Let $G$ be a locally compact group and let $H$ be a Haar measure over $G$. Show that if $f,g\in L^{1}(G)$ then $f\ast g\in L^{1}(G)$ I came across this exercise in Conway's "A Course in Functional Analysis". The complete exercise has a total of four parts.
(1) If $\phi\in C_{c}(G)$ and $\epsilon>0$, show that there is an open neighbourhood $U$ of $e$ in $G$ such that $\Vert\phi_{x}-\phi_{y}\Vert<\epsilon$ whenever $xy^{-1}\in U$ ($\phi_{x}(z)=\phi(xz)$).
(2) Show that if $f\in L^{p}(G)$ ($1\leq p<\infty$), show that there is an open neighbourhood $U$ of $e$ in $G$ such that $\Vert f_{x}-f_{y}\Vert<\epsilon$ whenever $xy^{-1}\in U$.
(3) Show that if $f\in L^{1}(G)$ and $g\in L^{\infty}(G)$, then $h(x)=\int_{G}f(xy^{-1})g(y)dh(y)$ (in other words $h=f\ast g$) defines a bounded continuous function on $G$.
The fourth item is precisely what is stated in the title of the post. I have a pretty good idea how I could prove (1), from which I should be able to prove (2) using the density of $C_{c}(G)$ in $L^{p}(G)$. From here (3) should also be pretty straightforward. But I have no idea how to use all of the above to prove (4). At least I assume that's what I'm supposed to do.
I would be grateful if you could give me a hand with this.
 A: The following is a full solution of (1).
If I have time, I will solve other parts.
(1) Firstly, we prove the lamma: Given an open neighborhood $U$ of
$e\in G$, there exits an open neighborhood $V$ of $e$ such that
$V=V^{-1}$ and $VV\subseteq U$.
Proof of the lemma: Let $\theta:G\times G\rightarrow G$, $\theta(x,y)=xy$.
Note that $\theta$ is continuous with respect to the product topology,
so $\theta^{-1}(U)$ is an open neighborhood of $(e,e)$. Choose open
neighborhoods $V_{1},V_{2}$ of $e$ such that $V_{1}\times V_{2}\subseteq\theta^{-1}(U)$.
Define $V=V_{1}\cap V_{1}^{-1}\cap V_{2}\cap V_{2}^{-1}$, then $V$
is an open neighborhod of $e$, satisfying the condition $V=V^{-1}$.
Moreover $V\times V\subseteq\theta^{-1}(U)\Rightarrow VV\subseteq U$.
We go back to the problem. Let $\varepsilon>0$. For each $x\in G$,
choose an open symmetric neighborhood $V_{x}$ of $e$ such that $|\phi(x)-\phi(y)|<\varepsilon$
whenever $y\in V_{x}V_{x}x$. (This is possible. For, choose an open
neighborhood $U$ of $x$ such that $|\phi(x)-\phi(y)|<\varepsilon$
whenever $y\in U$. Note that $Ux^{-1}$ is an open neighborhood of
$e$, so $Ux^{-1}\supseteq V_{x}V_{x}$ for some open symmetric neighborhood
$V_{x}$ of $e$. Now, $U\supseteq V_{x}V_{x}x$...)
Let $K$ be the compact support of $\phi$. Observe that $\{V_{x}x\mid x\in K\}$
is an open covering of $K$, so it admits a finite subcover $\{V_{x_i}x_{i}\mid i=1,\ldots,n\}$.
Let $V=\cap_{i=1}^{n}V_{x_{i}}$, which is an open symmetric neighborhood
of $e$. Let $x,y\in G$ be arbitrary such that $xy^{-1}\in V$, then
$yx^{-1}=(xy^{-1})^{-1}\in V^{-1}=V$. Let $t\in G$. If $xt\in K$,
then $xt\in V_{x_{i}}x_{i}$ for some $i$. We have that $yt=(yx^{-1})(xt)\in VV_{x_{i}}x_{i}\subseteq V_{x_{i}}V_{x_{i}}x_{i}$.
Moreover, we trivially have $xt\in V_{x_{i}}x_{i}\subseteq V_{x_{i}}V_{x_{i}}x_{i}$.
Therefore
\begin{eqnarray*}
 &  & |\phi(xt)-\phi(yt)|\\
 & \leq & |\phi(xt)-\phi(x_{i})|+|\phi(yt)-\phi(x_{i})|\\
 & < & 2\varepsilon.
\end{eqnarray*}
If $yt\in K$, then $yt\in V_{x_j}x_{j}$ for some $j$. We have that $xt=(xy^{-1})(yt)\in VV_{x_{j}}x_{j}\subseteq V_{x_{j}}V_{x_{j}}x_{j}$.
Trivially, we have $yt\in V_{x_j}x_{j}\subseteq V_{x_{j}}V_{x_{j}}x_{j}$.
We also have $|\phi(xt)-\phi(yt)|\leq|\phi(xt)-\phi(x_{j})|+|\phi(yt)-\phi(x_{j})|<2\varepsilon$.
If $xt\notin K$ and $yt\notin K$, then $\phi(xt)=\phi(yt)=0$.
This shows that $||\phi_{x}-\phi_{y}||\leq2\varepsilon$ whenever
$xy^{-1}\in V$.
Remarks:

*

*The proof shown in above holds for general topological groups. However, if we do not impose restriction on the topology, $C_c(G)$ can be very small.


*We work with locally compact groups instead of just topological group because in the former case Urysohn Lemma is applicable.


*Urysohn Lemma plays a crucial rule. It guarantees that there are sufficiently many continuous functions with compact support that separate compact set and closed set.
Moreover, the existence of Haar measure, which is essentially a modification of the Riesz Representation Theorem (the dual of $C_0(G)$ is the $M(G)$), relies heavily on Urysohn Lemma.
