$L^1(G)$ is a $C^*$-algebra $\iff$ $G$ is trivial

Question

Let $L^1(G)$ be the group Banach $*$-algebra of a locally compact Hausdorff group $G$. Then it is known that $L^1(G)$ is a $C^*$-algebra if and only if the group $G$ is trivial.

The proof of this fact can be found here and it is not elementary. To my surprise, I came across a seemingly much simpler and completely elementary proof in Vahid Shirbisheh's Lectures on $C^*$-algebras, Proposition 2.2.21., Page 24.

However, upon more careful reading, I think that the proof is wrong. It goes like this:

Assume that $G$ is nontrivial and pick $s \in G\setminus \{1\}$. Pick an open relatively compact neighbourhood $U$ of $1$ in $G$ such that $U \cap sU = \emptyset$. Then by properties of left Haar measure we have $0 < \mu(U) < +\infty$. If we define $$f : G \to \Bbb{C}, \quad f(x) := \frac1{\sqrt{\Delta(x)}}(\mathbf{1}_{U} - i \mathbf{1}_{sU})(x), \quad x \in G$$ it is easy to show that $f \in L^1(G)$, where $\Delta : G \to \langle0,+\infty\rangle$ is the modular function.

We have \begin{align} \|f * f^*\|_{L^1(G)} &= \int_G |(f * f^*)(x)|\,d\mu(x)\\ &=\int_G \left|\int_G f(y)\Delta(x^{-1}y)\overline{f(x^{-1}y)}\,d\mu(y)\right|d\mu(x)\\ &= \int_G \left|\int_G \Delta(x^{-1}y)\cdot \frac{\mathbf{1}_{U}(y) - i \mathbf{1}_{sU}(y)}{\sqrt{\Delta(y)}}\cdot \frac{\mathbf{1}_{U}(x^{-1}y) + i \mathbf{1}_{sU}(x^{-1}y)}{\sqrt{\Delta(x^{-1}y)}}\,d\mu(y)\right|d\mu(x)\\ &= \int_G \frac1{\sqrt{\Delta(x)}} \left|\int_G (\mathbf{1}_U(y)\mathbf{1}_{U}(x^{-1}y) + i\mathbf{1}_U(y)\mathbf{1}_{sU}(x^{-1}y) - i\mathbf{1}_U(x^{-1}y)\mathbf{1}_{sU}(y) + \mathbf{1}_{sU}(y)\mathbf{1}_{sU}(x^{-1}y))\,d\mu(y)\right|d\mu(x)\\ &= \int_G \frac1{\sqrt{\Delta(x)}} \left|\mu(U \cap x U) + i \mu(U \cap xsU) - i\mu(xU \cap sU) + \mu(sU \cap xsU)\right|d\mu(x). \end{align}

Now comes the dubious step. The author writes (paraphrased): "In the last integral each term of the integrand is non-zero at least for some values of $x\in G$ and when one of the terms is non-zero the other three terms are zero. This feature justifies the following steps of our argument:"

\begin{align} \|f * f^*\|_{L^1(G)} &< \int_G \frac1{\sqrt{\Delta(x)}} (\mu(U \cap x U) + \mu(U \cap xsU) +\mu(xU \cap sU) + \mu(sU \cap xsU))\,d\mu(x)\\ &= \int_G \frac1{\sqrt{\Delta(x)}} \left|\int_G (\mathbf{1}_U(y)\mathbf{1}_{U}(x^{-1}y) + \mathbf{1}_U(y)\mathbf{1}_{sU}(x^{-1}y) + \mathbf{1}_U(x^{-1}y)\mathbf{1}_{sU}(y) + \mathbf{1}_{sU}(y)\mathbf{1}_{sU}(x^{-1}y))\,d\mu(y)\right|d\mu(x)\\ &= \int_G \left(\int_G \Delta(x^{-1}y)\cdot \frac{\mathbf{1}_{U}(y) + \mathbf{1}_{sU}(y)}{\sqrt{\Delta(y)}}\cdot \frac{\mathbf{1}_{U}(x^{-1}y) + \mathbf{1}_{sU}(x^{-1}y)}{\sqrt{\Delta(x^{-1}y)}}\,d\mu(y)\right)d\mu(x)\\ &= \int_G \left(\int_G \Delta(x^{-1}y)\cdot \left|\frac{\mathbf{1}_{U}(y) -i \mathbf{1}_{sU}(y)}{\sqrt{\Delta(y)}}\right|\cdot \left|\frac{\mathbf{1}_{U}(x^{-1}y) -i \mathbf{1}_{sU}(x^{-1}y)}{\sqrt{\Delta(x^{-1}y)}}\right|\,d\mu(y)\right)d\mu(x)\\ &= \int_G \left(\int_G \Delta(x^{-1}y)|f(y)||f(x^{-1}y)|\,d\mu(y)\right)d\mu(x)\\ &= \int_G \Delta(y)|f(y)|\left(\int_G |f(xy)|\,d\mu(x)\right)\,d\mu(y)\\ &= \int_G \Delta(y)|f(y)|\Delta(y^{-1})\left(\int_G |f(x)|\,d\mu(x)\right)\,d\mu(y)\\ &= \|f\|_{L^1(G)}^2. \end{align}

So, the idea seems to be to notice that the supports of the functions $$x \mapsto \mu(U \cap x U), \quad x \mapsto \mu(U \cap xsU), \quad x \mapsto \mu(xU \cap sU), \quad x \mapsto \mu(sU \cap xsU)$$ are pairwise disjoint and nonempty. Hence we somehow get the strict inequality, but I argue that it should in fact be equality. Precisely because the supports are pairwise disjoint we have $$\left|\mu(U \cap x U) + i \mu(U \cap xsU) - i\mu(xU \cap sU) + \mu(sU \cap xsU)\right|=\mu(U \cap x U) + \mu(U \cap xsU) +\mu(xU \cap sU) + \mu(sU \cap xsU)$$ for all $x \in G$. It is irrelevant that the supports are nonempty.

My questions are:

1. Is the proof indeed wrong?
2. If yes, can it be easily fixed?

Answer 1

Yes, the proof is wrong, and I don't think it can be easily fixed. If $G$ is discrete, we can take $U=\{e\}$ and the modular function is identically $1$, so that $f$ is just $\delta_e-i\delta_s$, which has $L^1$ norm $2$. Then $f\ast f^\ast=2\delta_e-i\delta_s+i\delta_{s^{-1}}$, which has $L^1$ norm $4$. So at least for this choice of $f$ there is no contradiction to the $C^\ast$-identity.