Minimal C star unitization

For $a\in A$ one has $\mathrm{\parallel }{L}_a\mathrm{\parallel }=\mathrm{\parallel }a\mathrm{\parallel }$: the inequality $\mathrm{\parallel }{L}_{a}b\mathrm{\parallel }\le \mathrm{\parallel }a\mathrm{\parallel }\mathrm{\parallel }b\mathrm{\parallel }$ gives $\mathrm{\parallel }{L}_a\mathrm{\parallel }\le \mathrm{\parallel }a\mathrm{\parallel }$, while $\mathrm{\parallel }{L}_a{a}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }a{a}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }a{\mathrm{\parallel }}^2$ gives $\mathrm{\parallel }{L}_a\mathrm{\parallel }\ge \mathrm{\parallel }a{\mathrm{\parallel }}^2\mathrm{/}\mathrm{\parallel }{a}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }a\mathrm{\parallel }$ when $a\mathrm{\ne }0$, and the case $a=0$ is trivial; in particular $L$ is an injective linear isometry, so $L(A)$ is a closed subspace of $\mathcal{B}(A)$: a point in its closure admits approximants $L_{a_n}$ at distance less than $1\mathrm{/}(n+1)$ by AC, the isometry makes $(a_n)$ Cauchy, and completeness gives a limit in $A$ with that operator image.

The map $\phi :A^{+}\to \mathcal{B}(A)$, $\phi (a,\lambda ):=L_a+\lambda I$, is complex-linear and multiplicative with $\phi ((a,\lambda )(b,\mu ))=\phi (a,\lambda )\phi (b,\mu )$ for all $(a,\lambda ),(b,\mu )\in A^{+}$, and $\phi (0,1)=I$, where the multiplicativity is the computation $\phi (a,\lambda )\phi (b,\mu )(c)=a(bc+\mu c)+\lambda (bc+\mu c)=(ab+\lambda b+\mu a)c+\lambda \mu c=\phi ((a,\lambda )(b,\mu ))(c)$ [L2, algebra].

The map $\phi$ of [step 1.2] is injective when $A$ is genuinely nonunital: if $L_a+\lambda I=0$ with $\lambda \mathrm{\ne }0$ then $ab=-\lambda b$ for all $b$, so $e:=-a\mathrm{/}\lambda$ satisfies $eb=b$ for all $b$, that is, $e$ is a left identity. Taking adjoints in $eb=b$ gives $b^{\ast }{e}^{\ast }=b^{\ast }$ for every $b$, and every element of $A$ is of the form $b^{\ast }$, so $e^{\ast }$ is a right identity. Applying the left identity to $b=e^{\ast }$ gives $e{e}^{\ast }=e^{\ast }$, and applying the right identity to $b=e$ gives $e{e}^{\ast }=e$; hence $e=e^{\ast }$ is a two-sided identity of $A$, a contradiction, and if $\lambda =0$ then $L_a=0$ forces $a=0$ by [step 1.1].

Put $d={\inf }_{S\in L(A)}\mathrm{\parallel }I-S\mathrm{\parallel }$. By step 2.1, $I\mathrm{\notin }L(A)$, and by step 1.1 this subspace is closed, so some open ball about $I$ is disjoint from it and $d>0$. For $T=L_a+\lambda I$ one has $\mathrm{\mid }\lambda \mathrm{\mid }d\le \mathrm{\parallel }T\mathrm{\parallel }$: this is immediate for $\lambda =0$, and otherwise divide by $\lambda$ and use $-L_a\mathrm{/}\lambda \in L(A)$. If $T_n=L_{a_n}+{\lambda }_{n}I$ converges in $\mathcal{B}(A)$, applying this bound to differences makes $({\lambda }_n)$ Cauchy in $\mathbb{C}$, with limit $\lambda$ by [L5]. Then $L_{a_n}=T_n-{\lambda }_{n}I$ converges into the closed subspace $L(A)$. Its limit is $L_a$ for some $a$, and the limit of $T_n$ is $L_a+\lambda I$. Thus $M=L(A)+\mathbb{C}I$ is sequentially closed, hence closed: under AC any point in its closure has a sequence at distances less than $1\mathrm{/}(n+1)$. No compact-subsequence argument is needed.

On $M$, the map $\sigma (T):=L_{a^{\ast }}+\bar{\lambda }I$ for $T=L_a+\lambda I$ is well defined (by injectivity from [step 2.1]) and involutive with $\sigma (\sigma (T))=T$; and for every $T\in M$ one has $\mathrm{\parallel }T{\mathrm{\parallel }}^2\le \mathrm{\parallel }\sigma (T)T\mathrm{\parallel }$ and $\sigma (T)T=\phi (x^{\ast }x)$ if $T=\phi (x)$: for $b\in A$, $\mathrm{\parallel }Tb{\mathrm{\parallel }}^2=\mathrm{\parallel }(Tb{)}^{\ast }(Tb)\mathrm{\parallel }=\mathrm{\parallel }{b}^{\ast }\sigma (T)Tb\mathrm{\parallel }\le \mathrm{\parallel }b{\mathrm{\parallel }}^2\mathrm{\parallel }\sigma (T)T\mathrm{\parallel }$ using [L1], and the identity $\sigma (T)T=\phi (x^{\ast }x)$ is multiplicativity of $\phi$ from [step 1.2] combined with the involution of [L2].

The involution $\sigma$ is contractive for the operator norm. Write $T=\phi (x)$ and let $b\in A$; put $y:=\sigma (T)b=x^{\ast }b$. Then $\mathrm{\parallel }y{\mathrm{\parallel }}^2=\mathrm{\parallel }{y}^{\ast }y\mathrm{\parallel }=\mathrm{\parallel }{b}^{\ast }x{x}^{\ast }b\mathrm{\parallel }=\mathrm{\parallel }{b}^{\ast }Ty\mathrm{\parallel }\le \mathrm{\parallel }b\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ by [L1]. If $y\mathrm{\ne }0$, cancellation gives $\mathrm{\parallel }\sigma (T)b\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }b\mathrm{\parallel }$, while the same inequality is trivial for $y=0$. Taking the supremum over the unit ball gives $\mathrm{\parallel }\sigma (T)\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$.

The norm $\mathrm{\parallel }(a,\lambda ){\mathrm{\parallel }}_{A^{+}}:=\mathrm{\parallel }\phi (a,\lambda )\mathrm{\parallel }$ makes $\phi$ an isometry onto the closed subspace $M$ of the Banach space $\mathcal{B}(A)$, so $A^{+}$ is a Banach space with a submultiplicative norm (both transported along the isometric algebra isomorphism $\phi$); and the C*-identity holds: for $x=(a,\lambda )$, $\mathrm{\parallel }x{\mathrm{\parallel }}^2=\mathrm{\parallel }\phi (x){\mathrm{\parallel }}^2=\mathrm{\parallel }\sigma (\phi (x))\phi (x)\mathrm{\parallel }=\mathrm{\parallel }\phi (x^{\ast }x)\mathrm{\parallel }=\mathrm{\parallel }{x}^{\ast }x\mathrm{\parallel }$. Indeed, [step 3.2] supplies the first inequality $\mathrm{\parallel }\phi (x){\mathrm{\parallel }}^2\le \mathrm{\parallel }\sigma (\phi (x))\phi (x)\mathrm{\parallel }$, while submultiplicativity and [step 3.3] supply the reverse inequality $\mathrm{\parallel }\sigma (\phi (x))\phi (x)\mathrm{\parallel }\le \mathrm{\parallel }\sigma (\phi (x))\mathrm{\parallel }\mathrm{\parallel }\phi (x)\mathrm{\parallel }\le \mathrm{\parallel }\phi (x){\mathrm{\parallel }}^2$.

The norm of [step 4.1] extends the norm of $A$: $\mathrm{\parallel }(a,0){\mathrm{\parallel }}_{A^{+}}=\mathrm{\parallel }{L}_a\mathrm{\parallel }=\mathrm{\parallel }a\mathrm{\parallel }$ by [step 1.1]; the element $(0,1)$ is a unit of norm one, since $\phi (0,1)=I$ has operator norm one; and $A\oplus \{0\}$ is a closed two-sided $\ast$-ideal of codimension one by the algebra identities of [L2] and the isometry of [step 1.1].

Uniqueness of the norm: let $\mathrm{\parallel }\cdot {\mathrm{\parallel }}^{\mathrm{\prime }}$ be a C*-norm on $A^{+}$ extending the norm of $A$. The algebraic unit is self-adjoint, so its positive primed norm satisfies $\mathrm{\parallel }1{\mathrm{\parallel }}^{\mathrm{\prime }2}=\mathrm{\parallel }{1}^{\ast }1{\mathrm{\parallel }}^{\mathrm{\prime }}=\mathrm{\parallel }1{\mathrm{\parallel }}^{\mathrm{\prime }}$ and hence equals one. Thus the normalized unital hypothesis of [L4] holds for both norms. For $x=(a,\lambda )$ the element $y:=x^{\ast }x=(c,\mathrm{\mid }\lambda {\mathrm{\mid }}^2)$ with $c:=a^{\ast }a+\bar{\lambda }a+\lambda a^{\ast }$ is self-adjoint, hence normal, in the unital C*-algebra $(A^{+},\mathrm{\parallel }\cdot {\mathrm{\parallel }}^{\mathrm{\prime }})$, so $\mathrm{\parallel }y{\mathrm{\parallel }}^{\mathrm{\prime }}=r^{\mathrm{\prime }}(y)$ by [L4]; and the spectrum of $y$ in the unital algebra $A^{+}$ is independent of the norm, so $r^{\mathrm{\prime }}(y)=r(y)$, where $r$ is computed with the norm of [step 4.1]; applying the same identity with the operator norm gives $\mathrm{\parallel }x{\mathrm{\parallel }}^2=\mathrm{\parallel }y\mathrm{\parallel }=r(y)=r^{\mathrm{\prime }}(y)=\mathrm{\parallel }y{\mathrm{\parallel }}^{\mathrm{\prime }}=\mathrm{\parallel }x{\mathrm{\parallel }}^{\mathrm{\prime }2}$, hence $\mathrm{\parallel }\cdot {\mathrm{\parallel }}^{\mathrm{\prime }}=\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{A^{+}}$.

Claims 1, 2 and 3 are proved: [step 4.1] and [step 5.1] give the C*-algebra structure with $A$ as a closed ideal of codimension one, [step 5.2] gives uniqueness of the norm among C*-norms extending the norm of $A$, and the zero algebra case is separate: [L2] identifies $A^{+}$ with $\mathbb{C}$, whose modulus is complete and satisfies the C*-identity by [L5]. The operator norm on $\mathcal{B}(\{0\})$ is not used. If a C*-norm on this scalar algebra is required, complex homogeneity and the unit C*-identity force $\mathrm{\parallel }\lambda \mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }1\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }$, so its usual norm is unique.