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For $\lambda \ne 0$, suppose $R=(\lambda 1-ba{)}^{-1}$ exists. The identities $(\lambda 1-ab)a=a(\lambda 1-ba)$ and $b(\lambda 1-ab)=(\lambda 1-ba)b$ show, by multiplication on either side, thatFor example, multiplying on the left gives $1-{\lambda }^{-1}ab+{\lambda }^{-1}a(\lambda 1-ba)Rb=1$, and the right product gives the same cancellation. Thus invertibility of $\lambda 1-ba$ implies invertibility of $\lambda 1-ab$. Interchanging $a,b$ proves the converse. The nonzero spectra of products in opposite orders are therefore equal:In particular the requested inclusion follows. The exclusion of zero matters: for the unilateral shift operator $S$, $S^{\ast }S=1$ has Banach algebra spectrum $\{1\}$, whereas $S{S}^{\ast }=1-P_0$ has Banach algebra spectrum $\{0,1\}$.

Use the spectral definition of a Positive element of a C-star algebra: $h$ is positive if $h=h^{\ast }$ and ${\sigma }_A(h)\subseteq [0,\infty )$. We use the standard continuous functional calculus for a Hermitian element of a C-star algebra, including its real Banach algebra spectrum and the equality of its norm with its spectral radius. To prove $a^{\ast }a$ positive without assuming this as part of the definition, first establish two elementary cone facts.

If $p,q$ are positive and $s=\Vert p\Vert +\Vert q\Vert$, the continuous functional calculus gives $\Vert \Vert p\Vert 1-p\Vert \le \Vert p\Vert$ and the corresponding bound for $q$, henceSince $p+q$ is self-adjoint, every spectral value $t$ is real and satisfies $|s-t|\le s$, so $t\ge 0$. Thus the positive cone of a C-star algebra is closed under addition. It is proper: if both $h$ and $-h$ are positive, their Banach algebra spectra force $\sigma (h)\subseteq \{0\}$, and the self-adjoint norm equality gives $h=0$. It is also norm closed: if $p_n\ge 0$ converges to $p$, choose $r\ge {\sup }_n\Vert p_n\Vert$; then $\Vert r1-p_n\Vert \le r$ passes to the limit and forces every spectral value of $p$ to be nonnegative.

Now put $h=a^{\ast }a$, and form its negative part $d=\max (-h,0)$ by continuous functional calculus. Let $c=a{d}^{1/2}$. Because $d$ is a function of $h$, the scalar identity $t\max (-t,0)=-\max (-t,0{)}^2$ givesThis has nonpositive Banach algebra spectrum. The previously proved equality of nonzero product Banach algebra spectra shows that $c{c}^{\ast }$ also has nonpositive Banach algebra spectrum; it is self-adjoint, so both $c^{\ast }c$ and $c{c}^{\ast }$ are negative elements. Write $c=u+iv$, where $u=(c+c^{\ast })/2$ and $v=(c-c^{\ast })/(2i)$ are self-adjoint. Direct multiplication givesBy the spectral mapping theorem, the squares $u^2,v^2$ are positive, so their sum is positive by the cone fact above. But the same sum is negative, since both its original summands are negative. Properness forces $c^{\ast }c+c{c}^{\ast }=0$. Then $c^{\ast }c=-c{c}^{\ast }$ is both negative and positive, so $c^{\ast }c=0$, whence $d^2=0$. The C-star identity applied to the self-adjoint $d$ gives $\Vert d{\Vert }^2=\Vert d^2\Vert =0$. Thus $d=0$, provingThis establishes positivity of adjoint products from spectral positivity. It also proves positivity under congruence: for $p\ge 0$ and any $z$, write $p=r^2$ using its functional-calculus square root; then $z^{\ast }pz=(rz{)}^{\ast }(rz)\ge 0$.

For the comparison of squares, we first derive inversion reverses the order of strictly positive elements. If $0<x\le y$ with both positive and invertible, thenThe continuous functional calculus gives $k^{-1}\le 1$, because the scalar function $1-t^{-1}$ is nonnegative on $\sigma (k)\subseteq [1,\infty )$. Congruence then givesNo commutation of $x,y$ was assumed.

Next derive the square-root resolvent integral in a C-star algebra. For $s\ge 0$,For $s>0$, substituting $t=s{u}^2$ reduces the integral to $\sqrt{s}{\pi }^{-1}{\int }_0^{\infty }2/(1+u^2)du=\sqrt{s}$; for $s=0$ it is zero. For a positive element $x$, the norm of $x(t1+x{)}^{-1}$ is at most $\min (1,\Vert x\Vert /t)$. Thus the integral converges in norm, with integrable bounds $t^{-1/2}$ near zero and $\Vert x\Vert t^{-3/2}$ near infinity, uniform on $\sigma (x)$. The continuous functional calculus consequently givesPut $\alpha =a^2$ and $\beta =b^2$. The assumption $b^2-a^2\ge 0$ says $\alpha \le \beta$. For every $t>0$, inversion reverses the order of the strictly positive elements $t1+\alpha$ and $t1+\beta$. Subtracting the two square-root formulas therefore yieldsHere ${\alpha }^{1/2}=a$ and ${\beta }^{1/2}=b$ because $a,b$ themselves are positive. Every truncated integral has a positive integrand, so is positive; norm convergence and closedness of the positive cone of a C-star algebra make its limit positive. This proves order preservation by the positive square root in the required case:It is important that this argument does not factor $b^2-a^2$ as $(b-a)(b+a)$, which would be invalid for noncommuting elements.

For $p=b-a\ge 0$, apply the continuous function $t\mapsto \sqrt{t}$ on its nonnegative compact Banach algebra spectrum. The resulting element $p^{1/2}$ is self-adjoint, has nonnegative Banach algebra spectrum by continuous functional calculus, and satisfies $(p^{1/2}{)}^2=p$. Thus $b-a$ has a positive square root in a C-star algebra.

Finally, if the positive element $a$ is invertible, its compact Banach algebra spectrum is bounded away from zero. Let $\epsilon =\min \sigma (a)>0$; continuous functional calculus gives $a\ge \epsilon 1$. Since $b\ge a$, we also have $b\ge \epsilon 1$, so $\sigma (b)\subseteq [\epsilon ,\infty )$. The reciprocal function is continuous on this Banach algebra spectrum and supplies $b^{-1}$. Therefore $\boxed{a\text{ invertible}⟹b\text{ invertible}}$ under the given positivity assumptions.

Although $ab$ need not be self-adjoint, it is similar to a Positive element of a C-star algebra. Namely,By congruence preserves positivity in a C-star algebra, $d\ge 0$. In fact $b\ge 1$ givesso $d\ge a\ge 1$. Similar elements have the same spectrum of an element, because $s^{-1}(x-\lambda 1)s$ is invertible exactly when $x-\lambda 1$ is. Therefore the stronger conclusion isPositivity of the spectrum is being asserted, not positivity of the generally non-self-adjoint product itself.

For an element $a$ of a complex unital Banach algebra, its spectrum of an element isThe nonzero spectra of products in opposite orders agree:For $\lambda \ne 0$, if $R=(\lambda 1-ab{)}^{-1}$ exists, direct multiplication on both sides verifiesInterchanging $a,b$ proves the converse. Zero may differ. On the Hilbert space ${\ell }^2({\mathbb{N}}_0)$, let $S$ be the unilateral shift operator $S{e}_n=e_{n+1}$ and take $a=S^{\ast }$, $b=S$. Then $ab=S^{\ast }S=I$, whereas $ba=S{S}^{\ast }=I-P_0$, where $P_0$ is projection onto the first coordinate. HenceBoth values for the second operator occur on its two nonzero invariant summands; every other value has an inverse obtained separately on those summands.

We use the spectral definition of a Positive element of a C-star algebra: $h$ is positive when $h=h^{\ast }$ and $\sigma (h)\subseteq [0,\infty )$. The continuous functional calculus, supplied by the Commutative Gelfand--Naimark theorem for the algebra generated by a Normal element of a C-star algebra and its adjoint, sends $F\in C(\sigma (h))$ to $F(h)$. It preserves products, sums, conjugation and the identity, is isometric, and satisfies $\sigma (F(h))=F(\sigma (h))$. For a self-adjoint element the spectrum is real, so real-valued functions produce self-adjoint elements.

For example, $F(t)=\sqrt{t}$ gives the positive square root in a C-star algebra of a Positive element of a C-star algebra. It satisfies $(h^{1/2}{)}^2=h$. Its uniqueness follows from the composition rule: if $d\ge 0$ and $d^2=h$, then the continuous functional calculus in the algebra generated by $d$ gives $\sqrt{d^2}=d$. For positive invertible $h$, the functions $t^{-1}$ and $t^{-1/2}$ give positive inverses and inverse square roots. If $0\le h$ and $r\ge \Vert h\Vert$, the function $r-t$ gives $r1-h\ge 0$; the same calculus givesThese transfer elementary scalar inequalities for functions of one element to the algebra. They do not justify applying a scalar inequality to arbitrary noncommuting elements.

To prove closure under addition without assuming an order theorem, let $a,b\ge 0$, set $\alpha =\Vert a\Vert$, $\beta =\Vert b\Vert$ and $s=\alpha +\beta$. If $s=0$, both elements are zero. Otherwise,The element $a+b$ is self-adjoint, so any $\lambda \in \sigma (a+b)$ is real and satisfies $|s-\lambda |\le s$. Thus $\lambda \ge 0$, provingWe will also use that the positive cone is proper: if $h\ge 0$ and $-h\ge 0$, then $\sigma (h)\subseteq \{0\}$, and the norm formula for self-adjoint elements gives $h=0$.

We next prove $a^{\ast }a\ge 0$ without presupposing that positivity is defined by such products. Put $h=a^{\ast }a$, which is self-adjoint, and form its negative part by the continuous functional calculus:This is positive. Let $c=a{h}_{-}^{1/2}$. Since $h_{-}$ commutes with $h$,The nonzero product-spectrum identity proved above gives $\sigma (c{c}^{\ast })\setminus \{0\}=\sigma (c^{\ast }c)\setminus \{0\}$. As $c{c}^{\ast }$ is self-adjoint, it too has spectrum in $(-\infty ,0]$, so $c{c}^{\ast }\le 0$.

Write $c=u+iv$ with $u=u^{\ast }$ and $v=v^{\ast }$. The continuous functional calculus makes $u^2$ and $v^2$ positive, and the addition result givesBut both summands on the left are nonpositive, so their sum is also nonpositive. Properness of the cone forces $c^{\ast }c+c{c}^{\ast }=0$. Now $c^{\ast }c=-c{c}^{\ast }$ is positive as well as nonpositive, so $c^{\ast }c=0$. The C-star identity gives $c=0$, whence $h_{-}^2=0$ and $h_{-}=0$. Thus $h$ has no negative spectral part:This proves the positivity of adjoint products from spectral positivity and the equivalence with the usual definition by elements of the form $d^{\ast }d$: the reverse direction is the factorization $h=(h^{1/2}{)}^{\ast }{h}^{1/2}$ for $h\ge 0$.

Finally, if $a\ge 0$ and $c\in A$, writeThe preceding result therefore provesIn particular, congruence preserves positivity in a C-star algebra, and $a\le b$ implies $c^{\ast }ac\le c^{\ast }bc$. These facts will justify the noncommutative order manipulations in the numbered parts.

In a unital C-star algebra $A$, an element is Hermitian when $x=x^{\ast }$, unitary when $x^{\ast }x=x{x}^{\ast }=1$, and normal when $x^{\ast }x=x{x}^{\ast }$. It is a Positive element of a C-star algebra when $x=y^{\ast }y$ for some $y$, equivalently when it is Hermitian and its spectrum lies in $[0,\infty )$.

Suppose first that $\Vert \tau \Vert =\tau (1)$, and normalize so that both equal one. If $x=x^{\ast }$, the C-star identity givesWriting $\tau (x)=a+ib$, we obtainfor every real $t$, which forces $b=0$. Decomposing an arbitrary $x$ into its Hermitian real and imaginary parts now gives

Let $x=x^{\ast }$. Every character of an algebra on the commutative unital Banach algebra generated by $x$ has norm and value at one equal to one, so the preceding argument makes its value on $x$ real. The character description of the spectrum of an element therefore gives $\sigma (x)\subset \mathbb{R}$. Iterating the C-star identity,so $\Vert x\Vert =r(x)$. If $c=\Vert x\Vert$, then $x+c1$ has nonnegative spectrum and is positive, as is $c1$; henceis a difference of positive elements.

If $0\le x$ and $\Vert x\Vert \le 1$, then spectral translation givesThus $1-x$ is positive and, by the norm--spectral-radius equality for Hermitian elements, $\Vert 1-x\Vert \le 1$.

Return to a norm-one $\tau$ with $\tau (1)=1$. For $0\le x\le 1$, the preceding paragraph and reality on Hermitian elements giveScaling proves that $\tau$ is a positive functional on a C-star algebra. Every character has norm and value at one equal to one, so every character is positive. On $C[0,1]$, the functionalis positive but is not multiplicative, and hence is not a character.

Conversely, let $\tau$ be positive. Writing a Hermitian element as a difference of positive elements shows that $\tau$ is real on Hermitian elements. Positivity offor every $\lambda \in \mathbb{C}$ says that the associated quadratic polynomial is nonnegative. Minimizing it in $\lambda$ gives the Cauchy--Schwarz inequalityTaking $y=1$ yields $|\tau (x){|}^2\le \tau (x^{\ast }x)\tau (1)$. Since $0\le x^{\ast }x\le \Vert x{\Vert }^{2}1$, positivity givesTogether with $\tau (1)\le \Vert \tau \Vert$, this proves $\Vert \tau \Vert =\tau (1)$.

Convex combinations preserve positivity and value one at the identity, so the state space $S(A)$ is convex. If $x$ is normal, the unital C-star subalgebra $C^{\ast }(1,x)$ is commutative and its Gelfand transform identifies it with $C(\sigma (x))$. Choose $\lambda \in \sigma (x)$ with $|\lambda |=\Vert x\Vert$. Evaluation at $\lambda$ is a state taking $x$ to $\lambda$. Its norm-preserving Hahn--Banach extension to $A$ still takes $1$ to one, so the norm criterion makes the extension a state $\tau$ satisfying $|\tau (x)|=\Vert x\Vert$.

The state space is nonempty, convex, and weak-star compact by the Banach-Alaoglu theorem. The Krein-Milman theorem gives an extreme point, so a pure state on a C-star algebra exists. For positive $x$, the preceding norm-attainment result makesa nonempty weak-star compact convex set. It is a face: if a convex combination has the maximal possible value on $x$, each summand does. An extreme point of $F$ exists by Krein--Milman and, because $F$ is a face, is extreme in $S(A)$. It is the required pure state.

A bounded linear functional $\tau$ on a C-star algebra is positive when $\tau (x)\ge 0$ for every Positive element of a C-star algebra. Positivity impliesand $\Vert \tau \Vert =\tau (1)$; conversely, $\Vert \tau \Vert =\tau (1)$ implies positivity.

Every Positive element of a C-star algebra $a$ has a unique positive element $a^{1/2}$ satisfying $(a^{1/2}{)}^2=a$. It is obtained by applying the continuous functional calculus to $t\mapsto \sqrt{t}$ on $\sigma (a)\subseteq [0,\infty )$.

For a Positive element of a C-star algebra $x$, the displayed improper integral converges in norm. Near zero its integrand has norm at most $t^{-1/2}$; near infinity it has norm at most $\Vert x\Vert t^{-3/2}$. The scalar identity follows by setting $t=s{u}^2$ for $s>0$ and using ${\int }_0^{\infty }2/(1+u^2)du=\pi$; for $s=0$ both sides vanish. The bounds are uniform on the compact spectrum of an element, so continuous functional calculus gives the operator identity.