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For Positive elements of a C-star algebra $x\le y$, inversion reverses the order of strictly positive elements, so $(t1+x{)}^{-1}\ge (t1+y{)}^{-1}$ for $t>0$. Subtracting their square-root resolvent integral in a C-star algebra formulas givesThe integrand is positive, and the positive cone is closed, so the norm limit of these integrals is positive. No commutativity of $x$ and $y$ is required. In particular, $b^2\ge a^2$ for positive $a,b$ implies $b\ge a$, even though squaring itself is not generally order preserving.

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.