Proof. (1): Let $\seq{g_n}\subset L^{2}(G; \complex)$, $\seq{f * g_n}\subset L^{2}(G; \complex)$, and $g, h \in L^{2}(G; \complex)$ such that $g_{n} \to g$ and $f * g_{n} \to h$ in $L^{2}(G; \complex)$ as $n \to \infty$. After taking a subsequence, assume without loss of generality that:

1. (i)

   $f * g_{n} \to h$ almost everywhere.

2. (ii)

   $\sum_{n \in \natp}\norm{g_n - g}_{L^2(G; \complex)}< \infty$.

In which case, for almost every $x \in G$,

where $\int_{G}|f(y)|\sum_{n \in \natp}|(g_{n} - g)(y^{-1}x)| dy < \infty$ by (ii). As such, by the Dominated Convergence Theorem, $f * g_{N} \to f * g$ almost everywhere. By (i), $f * g = h$ almost everywhere. Therefore the Closed Graph Theorem implies that $L_{f}$ is continuous.

(2): Let $U \in \cn_{G}(1_{G})$ be relatively compact with $U = U^{-1}$, then $\one_{U} \in L^{2}(G; \complex)$, and $\int_{G} |f(y)|\one_{U}(y^{-1}x)dy < \infty$ for almost every $x \in G$. For each $x_{0} \in G$, by Proposition 31.2.7, there exists $x \in G$ such that $x_{0} \in xU$ and $\int_{G} |f(y)|\one_{U}(y^{-1}x)dy < \infty$. As such, for any $K \subset G$ compact, there exists $F \subset G$ finite such that $\int_{G} |f(y)|\one_{U}(y^{-1}x)dy < \infty$ for each $x \in F$ and $K \subset \bigcup_{x \in F}xU$. In which case,

(3): By Lemma 44.2.4, $f$ is locally integrable. Let $K \subset G$ be relatively compact. Using Lemma 5.20.2, assume without loss of generality that $K$ is also open.

Let $g \in C_{c}(G; \complex)$ with $\supp{g}\subset K^{-1}$, then there exists $V \in \cn_{G}(1_{G})$ such that $\supp{g}^{-1}V \subset K^{-1}$. In which case, for each $x \in V$,

so $f * g$ is continuous on $V$ by Young’s Inequality. In particular, $f * g|_{V}= 0$ almost everywhere implies that $\int_{G} f(y)g(y^{-1}) dy = 0$. As such, $\int_{G} fg = 0$ for all $g \in C_{c}(G; \complex)$ with $\supp{g}\subset K$. By uniqueness from the Riesz Representation Theorem, $f|_{K} = 0$ almost everywhere. Therefore $f = 0$ locally almost everywhere.$\square$