{(\ref{definitionofOmega})}{=} (\Gamma^{-1}\circ\Theta)^{-1}(\Psi_f) = \Theta^{-1}\circ\Gamma(\Psi_f) = \Theta^{-1}(\widehat{\Psi_f}) \stackrel{(\ref{inverseTheta})}{=} \widehat{\Psi_f}\circ \Psi_{\Delta}^{-1}\circ M.
\end{gather*}

\noindent
Furthermore, we have
\begin{gather*}
\forall_{\chi\in\widehat{G}}\ \widehat{\Psi_f}\circ\Psi_{\Delta}^{-1}\circ M(\chi) \stackrel{(\ref{definitionM})}{=} \widehat{\Psi_f}\circ\Psi_{\Delta}^{-1}(m_{\chi}) \stackrel{(\ref{inversePsi})}{=} \widehat{\Psi_f}(m_{\chi}\circ\Psi^{-1})\\
= m_{\chi}\circ\Psi^{-1}(\Psi_f) = m_{\chi}(f) = \widehat{f}(\chi),
\end{gather*}

\noindent
which leads to the conclusion that $\Omega^{-1}\circ\Psi(\Ffamily) = \widehat{\Ffamily}.$ By Theorem \ref{AAforC0X} the family $\Omega^{-1}\circ \Psi(\Ffamily)$ (equivalently $\Ffamily$) is relatively compact in $C_0(\widehat{G})$ (respectively in $L^1(G)$) if and only if \textbf{(P1)}, \textbf{(P2)} and \textbf{(P3)} are satisfied. This demonstrates the first part of the theorem. The second part easily follows from Theorem \ref{AASudakov}.
\end{proof}

\begin{cor}
A family $\Ffamily\subset L^1(\reals^N)$ is relatively compact if and only if
\begin{itemize}
	\item $\widehat{\Ffamily}$ is equicontinuous,
	\item $\widehat{\Ffamily}$ is equivanishing.
\end{itemize}
\end{cor}

\section*{Acknowledgements}

config: arxiv


TITLE: Find the point where $\tau(t)$ is minimal.
QUESTION [0 upvotes]: Question.
$\vec{r}(t)=\left[\frac{4}{9}(1+t)^{3/2},\frac{4}{9}(1-t)^{3/2},\frac{1}{3}t\right]$ is a parametrization of a curve $C$ ($t\in\left[-1,1\right]$)
find the point where $\tau(t)$ is minimal.
Attempt.
$$
\tau(t) =\frac{(\vec{r}\,'(t)\times\vec{r}\,''(t))\cdot\vec{r}'''(t)}{||\vec{r}\,'(t)\times\vec{r}\,''(t)||^{2}}=\frac{\left[-\frac{1}{9}(1-t)^{-1/2},\frac{1}{9}(1+t)^{-1/2},\frac{4}{9}(1-t^{2})^{-1/2}\right]\cdot\left[-\frac{1}{6}(1+t)^{-3/2},\frac{1}{6}(1-t)^{-3/2},0\right]}{\left(\frac{\sqrt{2}}{3}(1-t^{2})^{-1/2}\right)^{2}}
 =\frac{\frac{1}{54}(1-t)^{-1/2}(1+t)^{-3/2}+\frac{1}{54}(1+t)^{-1/2}(1-t)^{-3/2}}{\frac{2}{9}(1-t^{2})^{-1}}
 =\frac{1}{12}(1-t^{2})\left((1-t)^{-1/2}(1+t)^{-3/2}+(1+t)^{-1/2}(1-t)^{-3/2}\right)
 =\frac{1}{12}(1-t)(1+t)\left((1-t)^{-1/2}(1+t)^{-3/2}+(1+t)^{-1/2}(1-t)^{-3/2}\right)
 =\frac{1}{12}\left((1-t)^{1/2}(1+t)^{-1/2}+(1+t)^{1/2}(1-t)^{-1/2}\right)
 =\frac{1}{12}\left(\frac{\sqrt{1-t}}{\sqrt{1+t}}+\frac{\sqrt{1+t}}{\sqrt{1-t}}\right)=\frac{1}{12}\left(\frac{1-t+1+t}{\sqrt{1-t^{2}}}\right)=\frac{1}{6}(1-t^{2})^{-1/2}
$$
But I think my calculation is wrong, can you take a look?

REPLY [0 votes]: Your calculations are correct, as also checked by @ mathametics2X2life.
The $\pm$ plus and minus sign of parameterization admits both signs $(>0,<0) $
$$\vec{r_1}(t)=(+\frac49 (1+t)^{3/2},+\frac49 (1-t)^{3/2},\frac{1}{3}t)$$
$$\vec{r_2}(t)=(-\frac49 (1+t)^{3/2},-\frac49 (1-t)^{3/2},\frac{1}{3}t)$$
and acordingly we can have torsion $\tau$  at $t=0$ both as a  minimum and as a maximum respectively. A rough plot of torsions:


TITLE: how many ways can 5 prizes be given to a group of 20 people with restrictions
QUESTION [1 upvotes]: Sorry if this is super simple, I've just been stuck on this concept for way too long.
Each person can only win one prize.
How many ways can 5 prizes be handed out to a group of 20 people if:
(a)  all 5 prizes are identical, also Alex and Sally can't both get prizes.
(b)  the 5 prizes are ranked 1st-5th, also Alex and Sally can't both get prizes.
So I think I've tried to solve this but I think I've made a mistake.
New Attempt
a)
Now I'm trying to find the total number of ways that the 5 prizes could be handed out to 20 people.  I think this should be:
$$ {20 \choose 5} $$
now we need to find the situations in which both Sally and Alex win prizes and subtract that from the total.  I think this is wrong but it might just be:
$$  {18 \choose 3}  $$
So the final value would be:
$$ {20 \choose 5}   -  {18 \choose 3}  $$
b)
I think the trick here is still to multiply the answer from part (a) by '5!'.
$$ 5! ({20 \choose 5}   -  {18 \choose 3})  $$

REPLY [1 votes]: For a.  possibilities are neither Alex or Sally get a prize $\binom{18}{5}$ or one of them gets a prize and the other doesn't $2\binom{18}{4}$, so total is $\binom{18}{5}+2\binom{18}{4}$.
For b. Use a. result and multiply by permutations - net $5!\times(\binom{18}{5}+2\binom{18}{4})$.


TITLE: How to understand the Harish-Chandra isomorphism?
QUESTION [24 upvotes]: The Harish-Chandra isomorphism describe the center $Z(\mathfrak{g})$ of $U(\mathfrak{g})$ as invariants of $\text{Sym}^*\mathfrak{h}$ under the action of the Weyl group. (One need to twist the action or make a change of coordinate on the affine space $\mathfrak{h}^*$ for this isomorphism work.)
My question is, how does one understand this isomorphism, and what's the geometric context of it? Why should one expect something like this might be true?

REPLY [6 votes]: Depending what one means by "intuition", the computation of the behavior of the center $\mathfrak z$ of $U\mathfrak g$ on unramified principal series repns not only suggests the form of the Harish-Chandra isomorphism, but also suggests the $W$-invariance, since the intertwinings (generically isomorphisms) among principal series are exactly given by Weyl group elements (with the shift by the half-sum $\rho$ of positive roots as a result of a change-of-measure at the group level).


TITLE: Continuous function and IVT.
QUESTION [1 upvotes]: Let $f:[0,1]->\mathbb R$ be a continuous function which satisfies $f(0)=f(1)$. Prove that there exists a number $a \in [0,1/2]$ such that $f(a)=f(a+1/2)$.
I know that i'll use IVT theorem but i confused a bit.Can you help me ?

REPLY [2 votes]: Consider the function $g: [0, \frac{1}{2}] \to \mathbb{R}$ given by
$$
g(x) = f(x) - f\left(x + \frac{1}{2}\right).
$$
Then $g$ is continuous, and $g(0) = -g\left(\frac{1}{2}\right)$. It follows from the intermediate value theorem that $g$ has a zero somewhere.


TITLE: I want the pullback of a non-closed 1-form to be closed. Is that possible?
QUESTION [0 upvotes]: Say $\omega$ is any old meromorphic 1-form on a Riemann surface X which has residue 0 at each pole. Apparently it is possible to find a covering map $p: Y \to X$ (where $Y$ is another Riemann surface) such that $p^*\omega$ (i.e. the pullback of $\omega$) actually has a meromorphic primitive! And that is what I seek to prove. I have a feeling that the statement that a closed differential form has a primitive if and only if all of its periods are zero may be useful in trying to prove this claim...but if I want to use this statement, I need a closed 1-form, so I thought I might need to find a covering map $p: Y \to X$ such that $p^*\omega$ is closed, and then go ahead and try to use this statement.But am I just trying in vain? Is it even possible? I know that pullbacks of closed forms are closed but is the converse (i.e. if the pullback of a form is closed then the form is closed) also unfortunately true? More generally, am I proceeding in the right way? Any thoughts would be appreciated.

REPLY [3 votes]: This answer is just an expansion of my comments.
Every holomorphic $1$-form on a complex curve is closed. As being closed is a local property, this automatically means that every meromorphic $1$-form is closed as well. So, to begin with, your $\omega$ is already closed.
Now arrises the question: under what circumstances can we tell that a closed meromorphic $1$-form is exact?
The answer is that there are two kinds of obstructions preventing a closed meromorphic $1$-form $\omega$ from being exact, as follows.
a) The integral of $\omega$ along a curve around each pole has to vanish.
b) The integral of $\omega$ along any curve which represents a non-trivial class in the fundamental group of the surface has to vanish.
Now, by assumption, your $\omega$ satisfies condition a), and so does any pullback $p^*\omega$, where $p$ is a covering map. So, you only need to do something to make condition b) hold. But the universal cover of your surface is simply connected, which means that condition b) automatically holds on it. This means that the pullback of $\omega$ to the universal cover indeed has a meromorphic primitive.


TITLE: Find a maximal ideal in $F_7[x]/(x^2+5)$
QUESTION [0 upvotes]: $F_p=Z/(pZ)$
I'm stuck at this problem. I tried to find the irreducible factorisation of $x^2+5$, but that didn't help too much. Any ideas? and also could anyone explain the general philosophy with regard to solving this kind of problem?

REPLY [0 votes]: sorry, but the further explanation in the comments would actually give the solution, so I just write it here.]
We know for that $\mathbb{F}_7[x]/(x^2+5)$ some things:
1.$(x^2+5)=(x^2-9)=(x-3)(x+3)$
2.$\mathbb{F}_7[x]/(x^2+5)\cong \mathbb{F}_7 \oplus [x]\mathbb{F}_7$ as vectorspaces.
3.The above isomorphism makes $\mathbb{F}_7[x]/(x^2+5)$ into a $\mathbb{F}_7$ algebra.
Now we get that any ideal is a sub-$\mathbb{F}_7$-vectorspace of $\mathbb{F}_7[x]/(x^2+5)$ not containing $1$ in particular it cannot contain any element of $\mathbb{F}_7 \hookrightarrow \mathbb{F}_7[x]/(x^2+5)$. Now for dimensionreasons this means that our maximal ideal has maximal dimension one, which means that any nontrivial ideal is already maximal, so it suffices to find a nontrivial nonunit element and take the ideal generated by it. So a nontrivial zero-devisor has to do the trick of which we clearly already have 2 obvious possible choices: $(x-3)$ and $(x+3)$.
And we are done, actually no proper algebra involved, only linear algebra.


TITLE: Proving the sharper form of the Lebesgue Differentiation Theorem
QUESTION [0 upvotes]: I'm following the book- 'Fourier Analysis' by Javier Duoandikoetxea. In the chapter named- 'The Hardy-Littlewood Maximal Function', the Lebesgue Differentiation theorem was proved as a corollary to the weak $(1,1)$ inequality for the maximal function. The proved theorem states-

If $f\in L^1_{\text{loc}}(\mathbb{R}^n)$ then
  $$ \lim_{x\rightarrow 0^+}\frac{1}{|B_r|}\int_{B_r}f(x-y)dy=f(x) \text{  a.e}$$

The book says that the corollary can be made sharper, i.e.

If $f\in L^1_{\text{loc}}(\mathbb{R}^n)$ then
  $$ \lim_{x\rightarrow 0^+}\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy=0 \text{  a.e}$$

The given hint states that
$$\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy\leq Mf(x) + |f(x)|$$
where $M$ denotes the Hardy Littlewood maximal function, and $|B_r|$ denotes the Lebesgue measure of the ball of radius $r$ centered at the origin. I am unable to understand how to prove the sharper theorem using the hint.
I have been trying to find another function and use the corollary on that function to get the above result but have not succeeded. A proof which I have read earlier approximates $f(x)$ by rationals and uses the countability of rationals along with using the corollary on the function $|f(x-y)-r|$ to obtain the result. However, this does not use the hint. Could someone give another hint on how to proceed using the given hint?

REPLY [3 votes]: There's nothing to it. Define $$M'f(x)=\sup_{r>0}\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy.$$
Since $M'f(x)\le Mf(x)+|f(x)|$ it follows that $$m(\{M'f(x)>\lambda\})
\le\frac c\lambda||f||_1.$$
(Hint for that: $a+b>\lambda$ implies either $a>\lambda/2$ or $b>\lambda/2$.)
And you certainly have $$\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy\to0$$almost  everywhere for $f$ is a dense subset of $L^1$; now exactly the proof of the differentiation theorem from the H-L maximal theorem works.
OK, here's the proof. Define $$\omega f(x) =  \limsup_{r\to0}\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy.$$We need to show that $\omega f=0$ almost everywhere. So  it's enough to show that $$m\{\omega f>\lambda\}=0$$for every $f\in L^1$ and $\lambda>0$.
Note that $$\omega(f+g)\le\omega f+\omega g,$$also $$\omega f\le M'f.$$
Say $f\in L^1$ and $\epsilon>0$. Let $g\in C_c(\Bbb R^n)$ with $||f-g||_1<\epsilon$. Then $\omega g=0$, so $$\omega f\le\omega g+\omega(f-g)
=\omega(f-g)\le M'(f-g),$$so $$m\{\omega f>\lambda\}\le\frac c\lambda||f-g||_1<\frac c\lambda\epsilon.$$