The Spectral Scale of a Self-Adjoint Operator in a Semifinite von Neumann Algebra

Abstract

We extend Akemann, Anderson, and Weaver's Spectral Scale definition to include selfadjoint operators from semifinite von Neumann algebras. New illustrations of spectral scales in both the finite and semifinite von Neumann settings are presented. A counterexample to a conjecture made by Akemann concerning normal operators and the geometry of the their perspective spectral scales (in the finite setting) is offered.

1. Introduction

The notion of a “joint spectrum” for an 𝑛-tuple of operators was created and developed through the 1970s and work continues today. For an account of the various approaches taken and the achievements made see [1, page 3]. One attempt at a geometric interpretation of the “joint spectrum” of an 𝑛-tuple of self-adjoint operators is developed in spectral scale [2–6].

Definition 1.1. Let ℳ be a finite von Neumann algebra equipped with a finite, faithful, normal, trace 𝜏 such that 𝜏(1)=1, and let 𝑏=𝑏₁+𝑖𝑏₂∈ℳ with 𝑏₁ and 𝑏₂ self-adjoint. Define Ψ∶ℳ⟶ℝ³ (1.1) by Ψ(𝑐)=(𝜏(𝑐),𝜏(𝑏₁𝑐),𝜏(𝑏₂𝑐)) for all 𝑐∈ℳ. The spectral scale of 𝑏 with respect to 𝜏 is the set 𝐵(𝑏)=Ψℳ⁺₁=𝜏(𝑐),𝜏𝑏₁𝑐,𝜏𝑏₂𝑐∶0≤𝑐≤1. (1.2)

The spectral scale of an 𝑛-tuple of self-adjoint operators is defined analogously (in the canonical fashion), becoming a subset of ℝ^𝑛+1.

The set 𝐵(𝑏) has proven useful in gaining insight on problems in operator theory and operator algebras. In 2006, for a self-adjoint operator 𝑏 in a finite factor ℳ (i.e., 𝜏(1)<∞ and ℳ has trivial center), Akemann and Sherman [7] were able to use 𝐵(𝑏) to characterize the 𝑤^∗-closure of the unitary orbit 𝒰(𝑏)={𝑢𝑏𝑢^∗∶𝑢isunitaryinℳ} of 𝑏. More precisely, they showed that 𝒰(𝑏)^𝑤∗=𝒞(𝑏)={𝑑∈ℳ∶𝐵(𝑑)⊂𝐵(𝑏)}. Unfortunately, the spectral scale has only been developed for finite von Neumann algebras. In 2010, Wills [6] was able to extend the definition of 𝐵(𝑏) to include the class of unbounded operators.

The purpose of the work that follows to develop 𝐵(𝑏) for 𝑏=𝑏^∗∈ℳ, where ℳ is a semifinite von Neumann algebra, equipped with a semifinite, faithful, normal trace 𝜏 such that 𝜏(1)=∞. We will show that 𝐵(𝑏) is an unbounded subset of ℝ² which may or may not be closed, and that 𝐵(𝑏) gives a complete description of elements in the spectrum 𝜎(𝑏) of 𝑏 that lie outside the open interval determined by the minimum and maximum elements of the essential spectrum ℰ of 𝑏.

We begin with a summary of “how to read” 𝐵(𝑏) in the finite von Neumann algebra setting.

2. Geometric Interpretations of 𝐵(𝑏) When ℳ Is a Finite von Neumann Algebra

The following is a summary of the main results concerning the geometry of 𝐵(𝑏) [2–5].

Theorem 2.1. Let ℳ, 𝜏, 𝑏, Ψ, and 𝐵(𝑏) be as given in Definition 1.1. Then the following hold.(i)𝐵(𝑏) is a compact, convex subset of ℝ³ which is symmetric about the point (1/2,𝑏₁/2,𝑏₂/2). If 𝑏=𝑏^∗, then 𝐵(𝑏) can be identified with a subset of ℝ².(ii)𝐵(𝑏) gives a complete description of 𝜎(𝑡₁𝑏₁+𝑡₂𝑏₂) for all 𝑡₁,𝑡₂∈ℝ.(iii)𝜏(𝑏) can be read from the point Ψ(1).(iv)If 𝑏=𝑏^∗, then the set of all slopes of lines tangent to the boundary of 𝐵(𝑏) is equal to 𝜎(𝑏).(v)If 𝑏=𝑏^∗, then corners in the boundary of 𝐵(𝑏) correspond to gaps in 𝜎(𝑏).(vi)If 𝑏=𝑏^∗, then the extreme points of 𝐵(𝑏) are images under Ψ of spectral projections of 𝑏.(vii)If 𝑏≠𝑏^∗, then the extreme points of 𝐵(𝑏) are images under Ψ of spectral projections of the operators 𝑡₁𝑏₁+𝑡₂𝑏₂ for various 𝑡₁,𝑡₂∈ℝ.(viii)If 𝑏 is normal, then the set of all complex slopes of 1-dimensional faces of 𝐵(𝑏) (i.e., all the 1-dimensional “edges” in the boundary) is equal to 𝜎_𝑝(𝑏).(ix)The Numerical Range 𝑊(𝑏) is equal to the set of all slopes (complex when 𝑏≠𝑏^∗) of line segments emanating from the origin to points in 𝐵(𝑏).(x)If dim(ℳ)=𝑛, then for each 1≤𝑘≤𝑛 the cross-section of 𝐵(𝑏) taken at 𝑥=𝑘/𝑛 is isomorphic to the 𝑘-numerical range 𝑊_𝑘(𝑏)=1𝑘^𝑘_𝑖=1𝑏𝑥_𝑖,𝑥_𝑖∶𝑥_𝑖isanorthonormalset,1≤𝑘≤dim(ℳ). (2.1)

We finish this section with a few examples of spectral scales in the finite setting.

Example 2.2. Let ℳ=𝑀₅(ℂ) with 𝜏(𝑥)=(1/5)tr(𝑥) for all 𝑥∈ℳ, where tr is the canonical trace on 𝑀₅(ℂ). Let 𝑏=100000100000000000−100000−10. (2.2)𝐵(𝑏) is given in Figure 1. The MATLAB program SpecGUI [8] was used to generate Figure 1, and by Theorem 2.1 one can see that 𝜎(𝑏)={−10,−1,0,1}, ‖𝑏‖=10, 1 has multiplicity 2, 𝑊(𝑏)=[−10,1], and 𝜏(𝑏)=−9/5.

Figure 1 

𝐵(𝑏)—Example 2.2.

Example 2.3. Let ℳ=𝐿^∞([0,1],𝑚), where 𝑚 is Lebesgue measure on the interval [0,1], let 𝜏(𝑐)=∫¹₀𝑐𝑑𝑚 for all 𝑐∈ℳ, and let 𝑏∈ℳ be defined by 𝑏(𝑥)=𝑥 for all 𝑥∈[0,1]. Then 𝑏 is positive, with no eigenvalues, and 𝜎(𝑏)=[0,1]. 𝐵(𝑏) is given in Figure 2.

Figure 2 

𝐵(𝑏)—Example 2.3.

Example 2.4. Let ℳ=𝑀₃(ℂ) with 𝜏(𝑐)=(1/3)tr(𝑐) for all 𝑐∈ℳ, and let 𝑏=1000𝑖0002𝑖. (2.3)𝐵(𝑏) is given in Figure 3 [8].

Figure 3 

𝐵(𝑏)—Example 2.4.

Since 𝑏 is normal, by Theorem 2.1 we know that the eigenvalues of 𝑏 can be read from 𝐵(𝑏) as the slopes (complex) of 1-dimensional faces of 𝐵(𝑏) (i.e., slopes of 1-dimensional “edges” in the boundary of 𝐵(𝑏)). Alternatively, if we were given 𝐵(𝑏) without knowing what 𝑏 actually was, we could conclude by [3, Theorem 6.1] that 𝑏 is normal. The numerical range 𝑊(𝑏) of 𝑏 can be seen as the slopes (complex) of secant lines between the origin and points in 𝐵(𝑏).

Our last example is a counterexample to conjecture made by Akemann in 2005 which stated: 𝑏 is normal if the cross-section of 𝐵(𝑏) taken at 𝑥=1/2 is a polygon. The following is a spectral scale that has no 1-dimensional faces, and the cross-section taken 𝑥=1/2 is not a polygon, yet the operator it corresponds to is normal.

Example 2.5. Let 𝒞 be the unit circle, and let 𝑚 denote Lebesgue measure on 𝒞. Put ℳ=𝐿^∞(𝒞), and note that ℳ is a finite von Neumann algebra equipped with the finite, faithful, normal trace 𝜏∶𝐿^∞(𝒞)→𝐑 defined by 𝜏(𝑐)=(1/2𝜋)∫_𝒞𝑐𝑑𝑚 for all 𝑐∈𝐿^∞(𝒞).

Consider the function 𝑏(𝑧)=𝑧 on 𝒞. The spectral scale of 𝑏 with respect to 𝜏 is the collection of points𝐵(𝑏)=(𝜏(𝑐),𝜏(𝑏𝑐))∶𝑐∈𝐿^∞(𝒞)⁺₁ (2.4) in ℝ³. From [2, Theorem 2.3] we know that Ψ applied to spectral projections of the operators 𝑏_𝐭=𝑡₁Re(𝑏)+𝑡₂Im(𝑏) for various 𝐭=(𝑡₁,𝑡₂)∈ℝ² are extreme points of 𝐵(𝑏).

Let 𝜇∈[0,2𝜋), and let 𝜃∈[0,𝜋]. If 𝜒_{𝒞𝜇,𝜃} is the characteristic function on the arc 𝒞_𝜇,𝜃 from 𝑒^{𝑖(𝜇−𝜃)} to 𝑒^{𝑖(𝜇+𝜃)}, then, 𝜒_{𝒞𝜇,𝜃}(𝑧)=𝜒_{[cos(𝜃),∞)}Re𝑒^−𝑖𝜇𝑧=𝜒_{[cos(𝜃),∞)}cos𝜇𝑥−sin𝜇𝑦 (2.5) for every 𝑧=𝑥+𝑖𝑦∈ℂ (to see this think of the case where 𝜇=0). But 𝜒_{[cos(𝜃),∞)}cos𝜇𝑥−sin𝜇𝑦=𝜒_{[cos(𝜃),∞)}𝑡₁Re(𝑏)+𝑡₂Im(𝑏)=𝜒_{[cos(𝜃),∞)}𝑏_𝐭 (2.6) for 𝐭=(𝑡₁,𝑡₂)=(cos(𝜇),−sin(𝜇)). Therefore, every 𝜒_{𝒞𝜇,𝜃} is a spectral projection of some 𝑏_𝐭. Conversely, if 𝐭=(𝑡₁,𝑡₂)∈ℝ² and 𝑠∈ℝ, then for if 𝜇 is chosen so that cos(𝜇)=𝑡₁/𝑡²₁+𝑡²₂ and sin(𝜇)=−𝑡₂/𝑡²₁+𝑡²₂ and 𝜃 is chosen so that ‖𝐭‖cos(𝜃)=𝑠, then we have 𝜒_[𝑠,∞)𝑏_𝐭=𝜒_{[‖𝐭‖cos(𝜃),∞)}‖𝐭‖cos𝜇Re(𝑏)−‖𝐭‖sin𝜇Im(𝑏)=𝜒_{[cos(𝜃),∞)}cos𝜇Re(𝑏)−sin𝜇Im(𝑏)=𝜒_{𝒞𝜇,𝜃}. (2.7)

Since the extreme points of 𝐵(𝑏) come from the spectral projections described above, we can calculate the extreme points of 𝐵(𝑏) explicitly:𝜏𝜒_{𝒞𝜇,𝜃},𝜏𝑏𝜒_{𝒞𝜇,𝜃}=_{𝒞𝜇,𝜃}1𝑑𝑚,_{𝒞𝜇,𝜃}=𝜃𝑏𝑑𝑚𝜋,_{𝒞𝜇,𝜃}𝑒^𝑖𝑧=𝜃𝑑𝑧𝜋,1𝑒2𝜋𝑖^{𝑖(𝜇+𝜃)}−𝑒^{𝑖(𝜇−𝜃)}=𝜃𝜋,1𝜋𝑒^𝑖𝜇sin(𝜃). (2.8) Since 𝐵(𝑏) is the convex hull of its extreme points, we see that 𝐵(𝑏) can be recovered by rotating the function 𝑓(𝑥)=(1/𝜋)sin(𝜋𝑥)0≤𝑥≤1 about the 𝑥-axis (see Figure 4).

Figure 4 

𝐵(𝑏)—Example 2.5.

3. The Spectral Scale of a Self-Adjoint Operator in a Semifinite von Neumann Algebra

For the remainder of what follows, let ℳ be a semifinite von Neumann algebra, equipped with a faithful, semifinite, normal trace 𝜏, and let 𝑏∈ℳ be self-adjoint. In order to avoid the finite setting (for which the spectral scale has already been developed), we also assume that 𝜏(1)=∞.

Define 𝔫_𝜏={𝑥∈ℳ∶𝜏(𝑥^∗𝑥)<+∞} and 𝔪_𝜏={∑^𝑛_𝑖=1𝑥_𝑖𝑦_𝑖∶𝑥_𝑖,𝑦_𝑖∈𝔫_𝜏}. The by [9, Lemma 2.16, page 318], we can define the function Ψ∶𝔪_𝜏→ℝ² byΨ(𝑐)=(𝜏(𝑐),𝜏(𝑏𝑐))forevery𝑐∈𝔪_𝜏. (3.1)

Definition 3.1. The spectral scale of 𝑏 (with respect to 𝜏) is the set 𝐵(𝑏)=Ψ𝔪_𝜏⁺₁=(𝜏(𝑐),𝜏(𝑏𝑐))∶𝑐∈𝔪_𝜏⁺₁. (3.2)

Since (𝔪_𝜏)⁺₁ is convex, Ψ is linear, and 𝜏 is positive, we have that 𝐵(𝑏) is a convex subset of ℝ² contained in the right half-plane. Since 𝜏 is semifinite and 𝜏(1)=∞, 𝐵(𝑏) is an unbounded subset of ℝ². We will see that 𝐵(𝑏) may or may not be closed.

Note 1. If 𝑝 and 𝑞 are projections in ℳ and 𝑝≤𝑞 then we write 𝑝,𝑞=𝑎∶𝑝≤𝑎≤𝑞 (3.3) for the order interval determined by 𝑝 and 𝑞. Given a real number 𝑠, we let 𝑝⁺_𝑠 denote the spectral projection of 𝑏 corresponding to the interval (−∞,𝑠]. Similarly, we let 𝑝⁻_𝑠 denote the spectral projection of 𝑏 corresponding to the interval (−∞,𝑠). We use 𝑝^±_𝑠 to indicate either of these projections and we write 𝑝_𝑠 when 𝑝⁺_𝑠=𝑝⁻_𝑠. We let 𝑞⁺_𝑠 and 𝑞⁻_𝑠 denote the spectral projections corresponding to the intervals [𝑠,∞) and (𝑠,∞), respectively. We also use 𝑞^±_𝑠 to indicate either of these projections and we write 𝑞_𝑠 when 𝑞⁺_𝑠=𝑞⁻_𝑠. Note that 𝑞^±_𝑠=1−𝑝^∓_𝑠 for every 𝑠∈ℝ. The spectral projections 𝑝^±_𝑠 and 𝑞^±_𝑠 will be used throughout the remainder of this text, and will be instrumental in deciphering 𝐵(𝑏).

The following lemma was proven in [2, Lemma 1.2] for any type of von Neumann algebra.

Lemma 3.2. Let 𝑏 be a self-adjoint element in ℳ,𝑠∈ℝ, 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠], and 𝑎∈ℳ⁺₁. Then the following statements hold:(1)(𝑏−𝑠1)(1−𝑐)=(𝑏−𝑠1)(1−𝑝⁺_𝑠)≥0 and the range projection of (𝑏−𝑠1)(1−𝑝⁺_𝑠)is1−𝑝⁺_𝑠;(2)if 𝑎^1/2(𝑏−𝑠1)(1−𝑐)𝑎^1/2=0, then 𝑎≤𝑝⁺_𝑠;(3)(𝑠1−𝑏)𝑐=(𝑠1−𝑏)𝑝⁻_𝑠≥0 and the range projection of (𝑠1−𝑏)𝑝⁻_𝑠 is 𝑝⁻_𝑠;(4)if (1−𝑎)^1/2(𝑠1−𝑏)𝑐(1−𝑎)^1/2=0 then 𝑎≥𝑝⁻_𝑠.

The following two lemmas tell us that points of the form (𝜏(𝑐),𝜏(𝑐𝑏)) lie on the boundary when 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠] (or 𝑐∈[𝑞⁻_𝑠,𝑞⁺_𝑠]) for some 𝑠∈ℝ. A slight variation of these lemmas was proved in [2, Lemma 1.3].

Lemma 3.3. If 𝑏∈ℳ is self-adjoint and 𝑠∈ℝ, then the following statements hold for each 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠]∩𝔪_𝜏 and 𝑎∈(𝔪_𝜏)⁺₁.(1)If 𝜏(𝑎)=𝜏(𝑐), then 𝜏(𝑏𝑎)≥𝜏(𝑏𝑐).(2)If 𝜏(𝑎)=𝜏(𝑐) and 𝜏(𝑏𝑎)=𝜏(𝑏𝑐), then 𝑎∈[𝑝⁻_𝑠,𝑝⁺_𝑠]; moreover, if =𝑝^±_𝑠, then equality in the second equation obtains only for =𝑐.

Proof. (1) From the previous lemma we have that (𝑏−𝑠1)(1−𝑐)=(𝑏−𝑠1)1−𝑝⁺_𝑠≥0,sothat𝑏(1−𝑐)≥𝑠(1−𝑐). (3.4) Therefore 𝑎^1/2𝑏(1−𝑐)𝑎^1/2≥𝑎^1/2𝑠(1−𝑐)𝑎^1/2. (3.5) Note that 𝜏(𝑏(1−𝑐)𝑎)<∞ since 𝑎∈𝔪_𝜏 and 𝔪_𝜏 is an ideal. Since 𝑎^1/2∈𝔫_𝜏 and 𝜏 is order preserving on 𝔪_𝜏 [6], we have that 𝜏(𝑏(1−𝑐)𝑎)=𝜏𝑎^1/2𝑏(1−𝑐)𝑎^1/2≥𝜏𝑎^1/2𝑠(1−𝑐)𝑎^1/2=𝜏(𝑠(1−𝑐)𝑎). (∗) Since 𝑐∈𝔪_𝜏, an identical argument yields 𝜏(𝑏𝑐(1−𝑎))≤𝜏(𝑠𝑐(1−𝑎)). (∗∗) It follows from (*), (**), and our hypothesis that 𝜏()()()(𝑏()𝑎)(((𝑠()𝑎)((()()(𝜏(𝑎)(𝑐𝑏𝑎−𝜏𝑏𝑐=𝜏𝑏𝑎−𝑏𝑐𝑎−𝑏𝑐+𝑏𝑐𝑎=𝜏1−𝑐−𝜏𝑏𝑐1−𝑎))≥𝜏1−𝑐−𝜏𝑠𝑐1−𝑎))=𝜏𝑠𝑎−𝜏𝑠𝑐=𝑠−𝜏))=0, (3.6) and so 𝜏(𝑏𝑎)≥𝜏(𝑏𝑐).
To prove (2), suppose that 𝜏(𝑎)=𝜏(𝑐) and 𝜏(𝑏𝑎)=𝜏(𝑏𝑐). Then𝜏)𝑎)()()(𝑐(,𝜏(𝑏()𝑎)()()((.((1−𝑐=𝜏𝑎−𝑐𝑎=𝜏𝑐−𝑐𝑎=𝜏1−𝑎))1−𝑐=𝜏𝑏𝑎−𝑏𝑐𝑎=𝜏𝑏𝑐−𝑏𝑐𝑎=𝜏𝑏𝑐1−𝑎)) (3.7) Combining these two results with (*) and (**) we get that 𝜏(𝑏𝑐(1−𝑎))=𝜏(𝑏(1−𝑐)𝑎)≥𝑠𝜏((1−𝑐)𝑎)=𝑠𝜏(𝑐(1−𝑎))≥𝜏(𝑏𝑐(1−𝑎)). (3.8) Therefore, all the above terms must be equal, and so 𝜏(𝑏()𝑎))𝑎),𝜏(((𝑐(.1−𝑐=𝑠𝜏((1−𝑐𝑏𝑐1−𝑎))=𝑠𝜏1−𝑎)) (3.9)
From this it follows that𝜏𝑎^1/2(𝑏−𝑠1)(1−𝑐)𝑎^1/2𝜏()=0,1−𝑎^1/2(𝑏−𝑠1)𝑐(1−𝑎)^1/2=0. (3.10)
Since 𝜏 is faithful we get that𝑎^1/2(𝑏−𝑠1)(1−𝑐)𝑎^1/2()=0,1−𝑎^1/2(𝑏−𝑠1)𝑐(1−𝑎)^1/2=0. (3.11) By parts (2) and (4) of the previous lemma, 𝑎∈[𝑝⁻_𝑠,𝑝⁺_𝑠].
If it is the case that 𝑐=𝑝^±_𝑠, 𝜏(𝑎)=𝜏(𝑝^±_𝑠), and 𝜏(𝑏𝑎)=𝜏(𝑏𝑐) then by what we just showed, 𝑎∈[𝑝⁻_𝑠,𝑝⁺_𝑠], so 𝜏(𝑝^±_𝑠−𝑎)=0, which implies that 𝑎=𝑝^±_𝑠 since 𝜏 is faithful.

Remark 3.4. Notice the importance of our choice of 𝑎 and 𝑐 as elements of ideal 𝔪_𝜏 in the above proof. Without this assumption, many of the equalities and inequalities above would have been either trivial or undefined because 𝜏 is not necessarily finite on all of ℳ.

Lemma 3.5. If 𝑏∈ℳ is self-adjoint and 𝑡∈ℝ, then the following statements hold for each 𝑑∈[𝑞⁻_𝑡,𝑞⁺_𝑡]∩𝔪_𝜏 and 𝑎∈(𝔪_𝜏)⁺₁.(1)If 𝜏(𝑎)=𝜏(𝑑), then 𝜏(𝑏𝑎)≤𝜏(𝑏𝑑).(2)If 𝜏(𝑎)=𝜏(𝑑) and 𝜏(𝑏𝑎)=𝜏(𝑏𝑑), then 𝑎∈[𝑞⁻_𝑡,𝑞⁺_𝑡]; moreover, if =𝑞^±_𝑡, then equality in the second equation obtains only for 𝑎=𝑑.

Proof. (1) Similarly, by Lemma 3.2 (since 1−𝑑∈[𝑝⁻_𝑡,𝑝⁺_𝑡]), we have that 𝑡()().𝑏𝑑≥𝑡𝑑,1−𝑑≥𝑏1−𝑑 (3.12) Therefore (1−𝑎)^1/2𝑏𝑑(1−𝑎)^1/2≥(1−𝑎)^1/2𝑡𝑑(1−𝑎)^1/2, (∗)𝑎^1/2𝑡(1−𝑑)𝑎^1/2≥𝑎^1/2𝑏(1−𝑑)𝑎^1/2. (∗∗) Using the fact that the trace is faithful and finite on 𝔪_𝜏, we get that 𝜏(𝑏𝑑(1−𝑎))=𝜏(1−𝑎)^1/2𝑏𝑑(1−𝑎)^1/2≥𝜏(1−𝑎)^1/2𝑡𝑑(1−𝑎)^1/2((,𝜏(𝑡()𝑎)𝑎=𝜏𝑡𝑑1−𝑎))1−𝑑=𝜏^1/2𝑡(1−𝑑)𝑎^1/2≥𝜏𝑎^1/2𝑏(1−𝑑)𝑎^1/2=𝜏(𝑏(1−𝑑)𝑎). (3.13) It follows that 𝜏()()(((𝑏()𝑎)(((𝑡()𝑎)()()𝑏𝑑−𝜏𝑏𝑎=𝜏𝑏𝑑1−𝑎))−𝜏1−𝑑≥𝜏𝑡𝑑1−𝑎))−𝜏1−𝑑=𝜏𝑡𝑑−𝜏𝑡𝑎=0. (3.14)
To prove (2), suppose that 𝜏(𝑎)=𝜏(𝑑) and 𝜏(𝑏𝑎)=𝜏(𝑏𝑑). Then𝜏)𝑎)()()(𝑑(,𝜏(𝑏()𝑎)()()((.((1−𝑑=𝜏𝑎−𝑑𝑎=𝜏𝑑−𝑑𝑎=𝜏1−𝑎))1−𝑑=𝜏𝑏𝑎−𝑏𝑑𝑎=𝜏𝑏𝑑−𝑏𝑑𝑎=𝜏𝑏𝑑1−𝑎)) (3.15)
From these two observations, (*), and (**) we get that 𝜏(𝑏𝑑(1−𝑎))=𝜏(𝑏(1−𝑑)𝑎)≤𝑡𝜏((1−𝑑)𝑎)=𝑡𝜏(𝑑(1−𝑎))≤𝜏(𝑏𝑑(1−𝑎)). (3.16) But then all these terms must be equal. Therefore 𝜏(𝑏(1−𝑑)𝑎)=𝑡𝜏((1−𝑑)𝑎) and 𝜏(𝑏𝑑(1−𝑎))=𝑡𝜏(𝑑(1−𝑎)). It follows that 𝜏𝑎^1/2(𝑏−𝑡1)(1−𝑑)𝑎^1/2𝜏()=0,1−𝑎^1/2(𝑏−𝑡1)𝑑(1−𝑎)^1/2=0. (3.17) But since (𝑏−𝑡1)(1−𝑑)≥0 and (𝑏−𝑡1)𝑑≥0 we get 𝑎^1/2(𝑏−𝑡1)(1−𝑑)𝑎^1/2()=0,1−𝑎^1/2(𝑏−𝑡1)𝑑(1−𝑎)^1/2=0. (3.18) By Lemma 3.2 we get 𝑎∈[𝑞⁻_𝑡,𝑞⁺_𝑡]. If 𝑑=𝑞^±_𝑡, then since 𝜏 is faithful we must have 𝑎=𝑑=𝑞^±_𝑡.

Notice that this is one major difference between 𝐵(𝑏) in the case where ℳ is finite, and 𝐵(𝑏) when ℳ is semifinite. In the finite setting 𝐵(𝑏) was convex and compact. Next, we define the boundary of 𝐵(𝑏), and then study some of its properties.

Definition 3.6. Let 𝐵(𝑏) denote the closure of 𝐵(𝑏) in ℝ². The lower boundary of 𝐵(𝑏) is defined as 𝜕_ℒ𝐵(𝑏)=𝑥,𝑦∈𝐵(𝑏)∶𝑥,𝑦^′∈𝐵(𝑏)⟹𝑦^′≥𝑦. (3.19) The upper boundary of 𝐵(𝑏) is defined analogously as 𝜕_𝒰𝐵(𝑏)=𝑥,𝑦∈𝐵(𝑏)∶𝑥,𝑦^′∈𝐵(𝑏)⟹𝑦^′≤𝑦. (3.20) Since 𝐵(𝑏) is a closed convex set, and 𝜏 takes on every value in [0,∞], we let 𝑓∶[0,∞)⟶ℝ (3.21) defined by 𝑓(𝑥)=𝑦 for each (𝑥,𝑦)∈𝜕_ℒ𝐵(𝑏) denote the lower boundary function determined by b. Similarly, let 𝑔∶[0,∞)⟶ℝ (3.22) defined by 𝑔(𝑥)=𝑦 for all (𝑥,𝑦)∈𝜕_𝒰𝐵(𝑏) denote the upper boundary function determined by b.
It is clear that 𝑓 is a convex function and 𝑔 is a concave function since 𝐵(𝑏) is convex.

Theorems 3.7 and 3.8 describe the possibilities for the boundary of 𝐵(𝑏). These theorems describe the extreme points 𝐵(𝑏), the line segments in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) and 𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏), respectively, and the rays in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) and 𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏) (resp.). Rays contained in 𝜕_ℒ𝐵(𝑏)⧵𝐵(𝑏) and 𝜕_𝒰𝐵(𝑏)⧵𝐵(𝑏) will be described in Theorems 3.9 and 3.10 (resp.). Differentiability of the functions 𝑓 and 𝑔 will be discussed in Theorems 3.12 and 3.13 (resp.).

Note 2. For a convex set 𝒞 we let Ext(𝒞) denote the set of extreme points of 𝒞.

Theorem 3.7. Let 𝑎∈(𝔪_𝜏)⁺₁. Then one has the following. (1)If 𝑠∈𝜎(𝑏) with 𝑝^±_𝑠∈𝔪_𝜏, then Ψ(𝑝^±_𝑠)∈Ext(𝐵(𝑏))∩𝜕_ℒ𝐵(𝑏).(2)If 𝑠∈𝜎_𝑝(𝑏) with 𝑝⁺_𝑠∈𝔪_𝜏, then Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) is a line segment in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) having slope 𝑠.(3)If Ψ(𝑎)∈𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) and Ψ(𝑎)≠Ψ(𝑝^±_𝑠) for all 𝑠∈ℝ, then 𝑎∈[𝑝⁻_𝑠,𝑝⁺_𝑠] for some 𝑠∈𝜎(𝑏) with 𝜏(𝑝⁻_𝑠)<∞.(4)If Ψ(𝑎)∈Ext(𝐵(𝑏))∩𝜕_ℒ𝐵(𝑏), then 𝑎=𝑝^±_𝑠 for some 𝑠∈𝜎_𝑝(𝑏) with 𝑝^±_𝑠∈𝔪_𝜏.(5)ℛ is a ray in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) having slope 𝑠 if and only if ℛ=Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) and 𝑠∈𝜎_𝑝(𝑏) with 𝜏(𝑝⁻_𝑠)<𝜏(𝑝⁺_𝑠)=∞.

Proof. (1) Fix 𝑠∈𝜎(𝑏) such that 𝜏(𝑝^±_𝑠)<∞. Then by part (1) of Lemma 3.3 we have that Ψ𝑝^±_𝑠∈𝜕_ℒ𝐵(𝑏). (3.23) By part (2) of Lemma 3.3  we have that Ψ⁻¹Ψ𝑝^±_𝑠∩𝔪_𝜏⁺₁=𝑝^±_𝑠. (3.24) To see that Ψ(𝑝^±_𝑠) are extreme in 𝐵(𝑏), suppose that Ψ𝑝^±_𝑠=𝜆Ψ𝑎₁+(1−𝜆)Ψ𝑎₂ (3.25) for some 𝑎₁,𝑎₂∈(𝔪_𝜏)⁺₁and 0<𝜆<1. But then Ψ𝑝^±_𝑠=Ψ𝜆𝑎₁+(1−𝜆)𝑎₂, (3.26) and by part (2) of Lemma 3.3 we have that 𝑝^±_𝑠=𝜆𝑎₁+(1−𝜆)𝑎₂. Since projections are extreme points of ℳ⁺₁, we must have 𝑎₁=𝑎₂=𝑝^±_𝑠. Therefore, Ψ(𝑝^±_𝑠)∈Ext(𝐵(𝑏))∩𝜕_ℒ𝐵(𝑏).
(2) Fix 𝑠∈𝜎_𝑝(𝑏) so that 𝑝⁺_𝑠∈𝔪_𝜏. Then we have that 𝑝⁻_𝑠<𝑝⁺_𝑠,Ψ(𝑝⁻_𝑠)≠Ψ(𝑝⁺_𝑠), and 𝑏(𝑝⁺_𝑠−𝑝⁻_𝑠)=𝑠(𝑝⁺_𝑠−𝑝⁻_𝑠). Note that if 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠], then Ψ(𝑐)∈𝜕_ℒ𝐵(𝑏) by part (1) of Lemma 3.3, soΨ𝑝⁻_𝑠,𝑝⁺_𝑠⊂𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏). (3.27) For each 𝜆∈(0,1) let 𝑝_𝜆=𝜆𝑝⁻_𝑠+(1−𝜆)𝑝⁺_𝑠. Then, each 𝑝_𝜆 is in [𝑝⁻_𝑠,𝑝⁺_𝑠], and so Ψ(𝑝_𝜆)∈𝜕𝐵_ℒ(𝑏)∩𝐵(𝑏). Furthermore, Ψ(𝑝_𝜆) is a typical point on the line segment between Ψ(𝑝⁻_𝑠) and Ψ(𝑝⁺_𝑠) and the slope of this line segment is 𝜏𝑏𝑝⁺_𝑠−𝜏𝑏𝑝⁻_𝑠𝜏𝑝⁺_𝑠−𝜏𝑝⁻_𝑠=𝜏𝑏𝑝⁺_𝑠−𝑏𝑝⁻_𝑠𝜏𝑝⁺_𝑠−𝑝⁻_𝑠=𝜏𝑠𝑝⁺_𝑠−𝑝⁻_𝑠𝜏𝑝⁺_𝑠−𝑝⁻_𝑠=𝑠. (3.28) Therefore Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) is a line segment in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) with slope 𝑠. Note that since Ψ(𝑝^±_𝑠) are extreme points of 𝐵(𝑏) contained in 𝜕_ℒ𝐵(𝑏), we know that the endpoints of any line segment containing Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) must be Ψ(𝑝⁻_𝑠) and Ψ(𝑝⁺_𝑠).
(3) Suppose Ψ(𝑎)∈𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) and Ψ(𝑎)≠Ψ(𝑝^±_𝑠) for all 𝑠∈𝜎(𝑏). We will show that 𝑝⁻_𝑠<𝑎<𝑝⁺_𝑠 for some 𝑠∈𝜎(𝑏) with 𝜏(𝑝⁻_𝑠)<∞. Let𝑟₁=sup𝑠∈𝜎(𝑏)∶𝜏𝑝⁻_𝑠(𝑎,𝑟≤𝜏)₂=inf𝑠∈𝜎(𝑏)∶𝜏(𝑎)≤𝜏𝑝⁺_𝑠. (3.29) We want to show that 𝑟₁=𝑟₂. Note that 𝑟₁ and 𝑟₂ are in 𝜎(𝑏) since 𝜎(𝑏) is closed, and 𝜏(𝑝⁻_𝑟1)≤𝜏(𝑎) by lower semicontinuity of 𝜏.
We first show that 𝑟₁≤𝑟₂. If 𝑠,𝑡∈𝜎(𝑏) with 𝜏(𝑝⁻_𝑠)≤𝜏(𝑎)≤𝜏(𝑝⁺_𝑡), and 𝑡<𝑠, then by the positivity of 𝜏 we must also have 𝜏(𝑝⁺_𝑡)≤𝜏(𝑝⁻_𝑠), so 𝜏(𝑝⁻_𝑠)=𝜏(𝑎)=𝜏(𝑝⁺_𝑡). This fact, along with Lemma 3.3, contradicts our assumption about Ψ(𝑎). Thus, everything in {𝑠∈𝜎(𝑏)∶𝜏(𝑝⁻_𝑠)≤𝜏(𝑎)} is less than or equal to everything in {𝑠∈𝜎(𝑏)∶𝜏(𝑎)≤𝜏(𝑝⁺_𝑠)}, hence 𝑟₁≤𝑟₂.
Now, either 𝜏(𝑎)≤𝜏(𝑝⁺_𝑟2) or not. Suppose 𝜏(𝑝⁺_𝑟2)<𝜏(𝑎). Since 𝜏(𝑝⁻_𝑟2)≤𝜏(𝑝⁺_𝑟2)<𝜏(𝑎), and since 𝑟₁ is an upper bound for {𝑠∈𝜎(𝑏)∶𝜏(𝑝⁻_𝑠)≤𝜏(𝑎)}, we must have 𝑟₂≤𝑟₁. But 𝑟₁≤𝑟₂. Therefore 𝑟₁=𝑟₂ when 𝜏(𝑝⁺_𝑟2)<𝜏(𝑎).
Assume that 𝜏(𝑎)≤𝜏(𝑝⁺_𝑟2). Then, 𝜏(𝑝⁻_𝑟1)≤𝜏(𝑎)≤𝜏(𝑝⁺_𝑟2). But we are assuming that Ψ(𝑎)≠Ψ(𝑝^±_𝑠) for all 𝑠∈𝜎(𝑏), and that Ψ(𝑎)∈𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏). Therefore, by Lemma 3.3 we get𝜏𝑝⁻_𝑟1<𝜏(𝑎)<𝜏𝑝⁺_𝑟2. (3.30) Observe that, since 𝑟₂ is a lower bound for the set {𝑠∈𝜎(𝑏)∶𝜏(𝑎)≤𝜏(𝑝⁺_𝑠)}, if 𝑠<𝑟₂ and 𝑠∈𝜎(𝑏) then 𝜏𝑝⁺_𝑠<𝜏(𝑎)<𝜏𝑝⁺_𝑟2. (3.31) Similarly, if 𝑠∈𝜎(𝑏) with 𝑟₁<𝑠, then 𝜏𝑝⁻_𝑟1<𝜏(𝑎)<𝜏𝑝⁻_𝑠. (3.32) Suppose that 𝑟₁<𝑟₂. Then since 𝑟₁,𝑟₂∈𝜎(𝑏) we have that 𝜏𝑝⁺_𝑟1<𝜏(𝑎)<𝜏𝑝⁻_𝑟2. (3.33) On the other hand if 𝑠∈(𝑟₁,𝑟₂)∩𝜎(𝑏) then by (3.32) and (3.33) we would have 𝜏(𝑝⁺_𝑠)<𝜏(𝑎)<𝜏(𝑝⁻_𝑠). This is not possible, so it must be that (𝑟₁,𝑟₂)∩𝜎(𝑏)=∅. Therefore 𝑝⁺_𝑟1=𝑝⁻_𝑟2, and we have 𝜏(𝑝⁺_𝑟1)=𝜏(𝑝⁻_𝑟2), contradicting (3.33). Hence, it must be that 𝑟₁=𝑟₂=𝑟 and so 𝜏(𝑝⁻_𝑟)<𝜏(𝑎)<𝜏(𝑝⁺_𝑟). Therefore 𝑟∈𝜎_𝑝(𝑏) and 𝑎∈[𝑝⁻_𝑟,𝑝⁺_𝑟] by Lemma 3.3.
(4) Since 𝐵(𝑏) is convex, every point in the boundary of 𝐵(𝑏) must either be an extreme point, or lie on a flat spot (face). Therefore, (1), and (2), (3), imply (4).
(5) (⇐) Suppose 𝜏(𝑝⁻_𝑠)<𝜏(𝑝⁺_𝑠)=∞ for some 𝑠∈𝜎_𝑝(𝑏). We claim that ℛ=Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) is a ray in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) with slope 𝑠, emanating from the point Ψ(𝑝⁻_𝑠)=(𝜏(𝑝⁻_𝑠),𝜏(𝑏𝑝⁻_𝑠)). We will first show that every point in Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) lies on the ray through Ψ(𝑝⁻_𝑠) with slope 𝑠.
Let 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠]∩𝔪_𝜏, and write 𝑐=𝑝⁻_𝑠+𝑑 for some 𝑑∈(𝑝⁺_𝑠−𝑝⁻_𝑠)ℳ(𝑝⁺_𝑠−𝑝⁻_𝑠). Then 𝑏(𝑐−𝑝⁻_𝑠)=𝑏𝑑=𝑠𝑑=𝑠(𝑐−𝑝⁻_𝑠), so 𝜏(𝑏(𝑐−𝑝⁻_𝑠))=𝜏(𝑠(𝑐−𝑝⁻_𝑠)). This implies that𝜏(𝑏𝑐)=𝜏𝑠𝑐−𝑝⁻_𝑠+𝜏𝑏𝑝⁻_𝑠=𝑠⋅𝜏(𝑐)+𝜏(𝑏−𝑠1)𝑝⁻_𝑠. (3.34) Therefore Ψ(𝑐) lies on the ray with slope 𝑠 passing through the point (𝜏(𝑝⁻_𝑠),𝜏(𝑏𝑝⁻_𝑠)). Furthermore, by the convexity of 𝐵(𝑏), we know there can only be one ray having slope 𝑠 in the lower boundary of 𝐵(𝑏), and so if Ψ(𝑥) is any other point in 𝐵(𝑏) that lies on a ray with slope 𝑠 in the lower boundary, then we must have Ψ(𝑥)∈ℛ=Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]).
(5) (⇒) If ℛ is a ray in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) with slope 𝑠, emanating from Ψ(𝑥), then Ψ(𝑥) is an extreme point of 𝐵(𝑏) (if not, then ℛ does not emanate from Ψ(𝑥) since Ψ(𝑥) would then lie on a flat portion of 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏)), so we must have 𝑥=𝑝^±_𝑟∈𝔪_𝜏 for some 𝑟∈𝜎(𝑏) by part (4). But ℛ is a ray, so ℛ cannot contain any extreme points of 𝐵(𝑏). So by part (3) we know that if 𝑧∈(𝔪_𝜏)⁺₁ with Ψ(𝑧)∈ℛ, then 𝑧∈[𝑝⁻_𝑠𝑧,𝑝⁺_𝑠𝑧] for some 𝑠_𝑧∈𝜎_𝑝(𝑏) with 𝜏(𝑝⁻_𝑠𝑧)<∞. Furthermore, since any such Ψ([𝑝⁻_𝑠𝑧,𝑝⁺_𝑠𝑧]) is a line segment (or ray) with slope 𝑠_𝑧, and since Ψ([𝑝⁻_𝑠𝑧,𝑝⁺_𝑠𝑧])⊂ℛ, we must have 𝑠_𝑧=𝑠. Since there can only be one ray with slope 𝑠 in the lower boundary of 𝐵(𝑏), we may conclude that 𝑟=𝑠 and ℛ=Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]).

Theorem 3.8. Let 𝑎∈(𝔪_𝜏)⁺₁. Then one has the following.(1)If 𝑡∈𝜎(𝑏) with 𝑞^±_𝑡∈𝔪_𝜏, then Ψ(𝑞^±_𝑡)∈Ext(𝐵(𝑏))∩𝜕_𝒰𝐵(𝑏).(2)If 𝑡∈𝜎_𝑝(𝑏) with 𝑞⁺_𝑡∈𝔪_𝜏, then Ψ([𝑞⁻_𝑡,𝑞⁺_𝑡]) is a line segment in 𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏) having slope 𝑡.(3)If Ψ(𝑎)∈𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏) and Ψ(𝑎)≠Ψ(𝑞^±_𝑡) for all 𝑡∈ℝ, then 𝑎∈[𝑞⁻_𝑡,𝑞⁺_𝑡] for some 𝑡∈𝜎(𝑏) with 𝜏(𝑞⁻_𝑡)<∞.(4)If Ψ(𝑎)∈Ext(𝐵(𝑏))∩𝜕_𝒰𝐵(𝑏) then 𝑎=𝑞^±_𝑡 for some 𝑡∈𝜎_𝑝(𝑏) with 𝑞^±_𝑡∈𝔪_𝜏.(5)ℛ is a ray in 𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏) having slope 𝑡 if and only if ℛ=Ψ([𝑞⁻_𝑡,𝑞⁺_𝑡]) and 𝑡∈𝜎_𝑝(𝑏) with 𝜏(𝑞⁻_𝑡)<𝜏(𝑞⁺_𝑡)=∞.

Proof. Note that for each 𝑠∈ℝ we have that 𝑞⁻_𝑠=𝜒_(𝑠,∞)(𝑏)=𝜒_{(−∞,−𝑠)}(−𝑏). (3.35) Since −𝑏 is self-adjoint, the previous theorem applies. Alternatively, substitute Lemma 3.5 for Lemma 3.3 in the proof of Theorem 3.7, the claim follows immediately.

Until this point we have focused on points that lie in 𝐵(𝑏). As mentioned earlier, it may be the case that 𝐵(𝑏) is not closed. The next four results describe this phenomenon, and tell us exactly when 𝐵(𝑏) is closed, when it is not, and, when we get a “dotted” (open) portion in the boundary of 𝐵(𝑏).

Theorem 3.9. Let ℰ_ℒ={𝑠∈𝜎(𝑏)∶𝜏(𝑝⁺_𝑠)=∞} and let 𝜆=infℰ_ℒ. Then 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏)=𝜕_ℒ𝐵(𝑏) if and only if 𝜏(𝑝⁺_𝜆)=∞. Furthermore, if 𝜏(𝑝⁺_𝜆)<∞ then 𝑦=𝜆𝑥+𝜏(𝑏−𝜆)𝑝⁺_𝜆,𝑥>𝜏𝑝⁺_𝜆 (3.36) is a ray in 𝜕_ℒ𝐵(𝑏)⧵𝐵(𝑏) emanating from Ψ(𝑝⁺_𝜆).

Proof. Suppose that 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏)=𝜕_ℒ𝐵(𝑏). Theorem 3.7 tells us that every Ψ(𝑐)∈𝜕_ℒ𝐵(𝑏) has the property that either 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠] for some 𝑠∈𝜎_𝑝(𝑏) with 𝜏(𝑝⁻_𝑠)<∞, or 𝑐=𝑝^±_𝑠 for some 𝑠∈𝜎(𝑏) with 𝑝^±_𝑠∈𝔪_𝜏. Note that by the definition of 𝜆, 𝜏(𝑝⁺_𝑠)=∞ for all 𝑠>𝜆. Furthermore, for every 𝑠>𝜆 we can choose 𝜆<𝑠^′<𝑠, so ∞=𝜏(𝑝⁺_𝑠′)≤𝜏(𝑝⁻_𝑠). Therefore, for every 𝑠>𝜆 we have 𝜏(𝑝⁻_𝑠)=∞. Suppose, for the sake of contradiction, that 𝜏(𝑝⁺_𝜆)<∞. Then, if 𝑑∈(𝔪_𝜏)⁺₁ with Ψ(𝑑)∈𝜕_ℒ𝐵(𝑏) and 𝜏(𝑑)>𝜏(𝑝⁺_𝜆), then 𝑑∈[𝑝⁻_𝑟,𝑝⁺_𝑟] for some 𝑟∈𝜎(𝑏) with 𝜏(𝑝⁻_𝑟)<∞. Therefore it must be that 𝑟≤𝜆. Clearly this is a contradiction since 𝜏(𝑑)>𝜏(𝑝⁺_𝜆). Hence, it must be that 𝜏(𝑝⁺_𝜆)=∞.
Suppose 𝜏(𝑝⁺_𝜆)=∞, and let (𝑥,𝑦)∈𝜕_ℒ𝐵(𝑏). We want to show that (𝑥,𝑦)∈𝐵(𝑏). Now, either 𝜏(𝑝⁻_𝜆)=∞ or not. If 𝜏(𝑝⁻_𝜆)=∞ then since 𝜏(𝑝⁺_𝑠)<∞ for all 𝑠<𝜆, we get that 𝜏(𝜒_{(𝜆−𝜖,𝜆)}(𝑏))=∞ for every 𝜖>0. For each 𝑛∈ℕ let 𝜆_𝑛=𝜆−1/𝑛. Then 𝜏(𝑝⁻_𝜆𝑛)≤𝜏(𝑝⁺_𝜆𝑛)<∞ for every 𝑛, and 𝜏(𝑝⁺_𝜆𝑛)→∞ as 𝑛→∞. Therefore, we can find 𝑠∈𝜎(𝑏) such that 𝜏(𝑝⁺_𝑠)>𝑥. Let𝑟₁=sup𝑠∈𝜎(𝑏)∶𝜏𝑝⁻_𝑠,𝑟≤𝑥₂=inf𝑠∈𝜎(𝑏)∶𝑥≤𝜏𝑝⁺_𝑠. (3.37) To show that (𝑥,𝑦)∈𝐵(𝑏), we will show that 𝑟₁=𝑟₂, which will force (𝑥,𝑦)=Ψ(𝑐) for some 𝑐∈[𝑝⁻_𝑟1,𝑝⁺_𝑟1] by Theorem 3.7. Note that 𝜏(𝑝⁻_𝑟1)≤𝑥 by lower semicontinuity of 𝜏. We will first show that 𝑟₁≤𝑟₂.
If 𝑠,𝑡∈𝜎(𝑏) with 𝜏(𝑝⁻_𝑠)≤𝑥≤𝜏(𝑝⁺_𝑡), and 𝑡<𝑠, then by the positivity of 𝜏 we must also have 𝜏(𝑝⁺_𝑡)≤𝜏(𝑝⁻_𝑠), which implies 𝜏(𝑝⁻_𝑠)=𝜏(𝑝⁺_𝑡). Thus, everything in {𝑠∈𝜎(𝑏)∶𝜏(𝑝⁻_𝑠)≤𝑥} is less than or equal to everything in {𝑠∈𝜎(𝑏)∶𝑥≤𝜏(𝑝⁺_𝑠)}, hence 𝑟₁≤𝑟₂.
Suppose 𝑟₁<𝑟₂. Either 𝑥≤𝜏(𝑝⁺_𝑟2) or not. Suppose 𝜏(𝑝⁺_𝑟2)<𝑥. Since 𝜏(𝑝⁻_𝑟2)≤𝜏(𝑝⁺_𝑟2)<𝑥, and since 𝑟₁ is an upper bound for {𝑠∈𝜎(𝑏)∶𝜏(𝑝⁻_𝑠)≤𝑥}, we get 𝑟₂≤𝑟₁. But this can not be, so it must be that 𝑥≤𝜏(𝑝⁺_𝑟2). This implies 𝜏(𝑝⁻_𝑟1)≤𝑥≤𝜏(𝑝⁺_𝑟2). If equality holds on either side, then we would get (𝑥,𝑦)∈𝐵(𝑏) (again, by Theorem 3.7).
Suppose 𝜏(𝑝⁻_𝑟1)<𝑥<𝜏(𝑝⁺_𝑟2). Observe that, since 𝑟₂ is a lower bound for the set {𝑠∈𝜎(𝑏)∶𝑥≤𝜏(𝑝⁺_𝑠)}, if 𝑠<𝑟₂ and 𝑠∈𝜎(𝑏) then𝜏𝑝⁺_𝑠<𝑥<𝜏𝑝⁺_𝑟2. (3.38) Similarly, if 𝑠∈𝜎(𝑏) with 𝑟₁<𝑠 then 𝜏𝑝⁻_𝑟1<𝑥<𝜏𝑝⁻_𝑠. (3.39) Since 𝑟₁<𝑟₂, we have 𝜏𝑝⁺_𝑟1<𝑥<𝜏𝑝⁻_𝑟2. (3.40) Thus, if 𝑠∈(𝑟₁,𝑟₂)∩𝜎(𝑏) then by (3.32) and (3.33) we would have 𝜏(𝑝⁺_𝑠)<𝑥<𝜏(𝑝⁻_𝑠). This is not possible, so it must be that (𝑟₁,𝑟₂)∩𝜎(𝑏)=∅. Therefore 𝑝⁺_𝑟1=𝑝⁻_𝑟2, and we have 𝜏(𝑝⁺_𝑟1)=𝜏(𝑝⁻_𝑟2), contradicting (3.33). Hence, it must be that 𝑟₁=𝑟₂=𝑟 and so 𝜏(𝑝⁻_𝑟)<𝑥<𝜏(𝑝⁺_𝑟), and our desired result holds. Therefore, in this case we have 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏)=𝜕_ℒ𝐵(𝑏).
If 𝜏(𝑝⁻_𝜆)<∞ then by Theorem 3.7  Ψ([𝑝⁻_𝜆,𝑝⁺_𝜆]) will be a ray in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏). In both cases we have 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏)=𝜕_ℒ𝐵(𝑏).
For the second conclusion of the theorem, suppose 𝜏(𝑝⁺_𝜆)<∞. Plugging 𝑥=𝜏(𝑝⁺_𝜆) into 𝑦=𝜆𝑥+𝜏((𝑏−𝜆)𝑝⁺_𝜆) we get 𝑦=𝜏(𝑏𝑝⁺_𝜆), so the point Ψ(𝑝⁺_𝜆) lies on the line 𝑦=𝜆𝑥+𝜏((𝑏−𝜆)𝑝⁺_𝜆). We will now show that, for 𝑥>𝜏(𝑝⁺_𝜆), the ray 𝑦=𝜆𝑥+𝜏((𝑏−𝜆)𝑝⁺_𝜆) forms the lower boundary of 𝐵(𝑏), but no point on the ray lies in 𝐵(𝑏).
Suppose 𝑑∈(𝔪_𝜏)⁺₁ with 𝜏(𝑑)>𝜏(𝑝⁺_𝜆) (see Figure 5).
We will show that Ψ(𝑑) lies above the line 𝑦=𝜆𝑥+𝜏((𝑏−𝜆)𝑝⁺_𝜆), and, for any 𝜖>0 there exists 𝑑^′∈(𝔪_𝜏)⁺₁ such that 𝜏(𝑑^′)=𝜏(𝑑) and 𝜏(𝑏𝑑^′) is within 𝜖 of 𝑦=𝜆𝜏(𝑑^′)+𝜏((𝑏−𝜆)𝑝⁺_𝜆).
We will first show that Ψ(𝑑) lies above 𝑦=𝜆𝑥+𝜏((𝑏−𝜆)𝑝⁺_𝜆) when 𝜆=0, then prove the general case (𝜆≠0). Suppose 𝜆=0. We want to show that𝜏(𝑏𝑑)>𝜆𝜏(𝑑)+𝜏(𝑏−𝜆)𝑝⁺_𝜆=𝜏𝑏𝑝⁺₀. (3.41) Suppose for the sake of contradiction that 𝜏(𝑏𝑑)≤𝜏(𝑏𝑝⁺₀). Since 𝐵(𝑏) is convex there must be 𝑑^′∈(𝔪_𝜏)⁺₁ such that 𝜏(𝑑^′)=𝜏(𝑑) and 𝜏(𝑏𝑑^′)=𝜏(𝑏𝑝⁺₀) (see Figure 6).
Write 𝑏=𝑏⁺−𝑏⁻ with 𝑏⁺ and 𝑏⁻ positive and 𝑏⁺𝑏⁻=0. Note that 𝜏(𝑏𝑝⁺₀)=−𝜏(𝑏⁻), so we have 𝜏(𝑏𝑑^′)=−𝜏(𝑏⁻). On the other hand we have that −𝜏(𝑏⁻)=𝜏𝑏𝑑^′=𝜏𝑏⁺𝑑^′−𝜏𝑏⁻𝑑^′=𝜏𝑏⁺𝑑^′−𝜏𝑝⁻₀𝑏⁻𝑝⁻₀𝑑^′=𝜏𝑏⁺𝑑^′−𝜏(𝑏⁻)^1/2𝑝⁻₀𝑑^′𝑝⁻₀(𝑏⁻)^1/2≥𝜏𝑏⁺𝑑^′−𝜏(𝑏⁻)^1/2𝑝⁻₀(𝑏⁻)^1/2=𝜏𝑏⁺𝑑^′−𝜏(𝑏⁻). (3.42) Therefore, adding 𝜏(𝑏⁻) to both sides we get 𝜏(𝑏⁺𝑑^′)≤0. This implies 0=𝜏(𝑏⁺𝑑^′)=𝜏((𝑑^′)^1/2𝑏⁺(𝑑^′)^1/2). Therefore, since 𝜏 is faithful, 0=𝑑^′1/2𝑏⁺𝑑^′1/2=𝑑^′1/2𝑏1−𝑝⁺₀𝑑^′1/2. (3.43) By part (2) of Lemma 3.2 we have 𝑑^′≤𝑝⁺₀. (3.44) But our assumption is that 𝜏(𝑑^′)=𝜏(𝑑)>𝜏(𝑝⁺_𝜆)=𝜏(𝑝⁺₀). Therefore, we have reached a contradiction to our assumption that 𝜏(𝑏𝑑)≤𝜏(𝑏𝑝⁺₀), so it must be that 𝜏(𝑏𝑑)>𝜏(𝑏𝑝⁺₀).
Now suppose that 𝜆≠0. As before, we want to show that𝜏(𝑏𝑑)>𝜆𝜏(𝑑)+𝜏(𝑏−𝜆)𝑝⁺_𝜆. (3.45) Suppose that is not. Then as before we can find 𝑑^′∈(𝔪_𝜏)⁺₁ such that 𝜏𝑏𝑑^′=𝜆𝜏𝑑^′+𝜏(𝑏−𝜆)𝑝⁺_𝜆. (3.46)
Note that for every 𝑠∈ℝ we have that 𝑝⁺_𝑠=𝜒_{(−∞,𝑠]}(𝑏)=𝜒_(−∞,0](𝑏−𝑠1). (3.47) Let 𝑎=𝑏−𝜆1 and consider 𝐵(𝑎). From (3.46) and (3.47) we have that 𝜏𝑎𝑑^′=𝜏𝑎𝜒_(−∞,0](𝑎). (3.48) Using (3.47) and the argument concluded at (3.44) we get 𝑑^′≤𝜒_(−∞,0](𝑎)=𝑝⁺_𝜆. But this is a contradiction since 𝜏(𝑑^′)=𝜏(𝑑)>𝜏(𝑝⁺_𝜆). Therefore 𝜏(𝑏𝑑)>𝜆𝜏(𝑑)+𝜏((𝑏−𝜆)𝑝⁺_𝜆).
To prove 𝐵(𝑏) comes arbitrarily close to the line 𝑦=𝜆𝑥+𝜏((𝑏−𝜆)𝑝⁺_𝜆), let 𝜖>0. Using the definition of 𝜆 and our assumption that 𝜏(𝑝⁺_𝜆)<∞ we have 𝜏(𝑝⁺_𝜆+𝜖/2−𝑝⁺_𝜆)=∞. Since 𝜏 is semifinite we can find a nonzero, positive 𝑒<𝑝⁺_𝜆+𝜖/2−𝑝⁺_𝜆 such that 𝜏(𝑒)<∞. Furthermore, we may scale 𝑒 so that 𝜏(𝑝⁺_𝜆+𝑒)=𝜏(𝑑). Let 𝑑^′=𝑝⁺_𝜆+𝑒. Then since 𝑒<𝑝⁺_𝜆+𝜖/2−𝑝⁺_𝜆≤1−𝑝⁺_𝜆=𝑞⁻_𝜆, we have that 𝜏𝑏𝑑^′−𝜆𝜏𝑑^′+𝜏(𝑏−𝜆)𝑝⁺_𝜆=𝜏𝑏𝑑^′−𝜆𝜏𝑑^′−𝜏(𝑏−𝜆)𝑝⁺_𝜆=𝜏𝑏𝑝⁺_𝜆+𝑒−𝜆𝜏𝑝⁺_𝜆+𝑒−𝜏(𝑏−𝜆)𝑝⁺_𝜆)𝑒)()𝑞=𝜏((𝑏−𝜆=𝜏𝑏−𝜆⁻_𝜆𝑒𝑞⁻_𝜆=𝜏𝑞⁻_𝜆(𝑏−𝜆)𝑞⁻_𝜆𝑒=𝜏𝑒^1/2𝑞⁻_𝜆(𝑏−𝜆)𝑞⁻_𝜆𝑒^1/2≤𝜏𝑒^1/2𝑞⁻_𝜆𝑒^1/2=𝜏𝑞⁻_𝜆𝑒(𝑒)<𝜖=𝜏2. (3.49)

Figure 5 

Ray in 𝜕ℒ𝐵(𝑏)⧵𝐵(𝑏)—Theorem 3.9.

Figure 6 

Ray with slope 0 in 𝜕ℒ𝐵(𝑏)⧵𝐵(𝑏) —Theorem 3.9.

Theorem 3.10. Let ℰ_𝑈={𝑡∈𝜎(𝑏)∶𝜏(𝑞⁺_𝑡)=∞} and let 𝜌=supℰ_𝒰. Then 𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏)=𝜕_𝒰𝐵(𝑏) if and only if 𝜏(𝑞⁺_𝜌)=∞. Furthermore, if 𝜏(𝑞⁺_𝜌)<∞ then 𝑦=𝜌𝑥+𝜏𝑏−𝜌𝑞⁺_𝜌,𝑥>𝜏𝑞⁺_𝜌 (3.50) is a ray in 𝜕_𝒰𝐵(𝑏)⧵𝐵(𝑏) emanating from Ψ(𝑞⁺_𝜌).

Proof. Substitute Theorem 3.8 for Theorem 3.7 in the proof of Theorem 3.9.

Corollary 3.11. Let 𝜆 and 𝜌 be as in Theorems 3.9 and 3.10 (resp.). Then 𝐵(𝑏)=𝐵(𝑏) if and only if 𝜏(𝑝⁺_𝜆)=𝜏(𝑞⁺_𝜌)=∞.

Proof. Note that 𝜏(1)=∞ if and only if ℰ_𝒰≠∅ if and only if ℰ_ℒ≠∅ (since 𝑞^±_𝑠=1−𝑝^∓_𝑠 for each 𝑠∈ℝ). Thus, the claim follows immediately from Theorems 3.9 and 3.10.

Note that Theorems 3.7, 3.8, 3.9, and 3.10 tell us that the boundary of 𝐵(𝑏) (which contains information about 𝜎(𝑏)) is completely determined by spectral projections of the form 𝑝^±_𝑠 and 𝑞^±_𝑡=1−𝑝^∓_𝑡 where 𝑠,𝑡∈ℝ. To summarize, the following information is contained in the boundary of 𝐵(𝑏). (i)Extreme points contained in the lower boundary are of the form Ψ(𝑝^±_𝑠) with 𝑠∈ℝ.(ii)Line segments having slope 𝑠 in the lower boundary are of the form Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) with 𝑠∈ℝ and 𝜏(𝑝⁺_𝑠)<∞.(iii)Rays having slope 𝑠 in the lower boundary are of the form Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) with 𝑠∈ℝ and 𝜏(𝑝⁻_𝑠)<∞ and 𝜏(𝑝⁺_𝑠)=∞.(iv)A “dotted” ray in the lower boundary occurs if 𝜏(𝑝⁺_𝜆)<∞. Moreover, this ray emanates from Ψ(𝑝⁺_𝜆) and has as its equation 𝑦=𝜆𝑥+𝜏((𝑏−𝜆)𝑝⁺_𝜆).(v)Extreme points contained in the upper boundary are of the form Ψ(𝑞^±_𝑡) with 𝑡∈ℝ.(vi)Line segments having slope 𝑡 in the upper boundary are sets of the form Ψ([𝑞⁻_𝑡,𝑞⁺_𝑡]) with 𝑡∈ℝ and 𝜏(𝑞⁺_𝑡)<∞.(vii)Rays having slope 𝑡 in the upper boundary are sets of the form Ψ([𝑞⁻_𝑡,𝑞⁺_𝑡]) with 𝑡∈ℝ and 𝜏(𝑞⁻_𝑡)<∞ and 𝜏(𝑞⁺_𝑡)=∞.(viii)A “dotted” ray in the upper boundary occurs if 𝜏(𝑞⁺_𝜌)<∞. Moreover, this ray emanates from Ψ(𝑞⁺_𝜌) and has as its equation 𝑦=𝜌𝑥+𝜏((𝑏−𝜌)𝑞⁺_𝜌).

Recall that for each (𝑥,𝑦)∈𝜕_ℒ𝐵(𝑏) we let 𝑓∶[0,∞)→ℝ defined by 𝑓(𝑥)=𝑦 denote the lower boundary function generated by 𝑏. Similarly, 𝑔 denotes the upper boundary function generated by 𝑏. For each 𝑥∈[0,∞), let 𝑓^′₋(𝑥) (𝑔^′₋(𝑥)) and 𝑓^′₊(𝑥) (𝑔^′₊(𝑥)) denote the left-sided and right-sided, (respectively) derivatives of 𝑓 and 𝑔 at 𝑥. Note that by Theorem 3.7, every Ψ(𝑐) in 𝜕_ℒ∩𝐵(𝑏) has the property that Ψ(𝑐)∈Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) for some 𝑠∈𝜎(𝑏) with 𝜏(𝑝⁻_𝑠)<∞.

Theorem 3.12. Let 𝑠∈𝜎(𝑏), and let 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠]∩(𝔪_𝜏)⁺₁. Then(1)𝑓^′₋(𝜏(𝑐)) and 𝑓^′₊(𝜏(𝑐)) exist,(2)𝑓^′₋(𝜏(𝑝⁻_𝑠))=sup{𝑠^′∈𝜎(𝑏)∶𝑠^′<𝑠},(3)𝑓^′₊(𝜏(𝑝⁺_𝑠))=inf{𝑠^′∈𝜎(𝑏)∶𝑠<𝑠^′},(4)𝑓 is not differentiable at 𝜏(𝑝^±_𝑠) if and only if (𝑠−𝜖,𝑠)∩𝜎(𝑏)=∅(or(𝑠,𝑠+𝜖)∩𝜎(𝑏)=∅) for some 𝜖>0.

Proof. (1) Since 𝐵(𝑏) is convex, 𝑓 is a convex function, and so for any 𝑥∈(0,+∞)𝑓(𝑥)−𝑓(𝑥−ℎ)ℎ (3.51) increases as ℎ decreases to zero. Furthermore, if 𝑥=𝜏(𝑐) with 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠]∩𝔪_𝜏 for some 𝑠∈𝜎(𝑏), then since Ψ([𝑝⁻_𝑠,𝑝⁺_𝑠]) is a line segment (or ray) in the lower boundary with slope equal to 𝑠 (by Theorem 3.7), we know (𝑓(𝑥)−𝑓(𝑥−ℎ))/ℎ is bounded above by 𝑠. Since bounded increasing sequences converge, 𝑓^′₋(𝑥) exists at each 𝑥=𝜏(𝑐) with 𝑐∈[𝑝⁻_𝑠,𝑝⁺_𝑠]∩𝔪_𝜏 for some 𝑠∈𝜎(𝑏). Using a similar argument considering the sequence (𝑓(𝑥+ℎ)−𝑓(𝑥))/ℎ for some fixed 𝑥=𝜏(𝑐) as ℎ decreases to 0, and Theorem 3.7, we get that 𝑓^′₊(𝑥) also exists at every 𝑥∈[0,+∞).
(2) Fix 𝑠∈𝜎(𝑏) with 𝑠_min<𝑠, where 𝑠_min is the minimum of 𝜎(𝑏), and suppose 𝑥=𝜏(𝑝⁻_𝑠)<∞. Let𝑟=sup𝑠^′∈𝜎(𝑏)∶𝑠^′<𝑠. (3.52) Since 𝜎(𝑏) is closed, 𝑟∈𝜎(𝑏). We want to show that 𝑟=𝑓^′₋(𝜏(𝑝⁻_𝑠)). Clearly 𝑟≤𝑠.
First suppose that (𝑟−𝜖,𝑟)∩𝜎(𝑏)≠∅ for some 𝜖>0. If we had 𝑟=𝑠 then we would have (𝑠−𝜖,𝑠)∩𝜎(𝑏)≠∅, which contradicts the definition of 𝑟. Therefore, in this case we must have 𝑟<𝑠. But then again, by the definition of 𝑟, we have (𝑟,𝑠)∩𝜎(𝑏)≠∅. Therefore 𝑟∈𝜎_𝑝(𝑏), and by Theorem 3.7 we know that Ψ(𝑝⁻_𝑠)=Ψ(𝑝⁺_𝑟) is the right endpoint of a line segment in 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏) with slope equal to 𝑟. Thus, 𝑟=𝑓^′₋(𝜏(𝑝⁻_𝑠)).
If (𝑟−𝜖,𝑟)∩𝜎(𝑏)=∅ for every 𝜖>0, then we can choose an increasing sequence (𝑠_𝑛) of distinct points in 𝜎(𝑏) that converge to 𝑟. But then since 𝜏 is normal, we know that 𝛼_𝑛=𝜏(𝑝⁻_𝑠𝑛) converges in the 𝑤^∗-topology to 𝛼=𝜏(𝑝⁻_𝑟). Since all the elements under consideration are distinct spectral projections of 𝑏 we have𝑠_𝑛𝑝⁻_𝑟−𝑝⁻_𝑠𝑛≤𝑏𝑝⁻_𝑟−𝑝⁻_𝑠𝑛≤𝑟𝑝⁻_𝑟−𝑝⁻_𝑠𝑛 (3.53) for each 𝑛∈ℕ. Applying 𝜏 to both sides of this we get 𝑠_𝑛𝛼−𝛼_𝑛≤𝑓(𝛼)−𝑓𝛼_𝑛≤𝑟𝛼−𝛼_𝑛 (3.54) for every 𝑛∈ℕ. Therefore 𝑠_𝑛≤𝑓(𝛼)−𝑓𝛼_𝑛𝛼−𝛼_𝑛≤𝑟 (3.55) for every 𝑛. Since (𝑠_𝑛) converges to 𝑟, we get lim_𝑛𝑓(𝛼)−𝑓𝛼_𝑛𝛼−𝛼_𝑛=𝑟. (3.56)
Since the function defined byℎ(𝑥)=𝑓(𝛼)−𝑓(𝑥)[]𝛼−𝑥∀𝑥∈0,∞ (3.57) is an increasing function. From this and (3.55) above, we see that if (𝑟_𝑛) is any sequence in ℝ increasing to 𝜏(𝑝⁻_𝑠), then lim_𝑛𝑓(𝛼)−𝑓𝑟_𝑛𝛼−𝑟_𝑛=𝑟. (3.58) Thus, 𝑓^′₋(𝜏(𝑝⁻_𝑠))=𝑟.
(3) Let 𝑟=inf{𝑠^′∈𝜎(𝑏)∶𝑠<𝑠^′}. Clearly, by the definition of 𝑟, we have 𝑠≤𝑟. Using the same argument as in (2), we get𝑓^′₊𝜏𝑝⁺_𝑟=𝑟=inf𝑠^′∈𝜎(𝑏)∶𝑠<𝑠^′. (3.59)
 (4) From parts (2) and (3) above, we have𝑓^′₋𝜏𝑝⁻_𝑠=sup𝑠^′∈𝜎(𝑏)∶𝑠^′<𝑠,𝑓^′₊𝜏𝑝⁺_𝑠=inf𝑠^′∈𝜎(𝑏)∶𝑠<𝑠^′. (3.60) But then, (𝑠,𝑠+𝜖)∩𝜎(𝑏)=∅ for some 𝜖>0 if and only if 𝑥<𝑦 whenever 𝑥∈{𝑠^′∈𝜎(𝑏)∶𝑠^′<𝑠} and 𝑦∈{𝑠^′∈𝜎(𝑏)∶𝑠^′>𝑠} if and only if 𝑓^′₋(𝜏(𝑝⁻_𝑠)<𝑓^′₊(𝜏(𝑝⁺_𝑠) if and only if 𝑓 is not differentiable at 𝜏(𝑝⁻_𝑠) or 𝑓 is not differentiable at 𝜏(𝑝⁺_𝑠). An identical argument shows that (𝑠−𝜖,𝑠)∩𝜎(𝑏)=∅ for some 𝜖>0 if and only if 𝑓 is not differentiable at 𝜏(𝑝⁻_𝑠) or 𝑓 is not differentiable at 𝜏(𝑝⁺_𝑠).

Theorem 3.13. Let 𝑡∈𝜎(𝑏), and let 𝑐∈[𝑞⁻_𝑡,𝑞⁺_𝑡]∩(𝔪_𝜏)⁺₁. Then(1)𝑔^′₋(𝜏(𝑐)) and 𝑔^′₊(𝜏(𝑐)) exist,(2)𝑔^′₋(𝜏(𝑞⁻_𝑡))=sup{𝑡^′∈𝜎(𝑏)∶𝑡^′<𝑡},(3)𝑔^′₊(𝜏(𝑞⁺_𝑡))=inf{𝑡^′∈𝜎(𝑏)∶𝑡<𝑡^′},(4)𝑔 is not differentiable at 𝜏(𝑞^±_𝑡) if and only if (𝑡−𝜖,𝑡)∩𝜎(𝑏)=∅(or(𝑡,𝑡+𝜖)∩𝜎(𝑏)=∅) for some 𝜖>0.

Proof. Substitute (3.34) for (3.33) into the proof of Theorem 3.12.

Theorems 3.12 and 3.13 tell us that if 𝑝⁺_𝑠=𝑝⁻_𝑠 and 𝑠 does not lie in a gap of the spectrum, then 𝑓 is differentiable at 𝜏(𝑝_𝑠) and 𝑓^′(𝜏(𝑝_𝑠))∈𝜎(𝑏). Part (4) of these theorems also confirm that corners in the boundary of 𝐵(𝑏) correspond to gaps in 𝜎(𝑏).

We finish with a few examples of spectral scales in the semifinite von Neumann algebra setting. Recall from Theorems 3.9 and 3.10 that we letℰ_ℒ=𝑠∈𝜎(𝑏)∶𝜏𝑝⁺_𝑠=∞,with𝜆=infℰ_ℒ,ℰ_𝒰=𝑡∈𝜎(𝑏)∶𝜏𝑞⁺_𝑡=∞,with𝜌=supℰ_𝒰. (3.61) We also letℰ=𝜆∈ℂ∶𝜏𝜒_{(𝜆−𝜖,𝜆+𝜖)}(𝑏)=∞∀𝜖>0 (3.62) denote the essential spectrum of 𝑏. We will see in the next section that 𝐵(𝑏) describes the minimum and maximum elements of ℰ. Note that all of the figures that follow are (supposed to resemble) unbounded subsets of ℝ².

Example 3.14. Let 𝑝∈ℳ be a projection. Then either 𝜏(𝑝)<∞ and 𝜏(1−𝑝)=∞, or 𝜏(𝑝)=∞ and 𝜏(1−𝑝)<∞, or 𝜏(𝑝)=∞ and 𝜏(1−𝑝)=∞. In any case we have that 𝜏(𝑝⁺_𝜆)=𝜏(𝑞⁺_𝜌)=∞. Therefore by Corollary 3.11 we have 𝐵(𝑝)=𝐵(𝑝). Furthermore, 𝐵(𝑝) must be one of the following (see Figure 7).

(a)

(b)

(c)

Figure 7 

𝐵(𝑝)—Example 3.14.

Example 3.15. Let ℳ=𝐵(ℋ) for some infinite-dimensional separable Hilbert Space ℋ with 𝜏 the standard canonical trace on ℳ, and let 1𝑏=1000000⋯0−3100000⋯00210000⋯000−32000⋯00003100⋯00000−330⋯0000004⋯⋮⋮⋮⋮⋮⋮⋮⋱. (3.63)𝐵(𝑏) is pictured in Figure 8.

Figure 8 

𝐵(𝑏)—Example 3.15.

Notice that in this case we have 𝜆=−1/3 and 𝜌=1 are the minimum and maximum of the essential spectrum ℰ of 𝑏. Moreover, 𝜏(𝑝⁺_𝜆)=∞ and 𝜏(𝑞⁺_𝜌)<∞. By Theorem 3.9, 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏)=𝜕_ℒ𝐵(𝑏), and by Theorem 3.10, 𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏)≠𝜕_𝒰𝐵(𝑏). Corollary 3.11 then tells us that 𝐵(𝑏)≠𝐵(𝑏). Note the closed portion of 𝜕_𝒰𝐵(𝑏) between 𝑥=0 and 𝑥=1, which tells us that 𝜌=1 is an eigenvalue with multiplicity equal to 1.

Example 3.16. Suppose ℳ=𝐵(ℋ) for some infinite-dimensional separable Hilbert Space ℋ, with 𝜏 the standard canonical trace, and let 𝑏∈ℳ be the diagonal operator with diag(𝑏)=ℚ∩(0,1). In this case we have 𝜆=0 and 𝜌=1. Furthermore, 𝜆∉ℰ_ℒ and 𝜌∉ℰ_𝒰, so 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏)≠𝜕_ℒ𝐵(𝑏) and 𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏)≠𝜕_𝒰𝐵(𝑏). Again, by Corollary 3.11  𝐵(𝑏)≠𝐵(𝑏).  𝐵(𝑏) is pictured in Figure 9.

Figure 9 

𝐵(𝑏)—Example 3.16.

Example 3.17. Let ℳ=𝐿^∞([0,∞),𝑚) with 𝜏(𝑐)=∫^∞₀𝑐𝑑𝑚 for all 𝑐∈ℳ, where 𝑚 is Lebesgue measure on [0,∞). Let 𝑏∈ℳ be defined by 𝑏(𝑥)=𝑒^−𝑥 on [0,∞). Then 𝜎(𝑏) is equal to the essential range of 𝑏, which equals [0,1]. For each 𝑠∈[0,1] we have 𝑝⁻_𝑠=𝑝⁺_𝑠=𝜒_{(−∞,𝑠]}(𝑏)=𝜒_{[−ln(𝑠),∞)},𝑞⁺_𝑠=𝑞⁻_𝑠=1−𝑝⁺_𝑠=1−𝜒_{[−ln(𝑠),∞)}=𝜒_{[0,−ln(𝑠))}. (3.64) Since points of the form (𝜏(𝑝^±_𝑠),𝜏(𝑏𝑝^±_𝑠)) and (𝜏(𝑞^±_𝑠),𝜏(𝑏𝑞^±_𝑠)) determine the extreme points of 𝐵(𝑏) (see Theorems 3.7 and 3.8), by performing a couple quick integrals we find 𝑔 and 𝑓 (the upper and lower boundary curves (resp.) generated by 𝑏): 𝑔(𝑥)=1−𝑒^−𝑥,𝑓(𝑥)=0. (3.65) Note that we have 𝜆=0∉ℰ_ℒ and 𝜌=0∈ℰ_𝒰. Therefore 𝜕_ℒ𝐵(𝑏)∩𝐵(𝑏)≠𝐵(𝑏) and 𝜕_𝒰𝐵(𝑏)∩𝐵(𝑏)=𝐵(𝑏). Thus 𝐵(𝑏)≠𝐵(𝑏) by Corollary 3.11. 𝐵(𝑏) is pictured in Figure 10.

Figure 10 

𝐵(𝑏)—Example 3.17.

Since 𝑔(𝑥) has a horizontal asymptote at 𝑦=1, we know that 𝜏(𝑏)<∞. The spectrum of 𝑏 can be read from Figure 10 as slopes of lines tangent to 𝑔 and 𝑓. Since 𝜆=0∈𝜎(𝑏)⧵𝜎_𝑝(𝑏) and 𝜏(𝑝⁺₀)<∞, 𝑓(𝑥) is “dotted” (or “open”).

Example 3.18. Suppose ℳ=𝐵(ℋ) for some infinite-dimensional separable Hilbert Space ℋ, with 𝜏 the standard canonical trace, and let 1𝑏=10000000⋯01000000⋯00100000⋯000−10000⋯000021000⋯00000−2100⋯000000310⋯0000000−3⋯⋮⋮⋮⋮⋮⋮⋮⋮⋱. (3.66)𝐵(𝑏) is pictured in Figure 11 (the black dots in the boundary are meant to emphasize the extreme points).

Figure 11 

𝐵(𝑏)—Example 3.18.

From Figure 11 we can see that the eigenvalue 1 has multiplicity three by noticing that the line segment Ψ([𝑝⁻₁,𝑝⁺₁]) extends from 𝑥=0 to 𝑥=3. We can also see that 𝑏 is trace class since the upper and lower boundary curves 𝑔 and 𝑓 (resp.) have asymptotes. In fact, if we write 𝑏=𝑏⁺−𝑏⁻ with 𝑏⁺≥0 and 𝑏⁻≥0, we see that the asymptote for 𝑔 is 𝑦=𝜏(𝑏⁺) and the asymptote for 𝑓 is 𝑦=−𝜏(𝑏⁻). Also note that we have 𝜆=0∈ℰ_ℒ and 𝜌=0∈ℰ_𝒰, so by Corollary 3.11  𝐵(𝑏) is closed. Since the slopes of the line segments in the boundary of 𝐵(𝑏) tend to 0 as 𝑥→∞, we know that 𝑏 is compact.