Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 6 (Introduction to Functional Analysis)

We first prove the algebraic extension step needed for Hahn-Banach separation theorem. Suppose $p$ is a finite sublinear functional on a real vector space and $g$ is a linear functional on a vector subspace $W$, with $g\le p$ there. For $v\notin W$, choose a real number $c$ betweenThis interval is nonempty: for any $w,z\in W$,Taking $w=0$ or $z=0$ also bounds the endpoints by finite numbers. Define $\tilde{g}(w+tv)=g(w)+tc$. For $t>0$, the upper bound for $c$, applied to $w/t$, proves $\tilde{g}(w+tv)\le p(w+tv)$. For $t<0$, writing $t=-s$ and using the lower bound at $w/s$ proves the same inequality. The case $t=0$ is the original domination. Thus a dominated linear functional extends over one further direction. Order all dominated extensions by extension of their domains. A chain has its union as an upper bound, so Zorn's lemma supplies a maximal extension. The one-direction argument shows its domain is the whole vector space. This proves the required algebraic Hahn-Banach theorem, rather than assuming the separation result.

Choose $a\in A$, and put $C=A-a$. This is a radially open convex set containing zero. It is absorbing: along each line through zero, sufficiently small multiples of any vector lie in $C$. Its Minkowski functionalis finite and nonnegative. Positive homogeneity follows by rescaling $t$. If $x\in sC$ and $y\in tC$, convexity gives $x+y\in (s+t)C$, proving subadditivity by taking infima. Moreover,Indeed, $p(x)<1$ gives $x\in sC$ for some $s<1$, hence $x\in C$ by convexity and $0\in C$. Conversely, radial openness at $x\in C$ permits $(1+ϵ)x\in C$ for some $ϵ>0$, giving $p(x)<1$.

Since $A\cap U=\varnothing$, $a\notin U$ and $u-a\notin C$ for every $u\in U$. Define a linear functional on $U+\mathbb{R}a$ byIt is well defined because $a\notin U$. For $t<0$, write $s=-t$; then $p(u+ta)=sp(u/s-a)\ge s=g(u+ta)$. For $t\ge 0$, domination follows from $g(u+ta)\le 0\le p(u+ta)$. The extension step therefore gives $L:V\to \mathbb{R}$ with $L\le p$, $L{|}_U=0$ and $L(a)=-1$. If $x\in A$, thenso $L(x)<0$. ConsequentlyBecause $L(a)\ne 0$, this kernel of a linear map has codimension one and is a hyperplane. This establishes the geometric Hahn-Banach theorem for radially open sets without assuming a norm or a topology on $V$.

For the convex function, use its strict epigraph in $V\times \mathbb{R}$. It is nonempty and convex, and does not contain $(0,0)$. It is also radially open: on every affine line, a finite convex function is continuous. To justify this last fact, convexity orders secant slopes; on a smaller interval the slopes are bounded above and below by slopes to two fixed exterior endpoints, giving a local Lipschitz bound. Thus $t\mapsto f(x+ty)-\beta -ts$ is continuous, and a strict negative value persists near $t=0$.

Apply the separation theorem with the zero subspace. A nonzero separating linear functional has form $L(x,\beta )=g(x)+c\beta$. Its values on the convex epigraph cannot have both signs, since a line segment would then meet its kernel. Choose the sign so that $L>0$ on the epigraph. At $x=0$ and $\beta >0$ this gives $c>0$. Letting $\beta$ decrease to $f(x)$ yields $g(x)+cf(x)\ge 0$. ThereforeThe resulting linear functional need not be nonzero: for example, the zero functional is the only linear minorant of $f(x)=x^2$ on $\mathbb{R}$.

Now suppose the norm and continuity hypotheses hold. Domination at $x$ and $-x$ givesIf $x_n\in \ker l$ and $x_n\to x$ in norm, apply this inequality to $x-x_n$. Since $l(x-x_n)=l(x)$ and both bounding values tend to $f(0)=0$, we obtain $l(x)=0$. The kernel of a linear map is therefore closed.

Finally, a linear functional with closed kernel on a normed vector space is continuous. If $l=0$, this is immediate. Otherwise choose $v$ with $l(v)=1$ and put $M=\ker l$. Closedness gives $d=\mathrm{dist}(v,M)>0$. For $l(x)\ne 0$, $x/l(x)-v\in M$, so $d\le \Vert x\Vert /|l(x)|$. The same resulting bound holds when $l(x)=0$:This proves continuity and explains exactly how closedness of the kernel is used.

False. We construct a compact Hausdorff space with a countable dense discrete subspace but no compatible metric. Letfor each subset $S\subseteq \mathbb{N}$. Give $P$ its product topology, and let $X=\overline{j(\mathbb{N})}$ in $P$. The Tychonoff theorem makes $P$ compact, and the product is Hausdorff; hence $X$ is a compact Hausdorff space. By construction, $Y=j(\mathbb{N})$ is countable and dense in $X$, so $X$ is a separable topological space. The coordinate $S=\{n\}$ isolates $j(n)$ within $Y$. Thus the subspace topology on $Y$ is discrete and is induced by the discrete metric.

Suppose $X$ were metrizable. An infinite sequence of distinct points of $Y$ would have a convergent subsequence, by sequential compactness of a compact metric space. Write this subsequence as $j(n_k)$ with distinct $n_k$. For $S=\{n_2,n_4,n_6,\dots \}$, its $S$ coordinate alternates between zero and one and therefore does not converge. But coordinate projections are continuous in the product topology, so every coordinate of a convergent sequence must converge. This contradiction proves that $X$ is not metrizable. This is a separable compactification need not be metrizable counterexample.

True. In the norm topology on the unit ball, the set $\{x\in B:\Vert x\Vert <1/2\}$ is a neighbourhood of zero. If the relative weak topology agrees with it, there are finitely many continuous linear functionals $f_1,\dots ,f_m$ and positive ${ϵ}_j$ such thatIf ${\bigcap }_j\ker f_j$ contained a nonzero vector, rescaling it to have norm $3/4$ would put it in the left-hand set but not the right-hand set. Therefore ${\bigcap }_j\ker f_j=\{0\}$. The linear mapis injective into a finite-dimensional scalar space, soThis proof works for either the open or closed unit-ball convention and does not require completeness.

False. Take the separable Banach space $E={\ell }^1$, whose continuous dual space is ${\ell }^{\infty }$. We prove that first countability of the weak unit ball would force this dual to be norm separable, producing a contradiction.

Suppose that $B$ had a countable weak neighbourhood base $(U_n)$ at zero. Inside each $U_n$, choose a basic relative weak neighbourhood specified by finitely many continuous linear functionals $F_n$. The union $F={\bigcup }_n{F}_n$ is countable. For any $f\in E^{\prime }$ and $ϵ>0$, some $U_n$ lies in $\{x\in B:|f(x)|<ϵ\}$. On the vector subspacethis implies $\Vert f{|}_{M_n}\Vert \le ϵ$, by scaling vectors into $B$. The Hahn-Banach theorem extends $f{|}_{M_n}$ to $h\in E^{\prime }$ with $\Vert h\Vert \le ϵ$. The functional $f-h$ vanishes on $M_n$ and is therefore a linear combination of $F_n$: it factors through the finite-dimensional coordinate map $x\mapsto (g(x){)}_{g\in F_n}$, and a linear functional on that map's image extends to the finite-dimensional coordinate space. HenceEvery $f\in E^{\prime }$ thus belongs to the norm closure of $\mathrm{span}F$. Taking rational coefficients, or rational real and imaginary parts, gives a countable norm-dense subset of $E^{\prime }$. This proves that weak-ball metrizability requires a norm-separable dual.

For ${\ell }^1$, every bounded sequence $a=(a_j)$ defines $f_a(x)={\sum }_j{a}_j{x}_j$ with $\Vert f_a\Vert =\Vert a{\Vert }_{\infty }$. Conversely, the values $f(e_j)$ of any bounded linear functional determine such a sequence and recover $f$ by density of finitely supported vectors. Thus $E^{\prime }={\ell }^{\infty }$. The uncountable family $\{0,1{\}}^{\mathbb{N}}$ in ${\ell }^{\infty }$ has distance one between distinct members, so this dual is not norm separable: disjoint balls of radius less than $1/2$ would require distinct members of any countable dense set. Since every metrizable space is first countable, the weak unit ball of ${\ell }^1$ is not metrizable.

True. Choose a norm-dense sequence $(x_n)$ in the unit ball of the separable normed vector space $E$. On the dual unit ball $B^{\prime }$, defineEach summand is at most $2^{1-n}$, so the series converges uniformly. Symmetry and the triangle inequality follow term by term. If $d(\varphi ,\psi )=0$, the two continuous functionals agree on the dense sequence, hence on the unit ball and then on all of $E$. Thus $d$ is a metric.

A relative weak-star neighbourhood controls finitely many evaluations. Conversely, to make $d$ small, first choose a tail whose sum is small and then control the finitely many evaluations at $x_1,\dots ,x_N$. This proves that each metric ball is weak-star open. For the other direction, let $x\in E$ and $ϵ>0$. Approximate $x/\Vert x\Vert$ by $x_n$ when $x\ne 0$. Since $\Vert \varphi -\psi \Vert \le 2$ on $B^{\prime }$,The second term can be made smaller than $ϵ/2$ by choosing $n$; the first is smaller than $ϵ/2$ whenever $d(\varphi ,\psi )$ is sufficiently small, because $|\varphi (x_n)-\psi (x_n)|\le 2^{n}d(\varphi ,\psi )$. Every evaluation is therefore metric-continuous. These two inclusions show that $d$ induces exactly the weak-star topology, proving weak-star metrizability of the dual ball without a completeness assumption on $E$.

We construct a normalized positive translation-invariant functional on $C(G)$ and then represent it by a measure. The main step is a finite matching in a graph argument, for which we establish all the covering-net properties.

Let $V$ be an open neighbourhood of the identity in the compact topological group $G$. The sets $xV$, $x\in G$, form an open cover. Compactness supplies a finite set $F$ with $G=FV$. Among all such finite sets choose one of smallest cardinality, denoted $m(V)$. This is a minimal left covering net of a compact group. If $g\in G$, then $G=gFV$, and $gF$ has the same cardinality. Hence $gF$ is also minimal for the same $V$.

Let $F,F^{\prime }$ be any two minimal left covering nets of a compact group for $V$. They both have cardinality $m(V)$. Form a bipartite graph by joining $x\in F$ to $y\in F^{\prime }$ whenever $xV\cap yV\ne \varnothing$. For a subset $S\subseteq F$, let $N(S)$ be its set of neighbours. Every point of $SV$ lies in some $yV$ with $y\in F^{\prime }$; that $y$ is a neighbour of a member of $S$. Thereforeand $(F\setminus S)\cup N(S)$ is still a left covering set. By minimality,Thus $|N(S)|\ge |S|$. Hall's marriage theorem gives a bijection $\pi :F\to F^{\prime }$ with $xV\cap \pi (x)V\ne \varnothing$ for every $x$. If $xv=\pi (x)v^{\prime }$ at an intersection point, thenThis proves the matching minimal covering nets of a compact group property, including the estimate needed below.

For a continuous complex-valued function $f$ on $G$, defineContinuity and compactness imply ${\omega }_f(W)\to 0$ as identity neighbourhoods $W$ shrink. Here is the uniformity argument: continuity of $(x,w)\mapsto f(xw)-f(x)$ at each $(x,e)$ gives a product neighbourhood where its absolute value is small; finitely many of the first-coordinate neighbourhoods cover $G$, and intersecting the corresponding second-coordinate neighbourhoods gives one $W$ that works for every $x$.

Matched pairs therefore giveIn particular, with $F^{\prime }=gF$,The estimate holds for every $g\in G$; no commutativity of $G$ is being assumed.

Direct the open identity neighbourhoods by reverse inclusion and choose a minimal left covering net of a compact group $F_V$ for each. The values $I_{F_V}(f)$ lie in the closed complex disk of radius $\Vert f{\Vert }_{\infty }$. The Tychonoff theorem makes the product of these disks over all $f\in C(G)$ compact. Consequently the net of vectors $(I_{F_V}(f){)}_{f\in C(G)}$ has a convergent subnet. Call its coordinatewise limit $I(f)$. Since each finite average is linear, positive on nonnegative real functions, and equal to one at the constant function one, the limit satisfiesand $|I(f)|\le \Vert f{\Vert }_{\infty }$. Thus $I$ is a bounded positive linear functional.

Given an identity neighbourhood $W$, continuity of multiplication and inversion supplies a sufficiently small $V$ with $V{V}^{-1}\subseteq W$. The translation estimate consequently tends to zero along the subnet. For every $f\in C(G)$ and every $g\in G$,The Riesz-Markov-Kakutani representation theorem now represents $I$ by a regular positive Borel measure $\mu$ with $\mu (G)=1$. The preceding equality and uniqueness in that representation imply that the pushforward under every left translation equals $\mu$. HenceThis is the required normalized Haar measure. It is nonzero because its total mass is one. In fact every nonempty open set has positive measure: its left translates cover $G$, a finite subcover exists, and a zero measure for that open set would force $\mu (G)=0$.

For completeness, the constructed probability measure is also right invariant. Apply Fubini's theorem to $f(x^{-1}y)$ with both variables distributed according to $\mu$. Integrating first in $y$ gives $\int fd\mu$ by left invariance. Integrating first in $x$ and substituting $x=yz$ gives $\int f(z^{-1})d\mu (z)$. Thus $\mu$ is invariant under inversion. Inversion turns a right translation into a left translation, so right invariance follows. This last observation is not needed for the existence of a left Haar measure, but confirms the usual two-sided normalization for compact groups.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra $A\subseteq C(K)$ containing the constants and separating points of $K$, with the supremum norm. It is therefore a commutative unital Banach algebra. Its character space of an algebra ${\Phi }_A$, also called its carrier space, consists of nonzero multiplicative linear functionals, equipped with the Gelfand topology.

For $x\in K$, evaluation gives the character of an algebra ${\delta }_x(f)=f(x)$. The map $\delta :K\to {\Phi }_A$ is continuous: composing it with the coordinate $\phi \mapsto \phi (f)$ gives the continuous function $f$ for every $f\in A$. It is injective because $A$ separates points. The Gelfand topology is Hausdorff because distinct algebra characters differ on some element of $A$. A continuous injection from compact $K$ into a Hausdorff space is a homeomorphism onto its image: closed subsets of $K$ have compact, hence closed, images. Its image is also compact and closed. Thusis the asserted homeomorphism. A natural uniform algebra is one for which this image is all of ${\Phi }_A$.

We prove the forward direction of the finite-generator criterion for a natural uniform algebra. Suppose that $A$ is a natural uniform algebra, and let $f_1,\dots ,f_n$ have no common zero on $K$. The idealis all of $A$. Otherwise it is contained in a maximal ideal $M$, whose associated character of an algebra vanishes on all the $f_j$. Since every algebra character is an evaluation character under naturality, this would give a common zero, a contradiction. Therefore $1\in J$, meaning that

Here are the Banach-algebra details behind the algebra character associated with $M$. A proper ideal cannot have dense closure: if it contained an element within distance less than one of $1$, the Neumann series would make that element invertible, forcing the ideal to contain $1$. Thus the closure of a proper ideal is proper. A maximal ideal, obtained from Zorn's lemma, is therefore closed. Its quotient is a complex unital Banach algebra in which every nonzero element is invertible. The Gelfand-Mazur theorem identifies this quotient with $\mathbb{C}$, so the quotient map is the required algebra character. No closedness of the originally generated ideal $J$ is needed.

Assume the finite-generator identity holds whenever there is no common zero. Let $\phi \in {\Phi }_A$. For every finite family $f_1,\dots ,f_n\in \ker \phi$, their zero sets must have nonempty intersection: otherwise an identity $\sum f_j{g}_j=1$ would give $0=\phi (1)=1$. These zero sets are closed in compact $K$. The finite intersection property therefore supplies $x\in K$ at which every member of $\ker \phi$ vanishes. For any $f\in A$, the element $f-\phi (f)1$ is in that kernel, so $f(x)=\phi (f)$. Thus $\phi ={\delta }_x$, proving

For $C(K)$, the identity is explicit. If the $f_j$ have no common zero, the continuous function $q={\sum }_j|f_j{|}^2$ is strictly positive on compact $K$. SetThen $\sum f_j{g}_j=1$. The equivalence just proved shows that $C(K)$ is natural.

Next consider the disc algebra $A(\overline{\mathbb{D}})$, the continuous functions on the closed unit disk that are holomorphic in its interior. We show directly that each algebra character is evaluation. Polynomials are uniformly dense in this algebra: for $0<r<1$, $f_r(z)=f(rz)$ tends uniformly to $f$ as $r\uparrow 1$, by uniform continuity. The function $f_r$ is holomorphic on $|z|<1/r$, so its Taylor polynomials converge uniformly on $|z|\le 1$.

Every algebra character of a unital Banach algebra is bounded, with $|\phi (f)|\le \Vert f\Vert$: if $f-\phi (f)1$ were invertible, applying $\phi$ would contradict invertibility, so $\phi (f)$ belongs to the spectrum of an element, which is contained in the disk of radius $\Vert f\Vert$ by the Neumann series. For the coordinate function $z$, put $\lambda =\phi (z)$; then $|\lambda |\le 1$. Multiplicativity gives $\phi (p)=p(\lambda )$ for every polynomial. By polynomial density and continuity, $\phi (f)=f(\lambda )$ for every member of the disc algebra. Hence

Finally consider the algebra $C$ in the last request. For each $f\in C$, its holomorphic extension $g$ from the boundary is unique by the maximum modulus principle. Define $T(f)=g\in A(\overline{\mathbb{D}})$. This is a surjective unital algebra homomorphism, since it restricts to the identity on the disc algebra. AlsoThe algebra $C$ is closed: if $f_n\to f$ uniformly, the associated $T(f_n)$ are uniformly Cauchy by this bound, so their limit belongs to the closed disc algebra and agrees with $f$ on the boundary. Constants and the coordinate function belong to $C$, so it is itself a uniform algebra on the disk. Its idealis contained in $C$, and $C/I$ is the disc algebra.

If a algebra character $\chi$ on $C$ vanishes on $I$, it factors through $T$ and the preceding result givesIf $\chi$ does not vanish on $I$, choose $h\in I$ with $\chi (h)\ne 0$. For $F\in C(\overline{\mathbb{D}})$, defineThe product $hF$ is in $I$, so this is well defined. It is unital and linear, andshows multiplicativity. Since $C(\overline{\mathbb{D}})$ is natural, $\tilde{\chi }$ is evaluation at a point $x$ of the closed disk. The equality $h(x)=\tilde{\chi }(h)=\chi (h)\ne 0$ places $x$ in the open disk. On $C$, $\tilde{\chi }(f)=\chi (f)$, so $\chi$ itself is evaluation at that interior point.

We therefore obtain two families: actual evaluations $f\mapsto f(x)$ on one closed disk, and analytic-extension evaluations $f\mapsto T(f)(\lambda )$ on a second closed disk. They agree on the boundary. They are distinct at interior points: functions in $I$ distinguish actual interior evaluations from every analytic-extension evaluation, and the coordinate function distinguishes points within each family. ThusThis is also a topological identification. Each disk family is continuous in the Gelfand topology, and they agree along their boundary, giving a continuous bijection from the glued disks to ${\Phi }_C$. The glued disks form a compact space, explicitly homeomorphic to the sphere by sending $(x,y)$ in the two copies to $(x,y,\pm \sqrt{1-x^2-y^2})$. Since the character space is Hausdorff, the continuous bijection is a homeomorphism. This boundary-analytic disk algebra with doubled character space has more algebra characters than evaluations on its original disk.

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.

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.

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.

Take the C-star algebra $M_2(\mathbb{C})$ and chooseThen $a-1=\mathrm{diag}(0,1)$ is positive, whileis positive, since its quadratic form at $(x,y)$ is $|x+y{|}^2$. Thus $1\le a\le b$, as required. However,has determinant $-1$ and eigenvalues $5\pm \sqrt{26}$, one of which is negative. Equivalently, its quadratic form at the real vector $(5,-4)$ is $-4$. It is therefore not positive. This proves that squaring is not order preserving in a C-star algebra, even for invertible elements bounded below by the identity.