Moved harmonic to its own folder.
This commit is contained in:
Bokuan Li
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\section{The $L^1$ Group Algebra}
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\label{section:group-algebra-l1}
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\begin{definition}[Convolution of Measures]
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\label{definition:lcg-convolution-measures}
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Let $G$ be a locally compact group, and $\mu, \nu \in M_R(G; \complex)$ be finite Radon measures, then the \textbf{convolution} of $\mu$ and $\nu$ is the Radon measure $\mu * \nu$ defined by
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\[
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\int_{G}\phi(x)(\mu * \nu)(dx) = \iint_{G \times G} \phi(xy)\mu(dx)\nu(dy)
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\]
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for all $\phi \in C_c(G; \complex)$\footnote{There seem to be some subtleties with having to take the Radon product here.}.
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\end{definition}
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\begin{proposition}
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\label{proposition:convolution-measures-property}
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Let $G$ be a locally compact group, then:
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\begin{enumerate}
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\item For any $\mu, \nu, \sigma \in M_R(G; \complex)$, $(\mu * \nu) * \sigma = \mu * (\nu * \sigma)$.
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\item For any $\mu, \nu \in M_R(G; \complex)$, $\norm{\mu * \nu}_{\text{var}} \le \norm{\mu}_{\text{var}}\norm{\nu}_{\text{var}}$.
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\item Convolution on $M_R(G; \complex)$ is commutative if and only if $G$ is commutative.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $\phi \in C_c(G; \complex)$, then
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\begin{align*}
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\int_G \phi d(\mu * (\nu * \sigma)) &= \iint_{G \times G} \phi(xy)\mu(dx)(\nu * \sigma)(dy) \\
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&= \iiint_{G \times G \times G}\phi(xyz) \mu(dx) \nu(dy)\sigma(dz) \\
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&= \iint_{G \times G} \phi(yz) (\mu * \nu)(dy) \sigma(dz) = \int_G \phi d((\mu * \nu) * \sigma)
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\end{align*}
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(2): Let $\phi \in C_c(G; \complex)$, then
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\begin{align*}
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\abs{\int_G \phi d(\mu * \nu)} &= \abs{\iint_{G \times G}\phi(xy)\mu(dx)\nu(dy)} \\
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&\le \abs{\int_{G}\norm{\phi}_u\norm{\mu}_{\text{var}}d|\nu|} \le \norm{\phi}_u\norm{\mu}_{\text{var}}\norm{\nu}_{\text{var}}
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\end{align*}
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(3): For any $g, h \in G$ and $\phi \in C_c(G; \complex)$,
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\[
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\int_G \phi d(\delta_g * \delta_h) = \iint_{G \times G} \phi(xy) \delta_g(dx)\delta_h(dy) = \phi(gh)
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\]
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so $\delta_g * \delta_h = \delta_{gh}$, and $\delta_g * \delta_h = \delta_h * \delta_g$ if and only if $gh = hg$. Therefore the convolution in $M_R(G; \complex)$ is commutative if and only if $G$ is commutative.
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\end{proof}
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\begin{definition}[Measure Algebra]
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\label{definition:group-measure-algebra}
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Let $G$ be a locally compact group. For each $\mu \in M_R(G; \complex)$, let
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\[
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\mu^*: \cb_G \to \complex \quad \mu^*(A) = \ol{\mu(A^{-1})}
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\]
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then $M_R(G; \complex)$ equipped with the convolution product and the above involution is an involutive unital Banach algebra with identity $\delta_1$. The algebra $M_R(G; \complex)$ is the \textbf{measure algebra} of $G$.
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\end{definition}
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\begin{proof}
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By (2) of \autoref{lemma:convolution-measures-property}, $M_R(G; \complex)$ with the convolution product is a Banach algebra. For any $\mu \in M_R(G; \complex)$ and $\phi \in C_c(G; \complex)$,
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\begin{align*}
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\int_G \phi d(\mu * \delta_1) &= \iint_{G \times G} \phi(xy)\mu(dx)\delta_1(dy) \\
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&= \iint_{G \times G} \phi(xy)\delta_1(dy)\mu(dx) = \int_G \phi(x)\mu(dx) \\
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\int_G \phi d(\delta_1 * \mu) &= \iint_{G \times G}\phi(xy)\delta_1(dx)\mu(dy) = \int_G \phi(y)\mu(dy)
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\end{align*}
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so $\mu * \delta_1 = \mu = \delta_1 * \mu$, and $\delta_1$ is the identity.
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It remains to show that $\mu \mapsto \mu^*$ is an involution. Let $\mu, \nu \in M_R(G; \complex)$, then for any $\phi \in C_c(G; \complex)$,
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\begin{align*}
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\int_G \phi d(\mu * \nu)^* &= \int_G \phi(x^{-1})d(\ol{\mu * \nu}) = \iint_{G \times G} \phi((xy)^{-1})\ol{\mu}(dx)\ol{\nu}(dy) \\
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&= \iint_{G \times G} \phi(y^{-1}x^{-1}) \ol{\mu}(dx)\ol{\nu}(dy) = \iint_{G \times G}\phi(yx)\mu^*(dx)\nu^*(dy) \\
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&= \iint_{G \times G}\phi(yx)\nu^*(dy)\mu^*(dx) = \int_{G}\phi d(\nu^* * \mu^*)
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\end{align*}
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Therefore $(\mu * \nu)^* = \nu^* * \mu^*$.
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\end{proof}
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\begin{definition}[Convolution]
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\label{definition:group-convolution-functions}
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Let $G$ be a locally compact group and $f, g: G \to \complex$ be measurable functions, then the \textbf{convolution} of $f$ and $g$ is the function
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\[
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f * g(x) = \int_G f(y)g(y^{-1}x)dy
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\]
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defined wherever the integral exists.
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\end{definition}
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\begin{theorem}[Young's Inequality]
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\label{theorem:young-fourier}
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Let $G$ be a locally compact group, $f \in L^1(G; \complex)$, $p \in [1, \infty]$, and $g \in L^p(G; \complex)$, then
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\begin{enumerate}
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\item If $p = \infty$, then $(f * g)(x) = \int_G f(y)g(y^{-1}x)dy$ exists for all $x \in G$, with
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\[
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\norm{f * g}_{L^\infty(G; \complex)}\le \norm{f}_{L^1(G; \complex)} \norm{g}_{L^\infty(G; \complex)}
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\]
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\item If $p \in [1, \infty)$, then $(f * g)(x)$ exists for almost every $x \in G$, and $f * g \in L^p(G; \complex)$ with
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\[
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\norm{f * g}_{L^p(G; \complex)} \le \norm{f}_{L^1(G; \complex)} \norm{g}_{L^p(G; \complex)}
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\]
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\end{enumerate}
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Proposition 2.40]{Folland}}}. ]
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(1): By \autoref{definition:modular-function}, the measures $dx$ and $dx^{-1}$ are equivalent. Thus for each $x \in G$, $\normn{y \mapsto g(y^{-1}x)}_{L^\infty(G; \complex)} = \norm{g}_{L^\infty(G; \complex)}$. By \hyperref[Hölder's Inequality]{theorem:holder}
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\[
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|(f * g)(x)| = \abs{\int_G f(y)g(y^{-1}x)dy} \le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^\infty(G; \complex)}
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\]
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(2): Since $\bracsn{f \ne 0}$ and $\bracsn{g \ne 0}$ are $\sigma$-finite, assume without loss of generality that $dx$ is $\sigma$-finite. In which case, by \hyperref[Minkowski's Inequality for integrals]{theorem:minkowski-integral},
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\begin{align*}
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\norm{f * g}_{L^p(G; \complex)} &= \braks{\int_G \abs{\int_G f(y)g(y^{-1}x)dy}^pdx}^{1/p} \\
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&\le \int_{G} \braks{\int_G|f(y)g(y^{-1}x)|^p dx}^{1/p} dy \\
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&= \norm{g}_{L^p(G; \complex)}\int_G |f(y)|dy = \norm{f}_{L^1(G; \complex)} \norm{g}_{L^p(G; \complex)}
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\end{align*}
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\end{proof}
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\begin{lemma}
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\label{lemma:convolution-measure-function-agree}
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Let $G$ be a locally compact group, $f, g \in L^1(G; \complex)$, $\mu = fdx$, and $\nu = gdx$, then $\mu * \nu = (f * g)dx$.
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\end{lemma}
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\begin{proof}
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Let $\phi \in C_c(G; \complex)$, then
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\begin{align*}
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\int_G \phi(x) (f * g)(x)dx &= \iint_{G \times G} \phi(x) f(y)g(y^{-1}x)dydx \\
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&= \iint_{G \times G} \phi(yx)g(x)dxf(y)dy \\
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&= \iint_{G \times G} \phi(yx) \mu(dy)\nu(dx) = \int_G \phi d(\mu * \nu)
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\end{align*}
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\end{proof}
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\begin{definition}[$L^1$ Group Algebra]
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\label{definition:l1-group-algebra}
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Let $G$ be a locally compact group, then $L^1(G; \complex)$ equipped with the convolution product is an involutive Banach subalgebra of $M_R(G; \complex)$, where for any $f \in L^1(G; \complex)$,
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\[
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f^*(x) = \Delta(x^{-1}) \ol{f(x^{-1})}
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\]
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The algebra $L^1(G; \complex)$ is the \textbf{$L^1$ group algebra} of $G$.
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\end{definition}
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\begin{proof}
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By \autoref{lemma:convolution-measure-function-agree}, the convolution operation on $L^1(G; \complex)$ agrees with the convolution on $M_R(G; \complex)$. By \hyperref[Young's Inequality]{theorem:young-fourier}, $L^1(G; \complex)$ is a Banach subalgebra of $M_R(G; \complex)$.
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For any $f \in L^1(G; \complex)$, let $\mu = f dx$, then for each $A \in \cb_G$,
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\[
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\mu^*(A) = \ol{\mu(A^{-1})} = \ol{\int_{A^{-1}}d\mu} = \int_{A^{-1}} \ol{f(x)}dx = \int_A \ol{f(x^{-1})}\Delta_G(x^{-1})dx
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\]
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so $f^*dx = d\mu^* = \Delta(x^{-1}) \ol{f(x^{-1})} dx$, and $f^* = \Delta(x^{-1}) \ol{f(x^{-1})}$.
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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, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then
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\begin{enumerate}
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\item The mapping $G \times L^p(G; E) \to L^p(G; E)$ defined by $(x, f) \mapsto L_xf$ is jointly continuous.
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\item The mapping $G \times L^p(G; E) \to L^p(G; E)$ defined by $(x, f) \mapsto R_xf$ is jointly continuous.
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\end{enumerate}
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition 2.42]{FollandHarmonic}}}. ]
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(1): 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 $\phi \in C_c(G; E)$ 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} \cup x^{-1}y\bracs{\phi \ne 0})^{1/p}\\
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&\le 2\eps + 2\normn{L_{x^{-1}y}\phi - \phi}_u\mu\bracs{\phi \ne 0}^{1/p}
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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 $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/(2\mu\bracs{\phi \ne 0}^{1/p})$. Thus if $x^{-1}y \in V$, then
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\[
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\norm{L_xf - L_yg}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)}
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\]
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(2): Let $V \in \cn_G(1)$ be compact, then since $\Delta_G: G \to (0, \infty)$ is a continuous homomorphism, $C := \sup_{y \in V}\Delta_G(y^{-1}) < \infty$ by \autoref{proposition:compact-extensions}. For any $f \in L^p(\mu; E)$ and $x \in V$,
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\begin{align*}
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\int_G \norm{R_xf(y)}_E^p dy &= \int_G \norm{f(yx)}_E^p dy = \int_G \Delta_G(x^{-1}) \norm{f(y)}_E^pdy \\
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\norm{R_xf}_{L^p(G; E)} &= \Delta_G(x^{-1})^{1/p}\norm{f}_{L^p(G; E)} \le C^{1/p}\norm{f}_{L^p(G; E)}
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\end{align*}
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Let $f, g \in L^p(G; E)$ and $x \in V$, then
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\begin{align*}
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\norm{R_xg - f}_{L^p(G; E)} &\le \norm{R_xf - f}_{L^p(G; E)} + \norm{R_xg - R_xf}_{L^p(G; E)} \\
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&\le \norm{R_xf - f}_{L^p(G; E)} + C^{1/p}\norm{f - g}_{L^p(G; E)}
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\end{align*}
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By \autoref{proposition:radon-cc-dense}, there exists $\phi \in C_c(G; E)$ with $\norm{f - \phi}_{L^p(G; E)} < \eps$. Thus
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\begin{align*}
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\norm{R_xf - f}_{L^p(G; E)} &\le \norm{R_x(f - \phi)}_{L^p(G; E)} + \norm{f - \phi}_{L^p(G; E)} + \norm{R_x\phi - \phi}_{L^p(G; E)} \\
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&\le (1 + C^{1/p})\eps + \norm{R_x\phi - \phi}_{L^p(G; E)}
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\end{align*}
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Since $V$ is compact, $\bracsn{\phi \ne 0}V^{-1}$ is relatively compact, and of finite measure. By \autoref{proposition:lcg-cc-uc}, $\phi$ is left and right uniformly continuous, so there exists $W \in \cn_G(1)$ such that $W \subset V$ and $\norm{R_x\phi - \phi}_u < \eps/\normn{\one_{\bracsn{\phi \ne 0}V^{-1}}}_{L^p(G; \real)}$. In which case, for any $x \in W$,
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\begin{align*}
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\norm{R_x\phi - \phi}_{L^p(G; E)} &\le \norm{R_x\phi - \phi}_u \cdot \normn{\one_{\bracsn{\phi \ne 0} \cup \bracsn{\phi \ne 0}x^{-1}}}_{L^p(G; \real)} \\
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&\le \norm{R_x\phi - \phi}_u \cdot \normn{\one_{\bracsn{\phi \ne 0}V^{-1}}}_{L^p(G; \real)} \le \eps
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\end{align*}
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and
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\[
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\norm{R_xg - f}_{L^p(G; E)} \le (2 + C^{1/p})\eps + C^{1/p}\norm{f - g}_{L^p(G; E)}
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\]
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\end{proof}
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\begin{proposition}
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\label{proposition:convolution-function-gymnastics}
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Let $G$ be a locally compact group and $f, g: G \to \complex$ be Borel measurable, then:
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\begin{enumerate}
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\item For any $x \in G$, $(f * g)(x) = \int_G f(y)L_yg(x) dy = \int_G R_yf(x) g(y^{-1}) dy$.
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\item For any $z \in G$, $L_z(f * g) = (L_zf) * g$, and $R_z(f * g) = f * (R_z g)$.
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\item If $f \in L^1(G; \complex)$ and $g \in L^\infty(G; \complex)$, then $f * g$ is left uniformly continuous\footnote{Folland also claims that $g * f$ is right uniformly continuous \cite[Proposition 2.41, Proposition 2.43]{FollandHarmonic}. This is not true unless $G$ is unimodular. A counterexample is given by the $ax + b$ group, where $f = \one_{[1, \infty) \times [0, 1]}$ and $g = \one$.}.
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\item If $G$ is unimodular, $p, q \in (1, \infty)$ are Hölder conjugates, $f \in L^p(G; \complex)$, and $g \in L^q(G; \complex)$, then $f * g \in C_0(G; \complex)$ with
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\[
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\norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{g}_{L^q(G; \complex)}
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\]
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\end{enumerate}
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition 2.41, Proposition 2.43]{FollandHarmonic}}}. ]
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(1): For any $x \in G$, by translation-invariance,
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\begin{align*}
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(f * g)(x) &= \int_G f(y)g(y^{-1}x) dy = \int_G f(y)L_yg(x)dy \\
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&= \int_G f(xy)g(y^{-1})dy = \int_G R_yf(x)g(y^{-1})dy
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\end{align*}
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(2): Let $x, z \in G$, then by translation-invariance,
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\begin{align*}
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L_z(f * g)(x) &= (f * g)(z^{-1}x) = \int_G f(y)g(y^{-1}z^{-1}x) dy \\
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&= \int_G f(z^{-1}y)g(y^{-1}x)dy = \int_G L_zf(y)g(y^{-1}x)dy \\
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&= (L_zf * g)(x) \\
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R_z(f * g)(x) &= (f * g)(xz) = \int_G f(y)g(y^{-1}xz)dy = (f * R_zg)(x)
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\end{align*}
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(3): By (2), $L_z(f * g) - f * g = (L_z f - f) * g$. By \hyperref[continuity of translation]{proposition:haar-translation} and \hyperref[Young's Inequality]{theorem:young-fourier},
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\[
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\lim_{z \to 1}\norm{L_z(f * g) - (f * g)}_{L^\infty(G; \complex)} \le \lim_{z \to 1}\norm{L_zf - f}_{L^1(G; \complex)} \cdot \norm{g}_{L^\infty(G; \complex)} = 0
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\]
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Therefore $(f * g)$ is left uniformly continuous.
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(4): Let $h: G \to \complex$ be defined by $h(y) = g(y^{-1})$, then for each $x \in G$, \hyperref[Hölder's Inequality]{theorem:holder} implies that
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\begin{align*}
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(f * g)(x) &= \int_G f(y)g(y^{-1}x) dy = \int_G f(xy)g(y^{-1}) dy \\
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|(f * g)(x)| &\le \int_G |f(xy)h(y)| dy \le \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)}
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\end{align*}
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Since $G$ is unimodular, $\norm{h}_{L^q(G; \complex)} = \norm{g}_{L^q(G; \complex)}$, so
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\[
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\norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)}
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\]
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For any $f, g \in C_c(G; \complex)$, $\supp{f * g} \subset \supp{f}\supp{g}$, so $f * g \in C_c(G; \complex)$. By \autoref{proposition:radon-cc-dense}, $C_c(G; \complex)$ is dense in $L^p(G; \complex)$ and $L^q(G; \complex)$. By \autoref{proposition:closure-of-image},
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\begin{align*}
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L^p(G; \complex) * L^q(G; \complex) &= \ol{C_c(G; \complex)}^{L^p(G; \complex)} * \ol{C_c(G; \complex)}^{L^q(G; \complex)} \\
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&\subset \ol{C_c(G; \complex) * C_c(G; \complex)}^{C_0(G; \complex)} = C_0(G; \complex)
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\end{align*}
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\end{proof}
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Unfortunately, $L^1(G)$ is not a $C^*$-algebra, and its approximate identity requires a manual construction. Fortunately, the candidate is clear.
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|
||||
\begin{proposition}[Existence of Approximate Identity]
|
||||
\label{proposition:group-algebra-approximate-identity}
|
||||
Let $G$ be a locally compact group and $\fB \subset \cn_G(1)$ be a fundamental system of neighbourhoods at $1$, directed under reverse inclusion. For each $U \in \fB$, let $\phi_U \in L^+(G)$ with $\ol{\bracsn{\phi_U \ne 0}} \subset U$ and $\int_G \phi_U = 1$, then:
|
||||
\begin{enumerate}
|
||||
\item For each $p \in [1, \infty)$ and $f \in L^p(G; \complex)$, $\phi_U * f \to f$ in $L^p(G; \complex)$.
|
||||
\item For each left uniformly continuous $f \in C(G; \complex)$, $\phi_U * f \to f$ uniformly.
|
||||
\end{enumerate}
|
||||
|
||||
If in addition, $\phi_U(x^{-1}) = \phi_U(x)$ for all $x \in G$, then:
|
||||
\begin{enumerate}[start=2]
|
||||
\item For each $p \in [1, \infty)$ and $f \in L^p(G; \complex)$, $f * \phi_U \to f$ in $L^p(G; \complex)$.
|
||||
\item For each right uniformly continuous $f \in C(G; \complex)$, $f * \phi_U \to f$ uniformly.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 2.44]{FollandHarmonic}}}. ]
|
||||
(1): Let $U \in \fB$, $f \in L^p(G; \complex)$, and $x \in G$, then since $\int_G \phi_U = 1$,
|
||||
\[
|
||||
(\phi_U * f)(x) - f(x) = \int_G \phi_U(y)[L_yf(x) - f(x)]dy
|
||||
\]
|
||||
|
||||
By \hyperref[Minkowski's Inequality for integrals]{theorem:minkowski-integral},
|
||||
\begin{align*}
|
||||
\norm{\phi_U * f - f}_{L^p(G; \complex)} &= \braks{\int_G \abs{\int_G \phi_U(y)[L_yf(x) - f(x)]dy}^p dx}^{1/p} \\
|
||||
&\le \braks{\int_G \phi_U(y)\norm{L_yf - f}_{L^p(G; \complex)}^p dy }^{1/p} \\
|
||||
&\le \sup_{y \in U}\norm{L_yf - f}_{L^p(G; \complex)}
|
||||
\end{align*}
|
||||
|
||||
Therefore by \hyperref[continuity of translation]{proposition:haar-translation}, $\phi_U * f \to f$ in $L^p(G; \complex)$.
|
||||
|
||||
(3): Let $U \in \fB$, $f \in L^p(G; \complex)$, and $x \in G$, then since $\phi_U(y) = \phi_U(y^{-1})$ for all $y \in G$,
|
||||
\begin{align*}
|
||||
(f * \phi_U)(x) - f(x) &= \int_G [R_yf(x) - f(x)]\phi_U(y^{-1})dy \\
|
||||
&= \int_G \phi_U(y)[R_yf(x) - f(x)]dy
|
||||
\end{align*}
|
||||
|
||||
By \hyperref[Minkowski's Inequality for integrals]{theorem:minkowski-integral},
|
||||
\[
|
||||
\norm{f * \phi_U - f}_{L^p(G; \complex)} \le \sup_{y \in U}\norm{R_yf - f}_{L^p(G; \complex)}
|
||||
\]
|
||||
|
||||
and $f * \phi_U \to f$ in $L^p(G; \complex)$ by \hyperref[continuity of translation]{proposition:haar-translation}.
|
||||
\end{proof}
|
||||
|
||||
@@ -0,0 +1,230 @@
|
||||
\section{Haar Measures}
|
||||
\label{section:haar}
|
||||
|
||||
\begin{definition}[Haar Measure]
|
||||
\label{definition:haar-measure}
|
||||
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
|
||||
\begin{enumerate}
|
||||
\item[(LH)] For each $g \in G$ and $A \in \cb_G$, $\mu(gA) = \mu(A)$.
|
||||
\end{enumerate}
|
||||
|
||||
and a \textbf{right Haar measure} if
|
||||
\begin{enumerate}
|
||||
\item[(RH)] For each $g \in G$ and $A \in \cb_G$, $\mu(Ag) = \mu(A)$.
|
||||
\end{enumerate}
|
||||
\end{definition}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:lc-sigma-compact}
|
||||
Let $G$ be a locally compact group, then there exists an open and closed subgroup $H$ that is $\sigma$-compact.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
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.
|
||||
|
||||
Finally, since
|
||||
\[
|
||||
G \setminus H = G \setminus \bigcup_{n \in \natp} K^{(n)} = \bigcap_{n \in \natp}G \setminus K^{(n)}
|
||||
\]
|
||||
|
||||
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.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{definition}[Covering Ratio]
|
||||
\label{definition:lcg-covering-ratio}
|
||||
Let $G$ be a locally compact group and $f, g \in C_c^+(G)$, then
|
||||
\[
|
||||
(f: g) = \inf\bracs{\sum_{j = 1}^n c_j \bigg | \seqf{c_j} \subset [0, \infty), \seqf{x_j} \subset G, f \le \sum_{j = 1}^n c_j L_{x_j}g}
|
||||
\]
|
||||
|
||||
is the \textbf{covering ratio} of $f$ by $g$.
|
||||
\end{definition}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:covering-ratio-gymnastics}
|
||||
Let $G$ be a locally compact group and $f, h, g \in C_c^+(G)$, then:
|
||||
\begin{enumerate}
|
||||
\item If $g \ne 0$, then $(f: g) < \infty$.
|
||||
\item $(f + h: g) \le (f: g) + (h: g)$.
|
||||
\item For each $\lambda \ge 0$, $(\lambda f: g) = \lambda(f: g)$.
|
||||
\item If $f \le h$, then $(f: g) \le (h: g)$.
|
||||
\item $(f: g) \le (f: h)(h: g)$.
|
||||
\item $(f: g) \ge \norm{f}_u/\norm{g}_h$.
|
||||
\item For each $x \in G$, $(L_xf: g) = (f: g)$.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
% Proof omitted due to obviousness.
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:haar-approx}
|
||||
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 $g \ne 0$,
|
||||
\[
|
||||
(f: g) + (f': g) \le (f + f': g) + \eps
|
||||
\]
|
||||
\end{lemma}
|
||||
\begin{proof}[Proof, {{\cite[Lemma 2.18]{FollandHarmonic}}}. ]
|
||||
By \hyperref[Urysohn's Lemma]{lemma:lch-urysohn}, there exists $\eta \in C_c^+(G; [0, 1])$ such that $\eta|_{\supp{f} \cup \supp{f'}} = 1$.
|
||||
|
||||
Let $\delta > 0$, and define
|
||||
\[
|
||||
H = f + f' + \delta \eta \quad h = \frac{f}{H} \quad h' = \frac{f'}{H}
|
||||
\]
|
||||
|
||||
By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that for any $x, y \in G$ with $x^{-1}y \in V$,
|
||||
\[
|
||||
|h(x) - h(y)|, |h'(x) - h'(y)| < \delta
|
||||
\]
|
||||
|
||||
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 $x \in G$,
|
||||
\begin{align*}
|
||||
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) \\
|
||||
&\le \sum_{j = 1}^n c_j[h(x_j) + \delta] \cdot L_{x_j}g(x)
|
||||
\end{align*}
|
||||
|
||||
Likewise,
|
||||
\[
|
||||
f'(x) \le \sum_{j = 1}^n c_j[h'(x_j) + \delta] \cdot L_{x_j}g(x)
|
||||
\]
|
||||
|
||||
As $h + h' \le 1$,
|
||||
\[
|
||||
(f: g) + (f': g) \le \sum_{j = 1}^n c_j[h(x_j) + h'(x_j) + 2\delta]
|
||||
\]
|
||||
|
||||
Since the above holds for all such $\seqf{c_j} \subset [0, \infty)$ and $\seqf{x_j} \subset G$,
|
||||
\begin{align*}
|
||||
(f: g) + (f': g) &\le (1 + 2\delta)(H: g) \\
|
||||
&\le (1 + 2\delta)[(f + f': g) + \delta(\eta: g)]
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}[Haar]
|
||||
\label{theorem:haar}
|
||||
Let $G$ be a locally compact group, then:
|
||||
\begin{enumerate}
|
||||
\item There exists a left/right Haar measure on $G$.
|
||||
\item For any two left/right Haar measures $\mu$ and $\nu$ on $G$, there exists $\lambda > 0$ such that $\mu = \lambda \nu$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 2.10, 2.20]{FollandHarmonic}}}. ]
|
||||
(1): Fix $h \in C_c^+(G)$ with $h \ne 0$. For each $g \in C_c^+(G)$ with $g \ne 0$, let
|
||||
\[
|
||||
I_g: C_c^+(G) \to [0, \infty) \quad f \mapsto \frac{(f: g)}{(h: g)}
|
||||
\]
|
||||
|
||||
then by (5) of \autoref{proposition:covering-ratio-gymnastics}, for each $f \in C_c^+(G)$ with $f \ne 0$,
|
||||
\[
|
||||
\frac{1}{(h: f)} = \frac{(f: g)}{(h: f)(f: g)} \le I_g(f) \le \frac{(f: h)(h: g)}{(h: g)} = (f: h)
|
||||
\]
|
||||
|
||||
Thus $\mathcal{I}(f) = \bracs{I_g(f)|g \in C_c^+(G) \setminus \bracs{0}}$ is relatively compact for each $f \in C_c^+(G)$.
|
||||
|
||||
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 $\bigcap_{V \in \cn_G(1)}\ol{E_V} \ne \emptyset$.
|
||||
|
||||
Let $I \in \bigcap_{V \in \cn_G(1)}\ol{E_V}$, then by continuity,
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item For each $f \in C_c^+(G) \setminus \bracs{0}$, $I(f) \in [(h: f)^{-1}, (f: h)]$.
|
||||
\item For every $\lambda \ge 0$ and $f \in C_c^+(G)$, $I(\lambda f) = \lambda I(f)$.
|
||||
\item For any $x \in G$, $I(L_xf) = I(f)$.
|
||||
\item For each $f, f' \in C_c^+(G)$, $I(f + f') \le I(f) + I(f')$.
|
||||
\end{enumerate}
|
||||
|
||||
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 $g \in E_V$,
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item $|I_g(f) - I(f)|, |I_g(f') - I(f')| < \eps$.
|
||||
\item $I_g(f) + I_g(f') \le I_g(f + f') + \eps$.
|
||||
\end{enumerate}
|
||||
|
||||
In which case, $I(f) + I(f') \le I(f + f') + 3\eps$. Since this holds for all $\eps > 0$,
|
||||
\begin{enumerate}[start=4, label=(\roman*)]
|
||||
\item For each $f, f' \in C_c^+(G)$, $I(f + f') \ge I(f) + I(f')$.
|
||||
\end{enumerate}
|
||||
|
||||
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
|
||||
\begin{enumerate}
|
||||
\item[(LH)] For each $f \in C_c(G)$ and $x \in G$, $I(L_xf) = I(f)$.
|
||||
\end{enumerate}
|
||||
|
||||
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.
|
||||
|
||||
(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 $y \in V$,
|
||||
\[
|
||||
|f(xy) - f(yx)|, |g(xy) - g(yx)| < \eps
|
||||
\]
|
||||
|
||||
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},
|
||||
\begin{align*}
|
||||
\paren{\int f d\mu}\paren{\int h d\nu} &= \iint f(x)h(y) \mu(dx)\nu(dy) \\
|
||||
&= \iint f(yx)h(y) \nu(dx)\mu(dy) \\
|
||||
\end{align*}
|
||||
|
||||
Similarly, by symmetry of $h$,
|
||||
\begin{align*}
|
||||
\paren{\int h d\mu}\paren{\int f d\nu} &= \iint h(x)f(y) \mu(dx)\nu(dy) \\
|
||||
&= \iint h(y^{-1}x)f(y) \mu(dx)\nu(dy) \\
|
||||
&= \iint h(x^{-1}y)f(y) \nu(dy)\mu(dx) \\
|
||||
&= \iint h(y)f(xy) \nu(dy)\mu(dx) \\
|
||||
&= \iint h(y)f(xy) \mu(dx)\nu(dy)
|
||||
\end{align*}
|
||||
|
||||
Thus there exists $C > 0$ such that
|
||||
\begin{align*}
|
||||
&\abs{\paren{\int f d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int f d\nu}} \\
|
||||
&\le \abs{\iint h(y)[f(xy) - f(yx)]\mu(dx)\nu(dy)} \le C\eps
|
||||
\end{align*}
|
||||
|
||||
and
|
||||
\[
|
||||
\abs{\paren{\int g d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int g d\nu}}
|
||||
\le C\eps
|
||||
\]
|
||||
|
||||
so there exists $C' > 0$ such that
|
||||
\[
|
||||
\abs{\frac{\int h d\nu}{\int h d\mu} - \frac{\int g d\nu}{\int g d\mu}}
|
||||
\le C'\eps
|
||||
\]
|
||||
|
||||
and
|
||||
\[
|
||||
\abs{\frac{\int f d\nu}{\int f d\mu} + \frac{\int h d\nu}{\int h d\mu}}
|
||||
\le C'\eps
|
||||
\]
|
||||
|
||||
Therefore
|
||||
\[
|
||||
\abs{\frac{\int f d\nu}{\int f d\mu} - \frac{\int g d\nu}{\int g d\mu}} \le 2C' \eps
|
||||
\]
|
||||
|
||||
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 $\mu = \lambda \nu$.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:haar-measure-charge}
|
||||
Let $G$ be a locally compact group and $\mu: \cb_G \to [0, \infty]$ be a left/right Haar measure, then:
|
||||
\begin{enumerate}
|
||||
\item For each $U \subset G$ open with $U \ne \emptyset$, $\mu(U) > 0$.
|
||||
\item For each $f \in C_c^+(G) \setminus \bracs{0}$, $\int \phi d\mu > 0$.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, for left Haar measures. ]
|
||||
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:
|
||||
\[
|
||||
\supp{g} \subset \bigcup_{j = 1}^n x_j^{-1}U \quad g \le \sum_{j = 1}^n L_{x_j}f
|
||||
\]
|
||||
|
||||
Thus $0 < \int g d\mu \le n\mu(U)$ and $0 < \int g d\mu \le n \int f d\mu$.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}
|
||||
\label{definition:haar-measure-lp-notation}
|
||||
Let $G$ be a locally compact group and assume that $G$ is equipped with a fixed left Haar measure $\mu: \cb_G \to [0, \infty]$, then
|
||||
\begin{enumerate}
|
||||
\item For any normed vector space $E$ over $K \in \RC$, $L^p(G; E)$ denotes $L^p(G, \mu; E)$.
|
||||
\item For any $f \in L^1(G; E) \cup L^+(G)$, $\int_G f(x)dx$ denotes integration with respect to $\mu$.
|
||||
\end{enumerate}
|
||||
\end{definition}
|
||||
|
||||
|
||||
|
||||
|
||||
@@ -0,0 +1,7 @@
|
||||
\chapter{Integration on Locally Compact Groups}
|
||||
\label{chap:locally-compact-integration}
|
||||
|
||||
\input{./lcg.tex}
|
||||
\input{./haar.tex}
|
||||
\input{./modular.tex}
|
||||
\input{./algebra.tex}
|
||||
@@ -0,0 +1,17 @@
|
||||
\section{Locally Compact Groups}
|
||||
\label{section:lcg}
|
||||
|
||||
\begin{definition}[Locally Compact Group]
|
||||
\label{definition:lcg}
|
||||
Let $G$ be a topological group, then $G$ is \textbf{locally compact} if $G$ is an LCH space.
|
||||
\end{definition}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:lcg-cc-uc}
|
||||
Let $G$ be a locally compact group, $E$ be a TVS over $K \in \RC$, and $\phi \in C_c(G; E)$, then $\phi$ is left and right uniformly continuous.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
By \autoref{lemma:lch-compact-neighbour}, there exists a compact neighbourhood $U \in \cn_G(\text{supp}(\phi))$. By \autoref{proposition:uniform-continuous-compact}, $\phi|_{U}$ and $\phi|_{\supp(\phi)^c}$ are both left and right uniformly continuous. Therefore $\phi$ is left and right uniformly continuous.
|
||||
\end{proof}
|
||||
|
||||
|
||||
@@ -0,0 +1,53 @@
|
||||
\section{The Modular Function}
|
||||
\label{section:modular-function}
|
||||
|
||||
\begin{definition}[Modular Function]
|
||||
\label{definition:modular-function}
|
||||
Let $G$ be a locally compact group and $\mu$ be a left Haar measure on $G$, then
|
||||
\begin{enumerate}
|
||||
\item For any $f, g \in C_c^+(G; \real) \setminus \bracs{0}$, $A, B \in \cb_G$ with $\mu(A), \mu(B) > 0$, and $y \in G$,
|
||||
\[
|
||||
\Delta_G(y) = \frac{\int R_{y^{-1}} f d\mu}{\int f d\mu} = \frac{\int R_{y^{-1}} g d\mu}{\int g d\mu} = \frac{\mu(Ay)}{\mu(A)} = \frac{\mu(By)}{\mu(B)} > 0
|
||||
\]
|
||||
\item For each $y \in G$ and $A \in \cb_G$, denote $\mu_y(A) = \mu(Ay)$, then $\mu_y(dx) = \Delta_G(y)\mu(dx)$.
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\item For any choice of $f \in C_c^+(G; \real) \setminus \bracs{0}$, the mapping $\Delta_G: G \to (0, \infty)$ defined by $y \mapsto \int R_{y^{-1}}f d\mu$ is a continuous group homomorphism.
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\item For each $A \in \cb_G$, let $\nu(A) = \mu(A^{-1})$, then $\nu(dx) = \Delta(x^{-1})\mu(dx)$.
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\end{enumerate}
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The homomorphism $\Delta_G: G \to (0, \infty)$ is the \textbf{modular function} of $G$.
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\end{definition}
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\begin{proof}[Proof, {{\cite[Proposition 2.24, Proposition 2.31]{FollandHarmonic}}}. ]
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(1), (2): For each $y \in G$, $\mu_y$ is also a left Haar measure. By \hyperef[Haar's Theorem]{theorem:haar}, there exists $\lambda > 0$ such that $\mu_y = \lambda \mu$. In which case,
|
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\[
|
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\Delta_G(y^{-1}) = \frac{\int R_y f d\mu}{\int f d\mu} = \frac{\int R_y g d\mu}{\int g d\mu} = \frac{\mu(Ay)}{\mu(A)} = \frac{\mu(By)}{\mu(B)}
|
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\]
|
||||
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(3): By \autoref{proposition:haar-translation}, the mapping $y \mapsto \int R_y f d\mu$ is continuous. For any $x, y \in G$ and $A \in \cb_G$ with $\mu(A) > 0$,
|
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\[
|
||||
\Delta_G(xy)\mu(A) = \mu(Axy) = \Delta_G(y)\mu(Ax) = \Delta_G(y)\Delta_G(x)\mu(A)
|
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\]
|
||||
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(4): Let $f \in C_c^+(G)$ and $y \in G$, then
|
||||
\begin{align*}
|
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\int R_{y}f(x) \Delta_G(x^{-1})\mu(dx) &= \Delta_G(y)\int f(xy)\Delta_G[(xy)^{-1}]\mu(dx) \\
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&= \int f(x)\Delta(x^{-1})\mu(dx)
|
||||
\end{align*}
|
||||
|
||||
so $\Delta_G(x^{-1})\mu(dx)$ is a right Haar measure. By Haar's Theorem, there exists $\lambda > 0$ such that $\nu(dx) = \lambda \Delta_G(x^{-1})\mu(dx)$.
|
||||
|
||||
If $\lambda \ne 1$, then there exists $U \in \cn_G(1)$ symmetric and compact such that $|\Delta_G(x^{-1}) - 1| \le 2^{-1}|\lambda - 1|$ on $U$. In which case, by symmetry, $\mu(U) = \nu(U)$, and
|
||||
\begin{align*}
|
||||
|\lambda - 1|\mu(U) &= |\lambda \nu(U) - \mu(U)| = \abs{\int_U \Delta_G(x^{-1})-1 \mu(dx)} \\
|
||||
&\le \frac{1}{2}|\lambda - 1|\mu(U)
|
||||
\end{align*}
|
||||
|
||||
which contradicts the fact that $\lambda \ne 1$. Therefore $\lambda = 1$, and $\nu(dx) = \Delta_G(x^{-1})\mu(dx)$.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Unimodular]
|
||||
\label{definition:lcg-unimodular}
|
||||
Let $G$ be a locally compact group, then $G$ is \textbf{unimodular} if $\Delta_G = 1$.
|
||||
\end{definition}
|
||||
|
||||
|
||||
|
||||
Reference in New Issue
Block a user