Lemma 31.4.11 (Riemann Sums).label Let $G$ be a locally compact group and $\phi \in L^{1}(G; \complex)^{*}$, then for any $f, g \in L^{1}(G; \complex)$,
\[\dpn{f * g, \phi}{L^1(G; \complex)}= \int_{G} f(y)\dpn{L_yg, \phi}{L^1(G; \complex)}dy\]
Proof. Let $K \subset G$ be compact and $\eps > 0$. By continuity of translation, there exists a partition $K = \bigsqcup_{j = 1}^{n} U_{j}$ of $K$ and $\seqf{y_j}$ such that:
(1)
For each $1 \le j \le n$, $y_{j} \in U_{j}$.
(2)
For each $1 \le j \le n$ and $y \in U_{j}$,
\[\normn{L_yg - L_{y_j}g}_{L^1(G; \complex)}\vee \normn{\dpn{L_yg, \phi}{L^1(G; \complex)} - \dpn{L_{y_j}g, \phi}{L^1(G; \complex)}}< \eps\]
Since $g \in L^{1}(G; \complex)$, $\bracsn{g \ne 0}$ is $\sigma$-finite. By Tonelli’s Theorem,
\begin{align*}\norm{\one_K * g - \sum_{j = 1}^n|U_j| L_{y_j}g}_{L^1(G)}&= \int_{G}\abs{\one_K * g(x) - \sum_{j = 1}^n|U_j| L_{y_j}g(x)}dx \\&\le \iint_{G \times G}\abs{\one_K(y)L_yg(x) - \sum_{j = 1}^n\one_{U_j}(y) L_{y_j}g(x)}dy dx \\&\le \sum_{j = 1}^{n} \int_{G}\int_{U_j}|L_{y}g(x) - L_{y_j}g(x)|dydx \\&\le \sum_{j = 1}^{n} \eps |U_{j}| = \eps |K|\end{align*}
Similarly,
\[\abs{\int_G \one_K(y)\dpn{L_yg, \phi}{L^1(G; \complex)}dy - \sum_{j = 1}^n |U_j|\dpn{L_{y_j}g, \phi}{L^1(G; \complex)}}\le \eps |K|\]
By linearity,
\[\angles{\sum_{j = 1}^n |U_j|L_{y_j}g, \phi}_{L^1(G; \complex)}= \sum_{j = 1}^{n} |U_{j}| \dpn{L_{y_j}g, \phi}{L^1(G; \complex)}\]
so
\[\abs{\dpn{\one_K * g, \phi}{L^1(G; \complex)} - \int \one_K(y)\dpn{L_y g, \phi}{L^1(G; \complex)}dy}\le 2\eps |K| (1 \vee \norm{\phi}_{L^1(G; \complex)^*})\]
As the above holds for all $\eps > 0$, $\dpn{\one_K * g, \phi}{L^1(G; \complex)}= \int_{G}\one_{K}(y)\dpn{L_y g, \phi}{L^1(G; \complex)}dy$.
By inner regularity of the Haar measure,
\[\dpn{\one_B * g, \phi}{L^1(G; \complex)}= \int_{G}\one_{B}(y)\dpn{L_y g, \phi}{L^1(G; \complex)}dy\]
for all $B \in \cb_{\complex}$. By density of simple functions,
\[\dpn{f * g, \phi}{L^1(G; \complex)}= \int_{G} f(y)\dpn{L_y g, \phi}{L^1(G; \complex)}dy\]
for all $f \in L^{1}(G; \complex)$.$\square$
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