The unitary dual need not be Hausdorff

$G$ is a second-countable locally compact Hausdorff group: the product ${\mathbb{R}}_{>0}\times \mathbb{R}$ is second-countable and locally compact, the multiplication $(a,b)(a^{\mathrm{\prime }},b^{\mathrm{\prime }})=(a{a}^{\mathrm{\prime }},b+a{b}^{\mathrm{\prime }})$ and the inverse $(a,b{)}^{-1}=(a^{-1},-b\mathrm{/}a)$ are continuous, so $G$ is a topological group.

$\pi$ is a unitary representation. Each $\pi (a,b)$ is the composition of the translation $\eta \mapsto \eta (\cdot +\log a)$ and the multiplication by the modulus-one function $u\mapsto e^{ib{e}^u}$, hence unitary by [F2]; and $\pi (a,b)\pi (a^{\mathrm{\prime }},b^{\mathrm{\prime }})\eta (u)=e^{ib{e}^u}{e}^{i{b}^{\mathrm{\prime }}{e}^{u+\log a}}\eta (u+\log a+\log a^{\mathrm{\prime }})=e^{i(b+a{b}^{\mathrm{\prime }})e^u}\eta (u+\log (a{a}^{\mathrm{\prime }}))=\pi (a{a}^{\mathrm{\prime }},b+a{b}^{\mathrm{\prime }})\eta (u)$ by $e^{u+\log a}=a{e}^u$.

$\pi$ is strongly continuous. Write ${\tau }_h\eta (u)=\eta (u+h)$ and $M_b\eta (u)=e^{ib{e}^u}\eta (u)$. Since $M_b$ is unitary, $\mathrm{\parallel }\pi (a,b)\eta -\pi (a_0,b_0)\eta \mathrm{\parallel }\le \mathrm{\parallel }{\tau }_{\log a}\eta -{\tau }_{\log a_0}\eta \mathrm{\parallel }+\mathrm{\parallel }(M_b-M_{b_0}){\tau }_{\log a_0}\eta \mathrm{\parallel }$. The first term tends to zero by continuity of translations. In the second, the vector is fixed, the multiplier converges pointwise as $b\to b_0$, and its squared modulus is bounded by $4\mathrm{\mid }{\tau }_{\log a_0}\eta {\mathrm{\mid }}^2\in L^1$. Dominated convergence gives the second limit; its sequential form suffices because the parameter space is metrizable.

Every bounded operator commuting with $\pi$ is a multiplication $M_m$. For $n\ge 1$, $\pi (1,\pm 1\mathrm{/}n)$ is multiplication by $e^{\pm i{e}^u\mathrm{/}n}$, so a commuting $T$ commutes with their real and imaginary parts, the multiplications $M_{{\phi }_n}$, $M_{{\psi }_n}$ for ${\phi }_n(u)=\cos (e^u\mathrm{/}n)$ and ${\psi }_n(u)=\sin (e^u\mathrm{/}n)$. The family $\mathcal{F}=\{{\phi }_n,{\psi }_n:n\ge 1\}$ consists of bounded real Borel functions with $\sigma (\mathcal{F})=$ Borel: indeed $n\sin (e^u\mathrm{/}n)\to e^u$ pointwise, so $e^u$, hence $u=\log e^u$, is $\sigma (\mathcal{F})$-measurable, and the sets $\{u\le c\}$ generate the Borel $\sigma$-algebra. By [F4], $T=M_m$ for some bounded measurable $m$.

Coefficient estimate. For $t\in \mathbb{R}$ and $R>0$ put $I_R=[-R,-R\mathrm{/}2]$ and ${\eta }_R:=(R\mathrm{/}2{)}^{-1\mathrm{/}2}{e}^{itu}{\mathbf{1}}_{I_R}$, a unit vector of $L^2(\mathbb{R})$. For $(a,b)\in G$ with $h:=\log a$, the change of variable $u\mapsto u+h$ and $\mathrm{\mid }{e}^{it(u+h)}\mathrm{\mid }=1$ give $⟨\pi (a,b){\eta }_R,{\eta }_R⟩=(2\mathrm{/}R){\int }_{I_R\cap (I_R-h)}{e}^{ib{e}^u}{e}^{ith}du=a^{it}(2\mathrm{/}R){\int }_{I_R\cap (I_R-h)}{e}^{ib{e}^u}du$. If $\mathrm{\mid }h\mathrm{\mid }\le R\mathrm{/}2$, the intersection has measure $R\mathrm{/}2-\mathrm{\mid }h\mathrm{\mid }$ by [F8], and $\mathrm{\mid }{e}^{ib{e}^u}-1\mathrm{\mid }\le \min (2,\mathrm{\mid }b\mathrm{\mid }{e}^{-R\mathrm{/}2})$ on $I_R$ because $u\le -R\mathrm{/}2$ there; therefore $\mid ⟨\pi (a,b){\eta }_R,{\eta }_R⟩-{\chi }_t(a,b)\mid \le (2\mathrm{/}R){\int }_{I_R\cap (I_R-h)}\mathrm{\mid }{e}^{ib{e}^u}-1\mathrm{\mid }du+2\mathrm{\mid }h\mathrm{\mid }\mathrm{/}R\le 2\mathrm{\mid }h\mathrm{\mid }\mathrm{/}R+\mathrm{\mid }b\mathrm{\mid }{e}^{-R\mathrm{/}2}$.

The operator $M_m$ commutes with all translations $U_h\eta (u)=\eta (u+h)$, $h\in \mathbb{R}$, because $\pi (e^h,0)=U_h$. The family of translations is ergodic for Lebesgue measure: if a Lebesgue measurable set $E$ is invariant under every translation up to null sets, replace it by a Borel set equal to it a.e. using [F5]. Translation preserves null sets, so the new indicator is still invariant a.e. under each translation, and its composition with $(u,t)\mapsto u+t$ is Borel, hence product-measurable. Then ${\int }_{\mathbb{R}}{\int }_{\mathbb{R}}\mathrm{\mid }{1}_E(u+t)-1_E(u){\mathrm{\mid }}^{2}dudt=0$ by Tonelli [F5], since every $t$-slice vanishes; hence $(u,t)\mapsto \mathrm{\mid }{1}_E(u+t)-1_E(u)\mathrm{\mid }$ is $0$ a.e., so for some $u_0$ one has $1_E(u_0+t)=1_E(u_0)$ for a.e. $t$, and $1_E$ is a.e. equal to the constant $1_E(u_0)$; being an indicator, $E$ is null or conull. Applying the moreover clause of [F4] to the commuting translation unitaries gives that $m$ is constant a.e., so $T=\lambda I$.

Hence ${\chi }_t\prec \pi$. Let $K\subseteq G$ be compact and $ϵ>0$, and let $z\in \mathbb{C}$; since $\log a$ and $b$ stay in bounded ranges $\mathrm{\mid }h\mathrm{\mid }\le L$, $\mathrm{\mid }b\mathrm{\mid }\le B$ on $K$, choose $R>2L$ so large that $2L\mathrm{/}R+B{e}^{-R\mathrm{/}2}<ϵ\mathrm{/}(1+\mathrm{\mid }z{\mathrm{\mid }}^2)$. Then step 1.5 gives ${\sup }_{(a,b)\in K}\mid \mathrm{\mid }z{\mathrm{\mid }}^2{\chi }_t(a,b)-⟨\pi (a,b)(z{\eta }_R),z{\eta }_R⟩\mid \le \mathrm{\mid }z{\mathrm{\mid }}^2(2L\mathrm{/}R+B{e}^{-R\mathrm{/}2})<ϵ$, so every diagonal coefficient of the one-dimensional representation ${\chi }_t$, hence every function of positive type associated to ${\chi }_t$, is a compact-uniform limit of coefficients of $\pi$; by [F9], ${\chi }_t\prec \pi$.

$\pi$ is irreducible. If $M\subseteq L^2(\mathbb{R})$ were a closed invariant subspace different from $0$ and $H$, its orthogonal projection $P_M$ would commute with every $\pi (g)$ by [F6] and would satisfy $P_M\mathrm{\ne }0,I$, contradicting step 2.1. Hence $H$ has no nontrivial closed invariant subspace, and $\pi$ is irreducible; its carrier $L^2(\mathbb{R})$ is infinite dimensional, since the indicators of $[n,n+1)$, $n\in \mathbb{Z}$, are an infinite orthonormal family.

Each ${\chi }_t(a,b)=a^{it}$ is a continuous unitary character, i.e. a one-dimensional strongly continuous unitary representation; its diagonal coefficients at $z\in \mathbb{C}$ are $\mathrm{\mid }z{\mathrm{\mid }}^2{\chi }_t$. No ${\chi }_t$ is unitarily equivalent to $\pi$, since unitarily equivalent representations have isomorphic carriers and $\dim L^2(\mathbb{R})=\mathrm{\infty }$ while $\dim \mathbb{C}=1$; in particular $\pi \simeq ̸1_G={\chi }_0$, and $[\pi ]\mathrm{\ne }[1_G]$.

Fell closure and failure of Hausdorffness. Since ${\chi }_t$ and $\pi$ are irreducible, [F10] gives ${\chi }_t\in \overline{\{[\pi ]\}}$ in the Fell topology; in particular $[1_G]=[{\chi }_0]\in \overline{\{[\pi ]\}}$ with $[1_G]\mathrm{\ne }[\pi ]$ by step 4.1, so $\{[\pi ]\}$ is not closed. If the Fell topology were Hausdorff, the distinct points $[1_G]$ and $[\pi ]$ would have disjoint open neighbourhoods $U\ni [1_G]$, $V\ni [\pi ]$; since $[1_G]$ lies in the closure of $\{[\pi ]\}$, the neighbourhood $U$ meets $\{[\pi ]\}$ and therefore $[1_G]\in U$ and $[\pi ]\in U\cap V$, contradicting disjointness. Hence $\widehat{G}$ is not Hausdorff.

The Axiom of Choice is inherited from the multiplication lemma and the Fell suppliers; the explicit representation, the coefficient estimates and the ergodicity computation use no further choice (The Axiom of Choice).