Recovering a unitary group representation from a nondegenerate L one representation

If $K=0$, the unique representation on that Hilbert space has zero integrated operators and satisfies every assertion. Hence assume $K\mathrm{\ne }0$. The star-representation is contractive: $\mathrm{\parallel }\sigma (f)\mathrm{\parallel }\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$ for every $f\in L^1(G)$. Indeed, the extension $\tilde{\sigma }$ of [F5] is a unital homomorphism, so an invertible $b\in B$ has invertible image with inverse $\tilde{\sigma }(b^{-1})$; hence ${\sigma }_{\mathcal{B}(K)}(\sigma (f))\subseteq {\sigma }_B((0,f))$ and therefore $r_{\mathcal{B}(K)}(\sigma (f))\le r_B((0,f))\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$ by [F5] and [F6]. For $a=f^{\ast }\ast f$ the operator $\sigma (a)=\sigma (f{)}^{\ast }\sigma (f)$ is self-adjoint, hence normal, so $\mathrm{\parallel }\sigma (f){\mathrm{\parallel }}^2=\mathrm{\parallel }\sigma (f{)}^{\ast }\sigma (f)\mathrm{\parallel }=\mathrm{\parallel }\sigma (f^{\ast }\ast f)\mathrm{\parallel }=r_{\mathcal{B}(K)}(\sigma (f^{\ast }\ast f))\le r_B((0,f^{\ast }\ast f))\le \mathrm{\parallel }{f}^{\ast }\ast f{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1^2$ by [F1] and [F6].

For all $g\in G$ and $u,w\in L^1(G)$ one has $L_g(u\ast w)=(L_{g}u)\ast w$; consequently $\sigma (L_g(u\ast w))=\sigma (L_{g}u)\sigma (w)$. Indeed, for $u,w\in C_c(G)$ the pointwise formula of [F1] gives $(L_g(u\ast w))(x)=\int u(g^{-1}y)w(y^{-1}x)dy=\int u(z)w(z^{-1}{g}^{-1}x)dz=(L_{g}u\ast w)(x)$ after $y=gz$, and both sides are bounded bilinear maps $L^1(G)\times L^1(G)\to L^1(G)$ (left translation is isometric by [F2], convolution is bounded by [F1]) that agree on the dense subspace $C_c(G)\times C_c(G)$.

For $w,u\in C_c(G)$, $F(g)=w(g)L_{g}u$ has compact norm image and vanishes outside the compact set $\mathrm{supp}w$. For every integer $n\ge 1$, choose a finite $2^{-n}$-net in that image, and partition the compact support into measurable sets by the first net point within $2^{-n}$ of $F(g)$. The resulting finite-valued simple function $s_n$, zero off that support, satisfies $\mathrm{\parallel }F-s_n\mathrm{\parallel }\le 2^{-n}$ there, hence $\int \mathrm{\parallel }F-s_n\mathrm{\parallel }\le 2^{-n}\mu (\mathrm{supp}w)\to 0$. Thus $F$ is strongly measurable and Bochner integrable by [F7], directly for the given Borel Haar measure. Put $P=\int F(g)dg$. For EVERY bounded measurable complex function $h$, the map $a\mapsto \int a(x)h(x)dx$ is bounded linear on $L^1$, so [F7] and Fubini [F8] give $\int P(x)h(x)dx=\int w(g)\int u(g^{-1}x)h(x)dxdg=\int (w\ast u)(x)h(x)dx\mathrm{.}$ The integrands are absolutely integrable on compact support: after $x=gy$, their absolute value is bounded by $\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }w(g)\mathrm{\mid }\mathrm{\mid }u(y)\mathrm{\mid }$. Taking $h(x)=\overline{a(x)}\mathrm{/}\mathrm{\mid }a(x)\mathrm{\mid }$ where $a=P-w\ast u\mathrm{\ne }0$, and $h=0$ where $a=0$, gives $\int \mathrm{\mid }P-w\ast u\mathrm{\mid }=0$, proving $P=w\ast u$ in $L^1$.

For every $f\in L^1(G)$ and $g\in G$: $\sigma (L_{g}f)\xi ={\lim }_U\sigma (L_g{e}_U)\sigma (f)\xi$ for every $\xi \in K$, and $\mathrm{\parallel }\sigma (L_g{e}_U)\mathrm{\parallel }\le 1$. Indeed, $L_g(e_U\ast f)=(L_g{e}_U)\ast f$ by step 1.2, so $\sigma (L_g(e_U\ast f))=\sigma (L_g{e}_U)\sigma (f)$ by [F4]; moreover $L_g(e_U\ast f)\to L_{g}f$ in $L^1(G)$ because $e_U\ast f\to f$ by [F3] and $L_g$ is isometric, so boundedness of $\sigma$ gives convergence in operator norm. Finally $\mathrm{\parallel }\sigma (L_g{e}_U)\mathrm{\parallel }\le \mathrm{\parallel }{L}_g{e}_U{\mathrm{\parallel }}_1=\mathrm{\parallel }{e}_U{\mathrm{\parallel }}_1\le 1$ by step 1.1, [F2] and [F3].

For $w,u\in C_c(G)$ and $\xi ,\eta \in K$: $⟨\sigma (w\ast u)\xi ,\eta ⟩={\int }_{G}w(g)⟨\sigma (L_{g}u)\xi ,\eta ⟩dg$. Indeed, the map $a\mapsto \sigma (a)\xi$ is bounded linear by [F4], so it commutes with the Bochner integral of step 1.3: $\sigma ({\int }_{G}w(g)L_{g}udg)\xi ={\int }_{G}w(g)\sigma (L_{g}u)\xi dg$; taking the pairing with $\eta$ and substituting $\int w(g)L_{g}udg=w\ast u$ from the preceding step gives the claim.

Let $D_0$ be the linear span of $\{\sigma (f)\xi :f\in L^1(G),\xi \in K\}$, a dense subspace of $K$ by [F4]. For $g\in G$ define $U_0(g)({\sum }_i\sigma (f_i){\xi }_i):={\sum }_i\sigma (L_g{f}_i){\xi }_i$ on $D_0$. This is well defined: if ${\sum }_i\sigma (f_i){\xi }_i=0$, then applying the bounded operator $\sigma (L_g{e}_U)$ and passing to the limit with step 2.1 gives ${\sum }_i\sigma (L_g{f}_i){\xi }_i=0$. It is complex-linear and a contraction, because $\parallel {\sum }_i\sigma (L_g{f}_i){\xi }_i\parallel ={\lim }_U\parallel \sigma (L_g{e}_U){\sum }_i\sigma (f_i){\xi }_i\parallel \le \mathrm{\parallel }d\mathrm{\parallel }$ for $d={\sum }_i\sigma (f_i){\xi }_i$ by step 2.1. Hence $U_0(g)$ extends uniquely to a contraction $U(g)\in \mathcal{B}(K)$.

For $h\in C_c(G)$, $f\in L^1(G)$ and $\xi ,\eta \in K$: ${\int }_{G}h(g)⟨\sigma (L_{g}f)\xi ,\eta ⟩dg=⟨\sigma (h\ast f)\xi ,\eta ⟩\mathrm{.}$ Both sides are complex-linear in $f$ and bounded by $\mathrm{\parallel }h{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }$: on the left, $\mathrm{\mid }⟨\sigma (L_{g}f)\xi ,\eta ⟩\mathrm{\mid }\le \mathrm{\parallel }\sigma (L_{g}f)\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }\le \mathrm{\parallel }{L}_{g}f{\mathrm{\parallel }}_1\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }$ by steps 1.1 and 1.2, and $\int \mathrm{\mid }h(g)\mathrm{\mid }dg=\mathrm{\parallel }h{\mathrm{\parallel }}_1$; on the right, $\mathrm{\parallel }\sigma (h\ast f)\xi \mathrm{\parallel }\le \mathrm{\parallel }h\ast f{\mathrm{\parallel }}_1\mathrm{\parallel }\xi \mathrm{\parallel }\le \mathrm{\parallel }h{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }\xi \mathrm{\parallel }$ by [F1] and step 1.1. By step 2.2 the two sides agree whenever $f\in C_c(G)$, and $C_c(G)$ is dense in $L^1(G)$ by [F1].

For all $g,h\in G$ and $\xi \in K$ one has $U(g)U(h)\sigma (f)\xi =U(gh)\sigma (f)\xi$ and $U(e)=I$: indeed $U(g)U(h)\sigma (f)\xi =U(g)\sigma (L_{h}f)\xi =\sigma (L_g{L}_{h}f)\xi =\sigma (L_{gh}f)\xi =U(gh)\sigma (f)\xi$ by [F2], and $U(e)\sigma (f)\xi =\sigma (L_{e}f)\xi =\sigma (f)\xi$. Since $D_0$ is dense, $U(g)U(h)=U(gh)$ and $U(e)=I$; taking $h=g^{-1}$ shows that every $U(g)$ is bijective with inverse $U(g^{-1})$ and is therefore, being a contraction with contractive inverse, an isometry, i.e. a unitary operator.

Uniqueness of $U$: if $V$ is a unitary representation of $G$ with $V(g)\sigma (f)\xi =\sigma (L_{g}f)\xi$ for all $g,f,\xi$, then $V(g)$ and $U(g)$ agree on the dense subspace $D_0$ and both are bounded, so $V(g)=U(g)$ for every $g$.

$U$ is strongly continuous. For fixed $f\in L^1(G)$, $\xi \in K$ and $g\to g_0$, $\mathrm{\parallel }U(g)\sigma (f)\xi -U(g_0)\sigma (f)\xi \mathrm{\parallel }=\mathrm{\parallel }\sigma (L_{g}f-L_{g_0}f)\xi \mathrm{\parallel }\le \mathrm{\parallel }{L}_{g}f-L_{g_0}f{\mathrm{\parallel }}_1\mathrm{\parallel }\xi \mathrm{\parallel }\to 0$ by steps 1.1 and 2.1 and the continuity in [F2]. For arbitrary $\eta \in K$ and $ϵ>0$ choose $d\in D_0$ with $\mathrm{\parallel }\eta -d\mathrm{\parallel }<ϵ\mathrm{/}3$, which [F4] permits, and use $\mathrm{\parallel }U(g)\eta -U(g_0)\eta \mathrm{\parallel }\le \mathrm{\parallel }U(g)(\eta -d)\mathrm{\parallel }+\mathrm{\parallel }U(g)d-U(g_0)d\mathrm{\parallel }+\mathrm{\parallel }U(g_0)(d-\eta )\mathrm{\parallel }\le 2ϵ\mathrm{/}3+\mathrm{\parallel }U(g)d-U(g_0)d\mathrm{\parallel }$ together with the preceding convergence for $d$; since $U(g)$ is unitary by step 4.1, this gives continuity of every orbit map.

For $h\in C_c(G)$ one has ${\pi }_U(h)=\sigma (h)$. Indeed, for all $f\in L^1(G)$, $\xi ,\eta \in K$, $⟨{\pi }_U(h)\sigma (f)\xi ,\eta ⟩={\int }_{G}h(g)⟨U(g)\sigma (f)\xi ,\eta ⟩dg={\int }_{G}h(g)⟨\sigma (L_{g}f)\xi ,\eta ⟩dg=⟨\sigma (h\ast f)\xi ,\eta ⟩=⟨\sigma (h)\sigma (f)\xi ,\eta ⟩,$ using [F9], step 3.1, step 3.2 and multiplicativity [F4]. Thus ${\pi }_U(h)-\sigma (h)$ vanishes on the dense subspace $D_0$ of [F4] and is bounded, so ${\pi }_U(h)=\sigma (h)$.

For every $h\in L^1(G)$ one has ${\pi }_U(h)=\sigma (h)$: the assignments $h\mapsto {\pi }_U(h)$ and $h\mapsto \sigma (h)$ are complex-linear and bounded with operator norm at most one by [F9] and step 1.1, and they agree on the dense subspace $C_c(G)$ of $L^1(G)$ by step 6.1 and [F1]; a bounded linear map is determined by its restriction to a dense subspace.

AC is used only through the suppliers: the $L^1$ convolution and Haar integration theory of [F1]–[F3], the spectral and unitization inputs of [F5]–[F6] and the Bochner toolkit of [F7]–[F8], each of which states the choice principle it requires; the reconstruction itself involves no further selection (The Axiom of Choice).