Proof. ($\supset$): Let $\seqf{f_j}\subset L^{1}(G; \complex)$. By inner regularity of the Haar measure and hence of $|f_{j}(x)|dx$, there exists $K \subset G$ compact such that $\int_{G \setminus K}|f_{j}| < \eps$ for all $1 \le j \le n$. In which case, for any $\phi, \psi \in S(G)$,
\begin{align*}\max_{1 \le j \le n}\abs{\int_{G} f_j (\phi - \psi)}&\le \max_{1 \le j \le n}\int_{G}|f_{j}(\phi - \psi)| \\&\le \max_{1 \le j \le n}\braks{\int_{K} |f_j(\phi - \psi)| + \int_{G \setminus K}|f_j(\phi - \psi)|}\\&\le \sup_{x \in K}|\phi(x) - \psi(x)|\max_{1 \le j \le n}\norm{f_j}_{L^1(G; \complex)}+ 2\eps\end{align*}
($\subset$): Let $\phi \in S(G)$ and $V \in \cn_{G}(1_{G})$ be relatively compact, then for each $x \in G$,
\begin{align*}\abs{\phi(x) - \frac{\one_{V}}{|V|} * \phi(x)}&\le \frac{1}{|V|}\int_{V} |\phi(x) - \phi(y^{-1}x)|dy \\&\le \frac{1}{|V|}\int_{V} ({2 - 2\text{Re}(\phi(y))})^{1/2}dy \\&\le |V|^{-1/2}\braks{\int_V 2 - 2\text{Re}(\phi(y)) dy}^{1/2}\end{align*}
by Lemma 32.4.6 and the Cauchy-Schwarz inequality. As such, for any $\phi, \psi \in S(G)$ and $x \in G$,
\begin{align*}|\phi(x) - \psi(x)|&\le \abs{\phi(x) - \frac{\one_{V}}{|V|} * \phi(x)}+ \frac{1}{|V|}\abs{\one_V * (\phi - \psi)(x)}\\&+ \abs{\psi(x) - \frac{\one_{V}}{|V|} * \psi(x)}\\&\le |V|^{-1/2}\braks{\int_V 2 - 2\text{Re}(\phi(y)) dy}^{1/2}+ \abs{\angles{\frac{L_{x}\one_{V}}{|V|}, \phi - \psi}_{L^1(G; \complex)}}\\&+ |V|^{-1/2}\braks{\int_V 2 - 2\text{Re}(\psi(y)) dy}^{1/2}\\&\le 2|V|^{-1/2}\braks{\int_V 2 - 2\text{Re}(\phi(y)) dy}^{1/2}+ 2|V|^{-1/2}|\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2}\\&+ \abs{\angles{\frac{L_{x}\one_{V}}{|V|}, \phi - \psi}_{L^1(G; \complex)}}\\&\le 4\sup_{y \in V}|\phi(y) - 1|^{1/2}+ |2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2}\\&+ \abs{\angles{\frac{L_{x}\one_{V}}{|V|}, \phi - \psi}_{L^1(G; \complex)}}\\\end{align*}
For any $K \subset G$ compact,
\begin{align*}\sup_{x \in K}|\phi(x) - \psi(x)|&\le 4\sup_{y \in V}|\phi(y) - 1|^{1/2}+ |2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2}\\&+\sup_{x \in K}\abs{\angles{\frac{L_{x}\one_{V}}{|V|}, \phi - \psi}_{L^1(G; \complex)}}\end{align*}
Let $\eps > 0$. By Corollary 32.3.4, $\phi$ is continuous with $\phi(1_{G}) = 1$, so there exists a relatively compact neighbourhood $V \in \cn_{G}(1_{G})$ with $4\sup_{y \in V}|\phi(y) - 1|^{1/2}< \eps$.
Since $S(G)$ is a bounded subset of $L^{1}(G; \complex)^{*}$, it is equicontinuous. By the Arzelà-Ascoli Theorem, the weak* topology and the topology of uniform convergence on compact subsets of $G$ coincide on $S(G)$. In particular, given that $K$ is compact, $\bracsn{L_x\one_V|x \in K}$ is compact in $L^{1}(G; \complex)$, so there exists $W \in \cn_{S(G)}(\phi)$ such that
\[|2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2}+\sup_{x \in K}\abs{\angles{\frac{L_{x}\one_{V}}{|V|}, \phi - \psi}_{L^1(G; \complex)}}\le \eps\]
for all $\psi \in W$. Therefore for any $\psi \in W$, $\sup_{x \in K}|\phi(x) - \psi(x)| \le 2\eps$.$\square$
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