%From auela@reed.edu Mon Nov 12 23:23:08 2001
%Date: Mon, 29 Oct 2001 03:20:27 -0800 (PST)
%From: Asher Natan Auel <auela@reed.edu>
%To: asher@reed.edu
%Subject: tgnotes1.tex

\documentclass[12pt]{book}
\usepackage{amsthm}
\theoremstyle{definition}
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\newtheorem{defin}[theorem]{Definition}
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\newtheorem{corollary}[theorem]{Corollary}
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\newtheorem{remark}[theorem]{Remark}
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\newcommand{\osc}{{\rm osc}}
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\begin{document}

\section{Topological Groups}

\begin{defin}
A set $G$ is a {\it topological group} if $G$ is a group, $G$ is a
topological space, and the group operations in $G$ are continuous
in the topological space $G$.  Equivalently, for all $a,b \in G$,
and for every neighborhood $W$ of $ab^{-1}$ there exists
neighborhoods $U$ and $V$ of $a$ and $b$ respectively, such that
$UV^{-1} \subset W$.
\end{defin}

\begin{remark}
Let $G$ be a topological group, and let $f:G \to \R$.  Then $f$ is
continuous at $a \in G$ if and only if for every $\epsilon > 0$
there exists a neighborhood $U$ of $a$ such that
$$x \in U \Rightarrow |f(x)-f(a)| < \epsilon.$$
Also, $f$ is continuous at $a \in G$ if and only if for all
$\epsilon > 0$ there exists a neighborhood $U$ of the identity
such that for all $x \in G$,
$$x\,a^{-1} \in U \Rightarrow |f(x)-f(a)| < \epsilon.$$
\end{remark}

\begin{defin}
Let $G$ be a topological group, let $M \subset G$, and let $f:M
\to \R$.  Then f is {\it uniformly continuous} if for every
$\epsilon > 0$ there exists a neighborhood $U$ of the identity of
$G$ such that for all $x,y \in M$,
$$x\,y^{-1} \in U \Rightarrow |f(x)-f(y)| < \epsilon.$$
\end{defin}

\begin{theorem}
Let $G$ be a topological group, let $M \subset G$ be compact, and
let $f:G \to \R$ be continuous. Then $f$ is uniformly continuous.
(in both senses...)
\end{theorem}

\begin{defin}
Let $G$ be a topological group, let $M \subset G$, and let
$\Delta$ be a set of function defined on $M$. Then $\Delta$ is
{\it uniformly equicontinuous} if for every $\epsilon > 0$ there
exists a neighborhood $U$ of the identity of $G$ such that for all
$x,y \in M$ and for all $f \in \Delta$,
$$x\,y^{-1} \in U \Rightarrow |f(x)-f(y)| < \epsilon.$$
\smallskip
Also, $\Delta$ is {\it uniformly bounded}  if there exists a $B
\in \R$ such that for all $x \in M$ and for all $f \in \Delta$ we
have $|f(x)| < B$.
\end{defin}

\begin{theorem}[Generalization of Arzel$\grave{{\bf a}}$'s Theorem]
Let $G$ be a topological group, let $M \subset G$ be compact, and
let $\Delta$ be a uniformly bounded and uniformly equicontinuous
set of real function defined on $M$.  Then any sequence of
functions in $\Delta$ has a uniformly convergent subsequence.
\end{theorem}

\section{Invariant Integration}

\begin{theorem}
Let $G$ be a compact topological group.  Then there is a unique
mapping $f \mapsto \int{f(x)dx}$ from $C(G)$ to $\R$ satisfying
the following conditions:
\begin{enumerate}
\item For all $f \in C(G)$ and for all $\alpha \in \R$,
$$\int{\alpha f(x)\,dx} = \alpha \int{f(x)\,dx}.$$
\item For all $f,g \in C(G)$,
$$\int{(f(x)+g(x))\,dx} = \int{f(x)\,dx} + \int{g(x)\,dx}.$$
\item For all $f \in C(G)$,
$$ f(x) \geq 0 {\rm ~for~all}~x \in G \Rightarrow \int{f(x)\,dx}
\geq 0.$$
\item If $f(x)=1$ for all $x \in G$ then $\int{f(x)\,dx} = 1$.
\item For all $f \in C(G)$ and for all $a \in G$,
$$\int{f(xa)\,dx} = \int{f(x)\,dx}.$$
\item For all $f \in C(G)$ and for all $a \in G$,
$$\int{f(ax)\,dx} = \int{f(x)\,dx}.$$
\item For all $f \in C(G)$,
$$\int{f(x^{-1})\,dx} = \int{f(x)\,dx}.$$
\end{enumerate}
Furthermore, if there is on $G$ any mapping satisfying conditions
1-5, then the remaining conditions hold.  The unique
$\int{f(x)dx}$ will be called the {\it integral} of $f$.
\end{theorem}

The proof of this theorem proceeds in a number of steps.
Throughout, let $G$ be a topological group.

\begin{remark}
Conditions 1-3 are natural for any conception of an integral, and
permit the integration of inequalities and make it possible to
obtain the usual estimate concerning the integral of absolute
values.  Namely, let $f,g \in C(G)$, then
$$f(x) \leq g(x) {\rm ~for~all}~x \in G \Rightarrow \int{ f(x)\,dx } \leq
\int{ g(x)\,dx },$$
$$\left| \int{ f(x)\,dx } \right| \leq \int{ |f(x)|\,dx }.$$
\end{remark}

\begin{proof}
Indeed, we see that for all $x \in G$, $f(x) \leq g(x) \Rightarrow
g(x)-f(x) \geq 0$ thus by condition 3 we have $\int{
(g(x)-f(x))\,dx } \geq 0$, and by conditions 1 and 2 we have
$\int{ g(x)\,dx } - \int{ f(x)\,dx } \geq 0$, i.e.
$$\int{ f(x)\,dx } \leq \int{ g(x)\,dx }.$$

Moreover, for all $x \in G$ we have $-|f(x)| \leq f(x) \leq
|f(x)|$, and from what we have just shown we have $-\int{
|f(x)|\,dx } \leq \int{ f(x)\,dx } \leq \int{ |f(x)|\,dx }$, in
other words,
$$\left| \int{ f(x)\,dx } \right| \leq \int{ |f(x)|\,dx }.$$
\end{proof}

\begin{defin}
Let $f \in C(G)$ and let $A = \{a_1,a_2,\ldots,a_n\} \subset G$.
Then define $M(A,f):G \to \R$ given by
$$M(A,f;x) = \frac{1}{n}\sum_{i=1}^{n}{ f(xa_i) } \quad ~{\rm for~all}~x
\in G.$$
\end{defin}

\begin{remark}
For all $f \in C(G)$ and $A = \{a_1,a_2,\ldots,a_n\} \subset G$,
the function $M(A,f)$ is continuous and the following hold:
\begin{enumerate}
\item $\max (M(A,f)) \leq \max (f)$,
\item $\min (M(A,f)) \geq \min (f)$,
\item $\osc (M(A,f)) \geq \osc (f)$, and
\item If $A=\{a_1,a_2,\ldots,a_n\} \subset G$ and
$B=\{b_1,b_2,\ldots,b_m\} \subset
G$, then
$$M(A,M(B,f)) = M(AB,f),$$
where $AB = \{a_i b_j \in G : 0 \leq i \leq n, 0 \leq j \leq m\}$.
\end{enumerate}
\end{remark}

\begin{lemma}
Let $f \in C(G)$ be non-constant.  Then there exists an $A \subset
G$ such that
$$\osc (M(A,f)) < \osc (f).$$
\end{lemma}

\begin{proof}
Let $f \in C(G)$ be non-constant, let $k = \min (f)$, and let $l =
\max (f)$. Then since $f$ is continuous and $k < l$ there exists
an open set $U \subset G$ such that for every $x \in G$ we have
$$f(x) \leq h < l \quad {\rm ~for~some} ~h \in \R.$$
Now the collection of open sets of the form $Ua^{-1}$ for some $a
\in G$ covers $G$, so by the compactness of $G$ we can choose an
$A=\{a_1,a_2,\ldots,a_n\} \subset G$ such that the open sets
$Ua_i^{-1}$ for all $a_i \in A$ covers $G$.

Now for every $x \in G$ and for every $a_i \in A$ we have
$f(xa_i^) \leq l$, but also for every $x \in G$ there exists an
$a_j \in A$ such that $x \in Ua_j^{-1}$, thus $xa_j \in U$ so
$f(xa_j) \leq h$. Thus we have that for all $x \in G$,
$$M(A,f;x) = \frac{1}{n}\sum_{i=1}^{n}{ f(xa_i) } \leq
\frac{1}{n}((m-1)l + h) \leq h < l.$$ Also we have $k \leq
M(A,f;x)$ for all $x \in G$.  Thus we have that
\begin{eqnarray*}
\osc (M(A,f)) &=& \max (M(A,f)) - \min (M(A,f)) \\ & \leq & \max
(M(A,f)) - \min (f) \\ &<& \max (f) - \min (f) = \osc (M(A,f)).
\end{eqnarray*}
\end{proof}

\begin{defin}
Let $f \in C(G)$. Then $p \in \R$ is a {\it right mean} of $f$ if
for every $\epsilon > 0$ there exists an $A \subset G$ such that
$$|M(A,f;x) - p| < \epsilon \quad ~{\rm for~all}~x \in G.$$
\end{defin}

\begin{lemma}
Every $f \in C(G)$ has at least one right mean.
\end{lemma}

\begin{proof}
Fix $f \in C(G)$, and let $\Delta = \{M(A,f) \in C(G):
A=\{a_1,a_2,\ldots,a_n\} \subset G\}$.  From Remark 10, $\Delta$
is uniformly bounded, we will show that $\Delta$ is uniformly
equicontinuous.

To that end, note that by theorem 4, $f$ is uniformly continuous,
thus for any $\epsilon > 0$ there exists a neighborhood $U$ of the
identity of $G$ such that for all $x,y \in G$,
$$xy^{-1} \in U \Rightarrow |f(x)-f(y)| < \epsilon.$$
But now, for all $a_i \in A$, we have
$$(xa_i)(ya_i)^{-1} = xy^{-1} \in U \Rightarrow |f(xa_i)-g(xa_i)|
< \epsilon.$$ Thus for all $xy^{-1} \in U$, and for all
$A=\{a_1,a_2,\ldots,a_n\} \subset G\}$ we have
\begin{eqnarray*}
|M(A,f;x)-M(A,f;y)| &=& \left| \frac{1}{n} \sum_{i=1}^{n}{
(f(xa_i)-f(ya_i)) } \right| \\ &\leq& \frac{1}{n} \sum_{i=1}^{n}{
|f(xa_i)-f(ya_i)| } \\ &<& \frac{1}{n}(n \epsilon) = \epsilon.
\end{eqnarray*}
So $\Delta$ is uniformly equicontinuous.

Now, let $s = \inf (\{\osc (h) \in \R : h \in \Delta \})$.  Then
there exists a sequence $\{h_n\} \subset \Delta$ such that
$\{\osc(h_n)\} \to s$.  But since $\Delta$ is uniformly bounded
and uniformly equicontinuous we may select from $\{h_n\}$ a
uniformly convergent subsequence $\{g_n\}$.  Suppose $\{g_n\} \to
g$, then $\osc (g) = s$.  Now we will show that $g$ is constant,
so suppose the contrary.  Then by lemma 11, there exists an $A
\subset G$, such that
$$\osc (M(A,g)) = s' < s.$$
Let $\epsilon = \frac{s-s'}{3}$.  Since $\{g_n\}$ converges
uniformly there exists a $k$ such that $|g(x) - g_k(x)| <
\epsilon$ for all $x \in G$.  Thus $|g(xa_i) - g_k(xa_i)| <
\epsilon$ for all $x \in G$ and for all $a_i \in A$.  So as before
we have for all $x \in G$ and for all $k \in \Z^+$,
$$|M(A,g;x) - M(A,g_k;x)| < \epsilon. $$

Fill this in. Get a contradiction. So $g$ is constant and put
$g(x) = p$.

Now since $\{g_n\}$ converges to $g$ uniformly, for every
$\epsilon > 0$ there exists an $n$ such that $|g_n(x) - p| <
\epsilon$ for every $x \in G$.  But now since $g \in \Delta$ we
have that
$$|M(A,f;x) - p| < \epsilon$$
for some $A \subset G$.
\end{proof}

By analogy,

\begin{defin}
Let $f \in C(G)$ and let $B = \{a_1,a_2,\ldots,a_m\} \subset G$.
Then define $M'(B,f):G \to \R$ given by
$$M'(B,f;x) = \frac{1}{m}\sum_{i=1}^{n}{ f(a_i x) } \quad ~{\rm for~all}~x
\in G.$$
\end{defin}

\begin{remark}
Let $f \in C(G)$, let $A=\{a_1,a_2,\ldots,a_n\} \subset G$, and
let $B=\{b_1,b_2,\ldots,b_m\} \subset G$.  Then
$$M(A,M'(B,f)) = M'(B,M(A,f)).$$
\end{remark}

\begin{defin}
Let $f \in C(G)$. Then $q \in \R$ is a {\it left mean} of $f$ if
for every $\epsilon > 0$ there exists an $B \subset G$ such that
$$|M'(B,f;x) - q| < \epsilon \quad ~{\rm for~all}~x \in G.$$
\end{defin}

\begin{remark}
Every $f \in C(G)$ has at least one right mean.
\end{remark}

\begin{lemma}
Let $f \in C(G)$.  Then $f$ has only one right mean and only one
left mean, and furthermore, these two numbers are the same.  This
unique {\it mean} for $f$ with be denoted by $M(f)$.
\end{lemma}

\begin{proof}
Let $p$ be any right mean of $f \in C(G)$, and let $q$ be any left
mean of $f$.  Then for every $\epsilon > 0$ there exist $A,B
\subset G$ such that for all $x \in G$,
$$|M(A,f;x) - p| < \epsilon ~{\rm and}~ |M'(B,f;x) - q| <
\epsilon.$$ As seen before we then have
$$|M'(B,M(A,f);x) - p| = \left| \frac{1}{n}\sum_{i=1}^{n}{ (M(A,f;a_i x) -
p)
} \right| < \epsilon$$ And similarly,
$$|M(A,M'(B,f);x) - q| < \epsilon.$$
And by remark 15 we have
$$|p-q| < 2 \epsilon.$$
Thus we have that $p = q$.
\end{proof}

\begin{lemma}
Let $f,g \in C(G)$.  Then
$$M(f+g) = M(f) + M(g).$$
\end{lemma}

\begin{proof}
Skip it.
\end{proof}

\begin{lemma}
Let $f \in C(G)$, and let $a \in G$.  Define $f_a(x) = f(xa)$ and
$f^a(x) = f(ax)$ for all $x \in M$. Then $M(f_a) = M(f^a) = M(f)$.
\end{lemma}

\begin{proof}
Let $A=\{a_1,a_2,\ldots,a_n\} \subset G$.  For all $a,x \in G$
observe that
$$M(A,f_a;x) = \frac{1}{n}\sum_{i=1}^{n}{ f_a(xa_i) } =
\frac{1}{n}\sum_{i=1}^{n}{ f(xa_ia)
} = M(Aa,f)$$ and that
$$M'(A,f^a;x) = \frac{1}{n}\sum_{i=1}^{n}{ f^a(a_ix) } =
\frac{1}{n}\sum_{i=1}^{n}{ f(aa_ix) } =
M'(aA,f).$$ Thus the right mean for $f_a$ and $f$ is the same and
the left mean for $f^a$ and $f$ is the same.  Thus $M(f_a) =
M(f^a) = M(f)$.
\end{proof}

We can now proceed to,

\begin{proof}[Proof of theorem]
Let $f \in C(G)$.  Then define the mapping from $C(G) \to \R$ by
$$f \mapsto \int{f(x)\,dx} = M(f).$$

First notice that conditions 3 and 4 are clearly satisfied, while
conditions 2,3 and 4 have been proved in the above remarks and
lemmas.

To check condition 1, let $\alpha \in \R \setminus \{0\}$ ($\alpha
= 0$ holds trivially) and let $p$ be a right mean for $f$. Thus
for any $\epsilon > 0$, there exists an $A \subset G$ such that
for all $x \in G$,
$$|M(A,f;x) - p| < \epsilon/|\alpha| \Rightarrow |M(A,\alpha
f;x) - \alpha p| < \epsilon$$ so $\alpha p$ is a right mean for
$\alpha f$, i.e. $\int{\alpha f(x)\,dx} = \alpha \int{f(x)\,dx}.$

Now, let $\int^*{ f(x)\,dx }$ be any integral defined on $G$
satisfying conditions 1-5. Now let $p$ be a right mean of $f$,
then for all $\epsilon > 0$ there exists an $A \subset G$ such
that for all $x \in G$,
$$|M(A,f;x) - p| < \epsilon.$$
As noted earlier, we only need conditions 1-3 in order to
integrate this inequality,
$$\left| \int^*{ M(A,f;x)\,dx } - p \right| \leq \epsilon$$
thus by condition 6 we have
$$\left| \int^*{ f(x)\,dx } - p \right| \leq \epsilon.$$
Thus we have
$$\int^*{ f(x)\,dx } = p = M(f) = \int{ f(x)\,dx }$$
Thus any integral satisfying conditions 1-5 is unique.

Now to verify condition 7, define a new integral on $G$ given by
$$\int^*{ f(x)\,dx } = \int{ f(x^{-1})\,dx }.$$
Then we can easily verify conditions 1-5 for this other integral.
For instance, verification of 6 is as follows, for any $a \in G$,
\begin{eqnarray*}
 \int^*{ f(xa)\,dx } &=& \int{ f((xa)^{-1})\,dx } =
\int{ f(xa)\,dx } = \int{ f(x)\,dx } \\ &=& \int{ f(x^{-1})\,dx }
= \int^*{ f(x)\,dx }.
\end{eqnarray*}
Thus by the uniqueness just established we have
$$\int{f(x^{-1})\,dx} = \int{f(x)\,dx},$$
which completes the proof.
\end{proof}


\end{document}








