The universal Weyl norm is finite and has the C*-completion property

Bilinearity gives $m_{\lambda }(v,w)m_{\lambda }(v+w,z)=m_{\lambda }(w,z)m_{\lambda }(v,w+z)$, so the two parenthesizations of ${\delta }_v{\delta }_w{\delta }_z$ agree. Also $m_{\lambda }(0,v)=m_{\lambda }(v,0)=1$, so ${\delta }_0$ is a two-sided unit. Since $\omega (v,-v)=0$, ${\delta }_v^{\ast }{\delta }_v={\delta }_v{\delta }_v^{\ast }={\delta }_0$. Finally, $\overline{m_{\lambda }(v,w)}=m_{\lambda }(-w,-v)$, which gives $({\delta }_v{\delta }_w{)}^{\ast }={\delta }_w^{\ast }{\delta }_v^{\ast }$; by bilinearity and conjugate-linearity these identities extend to all finite sums.

For any positive unital $\phi$, the form $B_{\phi }(a,b)=\phi (b^{\ast }a)$ is positive semidefinite. Positivity of $B_{\phi }(a+tb,a+tb)$ for every complex $t$ gives $\mathrm{\mid }{B}_{\phi }(a,b){\mathrm{\mid }}^2\le B_{\phi }(a,a)B_{\phi }(b,b)$: if $B_{\phi }(b,b)>0$, minimize the quadratic in $t$; if it is zero, varying the phase and magnitude of $t$ forces $B_{\phi }(a,b)=0$. Therefore its null space $N_{\phi }=\{a:B_{\phi }(a,a)=0\}$ is orthogonal to every vector and the quotient has the inner product specified in [F1].

Write $a={\sum }_v{a}_v{\delta }_v$ with finite support and define ${\phi }_0(a)=a_0$. The coefficient of ${\delta }_0$ in $a^{\ast }a$ is ${\sum }_v\overline{a_v}{a}_v{m}_{\lambda }(-v,v)={\sum }_v\mathrm{\mid }{a}_v{\mathrm{\mid }}^2$, because $\omega (v,-v)=0$. Thus ${\phi }_0(a^{\ast }a)\ge 0$ and ${\phi }_0({\delta }_0)=1$; the positive unital functional set in [F1] is nonempty.

Left multiplication by ${\delta }_v$ is an isometry on this quotient: $B_{\phi }({\delta }_{v}a,{\delta }_{v}a)=\phi (a^{\ast }{\delta }_v^{\ast }{\delta }_{v}a)=\phi (a^{\ast }a)=B_{\phi }(a,a)$. Hence it preserves the null space. For a finite sum $a={\sum }_v{a}_v{\delta }_v$, the triangle inequality gives $\mathrm{\parallel }{L}_a[b]\mathrm{\parallel }\le ({\sum }_v\mathrm{\mid }{a}_v\mathrm{\mid })\mathrm{\parallel }[b]\mathrm{\parallel }$. Left multiplication therefore extends to a bounded operator on the Hilbert completion of the quotient, and its adjoint is $L_{a^{\ast }}$ because $B_{\phi }(ab,c)=B_{\phi }(b,a^{\ast }c)$. Associativity from step 1.1 makes ${\pi }_{\phi }(a):=L_a$ a unital $\ast$-representation.

The supremum in [F1] is over a set of functionals (a subset of the function set ${\mathbb{C}}^{A_0}$), is nonempty by step 2.1, and is finite since step 2.2 gives $\mathrm{\parallel }{\pi }_{\phi }(a)\mathrm{\parallel }\le {\sum }_v\mathrm{\mid }{a}_v\mathrm{\mid }$ for every $\phi$. Each GNS Hilbert space is nonzero because $\mathrm{\parallel }[{\delta }_0]{\mathrm{\parallel }}^2=\phi ({\delta }_0)=1$, so [F3] applies to every ${\pi }_{\phi }$. Since each ${\pi }_{\phi }$ is a $\ast$-representation, $\mathrm{\parallel }{a}^{\ast }a{\mathrm{\parallel }}_u={\sup }_{\phi }\mathrm{\parallel }{\pi }_{\phi }(a{)}^{\ast }{\pi }_{\phi }(a)\mathrm{\parallel }={\sup }_{\phi }\mathrm{\parallel }{\pi }_{\phi }(a){\mathrm{\parallel }}^2=\mathrm{\parallel }a{\mathrm{\parallel }}_u^2$. Submultiplicativity and $\mathrm{\parallel }{a}^{\ast }{\mathrm{\parallel }}_u=\mathrm{\parallel }a{\mathrm{\parallel }}_u$ follow by the same operator-norm calculation. Thus $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_u$ is a nonzero C*-seminorm; its kernel is a $\ast$-ideal, and the induced norm on $A_0\mathrm{/}\ker \mathrm{\parallel }\cdot {\mathrm{\parallel }}_u$ is a C*-norm.

Let $\rho :A_0\to \mathcal{B}(\mathcal{K})$ be any unital $\ast$-representation. If $\mathcal{K}=\{0\}$, the unique zero representation extends; assume otherwise and choose a unit vector $\xi \in \mathcal{K}$. The vector functional ${\phi }_{\xi }(a)=⟨\rho (a)\xi ,\xi ⟩$ is positive and unital. The map $[a]\mapsto \rho (a)\xi$ preserves the inner product on the GNS quotient, extends to a unitary from its completion onto the cyclic subspace $\overline{\rho (A_0)\xi }$, and intertwines ${\pi }_{{\phi }_{\xi }}$ with the restriction of $\rho$. Therefore $\mathrm{\parallel }\rho (a)\xi \mathrm{\parallel }\le \mathrm{\parallel }{\pi }_{{\phi }_{\xi }}(a)\mathrm{\parallel }\le \mathrm{\parallel }a{\mathrm{\parallel }}_u$; taking the supremum over unit vectors gives $\mathrm{\parallel }\rho (a)\mathrm{\parallel }\le \mathrm{\parallel }a{\mathrm{\parallel }}_u$.

The quotient completion in [F1] is a C*-algebra: multiplication and $\ast$ extend continuously by submultiplicativity and isometry, and the C*-identity passes to limits. Step 4.1 makes every unital $\ast$-representation contractive for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_u$, so it extends uniquely to the completion; conversely each representation of the completion restricts to one of $A_0$. The state ${\phi }_0$ from step 2.1 sends ${\delta }_0$ to a unit vector in its GNS space, so $\mathrm{\parallel }{\delta }_0{\mathrm{\parallel }}_u=1$ and the completion is nonzero. This proves the universal property and all claims.