If $G$ is a locally compact group and $L^1(G)$ is unital, then $G$ is discrete.

Question

The book Principles of Harmonic Analysis by Anton Deitmar and Siegfried Echterhoff outlines the following proof to show that if $G$ is a locally compact group and $L^1(G)$ is unital, then $G$ is discrete. I have questions about certain steps.

Proof. Assume that $A = L^1(G)$ possesses a unit $\phi$ and $G$ is non-discrete. The latter fact implies that any unit neighborhood $U$ has at least two points. This implies by Urysohn’s Lemma (A.8.1) that for every unit-neighborhood $U$, there are two Dirac functions $\phi_U$ and $ψ_U$, both with support in $U$, such that the supports of $\phi_U$ and $ψ_U$ are disjoint, hence in particular, $\|\phi_U - \psi_U\|_1 = 2$ for every $n\in \mathbb N$.

1. How does Urysohn's lemma guarantee the existence of the required Dirac functions? Details would be great.

2. The statement "$\|\phi_U - \psi_U\|_1 = 2$ for every $n\in \mathbb N$" doesn't seem to make sense, there is no $n$ involved at all! What does the author mean?

The function $\phi$ being a unit means that we have $\phi ∗ f = f ∗ \phi = f$ for every $f ∈ L^1(G)$. There exists a unit-neighborhood $U$, such that one has $\|\phi_U \ast \phi - \phi\|_1 < 1$ and $\|\psi_U \ast \phi - \phi\|_1 < 1$. Hence $2 = \|\phi_U − ψ_U\|_1 ≤ \|\phi_U − \phi\|_1 + \|\phi − ψ_U\|_1 < 2$, a contradiction! Hence the assumption is false, and $G$ must be discrete. $\square$

3. Why does there exist a unit-neighborhood $U$ such that $\|\phi_U \ast \phi - \phi\|_1 < 1$ and $\|\psi_U \ast \phi - \phi\|_1 < 1$? $\color{blue}{\text{(Resolved.)}}$

Thanks a lot!

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Note:

• All groups in this discussion are Hausdorff.
• By a Dirac function we mean a function $f\in C_c(G)$ satisfying $f\ge 0$, $\int_G f(x)\, dx = 1$, and $f(x) = f(x^{-1})$ for all $x\in G$.

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Update $1$: $Q3.$ is resolved by the following lemma from the text (take $\epsilon = 1$):

Lemma $1.6.6.$ Let $\epsilon > 0$. For every $f\in L^1(G)$ there exists a unit-neighborhood $U$ such that for every Dirac function $\phi_U$ with support in $U$ one has $$\|f \ast \phi_U - f\|_1 < \epsilon, \quad\quad \|\phi_U \ast f - f\|_1 < \epsilon$$

Answer 1

The idea is to take symmetric neighborhoods of the unit, which I denote by $e$. Assume then that $G$ is a l.c.H. too group and that the topology is not discrete. Suppose $U$ is an open neighborhood of $e$ and choose a on open relative compact symmetric neighborhood $W$ of $e$ such that $\overline{W}\overline{W}\subset U$. There is $x\in W\setminus\{e\}$. Notice that $x^{-1}\neq e$. There is an open relatively compact symmetric neighborhood $V$ of $e$ such that $\overline{V}\subset W$ and $$V\cap (xV)=V\cap(Vx)=V\cap (x^{-1}V)=V\cap (Vx^{-1})=\emptyset$$ By Urysohn's Lemma, there are continuous functions $0\leq \phi_U\leq 1$ and $0\leq \psi_U\leq 1$ such that

1. $\operatorname{supp}(\phi_U)\subset \overline{V}$, $\phi_U(e)=1$,
2. $\operatorname{supp}(\psi_U)\subset x\overline{V}\cap(\overline{V}x)$ and $\psi_U(x)=1$.

Let $$\phi'_U(y)=\frac{c}2(\phi_U(y)+\phi_U(y^{-1}))\qquad\psi'_U(y)=\frac{d}2(\psi_U(y)+\psi_U(y^{-1}))$$ with $c,d>0$ chosen so that $\int\phi'_U=\int\psi'_U=1$. The supports of $\phi'_U$ and $\psi'_U$ are disjoint, they are included in $U$ and $\|\phi'_U-\psi'_U\|_{L_1(G)}=2$.

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The rest of the proof I gather is based on the fact that $$\lim_{U\rightarrow x}\|\phi_U*f-f\|_{L_1(G)}$$ for any net $\{\phi_U\in C^+_{00}(G): U\in \mathcal{N}_x\}$ where $\mathcal{N}_x$ is the collection of all open neighborhoods of $x$, and the $\phi_U$ are such that $\operatorname{supp}(\phi_U)\subset U$ for all $U\in\mathcal{N}_x$, and $\int\phi_U=1$ for all $U\in\mathcal{N}_x$. For this, symmetry of the $\phi_U$ not necessary). Symmetry becomes handy to also obtain that $$\lim_{U\rightarrow x}\|f*\phi_U-f\|_{L_1(G)}$$

Answer 2

$\bf{Correction:}$ The same result appears in Bourbaki, Integration as an exercise with hints. Let's see how it works.

First, assume that $G$ is not discrete. Then $\mu(\{e\}) = 0$. Why is that? If $e$ had an open neighborhood $U$ that is a finite set, then $U\backslash\{e\}$ is a finite union of closed sets ( $G$ is $T_1$), so closed. We conclude that $$\{e\} = U \backslash ( U\backslash\{e\}) $$ is open, contradiction.

Now, we also know that $G$ has a compact neighborhood $K$ of $e$. If we had $\mu(\{e\})> 0$, then from the invariance of $G$ we got $\mu(\{g\})=\mu(\{e\}) >0$ for all $G$. We know from the above that $K$ is infinite. But that would imply $\mu(K) = \infty$, contradiction. Therefore, $\mu(\{e\}) = 0$.

Now, let $f \in L^{1}(G, \mu)$. The map

$$A\mapsto \int_A |f| d \mu$$

is a measure, and is absolutely continuous with respect to $\mu$.

Claim: there exists $U$ neighborhood of $e$ such that

$$\int_U |f| < \frac{1}{2}$$

(we'll get back to this later)

Let $V$ a neighborhood of $e$ such that $V = V^{-1}$, and $V\cdot V \cdot V\subset U$. Consider $\phi$ with values in $[0,1]$, $1$ on $V$, and $0$ outside $V$ ( take it continuous, although it is not very important).

We calculate

$$\phi\star f(x) = \int_G \phi(x h^{-1}) f(h)$$

We see that $|\phi\star f|\le \frac{1}{2}$ on $V$. Hence $\phi\star f\ne \phi$. Conclude that $f$ is not a unit. Hence there are no units $f$.

Answer 3

Urysohn's lemma (in locally compact, Hausdorff spaces) implies that, given two disjoint compact sets $E,F$, there are $[0,1]$-valued continuous functions $f,g$ which are $1$ on $E,F$, respectively, and $0$ off neighborhoods $U,V$ of $E,F$, respectively... and we can take $U,V$ disjoint.