From d866e3e448a95d5b11c0fff355aa95e975969e00 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 21 Sep 2026 11:27:48 -0400 Subject: [PATCH] The $L^1$ Group Algebra is past revisions! --- src/measure/lcg/algebra.tex | 72 +++++++++++++++++++------------------ src/measure/lcg/index.tex | 1 + 2 files changed, 38 insertions(+), 35 deletions(-) diff --git a/src/measure/lcg/algebra.tex b/src/measure/lcg/algebra.tex index 7144f8b..34e5dcd 100644 --- a/src/measure/lcg/algebra.tex +++ b/src/measure/lcg/algebra.tex @@ -3,7 +3,7 @@ \begin{definition}[Convolution of Measures] \label{definition:lcg-convolution-measures} - Let $G$ be a locally compact group, $\mu, \nu \in M_R(G; \complex)$ be finite Radon measures, then the \textbf{convolution} of $\mu$ and $\nu$ is the Radon measure $\mu * \nu$ defined by + 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 \[ \int_{G}\phi(x)(\mu * \nu)(dx) = \iint_{G \times G} \phi(xy)\mu(dx)\nu(dy) \] @@ -32,7 +32,7 @@ (2): Let $\phi \in C_c(G; \complex)$, then \begin{align*} \abs{\int_G \phi d(\mu * \nu)} &= \abs{\iint_{G \times G}\phi(xy)\mu(dx)\nu(dy)} \\ - &\le \abs{\iint_{G \times G}\norm{\phi}_u\norm{\mu}_{\text{var}}\nu(dy)} \le \norm{\phi}_u\norm{\mu}_{\text{var}}\norm{\nu}_{\text{var}} + &\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}} \end{align*} (3): For any $g, h \in G$ and $\phi \in C_c(G; \complex)$, @@ -50,17 +50,17 @@ \mu^*: \cb_G \to \complex \quad \mu^*(A) = \ol{\mu(A^{-1})} \] - then $M_R(G; \complex)$ equipped with the convolution product and the above involution is an involutive unital Banach algebra with identity $\delta_e$. The algebra $M_R(G; \complex)$ is the \textbf{measure algebra} of $G$. + 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 $G$. \end{definition} \begin{proof} 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 $\phi \in C_c(G; \complex)$, \begin{align*} - \int_G \phi d(\mu * \delta_e) &= \iint_{G \times G} \phi(xy)\mu(dx)\delta_e(dy) \\ - &= \iint_{G \times G} \phi(xy)\delta_e(y)\mu(dx) = \int_G \phi(x)\mu(dx) \\ - \int_G \phi d(\delta_e * \mu) &= \iint_{G \times G}\phi(xy)\delta_e(dx)\mu(dy) = \int_G \phi(y)\mu(dy) + \int_G \phi d(\mu * \delta_1) &= \iint_{G \times G} \phi(xy)\mu(dx)\delta_1(dy) \\ + &= \iint_{G \times G} \phi(xy)\delta_1(dy)\mu(dx) = \int_G \phi(x)\mu(dx) \\ + \int_G \phi d(\delta_1 * \mu) &= \iint_{G \times G}\phi(xy)\delta_1(dx)\mu(dy) = \int_G \phi(y)\mu(dy) \end{align*} - so $\mu * \delta_e = \mu = \delta_e * \mu$, and $\delta_e$ is the identity. + so $\mu * \delta_1 = \mu = \delta_1 * \mu$, and $\delta_1$ is the identity. It remains to show that $\mu \mapsto \mu^*$ is an involution. Let $\mu, \nu \in M_R(G; \complex)$, then for any $\phi \in C_c(G; \complex)$, \begin{align*} @@ -88,7 +88,7 @@ \begin{enumerate} \item If $p = \infty$, then $(f * g)(x) = \int_G f(y)g(y^{-1}x)dy$ exists for all $x \in G$, with \[ - \norm{f * g}_{L^\infty(G; \complex)}, \norm{g * f}_{L^\infty(G; \complex)} \le \norm{f}_{L^1(G; \complex)} \norm{g}_{L^\infty(G; \complex)} + \norm{f * g}_{L^\infty(G; \complex)}\le \norm{f}_{L^1(G; \complex)} \norm{g}_{L^\infty(G; \complex)} \] \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 @@ -97,7 +97,7 @@ \] \end{enumerate} \end{theorem} -\begin{proof} +\begin{proof}[Proof, {{\cite[Proposition 2.40]{Folland}}}. ] (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} \[ |(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)} @@ -105,8 +105,8 @@ (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}, \begin{align*} - \int_G |(f*g)(x)|^p dx &= \int_G \abs{\int_G f(y)g(y^{-1}x)dy}^pdx \\ - \braks{\int_G |(f*g)(x)|^p dx}^{1/p} &\le \int_{G} \braks{\int_G|f(y)g(y^{-1}x)|^p dx}^{1/p} dy \\ + \norm{f * g}_{L^p(G; \complex)} &= \braks{\int_G \abs{\int_G f(y)g(y^{-1}x)dy}^pdx}^{1/p} \\ + &\le \int_{G} \braks{\int_G|f(y)g(y^{-1}x)|^p dx}^{1/p} dy \\ &= \norm{g}_{L^p(G; \complex)}\int_G |f(y)|dy = \norm{f}_{L^1(G; \complex)} \norm{g}_{L^p(G; \complex)} \end{align*} \end{proof} @@ -167,18 +167,19 @@ \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)} \\ &+ \norm{L_yf - L_y \phi}_{L^p(\mu; E)} \\ &= 2\norm{f - \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\ - &\le 2\eps + \normn{L_{x^{-1}y}\phi - \phi}_u\mu\bracs{\phi \ne 0}^{1/p} + &\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}\\ + &\le 2\eps + 2\normn{L_{x^{-1}y}\phi - \phi}_u\mu\bracs{\phi \ne 0}^{1/p} \end{align*} - By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that if $x^{-1}y \in V$, then $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/(\mu\bracs{\phi \ne 0}^{1/p})$. Thus if $x^{-1}y \in V$, then + 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 \[ - \norm{L_xf - L_yf}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)} + \norm{L_xf - L_yg}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)} \] - (2): Let $V \in \cn_G(1)$ be compact, then since $\Delta_G: G \to (0, \infty)$ is a continuous homomorphism, $C := \sup_{y \in V}\Delta_G(y) < \infty$ by \autoref{proposition:compact-extensions}. For any $f \in L^p(\mu; E)$ and $x \in V$, + (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 $x \in V$, \begin{align*} - \int_G \norm{R_xf(y)}_E^p dy &= \int_G \norm{f(yx)}_E^p dy = \int_G \Delta_G(x) \norm{f(y)}_E^pdy \\ - \norm{R_xf}_{L^p(G; E)} &= \Delta_G(x)^{1/p}\norm{f}_{L^p(G; E)} \le C^{1/p}\norm{f}_{L^p(G; E)} + \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 \\ + \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)} \end{align*} Let $f, g \in L^p(G; E)$ and $x \in V$, then @@ -190,13 +191,15 @@ By \autoref{proposition:radon-cc-dense}, there exists $\phi \in C_c(G; E)$ with $\norm{f - \phi}_{L^p(G; E)} < \eps$. Thus \begin{align*} \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)} \\ - &\le (1 + C^{1/p})\epsilon + \norm{R_x\phi - \phi}_{L^p(G; E)} + &\le (1 + C^{1/p})\eps + \norm{R_x\phi - \phi}_{L^p(G; E)} \end{align*} - By \autoref{proposition:lcg-cc-uc}, there exists $W \in \cn_G(1)$ such that $W \subset V$ and $\norm{R_x\phi - \phi}_u < \eps/\normn{\one_{\bracsn{\phi \ne 0}}}_{L^p(G; \real)}$. In which case, for any $x \in W$, - \[ - \norm{R_x\phi - \phi}_{L^p(G; E)} \le \norm{R_x\phi - \phi}_u \cdot \normn{\one_{\bracsn{\phi \ne 0}}}_{L^p(G; \real)} \le \epsilon - \] + 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 $x \in W$, + \begin{align*} + \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)} \\ + &\le \norm{R_x\phi - \phi}_u \cdot \normn{\one_{\bracsn{\phi \ne 0}V^{-1}}}_{L^p(G; \real)} \le \eps + \end{align*} + and \[ @@ -211,14 +214,14 @@ \begin{enumerate} \item For any $x \in G$, $(f * g)(x) = \int_G f(y)L_yg(x) dy = \int_G R_yf(x) g(y^{-1}) dy$. \item For any $z \in G$, $L_z(f * g) = (L_zf) * g$, and $R_z(f * g) = f * (R_z g)$. - \item If $f \in L^1(G; \complex)$ and $g \in L^\infty(G; \complex)$, then $f * g$ is left uniformly continuous, and $g * f$ is right uniformly continuous. + \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 $g = \one$.}. \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 \[ \norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{g}_{L^q(G; \complex)} \] \end{enumerate} \end{proposition} -\begin{proof} +\begin{proof}[Proof, {{\cite[Proposition 2.41, Proposition 2.43]{FollandHarmonic}}}. ] (1): For any $x \in G$, by translation-invariance, \begin{align*} (f * g)(x) &= \int_G f(y)g(y^{-1}x) dy = \int_G f(y)L_yg(x)dy \\ @@ -233,18 +236,17 @@ R_z(f * g)(x) &= (f * g)(xz) = \int_G f(y)g(y^{-1}xz)dy = (f * R_zg)(x) \end{align*} - (3): By (2), $L_z(f * g) - f * g = (L_z f - f) * g$ and $R_z(g * f) - g * f = g * (R_zf - f)$. By \hyperref[continuity of translation]{proposition:haar-translation} and \hyperref[Young's Inequality]{theorem:young-fourier}, - \begin{align*} - \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 \\ - \lim_{z \to 1}\norm{R_z(g * f) - (g * f)}_{L^\infty(G; \complex)} &\le \lim_{z \to 1}\norm{R_zf - f}_{L^1(G; \complex)} \cdot \norm{g}_{L^\infty(G; \complex)} = 0 \\ - \end{align*} + (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}, + \[ + \lim_{z \to 1}\norm{L_z(f * g) - (f * g)}_{L^\infty(G; \complex)} \le \lim_{z \to 1}\norm{L_zf - f}_{L^1(G; \complex)} \cdot \norm{g}_{L^\infty(G; \complex)} = 0 + \] - Therefore $(f * g)$ is left uniformly continuous, and $(g * f)$ is right uniformly continuous. + Therefore $(f * g)$ is left uniformly continuous. (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 \begin{align*} (f * g)(x) &= \int_G f(y)g(y^{-1}x) dy = \int_G f(xy)g(y^{-1}) dy \\ - |(f * g)(x)| &\le \int_G |f(xy)h(y)| dy = \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)} + |(f * g)(x)| &\le \int_G |f(xy)h(y)| dy \le \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)} \end{align*} Since $G$ is unimodular, $\norm{h}_{L^q(G; \complex)} = \norm{g}_{L^q(G; \complex)}$, so @@ -252,18 +254,18 @@ \norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)} \] - 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}, + 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}, \begin{align*} 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)} \\ &\subset \ol{C_c(G; \complex) * C_c(G; \complex)}^{C_0(G; \complex)} = C_0(G; \complex) \end{align*} \end{proof} -Unfortunately, $L^1(G)$ is not a $C^*$-algebra, and its approximate identity requires a manual construction. +Unfortunately, $L^1(G)$ is not a $C^*$-algebra, and its approximate identity requires a manual construction. Fortunately, the candidate is clear. \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 $\phi_U \in C_c(G; \real)$ with $\phi_U \ge 0$ and $\int_G \phi_U = 1$, then: + 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. @@ -275,7 +277,7 @@ Unfortunately, $L^1(G)$ is not a $C^*$-algebra, and its approximate identity req \item For each right uniformly continuous $f \in C(G; \complex)$, $f * \phi_U \to f$ uniformly. \end{enumerate} \end{proposition} -\begin{proof} +\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 diff --git a/src/measure/lcg/index.tex b/src/measure/lcg/index.tex index 32d87f2..ee593da 100644 --- a/src/measure/lcg/index.tex +++ b/src/measure/lcg/index.tex @@ -4,3 +4,4 @@ \input{./lcg.tex} \input{./haar.tex} \input{./modular.tex} +\input{./algebra.tex}