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Bokuan Li
2026-09-20 14:17:23 -04:00
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src/measure/lcg/algebra.tex
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@@ -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*}