The L1 group algebra has a unit exactly when the group is discrete

Discrete case. Let $G$ be discrete. By [F2] there is $c>0$ with $\mu =c\cdot \text{counting}$, so ${\int }_{G}Hd\mu =c{\sum }_{y\in G}H(y)$ and the class $u:=c^{-1}{\mathbf{1}}_{\{e\}}$ lies in $L^1(G)$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_1=1$. For $f\in C_c(G)$ and $x\in G$, [F3] gives $(u\ast f)(x)=c{\sum }_{y\in G}{c}^{-1}{\mathbf{1}}_{\{e\}}(y)f(y^{-1}x)=f(x)$ and $(f\ast u)(x)=c{\sum }_{y\in G}f(y)c^{-1}{\mathbf{1}}_{\{e\}}(y^{-1}x)=f(x)$, since the only nonzero term has $y=e$, respectively $y=x$. Thus $u\ast f=f\ast u=f$ for all $f\in C_c(G)$, and the identities extend to every $f\in L^1(G)$ by [F5] and the joint continuity of [F4]: $\mathrm{\parallel }u\ast f-f{\mathrm{\parallel }}_1\le 2\mathrm{\parallel }f-f_n{\mathrm{\parallel }}_1\to 0$ for $C_c$ functions $f_n\to f$. So $u$ is a two-sided identity.

The functions $u\ast f$ are continuous for $u\in L^1(G)$ and $f\in C_c(G)$, with $\mathrm{\parallel }u\ast f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }u{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Choose $u_n\in C_c(G)$ with $\mathrm{\parallel }{u}_n-u{\mathrm{\parallel }}_1\to 0$ by [F5]. Each $u_n\ast f$ lies in $C_c(G)$ by [F3] and satisfies $\mathrm{\mid }(u_n\ast f)(x)\mathrm{\mid }\le {\int }_G\mathrm{\mid }{u}_n(y)\mathrm{\mid }\mathrm{\mid }f(y^{-1}x)\mathrm{\mid }d\mu (y)\le \mathrm{\parallel }{u}_n{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$ for a constant $M$ independent of $n$ and $x$, and also $\mathrm{\mid }(u_n\ast f)(x)-(u_m\ast f)(x)\mathrm{\mid }\le \mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $x$. Hence $(u_n\ast f)$ is uniformly Cauchy and converges uniformly to a continuous function $h$. By [F4] it also converges in $L^1$ to $u\ast f$; Fatou's lemma (Fatou's lemma) applied to $\mathrm{\mid }{u}_n\ast f-k\mathrm{\mid }$, where $k$ is a measurable representative of the $L^1$ limit $u\ast f$, gives $\int \mathrm{\mid }h-k\mathrm{\mid }d\mu \le {\mathrm{lim\hspace{0.17em}inf}}_n\mathrm{\parallel }{u}_n\ast f-k{\mathrm{\parallel }}_1=0$, so $h=k=u\ast f$ almost everywhere, so $h$ is a continuous representative of $u\ast f$ and inherits the bound $\mathrm{\parallel }u\ast f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }u{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ by passing to the limit.

In a non-discrete $G$ the point $e$ has measure zero. If some identity neighbourhood $U$ of $G$ were finite, then every subset of the subspace $U$ would be closed there — each singleton is closed in $G$ by [F6] and a finite union of closed sets is closed — hence every subset of $U$ would be open in $U$, and in particular $\{e\}=U\cap W$ for some open $W\subseteq G$. Since $U$ is open in $G$, that makes $\{e\}$ open in $G$, so $G$ would be discrete. Thus, if $G$ is not discrete, every identity neighbourhood is infinite. Choose a compact neighbourhood $K$ of $e$ by [F7]. For each $n\ge 1$, choose distinct points $x_1,\dots ,x_n$ in the interior of $K$, which is an identity neighbourhood and hence infinite as just shown, and pairwise disjoint open neighbourhoods $V_i\subseteq K$ of $x_i$, which exist by the Hausdorff property [F6] applied to the finitely many points. Then $n\mu (\{e\})={\sum }_i\mu (\{x_i\})\le {\sum }_i\mu (V_i)=\mu \left({\bigcup }_i{V}_i\right)\le \mu (K)<\mathrm{\infty }$ by finite additivity and left invariance of $\mu$ and [F1], for every $n$; hence $\mu (\{e\})=0$.

If a continuous complex function $h$ on $G$ vanishes $\mu$-a.e., then $h=0$ everywhere: the set $\{x:h(x)\mathrm{\ne }0\}$ is open by continuity and has measure zero, so it is empty by the positivity of $\mu$ on nonempty open sets [F1].

Let $u\in L^1(G)$ satisfy $u\ast f=f$ for every $f\in L^1(G)$, and let $f\in C_c(G)$. The class $u\ast f$ equals the class of $f$, while the continuous function $u\ast f$ of step 1.2 represents its own class and $f$ is continuous; so $u\ast f-f$, continuous by step 1.2, vanishes $\mu$-a.e. and therefore vanishes everywhere by step 2.1. Consequently $(u\ast f)(e)=f(e)$ for every $f\in C_c(G)$.

Evaluate at $e$: $f(e)={\int }_{G}u(y)f(y^{-1})d\mu (y)$ for every $f\in C_c(G)$. Choose $u_n\in C_c(G)$ with $u_n\to u$ in $L^1(G)$, by [F5]. By [F3], $(u_n\ast f)(e)={\int }_G{u}_n(y)f(y^{-1})d\mu (y)$, and $\mathrm{\mid }{\int }_G(u_n-u)(y)f(y^{-1})d\mu (y)\mathrm{\mid }\le \mathrm{\parallel }{u}_n-u{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, so the right-hand integrals converge to ${\int }_{G}u(y)f(y^{-1})d\mu (y)$; the left-hand values converge to $(u\ast f)(e)=f(e)$ by the uniform bound of step 1.2. Hence $f(e)={\int }_{G}u(y)f(y^{-1})d\mu (y)$.

No unit exists when $G$ is not discrete. Suppose $G$ is not discrete and $u\in L^1(G)$ is such that $u\ast f=f$ for every $f\in L^1(G)$. Then $\mu (\{e\})=0$ by step 1.3. Choose $v\in C_c(G)$ with $\mathrm{\parallel }u-v{\mathrm{\parallel }}_1<1\mathrm{/}2$, possible by [F5]. Since $\mu$ is outer regular and $\mu (\{e\})=0$, there is an open identity neighbourhood $V$ with $\mu (V)\le 1\mathrm{/}(2\max (1,\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}))$; then ${\int }_V\mathrm{\mid }u\mathrm{\mid }d\mu \le {\int }_V\mathrm{\mid }u-v\mathrm{\mid }d\mu +{\int }_V\mathrm{\mid }v\mathrm{\mid }d\mu \le \mathrm{\parallel }u-v{\mathrm{\parallel }}_1+\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\mu (V)<1\mathrm{/}2+1\mathrm{/}2=1$. By [F8] under the Dependent Choice derived from [A1] there is $f\in C_c(G)$ with $0\le f\le 1$, $f(e)=1$ and $\mathrm{supp}f\subseteq V^{-1}$. Since $f(y^{-1})\mathrm{\ne }0$ forces $y^{-1}\in V^{-1}$, hence $y\in V$, step 4.1 gives $1=f(e)={\int }_{G}u(y)f(y^{-1})d\mu (y)$, so $1\le {\int }_V\mathrm{\mid }u(y)\mathrm{\mid }d\mu (y)<1$, a contradiction. Therefore no $u$ with $u\ast f=f$ for all $f$ exists, and a fortiori no two-sided identity exists.

Combining step 1.1 and step 5.1: $L^1(G)$ has a two-sided convolution identity exactly when $G$ is discrete, in which case $\mu =c\cdot \text{counting}$ and the identity is $c^{-1}{\mathbf{1}}_{\{e\}}$. ∎