255 lines
11 KiB
TeX
255 lines
11 KiB
TeX
\section{Haar Measures}
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\label{section:haar}
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\begin{definition}[Haar Measure]
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\label{definition:haar-measure}
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Let $G$ be a locally compact group and $\mu: \cb_G \to [0, \infty]$ be a non-zero Radon measure, then $\mu$ is a \textbf{left Haar measure} if
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\begin{enumerate}
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\item[(LH)] For each $g \in G$ and $A \in \cb_G$, $\mu(gA) = \mu(A)$.
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\end{enumerate}
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and a \textbf{right Haar measure} if
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\begin{enumerate}
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\item[(RH)] For each $g \in G$ and $A \in \cb_G$, $\mu(Ag) = \mu(A)$.
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\end{enumerate}
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\end{definition}
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\begin{lemma}
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\label{lemma:lc-sigma-compact}
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Let $G$ be a locally compact group, then there exists an open and closed subgroup $H$ that is $\sigma$-compact.
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\end{lemma}
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\begin{proof}
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Let $K \in \cn_G(1)$ and $K^{(1)} = K$. For each $n \in \natp$, let $K^{(n+1)} = KK^{(n)}$, then $K^{(n+1)}$ is compact by \autoref{proposition:compact-extensions} with $K^{(n+1)} \in \cn_G(K^{(n)})$. Let $H = \bigcup_{n \in \natp}K^{(n)}$, then $H$ is a subgroup of $G$, which is open by \autoref{lemma:openneighbourhood}. Since $H$ admits an exhaustion by compact sets, it is $\sigma$-compact.
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Finally, since
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\[
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G \setminus H = G \setminus \bigcup_{n \in \natp} K^{(n)} = \bigcap_{n \in \natp}G \setminus K^{(n)}
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\]
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and $K^{(n)}$ is closed for each $n \in \natp$ by \autoref{proposition:compact-closed}, $G \setminus H$ is closed, and hence $H$ is open.
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\end{proof}
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\begin{definition}[Covering Ratio]
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\label{definition:lcg-covering-ratio}
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Let $G$ be a locally compact group and $f, g \in C_c^+(G)$, then
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\[
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(f: g) = \inf\bracs{\sum_{j = 1}^n c_j \bigg | \seqf{c_j} \subset [0, \infty), \seqf{x_j} \subset G, f \le \sum_{j = 1}^n c_j L_{x_j}g}
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\]
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is the \textbf{covering ratio} of $f$ by $g$.
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\end{definition}
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\begin{proposition}
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\label{proposition:covering-ratio-gymnastics}
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Let $G$ be a locally compact group and $f, h, g \in C_c^+(G)$, then:
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\begin{enumerate}
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\item If $g \ne 0$, then $(f: g) < \infty$.
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\item $(f + h: g) \le (f: g) + (h: g)$.
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\item For each $\lambda \ge 0$, $(\lambda f: g) = \lambda(f: g)$.
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\item If $f \le h$, then $(f: g) \le (h: g)$.
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\item $(f: g) \le (f: h)(h: g)$.
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\item $(f: g) \ge \norm{f}_u/\norm{g}_h$.
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\item For each $x \in G$, $(L_xf: g) = (f: g)$.
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\end{enumerate}
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\end{proposition}
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% Proof omitted due to obviousness.
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\begin{lemma}
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\label{lemma:haar-approx}
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Let $G$ be a locally compact group, $f, f'\in C_c^+(G)$, and $\eps > 0$, then there exists $V \in \cn_G(1)$ such that for any $g \in C_c^+(V)$ with $g \ne 0$,
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\[
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(f: g) + (f': g) \le (f + f': g) + \eps
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\]
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\end{lemma}
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\begin{proof}[Proof, {{\cite[Lemma 2.18]{FollandHarmonic}}}. ]
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By \hyperref[Urysohn's Lemma]{lemma:lch-urysohn}, there exists $\eta \in C_c^+(G; [0, 1])$ such that $\eta|_{\supp{f} \cup \supp{f'}} = 1$.
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Let $\delta > 0$, and define
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\[
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H = f + f' + \delta \eta \quad h = \frac{f}{H} \quad h' = \frac{f'}{H}
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\]
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By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that for any $x, y \in G$ with $x^{-1}y \in V$,
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\[
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|h(x) - h(y)|, |h'(x) - h'(y)| < \delta
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\]
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Let $g \in C_c^+(V)$, $\seqf{c_j} \subset [0, \infty)$, and $\seqf{x_j} \subset G$ such that $H \le \sum_{j = 1}^n c_j L_{x_j}\phi$, then for each $x \in G$,
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\begin{align*}
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f(x) &= H(x)h(x) \le \sum_{j = 1}^n c_j L_{x_j}g(x)h(x) = \sum_{j = 1}^n c_jg(x_j^{-1}x)h(x) \\
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&\le \sum_{j = 1}^n c_j[h(x_j) + \delta] \cdot L_{x_j}g(x)
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\end{align*}
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Likewise,
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\[
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f'(x) \le \sum_{j = 1}^n c_j[h'(x_j) + \delta] \cdot L_{x_j}g(x)
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\]
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As $h + h' \le 1$,
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\[
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(f: g) + (f': g) \le \sum_{j = 1}^n c_j[h(x_j) + h'(x_j) + 2\delta]
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\]
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Since the above holds for all such $\seqf{c_j} \subset [0, \infty)$ and $\seqf{x_j} \subset G$,
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\begin{align*}
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(f: g) + (f': g) &\le (1 + 2\delta)(H: g) \\
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&\le (1 + 2\delta)[(f + f': g) + \delta(\eta: g)]
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\end{align*}
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\end{proof}
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\begin{theorem}[Haar]
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\label{theorem:haar}
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Let $G$ be a locally compact group, then:
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\begin{enumerate}
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\item There exists a left/right Haar measure on $G$.
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\item For any two left/right Haar measures $\mu$ and $\nu$ on $G$, there exists $\lambda > 0$ such that $\mu = \lambda \nu$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 2.10, 2.20]{FollandHarmonic}}}. ]
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(1): Fix $h \in C_c^+(G)$ with $h \ne 0$. For each $g \in C_c^+(G)$ with $g \ne 0$, let
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\[
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I_g: C_c^+(G) \to [0, \infty) \quad f \mapsto \frac{(f: g)}{(h: g)}
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\]
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then by (5) of \autoref{proposition:covering-ratio-gymnastics}, for each $f \in C_c^+(G)$ with $f \ne 0$,
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\[
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\frac{1}{(h: f)} = \frac{(f: g)}{(h: f)(f: g)} \le I_g(f) \le \frac{(f: h)(h: g)}{(h: g)} = (f: h)
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\]
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Thus $\mathcal{I}(f) = \bracs{I_g(f)|g \in C_c^+(G) \setminus \bracs{0}}$ is relatively compact for each $f \in C_c^+(G)$.
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For each $V \in \cn_G(1)$, let $E_V = \bracs{I_g|g \in C_c^+(V) \setminus \bracs{0}}$, then $\fF = \bracs{E_V|V \in \cn_G(1)}$ is a filter on the product space $\prod_{f \in C_c^+(G)}\ol{\mathcal{I}(f)}$. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $\prod_{f \in C_c^+(G)}\ol{\mathcal{I}(f)}$ is compact, and $\bigcap_{V \in \cn_G(1)}\ol{E_V} \ne \emptyset$.
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Let $I \in \bigcap_{V \in \cn_G(1)}\ol{E_V}$, then by continuity,
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\begin{enumerate}[label=(\roman*)]
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\item For each $f \in C_c^+(G) \setminus \bracs{0}$, $I(f) \in [(h: f)^{-1}, (f: h)]$.
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\item For every $\lambda \ge 0$ and $f \in C_c^+(G)$, $I(\lambda f) = \lambda I(f)$.
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\item For any $x \in G$, $I(L_xf) = I(f)$.
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\item For each $f, f' \in C_c^+(G)$, $I(f + f') \le I(f) + I(f')$.
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\end{enumerate}
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Let $f, f' \in C_c^+(G)$ and $\eps > 0$. By \autoref{lemma:haar-approx}, there exists $V \in \cn_G(1)$ such that for each $g \in E_V$,
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\begin{enumerate}[label=(\alph*)]
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\item $|I_g(f) - I(f)|, |I_g(f') - I(f')| < \eps$.
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\item $I_g(f) + I_g(f') \le I_g(f + f') + \eps$.
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\end{enumerate}
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In which case, $I(f) + I(f') \le I(f + f') + 3\eps$. Since this holds for all $\eps > 0$,
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\begin{enumerate}[start=4, label=(\roman*)]
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\item For each $f, f' \in C_c^+(G)$, $I(f + f') \ge I(f) + I(f')$.
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\end{enumerate}
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Using \autoref{lemma:positive-functional-extension}, $I$ extends to a positive linear functional on $C_c(G; \real)$, with $I(f) > 0$ for all $f \in C_c^+(G) \setminus \bracs{0}$, and
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\begin{enumerate}
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\item[(LH)] For each $f \in C_c(G)$ and $x \in G$, $I(L_xf) = I(f)$.
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\end{enumerate}
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By (i) and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, there exists a unique non-zero Radon measure $\mu: \cb_G \to [0, \infty]$ such that for each $f \in C_c^+(G)$, $I(f) = \int_G f d\mu$. Finally, by \hyperref[density of $C_c(G; \real)$ in $L^1(\mu; \real)$]{proposition:radon-cc-dense} and (LH), $\mu$ is a left Haar measure.
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(2): Let $f, g \in C_c^+(G) \setminus \bracs{0}$ and $\eps > 0$, then by \autoref{proposition:lcg-cc-uc}, there exists a symmetric neighbourhood $V \in \cn_G(1)$ such that for any $x \in G$ and $y \in V$,
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\[
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|f(xy) - f(yx)|, |g(xy) - g(yx)| < \eps
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\]
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By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $h \in C_c^+(V) \setminus \bracs{0}$ such that $h(x) = h(x^{-1})$ for all $x \in V$. Since $f$ and $h$ are both compactly supported and $\mu, \nu$ are locally finite, by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli},
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\begin{align*}
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\paren{\int f d\mu}\paren{\int h d\nu} &= \iint f(x)h(y) \mu(dx)\nu(dy) \\
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&= \iint f(yx)h(y) \nu(dx)\mu(dy) \\
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\end{align*}
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Similarly, by symmetry of $h$,
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\begin{align*}
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\paren{\int h d\mu}\paren{\int f d\nu} &= \iint h(x)f(y) \mu(dx)\nu(dy) \\
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&= \iint h(y^{-1}x)f(y) \mu(dx)\nu(dy) \\
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&= \iint h(x^{-1}y)f(y) \nu(dy)\mu(dx) \\
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&= \iint h(y)f(xy) \nu(dy)\mu(dx) \\
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&= \iint h(y)f(xy) \mu(dx)\nu(dy)
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\end{align*}
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Thus there exists $C > 0$ such that
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\begin{align*}
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&\abs{\paren{\int f d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int f d\nu}} \\
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&\le \abs{\iint h(y)[f(xy) - f(yx)]\mu(dx)\nu(dy)} \le C\eps
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\end{align*}
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and
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\[
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\abs{\paren{\int g d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int g d\nu}}
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\le C\eps
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\]
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so there exists $C' > 0$ such that
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\[
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\abs{\frac{\int h d\nu}{\int h d\mu} - \frac{\int g d\nu}{\int g d\mu}}
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\le C'\eps
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\]
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and
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\[
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\abs{\frac{\int f d\nu}{\int f d\mu} + \frac{\int h d\nu}{\int h d\mu}}
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\le C'\eps
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\]
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Therefore
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\[
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\abs{\frac{\int f d\nu}{\int f d\mu} - \frac{\int g d\nu}{\int g d\mu}} \le 2C' \eps
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\]
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As the above holds for all $\eps > 0$, $\int f d\nu/\int f d\mu = \int g d\nu/\int g d\mu$. By uniqueness from the \hyperref[Riesz Representation Theorem]{section:riesz-radon}, there exists $\lambda > 0$ such that $\mu = \lambda \nu$.
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\end{proof}
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\begin{proposition}
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\label{proposition:haar-measure-charge}
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Let $G$ be a locally compact group and $\mu: \cb_G \to [0, \infty]$ be a left/right Haar measure, then:
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\begin{enumerate}
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\item For each $U \subset G$ open with $U \ne \emptyset$, $\mu(U) > 0$.
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\item For each $f \in C_c^+(G) \setminus \bracs{0}$, $\int \phi d\mu > 0$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}[Proof, for left Haar measures. ]
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Since $\mu \ne 0$, there exists $g \in C_c^+(G) \setminus \bracs{0}$ such that $\int g d\mu > 0$. By compactness of $\supp{g}$, there exists $\seqf{x_j} \subset G$ such that:
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\[
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\supp{g} \subset \bigcup_{j = 1}^n x_j^{-1}U \quad g \le \sum_{j = 1}^n L_{x_j}f
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\]
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Thus $0 < \int g d\mu \le n\mu(U)$ and $0 < \int g d\mu \le n \int f d\mu$.
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\end{proof}
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\begin{proposition}
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\label{proposition:haar-translation}
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Let $G$ be a locally compact group, $\mu: \cb_G \to [0, \infty]$ be a left Haar measure, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then the mapping
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\[
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G \times L^p(\mu; E) \quad (x, f) \mapsto L_xf
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\]
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is jointly continuous. Similarly, if $\nu: \cb_G \to [0, \infty]$ is a right Haar measure, then
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\[
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G \times L^p(\mu; E) \quad (x, f) \mapsto R_xf
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\]
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is also jointly continuous.
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\end{proposition}
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\begin{proof}[Proof of the left case. ]
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Let $\eps > 0$, $x, y \in G$, and $f, g \in L^p(\mu; E)$, then
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\begin{align*}
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\norm{L_xf - L_yg}_{L^p(\mu; E)} &\le \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{L_yf - L_y g}_{L^p(\mu; E)} \\
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&= \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{f - g}_{L^p(\mu; E)}
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\end{align*}
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By \autoref{proposition:radon-cc-dense}, there exists $\phi \in C_c(G)$ such that $\norm{\phi - f}_{L^p(\mu; E)} < \eps$. In which case,
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\begin{align*}
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\norm{L_xf - L_yf}_{L^p(\mu; E)} &\le \norm{L_xf - L_x \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
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&+ \norm{L_yf - L_y \phi}_{L^p(\mu; E)} \\
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&= 2\norm{f - \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
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&\le 2\eps + \normn{L_{x^{-1}y}\phi - \phi}_u\mu\bracs{\phi \ne 0}
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\end{align*}
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By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that if $x^{-1}y \in V$, then $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/\mu\bracs{\phi \ne 0}$. Thus if $x^{-1}y \in V$, then
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\[
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\norm{L_xf - L_yf}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)}
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\]
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\end{proof}
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