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Here strictly positive means positive and invertible. Set $k=a^{-1/2}b{a}^{-1/2}\ge 1$ using congruence preserves positivity in a C-star algebra. The continuous functional calculus gives $k^{-1}\le 1$, since $1-t^{-1}\ge 0$ for $t\ge 1$. Then $a^{-1}-b^{-1}=a^{-1/2}(1-k^{-1})a^{-1/2}\ge 0$. The proof works without commutativity. If $1\le a$, functional calculus also gives $a^{-1}\le 1$.

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.

Since $a-1$ and $b-1=(b-a)+(a-1)$ are positive, their spectra are nonnegative. Translation of the spectrum of an element gives $\sigma (a),\sigma (b)\subseteq [1,\infty )$, so both elements are invertible. Their inverses and inverse square roots are positive by the continuous functional calculus. Also $1-a^{-1}$ is positive because the scalar function $1-t^{-1}$ is nonnegative on $\sigma (a)$.

For the order reversal, setThe inequality uses congruence preserves positivity in a C-star algebra. continuous functional calculus gives $k^{-1}\le 1$. Sinceanother congruence showsConsequentlyHere $b^{-1}$ is strictly positive in the stronger invertible sense: $b^{-1}\ge \Vert b{\Vert }^{-1}1>0$, because its spectrum is the set of reciprocals of the positive spectral values of $b$. This is inversion reverses the order of strictly positive elements; no commutativity of $a,b$ is required.

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.