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
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)=Ψ(M1+)={(τ(c),τ(b1c),τ(b2c)):0≤c≤1}.

The spectral scale of an n-tuple of self-adjoint operators is defined analogously (in the canonical fashion), becoming a subset of ℝn+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 <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M41"><mml:mi>B</mml:mi><mml:mo stretchy="false">(</mml:mo><mml:mi>b</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula> When <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M42"><mml:mrow><mml:mi>ℳ</mml:mi></mml:mrow></mml:math></inline-formula> 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.

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.

B(b) gives a complete description of σ(t1b1+t2b2) for all t1,t2∈ℝ.

τ(b) can be read from the point Ψ(1).

If b=b*, then the set of all slopes of lines tangent to the boundary of B(b) is equal to σ(b).

If b=b*, then corners in the boundary of B(b) correspond to gaps in σ(b).

If b=b*, then the extreme points of B(b) are images under Ψ of spectral projections of b.

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.

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 σp(b).

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).

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
Wk(b)={1k∑i=1k〈bxi,xi〉:{xi}isanorthonormalset},1≤k≤dim(M).

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).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.

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)=∫01cdm 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.

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).B(b) is given in Figure 3 [8].

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+}
in ℝ3. From [2, Theorem 2.3] we know that Ψ applied to spectral projections of the operators bt=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 χ𝒞μ,θ is the characteristic function on the arc 𝒞μ,θ from ei(μ-θ) to ei(μ+θ), then, χCμ,θ(z)=χ[cos(θ),∞)(Re(e-iμz))=χ[cos(θ),∞)((cos(μ))x-(sin(μ))y)
for every z=x+iy∈ℂ (to see this think of the case where μ=0). But χ[cos(θ),∞)((cos(μ))x-(sin(μ))y)=χ[cos(θ),∞)(t1Re(b)+t2Im(b))=χ[cos(θ),∞)(bt)
for t=(t1,t2)=(cos(μ),-sin(μ)). Therefore, every χ𝒞μ,θ is a spectral projection of some bt. Conversely, if t=(t1,t2)∈ℝ2 and s∈ℝ, then for if μ is chosen so that cos(μ)=t1/t12+t22 and sin(μ)=-t2/t12+t22 and θ is chosen so that ∥t∥cos(θ)=s, then we have χ[s,∞)(bt)=χ[‖t‖cos(θ),∞)(‖t‖cos(μ)Re(b)-‖t‖sin(μ)Im(b))=χ[cos(θ),∞)(cos(μ)Re(b)-sin(μ)Im(b))=χCμ,θ.

Since the extreme points of B(b) come from the spectral projections described above, we can calculate the extreme points of B(b) explicitly:(τ(χCμ,θ),τ(bχCμ,θ))=(∫Cμ,θ1dm,∫Cμ,θbdm)=(θπ,∫Cμ,θeizdz)=(θπ,12πi(ei(μ+θ)-ei(μ-θ)))=(θπ,1πeiμsin(θ)).
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).

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 𝔫τ={x∈ℳ:τ(x*x)<+∞} and 𝔪τ={∑i=1nxiyi:xi,yi∈𝔫τ}. The by [9, Lemma 2.16, page 318], we can define the function Ψ:𝔪τ→ℝ2 byΨ(c)=(τ(c),τ(bc))foreveryc∈mτ.

Definition 3.1.

The spectral scale of b (with respect to τ) is the set
B(b)=Ψ((mτ)1+)={(τ(c),τ(bc)):c∈(mτ)1+}.

Since (𝔪τ)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}
for the order interval determined by p and q. Given a real number s, we let ps+ denote the spectral projection of b corresponding to the interval (-∞,s]. Similarly, we let ps- denote the spectral projection of b corresponding to the interval (-∞,s). We use ps± to indicate either of these projections and we write ps when ps+=ps-. We let qs+ and qs- denote the spectral projections corresponding to the intervals [s,∞) and (s,∞), respectively. We also use qs± to indicate either of these projections and we write qs when qs+=qs-. Note that qs±=1-ps∓ for every s∈ℝ. The spectral projections ps± and qs± 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∈[ps-,ps+], and a∈ℳ1+. Then the following statements hold:

(b-s1)(1-c)=(b-s1)(1-ps+)≥0 and the range projection of (b-s1)(1-ps+)is1-ps+;

if a1/2(b-s1)(1-c)a1/2=0, then a≤ps+;

(s1-b)c=(s1-b)ps-≥0 and the range projection of (s1-b)ps- is ps-;

if (1-a)1/2(s1-b)c(1-a)1/2=0 then a≥ps-.

The following two lemmas tell us that points of the form (τ(c),τ(cb)) lie on the boundary when c∈[ps-,ps+] (or c∈[qs-,qs+]) 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∈[ps-,ps+]∩𝔪τ and a∈(𝔪τ)1+.

If τ(a)=τ(c), then τ(ba)≥τ(bc).

If τ(a)=τ(c) and τ(ba)=τ(bc), then a∈[ps-,ps+]; moreover, if =ps±, 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-ps+)≥0,sothatb(1-c)≥s(1-c).
Therefore
a1/2b(1-c)a1/2≥a1/2s(1-c)a1/2.
Note that τ(b(1-c)a)<∞ since a∈𝔪τ and 𝔪τ is an ideal. Since a1/2∈𝔫τ and τ is order preserving on 𝔪τ [6], we have that
τ(b(1-c)a)=τ(a1/2b(1-c)a1/2)≥τ(a1/2s(1-c)a1/2)=τ(s(1-c)a).
Since c∈𝔪τ, an identical argument yields
τ(bc(1-a))≤τ(sc(1-a)).
It follows from (*), (**), and our hypothesis that
τ(ba)-τ(bc)=τ(ba-bca-bc+bca)=τ(b(1-c)a)-τ(bc(1-a))≥τ(s(1-c)a)-τ(sc(1-a))=τ(sa)-τ(sc)=s(τ(a)-τ(c))=0,
and so τ(ba)≥τ(bc).

To prove (2), suppose that τ(a)=τ(c) and τ(ba)=τ(bc). Thenτ((1-c)a)=τ(a-ca)=τ(c-ca)=τ(c(1-a)),τ(b(1-c)a)=τ(ba-bca)=τ(bc-bca)=τ(bc(1-a)).
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)).
Therefore, all the above terms must be equal, and so
τ(b(1-c)a)=sτ((1-c)a),τ(bc(1-a))=sτ(c(1-a)).

From this it follows thatτ(a1/2(b-s1)(1-c)a1/2)=0,τ((1-a)1/2(b-s1)c(1-a)1/2)=0.

Since τ is faithful we get thata1/2(b-s1)(1-c)a1/2=0,(1-a)1/2(b-s1)c(1-a)1/2=0.
By parts (2) and (4) of the previous lemma, a∈[ps-,ps+].

If it is the case that c=ps±, τ(a)=τ(ps±), and τ(ba)=τ(bc) then by what we just showed, a∈[ps-,ps+], so τ(ps±-a)=0, which implies that a=ps± since τ is faithful.

Remark 3.4.

Notice the importance of our choice of a and c as elements of ideal 𝔪τ 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∈[qt-,qt+]∩𝔪τ and a∈(𝔪τ)1+.

If τ(a)=τ(d), then τ(ba)≤τ(bd).

If τ(a)=τ(d) and τ(ba)=τ(bd), then a∈[qt-,qt+]; moreover, if =qt±, then equality in the second equation obtains only for a=d.

Proof.

(1) Similarly, by Lemma 3.2 (since 1-d∈[pt-,pt+]), we have that
bd≥td,t(1-d)≥b(1-d).
Therefore
(1-a)1/2bd(1-a)1/2≥(1-a)1/2td(1-a)1/2,a1/2t(1-d)a1/2≥a1/2b(1-d)a1/2.
Using the fact that the trace is faithful and finite on 𝔪τ, 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(td(1-a)),τ(t(1-d)a)=τ(a1/2t(1-d)a1/2)≥τ(a1/2b(1-d)a1/2)=τ(b(1-d)a).
It follows that
τ(bd)-τ(ba)=τ(bd(1-a))-τ(b(1-d)a)≥τ(td(1-a))-τ(t(1-d)a)=τ(td)-τ(ta)=0.

To prove (2), suppose that τ(a)=τ(d) and τ(ba)=τ(bd). Thenτ((1-d)a)=τ(a-da)=τ(d-da)=τ(d(1-a)),τ(b(1-d)a)=τ(ba-bda)=τ(bd-bda)=τ(bd(1-a)).

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)).
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
τ(a1/2(b-t1)(1-d)a1/2)=0,τ((1-a)1/2(b-t1)d(1-a)1/2)=0.
But since (b-t1)(1-d)≥0 and (b-t1)d≥0 we get
a1/2(b-t1)(1-d)a1/2=0,(1-a)1/2(b-t1)d(1-a)1/2=0.
By Lemma 3.2 we get a∈[qt-,qt+]. If d=qt±, then since τ is faithful we must have a=d=qt±.

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}.
The upper boundary of B(b) is defined analogously as
∂UB(b)={(x,y)∈B(b)¯:(x,y′)∈B(b)⟹y′≤y}.
Since B(b)¯ is a closed convex set, and τ takes on every value in [0,∞], we let
f:[0,∞)⟶R
defined by f(x)=y for each (x,y)∈∂ℒB(b) denote the lower boundary function determined by b. Similarly, let
g:[0,∞)⟶R
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∈(𝔪τ)1+. Then one has the following.

If s∈σ(b) with ps±∈𝔪τ, then Ψ(ps±)∈Ext(B(b))∩∂ℒB(b).

If s∈σp(b) with ps+∈𝔪τ, then Ψ([ps-,ps+]) is a line segment in ∂ℒB(b)∩B(b) having slope s.

If Ψ(a)∈∂ℒB(b)∩B(b) and Ψ(a)≠Ψ(ps±) for all s∈ℝ, then a∈[ps-,ps+] for some s∈σ(b) with τ(ps-)<∞.

If Ψ(a)∈Ext(B(b))∩∂ℒB(b), then a=ps± for some s∈σp(b) with ps±∈𝔪τ.

ℛ is a ray in ∂ℒB(b)∩B(b) having slope s if and only if ℛ=Ψ([ps-,ps+]) and s∈σp(b) with τ(ps-)<τ(ps+)=∞.

Proof.

(1) Fix s∈σ(b) such that τ(ps±)<∞. Then by part (1) of Lemma 3.3 we have that
Ψ(ps±)∈∂LB(b).
By part (2) of Lemma 3.3 we have that
Ψ-1(Ψ(ps±))∩(mτ)1+={ps±}.
To see that Ψ(ps±) are extreme in B(b), suppose that
Ψ(ps±)=λΨ(a1)+(1-λ)Ψ(a2)
for some a1,a2∈(𝔪τ)1+and 0<λ<1. But then
Ψ(ps±)=Ψ(λa1+(1-λ)a2),
and by part (2) of Lemma 3.3 we have that ps±=λa1+(1-λ)a2. Since projections are extreme points of ℳ1+, we must have a1=a2=ps±. Therefore, Ψ(ps±)∈Ext(B(b))∩∂ℒB(b).

(2) Fix s∈σp(b) so that ps+∈𝔪τ. Then we have that ps-<ps+,Ψ(ps-)≠Ψ(ps+), and b(ps+-ps-)=s(ps+-ps-). Note that if c∈[ps-,ps+], then Ψ(c)∈∂ℒB(b) by part (1) of Lemma 3.3, soΨ([ps-,ps+])⊂∂LB(b)∩B(b).
For each λ∈(0,1) let pλ=λps-+(1-λ)ps+. Then, each pλ is in [ps-,ps+], and so Ψ(pλ)∈∂Bℒ(b)∩B(b). Furthermore, Ψ(pλ) is a typical point on the line segment between Ψ(ps-) and Ψ(ps+) and the slope of this line segment is
τ(bps+)-τ(bps-)τ(ps+)-τ(ps-)=τ(b(ps+-bps-))τ(ps+-ps-)=τ(s(ps+-ps-))τ(ps+-ps-)=s.
Therefore Ψ([ps-,ps+]) is a line segment in ∂ℒB(b)∩B(b) with slope s. Note that since Ψ(ps±) are extreme points of B(b) contained in ∂ℒB(b), we know that the endpoints of any line segment containing Ψ([ps-,ps+]) must be Ψ(ps-) and Ψ(ps+).

(3) Suppose Ψ(a)∈∂ℒB(b)∩B(b) and Ψ(a)≠Ψ(ps±) for all s∈σ(b). We will show that ps-<a<ps+ for some s∈σ(b) with τ(ps-)<∞. Letr1=sup{s∈σ(b):τ(ps-)≤τ(a)},r2=inf{s∈σ(b):τ(a)≤τ(ps+)}.
We want to show that r1=r2. Note that r1 and r2 are in σ(b) since σ(b) is closed, and τ(pr1-)≤τ(a) by lower semicontinuity of τ.

We first show that r1≤r2. If s,t∈σ(b) with τ(ps-)≤τ(a)≤τ(pt+), and t<s, then by the positivity of τ we must also have τ(pt+)≤τ(ps-), so τ(ps-)=τ(a)=τ(pt+). This fact, along with Lemma 3.3, contradicts our assumption about Ψ(a). Thus, everything in {s∈σ(b):τ(ps-)≤τ(a)} is less than or equal to everything in {s∈σ(b):τ(a)≤τ(ps+)}, hence r1≤r2.

Now, either τ(a)≤τ(pr2+) or not. Suppose τ(pr2+)<τ(a). Since τ(pr2-)≤τ(pr2+)<τ(a), and since r1 is an upper bound for {s∈σ(b):τ(ps-)≤τ(a)}, we must have r2≤r1. But r1≤r2. Therefore r1=r2 when τ(pr2+)<τ(a).

Assume that τ(a)≤τ(pr2+). Then, τ(pr1-)≤τ(a)≤τ(pr2+). But we are assuming that Ψ(a)≠Ψ(ps±) for all s∈σ(b), and that Ψ(a)∈∂ℒB(b)∩B(b). Therefore, by Lemma 3.3 we getτ(pr1-)<τ(a)<τ(pr2+).
Observe that, since r2 is a lower bound for the set {s∈σ(b):τ(a)≤τ(ps+)}, if s<r2 and s∈σ(b) then
τ(ps+)<τ(a)<τ(pr2+).
Similarly, if s∈σ(b) with r1<s, then
τ(pr1-)<τ(a)<τ(ps-).
Suppose that r1<r2. Then since r1,r2∈σ(b) we have that
τ(pr1+)<τ(a)<τ(pr2-).
On the other hand if s∈(r1,r2)∩σ(b) then by (3.32) and (3.33) we would have τ(ps+)<τ(a)<τ(ps-). This is not possible, so it must be that (r1,r2)∩σ(b)=∅. Therefore pr1+=pr2-, and we have τ(pr1+)=τ(pr2-), contradicting (3.33). Hence, it must be that r1=r2=r and so τ(pr-)<τ(a)<τ(pr+). Therefore r∈σp(b) and a∈[pr-,pr+] 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(ps-)<τ(ps+)=∞ for some s∈σp(b). We claim that ℛ=Ψ([ps-,ps+]) is a ray in ∂ℒB(b)∩B(b) with slope s, emanating from the point Ψ(ps-)=(τ(ps-),τ(bps-)). We will first show that every point in Ψ([ps-,ps+]) lies on the ray through Ψ(ps-) with slope s.

Let c∈[ps-,ps+]∩𝔪τ, and write c=ps-+d for some d∈(ps+-ps-)ℳ(ps+-ps-). Then b(c-ps-)=bd=sd=s(c-ps-), so τ(b(c-ps-))=τ(s(c-ps-)). This implies thatτ(bc)=τ(s(c-ps-))+τ(bps-)=s⋅τ(c)+τ((b-s1)ps-).
Therefore Ψ(c) lies on the ray with slope s passing through the point (τ(ps-),τ(bps-)). 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([ps-,ps+]).

(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=pr±∈𝔪τ 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∈(𝔪τ)1+ with Ψ(z)∈ℛ, then z∈[psz-,psz+] for some sz∈σp(b) with τ(psz-)<∞. Furthermore, since any such Ψ([psz-,psz+]) is a line segment (or ray) with slope sz, and since Ψ([psz-,psz+])⊂ℛ, we must have sz=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([ps-,ps+]).

Theorem 3.8.

Let a∈(𝔪τ)1+. Then one has the following.

If t∈σ(b) with qt±∈𝔪τ, then Ψ(qt±)∈Ext(B(b))∩∂𝒰B(b).

If t∈σp(b) with qt+∈𝔪τ, then Ψ([qt-,qt+]) is a line segment in ∂𝒰B(b)∩B(b) having slope t.

If Ψ(a)∈∂𝒰B(b)∩B(b) and Ψ(a)≠Ψ(qt±) for all t∈ℝ, then a∈[qt-,qt+] for some t∈σ(b) with τ(qt-)<∞.

If Ψ(a)∈Ext(B(b))∩∂𝒰B(b) then a=qt± for some t∈σp(b) with qt±∈𝔪τ.

ℛ is a ray in ∂𝒰B(b)∩B(b) having slope t if and only if ℛ=Ψ([qt-,qt+]) and t∈σp(b) with τ(qt-)<τ(qt+)=∞.

Proof.

Note that for each s∈ℝ we have that
qs-=χ(s,∞)(b)=χ(-∞,-s)(-b).
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 ℰℒ={s∈σ(b):τ(ps+)=∞} and let λ=infℰℒ. Then ∂ℒB(b)∩B(b)=∂ℒB(b) if and only if τ(pλ+)=∞. Furthermore, if τ(pλ+)<∞ then
y=λx+τ((b-λ)pλ+),x>τ(pλ+)
is a ray in ∂ℒB(b)∖B(b) emanating from Ψ(pλ+).

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∈[ps-,ps+] for some s∈σp(b) with τ(ps-)<∞, or c=ps± for some s∈σ(b) with ps±∈𝔪τ. Note that by the definition of λ, τ(ps+)=∞ for all s>λ. Furthermore, for every s>λ we can choose λ<s′<s, so ∞=τ(ps′+)≤τ(ps-). Therefore, for every s>λ we have τ(ps-)=∞. Suppose, for the sake of contradiction, that τ(pλ+)<∞. Then, if d∈(𝔪τ)1+ with Ψ(d)∈∂ℒB(b) and τ(d)>τ(pλ+), then d∈[pr-,pr+] for some r∈σ(b) with τ(pr-)<∞. Therefore it must be that r≤λ. Clearly this is a contradiction since τ(d)>τ(pλ+). Hence, it must be that τ(pλ+)=∞.

Suppose τ(pλ+)=∞, and let (x,y)∈∂ℒB(b). We want to show that (x,y)∈B(b). Now, either τ(pλ-)=∞ or not. If τ(pλ-)=∞ then since τ(ps+)<∞ for all s<λ, we get that τ(χ(λ-ϵ,λ)(b))=∞ for every ϵ>0. For each n∈ℕ let λn=λ-1/n. Then τ(pλn-)≤τ(pλn+)<∞ for every n, and τ(pλn+)→∞ as n→∞. Therefore, we can find s∈σ(b) such that τ(ps+)>x. Letr1=sup{s∈σ(b):τ(ps-)≤x},r2=inf{s∈σ(b):x≤τ(ps+)}.
To show that (x,y)∈B(b), we will show that r1=r2, which will force (x,y)=Ψ(c) for some c∈[pr1-,pr1+] by Theorem 3.7. Note that τ(pr1-)≤x by lower semicontinuity of τ. We will first show that r1≤r2.

If s,t∈σ(b) with τ(ps-)≤x≤τ(pt+), and t<s, then by the positivity of τ we must also have τ(pt+)≤τ(ps-), which implies τ(ps-)=τ(pt+). Thus, everything in {s∈σ(b):τ(ps-)≤x} is less than or equal to everything in {s∈σ(b):x≤τ(ps+)}, hence r1≤r2.

Suppose r1<r2. Either x≤τ(pr2+) or not. Suppose τ(pr2+)<x. Since τ(pr2-)≤τ(pr2+)<x, and since r1 is an upper bound for {s∈σ(b):τ(ps-)≤x}, we get r2≤r1. But this can not be, so it must be that x≤τ(pr2+). This implies τ(pr1-)≤x≤τ(pr2+). If equality holds on either side, then we would get (x,y)∈B(b) (again, by Theorem 3.7).

Suppose τ(pr1-)<x<τ(pr2+). Observe that, since r2 is a lower bound for the set {s∈σ(b):x≤τ(ps+)}, if s<r2 and s∈σ(b) thenτ(ps+)<x<τ(pr2+).
Similarly, if s∈σ(b) with r1<s then
τ(pr1-)<x<τ(ps-).
Since r1<r2, we have
τ(pr1+)<x<τ(pr2-).
Thus, if s∈(r1,r2)∩σ(b) then by (3.32) and (3.33) we would have τ(ps+)<x<τ(ps-). This is not possible, so it must be that (r1,r2)∩σ(b)=∅. Therefore pr1+=pr2-, and we have τ(pr1+)=τ(pr2-), contradicting (3.33). Hence, it must be that r1=r2=r and so τ(pr-)<x<τ(pr+), and our desired result holds. Therefore, in this case we have ∂ℒB(b)∩B(b)=∂ℒB(b).

If τ(pλ-)<∞ then by Theorem 3.7Ψ([pλ-,pλ+]) 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(pλ+)<∞. Plugging x=τ(pλ+) into y=λx+τ((b-λ)pλ+) we get y=τ(bpλ+), so the point Ψ(pλ+) lies on the line y=λx+τ((b-λ)pλ+). We will now show that, for x>τ(pλ+), the ray y=λx+τ((b-λ)pλ+) forms the lower boundary of B(b), but no point on the ray lies in B(b).

Suppose d∈(𝔪τ)1+ with τ(d)>τ(pλ+) (see Figure 5).

We will show that Ψ(d) lies above the line y=λx+τ((b-λ)pλ+), and, for any ϵ>0 there exists d′∈(𝔪τ)1+ such that τ(d′)=τ(d) and τ(bd′) is within ϵ of y=λτ(d′)+τ((b-λ)pλ+).

We will first show that Ψ(d) lies above y=λx+τ((b-λ)pλ+) when λ=0, then prove the general case (λ≠0). Suppose λ=0. We want to show thatτ(bd)>λτ(d)+τ((b-λ)pλ+)=τ(bp0+).
Suppose for the sake of contradiction that τ(bd)≤τ(bp0+). Since B(b) is convex there must be d′∈(𝔪τ)1+ such that τ(d′)=τ(d) and τ(bd′)=τ(bp0+) (see Figure 6).

Write b=b+-b- with b+ and b- positive and b+b-=0. Note that τ(bp0+)=-τ(b-), so we have τ(bd′)=-τ(b-). On the other hand we have that -τ(b-)=τ(bd′)=τ(b+d′)-τ(b-d′)=τ(b+d′)-τ(p0-b-p0-d′)=τ(b+d′)-τ((b-)1/2p0-d′p0-(b-)1/2)≥τ(b+d′)-τ((b-)1/2p0-(b-)1/2)=τ(b+d′)-τ(b-).
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′)1/2b(1-p0+)(d′)1/2.
By part (2) of Lemma 3.2 we have
d′≤p0+.
But our assumption is that τ(d′)=τ(d)>τ(pλ+)=τ(p0+). Therefore, we have reached a contradiction to our assumption that τ(bd)≤τ(bp0+), so it must be that τ(bd)>τ(bp0+).

Now suppose that λ≠0. As before, we want to show thatτ(bd)>λτ(d)+τ((b-λ)pλ+).
Suppose that is not. Then as before we can find d′∈(𝔪τ)1+ such that
τ(bd′)=λτ(d′)+τ((b-λ)pλ+).

Note that for every s∈ℝ we have that ps+=χ(-∞,s](b)=χ(-∞,0](b-s1).
Let a=b-λ1 and consider B(a). From (3.46) and (3.47) we have that
τ(ad′)=τ(aχ(-∞,0](a)).
Using (3.47) and the argument concluded at (3.44) we get d′≤χ(-∞,0](a)=pλ+. But this is a contradiction since τ(d′)=τ(d)>τ(pλ+). Therefore τ(bd)>λτ(d)+τ((b-λ)pλ+).

To prove B(b) comes arbitrarily close to the line y=λx+τ((b-λ)pλ+), let ϵ>0. Using the definition of λ and our assumption that τ(pλ+)<∞ we have τ(pλ+ϵ/2+-pλ+)=∞. Since τ is semifinite we can find a nonzero, positive e<pλ+ϵ/2+-pλ+ such that τ(e)<∞. Furthermore, we may scale e so that τ(pλ++e)=τ(d). Let d′=pλ++e. Then since e<pλ+ϵ/2+-pλ+≤1-pλ+=qλ-, we have that τ(bd′)-(λτ(d′)+τ((b-λ)pλ+))=τ(bd′)-λτ(d′)-τ((b-λ)pλ+)=τ(b(pλ++e))-λτ(pλ++e)-τ((b-λ)pλ+)=τ((b-λ)e)=τ((b-λ)qλ-eqλ-)=τ(qλ-(b-λ)qλ-e)=τ(e1/2qλ-(b-λ)qλ-e1/2)≤τ(e1/2qλ-e1/2)=τ(qλ-e)=τ(e)<ϵ2.

Ray in ∂ℒB(b)∖B(b)—Theorem 3.9.

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

Theorem 3.10.

Let ℰU={t∈σ(b):τ(qt+)=∞} and let ρ=supℰ𝒰. Then ∂𝒰B(b)∩B(b)=∂𝒰B(b) if and only if τ(qρ+)=∞. Furthermore, if τ(qρ+)<∞ then
y=ρx+τ((b-ρ)qρ+),x>τ(qρ+)
is a ray in ∂𝒰B(b)∖B(b) emanating from Ψ(qρ+).

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(pλ+)=τ(qρ+)=∞.

Proof.

Note that τ(1)=∞ if and only if ℰ𝒰≠∅ if and only if ℰℒ≠∅ (since qs±=1-ps∓ 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 ps± and qt±=1-pt∓ where s,t∈ℝ. To summarize, the following information is contained in the boundary of B(b).

Extreme points contained in the lower boundary are of the form Ψ(ps±) with s∈ℝ.

Line segments having slope s in the lower boundary are of the form Ψ([ps-,ps+]) with s∈ℝ and τ(ps+)<∞.

Rays having slope s in the lower boundary are of the form Ψ([ps-,ps+]) with s∈ℝ and τ(ps-)<∞ and τ(ps+)=∞.

A “dotted” ray in the lower boundary occurs if τ(pλ+)<∞. Moreover, this ray emanates from Ψ(pλ+) and has as its equation y=λx+τ((b-λ)pλ+).

Extreme points contained in the upper boundary are of the form Ψ(qt±) with t∈ℝ.

Line segments having slope t in the upper boundary are sets of the form Ψ([qt-,qt+]) with t∈ℝ and τ(qt+)<∞.

Rays having slope t in the upper boundary are sets of the form Ψ([qt-,qt+]) with t∈ℝ and τ(qt-)<∞ and τ(qt+)=∞.

A “dotted” ray in the upper boundary occurs if τ(qρ+)<∞. Moreover, this ray emanates from Ψ(qρ+) and has as its equation y=ρx+τ((b-ρ)qρ+).

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([ps-,ps+]) for some s∈σ(b) with τ(ps-)<∞.

Theorem 3.12.

Let s∈σ(b), and let c∈[ps-,ps+]∩(𝔪τ)1+. Then

f-′(τ(c)) and f+′(τ(c)) exist,

f-′(τ(ps-))=sup{s′∈σ(b):s′<s},

f+′(τ(ps+))=inf{s′∈σ(b):s<s′},

f is not differentiable at τ(ps±) 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
increases as h decreases to zero. Furthermore, if x=τ(c) with c∈[ps-,ps+]∩𝔪τ for some s∈σ(b), then since Ψ([ps-,ps+]) 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∈[ps-,ps+]∩𝔪τ 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 smin<s, where smin is the minimum of σ(b), and suppose x=τ(ps-)<∞. Letr=sup{s′∈σ(b):s′<s}.
Since σ(b) is closed, r∈σ(b). We want to show that r=f-′(τ(ps-)). 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∈σp(b), and by Theorem 3.7 we know that Ψ(ps-)=Ψ(pr+) is the right endpoint of a line segment in ∂ℒB(b)∩B(b) with slope equal to r. Thus, r=f-′(τ(ps-)).

If (r-ϵ,r)∩σ(b)=∅ for every ϵ>0, then we can choose an increasing sequence (sn) of distinct points in σ(b) that converge to r. But then since τ is normal, we know that αn=τ(psn-) converges in the w*-topology to α=τ(pr-). Since all the elements under consideration are distinct spectral projections of b we havesn(pr--psn-)≤b(pr--psn-)≤r(pr--psn-)
for each n∈ℕ. Applying τ to both sides of this we get
sn(α-αn)≤f(α)-f(αn)≤r(α-αn)
for every n∈ℕ. Therefore
sn≤f(α)-f(αn)α-αn≤r
for every n. Since (sn) converges to r, we get
limnf(α)-f(αn)α-αn=r.

Since the function defined byh(x)=f(α)-f(x)α-x∀x∈[0,∞]
is an increasing function. From this and (3.55) above, we see that if (rn) is any sequence in ℝ increasing to τ(ps-), then
limnf(α)-f(rn)α-rn=r.
Thus, f-′(τ(ps-))=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+′(τ(pr+))=r=inf{s′∈σ(b):s<s′}.

(4) From parts (2) and (3) above, we havef-′(τ(ps-))=sup{s′∈σ(b):s′<s},f+′(τ(ps+))=inf{s′∈σ(b):s<s′}.
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(ps-)<f+′(τ(ps+) if and only if f is not differentiable at τ(ps-) or f is not differentiable at τ(ps+). An identical argument shows that (s-ϵ,s)∩σ(b)=∅ for some ϵ>0 if and only if f is not differentiable at τ(ps-) or f is not differentiable at τ(ps+).

Theorem 3.13.

Let t∈σ(b), and let c∈[qt-,qt+]∩(𝔪τ)1+. Then

g-′(τ(c)) and g+′(τ(c)) exist,

g-′(τ(qt-))=sup{t′∈σ(b):t′<t},

g+′(τ(qt+))=inf{t′∈σ(b):t<t′},

g is not differentiable at τ(qt±) 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 ps+=ps- and s does not lie in a gap of the spectrum, then f is differentiable at τ(ps) and f′(τ(ps))∈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 letEL={s∈σ(b):τ(ps+)=∞},withλ=infEL,EU={t∈σ(b):τ(qt+)=∞},withρ=supEU.
We also letE={λ∈C:τ(χ(λ-ϵ,λ+ϵ)(b))=∞∀ϵ>0}
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(pλ+)=τ(qρ+)=∞. Therefore by Corollary 3.11 we have B(p)¯=B(p). Furthermore, B(p) must be one of the following (see 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
b=(1000000⋯0-1300000⋯00120000⋯000-13000⋯00002300⋯00000-130⋯00000034⋯⋮⋮⋮⋮⋮⋮⋮⋱).B(b) is pictured in 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(pλ+)=∞ and τ(qρ+)<∞. 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∉ℰℒ and ρ∉ℰ𝒰, so ∂ℒB(b)∩B(b)≠∂ℒB(b) and ∂𝒰B(b)∩B(b)≠∂𝒰B(b). Again, by Corollary 3.11B(b)¯≠B(b). B(b) is pictured in Figure 9.

B(b)—Example 3.16.

Example 3.17.

Let ℳ=L∞([0,∞),m) with τ(c)=∫0∞cdm 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
ps-=ps+=χ(-∞,s](b)=χ[-ln(s),∞),qs+=qs-=1-ps+=1-χ[-ln(s),∞)=χ[0,-ln(s)).
Since points of the form (τ(ps±),τ(bps±)) and (τ(qs±),τ(bqs±)) 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.
Note that we have λ=0∉ℰℒ and ρ=0∈ℰ𝒰. 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.

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)∖σp(b) and τ(p0+)<∞, 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
b=(10000000⋯01000000⋯00100000⋯000-10000⋯000012000⋯00000-1200⋯000000130⋯0000000-13⋯⋮⋮⋮⋮⋮⋮⋮⋮⋱).B(b) is pictured in Figure 11 (the black dots in the boundary are meant to emphasize the extreme points).

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([p1-,p1+]) 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∈ℰℒ and ρ=0∈ℰ𝒰, so by Corollary 3.11B(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.

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