From c7ab20155f914d8cfea48b6b286baeda7931a359 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sun, 20 Sep 2026 14:17:23 -0400 Subject: [PATCH] Revision 1. --- src/measure/lcg/algebra.tex | 44 ++++++++++++++++++------------------- 1 file changed, 22 insertions(+), 22 deletions(-) diff --git a/src/measure/lcg/algebra.tex b/src/measure/lcg/algebra.tex index f8f61cb..7144f8b 100644 --- a/src/measure/lcg/algebra.tex +++ b/src/measure/lcg/algebra.tex @@ -8,7 +8,7 @@ \int_{G}\phi(x)(\mu * \nu)(dx) = \iint_{G \times G} \phi(xy)\mu(dx)\nu(dy) \] - for all $\phi \in C_c(G; \complex)$\footnote{There seems to be some subtleties with having to take the Radon product here.}. + for all $\phi \in C_c(G; \complex)$\footnote{There seem to be some subtleties with having to take the Radon product here.}. \end{definition} @@ -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}\norm{\phi}_u\norm{\mu}_{\text{var}}\nu(dy)} \le \norm{\phi}_u\norm{\mu}_{\text{var}}\norm{\nu}_{\text{var}} + &\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}} \end{align*} (3): For any $g, h \in G$ and $\phi \in C_c(G; \complex)$, @@ -40,7 +40,7 @@ \int_G \phi d(\delta_g * \delta_h) = \iint_{G \times G} \phi(xy) \delta_g(dx)\delta_h(dy) = \phi(gh) \] - so $\delta_g * \delta_h = \delta_{gh}$, and $\delta_g * \delta_h = \delta_h * \delta_g$ if and only if $gh = gh$. Therefore the convolution in $M_R(G; \complex)$ is commutative if and only if $G$ is commutative. + 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. \end{proof} \begin{definition}[Measure Algebra] @@ -64,7 +64,7 @@ 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*} - \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) \\ + \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) \\ &= \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) \\ &= \iint_{G \times G}\phi(yx)\nu^*(dy)\mu^*(dx) = \int_{G}\phi d(\nu^* * \mu^*) \end{align*} @@ -79,7 +79,7 @@ f * g(x) = \int_G f(y)g(y^{-1}x)dy \] - defined on wherever the integral exists. + defined wherever the integral exists. \end{definition} \begin{theorem}[Young's Inequality] @@ -106,8 +106,8 @@ (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{g}_{L^p(X; \complex)}\int_G |f(y)|dy = \norm{f}_{L^1(G; \complex)} \norm{g}_{L^p(G; \complex)} + \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{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} @@ -162,23 +162,23 @@ &= \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{f - g}_{L^p(\mu; E)} \end{align*} - 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, + 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, \begin{align*} \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} + &\le 2\eps + \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 $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/\mu\bracs{\phi \ne 0}$. 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 $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/(\mu\bracs{\phi \ne 0}^{1/p})$. Thus if $x^{-1}y \in V$, then \[ \norm{L_xf - L_yf}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)} \] (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 $x \in V$, \begin{align*} - \int_G \norm{R_xf}_E^p &= \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(y)^{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) \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)} \end{align*} Let $f, g \in L^p(G; E)$ and $x \in V$, then @@ -193,7 +193,7 @@ &\le (1 + C^{1/p})\epsilon + \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 csae, for any $x \in W$, + 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 $x \in W$, \[ \norm{R_x\phi - \phi}_{L^p(G; E)} \le \norm{R_x\phi - \phi}_u \cdot \normn{\one_{\bracsn{\phi \ne 0}}}_{L^p(G; \real)} \le \epsilon \] @@ -207,14 +207,14 @@ \begin{proposition} \label{proposition:convolution-function-gymnastics} - Let $G$ be a locally compact group and $f, g: X \to \complex$ be Borel measurable, then: + Let $G$ be a locally compact group and $f, g: G \to \complex$ be Borel measurable, then: \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 $G$ 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 \[ - \norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)} + \norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{g}_{L^q(G; \complex)} \] \end{enumerate} \end{proposition} @@ -230,10 +230,10 @@ L_z(f * g)(x) &= (f * g)(z^{-1}x) = \int_G f(y)g(y^{-1}z^{-1}x) dy \\ &= \int_G f(z^{-1}y)g(y^{-1}x)dy = \int_G L_zf(y)g(y^{-1}x)dy \\ &= (L_zf * g)(x) \\ - R_z(f * g)(x) &= (f * g)(xz) = \int_G f(y)g(xz)dy = (f * R_zg)(x) + 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 (1), $L_z(f * g) - f * g = (L_z f - f) * g$ and $R_z(g * f) = g * (R_zf - f)$. By \hyperref[continuity of translation]{proposition:haar-translation} and \hyperref[Young's Inequality]{theorem:young-fourier}, + (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 \\ @@ -244,7 +244,7 @@ (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 = \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 @@ -266,13 +266,13 @@ Unfortunately, $L^1(G)$ is not a $C^*$-algebra, and its approximate identity req 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: \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 $f \in C(G; \complex)$, $\phi_U * f$ uniformly. + \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 $f \in C(G; \complex)$, $f * \phi_U$ uniformly. + \item For each right uniformly continuous $f \in C(G; \complex)$, $f * \phi_U \to f$ uniformly. \end{enumerate} \end{proposition} \begin{proof} @@ -283,8 +283,8 @@ Unfortunately, $L^1(G)$ is not a $C^*$-algebra, and its approximate identity req 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(X; \complex)}^p dy }^{1/p} \\ + \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*}