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[1].
- (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,
$\square$
- 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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