Positive calculus and order estimates in a C star algebra

Let $a\in B$ be normal. Then $C:=C^{\ast }(1,a)$ is commutative, since $a$ commutes with $a^{\ast }$, and unital. By [F2], its Gelfand transform $\mathrm{\Gamma }:C\to C(\mathrm{\Delta })$ is an isometric unital $\ast$-isomorphism. By [F3], the range of $\mathrm{\Gamma }(a)$ is ${\sigma }_B(a)$. For $f\in C({\sigma }_B(a))$ define $f(a):={\mathrm{\Gamma }}^{-1}(f\circ \mathrm{\Gamma }(a))$. This is a unital $\ast$-homomorphism in $f$, extends polynomials in $a,a^{\ast }$, and satisfies $\mathrm{\parallel }f(a)\mathrm{\parallel }={\sup }_{\sigma (a)}\mathrm{\mid }f\mathrm{\mid }$ and ${\sigma }_B(f(a))=f({\sigma }_B(a))$ by [F3]. If $a=h=h^{\ast }$ then $\mathrm{\Gamma }(h)$ is real-valued, so ${\sigma }_B(h)\subseteq \mathbb{R}$, and this construction gives the stated self-adjoint calculus. [F1, F2, F3].

For $u,v\in B$, put $c:=(1-vu{)}^{-1}$ when this inverse exists and $d:=1+ucv$. Then $(1-uv)d=1+u(c-1-vuc)v=1$ because $(1-vu)c=1$, and $d(1-uv)=1+u(c-1-cvu)v=1$ because $c(1-vu)=1$. Thus $1-vu$ invertible implies $1-uv$ invertible. Interchanging $u,v$ and replacing $u$ by $u\mathrm{/}\lambda$ gives $\sigma (uv)\setminus \{0\}=\sigma (vu)\setminus \{0\}$.

Let $\phi :E\to F$ be an algebraic star-homomorphism of C*-algebras. Give the forced algebraic unitizations $E^{†}=E\oplus \mathbb{C}$ and $F^{†}=F\oplus \mathbb{C}$ C*-norms extending the original norms: for genuinely nonunital algebras use [F1], including its zero case, and for a unital algebra $E$ use the algebraic $\ast$-isomorphism $(a,\lambda )\mapsto (a+\lambda 1_E,\lambda )$ onto $E\times \mathbb{C}$ with norm $\max (\mathrm{\parallel }a+\lambda 1_E\mathrm{\parallel },\mathrm{\mid }\lambda \mathrm{\mid })$; coordinatewise completeness, submultiplicativity and the C*-identity verify this norm, and $(a,0)$ has norm $\mathrm{\parallel }a\mathrm{\parallel }$. The map ${\phi }^{†}(a,\lambda )=(\phi (a),\lambda )$ is an algebraic unital $\ast$-homomorphism by the unitization formulas. If $z-\lambda 1$ is invertible, ${\phi }^{†}((z-\lambda 1{)}^{-1})$ is the inverse of ${\phi }^{†}(z)-\lambda 1$, so ${\sigma }_{F^{†}}({\phi }^{†}(z))\subseteq {\sigma }_{E^{†}}(z)$. Apply this to the self-adjoint element $z=(a^{\ast }a,0)$ and use [F4] in both unital C*-algebras to obtain $\mathrm{\parallel }\phi (a){\mathrm{\parallel }}^2=\mathrm{\parallel }\phi (a^{\ast }a)\mathrm{\parallel }=r_{F^{†}}({\phi }^{†}(z))\le r_{E^{†}}(z)=\mathrm{\parallel }{a}^{\ast }a\mathrm{\parallel }=\mathrm{\parallel }a{\mathrm{\parallel }}^2$. Thus every algebraic star-homomorphism is contractive, and in particular continuous.

Put $P:=\{h\in B:h=h^{\ast },\sigma (h)\subseteq [0,\mathrm{\infty })\}$. For a self-adjoint $h\in B$, one has $h\in P$ if and only if $\mathrm{\parallel }t1-h\mathrm{\parallel }\le t$ for some (equivalently, every) real $t\ge \mathrm{\parallel }h\mathrm{\parallel }$; in particular, $h\in P$ if and only if $\mathrm{\parallel }\mathrm{\parallel }h\mathrm{\parallel }1-h\mathrm{\parallel }\le \mathrm{\parallel }h\mathrm{\parallel }$. Indeed, step 1.1 and [F4] give $\mathrm{\parallel }t1-h\mathrm{\parallel }={\max }_{\lambda \in \sigma (h)}\mathrm{\mid }t-\lambda \mathrm{\mid }$. If $h\in P$ and $t\ge \mathrm{\parallel }h\mathrm{\parallel }$, every spectral value lies in $[0,t]$, so this maximum is at most $t$. Conversely, a negative spectral value $\lambda$ gives $\mathrm{\mid }t-\lambda \mathrm{\mid }=t-\lambda >t$ for every such $t$. Nonnegative scalar multiples preserve $P$ by step 1.1. If $h,k\in P$, choose $t>0$ with $t\ge \max (\mathrm{\parallel }h\mathrm{\parallel },\mathrm{\parallel }k\mathrm{\parallel })$ and put $w=(h+k)\mathrm{/}(2t)$. Then $w=w^{\ast }$, $\mathrm{\parallel }w\mathrm{\parallel }\le 1$, and $\mathrm{\parallel }1-w\mathrm{\parallel }\le (\mathrm{\parallel }1-h\mathrm{/}t\mathrm{\parallel }+\mathrm{\parallel }1-k\mathrm{/}t\mathrm{\parallel })\mathrm{/}2\le 1$, so the criterion with parameter $1$ gives $w\in P$ and hence $h+k\in P$. If $h_n\in P$ and $h_n\to h$, continuity of the involution gives $h=h^{\ast }$, and $\mathrm{\parallel }\mathrm{\parallel }h\mathrm{\parallel }1-h\mathrm{\parallel }={\lim }_n\mathrm{\parallel }\mathrm{\parallel }{h}_n\mathrm{\parallel }1-h_n\mathrm{\parallel }\le {\lim }_n\mathrm{\parallel }{h}_n\mathrm{\parallel }=\mathrm{\parallel }h\mathrm{\parallel }$, so $h\in P$. Finally, $h\in P\cap (-P)$ implies $\sigma (h)\subseteq \{0\}$, whence $\mathrm{\parallel }h\mathrm{\parallel }=r(h)=0$. Thus $P$ is a closed convex cone with $P\cap (-P)=\{0\}$.

Let $\phi :E\to F$ be a unital star-homomorphism of unital C*-algebras and let $a\in E$ be normal. Then ${\sigma }_F(\phi (a))\subseteq {\sigma }_E(a)$: invertibility of $a-\lambda 1$ implies invertibility of $\phi (a)-\lambda 1=\phi (a-\lambda 1)$ with inverse $\phi ((a-\lambda 1{)}^{-1})$. Moreover $\phi (f(a))=f(\phi (a))$ for every $f\in C({\sigma }_E(a))$: choose polynomials $p_n(z,\bar{z})$ with $\mathrm{\parallel }{p}_n-f{\mathrm{\parallel }}_{\mathrm{\infty },{\sigma }_E(a)}\to 0$, possible because the polynomials in $z$ and $\bar{z}$ form a point-separating self-adjoint unital complex function algebra on the compact set ${\sigma }_E(a)\subseteq \mathbb{C}$ and [F5] applies; then $\phi (p_n(a))=p_n(\phi (a))$ by multiplicativity, star preservation and unitality, while step 1.3 gives $\mathrm{\parallel }\phi (f(a)-p_n(a))\mathrm{\parallel }\le \mathrm{\parallel }f-p_n{\mathrm{\parallel }}_{\mathrm{\infty },{\sigma }_E(a)}$, and $\mathrm{\parallel }{p}_n(\phi (a))-f(\phi (a))\mathrm{\parallel }={\sup }_{\lambda \in {\sigma }_F(\phi (a))}\mathrm{\mid }{p}_n(\lambda )-f(\lambda )\mathrm{\mid }\le \mathrm{\parallel }{p}_n-f{\mathrm{\parallel }}_{\mathrm{\infty },{\sigma }_E(a)}$ by step 1.1 and the spectral inclusion just proved.

Suppose $A$ is nonunital, $h=h^{\ast }\in A$ and $f\in C(\sigma (h))$ vanishes at $0$. Then $f(h)\in A$. First $0\in {\sigma }_B(h)$: an element of the proper two-sided ideal $A$ of the unital algebra $B=A^{+}$ cannot be invertible, since an invertible element generates the unit ideal. Given $ϵ>0$, use [F5] to choose a polynomial $q$ with $\mathrm{\parallel }q-f{\mathrm{\parallel }}_{\mathrm{\infty },\sigma (h)}<ϵ\mathrm{/}2$; then $p:=q-q(0)$ satisfies $p(0)=0$ and $\mathrm{\parallel }p-f{\mathrm{\parallel }}_{\mathrm{\infty },\sigma (h)}\le \mathrm{\parallel }q-f{\mathrm{\parallel }}_{\mathrm{\infty },\sigma (h)}+\mathrm{\mid }q(0)-f(0)\mathrm{\mid }<ϵ$, since $f(0)=0$ and $\mathrm{\mid }q(0)-f(0)\mathrm{\mid }=\mathrm{\mid }q(0)\mathrm{\mid }\le \mathrm{\parallel }q-f{\mathrm{\parallel }}_{\mathrm{\infty },\sigma (h)}$. Every such $p$ is a finite combination of powers $h^k$ with $k\ge 1$, so $p(h)\in A$; by step 1.1, $\mathrm{\parallel }f(h)-p(h)\mathrm{\parallel }=\mathrm{\parallel }p-f{\mathrm{\parallel }}_{\mathrm{\infty },\sigma (h)}<ϵ$, and $A$ is closed in $B$. [F1, F5, step 1.1].

For every $a\in B$ one has $a^{\ast }a\in P$. Put $c:=a^{\ast }a=c^{\ast }$ and let $c_{+}:=\max (\cdot ,0)(c)$ and $c_{-}:=\max (-\cdot ,0)(c)$ be given by the calculus of step 1.1; then $c=c_{+}-c_{-}$ and $c_{+}{c}_{-}=0$, while $c_{+},c_{-}\in P$ by the spectral image formula of step 1.1. Put $d:=a{c}_{-}$. Then $d^{\ast }d=c_{-}{a}^{\ast }a{c}_{-}=c_{-}(c_{+}-c_{-})c_{-}=-c_{-}^3$, and $d^{\ast }d+d{d}^{\ast }=2x^2+2y^2=2(x^2+y^2)$, where $d=x+iy$ is the decomposition of $d$ into self-adjoint parts, so $d^{\ast }d+d{d}^{\ast }\in P$: indeed $x^2$ and $y^2$ lie in $P$ because their spectra are squares of the real spectra of $x$ and $y$ by step 1.1, and $P$ is a cone by step 2.1. Since $-d^{\ast }d=c_{-}^3\in P$, adding gives $d{d}^{\ast }\in P$; step 1.2 applied to $(u,v)=(d^{\ast },d)$ shows $\sigma (d^{\ast }d)\setminus \{0\}=\sigma (d{d}^{\ast })\setminus \{0\}\subseteq [0,\mathrm{\infty })$, so $d^{\ast }d\in P$ as well. Then $c_{-}^3=-d^{\ast }d\in P\cap (-P)=\{0\}$ by step 2.1; the calculus gives $\sigma (c_{-}{)}^3=\sigma (c_{-}^3)=\{0\}$, hence $\sigma (c_{-})=\{0\}$ and $\mathrm{\parallel }{c}_{-}\mathrm{\parallel }=r(c_{-})=0$ by [F4], so $c_{-}=0$ and $c=c_{+}\in P$. [F1, F4, step 1.1, step 1.2, step 2.1].

Every $h\in P$ is a square: the continuous function $\sqrt{\cdot }$ on $\sigma (h)\subseteq [0,\mathrm{\infty })$ has $\sqrt{\cdot }(h)$ self-adjoint by step 1.1 and $\sqrt{\cdot }(h{)}^2=h$. Hence $P\subseteq \{b^{\ast }b:b\in B\}$, and step 3.1 gives the reverse inclusion, so $P=\{a^{\ast }a:a\in B\}$ is exactly the set of algebraically positive elements. [step 1.1, step 3.1].

If $a\in P$ and $x\in B$, then $x^{\ast }ax\in P$: by step 4.1 write $a=y^2$ with $y=y^{\ast }$, so that $x^{\ast }ax=(yx{)}^{\ast }(yx)\in P$ by step 3.1. In particular the relation $\le$, defined by $a\le b$ iff $b-a\in P$, is compatible with conjugation, and it is reflexive and transitive because $P$ is a cone. [step 2.1, step 3.1, step 4.1].

If $0\le a\le b$ then $\mathrm{\parallel }a\mathrm{\parallel }\le \mathrm{\parallel }b\mathrm{\parallel }$. For $\delta >0$ the element $b+\delta 1$ has spectrum in $[\delta ,\mathrm{\infty })$, so it is invertible in $B$; with $z:=(b+\delta 1{)}^{1\mathrm{/}2}$ given by step 1.1 and $z^{-1}$ both self-adjoint, put $c:=z^{-1}a{z}^{-1}$. Then $c\in P$ by step 5.1, and $1-c=z^{-1}(b+\delta 1-a)z^{-1}\in P$ because $b-a\in P$ and $\delta 1\in P$; hence $\sigma (c)\subseteq [0,1]$ by step 2.1, so $\mathrm{\parallel }c\mathrm{\parallel }=r(c)\le 1$ by [F4]. Since $a=zcz$, the C*-identity and [F4] give $\mathrm{\parallel }a\mathrm{\parallel }\le \mathrm{\parallel }z{\mathrm{\parallel }}^2\mathrm{\parallel }c\mathrm{\parallel }\le \mathrm{\parallel }b+\delta 1\mathrm{\parallel }\le \mathrm{\parallel }b\mathrm{\parallel }+\delta$; letting $\delta \downarrow 0$ yields $\mathrm{\parallel }a\mathrm{\parallel }\le \mathrm{\parallel }b\mathrm{\parallel }$. [F1, F4, step 1.1, step 2.1, step 5.1].

AC is inherited from the unitization, Gelfand–Naimark, Stone–Weierstrass and spectral-radius suppliers of [F1]–[F5]; the cone, positivity, order and naturality arguments use no further choice (The Axiom of Choice).