Polar decomposition of functionals and weak compactness in preduals

Originally written by Claude Opus 5.5 (Anthropic), September 2026, with a separate historical AI spot-check. GPT-6.1 Sol (OpenAI), at the Ultra setting, read and replayed the entire lesson and all five solutions, compared the full programme prerequisite proofs, and completed the boundary and construction details, October 2026. Public domain (CC0).

An integrable complex function $h$ determines a functional $x\mapsto\int xh\,d\mu$, whose positive absolute value is $x\mapsto\int x|h|\,d\mu$; Example 2.6 proves this formula. A matrix $\rho$ can be written as $\rho=w|\rho|$, where $|\rho|=(\rho^*\rho)^{1/2}$ and $w$ is a partial isometry. Both are cases of one fact about von Neumann algebras: in any such algebra $M$, every normal functional $\varphi$ factors as $\varphi=v|\varphi|$, where $|\varphi|$ is a positive normal functional with $\||\varphi|\|=\|\varphi\|$ and $v\in M$ is a partial isometry. This lesson proves the factorization, identifies $|\varphi|$ by an inequality that does not mention $v$, and studies how $|\varphi|$ depends on $\varphi$. Then it uses the absolute value to study the predual $M_*$ as a Banach space.

Sections 1 to 6 are about the polar decomposition. Section 1 shows that when a projection cuts a functional without losing any of its norm, it cuts away nothing, and it defines the left and right supports of a functional. Section 2 proves the polar decomposition for normal functionals and, through the bidual, for all bounded functionals on a $C^*$-algebra. Section 3 characterizes $|\varphi|$ by the inequality $|\varphi(x)|^2\le\|\varphi\|\,|\varphi|(xx^*)$. Section 4 bounds $|a\varphi|$ by $\varphi$ for positive $\varphi$, and Section 5 shows that $\varphi\mapsto|\varphi|$ is continuous, with an explicit modulus. Section 6 describes the closed invariant subspaces of a predual by their positive parts, and through them the closed one-sided ideals of a $C^*$-algebra; every closed left ideal is an intersection of left kernels of pure states.

Sections 7 to 14 are about the predual as a Banach space. Section 7 proves Phillips's lemma and Schur's theorem on $\ell^1$. With them, Section 8 shows that the normal and singular parts of a weak$^*$ convergent sequence of functionals converge, so that preduals are weakly sequentially complete. Section 9 shows that the $\sigma$-strong topology on the unit ball comes from a complete metric exactly when there is a faithful normal state. Section 10 characterizes the relatively weakly compact subsets of a predual, and Section 11 deduces that on bounded sets the Mackey topology is the $\sigma$-strong$^*$ topology. Section 12 shows that in an atomic algebra, such as $B(H)$ or $\ell^\infty$, weak convergence of normal functionals becomes norm convergence once the absolute values are under control. Section 13 extends normal functionals from subalgebras without increasing the norm. Section 14 characterizes the $C^*$-algebras that are ideals in their biduals: they are exactly the $c_0$-direct sums of algebras of compact operators. Section 15 has exercises with solutions.

The absolute value reduces questions about normal functionals to questions about positive ones. For instance, every closed invariant subspace of a predual is generated by its positive part. The results on weak compactness are the noncommutative form of classical facts about $\ell^1$ and about spaces of measures. They are what one uses when compactness in a predual is needed, for example to average normal functionals, and they show where $B(H)_*$ behaves like $\ell^1$ and where it does not.

The full earlier programme proofs supply the bidual, positive supports, invariant predual subspaces, normal/singular splitting, operator polar decomposition, functional calculus and projection comparison. The next section gives exact proof locations, including the Banach-space inputs. The multiplication algebra is proved directly below. The freely readable comparison sources are identified at the end.

Nothing here assumes separability, a unit, or a faithful normal state unless this is said. Edge cases are worked out where they arise: functionals whose partial isometry lies outside the $C^*$-algebra (Example 3.3), the failure of the triangle inequality for absolute values (Example 3.5), weak$^*$ limits that are not norm limits (Example 5.4), nets in place of sequences (Example 8.4), uncountable families of singular functionals (Example 8.6), absolute values that escape from a weakly compact set (Example 10.7), and non-atomic algebras (Example 12.5).

The polar decomposition separates the size of a normal functional from its phase. Its support projections then control orthogonality, normal and singular parts, weak compactness and the Mackey topology. The complete arguments appear in the indicated sections below; the freely readable comparisons are listed at the end.

Conventions

Inner products are linear in the first variable. For vectors $\xi,\eta$ we write $\theta_{\xi,\eta}\zeta=\langle\zeta,\eta\rangle\xi$, $\omega_{\xi,\eta}(x)=\langle x\xi,\eta\rangle$ and $\omega_\xi=\omega_{\xi,\xi}$.
$M$ denotes a von Neumann algebra (or, what is the same, a $W^*$-algebra), $M_*$ its predual and $M_1=\{x\in M:\|x\|\le1\}$ its closed unit ball. The weak topology of the Banach space $M_*$ is $\sigma(M_*,M)$, because $(M_*)^*=M$. "Weakly convergent" and "relatively weakly compact" for subsets of $M_*$ refer to it. On $M^*$ we use the weak$^*$ topology $\sigma(M^*,M)$.
For $x\in M$, $|x|=(x^*x)^{1/2}$. For $\omega\in M_*^+$, $s(\omega)$ is its support: $\omega(x)=\omega(s(\omega)xs(\omega))$, $\omega$ is faithful on $s(\omega)Ms(\omega)$, and $1-s(\omega)$ is the largest projection on which $\omega$ vanishes.
Module actions. For a bounded functional $\varphi$ on a $C^*$-algebra $B$ and $a,b$ in $B$ (or in its unitization), $\begin{gathered} (a\varphi)(x)\\ =\varphi(xa),\\ (\varphi b)(x)\\ =\varphi(bx),\\ (a\varphi b)(x)\\ =\varphi(bxa). \end{gathered} \tag{0.1}$ Then $a(b\varphi)=(ab)\varphi$, $(\varphi a)b=\varphi(ab)$, $(a\varphi)b=a(\varphi b)$ and $\|a\varphi b\|\le\|a\|\|\varphi\|\|b\|$. The adjoint functional is $\varphi^*(x)=\overline{\varphi(x^*)}$; it satisfies $(a\varphi)^*=\varphi^*a^*$, $(\varphi b)^*=b^*\varphi^*$ and $\|\varphi^*\|=\|\varphi\|$. A functional is hermitian if $\varphi^*=\varphi$. We write $\psi\le\varphi$ if $\varphi-\psi$ is positive. For $\omega\in M_*^+$ the support satisfies $\omega=s(\omega)\omega=\omega s(\omega)$.
Representations of $C^*$-algebras are nondegenerate.
For a $C^*$-algebra $A$, $\tilde A=A^{**}$ is its universal enveloping von Neumann algebra. We regard $A$ as a $\sigma$-weakly dense $C^*$-subalgebra of $\tilde A$ and identify $A^*$ with the predual $\tilde A_*$: a functional $f\in A^*$ and its normal extension to $\tilde A$ are written with the same letter. So $\tilde A$ acts on $A^*$ by (0.1).
For $\varphi\in M^*$, $\varphi^{\rm n}$ and $\varphi^{\rm s}$ are its normal and singular parts: $\varphi=\varphi^{\rm n}+\varphi^{\rm s}$ and $\|\varphi\|=\|\varphi^{\rm n}\|+\|\varphi^{\rm s}\|$. $M_*^\perp$ is the space of singular functionals.
$r(a)$ is the spectral radius of $a$.

Full proof prerequisites

Banach spaces.

Baire's theorem. A nonempty complete metric space is not a countable union of closed sets with empty interior. Full programme proof: HB Theorem 3.1.
Uniform boundedness. A pointwise bounded family of bounded linear maps on a Banach space is norm bounded. Hence weak$^*$ convergent sequences in a dual Banach space, and weakly bounded subsets of a normed space, are norm bounded. Full programme proof: HB Theorem 4.2 and Corollary 4.3.
Hahn–Banach theorems. Two nonempty disjoint convex sets in a topological vector space, one of them open, are separated by a continuous real-linear functional; a point outside a closed convex set of a locally convex space is strictly separated from it. A continuous linear functional on a subspace of a locally convex space extends to a continuous linear functional on the whole space. The continuous functionals for a weak topology $\sigma(X,Y)$ are the elements of $Y$. Full programme proof: HB Section 6, separation and extension, and WT Theorem 1.2, the weak continuous dual.
Mazur's theorem. A convex subset of a normed space has the same closure in the weak and in the norm topology. Full programme proof: WT Theorem 4.1.
Banach–Alaoglu theorem. The closed unit ball of a dual Banach space is weak$^*$ compact. Full programme proof: WT Theorem 3.1.
Krein–Milman theorem. A nonempty compact convex subset of a Hausdorff locally convex space is the closed convex hull of its extreme points. Full programme proof: WT Theorem 6.1.
Annihilators. For a subspace $N$ of the dual $X^*$ of a Banach space, the weak$^*$ closure of $N$ is the set of functionals that vanish on $N_\perp=\{x:\psi(x)=0\ \forall\psi\in N\}$; for a subspace $L\subseteq X$, the norm closure of $L$ is the set of $x$ annihilated by $L^\perp$. Full programme proof: WT Theorem 5.1.
Finite dimension. On a finite-dimensional vector space all Hausdorff vector topologies coincide. Full programme proof: HB Theorem 7.1.
Eberlein–Šmulian theorem. For a subset $K$ of a Banach space $E$ the following are equivalent: $K$ is relatively weakly compact; every sequence in $K$ has a subsequence that converges weakly in $E$; every sequence in $K$ has a weak cluster point in $E$. Full programme proof: WT Lemma 7.1 and Theorem 7.2, the full bidual and subsequence proof.
Further reading, unproved and unused: the Mackey–Arens compatibility theorem identifies the topology defined directly in Section 11 with the finest locally convex topology having the given dual . No proof or solution here uses that compatibility characterization.

$C^*$-algebras. These facts are proved in the lessons on $C^*$-algebras, on Banach algebras and on the bidual, named above.

Continuous functional calculus, positive elements, and increasing approximate units $(u_\lambda)$ exist in every $C^*$-algebra (C*-algebras lesson). The spectral radius satisfies $r(a)=\lim_n\|a^n\|^{1/n}\le\|a\|$ (spectral radius formula). Exact full proofs: CF Section 5, CF Section 8, CF Theorem 11.1, and BN spectral-radius formula.
Positive functionals and GNS. Positivity implies hermiticity and the Cauchy–Schwarz inequality, by the full GN Proposition 3.2. The full GN Section 4 norm and continuity proofs and Section 5 GNS construction give a cyclic vector $\xi_\omega$ with $\omega(x)=\langle\pi_\omega(x)\xi_\omega,\xi_\omega\rangle$ and $\|\xi_\omega\|^2=\|\omega\|$, including the nonunital and zero cases. For an approximate identity $(u_\lambda)$ of positive contractions, $\pi_\omega(u_\lambda)\to1$ strongly: on $\pi_\omega(a)\eta$, use $\|u_\lambda a-a\|\to0$, then density and the uniform norm bound. Thus $\omega(u_\lambda)\to\|\omega\|$. For $0\le\psi\le\omega$, applying this limit to $\psi,\omega-\psi,\omega$ gives $\|\psi\|+\|\omega-\psi\|=\|\omega\|$.
Pure states. A state $\omega$ is pure exactly when $\pi_\omega$ is irreducible, and then $a+N_\omega\mapsto\pi_\omega(a)\xi_\omega$ is an isometry of $A/N_\omega$, with the quotient norm, onto $H_\omega$, where $N_\omega=\{x:\omega(x^*x)=0\}$ (Lemma 7.2 and Proposition 7.3). A minimal nonzero closed left ideal of $A$ is $Af$ for a projection $f\in A$ with $fAf=\mathbb Cf$ (Exercise 7.5). Exact full proofs: UE Lemma 7.2 and Proposition 7.3 and Exercise 7.5, complete solution.
The bidual. The closed unit ball of $A$ is $\sigma$-weakly dense in that of $\tilde A$. Every representation $\pi$ of $A$ has a unique normal extension $\bar\pi$ to $\tilde A$; its kernel is $\tilde A(1-z(\pi))$ for a central projection $z(\pi)$, the central support of $\pi$ (Lemma 2.1 and Theorem 3.3). For a Hilbert space $H$, the bidual of the compact operators $K(H)$ is $B(H)$, with $K(H)\subseteq B(H)$ as the canonical inclusion (Example 5.6). Exact full proofs: UE state-sum and predual-adjoint construction, UE Lemma 2.1, normal extension and central-kernel inverse, and UE Example 5.6. The bounded density input is KD Theorem 7.1.

von Neumann algebras.

Topologies. The $\sigma$-strong topology is defined by the seminorms $x\mapsto\omega(x^*x)^{1/2}$ and the $\sigma$-strong$^*$ topology by $x\mapsto\omega(x^*x+xx^*)^{1/2}$, $\omega\in M_*^+$. On bounded sets they agree with the strong and strong$^*$ operator topologies of any faithful normal representation. Normal functionals are $\sigma$-strongly continuous, and the involution is $\sigma$-strong$^*$ continuous (Lemma 1.2). Multiplication is jointly $\sigma$-strongly continuous on bounded sets (universal enveloping algebra lesson, background). Every $*$-isomorphism between von Neumann algebras is a homeomorphism for the $\sigma$-weak, $\sigma$-strong and $\sigma$-strong$^*$ topologies, in particular normal, so these topologies do not depend on the representation (Corollary 11.4). For a projection $e\in M$, $eMe$, acting on $eH$, is a von Neumann algebra whose $\sigma$-weak and $\sigma$-strong$^*$ topologies are the relative ones (reduced algebras). Exact full proofs: LT Sections 8–9, including Lemma 8.5 and relative weak topologies, UE Corollary 11.4, intrinsic topologies, and BI Theorem 5.8, reduced algebras.
Spectral theory. Self-adjoint elements of $M$ are norm limits of real combinations of their spectral projections, which lie in $M$; bounded Borel functions of self-adjoint elements lie in $M$. A bounded increasing net of self-adjoint elements converges $\sigma$-strongly to its least upper bound (universal enveloping algebra lesson, background). Exact full proofs: ST Theorem 3.1 and Proposition 5.1, ST Theorem 6.1, bounded increasing nets, and PT Fact 2.8, finite spectral approximation.
Polar decomposition of operators. Every $x\in M$ is $x=u|x|$ with $u\in M$ a partial isometry, $u^*u$ the projection onto $[x^*H]$ and $uu^*$ the projection onto $[xH]$ (Proposition 7.2). Exact full proof: BI Theorem 7.2.
Supports and left kernels. For $\omega\in M_*^+$, $\{x\in M:\omega(x^*x)=0\}=M(1-s(\omega))$ (Lemma 11.1). Exact full least-support and faithfulness proof: UE tools before Section 1; UE Lemma 11.1 identifies the left kernel.
One-sided ideals. A $\sigma$-weakly closed right ideal of $M$ is $pM$ for a unique projection $p$, and a $\sigma$-weakly closed left ideal is $Mp$ (Lemma 4.2). Exact full proof: UE Lemma 4.2, with the complete one-sided projection construction in BI Section 8.
Invariant subspaces. A norm-closed subspace $V\subseteq M_*$ with $aV\subseteq V$ for all $a$ in a $\sigma$-weakly total subset of $M$ is invariant under all of $M$, and has the form $V=M_*e=\{\varphi:\varphi=\varphi e\}$ for a unique projection $e$ (Theorem 4.3). Exact full proof: UE Theorem 4.3, including the Hahn–Banach annihilator argument passing from a total subset to all multipliers.
Positivity from a norming net. If $\omega\in A^*$ and $\omega(a_\lambda)\to\|\omega\|$ for a net of positive contractions $a_\lambda$, then $\omega$ is positive (Lemma 8.1). Exact full proof: UE Lemma 8.1.
Normal and singular parts (Section 10). There is a central projection $z_0$ of $M^{**}$ with $M_*=M^*z_0$ and $M_*^\perp=M^*(1-z_0)$. Both parts of a positive functional are positive, $M_*^\perp$ is norm closed, and it is spanned by its positive elements. Exact full proofs: UE Section 10 and Lemma 10.2, using its prior full Jordan construction.
Singular functionals (Theorem 11.2 and Corollary 11.5). A positive $\omega\in M^*$ is singular exactly when every nonzero projection majorizes a nonzero projection on which $\omega$ vanishes. A functional $\varphi\in M^*$ is normal exactly when it is completely additive: for every family $(p_i)$ of mutually orthogonal projections, the finite partial sums of $\sum_i\varphi(p_i)$ converge to $\varphi(\sum_ip_i)$. It suffices that the restriction of $\varphi$ to every abelian von Neumann subalgebra is normal (Remark 11.6). An infinite-dimensional von Neumann algebra has states that are not normal (Exercise 11.10). Exact full proofs: UE Theorem 11.2, Corollary 11.5, Remark 11.6 and Exercise 11.10. These include the complete-additivity converse, abelian restriction argument and non-normal state construction.
Normal representations. The cyclic representation of a positive normal functional is normal, and the image of a faithful normal representation is a von Neumann algebra (Proposition 12.1). Exact full proofs: UE Proposition 12.1 and UE Proposition 12.1, GNS normality paragraph.
Minimal projections (Section 6). A projection $f$ is minimal if $f\ne0$ and $fMf=\mathbb Cf$. Then $Mf$ is a Hilbert space for $\langle x,y\rangle f=y^*x$, whose norm is the norm of $M$; $\dim fMg\le1$ for minimal $f,g$; the join of finitely many minimal projections is a finite sum of mutually orthogonal minimal projections; and an infinite-dimensional von Neumann algebra contains an infinite sequence of mutually orthogonal nonzero projections, hence an isometric copy of $\ell^\infty$, so it is not reflexive. Exact full proofs: UE Lemmas 6.1, 6.2 and 6.5, including the sequence-space dualities and closed-subspace reflexivity argument.
Central supports and type I factors. The central support $c(e)$ of a projection $e$ is the least central projection majorizing $e$ (Definition 3.4). A von Neumann algebra is atomic if every nonzero projection majorizes a minimal projection; then $1$ is a sum of mutually orthogonal minimal projections (Example 7.6). A factor that contains a minimal projection is isomorphic to $B(L)$ for a Hilbert space $L$ (Lemma 6.2 and Corollary 10.4). Exact full proofs: PT Definition 3.4 and Proposition 3.5, Example 7.6, and Corollary 10.4.
Sums of vector functionals. If $M\subseteq B(H)$, every positive normal functional on $M$ is $x\mapsto\sum_n\langle x\xi_n,\xi_n\rangle$ with $\sum_n\|\xi_n\|^2<\infty$ (Theorem 10.1). Every $\sigma$-weakly continuous functional on $B(H)$ is a norm-convergent sum $\sum_n\omega_{\xi_n,\eta_n}$ with $\sum_n\|\xi_n\|\|\eta_n\|<\infty$ (universal enveloping algebra lesson, background). Exact full proofs: BI Theorem 10.1, LT Theorem 6.2(d), the vector series, and LT Theorem 9.1, equivalence with sigma-weak continuity.
Multiplication algebras. For a $\sigma$-finite measure space $(\Gamma,\mu)$, multiplication by $L^\infty(\Gamma,\mu)$ is a von Neumann algebra on $L^2(\Gamma,\mu)$. A full direct proof is given immediately below, so no decomposable-operator theorem is imported.

Elementary tools for the predual proofs

Sequence spaces and Hilbert balls. For any set $\Gamma$, $\ell^1(\Gamma)$ consists of the scalar families with finite sum of absolute values, where the sum is the supremum of finite subsums. Such a family has countable support, since each set $\{\gamma:|a_\gamma|\ge1/n\}$ is finite. Finite truncations are dense. An absolutely summable series of these families converges coordinatewise and in $\ell^1$: the nonnegative double sum bounds the norm of every tail by the sum of the norms of the series tail. A Cauchy sequence has a subsequence whose successive differences have summable norms; the preceding observation gives its limit, and then the whole sequence converges. Thus $\ell^1(\Gamma)$ is complete. A bounded functional is determined by its values $b_\gamma$ at the coordinate vectors, with $|b_\gamma|\le\|F\|$, and is $F(a)=\sum_\gamma a_\gamma b_\gamma$. Conversely every bounded family gives this functional, with norm $\sup_\gamma|b_\gamma|$, by coordinate tests. Hence $\ell^1(\Gamma)^*=\ell^\infty(\Gamma)$. The full Riesz theorem identifies the Hilbert dual isometrically with the Hilbert space; applying Banach–Alaoglu therefore makes its closed unit ball weakly compact. Complete and totally bounded metric spaces are compact by the full HS Lemma 5.0. These proofs supply the sequence-space, Hilbert-ball and metric inputs in Sections 7, 11 and 15.

Proof of fact 28. Let $D=\{M_h:h\in L^\infty(\Gamma,\mu)\}$. Multiplication is bounded with $\|M_h\|=\|h\|_\infty$. The upper bound follows by integration; for the lower bound, if $c<\|h\|_\infty$, the set $\{|h|>c\}$ has positive measure and meets a finite-measure set in positive measure, so test on its indicator. Null spaces give the zero algebra. Clearly $D\subseteq D'$. Conversely, take $T\in D'$, and partition $\Gamma$ into disjoint measurable sets $\Gamma_n$ of finite measure, using sigma-finiteness. On each non-null $\Gamma_n$ put $h_n=T1_{\Gamma_n}$; this is supported there because $T$ commutes with $M_{1_{\Gamma_n}}$. For every measurable $E\subseteq\Gamma_n$, $T1_E=1_Eh_n,\qquad \int_E|h_n|^2\,d\mu\le\|T\|^2\mu(E).$ Consequently $|h_n|\le\|T\|$ almost everywhere: a set where $|h_n|>\|T\|+\varepsilon$ of positive measure would contradict the inequality. The function $h$ obtained by joining the $h_n$ is measurable and bounded. The formulas give $T=M_h$ on simple functions supported in finitely many $\Gamma_n$. Those functions are dense in $L^2$, by truncation and simple approximation, whose full measure proofs are in the measure and Hilbert-space tools. Hence $T=M_h$, so $D'=D$ and $D''=D$. This proves the assertion on arbitrary sigma-finite measure spaces.

Choice and zero cases. Maximal orthogonal families use the Choice/Zorn convention of HB Section 1. For the zero algebra all functionals and projections are zero; the faithful positive functional is zero and the unit ball is a singleton. A faithful normal positive functional can be normalized to a state when the algebra is nonzero. This is the qualification in statements below about a faithful normal state.

1. Cutting a functional by a projection

Cutting a functional by a projection $e$ means passing from $\varphi$ to $e\varphi$, that is, to $x\mapsto\varphi(xe)$. The first result says how much norm can survive the two cuts $e\varphi$ and $(1-e)\varphi$. Its main use is the special case: if the cut by $e$ keeps all of the norm, it keeps all of the functional.

Lemma 1.1. Let $B$ be a $C^*$-algebra, $\varphi\in B^*$ and $e\in B$ a projection. Put $f=1-e$, computed in the unitization if $B$ has no unit, so that $xf=x-xe\in B$ for $x\in B$. Then $\begin{gathered} \|e\varphi\|^2+\|f\varphi\|^2\\ \le\|\varphi\|^2\\ \text{and}\\ \|\varphi e\|^2+\|\varphi f\|^2\\ \le\|\varphi\|^2 . \end{gathered} \tag{1.1}$ In particular, if $\|e\varphi\|=\|\varphi\|$ then $e\varphi=\varphi$, and if $\|\varphi e\|=\|\varphi\|$ then $\varphi e=\varphi$.

Proof. Put $\alpha=\|e\varphi\|$ and $\beta=\|f\varphi\|$, and let $\varepsilon>0$. Choose $a,b$ in the unit ball of $B$ with $\varphi(ae)\ge\alpha-\varepsilon$ and $\varphi(bf)\ge\beta-\varepsilon$; multiplying $a$ and $b$ by scalars of modulus one makes these numbers real. For $s,t\ge0$ with $s^2+t^2=1$, put $y=s\,ae+t\,bf\in B$. Since $ef=0$, the cross terms of $yy^*$ vanish: $\begin{gathered} yy^*\\ =s^2\,aea^*+t^2\,bfb^* ,\\ \text{so}\\ \|y\|^2\\ =\|yy^*\|\\ \le s^2+t^2\\ =1 . \end{gathered}$ Hence $\|\varphi\|\ge\operatorname{Re}\varphi(y)\ge s(\alpha-\varepsilon)+t(\beta-\varepsilon)$. If $\alpha^2+\beta^2>0$, take $(s,t)=(\alpha,\beta)/(\alpha^2+\beta^2)^{1/2}$ and let $\varepsilon\to0$: this gives $\|\varphi\|\ge(\alpha^2+\beta^2)^{1/2}$. The second inequality follows by applying the first to $\varphi^*$, since $(\varphi e)^*=e\varphi^*$ and adjoints keep norms. If $\|e\varphi\|=\|\varphi\|$, then $\|f\varphi\|=0$, so $\varphi=e\varphi+f\varphi=e\varphi$. $\square$

Example 1.2 (The two extremes). Let $B=B(H)$, $\xi,\eta\in H$ and $\varphi=\omega_{\xi,\eta}$. Then $\|\omega_{\xi,\eta}\|=\|\xi\|\|\eta\|$: the inequality $\le$ is Cauchy–Schwarz, and for nonzero $\xi,\eta$, $x=\theta_{\eta,\xi}/(\|\xi\|\|\eta\|)$ attains it. If either vector is zero, the functional and both sides are zero. Since $e\omega_{\xi,\eta}=\omega_{e\xi,\eta}$, $\begin{gathered} \|e\varphi\|^2+\|(1-e)\varphi\|^2\\ =\big(\|e\xi\|^2+\|(1-e)\xi\|^2\big)\|\eta\|^2\\ =\|\varphi\|^2 . \end{gathered}$ So (1.1) is an equality for vector functionals, and it cannot be improved. At the other extreme, for a central projection $e$ of a von Neumann algebra and a normal $\varphi$, the norm is additive: $\|\varphi\|=\|e\varphi\|+\|(1-e)\varphi\|$ (Lemma 10.2). Every functional lies between the two: $\begin{gathered} (\|e\varphi\|^2+\|(1-e)\varphi\|^2)^{1/2}\\ \le\|\varphi\|\\ \le\|e\varphi\|+\|(1-e)\varphi\|. \end{gathered}$

A normal functional determines two projections. Let $\varphi\in M_*$. The set $R_\varphi=\{y\in M:\varphi(yx)=0\text{ for all }x\in M\}$ is a $\sigma$-weakly closed right ideal, because $y\mapsto\varphi(yx)$ is $\sigma$-weakly continuous for each $x$. By background fact 19, $R_\varphi=pM$ for a unique projection $p$. Likewise $\begin{gathered} L_\varphi\\ =\{y\in M:\varphi(xy)=0\text{ for all }x\in M\}\\ =Mq \end{gathered}$ for a unique projection $q$.

Definition 1.3. The right support of $\varphi\in M_*$ is $s_r(\varphi)=1-p$ and the left support is $s_l(\varphi)=1-q$, with $p,q$ as above.

Lemma 1.4. Let $\varphi\in M_*$.

For a projection $e$, $\varphi=\varphi e$ exactly when $e\ge s_r(\varphi)$, and $\varphi=e\varphi$ exactly when $e\ge s_l(\varphi)$. So $s_r(\varphi)$ is the least projection $e$ with $\varphi(ex)=\varphi(x)$ for all $x$, and $s_l(\varphi)$ is the least projection $f$ with $\varphi(xf)=\varphi(x)$ for all $x$.
$\varphi(x)=\varphi(s_r(\varphi)\,x\,s_l(\varphi))$ for all $x\in M$.
$s_l(\varphi^*)=s_r(\varphi)$ and $s_r(\varphi^*)=s_l(\varphi)$.
If $\omega$ is positive, $s_l(\omega)=s_r(\omega)=s(\omega)$. If $\varphi$ is hermitian, $s_l(\varphi)=s_r(\varphi)$.

Proof. (1) $\varphi=\varphi e$ means $\varphi((1-e)x)=0$ for all $x$, that is $1-e\in R_\varphi=pM$. This holds exactly when $p(1-e)=1-e$, that is $1-e\le p$, that is $e\ge1-p$. The left case is the same with $L_\varphi=Mq$.

(2) With $e=s_r(\varphi)$ and $f=s_l(\varphi)$: $\varphi(x)=\varphi(ex)=\varphi(exf)$ by (1), applied twice.

(3) $(\varphi e)^*=e\varphi^*$, so $\varphi=\varphi e$ exactly when $\varphi^*=e\varphi^*$. Now use (1) for $\varphi$ and $\varphi^*$.

(4) From $\omega=\omega s(\omega)=s(\omega)\omega$ and (1), both supports lie under $s(\omega)$. If $\omega=\omega e$, then $\omega(1-e)=\omega(e(1-e))=0$, so $1-e\le1-s(\omega)$, that is $e\ge s(\omega)$; the left support is handled in the same way. For hermitian $\varphi$, (3) gives $s_l(\varphi)=s_r(\varphi^*)=s_r(\varphi)$. $\square$

For the vector functional $\omega_{\xi,\eta}$ on $B(H)$, with $\xi,\eta\ne0$: $\omega_{\xi,\eta}(ex)=\langle x\xi,e\eta\rangle$ and $\omega_{\xi,\eta}(xf)=\langle xf\xi,\eta\rangle$. So $s_r(\omega_{\xi,\eta})$ is the projection onto $\mathbb C\eta$ and $s_l(\omega_{\xi,\eta})$ the projection onto $\mathbb C\xi$.

2. The polar decomposition

We first record a comparison principle for positive functionals. It gives uniqueness in the polar decomposition, and in Section 3 it characterizes the absolute value.

Lemma 2.1 (Comparison of positive functionals). Let $A$ be a $C^*$-algebra and $\omega,\omega_1\in A^*_+$ with $\|\omega\|=\|\omega_1\|$ and $\begin{gathered} |\omega(x)|^2\\ \le\|\omega_1\|\,\omega_1(x^*x)\\ (x\in A). \end{gathered} \tag{2.1}$ Then $\omega=\omega_1$.

Proof. Put $c=\|\omega_1\|$; if $c=0$ both functionals vanish. Let $(\pi,H,\xi)$ be the cyclic representation of $\omega_1$, so $\omega_1(x^*x)=\|\pi(x)\xi\|^2$ and $\|\xi\|^2=c$. By (2.1), $\pi(x)\xi\mapsto\omega(x)$ is a well-defined linear functional on the dense subspace $\pi(A)\xi$, of norm at most $c^{1/2}$. The Riesz theorem gives $\eta\in H$ with $\|\eta\|\le c^{1/2}$ and $\omega(x)=\langle\pi(x)\xi,\eta\rangle$ for $x\in A$. For an approximate unit $(u_\lambda)$, $\omega(u_\lambda)\to\|\omega\|=c$ and $\pi(u_\lambda)\xi\to\xi$ (background fact 12). Hence $\langle\xi,\eta\rangle=c$, and $\begin{gathered} \|\xi-\eta\|^2\\ =\|\xi\|^2-2\operatorname{Re}\langle\xi,\eta\rangle+\|\eta\|^2\\ \le c-2c+c\\ =0 . \end{gathered}$ So $\eta=\xi$ and $\omega(x)=\langle\pi(x)\xi,\xi\rangle=\omega_1(x)$. $\square$

Theorem 2.2 (Polar decomposition). Let $M$ be a von Neumann algebra and $\varphi\in M_*$. There is exactly one pair $(v,\omega)$ of a partial isometry $v\in M$ and a positive normal functional $\omega$ with $\varphi=v\omega\qquad\text{and}\qquad v^*v=s(\omega). \tag{2.2}$ For this pair:

$\omega=v^*\varphi$, and $\|\omega\|=\|\varphi\|=\varphi(v^*)$;
$|\varphi(x)|^2\le\|\varphi\|\,\omega(xx^*)$ for all $x\in M$;
$v^*v=s_r(\varphi)$ and $vv^*=s_l(\varphi)$;
$\varphi^*=v^*\psi$ with $\psi=v\omega v^*$, that is $\psi(x)=\omega(v^*xv)$, and $(v^*,\psi)$ is the pair (2.2) for $\varphi^*$. Moreover $\varphi=\psi v$.

Proof. Existence. For $\varphi=0$, take $\omega=0$ and $v=0$; any pair with the stated support condition must also have both entries zero, since $\|\omega\|=\|\varphi\|$ by the argument below and $v^*v=0$. We may therefore assume $\varphi\ne0$. The unit ball $M_1$ is $\sigma$-weakly compact and $\varphi$ is $\sigma$-weakly continuous, so $|\varphi|$ attains its supremum on $M_1$; after a rotation there is $a\in M_1$ with $\varphi(a)=\|\varphi\|$. Let $a^*=u|a^*|$ be the polar decomposition of $a^*$ (background fact 17), where $|a^*|=(aa^*)^{1/2}$ and $u\in M$ is a partial isometry with $u^*u=s(|a^*|)$, the range projection of $|a^*|$. Then $a=|a^*|u^*$ and $0\le|a^*|\le1$.

Finally $(u^*u)\omega_0=u^*(uu^*\varphi)=u^*\varphi=\omega_0$, that is $\omega_0(xu^*u)=\omega_0(x)$ for all $x$. At $x=1$ this gives $\omega_0(1-u^*u)=0$, so $s(\omega_0)\le u^*u$. Put $v=us(\omega_0)$. It is a partial isometry with $v^*v=s(\omega_0)u^*us(\omega_0)=s(\omega_0)$, and $v\omega_0=u(s(\omega_0)\omega_0)=u\omega_0=\varphi$. So $(v,\omega_0)$ satisfies (2.2).

Properties (1) and (2) for any pair (2.2). Let $(v,\omega)$ satisfy (2.2) and put $s=s(\omega)$. Then $v^*\varphi=(v^*v)\omega=s\omega=\omega$. Also $\varphi(v^*)=\omega(v^*v)=\omega(s)=\omega(1)=\|\omega\|$, and $\|\varphi\|\le\|v\|\|\omega\|\le\|\omega\|=\varphi(v^*)\le\|\varphi\|$. So $\|\omega\|=\|\varphi\|=\varphi(v^*)$. By the Cauchy–Schwarz inequality, $\begin{gathered} |\varphi(x)|^2\\ =|\omega(xv)|^2\\ \le\omega(xx^*)\,\omega(v^*v)\\ =\|\varphi\|\,\omega(xx^*) . \end{gathered} \tag{2.3}$

Uniqueness. Let $(v,\omega)$ and $(v_1,\omega_1)$ both satisfy (2.2). By (1) and (2) for the second pair, $\|\omega_1\|=\|\varphi\|$ and $|\varphi(y)|^2\le\|\varphi\|\,\omega_1(yy^*)$. Since $\omega(x)=(v^*\varphi)(x)=\varphi(xv^*)$, $\begin{gathered} |\omega(x)|^2\\ =|\varphi(xv^*)|^2\\ \le\|\varphi\|\,\omega_1(xv^*vx^*)\\ \le\|\varphi\|\,\omega_1(xx^*) . \end{gathered}$ Replacing $x$ by $x^*$ and using that $\omega$ is hermitian, $|\omega(x)|^2\le\|\omega_1\|\,\omega_1(x^*x)$. Lemma 2.1 gives $\omega=\omega_1$. Then $(v-v_1)\omega=0$, so $\begin{gathered} \omega((v-v_1)^*(v-v_1))\\ =((v-v_1)\omega)((v-v_1)^*)\\ =0. \end{gathered}$ By background fact 18, $(v-v_1)s(\omega)=0$. As $v=vs(\omega)$ and $v_1=v_1s(\omega)$, we get $v=v_1$.

(3). Let $s=s(\omega)=v^*v$. Since $\omega(y)=\omega(sys)$ and $vs=v$, we get $\varphi(sx)=\omega(sxv)=\omega(sxvs)=\omega(xv)=\varphi(x)$, so $\varphi=\varphi s$ and $s_r(\varphi)\le s$ (Lemma 1.4). Conversely, if $\varphi=\varphi e$, then $\begin{gathered} \omega(x)\\ =\varphi(xv^*)\\ =\varphi(exv^*)\\ =\omega(exv^*v)\\ =\omega(exs)\\ =\omega(ex), \end{gathered}$ so $\omega=\omega e$ and $e\ge s$ by Lemma 1.4(4). Hence $s_r(\varphi)=v^*v$. Now put $q=vv^*$. Since $qv=v$, $\varphi(xq)=\omega(xqv)=\omega(xv)=\varphi(x)$, so $\varphi=q\varphi$ and $s_l(\varphi)\le q$. Conversely, let $\varphi=f\varphi$ for a projection $f$. Then $\omega(x(1-f)v)=\varphi(x(1-f))=0$ for all $x$. Taking $x=v^*(1-f)$ gives $\omega\big(((1-f)v)^*(1-f)v\big)=0$, so $(1-f)vs=0$ by background fact 18. Thus $(1-f)v=0$, $fv=v$, and $f\ge vv^*$. Hence $s_l(\varphi)=vv^*$.

(4). $\psi=v\omega v^*$ is positive and normal. Since $\omega$ is hermitian and $\omega(y)=\omega(ys)$, $\begin{gathered} \varphi^*(x)\\ =\overline{\omega(x^*v)}\\ =\omega(v^*x)\\ =\omega(v^*xs)\\ =\omega(v^*xv^*v)\\ =\psi(xv^*)\\ =(v^*\psi)(x). \end{gathered}$ The support of $\psi$ is $vv^*=(v^*)^*v^*$: first $\psi(1-vv^*)=\omega(v^*v-v^*vv^*v)=0$; and if $g$ is a projection with $\psi(1-g)=0$, then $\omega(((1-g)v)^*(1-g)v)=0$, so $(1-g)v=(1-g)vs=0$ and $g\ge vv^*$. So $(v^*,\psi)$ satisfies (2.2) for $\varphi^*$. Finally $\begin{gathered} (\psi v)(x)\\ =\psi(vx)\\ =\omega(v^*vxv)\\ =\omega(sxv)\\ =\omega(xv)\\ =\varphi(x), \end{gathered}$ using $\omega=\omega s$. $\square$

Definition 2.3. The functional $\omega$ of Theorem 2.2 is the absolute value $|\varphi|$ of $\varphi$, and $\varphi=v|\varphi|$ is the polar decomposition of $\varphi$. By Theorem 2.2, $\begin{gathered} \||\varphi|\|\\ =\||\varphi^*|\|\\ =\|\varphi\|,\\ |\varphi|\\ =v^*\varphi,\\ |\varphi^*|\\ =v|\varphi|v^*,\\ \varphi\\ =v|\varphi|\\ =|\varphi^*|v . \end{gathered} \tag{2.4}$ Also $v^*|\varphi^*|v=|\varphi|$, since $\omega(v^*vxv^*v)=\omega(x)$. If $\varphi$ is positive, then $\varphi=s(\varphi)\varphi$ is its polar decomposition, so $|\varphi|=\varphi$ and $v=s(\varphi)$.

Three examples show what the absolute value is in familiar cases.

Example 2.4 (Matrices). Let $M=M_n(\mathbb C)$. Every functional is $\varphi(x)=\operatorname{Tr}(\rho x)$ for a unique matrix $\rho$. Let $\rho=w|\rho|$ be the polar decomposition of the matrix $\rho$. Then $|\varphi|=\operatorname{Tr}(|\rho|\,\cdot\,)$ and $v=w$. Indeed, $(w|\varphi|)(x)=\operatorname{Tr}(|\rho|xw)=\operatorname{Tr}(w|\rho|x)=\varphi(x)$. And the support of $\operatorname{Tr}(\sigma\,\cdot\,)$, for $\sigma\ge0$, is the range projection of $\sigma$: $\operatorname{Tr}(\sigma(1-p))=\operatorname{Tr}((1-p)\sigma(1-p))$ vanishes exactly when $\sigma^{1/2}(1-p)=0$, that is when $p$ majorizes the range projection of $\sigma$. For $\sigma=|\rho|$ this range projection is $w^*w$. So (2.2) holds. By Theorem 2.2(3), $s_l(\varphi)=ww^*$ is the range projection of $\rho$, and $s_r(\varphi)=w^*w$ is the projection onto $(\ker\rho)^\perp$.

Example 2.5 (Vector functionals). Let $M=B(H)$ and $\xi,\eta\ne0$. Then $\begin{gathered} |\omega_{\xi,\eta}|\\ =\frac{\|\xi\|}{\|\eta\|}\,\omega_\eta,\\ v\\ =\frac{\theta_{\xi,\eta}}{\|\xi\|\|\eta\|} . \end{gathered} \tag{2.5}$ Indeed, $v^*v=\theta_{\eta,\eta}/\|\eta\|^2$ is the projection onto $\mathbb C\eta$, which is the support of $\omega_\eta$; and $v\eta=(\|\eta\|/\|\xi\|)\xi$, so $\begin{gathered} (\|\xi\|/\|\eta\|)\,\omega_\eta(xv)\\ =(\|\xi\|/\|\eta\|)\langle xv\eta,\eta\rangle\\ =\langle x\xi,\eta\rangle. \end{gathered}$ So the absolute value of $\omega_{\xi,\eta}$ lives on $\eta$, and that of $\omega_{\xi,\eta}^*=\omega_{\eta,\xi}$ lives on $\xi$.

Example 2.6 (Commutative algebras). Let $(\Gamma,\mu)$ be a $\sigma$-finite measure space and $M=L^\infty(\Gamma,\mu)$ acting on $L^2(\Gamma,\mu)$ (background fact 28). The normal functionals are exactly $\varphi_h(x)=\int xh\,d\mu$ with $h\in L^1(\Gamma,\mu)$, and $\|\varphi_h\|=\|h\|_1$. Indeed, $\varphi_h=\omega_{f,g}$ for $f=|h|^{1/2}\operatorname{sgn}h$ and $g=|h|^{1/2}$, so $\varphi_h$ is normal. Conversely, a $\sigma$-weakly continuous functional on $M$ extends to a $\sigma$-weakly continuous functional on $B(L^2)$ (background fact 3, since the $\sigma$-weak topology of $M$ is the relative one, and the $\sigma$-weak topology is locally convex). That extension is $\sum_n\omega_{\xi_n,\eta_n}$ with $\sum_n\|\xi_n\|\|\eta_n\|<\infty$ (background fact 27), so on $M$ it is $\varphi_h$ with $h=\sum_n\xi_n\bar\eta_n\in L^1$. Finally $\|\varphi_h\|\le\|h\|_1$, and $x=\overline{\operatorname{sgn}h}$ gives equality. Now put $v=\operatorname{sgn}h$, with $\operatorname{sgn}h=h/|h|$ where $h\ne0$ and $0$ elsewhere. Then $v\varphi_{|h|}=\varphi_h$, and $v^*v=1_{\{h\ne0\}}$ is the support of $\varphi_{|h|}$. So $|\varphi_h|=\varphi_{|h|} ,$ the familiar total variation. The same argument works for counting measure on any set $\Gamma$, even when $\Gamma$ is uncountable: $\ell^\infty(\Gamma)$, acting diagonally on $\ell^2(\Gamma)$, is a von Neumann algebra, because an operator that commutes with every coordinate projection is diagonal, so $\ell^\infty(\Gamma)'=\ell^\infty(\Gamma)$. Thus $\ell^\infty(\Gamma)_*=\ell^1(\Gamma)$, and the absolute value is taken coordinatewise.

For a $C^*$-algebra the partial isometry lives in the bidual.

Theorem 2.7 (Polar decomposition on a $C^*$-algebra). Let $A$ be a $C^*$-algebra and $f\in A^*$. There is exactly one pair of a partial isometry $v\in\tilde A$ and a positive $\omega\in A^*$ with $f=v\omega$ and $v^*v=s(\omega)$, the support of $\omega$ in $\tilde A$. We write $|f|=\omega$. Then $\||f|\|=\|f\|$ and $|f(x)|^2\le\|f\|\,|f|(xx^*)$ for all $x\in\tilde A$, in particular for all $x\in A$.

Proof. Apply Theorem 2.2 to the von Neumann algebra $\tilde A$, whose predual is $A^*$. $\square$

Examples in Section 3 show that $v$ need not lie in $A$, even when $f$ is positive.

The hermitian case of the polar decomposition is the norm-additive splitting of a hermitian functional into positive parts. The lesson on the universal enveloping algebra proves existence before the bidual construction and proves uniqueness directly (Proposition 3.4); here both drop out of Theorem 2.2.

Corollary 2.8 (Hermitian functionals). Let $\varphi\in M_*$ be hermitian, with polar decomposition $\varphi=v|\varphi|$, and put $s=s(|\varphi|)$.

$v=v^*$, $|\varphi^*|=|\varphi|$, and $v|\varphi|=|\varphi|v$.
$e=\tfrac12(s+v)$ and $f=\tfrac12(s-v)$ are orthogonal projections with $e+f=s$ and $e-f=v$. The functionals $\varphi_+=e|\varphi|=\tfrac12(|\varphi|+\varphi)$ and $\varphi_-=f|\varphi|=\tfrac12(|\varphi|-\varphi)$ are positive and normal, $s(\varphi_+)=e$, $s(\varphi_-)=f$, $\varphi=\varphi_+-\varphi_-$ and $\|\varphi\|=\|\varphi_+\|+\|\varphi_-\|$.
If $\varphi=\psi_1-\psi_2$ with $\psi_1,\psi_2\in M_*^+$ and $\|\varphi\|=\|\psi_1\|+\|\psi_2\|$, then $\psi_1=\varphi_+$ and $\psi_2=\varphi_-$.
For $\psi_1,\psi_2\in M_*^+$: $\|\psi_1-\psi_2\|=\|\psi_1\|+\|\psi_2\|$ exactly when $s(\psi_1)s(\psi_2)=0$.

(2) $v$ is a self-adjoint partial isometry with $v^2=v^*v=s$ and $vs=sv=v$. So $e^2=\tfrac14(s+2v+s)=e$, $f^2=f$, $ef=\tfrac14(s-v^2)=0$, $e+f=s$ and $e-f=v$. By (1), $e$ commutes with $|\varphi|$: $e|\varphi|=|\varphi|e$. Hence $\begin{gathered} (e|\varphi|)(x)\\ =|\varphi|(xe)\\ =|\varphi|(xe\cdot e)\\ =(|\varphi|e)(xe)\\ =|\varphi|(exe), \end{gathered}$ which is positive in $x$; likewise for $f$. Next $\varphi_+-\varphi_-=(e-f)|\varphi|=v|\varphi|=\varphi$ and $\varphi_++\varphi_-=s|\varphi|=|\varphi|$, which give the formulas $\tfrac12(|\varphi|\pm\varphi)$ and $\begin{gathered} \|\varphi_+\|+\|\varphi_-\|\\ =|\varphi|(e)+|\varphi|(f)\\ =|\varphi|(s)\\ =\|\varphi\|. \end{gathered}$ Since $\varphi_+(1-e)=|\varphi|(e(1-e)e)=0$, $s(\varphi_+)\le e$; and if $g\le e$ is a projection with $\varphi_+(e-g)=0$, then $|\varphi|(e-g)=0$ with $e-g\le s$, so $e=g$ by faithfulness of $|\varphi|$ on $sMs$. Thus $s(\varphi_+)=e$, and similarly $s(\varphi_-)=f$.

(3) By Theorem 2.2(1), $\|\varphi\|=\varphi(v^*)=\varphi(e-f)$, so $\begin{gathered} \psi_1(1)+\psi_2(1)\\ =\psi_1(e)-\psi_1(f)-\psi_2(e)+\psi_2(f). \end{gathered}$ Here $\psi_1(e)\le\psi_1(1)$, $\psi_2(f)\le\psi_2(1)$ and $\psi_1(f),\psi_2(e)\ge0$. So all four inequalities are equalities: $\psi_1(1-e)=0$ and $\psi_2(1-f)=0$, that is $s(\psi_1)\le e$ and $s(\psi_2)\le f$. Then $e\psi_1=\psi_1$, because $\psi_1(xe)=\psi_1(xes(\psi_1))=\psi_1(xs(\psi_1))=\psi_1(x)$; and $e\psi_2=0$, because $\psi_2(xe)=\psi_2(xes(\psi_2))=0$ as $es(\psi_2)=efs(\psi_2)=0$. Therefore $\psi_1=e(\psi_1-\psi_2)=e\varphi=ev|\varphi|=e|\varphi|=\varphi_+$, since $ev=e(e-f)=e$. And $\psi_2=\psi_1-\varphi=\varphi_-$.

(4) If $s(\psi_1)s(\psi_2)=0$, then $x=s(\psi_1)-s(\psi_2)$ has norm at most one and $(\psi_1-\psi_2)(x)=\psi_1(1)+\psi_2(1)$, so $\|\psi_1-\psi_2\|\ge\|\psi_1\|+\|\psi_2\|$; the other inequality is the triangle inequality. Conversely, if the norms add, the proof of (3), applied to $\varphi=\psi_1-\psi_2$, gives $s(\psi_1)\le e$ and $s(\psi_2)\le f$, which are orthogonal. $\square$

3. How the absolute value is determined

The absolute value was defined through the partial isometry $v$. It is also the only positive functional of the right size that dominates $\varphi$ in the sense of (2.3). This description does not mention $v$, does not require normality, and works for every $C^*$-algebra.

Theorem 3.1 (Characterization of the absolute value). Let $A$ be a $C^*$-algebra and $f\in A^*$. Then $|f|$ is the only positive functional $\omega$ on $A$ with $\|\omega\|\le\|f\|$ and $\begin{gathered} |f(x)|^2\\ \le\|f\|\,\omega(xx^*)\\ (x\in A). \end{gathered} \tag{3.1}$ If $A=M$ is a von Neumann algebra and $f\in M_*$, then the $|f|$ of Theorem 2.2 is the only positive functional on $M$, normal or not, with these two properties. In both cases such an $\omega$ has $\|\omega\|=\|f\|$.

Proof. The von Neumann case. By Theorem 2.2, $|f|$ has the two properties. Let $\omega$ be any positive functional on $M$ with the two properties, and let $f=v|f|$. If $f=0$ then $\omega=0$. Otherwise, (3.1) at $x=v^*$ gives $\|f\|^2=f(v^*)^2\le\|f\|\,\omega(v^*v)\le\|f\|\,\|\omega\|$, so $\|\omega\|=\|f\|$. Since $|f|(x)=f(xv^*)$, $\begin{gathered} ||f|(x)|^2\\ =|f(xv^*)|^2\\ \le\|f\|\,\omega(xv^*vx^*)\\ \le\|f\|\,\omega(xx^*). \end{gathered}$ Replacing $x$ by $x^*$, $||f|(x)|^2\le\|\omega\|\,\omega(x^*x)$. Lemma 2.1, applied in the $C^*$-algebra $M$, gives $|f|=\omega$. No normality of $\omega$ was used.

The general case. By Theorem 2.7, $|f|$ has the two properties. Let $\omega\in A^*_+$ have them, and let $(\pi,H,\xi)$ be its cyclic representation. By (3.1) with $x=y^*$, $|f(y^*)|\le\|f\|^{1/2}\|\pi(y)\xi\|$. So $\pi(y)\xi\mapsto\overline{f(y^*)}$ is a bounded linear functional on $\pi(A)\xi$, and the Riesz theorem gives $\eta\in H$ with $\|\eta\|\le\|f\|^{1/2}$ and $f(y^*)=\overline{\langle\pi(y)\xi,\eta\rangle}=\langle\pi(y^*)\eta,\xi\rangle$. Thus $f(x)=\langle\pi(x)\eta,\xi\rangle$ for $x\in A$. Let $\bar\pi$ be the normal extension of $\pi$ to $\tilde A$ (background fact 14). The normal functionals $X\mapsto\langle\bar\pi(X)\eta,\xi\rangle$ and $X\mapsto\langle\bar\pi(X)\xi,\xi\rangle$ agree with $f$ and $\omega$ on the dense subalgebra $A$, so they are the normal extensions. Hence, for $X\in\tilde A$, $\begin{gathered} |f(X)|^2\\ =|\langle\eta,\bar\pi(X^*)\xi\rangle|^2\\ \le\|f\|\,\|\bar\pi(X^*)\xi\|^2\\ =\|f\|\,\omega(XX^*) . \end{gathered}$ So the normal extension of $\omega$ has the two properties in $\tilde A$, and the von Neumann case gives $\omega=|f|$. $\square$

Corollary 3.2. Let $M$ be a von Neumann algebra and $\varphi\in M_*$. The absolute value of $\varphi$ as a functional on the $C^*$-algebra $M$ (Theorem 2.7, through $M^{**}$) is the absolute value of Theorem 2.2. In particular it is normal.

Proof. Both are positive functionals on $M$ of norm $\|\varphi\|$ that satisfy (3.1) on $M$. By the general case of Theorem 3.1 they coincide. $\square$

Example 3.3 (Where the partial isometry lives). Let $A=C[0,1]$ and $\lambda(x)=\int_0^1x\,dt$.

For $s\ne t$ in $[0,1]$, $|\delta_s-\delta_t|=\delta_s+\delta_t$. Indeed, $\omega=\delta_s+\delta_t$ is positive with $\|\omega\|=2=\|\delta_s-\delta_t\|$ (test on a continuous $x$ with $|x|\le1$, $x(s)=1$, $x(t)=-1$), and $\begin{gathered} |x(s)-x(t)|^2\\ \le2(|x(s)|^2+|x(t)|^2)\\ =2\,\omega(xx^*). \end{gathered}$ Theorem 3.1 applies.
Let $\sigma=1_{[0,1/2)}-1_{[1/2,1]}$ and $f(x)=\int_0^1x\sigma\,dt$. Then $\|f\|=1$: the bound $\le$ is clear, and continuous functions $x_n$ with $|x_n|\le1$ that equal $\sigma$ outside an interval of length $1/n$ give $f(x_n)\ge1-2/n$. Also $\|\lambda\|=1$ and $|f(x)|^2\le\big(\int|x|\big)^2\le\int|x|^2=\lambda(xx^*)$. By Theorem 3.1, $|f|=\lambda$. Now suppose the partial isometry $v$ of $f$ lay in $A$, say $v=a$. Then $f(x)=\lambda(xa)=\int_0^1xa\,dt$ for all $x\in A$. Take $x\ge0$ continuous, supported in $[\tfrac12-\delta,\tfrac12]$, with $\int x=1$. Then $1=f(x)=\int xa$, and as $\delta\to0$ this tends to $a(\tfrac12)$, by continuity of $a$. So $a(\tfrac12)=1$. The same test with support in $[\tfrac12,\tfrac12+\delta]$ gives $a(\tfrac12)=-1$, a contradiction. So $v\notin A$.
Even for positive functionals, $v=s(f)$ need not lie in $A$. Let $A=K(H)$, whose bidual is $B(H)$ (background fact 14), let $(e_n)$ be an infinite orthonormal sequence, and let $f=\sum_n2^{-n}\omega_{e_n}$. Its normal extension to $B(H)$ is given by the same formula. Its support is the projection $P$ onto the closed span of the $e_n$: $f(1-P)=0$, and a projection $q\le P$ with $f(q)=0$ has $qe_n=0$ for all $n$, so $q=0$. So $v=P$ has infinite rank and is not compact.

The next result replaces the triangle inequality, which fails for absolute values (Example 3.5).

Proposition 3.4 (A weak triangle inequality). Let $A$ be a $C^*$-algebra and $(f_k)$ a finite or infinite sequence in $A^*$ with $\sum_k\|f_k\|<\infty$. Put $f=\sum_kf_k$. Then for every $x\in\tilde A$, $\begin{gathered} ||f|(x)|^2\\ \le\Big(\sum_k\|f_k\|\Big)\Big(\sum_k|f_k|(xx^*)\Big). \end{gathered} \tag{3.2}$

Proof. Let $f=w|f|$ and $f_k=u_k|f_k|$ be polar decompositions in $\tilde A$. The series $f=\sum_kf_k$ converges in norm, so $|f|=w^*f=\sum_kw^*u_k|f_k|$, that is $|f|(x)=\sum_k|f_k|(xw^*u_k)$. By the Cauchy–Schwarz inequality for $|f_k|$, $\begin{gathered} ||f_k|(xw^*u_k)|^2\\ \le|f_k|(xx^*)\,|f_k|(u_k^*ww^*u_k)\\ \le|f_k|(xx^*)\,\|f_k\| . \end{gathered}$ So $||f|(x)|\le\sum_k\|f_k\|^{1/2}|f_k|(xx^*)^{1/2}$, and the Cauchy–Schwarz inequality for sequences gives (3.2). $\square$

For commutative algebras the absolute value is the total variation of a measure (Example 2.6), and then $|f+g|\le|f|+|g|$. This fails in general.

Example 3.5 (No triangle inequality). In $M_2(\mathbb C)$ let $\varphi(x)=\operatorname{Tr}(E_{11}x)$ and $\psi(x)=\operatorname{Tr}(E_{12}x)$, with matrix units $E_{ij}$. By Example 2.4, $|\varphi|=\operatorname{Tr}(E_{11}\,\cdot\,)$ and $|\psi|=\operatorname{Tr}(E_{22}\,\cdot\,)$, because $(E_{12}^*E_{12})^{1/2}=E_{22}$. For $\rho=E_{11}+E_{12}$ we have $\rho^*\rho=\begin{pmatrix}1&1\\1&1\end{pmatrix}$, so $|\rho|=2^{-1/2}\begin{pmatrix}1&1\\1&1\end{pmatrix}$ and $|\varphi+\psi|=\operatorname{Tr}(|\rho|\,\cdot\,)$. Let $P$ be the projection onto $2^{-1/2}(1,1)$. Then $|\varphi+\psi|(P)=\sqrt2$, while $(|\varphi|+|\psi|)(P)=\operatorname{Tr}(P)=1$. So $|\varphi+\psi|\le|\varphi|+|\psi|$ fails. Inequality (3.2) holds here with equality at $x=P$: $2\le(1+1)\cdot1$.

4. Domination by a positive functional

For a positive functional $\varphi$ and an element $a$, the functional $a\varphi$ need not be positive. But when $a\varphi$ is hermitian it is squeezed between $\pm r(a)\varphi$. The proof squares $a$ repeatedly, which is why the spectral radius, not the norm, appears.

Proposition 4.1. Let $B$ be a $C^*$-algebra, $\varphi\in B^*_+$ and $a\in B$, and suppose that $a\varphi$ is hermitian, that is $\varphi(ya)=\varphi(a^*y)$ for all $y\in B$. Then $\begin{gathered} |\varphi(ha)|\\ \le r(a)\,\varphi(h)\\ (h\in B_+), \end{gathered} \tag{4.1}$ that is, $-r(a)\varphi\le a\varphi\le r(a)\varphi$.

Proof. Step 1. By induction on $m$, $\varphi(ya^m)=\varphi((a^*)^my)$ for all $y$: $\begin{gathered} \varphi(ya^{m+1})\\ =\varphi((ya^m)a)\\ =\varphi(a^*ya^m)\\ =\varphi((a^*)^{m}a^*y). \end{gathered}$ Hence for $m\ge1$ and $c=a^{m}$, the functional $c\varphi$ is hermitian, and $\varphi(yc^2)=\varphi((ya^m)a^m)=\varphi((a^m)^*ya^m)$, so $c^2\varphi$ is positive.

Step 2. Let $h\in B_+$ and let $c$ be an element with $c\varphi$ hermitian. By the Cauchy–Schwarz inequality and $\varphi(c^*(hc))=\varphi((hc)c)$, $\begin{gathered} |\varphi(hc)|^2\\ =|\varphi(h^{1/2}\cdot h^{1/2}c)|^2\\ \le\varphi(h)\,\varphi(c^*hc)\\ =\varphi(h)\,\varphi(hc^2), \end{gathered}$ and $\varphi(hc^2)=\varphi(c^*hc)\ge0$.

Step 3. Apply Step 2 with $c=a,a^2,a^4,\dots$. By induction, $\begin{gathered} |\varphi(ha)|\\ \le\varphi(h)^{1-2^{-n}}\varphi(ha^{2^n})^{2^{-n}}\\ \le\varphi(h)^{1-2^{-n}}\big(\|\varphi\|\|h\|\big)^{2^{-n}}\|a^{2^n}\|^{2^{-n}} . \end{gathered}$ If $\varphi(h)=0$, Step 2 already gives $\varphi(ha)=0$. Otherwise let $n\to\infty$: the first factor tends to $\varphi(h)$, the second to $1$, and the third to $r(a)$ by the spectral radius formula. $\square$

Corollary 4.2. Let $M$ be a von Neumann algebra, $\varphi\in M_*^+$ and $a\in M$. Then $|a\varphi|\le\|as(\varphi)\|\,\varphi\le\|a\|\,\varphi . \tag{4.2}$ The same holds for $\varphi\in A^*_+$ and $a\in\tilde A$, for any $C^*$-algebra $A$.

Proof. Put $s=s(\varphi)$. Since $\varphi=s\varphi$, $a\varphi=(as)\varphi$. Let $a\varphi=u|a\varphi|$. Then $|a\varphi|=u^*a\varphi=(u^*as)\varphi$ is positive, hence hermitian, so Proposition 4.1 applies to $c=u^*as$: for $h\ge0$, $\begin{gathered} |a\varphi|(h)\\ =\varphi(hu^*as)\\ \le r(u^*as)\varphi(h)\\ \le\|as\|\varphi(h). \end{gathered}$ For a $C^*$-algebra, work in $\tilde A$. $\square$

Example 4.3. Let $M=B(H)$, $\xi$ a unit vector and $\varphi=\omega_\xi$.

Left multiplication: $a\omega_\xi=\omega_{a\xi,\xi}$, so $|a\omega_\xi|=\|a\xi\|\,\omega_\xi$ by (2.5). Since $s(\omega_\xi)=\theta_{\xi,\xi}$, $\|as(\omega_\xi)\|=\|a\xi\|$: the first inequality in (4.2) is an equality.
Right multiplication has no such bound. $\omega_\xi a=\omega_{\xi,a^*\xi}$, so $|\omega_\xi a|=\|a^*\xi\|^{-1}\omega_{a^*\xi}$ when $a^*\xi\ne0$. If $a^*\xi\notin\mathbb C\xi$, the support of $|\omega_\xi a|$ is not under $s(\omega_\xi)$, so $|\omega_\xi a|\le C\omega_\xi$ fails for every $C$.
The spectral radius matters. In $M_2(\mathbb C)$ let $a=E_{12}$, which has $r(a)=0$, and $\varphi=\operatorname{Tr}(\rho\,\cdot\,)$ with $\rho=\begin{pmatrix}p&q\\\bar q&t\end{pmatrix}\ge0$. Since $(a\varphi)(x)=\operatorname{Tr}(a\rho x)$, $a\varphi$ is hermitian exactly when $a\rho=\rho a^*$, that is $\begin{pmatrix}\bar q&t\\0&0\end{pmatrix}=\begin{pmatrix}q&0\\t&0\end{pmatrix}$. This forces $t=0$, hence $q=0$ by positivity, and then $a\rho=0$. So $a\varphi=0$, as (4.1) predicts.

5. Continuity of the absolute value

Theorem 5.1. Let $A$ be a $C^*$-algebra and $\varphi,\psi\in A^*$ (for instance $A=M$ and $\varphi,\psi\in M_*$). Then $\begin{gathered} \||\varphi|-|\psi|\|\\ \le\|\varphi-\psi\|+2\big(\|\varphi\|\,\|\varphi-\psi\|\big)^{1/2}. \end{gathered} \tag{5.1}$ The same bound holds for $\||\varphi^*|-|\psi^*|\|$. So $\varphi\mapsto|\varphi|$ is norm continuous, uniformly on bounded sets.

Proof. Work in $\tilde A$ (or in $M$), with $\varphi=v|\varphi|$ and $\psi=w|\psi|$. Since $|\psi|=w^*\psi$, $|\psi|(x)=\psi(xw^*)$. Also $\varphi(xw^*)=|\varphi|(xw^*v)$. So $\begin{gathered} |\varphi|(x)-|\psi|(x)\\ =|\varphi|\big(x(1-w^*v)\big)+(\varphi-\psi)(xw^*). \end{gathered}$ The second term has modulus at most $\|\varphi-\psi\|\|x\|$. For the first, the Cauchy–Schwarz inequality gives $\begin{gathered} \big||\varphi|(x(1-w^*v))\big|^2\\ \le|\varphi|(xx^*)\;|\varphi|\big((1-v^*w)(1-w^*v)\big). \end{gathered}$ Here $|\varphi|(xx^*)\le\|\varphi\|\|x\|^2$. Expanding, and using $|\varphi|(v^*ww^*v)\le|\varphi|(v^*v)=\|\varphi\|$ and $|\varphi|(w^*v)=\varphi(w^*)$, $\begin{gathered} |\varphi|\big((1-v^*w)(1-w^*v)\big)\\ \le2\|\varphi\|-2\operatorname{Re}\varphi(w^*). \end{gathered}$ Since $\psi(w^*)=|\psi|(w^*w)=\|\psi\|$, $\operatorname{Re}\varphi(w^*)\ge\|\psi\|-\|\varphi-\psi\|$. So the last quantity is at most $2(\|\varphi\|-\|\psi\|+\|\varphi-\psi\|)\le4\|\varphi-\psi\|$. Altogether $\begin{gathered} ||\varphi|(x)-|\psi|(x)|\\ \le\big(\|\varphi-\psi\|+2(\|\varphi\|\|\varphi-\psi\|)^{1/2}\big)\|x\|, \end{gathered}$ which is (5.1). Apply this to $\varphi^*,\psi^*$ for the second claim. $\square$

Remark 5.2. By symmetry $\|\varphi\|$ can be replaced by $\min(\|\varphi\|,\|\psi\|)$ in (5.1). In the commutative case of Example 2.6, the pointwise inequality $||h|-|k||\le|h-k|$ gives the Lipschitz bound $\||\varphi_h|-|\varphi_k|\|\le\|\varphi_h-\varphi_k\|$.

Weak$^*$ convergence alone does not move the absolute values along (Example 5.4), but convergence of the norms is enough.

Theorem 5.3. Let $A$ be a $C^*$-algebra and $(f_i)$ a net in $A^*$ with $f_i\to f$ in $\sigma(A^*,A)$ and $\|f_i\|\to\|f\|$. Then $|f_i|\to|f|$ in $\sigma(A^*,A)$. Likewise, if $(\varphi_i)$ is a net in the predual of a von Neumann algebra $M$ with $\varphi_i\to\varphi$ weakly and $\|\varphi_i\|\to\|\varphi\|$, then $|\varphi_i|\to|\varphi|$ weakly.

Proof. Eventually $\||f_i|\|=\|f_i\|\le\|f\|+1$, so the tail of the net $(|f_i|)$ lies in a weak$^*$ compact ball. It is therefore enough to show that every weak$^*$ cluster point $\omega$ of $(|f_i|)$ equals $|f|$. Let a subnet $|f_j|$ converge to $\omega$. Then $\omega$ is positive, and $\|\omega\|\le\liminf_j\||f_j|\|=\|f\|$, because the norm is weak$^*$ lower semicontinuous. By Theorem 2.7, $|f_j(x)|^2\le\|f_j\|\,|f_j|(xx^*)$ for $x\in A$, and in the limit $|f(x)|^2\le\|f\|\,\omega(xx^*)$. Theorem 3.1 gives $\omega=|f|$.

Example 5.4. On $A=C[0,1]$, $f_n=\delta_{1/n}-\delta_0$ tends to $0$ weak$^*$. By Example 3.3(1), $|f_n|=\delta_{1/n}+\delta_0$, which tends to $2\delta_0\ne|0|$. Here $\|f_n\|=2$ does not tend to $\|0\|$.

6. Invariant subspaces, hereditary cones and one-sided ideals

A subset $C\subseteq M_*^+$ is a hereditary cone if it is a convex cone (closed under sums and multiplication by nonnegative scalars) and $0\le\psi\le\omega\in C$ implies $\psi\in C$. The positive part of a closed invariant subspace is such a cone, and the cone determines the subspace.

Theorem 6.1. Let $M$ be a von Neumann algebra.

Let $V\subseteq M_*$ be a norm-closed subspace with $MV\subseteq V$, and $V=M_*e$ as in background fact 20. Then $\begin{gathered} V_+\\ =V\cap M_*^+\\ =\{\omega\in M_*^+:\omega(1-e)=0\} \end{gathered}$ is a norm-closed hereditary cone, and $V=MV_+=\{a\omega:a\in M,\ \omega\in V_+\}$.
Let $C\subseteq M_*^+$ be a norm-closed hereditary cone. Then $V=MC$ is a norm-closed subspace with $MV\subseteq V$ and $V\cap M_*^+=C$.
Hence $V\mapsto V\cap M_*^+$ is a bijection from the norm-closed left invariant subspaces of $M_*$ onto the norm-closed hereditary cones in $M_*^+$, with inverse $C\mapsto MC$. Both correspond to projections $e$: $V=M_*e$ and $C=\{\omega\ge0:\omega(1-e)=0\}$.

The right-handed statements (subspaces with $VM\subseteq V$, and $CM$) follow by applying $\varphi\mapsto\varphi^*$.

Proof. (1) For $\omega\ge0$: if $\omega=\omega e$, then $\omega(1-e)=\omega(e(1-e))=0$; conversely, if $\omega(1-e)=0$ then $s(\omega)\le e$ and $\omega(ex)=\omega(s(\omega)ex)=\omega(s(\omega)x)=\omega(x)$. This gives the formula for $V_+$, which is visibly a norm-closed hereditary cone. If $\varphi\in V$ has polar decomposition $\varphi=v|\varphi|$, then $|\varphi|=v^*\varphi\in V$, so $|\varphi|\in V_+$ and $\varphi\in MV_+$. The inclusion $MV_+\subseteq V$ is invariance.

(2) Step 1: $MC\cap M_*^+=C$. Clearly $C\subseteq MC$. If $\varphi=a\omega\ge0$ with $\omega\in C$, then $\varphi=|a\omega|\le\|a\|\omega$ by Corollary 4.2, so $\varphi\in C$.

Step 3: for $\chi\in M_*^+$, the norm closure of $M\chi$ is $M_*s(\chi)=\{\psi:\psi=\psi s(\chi)\}$. The set $M_*s(\chi)$ is norm closed and contains $M\chi$, since $(a\chi)(s(\chi)x)=\chi(s(\chi)xa)=\chi(xa)$. If some $\psi\in M_*s(\chi)$ were not in the closure of the subspace $M\chi$, the Hahn–Banach theorem would give $y\in M=(M_*)^*$ with $\chi(ya)=0$ for all $a$ and $\psi(y)\ne0$. With $a=y^*$, $\chi(yy^*)=0$, so $y^*s(\chi)=0$ (background fact 18) and $s(\chi)y=0$. Then $\psi(y)=\psi(s(\chi)y)=0$, a contradiction.

Step 4: $MC$ is a subspace. It is closed under scalar multiples. Let $a\omega,b\psi\in MC$ and $\chi=\omega+\psi\in C$. Since $\omega\le\chi$, $\omega(1-s(\chi))=0$ and $s(\omega)\le s(\chi)$; so $a\omega\in M_*s(\omega)\subseteq M_*s(\chi)$, and likewise $b\psi\in M_*s(\chi)$. By Step 3, $a\omega+b\psi$ lies in the closure of $M\chi\subseteq MC$, which is contained in $MC$ by Step 2.

Invariance $M(MC)\subseteq MC$ is clear.

(3) By (1), $V=M(V\cap M_*^+)$; by (2), $(MC)\cap M_*^+=C$. $\square$

Example 6.2 (Convexity is needed). In $M=\mathbb C^2$, $M_*^+$ is the quadrant $[0,\infty)^2$. The union of the two axes is a closed cone, and it is hereditary, but it is not convex. It is not the positive part of any subspace, because such a positive part is convex.

For a $C^*$-algebra $A$, a subset $V\subseteq A^*$ is left invariant if $aV\subseteq V$ for all $a\in A$, where $(af)(x)=f(xa)$.

Corollary 6.3. Let $A$ be a $C^*$-algebra and $V\subseteq A^*$ a norm-closed left invariant subspace, and put $V_+=V\cap A^*_+$. Then $V$ is invariant under $\tilde A$, $V_+$ is a norm-closed hereditary cone, $V=\tilde AV_+$, and $V$ is the norm closure of the set $AV_+=\{af:a\in A,\ f\in V_+\}$. In particular a nonzero $V$ contains a nonzero positive functional, and $V$ is determined by $V_+$.

Proof. By background fact 20 with the total set $A\subseteq\tilde A$, $V$ is invariant under $\tilde A$, and Theorem 6.1 applies to $\tilde A$ and $V\subseteq\tilde A_*=A^*$. Let $f\in V$ with $f=u|f|$; then $|f|=u^*f\in V_+$. The norm closure of $A|f|$ is a norm-closed subspace invariant under $A$, hence under $\tilde A$; so it contains $u|f|=f$. Thus $V\subseteq\overline{AV_+}\subseteq V$. If $f\ne0$, then $|f|\in V_+$ is nonzero. $\square$

To pass from norm-closed to weak$^*$ closed subspaces we need a closedness test on the unit ball.

Lemma 6.4 (Krein–Šmulian, subspace case). Let $X$ be a Banach space and $V\subseteq X^*$ a linear subspace whose intersection with the closed unit ball $B$ of $X^*$ is weak$^*$ closed. Then $V$ is weak$^*$ closed.

Proof. Let $\varphi_0\in X^*\setminus V$. We find $x\in X$ that is annihilated by $V$ but not by $\varphi_0$; by background fact 7 this proves the lemma.

Step 1. For $r>0$, $V\cap rB=r(V\cap B)$ is weak$^*$ closed. Hence $V$ is norm closed: a norm-convergent sequence in $V$ is bounded, so it lies in some $V\cap rB$, and norm limits are weak$^*$ limits. So $d=\operatorname{dist}(\varphi_0,V)>0$; replacing $\varphi_0$ by a multiple, we may assume $d>1$. Put $C=\varphi_0+V$, a convex set with $C\cap B=\emptyset$. For $r>0$, $C\cap rB=rB\cap\big(\varphi_0+V\cap(r+\|\varphi_0\|)B\big)$ is weak$^*$ closed.

Step 2. For a finite set $F\subseteq X$ write $F^\circ=\{\varphi:|\varphi(x)|\le1\ \forall x\in F\}$, a weak$^*$ closed set; $\emptyset^\circ=X^*$. We choose finite sets $F_1,F_2,\dots$ with $F_n\subseteq n^{-1}B_X$ (where $B_X$ is the unit ball of $X$) such that, with $G_n=F_1\cup\dots\cup F_n$ and $G_0=\emptyset$, $\begin{gathered} C\cap(n+1)B\cap G_n^\circ\\ =\emptyset\\ (n\\ \ge0). \end{gathered} \tag{6.1}$ For $n=0$ this is $C\cap B=\emptyset$. Suppose (6.1) holds for $n-1$ and fails for every choice of $F_n$. Then the weak$^*$ closed sets $C\cap(n+1)B\cap G_{n-1}^\circ\cap F^\circ$, for finite $F\subseteq n^{-1}B_X$, are nonempty. Any finitely many of them contain a common one (take the union of the $F$'s), and they lie in the weak$^*$ compact set $(n+1)B$. So they have a common point $\varphi$. Then $|\varphi(x)|\le1$ for all $x\in n^{-1}B_X$, that is $\|\varphi\|\le n$, and $\varphi\in C\cap nB\cap G_{n-1}^\circ$, which contradicts (6.1) for $n-1$.

Step 3. List the elements of $F_1,F_2,\dots$ in order as a sequence $(x_k)$; then $x_k\to0$, since each $F_n$ is finite and lies in $n^{-1}B_X$ (pad with zeros if the list is finite). Every $\varphi\in C$ lies in some $(n+1)B$, so by (6.1) it is not in $G_n^\circ$: $\sup_k|\varphi(x_k)|>1$.

Step 4. Let $T:X^*\to c_0$, $T\varphi=(\varphi(x_k))_k$. Then $T(C)$ is convex and misses the open unit ball $U$ of $c_0$. By background fact 3 there are a nonzero bounded functional $\Lambda$ on $c_0$ and a real $\gamma$ with $\operatorname{Re}\Lambda(u)<\gamma\le\operatorname{Re}\Lambda(y)$ for $u\in U$, $y\in T(C)$. Every bounded functional on $c_0$ is $\Lambda(y)=\sum_k\lambda_ky_k$ with $\sum_k|\lambda_k|=\|\Lambda\|$: put $\lambda_k=\Lambda(\delta_k)$; testing $\Lambda$ on $\sum_{k\le n}\overline{\operatorname{sgn}\lambda_k}\,\delta_k$ gives $\sum_{k\le n}|\lambda_k|\le\|\Lambda\|$, and $y=\lim_n\sum_{k\le n}y_k\delta_k$ in $c_0$. Normalize $\|\Lambda\|=1$; then $\sup_U\operatorname{Re}\Lambda=1$, so $\gamma\ge1$. Put $x=\sum_k\lambda_kx_k$, an absolutely convergent series in $X$. Then $\operatorname{Re}\varphi(x)=\operatorname{Re}\Lambda(T\varphi)\ge1$ for all $\varphi\in C$.

Step 5. For $\psi\in V$ and real $t$, $\varphi_0+t\psi\in C$, so $\operatorname{Re}\varphi_0(x)+t\operatorname{Re}\psi(x)\ge1$ for all $t$; hence $\operatorname{Re}\psi(x)=0$, and with $i\psi$ also $\operatorname{Im}\psi(x)=0$. So $V$ annihilates $x$, while $\operatorname{Re}\varphi_0(x)\ge1$. $\square$

Theorem 6.5. Let $A$ be a $C^*$-algebra.

If $V\subseteq A^*$ is a weak$^*$ closed left invariant subspace, then $V\cap A^*_+$ is a weak$^*$ closed hereditary cone.
If $C\subseteq A^*_+$ is a weak$^*$ closed hereditary cone, then $\tilde AC$ is a weak$^*$ closed left invariant subspace with $\tilde AC\cap A^*_+=C$.
The map $\mathfrak r\mapsto\mathfrak r^\perp\cap A^*_+$, where $\mathfrak r^\perp=\{f\in A^*:f(\mathfrak r)=0\}$, is a bijection from the closed right ideals of $A$ onto the weak$^*$ closed hereditary cones in $A^*_+$. Its inverse is $C\mapsto\{x\in A:\omega(xx^*)=0\ \forall\omega\in C\}$, and $\mathfrak r^\perp=\tilde A(\mathfrak r^\perp\cap A^*_+)$.

Proof. (1) The positive cone $A^*_+$ is weak$^*$ closed, and Corollary 6.3 applies to $V$, which is norm closed.

(2) $C$ is norm closed, so by Corollary 6.3 (through Theorem 6.1 for $\tilde A$) $V=\tilde AC$ is a left invariant subspace, closed in norm, with $V\cap A^*_+=C$, and $V=A^*e$ for a projection $e\in\tilde A$ with $s(\rho)\le e$ for every $\rho\in C$. By Lemma 6.4 it suffices to show that $V\cap B$ is weak$^*$ closed, $B$ being the unit ball of $A^*$. Let $(f_i)$ be a net in $V\cap B$ with weak$^*$ limit $f$. The functionals $|f_i|$ lie in $V\cap A^*_+=C$ and in $B$. Passing to a subnet, $|f_i|\to\rho$ weak$^*$, and $\rho\in C$ since $C$ is weak$^*$ closed. From $|f_i(x)|^2\le|f_i|(xx^*)$ we get $|f(x)|^2\le\rho(xx^*)$ for $x\in A$. The argument in the general case of Theorem 3.1 extends this inequality to $\tilde A$: $|f(X)|^2\le\rho(XX^*)$ for all $X\in\tilde A$. Let $f=w|f|$. Then $\begin{gathered} ||f|(X)|^2\\ =|f(Xw^*)|^2\\ \le\rho(Xw^*wX^*)\\ \le\rho(XX^*). \end{gathered}$ At $X=1-s(\rho)$ this gives $|f|(1-s(\rho))=0$, so $s(|f|)\le s(\rho)\le e$. Hence $|f|=|f|e\in V$, so $|f|\in C$ and $f=w|f|\in V$.

(3) Let $\mathfrak r$ be a closed right ideal. Then $\mathfrak r^\perp$ is weak$^*$ closed and left invariant, since $(af)(x)=f(xa)$ and $xa\in\mathfrak r$ for $x\in\mathfrak r$. So $C_{\mathfrak r}=\mathfrak r^\perp\cap A^*_+$ is a weak$^*$ closed hereditary cone, and $\mathfrak r^\perp=\tilde AC_{\mathfrak r}$ by Corollary 6.3. We recover $\mathfrak r$. If $x\in\mathfrak r$ then $xx^*\in\mathfrak r$, so $\omega(xx^*)=0$ for $\omega\in C_{\mathfrak r}$. Conversely, if $\omega(xx^*)=0$ for all $\omega\in C_{\mathfrak r}$, then for $a\in A$, $|(a\omega)(x)|^2=|\omega(xa)|^2\le\omega(xx^*)\omega(a^*a)=0$. So $x$ is annihilated by $AC_{\mathfrak r}$, hence by its norm closure $\mathfrak r^\perp$ (Corollary 6.3), and $x\in\mathfrak r$ by background fact 7.

Conversely, let $C$ be a weak$^*$ closed hereditary cone and $V=\tilde AC$, weak$^*$ closed by (2). Then $\mathfrak r=V_\perp$ is a closed right ideal: $f(xa)=(af)(x)=0$ for $x\in\mathfrak r$, $a\in A$, $f\in V$. By background fact 7, $\mathfrak r^\perp=V$, so $\mathfrak r^\perp\cap A^*_+=C$, and the formula for $\mathfrak r$ follows from the first part. $\square$

Applying the involution, closed left ideals $\mathfrak m$ correspond in the same way to weak$^*$ closed hereditary cones, through $\mathfrak m\mapsto\{\omega\ge0:\omega(\mathfrak m)=0\}$ and $C\mapsto\{x:\omega(x^*x)=0\ \forall\omega\in C\}$. Extreme points turn this into a statement about pure states. For a state $\omega$ let $N_\omega=\{x\in A:\omega(x^*x)=0\}$, its left kernel.

Corollary 6.6 (Left ideals and pure states). Let $A$ be a $C^*$-algebra.

Every closed left ideal $\mathfrak m$ is the intersection of the left kernels $N_\omega$ of the pure states $\omega$ with $\mathfrak m\subseteq N_\omega$. (For $\mathfrak m=A$ the family is empty and the intersection is $A$.)
For a pure state $\omega$, $N_\omega$ is a maximal closed left ideal. Every proper closed left ideal is contained in some $N_\omega$, and every maximal closed left ideal is some $N_\omega$.

The same holds for closed right ideals with right kernels $\{x:\omega(xx^*)=0\}$.

Proof. (1) Let $C=\{\omega\in A^*_+:\omega(\mathfrak m)=0\}$, so that $\mathfrak m=\{x:\omega(x^*x)=0\ \forall\omega\in C\}$ by the left-ideal form of Theorem 6.5(3). The set $K=\{\omega\in C:\|\omega\|\le1\}$ is convex and weak$^*$ compact. Let $\omega$ be a nonzero extreme point of $K$. Then $\|\omega\|=1$, for otherwise $\omega=\|\omega\|(\omega/\|\omega\|)+(1-\|\omega\|)\cdot0$. If $0\le\psi\le\omega$ and $\psi\ne0,\omega$, then $\psi$ and $\omega-\psi$ lie in $C$ (heredity), and $\begin{gathered} \omega\\ =\|\psi\|(\psi/\|\psi\|)+\|\omega-\psi\|\big((\omega-\psi)/\|\omega-\psi\|\big) \end{gathered}$ with $\|\psi\|+\|\omega-\psi\|=1$ (background fact 12); extremality forces $\psi=\|\psi\|\omega$. So $\omega$ is a pure state. By the Krein–Milman theorem, $K$ is the weak$^*$ closed convex hull of $0$ and the pure states in $C$. For $x\in A$, $\omega\mapsto\omega(x^*x)$ is affine and weak$^*$ continuous; so if it vanishes at every pure state in $C$, it vanishes on $K$, hence on $C$, and $x\in\mathfrak m$. Finally a pure state $\omega$ lies in $C$ exactly when $\mathfrak m\subseteq N_\omega$. If $\omega(\mathfrak m)=0$, then $\omega(y^*y)=0$ for $y\in\mathfrak m$, since $y^*y\in\mathfrak m$. Conversely, if $\omega(y^*y)=0$ for $y\in\mathfrak m$, then $|\omega(u_\lambda y)|^2\le\omega(u_\lambda^2)\omega(y^*y)=0$ and $\omega(y)=\lim_\lambda\omega(u_\lambda y)=0$.

(2) $N_\omega\ne A$, since $\omega\ne0$. By background fact 13, $a+N_\omega\mapsto\pi_\omega(a)\xi_\omega$ identifies $A/N_\omega$ isometrically with $H_\omega$, and it carries left multiplication by $b$ to $\pi_\omega(b)$. If $\mathfrak m\supsetneq N_\omega$ is a closed left ideal, its image is a closed, nonzero, $\pi_\omega(A)$-invariant subspace of $H_\omega$ (the image of a closed subspace containing the kernel of a quotient map is closed). As $\pi_\omega$ is irreducible, the image is $H_\omega$, and $\mathfrak m=A$. So $N_\omega$ is maximal. If $\mathfrak m$ is proper, the family in (1) is not empty, so $\mathfrak m\subseteq N_\omega$ for some pure $\omega$; if $\mathfrak m$ is maximal, $\mathfrak m=N_\omega$. The right-handed version follows by taking adjoints. $\square$

7. Phillips's lemma and Schur's theorem

For a set $\Gamma$, a bounded functional $\mu$ on $\ell^\infty(\Gamma)$ is a finitely additive set function $E\mapsto\mu(E)=\mu(1_E)$. The next lemma says that if such functionals tend to zero on every set, their parts on points tend to zero in $\ell^1$-norm. It is the tool behind Section 8.

Lemma 7.1 (Phillips's lemma). Let $(\mu_n)$ be a bounded sequence in $\ell^\infty(\Gamma)^*$ with $\mu_n(E)\to0$ for every $E\subseteq\Gamma$. Then $\sum_{\gamma\in\Gamma}|\mu_n(\{\gamma\})|\to0 .$

Proof. For $\mu\in\ell^\infty(\Gamma)^*$ and $E\subseteq\Gamma$ put $|\mu|(E)=\sup\{|\mu(x)|:\|x\|\le1,\ x=0\text{ off }E\}$. Three facts:

(a) If $E_1,\dots,E_m$ are disjoint, $\sum_j|\mu|(E_j)\le\|\mu\|$. Choose $x_j$ supported in $E_j$ with $\|x_j\|\le1$ and $\mu(x_j)\ge|\mu|(E_j)-\varepsilon$; then $x=\sum_jx_j$ has $\|x\|\le1$.
(b) $\sum_{\gamma\in E}|\mu(\{\gamma\})|\le|\mu|(E)$, by testing on $\sum_{\gamma\in F}\overline{\operatorname{sgn}\mu(\{\gamma\})}\,1_{\{\gamma\}}$ for finite $F\subseteq E$. In particular $a_\mu(\gamma)=\mu(\{\gamma\})$ defines $a_\mu\in\ell^1(\Gamma)$.
(c) For finite $F$ and any $x$, $\mu(x1_F)=\sum_{\gamma\in F}x(\gamma)\mu(\{\gamma\})$, by finite additivity.

Let $c=\sup_n\|\mu_n\|$ and $a_n=a_{\mu_n}$. Suppose the lemma fails. Passing to a subsequence, which keeps the hypotheses, we may assume $\|a_n\|_1\ge5\varepsilon$ for all $n$, for some $\varepsilon>0$. By hypothesis $a_n(\gamma)\to0$ for each $\gamma$, so $\sum_{\gamma\in D}|a_n(\gamma)|\to0$ for every finite $D$.

Humps. Choose indices $m_1<m_2<\cdots$ and disjoint finite sets $E_1,E_2,\dots$ as follows. Given $E_1,\dots,E_{k-1}$, with union $D$, choose $m_k>m_{k-1}$ with $\sum_{\gamma\in D}|a_{m_k}(\gamma)|<\varepsilon/2$; then $\sum_{\gamma\notin D}|a_{m_k}(\gamma)|>4\varepsilon$, so there is a finite $E_k$ disjoint from $D$ with $\sum_{\gamma\in E_k}|a_{m_k}(\gamma)|\ge4\varepsilon$.

Thinning. Fix an integer $p>c/\varepsilon$. If $L$ is an infinite set of indices and $\mu$ a functional, split $L$ into $p$ disjoint infinite sets $L_1,\dots,L_p$. The sets $\bigcup_{l\in L_r}E_l$ are disjoint, so by (a) some $r$ has $|\mu|(\bigcup_{l\in L_r}E_l)\le c/p<\varepsilon$. Starting from $L_0=\mathbb N$, define $k_j=\min L_{j-1}$ and apply this to $\mu=\mu_{m_{k_j}}$ and $L=L_{j-1}\setminus\{k_j\}$, to get an infinite $L_j\subseteq L_{j-1}\setminus\{k_j\}$ with $|\mu_{m_{k_j}}|\big(\bigcup_{l\in L_j}E_l\big)<\varepsilon$. Then $k_1<k_2<\cdots$, and $k_i\in L_j$ for $i>j$.

Test function. Let $\nu_j=\mu_{m_{k_j}}$ and define $x\in\ell^\infty(\Gamma)$ by $x(\gamma)=\overline{\operatorname{sgn}a_{m_{k_i}}(\gamma)}$ for $\gamma\in E_{k_i}$, $i\ge1$, and $x=0$ elsewhere; $\|x\|\le1$. Fix $j$ and split $x=x1_{P}+x1_{E_{k_j}}+x1_{T}$, where $P=\bigcup_{i<j}E_{k_i}$ and $T=\bigcup_{i>j}E_{k_i}$. By (c) and the choice of $m_{k_j}$ (the set $P$ lies in the $D$ used at that step), $|\nu_j(x1_P)|<\varepsilon/2$. By (c), $\nu_j(x1_{E_{k_j}})=\sum_{\gamma\in E_{k_j}}|a_{m_{k_j}}(\gamma)|\ge4\varepsilon$. And $T\subseteq\bigcup_{l\in L_j}E_l$, so $|\nu_j(x1_T)|\le|\nu_j|(T)<\varepsilon$. Hence $|\nu_j(x)|>2\varepsilon$ for every $j$.

Contradiction. The function $x$ is a uniform limit of finite linear combinations of indicator functions (cut the unit disc into finitely many small pieces). Each $\nu_j(1_E)\to0$ and $\sup_j\|\nu_j\|\le c$, so $\nu_j(x)\to0$. $\square$

Corollary 7.2 (Schur's theorem). In $\ell^1(\Gamma)$ every weakly convergent sequence converges in norm.

Proof. Let $g_n\to g$ weakly; replacing $g_n$ by $g_n-g$, let $g=0$. The sequence is bounded (background fact 2). Put $\mu_n(x)=\sum_\gamma x(\gamma)g_n(\gamma)$ on $\ell^\infty(\Gamma)$. For $E\subseteq\Gamma$, $g\mapsto\sum_{\gamma\in E}g(\gamma)$ is a bounded functional on $\ell^1(\Gamma)$, so $\mu_n(E)\to0$. Lemma 7.1 gives $\|g_n\|_1=\sum_\gamma|\mu_n(\{\gamma\})|\to0$. $\square$

So in $\ell^1$ weak and norm convergence of sequences agree. Section 12 shows how much of this survives in $B(H)_*$ and other atomic preduals.

8. Normal and singular parts along weak$^*$ convergent sequences

For the rest of the lesson we attach to each functional one positive functional that controls it.

Lemma 8.1. For $\varphi\in M^*$ put $\varphi^\sharp=|\varphi|+|\varphi^*|$, the absolute values being taken in the $C^*$-algebra $M$ (Theorem 2.7).

$\varphi^\sharp\ge0$, $\|\varphi^\sharp\|=2\|\varphi\|$, and $|\varphi(x)|^2\le\|\varphi\|\varphi^\sharp(xx^*)$, $|\varphi(x)|^2\le\|\varphi\|\varphi^\sharp(x^*x)$.
If $\varphi$ is normal, so is $\varphi^\sharp$, and $\varphi(x)=\varphi(gxg)$ with $g=s(\varphi^\sharp)$. If $\varphi$ is singular, so is $\varphi^\sharp$.
If $p$ is a projection with $\varphi^\sharp(p)=0$, then $\varphi(p)=0$.

Proof. (1) follows from (2.4) and Theorem 2.7, the second inequality by applying the first to $\varphi^*$ and $x^*$. (2) If $\varphi$ is normal, $|\varphi|$ and $|\varphi^*|$ are normal by Corollary 3.2. Their supports are $s_r(\varphi)$ and $s_l(\varphi)$ (Theorem 2.2), both under $g$, so $\begin{gathered} \varphi(gxg)\\ =\varphi(s_r(\varphi)\,gxg\,s_l(\varphi))\\ =\varphi(s_r(\varphi)\,x\,s_l(\varphi))\\ =\varphi(x) \end{gathered}$ by Lemma 1.4(2). If $\varphi$ is singular, then $\varphi\in M^*(1-z_0)$ (background fact 22), and so is $\varphi^*$ since $z_0$ is central. The space $M^*(1-z_0)$ is invariant under $M^{**}$, so $|\varphi|=v^*\varphi$ and $|\varphi^*|$ lie in it. (3) By (1), $|\varphi(p)|^2\le\|\varphi\|\varphi^\sharp(p)=0$. $\square$

Theorem 8.2. Let $(\varphi_k)$ be a sequence in $M^*$ that converges to $\varphi$ in $\sigma(M^*,M)$. Then $\varphi_k^{\rm n}\to\varphi^{\rm n}$ and $\varphi_k^{\rm s}\to\varphi^{\rm s}$ in $\sigma(M^*,M)$.

Proof. The splitting is linear, so we may assume $\varphi=0$. By background fact 2, $c=\sup_k\|\varphi_k\|<\infty$, and $\|\varphi_k^{\rm n}\|,\|\varphi_k^{\rm s}\|\le c$. Since $\varphi_k^{\rm s}=\varphi_k-\varphi_k^{\rm n}$, it suffices to show $\varphi_k^{\rm n}\to0$. The sequence is bounded and the span of the projections is norm dense in $M$ (background fact 16), so it suffices to show $\varphi_k^{\rm n}(p)\to0$ for each projection $p\ne0$.

Let $\omega=\sum_k2^{-k}(\varphi_k^{\rm s})^\sharp$, a norm-convergent series of singular positive functionals (Lemma 8.1), so $\omega$ is singular and positive. By Zorn's lemma choose a maximal family $(p_i)_{i\in I}$ of mutually orthogonal nonzero projections under $p$ with $\omega(p_i)=0$. Then $\sum_ip_i=p$: otherwise $p-\sum_ip_i$ majorizes a nonzero projection on which $\omega$ vanishes (background fact 23), against maximality. For each $k$ and $i$, $(\varphi_k^{\rm s})^\sharp(p_i)\le2^k\omega(p_i)=0$, so $\varphi_k^{\rm s}(p_i)=0$ (Lemma 8.1(3)) and $\varphi_k(p_i)=\varphi_k^{\rm n}(p_i)$.

The map $\rho:\ell^\infty(I)\to M$, $\rho(x)=\sum_ix(i)p_i$ (a strong sum), is a contractive $*$-homomorphism. Put $\mu_k=\varphi_k\circ\rho$. Then $\|\mu_k\|\le c$, and $\mu_k(E)=\varphi_k(\sum_{i\in E}p_i)\to0$ for every $E\subseteq I$. By Phillips's lemma, $\sum_i|\varphi_k^{\rm n}(p_i)|=\sum_i|\mu_k(\{i\})|\to0 .$ As $\varphi_k^{\rm n}$ is normal, it is completely additive, so $|\varphi_k^{\rm n}(p)|=|\sum_i\varphi_k^{\rm n}(p_i)|\le\sum_i|\varphi_k^{\rm n}(p_i)|\to0$. $\square$

Corollary 8.3 (Weak sequential completeness).

The predual $M_*$ of a von Neumann algebra is weakly sequentially complete: every weakly Cauchy sequence in $M_*$ converges weakly to an element of $M_*$.
The dual $A^*$ of a $C^*$-algebra is weakly sequentially complete.

Proof. (1) If $(\varphi_k)$ is weakly Cauchy, $\varphi_k(x)$ converges for every $x\in M=(M_*)^*$. By uniform boundedness the limit $\varphi(x)$ defines $\varphi\in M^*$, and $\varphi_k\to\varphi$ in $\sigma(M^*,M)$. By Theorem 8.2, $\varphi^{\rm s}=\lim\varphi_k^{\rm s}=0$, so $\varphi\in M_*$. (2) $A^*=\tilde A_*$ and $(\tilde A_*)^*=\tilde A$. $\square$

Example 8.4 (Sequences cannot be replaced by nets). Let $M$ be infinite-dimensional. It has a non-normal state (background fact 23), whose singular part, normalized, is a singular state $\chi$. The normal states are weak$^*$ dense in the state space. Indeed, otherwise the separation theorem in $(M^*,\sigma(M^*,M))$, whose dual is $M$, gives a self-adjoint $k\in M$ and a state $\chi'$ with $\chi'(k)>\sup\{\omega(k):\omega\text{ a normal state}\}$. The right side is at least $\sup_{\|\zeta\|=1}\langle k\zeta,\zeta\rangle=\max\sigma(k)$, computed in a faithful normal representation; and $\chi'(k)\le\max\sigma(k)$ because $k\le\max\sigma(k)1$. This is a contradiction. So there is a net of normal states $\omega_i\to\chi$. Their normal parts are $\omega_i$, which tend to $\chi$, not to $\chi^{\rm n}=0$. In particular the net $(\omega_i)$ is weakly Cauchy in $M_*$ without a weak limit in $M_*$.

The singular functionals form a norm-closed subspace. Its weak$^*$ closure can be large, but countable subsets cannot escape from it.

Proposition 8.5. The $\sigma(M^*,M)$-closure of any countable set of singular functionals consists of singular functionals.

Proof. If the countable set contains zero, handle that singular point separately; a finite union with the closed singleton $\{0\}$ has the corresponding union of closures. An empty set needs no argument. Enumerate the remaining nonzero elements as $\{\varphi_k\}\subseteq M_*^\perp$, repeating elements if the nonempty set is finite, and put $\omega=\sum_k2^{-k}\varphi_k^\sharp/\|\varphi_k\|$, a singular positive functional. If $\omega(q)=0$ for a projection $q$, then $\varphi_k(q)=0$ for all $k$ (Lemma 8.1(3)). Let $\varphi$ lie in the weak$^*$ closure of $\{\varphi_k\}$, and let $p$ be a projection. The functional $\omega+(\varphi^{\rm s})^\sharp$ is singular and positive. As in the proof of Theorem 8.2, take a maximal orthogonal family $(p_i)$ of nonzero projections under $p$ on which it vanishes; then $\sum_ip_i=p$. For finite $F$ put $p_F=\sum_{i\in F}p_i$. Then $\omega(p_F)=0$, so $\varphi_k(p_F)=0$ for all $k$; since $\psi\mapsto\psi(p_F)$ is weak$^*$ continuous and vanishes on the set $\{\varphi_k\}$, $\varphi(p_F)=0$. Also $\varphi^{\rm s}(p_F)=0$. So $\varphi^{\rm n}(p_F)=0$, and normality gives $\varphi^{\rm n}(p)=\lim_F\varphi^{\rm n}(p_F)=0$. Thus $\varphi^{\rm n}$ vanishes on all projections, so $\varphi^{\rm n}=0$ and $\varphi$ is singular. $\square$

Example 8.6 (Countability is needed). Let $M=L^\infty[0,1]$ with Lebesgue measure. We show that the singular functionals are weak$^*$ dense in $M^*$. By background fact 7 it suffices to show that no $x\ne0$ in $M$ is annihilated by all singular functionals.

First, for every measurable $E$ of positive measure there is a singular state $\psi$ of $M$ with $\psi(1_E)=1$. The algebra $N=1_EM\cong L^\infty(E)$ is infinite-dimensional: $t\mapsto\mu(E\cap[0,t])$ is continuous because its increments have absolute value at most the length of the interval. Bisect its positive total mass repeatedly by the intermediate value property, obtaining infinitely many disjoint subsets of positive measure and hence independent indicator functions; so it has a singular state $\psi_1$ (as in Example 8.4). Put $\psi(y)=\psi_1(y1_E)$. This is a state of $M$ with $\psi(1_E)=1$. It is singular by background fact 23: if $q\ne0$ is a projection of $M$ and $q1_E=0$, then $\psi(q)=0$; otherwise $q1_E$ is a nonzero projection of $N$, which majorizes a nonzero projection $q_0$ of $N$ with $\psi_1(q_0)=0$, and $q_0\le q$, $\psi(q_0)=0$.

Now let $x\ne0$. Choose $\varepsilon>0$ such that $\{|x|\ge\varepsilon\}$ has positive measure. Cover the bounded annulus $\{\varepsilon\le|z|\le\|x\|_\infty\}$ by finitely many disks of radius $\varepsilon/4$, with centres $c$ in the annulus. At least one preimage of a disk meets that set in positive measure. For its nonzero centre $c$, $E=\{t:|x(t)-c|\le|c|/2\}$ therefore has positive measure. With $\psi$ as above, $\psi(x)=\psi(x1_E)$ (Cauchy–Schwarz, as $\psi(1-1_E)=0$), and $|\psi(x1_E)-c|=|\psi((x-c)1_E)|\le|c|/2$. So $\psi(x)\ne0$. Hence Lebesgue measure, a normal state, is a weak$^*$ limit of a net of singular functionals, and Proposition 8.5 fails for uncountable sets.

In atomic algebras, such as $\ell^\infty$ and $B(H)$, the singular functionals do form a weak$^*$ closed subspace.

Proposition 8.7. Let $M$ be atomic, that is, every nonzero projection majorizes a minimal projection, and let $I_0$ be the norm-closed two-sided ideal generated by the minimal projections. Then $\varphi\in M^*$ is singular exactly when $\varphi(I_0)=0$. So the singular functionals form a weak$^*$ closed subspace, and Proposition 8.5 holds for every set of singular functionals. For $M=\ell^\infty$, $I_0=c_0$; for $M=B(H)$, $I_0$ is the algebra of compact operators.

Proof. Let $\psi\in M^*$ be singular and positive, and $f$ a minimal projection. The only nonzero projection under $f$ is $f$, so $\psi(f)=0$ by background fact 23. Let $x\in M$. Since $fx^*xf\in fMf=\mathbb Cf$, $fx^*xf=\mu f$ with $\mu=\|xf\|^2$. If $\mu>0$, $w=\mu^{-1/2}xf$ is a partial isometry with $w^*w=f$, and $f'=ww^*$ is minimal, since $f'Mf'=wfMfw^*=\mathbb Cf'$; so $\psi(xfx^*)=\mu\psi(f')=0$. This also holds if $\mu=0$. For $y\in M$, $(xfy)(xfy)^*\le\|y\|^2xfx^*$, and the Cauchy–Schwarz inequality gives $|\psi(xfy)|^2\le\psi(1)\,\psi((xfy)(xfy)^*)=0$. So $\psi$ vanishes on $I_0$. Singular functionals are linear combinations of positive singular ones (background fact 22), so they all vanish on $I_0$.

Conversely, let $\varphi(I_0)=0$. Then $\varphi^{\rm s}(I_0)=0$ by the first part, so $\varphi^{\rm n}(I_0)=0$. Write $1=\sum_jf_j$ with mutually orthogonal minimal projections (background fact 26). The finite partial sums $e_F=\sum_{j\in F}f_j$ lie in $I_0$ and increase to $1$, so for $x\in M$, $e_Fxe_F\in I_0$ and $e_Fxe_F\to x$ $\sigma$-weakly (background facts 15 and 16). Hence $\varphi^{\rm n}(x)=\lim_F\varphi^{\rm n}(e_Fxe_F)=0$, and $\varphi$ is singular. So the singular functionals are the annihilator of $I_0$, which is weak$^*$ closed.

In $\ell^\infty$ the minimal projections are the coordinate projections, and they generate $c_0$. In $B(H)$ they are the rank-one projections, and they generate the closure of the finite-rank operators, which is the algebra of compact operators. $\square$

9. The unit ball as a complete metric space

When $M$ has a faithful normal state, the $\sigma$-strong topology on the unit ball comes from a complete metric. This makes the Baire category theorem available in Section 10. The converse holds as well.

Proposition 9.1. Let $M$ be a von Neumann algebra.

Let $\omega$ be a faithful positive normal functional. Then $\begin{gathered} d(x,y)\\ =\omega\big((x-y)^*(x-y)\big)^{1/2},\\ d^\#(x,y)\\ =\omega\big((x-y)^*(x-y)\\ +(x-y)(x-y)^*\big)^{1/2} \end{gathered}$ are metrics on $M_1$ that define its $\sigma$-strong and $\sigma$-strong$^*$ topologies, and $M_1$ is complete for both.
Conversely, if the $\sigma$-strong topology, or the $\sigma$-strong$^*$ topology, on $M_1$ is metrizable, then $M$ has a faithful positive normal functional.

Proof. (1) Let $(\pi,H,\xi)$ be the cyclic representation of $\omega$. It is faithful, since $\pi(x)=0$ gives $\omega(x^*x)=0$, and normal (background fact 24). So $\pi(M)$ is a von Neumann algebra, and by background fact 15 we may assume $M\subseteq B(H)$, $\omega=\omega_\xi$ and $[M\xi]=H$. The vector $\xi$ is separating: $x\xi=0$ gives $\omega(x^*x)=0$, so $x=0$. Hence $\xi$ is cyclic for $M'$: the projection $p$ onto $[M'\xi]$ lies in $M''=M$, and $\omega(1-p)=\|(1-p)\xi\|^2=0$ forces $p=1$. Now $d(x,y)=\|(x-y)\xi\|$ is a metric on $M$, as $\xi$ is separating.

If $x_i\to x$ $\sigma$-strongly then $d(x_i,x)\to0$, since $d(\cdot,x)$ is one of the defining seminorms. Conversely let $x_i,x\in M_1$ with $d(x_i,x)\to0$. For $a'\in M'$, $\begin{gathered} \|(x_i-x)a'\xi\|\\ =\|a'(x_i-x)\xi\|\\ \le\|a'\|\,d(x_i,x)\to0. \end{gathered}$ The vectors $a'\xi$ are dense and $\|x_i-x\|\le2$, so $x_i\to x$ strongly, hence $\sigma$-strongly (background fact 15). The same argument applied to $x_i$ and $x_i^*$ shows that $d^\#$ defines the $\sigma$-strong$^*$ topology on $M_1$.

Completeness: $x\mapsto x\xi$ is an isometry of $(M_1,d)$ onto $M_1\xi\subseteq H$. The set $M_1$ is $\sigma$-weakly compact and $x\mapsto x\xi$ is continuous from the $\sigma$-weak topology to the weak topology of $H$. So $M_1\xi$ is weakly compact and convex, hence norm closed, hence complete. For $d^\#$ use $x\mapsto(x\xi,x^*\xi)\in H\oplus H$, noting that the involution is $\sigma$-weakly continuous.

(2) If $M=0$, the zero functional is faithful and the assertion holds. Otherwise suppose the $\sigma$-strong topology on $M_1$ has a countable base $B_1,B_2,\dots$ of neighbourhoods of $0$. Each $B_n$ contains a basic set $\{x\in M_1:\sum_{j\le m_n}\omega_{n,j}(x^*x)<\delta_n\}$ with $\omega_{n,j}\in M_*^+$, which we may take nonzero. Enumerate all $\omega_{n,j}$ as $\omega_1,\omega_2,\dots$ and put $\omega=\sum_l2^{-l}\omega_l/\|\omega_l\|$. If $p$ is a projection with $\omega(p)=0$, then $\omega_l(p^*p)=0$ for all $l$, so $p$ lies in every $B_n$. The topology is Hausdorff, so $\bigcap_nB_n=\{0\}$ and $p=0$. A positive normal functional that vanishes on no nonzero projection is faithful: if $\omega(y)=0$ with $y\ge0$, the spectral projections $q=1_{[\varepsilon,\infty)}(y)\le\varepsilon^{-1}y$ satisfy $\omega(q)=0$, so $q=0$ for every $\varepsilon>0$ and $y=0$. For the $\sigma$-strong$^*$ topology use the seminorms $\omega(x^*x+xx^*)^{1/2}$ in the same way. $\square$

10. Weak compactness in the predual

We now characterize the relatively weakly compact subsets of a predual. For $M=L^\infty(\Gamma,\mu)$, whose predual is $L^1(\Gamma,\mu)$ (Example 2.6), condition (4) of the theorem below, tested on indicator functions $a=1_E$, says that there is one $g\in L^1_+$ such that the integrals $\int_Eh\,d\mu$, $h\in K$, are uniformly small whenever $\int_Eg\,d\mu$ is small. In general the control is by one positive normal functional, applied to $a^*a+aa^*$.

We use the following Banach-space fact twice, here and in Section 14.

Lemma 10.1. A bounded subset $K$ of a Banach space $X$ is relatively weakly compact exactly when its $\sigma(X^{**},X^*)$-closure in $X^{**}$ lies in $X$.

Proof. The weak topology of $X$ is the restriction of $\sigma(X^{**},X^*)$. If $K$ is relatively weakly compact, its weak closure $K'$ in $X$ is weakly compact, hence $\sigma(X^{**},X^*)$-compact and closed in $X^{**}$; so the closure of $K$ in $X^{**}$ lies in $K'\subseteq X$. Conversely, if the closure $\bar K$ of $K$ in $X^{**}$ lies in $X$, it is $\sigma(X^{**},X^*)$-compact by the Banach–Alaoglu theorem, since $K$ is bounded; so $\bar K$ is a weakly compact subset of $X$ containing $K$. $\square$

For $X=M_*$ we have $X^*=M$ and $X^{**}=M^*$, so a bounded $K\subseteq M_*$ is relatively weakly compact exactly when its $\sigma(M^*,M)$-closure consists of normal functionals. Relatively weakly compact sets are bounded (background fact 2). For a projection $p$ we write $(1-p)\varphi(1-p)$ for $x\mapsto\varphi((1-p)x(1-p))$, as in (0.1).

Theorem 10.2 (Weak compactness in $M_*$). For a subset $K$ of the predual of a von Neumann algebra $M$, the following are equivalent.

$K$ is relatively weakly compact.
For every abelian von Neumann subalgebra $\mathcal A\subseteq M$, the set $K|_{\mathcal A}=\{\varphi|_{\mathcal A}:\varphi\in K\}$ is relatively weakly compact in $\mathcal A_*$.
$K$ is bounded, and $\varphi(p_n)\to0$ uniformly in $\varphi\in K$ whenever the projections $p_n$ decrease to $0$.
$K$ is bounded, and there is $\omega\in M_*^+$ with this property: for every $\varepsilon>0$ there is $\delta>0$ such that $|\varphi(a)|<\varepsilon$ for all $\varphi\in K$ whenever $a\in M_1$ and $\omega(a^*a+aa^*)<\delta$.
$K$ is bounded, and for every increasing net of projections $(p_i)$, $\varphi(p_i)$ converges uniformly in $\varphi\in K$.
$K$ is bounded, and $\|(1-p_i)\varphi(1-p_i)\|\to0$ uniformly in $\varphi\in K$ for every increasing net of projections $(p_i)$ with $\sup_ip_i=1$.
$K$ is bounded, and $\varphi(q_n)\to0$ uniformly in $\varphi\in K$ for every sequence $(q_n)$ of mutually orthogonal projections.

We prove the easy implications first; the implication (1)$\Rightarrow$(4) needs two lemmas.

Proof of (1)$\Rightarrow$(2). Restriction $\varphi\mapsto\varphi|_{\mathcal A}$ is continuous from $\sigma(M_*,M)$ to $\sigma(\mathcal A_*,\mathcal A)$, and continuous images of compact sets are compact. $\square$

Proof of (2)$\Rightarrow$(1). Every self-adjoint $h\in M$ lies in the abelian von Neumann algebra generated by $h$ and $1$. So $\{\varphi(h):\varphi\in K\}$ is bounded, hence so is $\{\varphi(x):\varphi\in K\}$ for every $x=h+ik$, and $K$ is bounded by uniform boundedness. Let $\varphi$ lie in the $\sigma(M^*,M)$-closure of $K$. For an abelian von Neumann subalgebra $\mathcal A$, $\varphi|_{\mathcal A}$ lies in the $\sigma(\mathcal A^*,\mathcal A)$-closure of $K|_{\mathcal A}$, which consists of normal functionals by Lemma 10.1. So $\varphi$ is normal on every abelian von Neumann subalgebra, hence normal (background fact 23). By Lemma 10.1, $K$ is relatively weakly compact. $\square$

Proof of (4)$\Rightarrow$(6)$\Rightarrow$(5)$\Rightarrow$(3). (4)$\Rightarrow$(6): let $\varepsilon,\delta$ be as in (4) and $p_i\uparrow1$. By normality of $\omega$ there is $i_0$ with $\omega(1-p_i)<\delta/2$ for $i\ge i_0$. If $a\in M_1$ and $a=(1-p_i)a(1-p_i)$, then $a^*a\le1-p_i$ and $aa^*\le1-p_i$, so $\omega(a^*a+aa^*)<\delta$ and $|\varphi(a)|<\varepsilon$. Hence $\begin{gathered} \|(1-p_i)\varphi(1-p_i)\|\\ =\sup_{a\in M_1}|\varphi((1-p_i)a(1-p_i))|\\ \le\varepsilon \end{gathered}$ for $i\ge i_0$ and all $\varphi\in K$.

(6)$\Rightarrow$(5): let $(p_i)$ increase to $p$. Then $q_i=p_i+(1-p)$ increases to $1$, and $1-q_i=p-p_i$. By (6), $\begin{gathered} |\varphi(p)-\varphi(p_i)|\\ =|\varphi(p-p_i)|\\ \le\|(p-p_i)\varphi(p-p_i)\|\to0 \end{gathered}$ uniformly in $\varphi\in K$.

(5)$\Rightarrow$(3): if $p_n\downarrow0$, then $1-p_n\uparrow1$. By (5), $\varphi(1-p_n)$ converges uniformly, and its pointwise limit is $\varphi(1)$ by normality. So $\varphi(p_n)\to0$ uniformly. $\square$

Proof of (3)$\Leftrightarrow$(7). (3)$\Rightarrow$(7): for orthogonal $(q_n)$, $r_m=\sum_{n\ge m}q_n$ decreases to $0$, and $\varphi(q_m)=\varphi(r_m)-\varphi(r_{m+1})$. (7)$\Rightarrow$(3): suppose $p_n\downarrow0$ but $\sup_{\varphi\in K}|\varphi(p_n)|\not\to0$. Then there are $\varepsilon>0$, indices $n_1<n_2<\cdots$ and $\varphi_j\in K$ with $|\varphi_j(p_{n_j})|\ge\varepsilon$. Since each $\varphi_j$ is normal, $\varphi_j(p_n)\to0$ as $n\to\infty$; choosing the indices and functionals inductively, we may assume $|\varphi_j(p_{n_{j+1}})|<\varepsilon/2$. The projections $q_j=p_{n_j}-p_{n_{j+1}}$ are mutually orthogonal and $|\varphi_j(q_j)|>\varepsilon/2$, which contradicts (7). $\square$

Proof of (4)$\Rightarrow$(3). If $p_n\downarrow0$, then $\omega(p_n^*p_n+p_np_n^*)=2\omega(p_n)\to0$. $\square$

Proof of (3)$\Rightarrow$(1). By the Eberlein–Šmulian theorem we only need a weak cluster point in $M_*$ for each sequence $(\varphi_n)$ in $K$. Let $\varphi\in M^*$ be a $\sigma(M^*,M)$-cluster point, which exists by the Banach–Alaoglu theorem. On $M_*$, $\sigma(M^*,M)$ is the weak topology, so it suffices to show that $\varphi$ is normal. Put $\omega=\sum_n2^{-n}\varphi_n^\sharp\in M_*^+$.

(a) If $q$ is a projection with $\omega(q)=0$, then $\varphi_n(q)=0$ for all $n$ (Lemma 8.1(3)), hence $\varphi(q)=0$.

(b) If projections $r_m$ decrease to $0$, then $|\varphi(r_m)|\le\sup_n|\varphi_n(r_m)|\to0$ by (3).

Let $(p_i)_{i\in I}$ be mutually orthogonal with sum $p$. Since $\omega$ is normal, $\sum_i\omega(p_i)=\omega(p)<\infty$, so $I_0=\{i:\omega(p_i)>0\}$ is countable. Every projection $q$ under $\sum_{i\notin I_0}p_i$ has $\omega(q)=0$, hence $\varphi(q)=0$ by (a). For $J\subseteq I_0$, enumerate $J=\{j_1,j_2,\dots\}$; the projections $r_m=\sum_{k\ge m}p_{j_k}$ decrease to $0$, so by (b), $\varphi(\sum_{i\in J}p_i)=\lim_m\sum_{k<m}\varphi(p_{j_k})$. Applying this to $J=\{i\in I_0:\operatorname{Re}\varphi(p_i)\ge0\}$ and to the three analogous sets shows that $\sum_{i\in I_0}|\varphi(p_i)|<\infty$. Now let $F\subseteq I$ be finite. Then $p-\sum_{i\in F}p_i$ is the sum of a projection under $\sum_{i\notin I_0}p_i$ and of $\sum_{i\in I_0\setminus F}p_i$, so $\begin{gathered} \Big|\varphi(p)-\sum_{i\in F}\varphi(p_i)\Big|\\ =\Big|\sum_{i\in I_0\setminus F}\varphi(p_i)\Big|\\ \le\sum_{i\in I_0\setminus F}|\varphi(p_i)|, \end{gathered}$ which tends to $0$ along the finite sets $F$. So $\varphi$ is completely additive, hence normal (background fact 23). $\square$

The remaining implication rests on the following two lemmas.

Lemma 10.3. Let $(\varphi_k)$ be a sequence in $M_*$ converging weakly to $\varphi_0\in M_*$, and let $(a_n)$ be a sequence in $M_1$ converging $\sigma$-strongly$^*$ to $0$. Then $\sup_k|\varphi_k(a_n)|\to0$ as $n\to\infty$.

Proof. Reduction. Let $c=\sup_k\|\varphi_k\|<\infty$. Put $\omega=\varphi_0^\sharp+\sum_k2^{-k}\varphi_k^\sharp\in M_*^+$ and $e=s(\omega)$. For every $k\ge0$, $s(\varphi_k^\sharp)\le e$, so $\varphi_k(x)=\varphi_k(exe)$ by Lemma 8.1(2). The restriction of $\omega$ to the von Neumann algebra $N=eMe$ is faithful; the restrictions $\varphi_k|_N$ converge weakly to $\varphi_0|_N$; the elements $ea_ne\in N_1$ converge $\sigma$-strongly$^*$ to $0$; and $\varphi_k(a_n)=\varphi_k(ea_ne)$. So it is enough to treat a faithful $\omega$. Let $d$ be the metric of Proposition 9.1 for $\omega$.

Baire. Put $\psi_k=\varphi_k-\varphi_0$, so $\psi_k\to0$ weakly and $\|\psi_k\|\le2c$. Fix $\varepsilon>0$ and let $S_m=\{a\in M_1:|\psi_k(a)|\le\varepsilon\text{ for all }k\ge m\}$. Each $\psi_k$ is $\sigma$-strongly continuous, hence $d$-continuous on $M_1$, so $S_m$ is $d$-closed; and $\bigcup_mS_m=M_1$. By Baire's theorem, some $S_{m_0}$ contains $\{a\in M_1:d(a,a_0)<\delta\}$ for some $a_0\in M_1$, $\delta>0$.

Cutting $a_n$. Let $h_n=a_n^*a_n+a_na_n^*$, so $0\le h_n\le2$ and $\omega(h_n)\to0$. Let $p_n=1_{[0,\varepsilon^2]}(h_n)$, a spectral projection. Then $\|p_nh_np_n\|\le\varepsilon^2$, which gives $\|a_np_n\|\le\varepsilon$ and $\|p_na_n\|\le\varepsilon$. Also $1-p_n\le\varepsilon^{-2}h_n$, so $d(p_n,1)^2=\omega(1-p_n)\le\varepsilon^{-2}\omega(h_n)\to0$, and $p_n\to1$ $\sigma$-strongly. Write $\begin{gathered} a_n\\ =p_na_n+(1-p_n)a_np_n+c_n,\\ c_n\\ =(1-p_n)a_n(1-p_n). \end{gathered}$ The first two terms have norm at most $\varepsilon$, so $|\psi_k(a_n)|\le4c\varepsilon+|\psi_k(c_n)|$.

Comparing with $a_0$. Put $b_n=p_na_0p_n+c_n$. It lies in $M_1$: $b_n^*b_n=p_na_0^*p_na_0p_n+c_n^*c_n$ is a sum of two positive contractions with orthogonal supports. As $n\to\infty$, $p_na_0p_n\to a_0$ $\sigma$-strongly (joint continuity of multiplication on bounded sets), and $d(c_n,0)^2=\omega(c_n^*c_n)\le\omega(1-p_n)\to0$. So $d(p_na_0p_n,a_0)\to0$ and $d(b_n,a_0)\to0$, and there is $n_0$ with both distances below $\delta$ for $n\ge n_0$. For $k\ge m_0$ and $n\ge n_0$, $\begin{gathered} |\psi_k(c_n)|\\ =|\psi_k(b_n)-\psi_k(p_na_0p_n)|\\ \le2\varepsilon ,\\ \text{so}\\ |\psi_k(a_n)|\\ \le(4c+2)\varepsilon . \end{gathered}$ For each of the finitely many $k<m_0$, $\psi_k(a_n)\to0$, because $\psi_k$ is normal and a bounded $\sigma$-strongly$^*$ null sequence is $\sigma$-weakly null. Hence $\limsup_n\sup_k|\psi_k(a_n)|\le(4c+2)\varepsilon$ for every $\varepsilon$. Finally $\varphi_k(a_n)=\psi_k(a_n)+\varphi_0(a_n)$ and $\varphi_0(a_n)\to0$. $\square$

Lemma 10.4. Let $K\subseteq M_*$ be relatively weakly compact and $\varepsilon>0$. There are a finite set $F\subseteq K$ and $\delta>0$ such that: if $a\in M_1$ and $\psi^\sharp(a^*a+aa^*)<\delta$ for all $\psi\in F$, then $|\varphi(a)|<\varepsilon$ for all $\varphi\in K$.

Proof. If $K=\varnothing$, take $F=\varnothing$ and any $\delta>0$. If $K=\{0\}$, the conclusion is also immediate. Otherwise $K$ is bounded; scaling, we may assume $\|\varphi\|\le1$ on $K$. Suppose the claim fails for some $\varepsilon$. Choose $\varphi_1\in K$. Given $\varphi_1,\dots,\varphi_n$, the failure for $F=\{\varphi_1,\dots,\varphi_n\}$ and $\delta=2^{-n}$ gives $a_n\in M_1$ and $\varphi_{n+1}\in K$ with $\begin{gathered} \varphi_k^\sharp(a_n^*a_n+a_na_n^*)<2^{-n}\ (k\\ \le n)\\ \text{and}\\ |\varphi_{n+1}(a_n)|\\ \ge\varepsilon . \end{gathered}$ Let $\omega=\sum_k2^{-k}\varphi_k^\sharp$ and $e=s(\omega)$. Since $\varphi_k^\sharp(a^*a+aa^*)\le2\|\varphi_k^\sharp\|\le4$ for $a\in M_1$, $\begin{gathered} \omega(a_n^*a_n+a_na_n^*)\\ \le2^{-n}+\sum_{k>n}4\cdot2^{-k}\\ =5\cdot2^{-n}\to0 . \end{gathered}$ In $N=eMe$ the functional $\omega$ is faithful, and $b_n=ea_ne$ has $\omega(b_n^*b_n+b_nb_n^*)\le\omega(a_n^*a_n+a_na_n^*)$, because $b_n^*b_n\le ea_n^*a_ne$, $b_nb_n^*\le ea_na_n^*e$ and $\omega(eye)=\omega(y)$. By Proposition 9.1, $b_n\to0$ $\sigma$-strongly$^*$ in $N$. By the Eberlein–Šmulian theorem, $(\varphi_n)$ has a weakly convergent subsequence $(\varphi_{n_j})$, and Lemma 10.3 in $N$ gives $\sup_j|\varphi_{n_j}(b_n)|\to0$ as $n\to\infty$. But $\varphi_{n_j}(b_n)=\varphi_{n_j}(a_n)$ (Lemma 8.1(2)), and $|\varphi_{n_j}(a_{n_j-1})|\ge\varepsilon$ for all $j$ with $n_j\ge2$. As $n_j-1\to\infty$, this is a contradiction. $\square$

Proof of (1)$\Rightarrow$(4). For $m\ge1$, Lemma 10.4 with $\varepsilon=1/m$ gives a finite $F_m\subseteq K$ and $\delta_m>0$. Put $\gamma_m=\big(1+\sum_{\psi\in F_m}\|\psi^\sharp\|\big)^{-1}$ and $\omega=\sum_m2^{-m}\gamma_m\sum_{\psi\in F_m}\psi^\sharp\in M_*^+ .$ Given $\varepsilon>0$, choose $m>1/\varepsilon$ and $\delta=2^{-m}\gamma_m\delta_m$. If $a\in M_1$ and $\omega(a^*a+aa^*)<\delta$, then $\psi^\sharp(a^*a+aa^*)<\delta_m$ for every $\psi\in F_m$, so $|\varphi(a)|<1/m<\varepsilon$ for all $\varphi\in K$. $\square$

This completes the proof of Theorem 10.2.

Remark 10.5. For a $C^*$-algebra $A$, apply Theorem 10.2 to $M=\tilde A$, whose predual is $A^*$. It characterizes the relatively weakly compact subsets of $A^*$, in terms of projections of the bidual.

Absolute values need not stay in a relatively weakly compact set (Example 10.7). One-sided multiples detect exactly when they do.

Proposition 10.6. Let $K\subseteq M_*$ and put $|K|=\{|\varphi|:\varphi\in K\}$, $|K^*|=\{|\varphi^*|:\varphi\in K\}$, $M_1K=\{a\varphi:a\in M_1,\ \varphi\in K\}$ and $KM_1=\{\varphi a:a\in M_1,\ \varphi\in K\}$.

$M_1K$ is relatively weakly compact exactly when $|K|$ is.
$KM_1$ is relatively weakly compact exactly when $|K^*|$ is.
In particular, if $K\subseteq M_*^+$ is relatively weakly compact, so are $M_1K$ and $KM_1$.

Proof. (1) If $\varphi=v|\varphi|$, then $|\varphi|=v^*\varphi\in M_1K$ and $a\varphi=(av)|\varphi|\in M_1|K|$. So it suffices to show that $M_1L$ is relatively weakly compact when $L\subseteq M_*^+$ is. It is bounded. If $p_n\downarrow0$, then for $\omega\in L$ and $a\in M_1$, by the Cauchy–Schwarz inequality, $\begin{gathered} |(a\omega)(p_n)|\\ =|\omega(p_na)|\\ \le\omega(p_n)^{1/2}\omega(a^*a)^{1/2}\\ \le\|\omega\|^{1/2}\omega(p_n)^{1/2}. \end{gathered}$ By Theorem 10.2 (1)$\Rightarrow$(3) for $L$, the right side tends to $0$ uniformly, and (3)$\Rightarrow$(1) applies to $M_1L$.

(2) The map $\psi\mapsto\psi^*$ is a conjugate-linear isometry of $M_*$ onto itself and a homeomorphism for the weak topology. Since $\varphi a=(a^*\varphi^*)^*$, $KM_1=(M_1K^*)^*$, and (1) for $K^*$ gives (2). (3) For positive $\varphi$, $|\varphi|=|\varphi^*|=\varphi$. $\square$

Example 10.7 (Absolute values can escape). Let $H$ have an orthonormal sequence $(\xi_n)$, $M=B(H)$, and $\varphi_n=\omega_{\xi_1,\xi_n}$, so $\varphi_n(x)=\langle x\xi_1,\xi_n\rangle$.

$\sum_n|\varphi_n(x)|^2\le\|x\xi_1\|^2$, so $\varphi_n\to0$ weakly, and $K=\{\varphi_n\}$ is relatively weakly compact. But $\|\varphi_n\|=1$ (Example 1.2).
By (2.5), $|\varphi_n|=\omega_{\xi_n}$. The set $|K|$ is not relatively weakly compact: $p_m=\sum_{j\ge m}\theta_{\xi_j,\xi_j}$ decreases to $0$ (strongly), while $\omega_{\xi_n}(p_m)=1$ for $n\ge m$, against Theorem 10.2(3). By Proposition 10.6, $M_1K$ is not relatively weakly compact either.
$\varphi_n^*=\omega_{\xi_n,\xi_1}$ and $|\varphi_n^*|=\omega_{\xi_1}$ for all $n$. So $|K^*|$ is a single point, and $KM_1$ is relatively weakly compact.
Two-sided multiples of a single positive functional need not form a relatively weakly compact set: $\omega_{\xi_n}=a_n\omega_{\xi_1}a_n^*$ with $a_n=\theta_{\xi_n,\xi_1}\in M_1$.

In the commutative case absolute values cannot escape.

Proposition 10.8. If $M$ is abelian and $K\subseteq M_*$ is relatively weakly compact, then so is $|K|$.

Proof. Let $\omega$ be as in Theorem 10.2(4). For $\varphi=v|\varphi|\in K$ and $a\in M_1$, $|\varphi|(a)=\varphi(av^*)$. In the abelian algebra $M$, with $b=av^*\in M_1$, $b^*b+bb^*=a^*a\,vv^*+aa^*\,v^*v\le a^*a+aa^* ,$ because $vv^*$ and $v^*v$ are projections that commute with $a^*a$ and $aa^*$. So $\omega(b^*b+bb^*)<\delta$ whenever $\omega(a^*a+aa^*)<\delta$, and then $||\varphi|(a)|=|\varphi(b)|<\varepsilon$. Thus $|K|$ satisfies Theorem 10.2(4) with the same $\omega$. $\square$

11. The Mackey topology on bounded sets

Definition 11.1. On a von Neumann algebra $M$, let $\tau$ be the locally convex topology defined by the seminorms $p_K(x)=\sup_{\varphi\in K}|\varphi(x)|,$ where $K$ runs over the relatively weakly compact subsets of $M_*$, with $p_\varnothing=0$. It is Hausdorff, since singletons are compact. We also define the Mackey topology $\tau(M,M_*)$ directly as uniform convergence on the absolutely convex weakly compact subsets of $M_*$. Thus its seminorms form a subfamily of these $p_K$. The finest-compatible-topology characterization is not needed.

Theorem 11.2. On every bounded subset of $M$, the topology $\tau$ coincides with the $\sigma$-strong$^*$ topology. More precisely, the $\sigma$-strong$^*$ topology is coarser than $\tau$ on all of $M$, and on bounded sets $\tau$ is coarser than the $\sigma$-strong$^*$ topology.

Proof. $\sigma$-strong$^*$ is coarser than $\tau$. Let $\omega\in M_*^+$ with cyclic representation $(\pi,H,\xi)$, which is normal. The sets $\begin{gathered} K_\omega\\ =\{x\mapsto\langle\pi(x)\xi,\zeta\rangle:\ \|\zeta\|\le1\},\\ K'_\omega\\ =\{x\mapsto\langle\pi(x)\zeta,\xi\rangle:\ \|\zeta\|\le1\} \end{gathered}$ are absolutely convex and weakly compact in $M_*$: they are images of the weakly compact unit ball of $H$ under maps that are continuous from the weak topology of $H$ to $\sigma(M_*,M)$. By Cauchy–Schwarz, with equality at $\zeta=\pi(x)\xi/\|\pi(x)\xi\|$ when the denominator is nonzero (and both sides zero otherwise), $\begin{gathered} \omega(x^*x)^{1/2}\\ =\|\pi(x)\xi\|\\ =p_{K_\omega}(x),\\ \omega(xx^*)^{1/2}\\ =\|\pi(x^*)\xi\|\\ =p_{K'_\omega}(x). \end{gathered}$ So every $\sigma$-strong$^*$ seminorm is dominated by $p_{K_\omega}+p_{K'_\omega}$.

On bounded sets $\tau$ is coarser. Let $D\subseteq M$ be bounded, with $\|x\|\le R$ on $D$, and let $x_i\to x$ $\sigma$-strongly$^*$ in $D$. If $R=0$, then $D\subseteq\{0\}$ and there is nothing to prove. For $R>0$, put $a_i=(x_i-x)/(2R)\in M_1$. Let $K$ be relatively weakly compact, $\varepsilon>0$, and $\omega,\delta$ as in Theorem 10.2(4). Since $\omega(a_i^*a_i+a_ia_i^*)\to0$, eventually $\omega(a_i^*a_i+a_ia_i^*)<\delta$, and then $p_K(x_i-x)=2Rp_K(a_i)\le2R\varepsilon$. So $x_i\to x$ in $\tau$. $\square$

Remark 11.3. By the direct definition above, $\tau(M,M_*)$ is coarser than $\tau$. The absolutely convex weakly compact sets $K_\omega,K'_\omega$ in the proof show directly that $\tau(M,M_*)$ is finer than the $\sigma$-strong$^*$ topology. Theorem 11.2 therefore proves, without an additional theorem, that all three topologies coincide on bounded sets. The Mackey–Arens theorem further identifies $\tau(M,M_*)$ as the finest locally convex topology whose continuous functionals are the normal ones. That compatibility characterization is unproved here, is only further reading (background item 10), and is not used by any proof or solution.

On unbounded sets the topologies differ, unless $M$ is finite-dimensional.

Proposition 11.4. Let $M$ contain an infinite sequence $(e_n)$ of mutually orthogonal nonzero projections, and put $D=\{\sqrt n\,e_n:n\ge1\}$. Then $0$ lies in the $\sigma$-strong$^*$ closure of $D$, but not in its closure for $\tau(M,M_*)$ or for $\tau$. Consequently the $\sigma$-strong$^*$ topology equals the Mackey topology on all of $M$ exactly when $M$ is finite-dimensional.

Proof. A basic $\sigma$-strong$^*$ neighbourhood of $0$ contains $\{x:\omega(x^*x+xx^*)<\delta\}$ for some $\omega\in M_*^+$ and $\delta>0$, since finitely many functionals can be added. As $\sum_n\omega(e_n)\le\omega(1)<\infty$, the numbers $n\,\omega(e_n)$ are not bounded below by a positive constant, so $2n\,\omega(e_n)<\delta$ for some $n$. Then $\begin{gathered} \omega((\sqrt ne_n)^*(\sqrt ne_n)+(\sqrt ne_n)(\sqrt ne_n)^*)\\ =2n\,\omega(e_n)<\delta. \end{gathered}$

For the other claim, choose normal states $\omega_n$ with $\omega_n(e_n)=1$ (for instance $e_n\psi e_n/\psi(e_n)$ for a normal state $\psi$ with $\psi(e_n)>0$), and put $\varphi_n=n^{-1/2}\omega_n$, so $\|\varphi_n\|\to0$. The operator $T:\ell^1\to M_*$, $T\lambda=\sum_n\lambda_n\varphi_n$, is compact. Indeed its truncations $T_N$ have finite-dimensional bounded images of the unit ball and $\|T-T_N\|\le\sup_{n>N}\|\varphi_n\|\to0$. Finite nets for the truncated images give finite nets for $T$ of the unit ball; its closure is complete, since $M_*$ is Banach, and is therefore compact by the metric lemma above. So the closure $K$ of $T(\text{unit ball of }\ell^1)$ is absolutely convex and norm compact, hence weakly compact, and it contains every $\varphi_n$. Now $p_K(\sqrt ne_n)\ge|\varphi_n(\sqrt ne_n)|=\omega_n(e_n)=1$, so the neighbourhood $\{x:p_K(x)<1\}$ of $0$ misses $D$, for $\tau(M,M_*)$ and hence for the finer $\tau$.

If $\dim M=\infty$, such a sequence $(e_n)$ exists (background fact 25), so the topologies differ. If $\dim M<\infty$, both are Hausdorff vector topologies on a finite-dimensional space and coincide (background fact 8). $\square$

12. Norm convergence in atomic algebras

The pairs $c_0\subseteq\ell^\infty$ with predual $\ell^1$, and compact operators inside $B(H)$ with predual $B(H)_*$, are the two basic atomic examples. Schur's theorem says that in $\ell^1$ weak and norm convergence of sequences agree. In $B(H)_*$ they do not (Example 10.7). The next theorem locates the difference: it disappears once the absolute values are under control.

Definition 12.1. A von Neumann algebra is atomic if every nonzero projection majorizes a minimal projection (background fact 26). A projection is of finite rank if it is a sum of finitely many mutually orthogonal minimal projections.

In an atomic algebra $1=\sum_jf_j$ for a family of mutually orthogonal minimal projections (background fact 26). The join of two finite-rank projections is of finite rank (background fact 25), so the finite-rank projections form an increasing net with supremum $1$. For a finite-rank $e=f_1+\dots+f_m$, $\dim eMe\le m^2$, since $\dim f_iMf_j\le1$.

Theorem 12.2. Let $M$ be atomic, and let $(\varphi_i)$ be a net in $M_*$ converging weakly to $\varphi\in M_*$ such that $\{|\varphi_i|\}$ and $\{|\varphi_i^*|\}$ are relatively weakly compact. Then $\|\varphi_i-\varphi\|\to0$.

Proof. The set $K=\{|\varphi|,|\varphi^*|\}\cup\{|\varphi_i|,|\varphi_i^*|\}$ is relatively weakly compact, and $c=\sup_{\psi\in K}\|\psi\|<\infty$; note $\|\varphi_i\|=\||\varphi_i|\|\le c$. Let $0<\varepsilon<1$. By Theorem 10.2(6) applied to the net of finite-rank projections, there is a finite-rank $e$ with $\|(1-e)\psi(1-e)\|<\varepsilon$, in particular $\psi(1-e)<\varepsilon$, for all $\psi\in K$.

Since $eMe$ is finite-dimensional, weak convergence of the restrictions to $eMe$ is norm convergence (background fact 8). The norm of $x\mapsto(\varphi_i-\varphi)(exe)$ on $M$ is the norm of the restriction of $\varphi_i-\varphi$ to $eMe$. So there is $i_0$ with $\|e(\varphi_i-\varphi)e\|<\varepsilon$ for $i\ge i_0$.

For $\psi\in M_*$ with $\psi=u|\psi|$ and $a\in M_1$, the Cauchy–Schwarz inequality gives $\begin{gathered} |\psi((1-e)a)|\\ =||\psi|((1-e)au)|\\ \le|\psi|(1-e)^{1/2}\,|\psi|(u^*a^*au)^{1/2}\\ \le\big(|\psi|(1-e)\,\|\psi\|\big)^{1/2}. \end{gathered} \tag{12.1}$ Also $|\psi(ea(1-e))|=|\psi^*((1-e)a^*e)|$, to which (12.1) applies with $\psi^*$ and $a^*e$. Now split $a=eae+ea(1-e)+(1-e)a$. For $i\ge i_0$ and $a\in M_1$, $\begin{gathered} |(\varphi-\varphi_i)(a)|\\ \le\varepsilon+|\varphi(ea(1-e))|\\ +|\varphi_i(ea(1-e))|\\ +|\varphi((1-e)a)|\\ +|\varphi_i((1-e)a)|\\ \le\varepsilon+4(c\varepsilon)^{1/2}, \end{gathered}$ since $|\varphi|,|\varphi^*|,|\varphi_i|,|\varphi_i^*|\in K$. So $\|\varphi-\varphi_i\|\le\varepsilon+4(c\varepsilon)^{1/2}$ for $i\ge i_0$. $\square$

For positive functionals no compactness hypothesis is needed.

Corollary 12.3. Let $M$ be atomic and $(\varphi_i)$ a net in $M_*^+$ converging weakly to $\varphi$. Then $\|\varphi_i-\varphi\|\to0$.

Proof. $\varphi$ is positive and $\|\varphi_i\|=\varphi_i(1)\to\varphi(1)=\|\varphi\|$, so eventually $\|\varphi_i\|\le c=\|\varphi\|+1$. Given $\varepsilon>0$, normality of $\varphi$ gives a finite-rank $e$ with $\varphi(1-e)<\varepsilon$; then eventually $\varphi_i(1-e)<\varepsilon$. Here $|\varphi_i|=|\varphi_i^*|=\varphi_i$, so the estimates in the proof of Theorem 12.2 apply and give $\|\varphi-\varphi_i\|\le\varepsilon+4(c\varepsilon)^{1/2}$ eventually. $\square$

Corollary 12.4 (Schur's theorem again). A weakly convergent sequence in $\ell^1(\Gamma)$ converges in norm.

Proof. $\ell^1(\Gamma)$ is the predual of the atomic abelian algebra $\ell^\infty(\Gamma)$ (Example 2.6). A weakly convergent sequence together with its limit is weakly compact. By Proposition 10.8 its absolute values form a relatively weakly compact set. Here $|\varphi^*|=|\varphi|$, since the adjoint of $g\in\ell^1(\Gamma)$ is $\bar g$. Theorem 12.2 applies. $\square$

Example 12.5. The hypotheses of Theorem 12.2 and Corollary 12.3 cannot be dropped.

For the first example, the Rademacher function $r_n$ has constant value $+1$ or $-1$ on each interval of length $2^{-n}$, alternately. Thus $\|r_n\|_2=1$. If $m>n$, each interval on which $r_n$ is constant contains an even number of the alternating intervals for $r_m$, so $\int r_nr_m\,dt=0$. Endpoint values affect only a null set. This proves the asserted orthonormality.

Atomicity. Let $M=L^\infty[0,1]$ and let $r_n(t)=\operatorname{sgn}\sin(2^n\pi t)$ be the Rademacher functions, an orthonormal sequence in $L^2[0,1]$. For $x\in L^\infty\subseteq L^2$, Bessel's inequality gives $\int xr_n\,dt\to0$. So $\varphi_n=\varphi_{r_n}\to0$ weakly, and by Example 2.6, $|\varphi_n|=|\varphi_n^*|=\varphi_{|r_n|}=\lambda$, Lebesgue measure, for every $n$. But $\|\varphi_n\|=\int|r_n|=1$. Likewise the positive functionals $\varphi_{1+r_n}$ tend weakly to $\lambda$, while $\|\varphi_{1+r_n}-\lambda\|=1$.
Both absolute values. In Example 10.7, $M=B(H)$ is atomic, $\varphi_n\to0$ weakly, and $\{|\varphi_n^*|\}$ is a single point, but $\|\varphi_n\|=1$. Passing to adjoints gives an example where $\{|\varphi_n|\}$ is compact and $\{|\varphi_n^*|\}$ is not.
Positivity. The same example shows that Corollary 12.3 fails for sequences that are not positive.

13. Extending normal functionals from subalgebras

A normal functional on a von Neumann subalgebra extends to a normal functional on the whole algebra with the same norm. For positive functionals this comes from writing them as sums of vector functionals; the polar decomposition then handles the general case.

Proposition 13.1. Let $M$ be a von Neumann algebra and $N\subseteq M$ a $\sigma$-weakly closed $*$-subalgebra.

Every $\varphi\in N_*^+$ extends to some $\tilde\varphi\in M_*^+$, and $\|\tilde\varphi\|=\|\varphi\|$.
Every $\varphi\in N_*$ extends to some $\tilde\varphi\in M_*$ with $\|\tilde\varphi\|=\|\varphi\|$.

Proof. Represent $M$ faithfully and normally on $H$. The subalgebra $N$ may have a smaller unit. Let $(u_\lambda)$ be its increasing approximate identity of positive contractions, and let $q$ be the projection onto $[NH]$. On $a\eta$, with $a\in N$, the estimate $\|(u_\lambda a-a)\eta\|\to0$, followed by density and the uniform bound, gives $u_\lambda\to q$ strongly on $[NH]$; on its orthogonal complement every element of $N$ acts as zero. Bounded strong convergence is sigma-weak convergence by fact 15, so the sigma-weak closedness gives $q\in N$. Thus $qa=aq=a$ for $a\in N$, and $N$ acts nondegenerately on $qH$. The double commutant theorem makes it a von Neumann algebra there, with unit $q$, and $N\subseteq qMq$. If $N=0$, take $q=0$ and the assertions are immediate.

The extension is not unique in general: for $N=\mathbb C1\subseteq M_2(\mathbb C)$ and $\varphi(\lambda1)=\lambda$, every state of $M_2(\mathbb C)$ is a norm-preserving extension.

14. Weakly compact and dual C*-algebras

For a $C^*$-algebra $A$ and $a\in A$, consider left multiplication $L_a:x\mapsto ax$ on $A$. When is it a weakly compact operator? We will see that this happens for all $a$ exactly when $A$ is an ideal of its bidual, exactly when every closed one-sided ideal is recovered from its annihilator, and exactly when $A$ is a $c_0$-direct sum of algebras of compact operators.

In this section a minimal projection of $A$ is a nonzero projection $f\in A$ with $fAf=\mathbb Cf$. For a closed left ideal $\mathfrak m$ and a closed right ideal $\mathfrak n$ of $A$ we write $\begin{gathered} \operatorname{ann}_r(\mathfrak m)\\ =\{x\in A\,:\,\mathfrak mx=0\},\\ \operatorname{ann}_l(\mathfrak n)\\ =\{y\in A\,:\,y\mathfrak n=0\}. \end{gathered}$ The first is a closed right ideal and the second a closed left ideal.

Definition 14.1. A $C^*$-algebra $A$ is weakly compact if every left multiplication $L_a$, $a\in A$, is a weakly compact operator, that is, $aA_1$ is relatively weakly compact in $A$. It is dual if $\operatorname{ann}_l(\operatorname{ann}_r(\mathfrak m))=\mathfrak m$ for every closed left ideal $\mathfrak m$ and $\operatorname{ann}_r(\operatorname{ann}_l(\mathfrak n))=\mathfrak n$ for every closed right ideal $\mathfrak n$.

Proposition 14.2. For a $C^*$-algebra $A$ the following are equivalent: (a) every left multiplication $L_a$ is weakly compact; (b) every right multiplication $R_a:x\mapsto xa$ is weakly compact; (c) $A$ is a two-sided ideal of $\tilde A$.

Proof. Fix $a\in A$. Left multiplication by $a$ on $\tilde A$ is $\sigma$-weakly continuous, and $A_1$ is $\sigma$-weakly dense in $\tilde A_1$ (background fact 14). So the $\sigma(\tilde A,A^*)$-closure of $aA_1$ contains $a\tilde A_1$; and $a\tilde A_1$ is $\sigma$-weakly compact and contains $aA_1$. Hence that closure is $a\tilde A_1$. By Lemma 10.1 with $X=A$, $X^{**}=\tilde A$, $L_a$ is weakly compact exactly when $a\tilde A_1\subseteq A$, that is $a\tilde A\subseteq A$. So (a) says $A\tilde A\subseteq A$, and in the same way (b) says $\tilde AA\subseteq A$. Since $A$ and $\tilde A$ are closed under the involution, each inclusion implies the other by taking adjoints, and together they say (c). $\square$

Proposition 14.3. Let $A$ be an ideal of $\tilde A$, and $\mathfrak m$ a closed left ideal of $A$ whose $\sigma$-weak closure in $\tilde A$ is $\tilde Ae$ (background fact 19). Then $\begin{gathered} \mathfrak m\\ =Ae\\ =\{x\in A:xe=x\},\\ \operatorname{ann}_r(\mathfrak m)\\ =(1-e)A,\\ \operatorname{ann}_l(\operatorname{ann}_r(\mathfrak m))\\ =\mathfrak m . \end{gathered}$ The symmetric statements hold for closed right ideals. In particular $A$ is dual.

Proof. The $\sigma$-weak closure of $\mathfrak m$ is a left ideal of $\tilde A$, since $A\mathfrak m\subseteq\mathfrak m$, multiplication is separately continuous and $A$ is dense; so it is $\tilde Ae$ for a projection $e$. If $x\in\mathfrak m$, then $x\in\tilde Ae$, so $x=xe$. If $x\in A$, then $xe\in A$ (as $A$ is an ideal) and $xe\in\tilde Ae$, the $\sigma(\tilde A,A^*)$-closure of $\mathfrak m$. A norm-closed convex subset of $A$ is weakly closed (Mazur), and the weak topology of $A$ is the restriction of $\sigma(\tilde A,A^*)$; so $xe\in\mathfrak m$. This proves $\mathfrak m=Ae=\{x\in A:xe=x\}$.

Next, $y\in\operatorname{ann}_r(\mathfrak m)$ means $Aey=0$; the set $\{X\in\tilde A:Xey=0\}$ is $\sigma$-weakly closed and contains $A$, hence contains $1$, so $ey=0$. Conversely $ey=0$ gives $\mathfrak my=Aey=0$. Thus $\operatorname{ann}_r(\mathfrak m)=\{y\in A:ey=0\}=(1-e)A$, using that $A$ is an ideal. In the same way $z\in\operatorname{ann}_l((1-e)A)$ exactly when $z(1-e)=0$, that is $z\in\mathfrak m$. $\square$

Lemma 14.4. Let $f$ be a minimal projection of $A$. Then $f$ is minimal in $\tilde A$, and $\tilde Af=Af$.

Proof. $f\tilde Af$ is the $\sigma$-weak closure of $fAf=\mathbb Cf$, so it is $\mathbb Cf$. By background fact 25, $\tilde Af$ is a Hilbert space for $\langle X,Y\rangle f=Y^*X$, with the norm of $\tilde A$. Let $\omega_f$ be the normal state with $fZf=\omega_f(Z)f$; then $\langle X,Y\rangle=\omega_f(Y^*X)$, so each $X\mapsto\langle X,Y\rangle$ is $\sigma$-weakly continuous. The subspace $Af=\{x\in A:xf=x\}$ is norm closed, and it is $\sigma$-weakly dense in $\tilde Af$, since $X\mapsto Xf$ is $\sigma$-weakly continuous. So $Af$ is weakly dense in the Hilbert space $\tilde Af$ and norm closed; closed subspaces of a Hilbert space are weakly closed, so $Af=\tilde Af$. $\square$

Proposition 14.5 (Structure of dual algebras). Let $A$ be dual.

$\mathfrak m\mapsto\operatorname{ann}_r(\mathfrak m)$ is an inclusion-reversing bijection from the closed left ideals onto the closed right ideals, with inverse $\operatorname{ann}_l$. A closed left ideal is maximal exactly when its right annihilator is a minimal nonzero closed right ideal; and symmetrically.
Every nonzero closed left ideal contains $Af$ for some minimal projection $f$ of $A$; symmetrically for right ideals.
$A$ is the closed linear span of the sets $Af$, and also of the sets $fA$, where $f$ runs over the minimal projections of $A$.
$\tilde A$ is atomic, and every finite-rank projection of $\tilde A$ lies in $A$.
The finite-rank projections of $\tilde A$, ordered by size, form an increasing approximate unit of $A$.
$A$ is an ideal of $\tilde A$.

Proof. (1) The maps reverse inclusions and are mutually inverse by duality. $\operatorname{ann}_r(0)=A$, and $\operatorname{ann}_r(A)=0$ because $Ax=0$ gives $x^*x=0$. So $\mathfrak m\ne A$ exactly when $\operatorname{ann}_r(\mathfrak m)\ne0$. An inclusion-reversing bijection between the proper closed left ideals and the nonzero closed right ideals matches maximal elements with minimal ones.

(2) Let $\mathfrak m\ne0$. Then $\mathfrak n=\operatorname{ann}_r(\mathfrak m)$ is a proper closed right ideal, since $\mathfrak n=A$ would give $\mathfrak m=\operatorname{ann}_l(A)=0$. By Corollary 6.6 (right-handed), $\mathfrak n$ lies in a maximal closed right ideal $\mathfrak n_0$. Then $\operatorname{ann}_l(\mathfrak n_0)\subseteq\operatorname{ann}_l(\mathfrak n)=\mathfrak m$, and $\operatorname{ann}_l(\mathfrak n_0)$ is a minimal nonzero closed left ideal by (1). By background fact 13 it is $Af$ for a minimal projection $f$ of $A$. The right-handed statement follows by taking adjoints.

(3) Let $\mathfrak l$ be the closed span of the $Af$, a closed left ideal. An element $x$ lies in $\operatorname{ann}_r(\mathfrak l)$ exactly when $fx=0$ for all such $f$ (use an approximate unit for one direction). If $\operatorname{ann}_r(\mathfrak l)\ne0$, by (2) it contains $fA$ for such an $f$, so $f=ff=0$, which is absurd. So $\operatorname{ann}_r(\mathfrak l)=0$ and $\mathfrak l=\operatorname{ann}_l(0)=A$. The right-handed statement is symmetric.

(4) Let $z$ be the join in $\tilde A$ of all minimal projections of $A$. For $y\in A$ and such $f$, $yfz=yf$; by (3), $xz=x$ for all $x\in A$, and by density $z=1$. Let $e\ne0$ be a projection of $\tilde A$. Some such $f$ has $ef\ne0$, since otherwise $e\le1-z=0$. As $f$ is minimal in $\tilde A$ (Lemma 14.4), $fef=\mu f$ with $\mu=\|ef\|^2>0$. So $w=\mu^{-1/2}ef$ is a partial isometry with $w^*w=f$, and $q=ww^*\le e$ is a projection with $q\tilde Aq=wf\tilde Afw^*=\mathbb Cq$. Thus $e$ majorizes the minimal projection $q$, and $\tilde A$ is atomic. If $e$ itself is minimal, then $q=e$, and $w\in\tilde Af=Af\subseteq A$ by Lemma 14.4, so $e=ww^*\in A$. Finite-rank projections are finite sums of minimal ones.

(5) The finite-rank projections form an upward directed set (background fact 25) of positive contractions in $A$, by (4). For $x=yf$ as in (3) and $e\ge f$ of finite rank, $xe=x$. As such $x$ span a dense subspace and $\|e\|\le1$, $\|xe-x\|\to0$ for all $x\in A$. Using the $fA$, also $\|ex-x\|\to0$.

(6) For $X\in\tilde A$ and $x=yf$ as in (3), $Xx=(Xy)f\in\tilde Af=Af\subseteq A$. By (3) and continuity, $XA\subseteq A$. So $A$ is a left ideal of $\tilde A$, and, being self-adjoint, a two-sided ideal. $\square$

So far: $A$ is weakly compact $\Leftrightarrow$ $A$ is an ideal of $\tilde A$ $\Leftrightarrow$ $A$ is dual (Propositions 14.2, 14.3 and 14.5(6)). It remains to identify these algebras.

Definition 14.6. For $C^*$-algebras $(A_j)_{j\in J}$, the $c_0$*-direct sum* $\bigoplus^0_jA_j$ is the set of families $x=(x_j)$ with $x_j\in A_j$ such that $\{j:\|x_j\|\ge\varepsilon\}$ is finite for every $\varepsilon>0$, with coordinatewise operations and $\|x\|=\sup_j\|x_j\|$.

Lemma 14.7. Let $B=\bigoplus^0_jA_j$, and let $\iota_j:A_j\to B$ put an element in the $j$-th coordinate.

$B$ is a $C^*$-algebra, and every $x\in B$ is the norm limit of its finite truncations $\sum_{j\in F}\iota_j(x_j)$.
The closed left ideals of $B$ are exactly the sets $\bigoplus^0_j\mathfrak m_j=\{x\in B:x_j\in\mathfrak m_j\ \forall j\}$, $\mathfrak m_j$ a closed left ideal of $A_j$. The same holds for right and two-sided ideals.
If every $A_j$ is dual, $B$ is dual.

Proof. (1) The bounded families form a $C^*$-algebra with the supremum norm, and $B$ is a $*$-subalgebra of it. It is closed: if $\|x-x'\|<\varepsilon/2$ with $x'\in B$, then $\{j:\|x_j\|\ge\varepsilon\}\subseteq\{j:\|x'_j\|\ge\varepsilon/2\}$ is finite. For $F=\{j:\|x_j\|\ge\varepsilon\}$, the truncation differs from $x$ by at most $\varepsilon$.

(2) Let $\mathfrak m$ be a closed left ideal and $\mathfrak m_j=\iota_j^{-1}(\mathfrak m)$, a closed left ideal of $A_j$. If $x\in\mathfrak m$ and $(u_\lambda)$ is an approximate unit of $A_j$, then $\iota_j(u_\lambda)x=\iota_j(u_\lambda x_j)\in\mathfrak m$ tends to $\iota_j(x_j)$, so $x_j\in\mathfrak m_j$. Conversely, if all $x_j\in\mathfrak m_j$, the truncations of $x$ lie in $\mathfrak m$ and converge to $x$. Right and two-sided ideals are handled in the same way.

(3) By (2), for $\mathfrak m=\bigoplus^0_j\mathfrak m_j$, $y\in\operatorname{ann}_r(\mathfrak m)$ exactly when $\mathfrak m_jy_j=0$ for all $j$, so $\operatorname{ann}_r(\mathfrak m)=\bigoplus^0_j\operatorname{ann}_r(\mathfrak m_j)$; likewise for $\operatorname{ann}_l$. So $\operatorname{ann}_l\operatorname{ann}_r(\mathfrak m)=\bigoplus^0_j\mathfrak m_j=\mathfrak m$, and symmetrically for right ideals. $\square$

Example 14.8. $K(H)$ is dual: its bidual is $B(H)$, in which it is an ideal (background fact 14 and Proposition 14.3). By Lemma 14.7(3), every $c_0$-direct sum of algebras $K(H_j)$ is dual; for instance $c_0(\Gamma)=\bigoplus^0_{\gamma\in\Gamma}\mathbb C$. An infinite-dimensional $C^*$-algebra with a unit, such as $C[0,1]$ or $B(H)$ with $\dim H=\infty$, is never dual. For if $A$ is dual, it is an ideal of $\tilde A$ (Proposition 14.5(6)), and if moreover $1\in A$, then $\tilde A=\tilde A\cdot1\subseteq A$, so $A$ would be a von Neumann algebra equal to its bidual, hence reflexive, which an infinite-dimensional von Neumann algebra is not (it contains a copy of $\ell^\infty$, background fact 25).

Representations $\pi_1,\pi_2$ of $A$ are disjoint if their central supports are orthogonal: $z(\pi_1)z(\pi_2)=0$. Then no nonzero operator $T$ satisfies $T\pi_1(a)=\pi_2(a)T$ for all $a$: such a $T$ also intertwines the normal extensions, so $\begin{gathered} T\\ =T\bar\pi_1(z(\pi_1))\\ =\bar\pi_2(z(\pi_1))T\\ =\bar\pi_2(z(\pi_1)z(\pi_2))T\\ =0. \end{gathered}$

Proposition 14.9. Let $A$ be dual and $(\pi_i)_{i\in I}$ a family of mutually disjoint representations, and $\pi=\bigoplus_i\pi_i$. Then $\pi(A)=\bigoplus^0_i\pi_i(A)$ as sets of operators on $\bigoplus_iH_i$. So $\pi(A)$ is isomorphic to the $c_0$-direct sum of the $\pi_i(A)$.

Proof. Every $\pi(a)$ lies in the $c_0$-sum. For a minimal projection $f$ of $\tilde A$, the central support $c(f)$ is a minimal central projection. Indeed, let $c\le c(f)$ be central and nonzero. Then $cf\ne0$, for otherwise $f\le1-c$, so $c(f)\le1-c$ and $c=cc(f)=0$. And $cf=fcf$ is a nonzero projection in $f\tilde Af=\mathbb Cf$, so $cf=f$, $f\le c$ and $c(f)\le c$. Hence for each $i$, either $c(f)\le z(\pi_i)$ or $c(f)z(\pi_i)=0$, and the first case occurs for at most one $i$. In the second case $\bar\pi_i(f)=\bar\pi_i(fz(\pi_i))=\bar\pi_i(fc(f)z(\pi_i))=0$. Now let $a\in A$ and $\varepsilon>0$. By Proposition 14.5(5) there is a finite-rank projection $e=f_1+\dots+f_m$ of $\tilde A$ with $\|a-ea\|<\varepsilon$. Then $\bar\pi_i(e)=0$ for all $i$ outside a set of at most $m$ indices, and for those $i$, $\|\pi_i(a)\|=\|\pi_i(a-ea)\|<\varepsilon$.

The $c_0$-sum lies in $\pi(A)$. The normal extension of $\pi$ is $\bigoplus_i\bar\pi_i$. For $i_0\in I$, $\bar\pi_i(z(\pi_{i_0}))$ is $1$ for $i=i_0$ and $0$ otherwise, by disjointness. Since $A$ is an ideal of $\tilde A$ (Proposition 14.5(6)), $az(\pi_{i_0})\in A$, and $\pi(az(\pi_{i_0}))$ is $\pi_{i_0}(a)$ in the $i_0$-th place and $0$ elsewhere. So $\pi(A)$ contains all finite sums of such elements. It is norm closed, being the image of a $C^*$-algebra, and these finite sums are dense in $\bigoplus^0_i\pi_i(A)$ by Lemma 14.7(1). $\square$

Theorem 14.10. Every dual $C^*$-algebra is isomorphic to a $c_0$-direct sum of algebras $K(H_j)$ of compact operators.

Proof. Let $A$ be dual. By Proposition 14.5(4), $\tilde A$ is atomic. Let $(c_j)_{j\in J}$ be the distinct central supports of minimal projections of $\tilde A$. They are minimal central projections (proof of Proposition 14.9), hence mutually orthogonal, and $\sum_jc_j=1$, because $1$ is a sum of minimal projections $f$, each under $c(f)$. Each $\tilde Ac_j$ is a factor containing a minimal projection, so there is an isomorphism $\theta_j:\tilde Ac_j\to B(H_j)$ (background fact 26). Put $\pi_j(a)=\theta_j(ac_j)$. Since $\theta_j$ is normal (every $*$-isomorphism between von Neumann algebras is, background fact 15), $\bar\pi_j(X)=\theta_j(Xc_j)$, whose kernel is $\tilde A(1-c_j)$. As $\bar\pi_j(1)=\theta_j(c_j)=1$, normality carries the increasing approximate identity of $A$, whose supremum in $\tilde A$ is $1$, to a net with supremum $1$. By fact 16 it converges strongly to $1$, and its ranges lie in $[\pi_j(A)H_j]$. Hence $\pi_j$ is nondegenerate; so $z(\pi_j)=c_j$, and the $\pi_j$ are mutually disjoint. The representation $\pi=\bigoplus_j\pi_j$ is faithful: if $ac_j=0$ for all $j$, then $a=\sum_jac_j=0$. By Proposition 14.9, $A\cong\pi(A)=\bigoplus^0_j\pi_j(A)$.

It remains to show $\pi_j(A)=K(H_j)$. First, $\pi_j(A)=\theta_j(Ac_j)$ is an ideal of $B(H_j)$, because $Ac_j$ is an ideal of $\tilde Ac_j$ (Proposition 14.5(6)). It contains $\theta_j(f)$ for a minimal projection $f\le c_j$, which lies in $A$ by Proposition 14.5(4); $\theta_j(f)$ is a minimal projection of $B(H_j)$, hence of rank one. An ideal of $B(H_j)$ containing a rank-one projection $\theta_{\zeta,\zeta}$ contains every rank-one operator $\theta_{\xi,\eta}=\theta_{\xi,\zeta}\theta_{\zeta,\zeta}\theta_{\zeta,\eta}$ (with $\|\zeta\|=1$), hence every finite-rank operator, hence $K(H_j)$, as $\pi_j(A)$ is closed. Conversely, by Proposition 14.5(3), $A$ is the closed span of elements $yf$ with $f\in A$ minimal in $\tilde A$. Either $c(f)=c_j$ and $fc_j=f$, or $fc_j=0$. So $\pi_j(yf)=\pi_j(y)\theta_j(fc_j)$ has rank at most one, and $\pi_j(A)\subseteq K(H_j)$. $\square$

Corollary 14.11. For a $C^*$-algebra $A$ the following are equivalent: $A$ is weakly compact; every right multiplication is weakly compact; $A$ is an ideal of $\tilde A$; $A$ is dual; $A$ is isomorphic to a $c_0$-direct sum of algebras of compact operators.

Proof. Proposition 14.2 gives the first three equivalences, Proposition 14.3 gives "ideal $\Rightarrow$ dual", Theorem 14.10 gives "dual $\Rightarrow$ $c_0$-sum", and Example 14.8 gives "$c_0$-sum $\Rightarrow$ dual $\Rightarrow$ ideal" (with Proposition 14.5(6)). $\square$

Corollary 14.12. Quotients of dual $C^*$-algebras by closed two-sided ideals are dual.

Proof. By Theorem 14.10 we may take $B=\bigoplus^0_{j\in J}K(H_j)$. Let $\mathcal I$ be a closed two-sided ideal of $B$. By Lemma 14.7(2), $\mathcal I=\bigoplus^0_j\mathcal I_j$ with $\mathcal I_j$ a closed ideal of $K(H_j)$. A nonzero closed ideal of $K(H)$ is all of $K(H)$: if $0\ne x\in\mathcal I_j$ and $x\zeta\ne0$, then $\theta_{\xi,x\zeta}\,x\,\theta_{\zeta,\eta}=\|x\zeta\|^2\theta_{\xi,\eta}$, so the ideal contains every rank-one operator, hence every compact operator. So $\mathcal I=\bigoplus^0\{K(H_j):j\in S\}$ for a set $S\subseteq J$, and $x+\mathcal I\mapsto(x_j)_{j\notin S}$ is an isomorphism of $B/\mathcal I$ onto $\bigoplus^0\{K(H_j):j\notin S\}$: it is a well-defined, surjective $*$-homomorphism with kernel $\mathcal I$, and injective $*$-homomorphisms of $C^*$-algebras are isometric. This algebra is dual by Example 14.8. $\square$

15. Exercises

Exercise 15.1. (easy) In $M_2(\mathbb C)$ let $\varphi(x)=x_{21}$, the $(2,1)$ entry. Find $|\varphi|$, the partial isometry $v$, $s_l(\varphi)$, $s_r(\varphi)$ and $|\varphi^*|$. Check inequality (3.1), and check that (1.1) is an equality for $e=E_{11}$ and for the projection $e$ onto $2^{-1/2}(1,1)$.

Solution. $\varphi=\omega_{\epsilon_1,\epsilon_2}$ for the standard basis $\epsilon_1,\epsilon_2$, since $\langle x\epsilon_1,\epsilon_2\rangle=x_{21}$. By (2.5), $|\varphi|=\omega_{\epsilon_2}$, that is $|\varphi|(x)=x_{22}$, and $v=\theta_{\epsilon_1,\epsilon_2}=E_{12}$. Check: $|\varphi|(xE_{12})=(xE_{12})_{22}=x_{21}$, and $E_{12}^*E_{12}=E_{22}=s(|\varphi|)$. By Theorem 2.2(3), $s_r(\varphi)=E_{22}$ and $s_l(\varphi)=E_{11}$. Also $\varphi^*(x)=x_{12}$ and $|\varphi^*|(x)=x_{11}$. Inequality (3.1) reads $|x_{21}|^2\le(xx^*)_{22}=|x_{21}|^2+|x_{22}|^2$. For $e=E_{11}=s_l(\varphi)$, $e\varphi=\varphi$ and $(1-e)\varphi=0$: $1+0=1$. For $e$ the projection onto $2^{-1/2}(1,1)$, $e\varphi=\omega_{e\epsilon_1,\epsilon_2}$ has norm $\|e\epsilon_1\|=2^{-1/2}$, and so does $(1-e)\varphi$: $\tfrac12+\tfrac12=1$. $\square$

Exercise 15.2. (medium) Let $\varphi\in M_*^+$ and let $u\in M$ be unitary. Show that $|u\varphi|=\varphi$ and $|\varphi u|=u^*\varphi u$, where $(u^*\varphi u)(x)=\varphi(uxu^*)$. Deduce that $\||\varphi u|-|u\varphi|\|$ can be as large as $2\|\varphi\|$.

Solution. $u\varphi=(us(\varphi))\varphi$, and $w=us(\varphi)$ is a partial isometry with $w^*w=s(\varphi)$. By uniqueness in Theorem 2.2, $|u\varphi|=\varphi$. Next, $\psi=u^*\varphi u$ is positive with support $u^*s(\varphi)u$, since $\psi(x)=\varphi(uxu^*)$ and conjugation by $u$ preserves the order of projections. Put $w'=s(\varphi)u$, so $w'^*w'=u^*s(\varphi)u=s(\psi)$. Then $\begin{gathered} (w'\psi)(x)\\ =\psi(xs(\varphi)u)\\ =\varphi(uxs(\varphi))\\ =\varphi(ux)\\ =(\varphi u)(x), \end{gathered}$ using $\varphi=s(\varphi)\varphi$. So $|\varphi u|=\psi$. For $M=B(H)$, $\varphi=\omega_\xi$ with a unit vector $\xi$, and a unitary $u$ with $u^*\xi\perp\xi$: $|\varphi u|=\omega_{u^*\xi}$ and $|u\varphi|=\omega_\xi$, and $\|\omega_{u^*\xi}-\omega_\xi\|=2$ by Corollary 2.8(4), as the supports are orthogonal. $\square$

Exercise 15.3. (medium) Show that a subset $K$ of $\ell^1(\Gamma)$ is relatively weakly compact exactly when it is bounded and for every $\varepsilon>0$ there is a finite $F\subseteq\Gamma$ with $\sum_{\gamma\notin F}|g(\gamma)|<\varepsilon$ for all $g\in K$. Deduce that in $\ell^1(\Gamma)$ the relatively weakly compact sets are the relatively norm compact sets.

Solution. Suppose $K$ is bounded with uniformly small tails. Given $\varepsilon$, take $F$ as stated. The truncations $g1_F$ form a bounded set in the finite-dimensional space $\mathbb C^F$, so finitely many $\varepsilon$-balls cover them, and the $2\varepsilon$-balls with the same centres cover $K$. So $K$ is totally bounded, hence relatively norm compact, hence relatively weakly compact. Conversely let $K$ be relatively weakly compact. By the Eberlein–Šmulian theorem every sequence in $K$ has a weakly convergent subsequence, which converges in norm by Schur's theorem (Corollary 7.2). If $K$ were not totally bounded, an inductively chosen sequence of points at mutual distance at least some $\varepsilon>0$ would have no norm-convergent subsequence, a contradiction. Its norm closure is complete and totally bounded, hence compact by the metric lemma above. Thus $K$ is relatively norm compact. Cover $K$ by finitely many $\varepsilon$-balls with centres $g_1,\dots,g_m$, and choose a finite $F$ with $\sum_{\gamma\notin F}|g_l(\gamma)|<\varepsilon$ for all $l$. Then $\sum_{\gamma\notin F}|g(\gamma)|<2\varepsilon$ for every $g\in K$. The two parts together prove the last statement. $\square$

Exercise 15.4. (medium) Show that every bounded subset of $M_*$ is relatively weakly compact exactly when $M$ is finite-dimensional.

Solution. If $\dim M<\infty$, then $M_*$ is finite-dimensional and bounded sets are relatively compact. If $\dim M=\infty$, take an infinite sequence $(e_n)$ of mutually orthogonal nonzero projections (background fact 25) and normal states $\omega_n$ with $\omega_n(e_n)=1$ (as in the proof of Proposition 11.4). The bounded set $K=\{\omega_n\}$ violates condition (7) of Theorem 10.2, since $\sup_{\varphi\in K}\varphi(e_n)\ge1$ for all $n$. $\square$

Exercise 15.5. (medium) Let $S$ be the unilateral shift on $\ell^2=\ell^2(\{0,1,2,\dots\})$, $S\delta_k=\delta_{k+1}$, and $x_n=(S^*)^n$. Show that $x_n\to0$ $\sigma$-strongly but not in the topology $\tau$ of Definition 11.1, by exhibiting a relatively weakly compact $K\subseteq B(\ell^2)_*$ with $p_K(x_n)\ge1$ for all $n$. Why does this not contradict Theorem 11.2?

Solution. $(S^*)^n\delta_k=\delta_{k-n}$ for $k\ge n$ and $0$ otherwise, so $\|x_n\xi\|^2=\sum_{k\ge n}|\xi_k|^2\to0$: $x_n\to0$ strongly, hence $\sigma$-strongly, as $\|x_n\|\le1$. Let $\psi_n=\omega_{\delta_n,\delta_0}$, so $\psi_n(x)=\langle x\delta_n,\delta_0\rangle$ and $\psi_n(x_n)=\langle\delta_0,\delta_0\rangle=1$. By Bessel's inequality, as in Example 10.7(1), $\sum_n|\langle x\delta_0,\delta_n\rangle|^2\le\|x\delta_0\|^2$, so the functionals $\omega_{\delta_0,\delta_n}$ tend weakly to $0$; their adjoints $\psi_n$ do too, as the adjoint map is weakly continuous. So $K=\{\psi_n\}\cup\{0\}$ is weakly compact, and $p_K(x_n)\ge1$. There is no contradiction: $x_n^*=S^n$ is an isometry, so $\|x_n^*\delta_0\|=1$ and $x_n\not\to0$ $\sigma$-strongly$^*$. Theorem 11.2 compares $\tau$ with the $\sigma$-strong$^*$ topology, not with the $\sigma$-strong one. $\square$

References

[Kostecki] R. P. Kostecki, W*-algebras and noncommutative integration, lecture notes, arXiv:1307.4818 (version 5, 2014), https://arxiv.org/abs/1307.4818v5.
[Blackadar] B. Blackadar, Operator Algebras: Theory of C*-Algebras and von Neumann Algebras, revised author edition, 8 February 2017.

Freely accessible reading: Jan Hamhalter; Ondřej F. K. Kalenda; Antonio M. Peralta; Hermann Pfitzner, Measures of weak non-compactness in preduals of von Neumann algebras and JBW*-triples, §11, Theorems 11.1 and 11.3 gives a route through modern weak-noncompactness results, including JBW*-triples; these pages cite the foundational criterion, which is proved here. The lesson includes its own complete proofs at the stated hypotheses; references to human sources do not imply permission to adapt their expression.

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