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First draft of L^1.

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\section{The $L^1$ Group Algebra}
\label{section:group-algebra-l1}
\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
\[
\int_{G}\phi(x)(\mu * \nu)(dx) = \iint_{G \times G} \phi(xy)\mu(dx)\nu(dy)
\]
for all $\phi \in C_c(G; \complex)$\footnote{There seems to be some subtleties with having to take the Radon product here.}.
\end{definition}
\begin{proposition}
\label{proposition:convolution-measures-property}
Let $G$ be a locally compact group, then:
\begin{enumerate}
\item For any $\mu, \nu, \sigma \in M_R(G; \complex)$, $(\mu * \nu) * \sigma = \mu * (\nu * \sigma)$.
\item For any $\mu, \nu \in M_R(G; \complex)$, $\norm{\mu * \nu}_{\text{var}} \le \norm{\mu}_{\text{var}}\norm{\nu}_{\text{var}}$.
\item Convolution on $M_R(G; \complex)$ is commutative if and only if $G$ is commutative.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $\phi \in C_c(G; \complex)$, then
\begin{align*}
\int_G \phi d(\mu * (\nu * \sigma)) &= \iint_{G \times G} \phi(xy)\mu(dx)(\nu * \sigma)(dy) \\
&= \iiint_{G \times G \times G}\phi(xyz) \mu(dx) \nu(dy)\sigma(dz) \\
&= \iint_{G \times G} \phi(yz) (\mu * \nu)(dy) \sigma(dz) = \int_G \phi d((\mu * \nu) * \sigma)
\end{align*}
(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}}
\end{align*}
(3): For any $g, h \in G$ and $\phi \in C_c(G; \complex)$,
\[
\int_G \phi d(\delta_g * \delta_h) = \iint_{G \times G} \phi(xy) \delta_g(dx)\delta_h(dy) = \phi(gh)
\]
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.
\end{proof}
\begin{definition}[Measure Algebra]
\label{definition:group-measure-algebra}
Let $G$ be a locally compact group. For each $\mu \in M_R(G; \complex)$, let
\[
\mu^*: \cb_G \to \complex \quad \mu^*(A) = \ol{\mu(A^{-1})}
\]
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 $G$.
\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)
\end{align*}
so $\mu * \delta_e = \mu = \delta_e * \mu$, and $\delta_e$ 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*}
\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*}
Therefore $(\mu * \nu)^* = \nu^* * \mu^*$.
\end{proof}
\begin{definition}[Convolution]
\label{definition:group-convolution-functions}
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
\[
f * g(x) = \int_G f(y)g(y^{-1}x)dy
\]
defined on wherever the integral exists.
\end{definition}
\begin{theorem}[Young's Inequality]
\label{theorem:young-fourier}
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
\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
\[
\norm{f * g}_{L^\infty(G; \complex)}, \norm{g * f}_{L^\infty(G; \complex)} \le \norm{f}_{L^1(G; \complex)} \norm{g}_{L^\infty(G; \complex)}
\]
\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
\[
\norm{f * g}_{L^p(G; \complex)} \le \norm{f}_{L^1(G; \complex)} \norm{g}_{L^p(G; \complex)}
\]
\end{enumerate}
\end{theorem}
\begin{proof}
(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}
\[
|(f * g)(x)| = \abs{\int_G f(y)g(y^{-1}x)dy} \le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^\infty(G; \complex)}
\]
(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)}
\end{align*}
\end{proof}
\begin{lemma}
\label{lemma:convolution-measure-function-agree}
Let $G$ be a locally compact group, $f, g \in L^1(G; \complex)$, $\mu = fdx$, and $\nu = gdx$, then $\mu * \nu = (f * g)dx$.
\end{lemma}
\begin{proof}
Let $\phi \in C_c(G; \complex)$, then
\begin{align*}
\int_G \phi(x) (f * g)(x)dx &= \iint_{G \times G} \phi(x) f(y)g(y^{-1}x)dydx \\
&= \iint_{G \times G} \phi(yx)g(x)dxf(y)dy \\
&= \iint_{G \times G} \phi(yx) \mu(dy)\nu(dx) = \int_G \phi d(\mu * \nu)
\end{align*}
\end{proof}
\begin{definition}[$L^1$ Group Algebra]
\label{definition:l1-group-algebra}
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 $f \in L^1(G; \complex)$,
\[
f^*(x) = \Delta(x^{-1}) \ol{f(x^{-1})}
\]
The algebra $L^1(G; \complex)$ is the \textbf{$L^1$ group algebra} of $G$.
\end{definition}
\begin{proof}
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 $M_R(G; \complex)$.
For any $f \in L^1(G; \complex)$, let $\mu = f dx$, then for each $A \in \cb_G$,
\[
\mu^*(A) = \ol{\mu(A^{-1})} = \ol{\int_{A^{-1}}d\mu} = \int_{A^{-1}} \ol{f(x)}dx = \int_A \ol{f(x^{-1})}\Delta_G(x^{-1})dx
\]
so $f^*dx = d\mu^* = \Delta(x^{-1}) \ol{f(x^{-1})} dx$, and $f^* = \Delta(x^{-1}) \ol{f(x^{-1})}$.
\end{proof}
\begin{proposition}
\label{proposition:haar-translation}
Let $G$ be a locally compact group, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then
\begin{enumerate}
\item The mapping $G \times L^p(G; E) \to L^p(G; E)$ defined by $(x, f) \mapsto L_xf$ is jointly continuous.
\item The mapping $G \times L^p(G; E) \to L^p(G; E)$ defined by $(x, f) \mapsto R_xf$ is jointly continuous.
\end{enumerate}
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition 2.42]{FollandHarmonic}}}. ]
(1): Let $\eps > 0$, $x, y \in G$, and $f, g \in L^p(\mu; E)$, then
\begin{align*}
\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)} \\
&= \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,
\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}
\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
\[
\norm{L_xf - L_yf}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)}
\]
(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)}
\end{align*}
Let $f, g \in L^p(G; E)$ and $x \in V$, then
\begin{align*}
\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)} \\
&\le \norm{R_xf - f}_{L^p(G; E)} + C^{1/p}\norm{f - g}_{L^p(G; E)}
\end{align*}
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)}
\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 $x \in W$,
\[
\norm{R_x\phi - \phi}_{L^p(G; E)} \le \norm{R_x\phi - \phi}_u \cdot \normn{\one_{\bracsn{\phi \ne 0}}}_{L^p(G; \real)} \le \epsilon
\]
and
\[
\norm{R_xg - f}_{L^p(G; E)} \le (2 + C^{1/p})\eps + C^{1/p}\norm{f - g}_{L^p(G; E)}
\]
\end{proof}
\begin{proposition}
\label{proposition:convolution-function-gymnastics}
Let $G$ be a locally compact group and $f, g: X \to \complex$ 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
\[
\norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)}
\]
\end{enumerate}
\end{proposition}
\begin{proof}
(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 \\
&= \int_G f(xy)g(y^{-1})dy = \int_G R_yf(x)g(y^{-1})dy
\end{align*}
(2): Let $x, z \in G$, then by translation-invariance,
\begin{align*}
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)
\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},
\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*}
Therefore $(f * g)$ is left uniformly continuous, and $(g * f)$ is right 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)}
\end{align*}
Since $G$ is unimodular, $\norm{h}_{L^q(G; \complex)} = \norm{g}_{L^q(G; \complex)}$, so
\[
\norm{f * g}_u \le \norm{f}_{L^p(G; \complex)}\norm{h}_{L^q(G; \complex)}
\]
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},
\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.
\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:
\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.
\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.
\end{enumerate}
\end{proposition}
\begin{proof}
(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(X; \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}
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Thus $0 < \int g d\mu \le n\mu(U)$ and $0 < \int g d\mu \le n \int f d\mu$.
\end{proof}
\begin{proposition}
\label{proposition:haar-translation}
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
\[
G \times L^p(\mu; E) \quad (x, f) \mapsto L_xf
\]
\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}
is jointly continuous. Similarly, if $\nu: \cb_G \to [0, \infty]$ is a right Haar measure, then
\[
G \times L^p(\mu; E) \quad (x, f) \mapsto R_xf
\]
is also jointly continuous.
\end{proposition}
\begin{proof}[Proof of the left case. ]
Let $\eps > 0$, $x, y \in G$, and $f, g \in L^p(\mu; E)$, then
\begin{align*}
\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)} \\
&= \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,
\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}
\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
\[
\norm{L_xf - L_yf}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)}
\]
\end{proof}
src/measure/lcg/modular.tex
+6 -2
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@@ -42,8 +42,12 @@
\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}