30.4 The $L^{1}$ Group Algebra
Definition 30.4.1 (Convolution of Measures).label Let $G$ be a locally compact group, and $\mu, \nu \in M_{R}(G; \complex)$ be finite Radon measures, then the convolution of $\mu$ and $\nu$ is the Radon measure $\mu * \nu$ defined by
for all $\phi \in C_{c}(G; \complex)$[1].
Proposition 30.4.2.label Let $G$ be a locally compact group, then:
- (1)
For any $\mu, \nu, \sigma \in M_{R}(G; \complex)$, $(\mu * \nu) * \sigma = \mu * (\nu * \sigma)$.
- (2)
For any $\mu, \nu \in M_{R}(G; \complex)$, $\norm{\mu * \nu}_{\text{var}}\le \norm{\mu}_{\text{var}}\norm{\nu}_{\text{var}}$.
- (3)
Convolution on $M_{R}(G; \complex)$ is commutative if and only if $G$ is commutative.
Proof. (1): Let $\phi \in C_{c}(G; \complex)$, then
(2): Let $\phi \in C_{c}(G; \complex)$, then
(3): For any $g, h \in G$ and $\phi \in C_{c}(G; \complex)$,
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.$\square$
Definition 30.4.3 (Measure Algebra).label Let $G$ be a locally compact group. For each $\mu \in M_{R}(G; \complex)$, let
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 measure algebra of $G$.
Proof. By (2) of Unknown (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 $\phi \in C_{c}(G; \complex)$,
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 $\phi \in C_{c}(G; \complex)$,
Therefore $(\mu * \nu)^{*} = \nu^{*} * \mu^{*}$.$\square$
Definition 30.4.4 (Convolution).label Let $G$ be a locally compact group and $f, g: G \to \complex$ be measurable functions, then the convolution of $f$ and $g$ is the function
defined wherever the integral exists.
Theorem 30.4.5 (Young’s Inequality).label 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
- (1)
If $p = \infty$, then $(f * g)(x) = \int_{G} f(y)g(y^{-1}x)dy$ exists for all $x \in G$, with
\[\norm{f * g}_{L^\infty(G; \complex)}\le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^\infty(G; \complex)}\] - (2)
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
\[\norm{f * g}_{L^p(G; \complex)}\le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^p(G; \complex)}\]
Proof, [Proposition 2.40, Fol99]. (1): By Definition 30.3.1, 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 Hölder’s Inequality
(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 Minkowski’s Inequality for integrals,
$\square$
Lemma 30.4.6.label Let $G$ be a locally compact group, $f, g \in L^{1}(G; \complex)$, $\mu = fdx$, and $\nu = gdx$, then $\mu * \nu = (f * g)dx$.
Proof. Let $\phi \in C_{c}(G; \complex)$, then
$\square$
Definition 30.4.7 ($L^{1}$ Group Algebra).label 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 $f \in L^{1}(G; \complex)$,
The algebra $L^{1}(G; \complex)$ is the $L^{1}$ group algebra of $G$.
Proof. By Lemma 30.4.6, the convolution operation on $L^{1}(G; \complex)$ agrees with the convolution on $M_{R}(G; \complex)$. By Young’s Inequality, $L^{1}(G; \complex)$ is a Banach subalgebra of $M_{R}(G; \complex)$.
For any $f \in L^{1}(G; \complex)$, let $\mu = f dx$, then for each $A \in \cb_{G}$,
so $f^{*}dx = d\mu^{*} = \Delta(x^{-1}) \ol{f(x^{-1})}dx$, and $f^{*} = \Delta(x^{-1}) \ol{f(x^{-1})}$.$\square$
Proposition 30.4.8.label Let $G$ be a locally compact group, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then
- (1)
The mapping $G \times L^{p}(G; E) \to L^{p}(G; E)$ defined by $(x, f) \mapsto L_{x}f$ is jointly continuous.
- (2)
The mapping $G \times L^{p}(G; E) \to L^{p}(G; E)$ defined by $(x, f) \mapsto R_{x}f$ is jointly continuous.
Proof, [Proposition 2.42, Fol16]. (1): Let $\eps > 0$, $x, y \in G$, and $f, g \in L^{p}(\mu; E)$, then
By Proposition 25.1.7, there exists $\phi \in C_{c}(G; E)$ such that $\norm{\phi - f}_{L^p(\mu; E)}< \eps$. In which case,
By Proposition 30.1.2, 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
(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 Proposition 5.16.3. For any $f \in L^{p}(\mu; E)$ and $x \in V$,
Let $f, g \in L^{p}(G; E)$ and $x \in V$, then
By Proposition 25.1.7, there exists $\phi \in C_{c}(G; E)$ with $\norm{f - \phi}_{L^p(G; E)}< \eps$. Thus
Since $V$ is compact, $\bracsn{\phi \ne 0}V^{-1}$ is relatively compact, and of finite measure. By Proposition 30.1.2, $\phi$ 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 $x \in W$,
and
$\square$
Proposition 30.4.9.label Let $G$ be a locally compact group and $f, g: G \to \complex$ be Borel measurable, then:
- (1)
For any $x \in G$, $(f * g)(x) = \int_{G} f(y)L_{y}g(x) dy = \int_{G} R_{y}f(x) g(y^{-1}) dy$.
- (2)
For any $z \in G$, $L_{z}(f * g) = (L_{z}f) * g$, and $R_{z}(f * g) = f * (R_{z} g)$.
- (3)
If $f \in L^{1}(G; \complex)$ and $g \in L^{\infty}(G; \complex)$, then $f * g$ is left uniformly continuous[2].
- (4)
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
\[\norm{f * g}_{u} \le \norm{f}_{L^p(G; \complex)}\norm{g}_{L^q(G; \complex)}\]
Proof, [Proposition 2.41, Proposition 2.43, Fol16]. (1): For any $x \in G$, by translation-invariance,
(2): Let $x, z \in G$, then by translation-invariance,
(3): By (2), $L_{z}(f * g) - f * g = (L_{z} f - f) * g$. By continuity of translation and Young’s Inequality,
Therefore $(f * g)$ is left uniformly continuous.
(4): Let $h: G \to \complex$ be defined by $h(y) = g(y^{-1})$, then for each $x \in G$, Hölder’s Inequality implies that
Since $G$ is unimodular, $\norm{h}_{L^q(G; \complex)}= \norm{g}_{L^q(G; \complex)}$, so
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 Proposition 25.1.7, $C_{c}(G; \complex)$ is dense in $L^{p}(G; \complex)$ and $L^{q}(G; \complex)$. By Proposition 5.5.3,
$\square$
Unfortunately, $L^{1}(G)$ is not a $C^{*}$-algebra, and its approximate identity requires a manual construction. Fortunately, the candidate is clear.
Proposition 30.4.10 (Existence of Approximate Identity).label 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:
- (1)
For each $p \in [1, \infty)$ and $f \in L^{p}(G; \complex)$, $\phi_{U} * f \to f$ in $L^{p}(G; \complex)$.
- (2)
For each left uniformly continuous $f \in C(G; \complex)$, $\phi_{U} * f \to f$ uniformly.
If in addition, $\phi_{U}(x^{-1}) = \phi_{U}(x)$ for all $x \in G$, then:
- (3)
For each $p \in [1, \infty)$ and $f \in L^{p}(G; \complex)$, $f * \phi_{U} \to f$ in $L^{p}(G; \complex)$.
- (4)
For each right uniformly continuous $f \in C(G; \complex)$, $f * \phi_{U} \to f$ uniformly.
Proof, [Proposition 2.44, Fol16]. (1): Let $U \in \fB$, $f \in L^{p}(G; \complex)$, and $x \in G$, then since $\int_{G} \phi_{U} = 1$,
By Minkowski’s Inequality for integrals,
Therefore by continuity of 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 $y \in G$,
By Minkowski’s Inequality for integrals,
and $f * \phi_{U} \to f$ in $L^{p}(G; \complex)$ by continuity of translation.$\square$
- There seem to be some subtleties with having to take the Radon product here.keyboard_return
- Folland also claims that $g * f$ is right uniformly continuous [Proposition 2.41, Proposition 2.43, Fol16]. 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 $g = \one$.keyboard_return
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