The GNS translation action is unitary and strongly continuous

By [A13] and [A14], AC supplies Countable Choice. Hence [A7] gives the Hilbert completion $H_{\phi }$ of $Q_{\phi }$ with a dense linear isometry $\kappa :Q_{\phi }\to H_{\phi }$. By [A8], this completion is Banach. The quotient norm used here is the norm induced by its inner product, as in [A4]–[A5].

For $g\in G$, set $T_g[f]=[L_{g}f]$ on $Q_{\phi }$. It is well-defined by [A2] and complex-linear by the pointwise formula for $L_g$ and [A10]. For finitely supported $f,h$, reindex the finite sum in [A1] by $a=gx$ and $b=gy$; since $(gy{)}^{-1}(gx)=y^{-1}x$, this gives $B_{\phi }(L_{g}f,L_{g}h)=B_{\phi }(f,h)\mathrm{.}$ Thus $T_g$ preserves the quotient inner product and norm. The group laws for left translation give $T_e=I$, $T_g{T}_h=T_{gh}$, and $T_{g^{-1}}=T_g^{-1}$.

Each $T_g$ is bounded with bound $1$. Apply [A9] to extend it uniquely to a bounded linear operator ${\pi }_{\phi }(g):H_{\phi }\to H_{\phi }$ with $\mathrm{\parallel }{\pi }_{\phi }(g)v\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }$. The extension of $T_{g^{-1}}$ is an inverse: both compositions extend the identity on the dense subspace $\kappa (Q_{\phi })$, so uniqueness in [A9] makes them the identity on $H_{\phi }$. The same dense-set uniqueness applied to $T_g{T}_h=T_{gh}$ gives ${\pi }_{\phi }(g){\pi }_{\phi }(h)={\pi }_{\phi }(gh)$ and ${\pi }_{\phi }(e)=I$. Applying the contraction bound also to the inverse shows $\mathrm{\parallel }{\pi }_{\phi }(g)v\mathrm{\parallel }=\mathrm{\parallel }v\mathrm{\parallel }$. Thus every ${\pi }_{\phi }(g)$ is bijective, complex-linear, and isometric, so belongs to $U(H_{\phi })$ by [A11].

Fix $x\in G$ and put $t=x^{-1}gx$. Since $T_g[{\delta }_x]=[{\delta }_{gx}]$, [A1] and sesquilinearity give $\mathrm{\parallel }{\pi }_{\phi }(g)\kappa [{\delta }_x]-\kappa [{\delta }_x]{\mathrm{\parallel }}^2=2\phi (e)-\phi (t)-\phi (t^{-1})\mathrm{.}$ The left side is a nonnegative real. By [A3], both maps $g\mapsto x^{-1}gx$ and $g\mapsto (x^{-1}gx{)}^{-1}$ are continuous at $e$ and take $e$ to $e$. Continuity of $\phi$ and the metric description in [A15] therefore let us choose a neighborhood of $e$ on which each of $\mathrm{\mid }\phi (t)-\phi (e)\mathrm{\mid }$ and $\mathrm{\mid }\phi (t^{-1})-\phi (e)\mathrm{\mid }$ is less than ${\epsilon }^2\mathrm{/}2$. On this neighborhood, [A16] and the nonnegativity above give $0\le \mathrm{\parallel }{\pi }_{\phi }(g)\kappa [{\delta }_x]-\kappa [{\delta }_x]{\mathrm{\parallel }}^2\le \mathrm{\mid }\phi (e)-\phi (t)\mathrm{\mid }+\mathrm{\mid }\phi (e)-\phi (t^{-1})\mathrm{\mid }<{\epsilon }^2\mathrm{.}$ Since $\epsilon >0$ and the norm is nonnegative, [A17] yields $\mathrm{\parallel }{\pi }_{\phi }(g)\kappa [{\delta }_x]-\kappa [{\delta }_x]\mathrm{\parallel }<\epsilon$. Thus this orbit is continuous at $e$.

Every element of $Q_{\phi }$ is a finite linear combination of the $[{\delta }_x]$. Write $w={\sum }_{j=1}^m{a}_j\kappa [{\delta }_{x_j}]$. For $m=0$, $w=0$ and its orbit is constant. For $m>0$, put $C={\sum }_{j=1}^m\mathrm{\mid }{a}_j\mathrm{\mid }$. If $C=0$, again $w=0$. If $C>0$, then for any $\epsilon >0$, step 2.2 gives a neighborhood for each $j$ on which the corresponding generator displacement is less than $\epsilon \mathrm{/}C$. Their finite intersection is a neighborhood of $e$, and on it linearity and [A5] give $\mathrm{\parallel }{\pi }_{\phi }(g)w-w\mathrm{\parallel }\le {\sum }_{j=1}^m\mathrm{\mid }{a}_j\mathrm{\mid }\mathrm{\parallel }{\pi }_{\phi }(g)\kappa [{\delta }_{x_j}]-\kappa [{\delta }_{x_j}]\mathrm{\parallel }<\epsilon \mathrm{.}$

Let $v\in H_{\phi }$ and $\epsilon >0$. By density choose $w\in \kappa (Q_{\phi })$ with $\mathrm{\parallel }v-w\mathrm{\parallel }<\epsilon \mathrm{/}4$. Step 3.1 supplies a neighborhood of $e$ where $\mathrm{\parallel }{\pi }_{\phi }(g)w-w\mathrm{\parallel }<\epsilon \mathrm{/}2$. Since ${\pi }_{\phi }(g)$ is an isometry by step 2.1, the triangle inequality gives $\mathrm{\parallel }{\pi }_{\phi }(g)v-v\mathrm{\parallel }\le 2\mathrm{\parallel }v-w\mathrm{\parallel }+\mathrm{\parallel }{\pi }_{\phi }(g)w-w\mathrm{\parallel }<\epsilon \mathrm{.}$ Every orbit map is therefore continuous at $e$, including when $H_{\phi }=\{0\}$.

For any $g_0\in G$ and $g\to g_0$, unitarity and the homomorphism law give $\mathrm{\parallel }{\pi }_{\phi }(g)v-{\pi }_{\phi }(g_0)v\mathrm{\parallel }=\mathrm{\parallel }{\pi }_{\phi }(g_0^{-1}g)v-v\mathrm{\parallel }\mathrm{.}$ The map $g\mapsto g_0^{-1}g$ is continuous by [A3] and sends $g_0$ to $e$; step 4.1 thus proves continuity of the orbit map at $g_0$. This holds for every $g_0$ and $v$, so the representation is strongly continuous.

The zero function has $B_{\phi }=0$, hence $Q_{\phi }=H_{\phi }=\{0\};$ the unique operator on this space is the identity and all orbit maps are constant. In all cases, AC is used only to obtain Countable Choice for the published Hilbert-completion theorem [A7] and extension theorem [A9]. The extensions are unique, so assembling them as $g$ varies requires no further choice; the finite sums and continuity arguments above are choice-free.