Ar Ve Son Spectral Solutions

SOLUTIONS TO A SHORT COURSE ON SPECTRAL THEORY

BY WILLIAM ARVESON

Contents

1. Spectral Theory and Banach Algebras 21.1. Origins of Spectral Theory 21.2. The Spectrum of an Operator 41.3. Banach Algebras: Examples 51.4. The Regular Representation 81.5. The General Linear Group of A 91.6. Spectrum of an Element of a Banach Algebra 111.7. Spectral Radius 121.8. Ideals and Quotients 141.9. Commutative Banach Algebras 151.10. Examples: C(X) and the Wiener Algebra 161.11. Spectral Permanence Theorem 171.12. Brief on the Analytic Functional Calculus 182. Operators on Hilbert Space 212.1. Operators and Their C∗-Algebras 212.2. Commutative C∗-Algebras 232.3. Continuous Functions of Normal Operators 262.4. The Spectral Theorem and Diagonalizations 272.5. Representations of Banach *-Algebras 312.6. Borel Functions of Normal Operators 332.7. Spectral Measures 342.8. Compact Operators 372.9. Adjoining a Unit to a C∗-Algebra 392.10. Quotients of C∗-Algebras 403. Asymptotics: Compact Perturbations and Fredholm Theory 423.1. The Calkin Algebra 423.2. Riesz Theory of Compact Operators 433.3. Fredholm Operators 433.4. The Fredholm Index 434. Methods and Applications 434.1. Maximal Abelian von Neumann Algebras 434.2. Toeplitz Matrices and Toeplitz Operators 434.3. The Toeplitz C∗-Algebra 434.4. Index Theorem for Continuous Symbols 434.5. Some H2 Function Theory 434.6. Spectra of Toeplitz Operators with Continuous Symbol 434.7. States and the GNS Construction 434.8. Existence of States: The Gelfand-Naimark Theorem 43References 43

1

2 ARVESON SPECTRAL THEORY SOLUTIONS

1. Spectral Theory and Banach Algebras

1.1. Origins of Spectral Theory.

(1) Fix a sequence an of numbers with 0 < ε ≤ an ≤M , and let A : `2 → `2

be the operator determined by (Ax)n = anxn.Show that A is a boundedoperator on `2, and exhibit a bounded operator B on `2 such that AB =BA = 1.

Solution. A is bounded because

‖Ax‖2 =∑n

(Ax)n(Ax)n =∑n

anxnanxn =∑n

|an|2|xn|2 ≤∑n

M2|xn|2 = M2‖x‖2

The inverse operator B is determined by (Bx)n = 1anxn; this operator is

bounded by 1/ε, by the same argument.

(2) Let an be a bounded increasing sequence of positive numbers and letDn = a1a2 . . . an. Show that the sequence Dn converges to a nonzero limitD(A) iff

∞∑n=1

(1− an) <∞.

Solution. To clarify, when we say∑

(1−an) <∞, we mean to say that thesum converges to a finite number. In particular, the case where it convergesto −∞ (for example, if all the an are equal to 2) is disallowed.

It is clear that either of the conditions whose equivalence we seek toestablish implies that an ↑ 1, so we will assume this to be the case.

The condition thatDn approaches a nonzero limit is equivalent to∑j(− ln aj) <

∞. We want to show this is equivalent to∑j(1− aj) <∞. One direction

is trivial, because − lnx ≥ 1−x for all positive x (just compare the deriva-tives to the right and left of the intersection point x = 1). For the otherdirection, we note that − lnx < 2(1−x) in a neighborhood of x = 1 (again,this follows by comparing the derivatives at x = 1); since an → 1, one has− ln aj < 2(1− aj) for sufficiently large j, so that the result follows by thecomparison test.

We note that this argument is standard in complex analysis in the con-text of the Weierstrass factorization theorem; see for instance ([Rud87])Theorem 15.5 or ([Con78]) Proposition 5.4.

(3) Let k(x, y) be a continuous function on (x, y) : 0 ≤ y ≤ x ≤ 1 and letf ∈ C([0, 1]). Show that the function g : [0, 1]→ C defined by

g(x) =

∫ x

0

k(x, y)f(y) dy

is continuous, and that the linear map K : f → g defines a boundedoperator on C([0, 1]).

Solution. Let M = ‖f‖sup and L = max |k(x, y)|. Fix any x ∈ [0, 1].Let ε > 0. Since k is uniformly continuous (its domain is compact), one

can choose δ1 > 0 such that |k(x1, y1)− k(x2, y2)| < ε whenever ‖(x1, y1)−

ARVESON SPECTRAL THEORY SOLUTIONS 3

(x2, y2)‖ < δ1. Let δ = min(δ, ε/L). For 0 ≤ h < δ one has

|g(x+ h)− g(x)| =

∣∣∣∣∣∫ x+h

0

k(x+ h, y)f(y) dy −∫ x

0

k(x, y)f(y) dy

∣∣∣∣∣=

∣∣∣∣∣∫ x

0

[k(x+ h, y)− k(x, y)]f(y) dy +

∫ x+h

x

k(x+ h, y)f(y) dy

∣∣∣∣∣≤∫ x

0

|k(x+ h, y)− k(x, y)||f(y)| dy +

∫ x+h

x

|k(x+ h, y)|f(y)| dy

≤M∫ x

0

|k(x+ h, y)− k(x, y)| dy +M

∫ x+h

x

|k(x+ h, y)| dy

≤Mε+MLh < 2Mε.

A similar calculation holds for −δ < h ≤ 0 (the second integral is reversed).Thus g(x+ h)→ g(x) as h→ 0, so that g is continuous at x.

For the boundedness of K, note that for each x ∈ [0, 1],

|g(x)| =∣∣∣∣∫ x

0

k(x, y)f(y) dy

∣∣∣∣≤∫ x

0

|k(x, y)||f(y)| dy

≤M∫ x

0

|k(x, y)|dy

≤ML = L‖f‖sup

Taking the supremum over x ∈ [0, 1], we have ‖g‖sup ≤ L‖f‖sup. Thus‖K‖op ≤ L.

(4) For the kernel k(x, y) = 1 for 0 ≤ y ≤ x ≤ 1 consider the correspondingVolterra operator V : C([0, 1])→ C([0, 1]), i.e.

(V f)(x) =

∫ x

0

f(y) dy, f ∈ C([0, 1]).

Given a function g ∈ C([0, 1]), show that the equation V f = g has asolution f ∈ C([0, 1]) iff g is continuously differentiable and g(0) = 0.

Solution. Clearly every element V f of the range is continuously differen-tiable and has (V f)(0) = 0. Conversely, suppose g satisfies those hypothe-ses; let f = g′. Then (V f)′ = f = g′, so that V f and g differ by an additiveconstant; but (V f)(0) = 0 = g(0), so that V f = g.

(5) Let k(x, y), 0 ≤ x, y ≤ 1 be a continuous function on the unit square, andconsider the bounded operator K defined on C([0, 1]) by

(Kf)(x) =

∫ 1

0

k(x, y)f(y) dy, 0 ≤ x ≤ 1.

Let B1 = f ∈ C([0, 1]) : ‖f‖ ≤ 1 be the closed unit ball in C([0, 1]).Show that K is a compact operator in the sense that the norm closure ofthe image KB1 under K is a compact subset of C([0, 1]).


Solution. Clearly KB1 is a pointwise bounded family of functions (since Kis a bounded operator, by a straightforward generalization of exercise 3).

Given ε > 0, use the uniform continuity of k to choose δ > 0 such that|k(x1, y1)− k(x2, y2)| < ε whenever ‖(x1, y1)− (x2, y2)‖ < δ. Then for anyf ∈ B1 and any x1, x2 ∈ [0, 1] satisfying |x1 − x2| < δ,

|(Kf)(x1)− (Kf)(x2)| =∣∣∣∣∫ 1

0

[k(x1, y)− k(x2, y)]f(y) dy

∣∣∣∣≤∫ 1

0

|k(x1, y)− k(x2, y)||f(y)| dy

≤∫ 1

0

ε · 1 dy = ε.

Thus, KB1 is uniformly equicontinuous; by the Arzela-Ascoli theorem, ithas compact closure.

Note: The hint is misleading, as KB1 is not uniformly Lipschitz ingeneral; indeed, it may contain functions which are not Lipschitz. Forexample, suppose k(x, y) =

√x and f is the constant function f(y) = 1;

then (Kf)(x) =√x is not Lipschitz. One needs to use uniform rather than

Lipschitz continuity.

1.2. The Spectrum of an Operator.

(1) Give explicit examples of bounded operators A,B on `2(N) such that AB =1 and BA is the projection onto a closed infinite-dimensional subspace ofinfinite codimension.

Solution. We define A and B on the standard basis vectors by Bek = e2k

and

Aek =

ek/2 k even

0 else.

Then AB = 1 and BA is the projection onto spane2, e4, e6, . . ..

(2) Let A and B be the operators defined on `2(N) be

A(x1, x2, . . . ) = (0, x1, x2, . . . )

B(x2, x2, . . . ) = (x2, x3, x4, . . . )

Show that ‖A‖ = ‖B‖ = 1, and compute both BA and AB. Deduce thatA is injective but not surjective, B is surjective but not injective, and thatσ(AB) 6= σ(BA).

Solution. ClearlyA is isometric andB is contractive; since ‖B(0, 1, 0, 0, . . . )‖ =‖(0, 1, 0, 0, . . . )‖ we see that ‖B‖ = 1.BA = 1, whereas AB is projection onto the complement of the span of

e1. The former of these implies the injectivity of A and surjectivity of B;if A were surjective or B injective, the composition AB would have to beas well, but it’s not.

The non-injectivity of AB implies that its spectrum includes 0, whichσ(BA) does not.


(3) Let E be a Banach space and let A and B be bounded operators on E.Show that 1−AB is invertible iff 1−BA is invertible.

Solution. Suppose 1− AB is invertible, and let C be its inverse. Let D =1 +BCA. Then

D(1−BA) = 1+BCA−BA−BCABA = 1+B(C−CAB)A−BA = 1+BA−BA = 1

and

(1−BA)D = 1−BA+BCA−BABCA = 1−BA+B(C−ABC)A = 1−BA+BA = 1

so that 1 − BA is invertible. Interchanging A and B yields the otherdirection of implication.

We note that norm properties enter nowhere into this calculation; theresult holds in any unital algebra.

(4) Use the result of the preceding exercise to show that for any two boundedoperators A,B acting on a Banach space, σ(AB) and σ(BA) agree exceptperhaps for 0: σ(AB) \ 0 = σ(BA) \ 0.

Solution. For λ 6= 0,

λ /∈ σ(AB)⇔ λ−AB invertible

⇔ λ

(1− A

λB

)invertible

⇔ λ

(1−BA

λ

)invertible

⇔ λ−BA invertible

⇔ λ /∈ σ(BA).

1.3. Banach Algebras: Examples.

(1) Let E be a normed linear space. Show that E is a Banach space iff forevery sequence of elements xn ∈ X satisfying

∑n ‖xn‖ < ∞, there is an

element y ∈ X such that

limn→∞

‖y − (x1 + · · ·+ xn)‖ = 0.

Solution. Suppose E is Banach. Let xn be an absolutely summable se-quence. Let yn =

∑ni=1 xi be the nth partial sum. Then for n ≥ m,

‖yn − ym‖ =

∥∥∥∥∥n∑

i=m+1

xi

∥∥∥∥∥ ≤n∑

i=m+1

‖xi‖ ≤∞∑

i=m+1

‖xi‖.

Since the tails of a convergent sequence tend to zero, this shows that ‖yn−ym‖ → 0 as n,m → ∞, so that yn is Cauchy. By completeness, itconverges to a limit y. Thus, every absolutely summable sequence in E issummable.

Conversely, suppose E is a normed space in which every absolutelysummable sequence is summable. Let xn be a Cauchy sequence in E.Let xnk be a subsequence such that ‖xnk+1

− xnk‖ < 2−k. Let yk =


xnk+1− xnk . Then yk is absolutely summable, hence summable. Let y

be its sum. For each k,

xnk+1= xn1

+

k∑i=1

yk;

since the right-hand side approaches a limit as k → ∞, the left-hand sidemust as well. Hence the sequence xnk converges. But then xn is a Cauchysequence with a convergent subsequence, so it converges as well.

(2) Prove that the convolution algebra L1(R) does not have an identity.

Solution. Suppose g is a convolution identity. Let fn = 1[0,1/n]

. Then

1 = fn(0) = (fn ∗ g)(0) =

∫ 0

−1/n

g(y) dy.

Since L1 functions generate absolutely continuous measures ([Fol99] Theo-rem 3.5), this is impossible.

Another way to prove impossibility is to cite the fact that the convolutionof L1 functions with L∞ functions is continuous ([Fol99] Proposition 8.8).But then the convolution of g with a characteristic function would have tobe a continuous characteristic function, which is impossible.

(3) For every n = 1, 2, . . . let φn be a nonnegative function in L1(R) such thatφn vanishes outside the interval [−1/n, 1/n] and∫ ∞

−∞φn(t) dt = 1.

Show that φ1, φ2, . . . is an approximate identity for the convolution algebraL1(R) in the sense that

limn→∞

‖f ∗ φn − f‖1 = 0

for every f ∈ L1(R).

Solution. For any f ∈ L1(R),

‖f ∗ φn − f‖ =

∫R|(f ∗ φn)(x)− f(x)| dx

=

∫R

∣∣∣∣∫Rf(x− y)φn(y) dy − f(x)

∫Rφn(y) dy

∣∣∣∣ dx≤∫R

∫ 1/n

−1/n

|f(x− y)− f(x)||φn(y)| dydx

=

∫ 1/n

−1/n

φn(y)

∫R|fy(x)− f(x)| dxdy

=

∫ 1/n

−1/n

φn(y)‖fy − f‖1 dy.

By the uniform continuity of translation on L1, this tends to 0 as n →∞.


(4) Let f ∈ L1(R). The Fourier transform of f is defined as follows:

f(ξ) =

∫ ∞−∞

eitξf(t) dt, ξ ∈ R.

Show that f belongs to the algebra C∞(R) of all continuous functions onR that vanish at ∞.

Solution. Fix ξ ∈ R. Let ε > 0 be given. Let K ⊆ R be a compact subsetsuch that

∫R\K |f | < ε. Now for any η ∈ R,

|f(ξ)− f(η)| =∣∣∣∣∫

Reiξtf(t) dt−

∫Reiηtf(t) dt

∣∣∣∣=

∣∣∣∣∫R

(eiξt − eiηt

)f(t) dt

∣∣∣∣≤∫R

∣∣eiξt − eiηt∣∣ |f(t)| dt

≤ 2ε+

∫K

∣∣eiξt − eiηt∣∣ |f(t)| dt.

As η → ξ, eiηt → eiξt uniformly on K, so that the above expression is less

than 3ε for η sufficiently close to ξ. This proves that f is continuous.For the Riemann-Lebesgue lemma, one could use the identification of R

with the maximal ideal space of L1(R), and the corresponding identificationof the Fourier transform with the Gelfand transform; my presentation inJorgensen’s class in spring 2011 goes into this. A more elementary proof,however (thank to [SS05]), is as follows: For any ξ 6= 0, the change ofvariables x′ = x+ π

ξ yields

f(ξ) = −∫Rf

(x′ − π

ξ

)eix′ξ dx′.

Averaging this with the original expression for f(ξ) yields

f(ξ) =1

2

∫R

(f(x)− f

(x− π

ξ

))eixξ dx

from which |f(ξ)| ≤ 12‖f − fπ/ξ‖. The result follows by the L1 continuity

of translation.

(5) Show that the Fourier transform is a homomorphism of the convolutionalgebra L1(R) onto a subalgebra A of C∞(R) which is closed under complexconjugation and separates points of R.

Solution. Linearity is obvious. Multiplicativity is a well-known consequenceof Fubini’s theorem. For self-adjointness, transform the function f∗(x) =

f(−x). For separation of points, suppose |ξ| < |η| and consider the Fouriertransform of the characteristic function of [0, 2π

|η| ].


1.4. The Regular Representation.

(1) Let E and F be normed linear spaces with E 6= 0. Show that B(E,F ) isa Banach space iff F is a Banach space.

Solution. Suppose F is Banach. Let Tn be a Cauchy sequence inB(E,F ).For each e ∈ E and each m,n ∈ N,

‖Tme− Tne‖ = ‖(Tm − Tn)e‖ ≤ ‖Tm − Tn‖‖e‖

so that Tne is a Cauchy sequence in F . Define the function T : E → Fby Te = limTne. Clearly this is linear; it is bounded because

‖Te‖ = ‖ limTne‖ = lim ‖Tne‖ ≤ (lim ‖Tn‖)‖e‖

so that ‖T‖ ≤ lim ‖Tn‖ (which limit exists because ‖Tn‖ is a Cauchysequence in R). Finally, we must prove that Tn → T in norm, not justpointwise. Given ε > 0, choose N ∈ N such that ‖Tm − Tn‖ < ε form,n ≥ N . Then for n ≥ N and e ∈ E,

‖Te− Tne‖ = ‖ limmTme− Tne‖ = lim

m‖Tme− Tne ≤ ε‖e‖

since ‖Tme− Tne‖ < ε‖e‖ for each m. Hence ‖T − Tn‖ → 0.Conversely, suppose B(E,F ) is Banach. Let fn be a Cauchy sequence

in F . Let e ∈ E be some nonzero vector. Define linear maps Tn : E → Fby Tne = fn and Tn = 0 on the (algebraic) complement of the span of e.

Clearly Tn is linear with norm ‖fn‖‖e‖ . Moreover, since

(Tm − Tn)x =

α(fm − fn) x = αe

0 else,

we see that ‖Tm−Tn‖ = ‖fm−fn‖‖e‖ , so that Tn is Cauchy in B(E,F ). Let

T be its limit, and let f = Te. Then

‖f − fn‖ = ‖Te− Tne‖ ≤ ‖T − Tn‖‖e‖ → 0.

Thus F is Banach.

(2) Let A ∈ B(E) be an operator with the property that there is a sequenceAm of finite-rank operators such that ‖A − An‖ → 0. Show that A iscompact.

Solution. Let xn be a sequence of points in the unit ball of E. Define

subsequences x(i)n recursively x as follows: x

(0)n = xn, while x

(i+1)n is a

subsequence of x(i)n such that Ai+1x

(i+1)n converges. (This is possible by

the compactness of the unit ball for any finite-dimensional Banach space,

such as the range of each Am.) Let yn = x(n)n be the diagonal subsequence,

so that Akyn converges for each fixed k; call the limit zk. Then

‖zj − zk‖ = ‖ limnAkyn − lim

nAjyn‖ = lim

n‖(Ak −Aj)yn‖ ≤ ‖Ak −Aj‖

since ‖(Ak − Aj)yn‖ ≤ ‖Ak − Aj‖ for each n. This proves that zk isCauchy; let its limit be z. Then for each k,

‖Ayn−z‖ = ‖Akyn−zk+(A−Ak)yn+(zk−z)‖ ≤ ‖Akyn−zk‖+‖A−Ak‖+‖zk−z‖.


Since all three terms on the right approach 0 as k → ∞, we see thatAyn → z. Thus, we have found a subsequence yn such that Aynconverges. So A is compact.

By the way, this diagonal subsequence trick is used in general topology(nets) to show that a cluster point of the cluster points of A is a clusterpoint of A; see 11.5 of my putative functional HW. It’s also used to proveArzela-Ascoli, etc.

(3) Let a1, a2, . . . be a bounded sequence of complex numbers and A the corre-sponding multiplication operator on `2. Show that A is compact iff an → 0.

Solution. Suppose an → 0. LetAn be the truncated multiplication operatorwhich replaces ak with 0 for k > n. Then ‖A − An‖ = supk>n |ak| →0. Since each An has finite rank, the previous exercise implies that A iscompact.

Conversely, suppose an 6→ 0. Let ε > 0 and ank a subsequence with|ank | > ε for each k. Then enk is a sequence of unit vectors such that

Aenk has no convergent subsequence (indeed, ‖Aenk −Aenj‖ > ε√

2), sothat A is not compact.

(4) Let k ∈ C([0, 1]× [0, 1]) and define A : C([0, 1])→ C([0, 1]) by

(Af)(x) =

∫ 1

0

k(x, y)f(y) dy.

Show that A is bounded with ‖A‖ ≤ ‖k‖sup.

Solution. Already done in exercise 1.1.5, where we showed A is in factcompact.

(5) Show that there is a sequence of finite-rank operators An such that ‖An −A‖ → 0.

Solution. As can easily be checked, if A′ is another integral operator withkernel k′, then ‖A − A′‖ ≤ ‖k − k′‖sup. Now by Stone-Weierstrass, there

exists a sequence kn of kernels of the form kn(x, y) =∑Nnj=1 p

(n)j (x)q

(n)j (y),

with p(n)j and q

(n)j polynomials, such that ‖kn − k‖ → 0, and therefore

‖A−An‖ → 0, where An is the operator with kernel kn. But each An has

finite rank; in particular, its range is spanned by the p(n)j .

1.5. The General Linear Group of A. Let A be a unital Banach algebra with‖1‖ = 1, and let G = A−1.

(1) Show that for every x ∈ A with ‖x‖ < 1, there exists a continuous functionf : [0, 1]→ G with f(0) = 1 and f(1) = (1− x)−1.

Solution. We define f(t) = (1− tx)−1. This is well-defined since ‖tx‖ < t.It is continuous because it is the composition of the continuous functionst 7→ 1− tx with y 7→ y−1.

(2) Show that for every element x ∈ G there is an ε > 0 with the followingproperty: For every element y ∈ G satisfying ‖y− x‖ < ε there is an arc inG connecting y to x.


Solution. Let ε = ‖x−1‖−1. If h ∈ A is such that x + h ∈ G and ‖h‖ < ε,then let f be a path from 1 to 1 + x−1h as in the previous problem; thent 7→ xf(t) is a path in G from x to x+ h.

In words, what we’ve just shown is that G is locally path connected.

(3) Let G0 be the set of all finite products of elements of G of the form 1− xor (1−x)−1, where x ∈ A satisfies ‖x‖ < 1. Show that G0 is the connectedcomponent of 1 in G.

Solution. Let I denote the connected component of the identity. Note thatthis is the same as the path component of the identity, because G is locallypath connected by problem 2.• By problem 1 above, each element of the form (1− x)−1 is in I.• I is closed under products because, if f is a path from 1 to x and g a

path from 1 to y, then

h(t) =

f(2t) 0 ≤ t ≤ 1/2

xg(2t− 1) 1/2 ≤ t ≤ 1

defines a path from 1 to xy.• I is closed under inverses because, if f is a path from 1 to x, theng(t) = f(t)−1 is a path from 1 to x−1.

• Thus, G0 ⊆ I.• G0 is open: Let a = a1 . . . an be an element of G0, where ai = 1−xi or

(1−xi)−1 for each i, with ‖xi‖ < 1. Then for h ∈ A with ‖h‖ < ‖a−1‖,one has

a− h = a(1− a−1h)

and since both a and 1 − a−1h are elements of G0, this product isagain in G0.

• G0 is clopen, because open subgroups of topological groups are alwaysclopen: Suppose G is a topological group and H a subgroup. Then

G \H =⋃

x∈G\H

xH

is a union of open sets (because left multiplication y 7→ xy is a home-omorphism), hence open, so that H is closed.

• Thus, I ⊆ G0, completing the proof.

(4) Deduce that G0 is a normal subgroup of G and that the quotient topologyon G/G0 makes it into a discrete group.

Solution. The connected component of the identity is a subgroup, as shownabove; it is normal because, if f is a path from 1 to x and y ∈ G, then t 7→yf(t)y−1 is a path from 1 to yxy−1, so that G0 is closed under conjugationby elements of G.

The cosets of G0 in G are precisely the connected components of G; to seethis, let Gx denote the component of some x ∈ G, and note that xG0 ⊆ Gxand x−1Gx ⊆ G0. As a result, the preimage of any subset of G/G0 will bea union of connected components of G, and therefore open.


1.6. Spectrum of an Element of a Banach Algebra.

(1) Give an example of a one-dimensional Banach algebra that is not isomorphicto the algebra of complex numbers.

Solution. Such an algebra would have to have the same linear structurebut a different multiplication. The simplest way to do this is to define allproducts to be zero.

Note that it is impossible to give a unital counterexample, because themultiplicative structure of a unital one-dimensional algebra is determinedby (λ1)(µ1) = λµ1.

(2) Let X be a compact Hausdorff space and let A = C(X) be the Banachalgebra of all complex-valued continuous functions on X. Show that forevery f ∈ C(X), σ(f) = f(X).

Solution. The invertible functions in C(X) are precisely the functions whichare never zero. Thus, for a given λ, f − λ1 is invertible iff λ /∈ f(X).

(3) Let T be the operator defined on L2([0, 1]) by (Tf)(x) = xf(x). What isthe spectrum of T? Does T have point spectrum?

Solution. T has no point spectrum, because if f is a function such thatxf(x) = f(x) then f(x) = 0 for x 6= 1, so that f = 0 in L2.

The spectrum is [0, 1]. It is at most this, because for λ /∈ [0, 1], theinverse of T − λ1 is the operator (Sf)(x) = 1

x−λf(x). On the other hand,

if λ ∈ [0, 1] then such an inverse S cannot exist; when f(x) is the constantfunction f(x) = 1, we must have (Sf)(x) = 1

x−λ a.e., but this is not squareintegrable.

We recall that, more generally, the spectrum of a multiplication operatorMf on L2 is the essential range of f .

For the remaining exercises, let an be a bounded sequence of complexnumbers and H a Hilbert space with ONB en.

(4) Show that there is a (necessarily unique) bounded operator A ∈ B(H) sat-isfying Aen = anen+1 for every n. Such an operator A is called a unilateralweighted shift.

Solution. This operator is just the composition of the multiplication oper-ator A′ determined by an with the unilateral shift S. These are bothbounded by previous work.

(5) Let A ∈ B(H) be a weighted shift as above. Show that for every complexnumber λ with |λ| = 1 there is a unitary operator Uλ ∈ B(H) such thatUAU−1 = λA.

Solution. Define U by Uen = λnen. It’s easy to see that this is unitary andthat the desired conjugation relation holds.

(6) Deduce that the spectrum of a weighted shift must be the union of (possiblydegenerate) concentric circles about z = 0.

Solution. For any µ, λ ∈ C with |λ| = 1, an operator T is the inverse forA−µ1 iff UTU−1 is the inverse for λA−µ1. Hence µ ∈ ρ(A) iff µ/λ ∈ ρ(A).It follows that ρ(A), and therefore σ(A), is radially symmetric.


(7) Let A be the weighted shift associated with a sequence an ∈ `∞.(a) Calculate ‖A‖ in terms of an.(b) Assuming that an → 0, show that ‖An‖1/n → 0.

Solution.(a) We have ‖A‖ = sup |an|. Clearly it is at least this, by considering only

its action on the ONB; on the other hand, it’s at most this, becauseof the factorization A = SA′ mentioned above.

(b) Because each An maps the ONB onto an orthonormal set, Parseval’sidentity implies that the norm may be calculated as

‖An‖ = supk‖Anek‖.

Now

Anek =

k+n−1∏j=k

aj

ek+n

so that

‖Anek‖1/n = supk

k+n−1∏j=k

aj

1/n

.

This approaches 0. Explicitly, given ε > 0, choose N such that |an| <ε/2 for n ≥ N ; letM be the product of all a1, . . . aN with absolute valuegreater than 1, if there are any; choose N2 such that M(ε/2)N2−N < ε;then the geometric mean of any N2 consecutive terms of the sequencean is less than ε.For amusement, one could apply AM-GM and do the above epsilonticswith arithmetic rather than geometric means.

1.7. Spectral Radius.

(1) Let an be a sequence with an → 0. Show that the associated weightedshift operator has spectrum 0.

Solution. This follows from exercise 1.6#7(b) and the spectral radius for-mula.

(2) Consider the simplex ∆n ⊆ [0, 1]n defined by

∆n = (x1, . . . , xn) ∈ [0, 1]n : x1 ≤ x2 ≤ · · · ≤ xn.

Show that the volume of ∆n is 1n! .

First Solution. We have the explicit expression

|∆n| =∫ 1

0

∫ xn−1

0

∫ xn−2

0

. . .

∫ x2

0

1dx1dx2 . . . dxn.

One can see inductively that performing the first k integrals yields the

integrandxkk+1

k! , so that one ends up integratingxn−1n

(n−1)! from 0 to 1, yielding

an answer of 1n! .


Second Solution. We proceed by induction. First, let us define more generalregions

∆n,r = (x1, . . . , xn) ∈ [0, r]n : x1 ≤ · · · ≤ xn ≤ rfor each r ≥ 0. By homothety, |∆n,r| = rn|∆n|.

Now the base case |∆1| = 11! is clear; for the induction, we note that

|∆n+1| =∫

∆n+1

1dVn+1 =

∫ 1

0

∫∆n,xn+1

1dVndxn+1 =

∫ 1

0

xnn+1|∆n| dxn+1 =|∆n|n+ 1

.

Third Solution. We note that the permutation group Sn acts isometricallyon the cube [0, 1]n. Also,⋃

π∈Sn

π(∆n) = [0, 1]n.

Finally, the images π(∆n) are almost disjoint. For this, it suffices to show∆n is almost disjoint from π(∆n) for π 6= id. Referring to the cycle decom-position of π, if (a1a2 . . . ) is a cycle in standard form (i.e. smallest elementfirst), then

∆n ∩ π(∆n) ⊆ (x1, . . . , xn) ∈ [0, 1]n : xa1 = xa2which has volume zero.

Thus, [0, 1]n is the union of n! almost-disjoint regions of equal volume,so each must have volume 1

n! .

(3) Let k(x, y) be a Volterra kernel as in Example 1.1.4, and let K be itscorresponding integral operator on the Banach space C([0, 1]). Estimatethe norms ‖Kn‖ by showing that there is a positive constant M such thatfor every f ∈ C([0, 1]) and every n = 1, 2, . . . ,

‖Knf‖ ≤ Mn

n!‖f‖.

Solution. By induction,

(Knf)(t) =

∫∆n,t

k(t, xn)k(xn, xn+1) . . . k(x2, x1)f(x1)dx1 . . . dxn

so that

|(Knf)(t)| ≤∫

∆n,t

|k(t, xn)| . . . |k(x2, x1)||f(x1)|dx1 . . . dxn ≤∫

∆n,t

Mn‖f‖dx1 . . . dxn =Mn‖f‖tn

n!≤ Mn‖f‖

n!

where M = ‖k‖C([0,1]2).

Hence, ‖Kn‖ ≤ Mn

n! .

(4) Let K be a Volterra operator as in the preceding exercise. Show that forevery complex number λ 6= 0 and every g ∈ C([0, 1]), the Volterra equationof the second kind Kf − λf = g has a unique solution f ∈ C([0, 1]).

Solution. Since (n!)1/n → ∞ (one can compare n! to any fixed an to getthis result, or use the fact that n

(n!)1/n→ e), the previous exercise implies

that the spectral radius of K is 0, so that the spectrum is 0. HenceK − λ1 is invertible for every λ 6= 0.


1.8. Ideals and Quotients.

(1) Let V and W be finite-dimensional vector spaces over C and let T : V →Wbe a linear map satisfying TV = W , and having kernel K = x ∈ V : Tx =0. Then we have a short exact sequence of vector spaces

0 −→ K −→ V −→W −→ 0.

Show that dimV = dimK + dimW .

Solution. Let k1, . . . , kn be a basis for K and w1, . . . , wm a basis forW . For each j = 1, . . . ,m let vj ∈ V be a vector such that Tvj = wj . Wewill show that B = k1, . . . , kn, v1, . . . , vm is a basis for V . It is linearlyindependent because, if

a1k1 + · · ·+ ankn + b1v1 + · · ·+ bmvm = 0,

then, applying T , we have b1w1 + · · · + bmwm = 0, which by the indepen-dence of w1, . . . , wm implies that b1 = · · · = bm = 0; we therefore havea1k1 + · · · + ankn = 0, which by the independence of k1, . . . , kn impliesa1 = · · · = an = 0.

Furthermore, B spans V , because if v ∈ V , there exist scalars c1, . . . , cmsuch that Tv = c1w1 + · · ·+cmwm; but then v−(c1v1 + · · ·+cmvm) ∈ K, sothat v− (c1v1 + · · ·+cmvm) = d1k1 + · · ·+dnkn for some scalars d1, . . . , dn.We thus have

v = d1k1 + · · ·+ dnkn + c1v1 + · · ·+ cmvm.

(2) For n = 1, 2, . . . let V1, V2, . . . , Vn be finite-dimensional vector spaces andset V0 = Vn+1 = 0. Suppose that for k = 0, . . . , n we have a linear mapTk : Vk → Vk+1 such that

0T0→ V1

T1→ V2T2→ . . .

Tn−1→ VnTn→ 0

is exact. Show that∑nk=1(−1)k dimVk = 0.

Solution. By exactness,R(Tk−1) = kerTk, so that dimR(Tk−1) = dim kerTk.By the Rank Theorem, dim kerTk + dimR(Tk) = dimVk. Substituting theformer equation into the latter,

dimVk = dimR(Tk−1) + dimR(Tk).

Thusn∑k=1

(−1)k dimVk =

n∑k=1

(−1)k[dimR(Tk−1)+dimR(Tk)] = − dimR(T0)+(−1)n dimR(Tn)

since the sum telescopes. But both T0 and Tn have 0-dimensional range.Hence the sum is zero.

(3) Show that every normed linear space E has a basis consisting of unit vec-tors, and deduce that every infinite-dimensional normed linear space has adiscontinuous linear functional f : E → C.


Solution. Any basis can be scaled to a unit basis. Alternatively, one canadapt the standard proof of the existence of a basis to find a maximal setof independent unit vectors.

If eα is a unit basis of the infinite-dimensional normed space E, leten be a countable subset. Define f(en) = n and f(eα) = 0 for all otherα. Then f is discontinuous.

(4) Let A be a complex algebra and let I be a proper ideal of A. Show that Iis a maximal ideal iff the quotient algebra A/I is simple.

Solution. Suppose A/I is not simple, so that it has a nontrivial ideal J .Then π−1(J) is an ideal strictly between A and I, where π : A → A/I isthe canonical quotient map, so that I is not maximal.

Conversely, suppose I is not maximal, and let J be a nontrivial ideal ofA containing I. Then π(J) is a nontrivial ideal of A/I, so that A/I is notsimple.

(5) Let A be a unital Banach algebra, let n be a positive integer, and letω : A → Mn be a homomorphism of complex algebras such that ω(A) =Mn. Show that ω is continuous. Deduce that every multiplicative linearfunctional f : A→ C is continuous.

Solution. Since Mn is simple, the previous problem implies that the kernelof ω would be a maximal ideal in A, and therefore closed. But a homomor-phism is continuous precisely when its kernel is closed (indeed, when it is,then the homomorphism factors into the contractive quotient map followedby an isomorphism of finite-dimensional algebras), so ω is continuous. Thefinal comment is simply the case n = 1.

1.9. Commutative Banach Algebras.

(1) Show that if A is nontrivial in the sense that A 6= 0 (equivalently, 1 6= 0),one has sp(A) 6= 0.

Solution. Apply Theorem 1.9.5 to the element 1.

(2) Show that ω 7→ kerω is a bijection of the Gelfand spectrum onto the set ofmaximal ideals in A.

Solution. This maps into the set of maximal ideals as claimed, becausekerω has codimension one. It is injective because if kerω1 = kerω2 thenker(ω1 − ω2) includes both kerω1 and 1 and hence is all of A. Finally, it issurjective, because if I is a maximal ideal, then A/I is a division algebraand hence isomorphic to C by Gelfand-Mazur; hence the quotient mapA→ A/I has the form a 7→ ω(a) + I for some complex homomorphism ω,from which it follows that I = kerω.

(3) Show that the Gelfand map is an isometry iff ‖x2‖ = ‖x‖2 for every x ∈ A.

Solution. One always has ‖x2‖ = ‖x‖2; hence, if the Gelfand map is anisometry, one must have ‖x2‖ = ‖x‖2.

Conversely, suppose ‖x2‖ = ‖x‖2. By induction, this implies that ‖x2n‖ =‖x‖2n for each n ∈ N. The spectral radius then implies that ‖x‖ = r(x).But one always has r(x) = ‖x‖ as well, so that ‖x‖ = ‖x‖.


(4) The radical of A is the set of quasinilpotent elements

rad(A) =x ∈ A : lim

n→∞‖xn‖1/n = 0

.

Show that rad(A) is a closed ideal in A with the property that A/rad(A) hasno nonzero quasinilpotents (such a commutative Banach algebra is calledsemisimple).

Solution. The quasinilpotent elements are precisely those with spectrum0, by the spectral radius formula. The set of all such is a closed ideal be-cause it is the kernel of the contractive homomorphism Γ : A→ C(sp(A)).Hence the induced map A/rad(A) → C(sp(A)) is injective; a quasinilpo-tent element of A/rad(A) would be in the kernel and therefore would bezero.

(5) Let A and B be commutative unital Banach algebras and let θ : A→ B bea unital algebra homomorphism.

(a) Show that θ induces a continuous map θ : sp(B) → sp(A) by θ(ω) =ω θ.

(b) Assuming that B is semisimple, show that θ is bounded.(c) Deduce that every automorphism of a commutative unital semisimple

Banach algebra is a topological automorphism.

Solution.(a) Clearly θ maps sp(B) to sp(A) as claimed. For continuity, suppose

ων → ω in sp(B), which means ων(b)→ ω(b) for every b ∈ B. Then

θ(ων)(a) = ων(θ(a))→ ω(θ(a)) = θ(ω)(a)

for every a ∈ A, so that θ(ων)→ θ(ω) in sp(A).(b) Suppose (aν , θ(aν)) → (a, b). Then for every ω ∈ sp(B), ω(θ(aν)) →

ω(b). On the other hand, because θ is continuous, we have

ω(θ(aν)) = θ(ω)(aν)→ θ(ω)(a) = ω(θ(a)).

It follows that ω(θ(a)) = ω(b) for every ω ∈ sp(B); by Theorem 1.9.5,this means θ(a) − b has spectrum 0, hence is quasinilpotent. Bysemisimplicity, θ(a) = b. By the Closed Graph Theorem, θ is continu-ous.

(c) This is a special case of the previous statement with A = B.

1.10. Examples: C(X) and the Wiener Algebra.

(1) Let B be the space of functions in C(D) which are represented by convergentpower series

f(z) =

∞∑n=0

anzn

with∑|an| <∞.

Prove the following analogue of Wiener’s theorem: If f ∈ B satisfiesf(z) 6= 0 for every z ∈ D, then 1

f ∈ B.


Solution. First, we show that the maximal ideal space of B (also known asthe disk algebra A(D)) is homeomorphic to D. Clearly point evaluationsare MLF’s on B. For the converse, suppose ω ∈ sp(B); let z0 = ω(id).Since ‖ω‖ = 1, z0 ∈ D. By linearity and multiplicativity, ω agrees withevaluation at z0 on all polynomials; but the latter are dense in B, so ω isevaluation at z0.

Now the Gelfand transform is just the inclusion map into C(D). Also,the Gelfand transform preserves invertibility, by Theorem 1.9.5. If f ∈ B,then Γ(f) has an inverse in C(D), and therefore f has an inverse in B.

(2) Let Z+ denote the semigroup of nonnegative integers. Let T be the rightshift on `1(Z+). Let a = (a0, a1, . . . ) ∈ `1(Z). Show that the set of trans-

lates a, Ta, T 2a, . . . spans `1(Z+) iff the power series

f(z) =

∞∑n=0

anzn, |z| ≤ 1

has no zeros in the closed unit disk.

Solution. Note that `1(Z+) is a Banach algebra; indeed, it is a closed sub-algebra of `1(Z).

A calculation like that on page 30 shows that the maximal ideal spaceof `1(Z+) is homeomorphic to D, with the Gelfand transform being theFourier transform a 7→

∑anz

n. By the previous exercise, f has no zerosin the disk iff f is invertible in C(D), which happens iff a is invertible in`1(Z+) since the Gelfand transform preserves invertibility.

Finally, note that the convolution of a with another element of `1 consistsof a (convergent infinite) sum of translates of a. Hence the span of suchtranslates includes the delta function; but, being translation-invariant, ittherefore includes translates of the delta function, hence everything.

1.11. Spectral Permanence Theorem.

(1) Let A be a unital Banach algebra, let x ∈ A, and let Ω∞ be the unboundedcomponent of C \ σA(x). Show that for every λ ∈ Ω∞ there is a sequenceof polynomials p1, p2, . . . such that

limn→∞

‖(x− λ1)−1 − pn(x)‖ = 0.

Solution. Let B be the unital subalgebra generated by x. Then σB(x) hasthe same unbounded component as σA(x); in particular, λ /∈ σB(x), sothat x − λ1 has an inverse in B; but this inverse must be a norm limit ofpolynomials, because everything in B is.

(2) Let A be a unital Banach algebra that is generated by 1, x for somex ∈ A. Show that σA(x) has no holes.

Solution. Let K denote the “filling in” of σA(x); that is, K is the comple-ment of the unbounded component of the complement. We wish to showthat K = σA(x). By Theorem 1.9.5, it suffices to show that for each ζ ∈ Kthere is a multiplicative linear function ω on A with ω(x) = ζ. Given ζ,define ω on polynomials p(x) by ω(p(x)) = p(ζ); this is contractive because,given a polynomial p, the Maximum Modulus Principle implies that there


exists y ∈ ∂K with |p(ζ)| ≤ |p(y)|. But by the Spectral Mapping Theorem,p(y) ∈ σA(p(x)), so that |p(y)| ≤ r(p(x)) ≤ ‖p(x)‖. We thus have

|ω(p(x))| = |p(ζ)| ≤ |p(y)| ≤ ‖p(x)‖

so that ω is contractive. We can therefore extend ω to a multiplicativelinear functional on the closure of the set of all polynomials in x; but thisis all of A.

(3) Deduce the following theorem of Runge: Let X ⊆ C be a compact setwhose complement is connected. Show that if f(z) = p(z)/q(z) is a rationalfunction with q(z) 6= 0 for z ∈ X, then there is a sequence of polynomialsf1, f2, . . . such that

supz∈X|f(z)− fn(z)| → 0.

Solution. Let A be the unital subalgebra of C(K) generated by the identityfunction—that is, the algebra of norm limits of polynomials. We wish toshow that f ∈ A. Clearly it suffices to show that q is invertible in A.Now σC(K)(id) = K, which has no holes, so that σA(id) = K as well. BySpectral Mapping, σA(q) = σA(q(id)) = q(K), which by hypothesis doesnot contain 0. Hence q is invertible in A.

I’m not sure why the previous problem is needed for this; Corollary 2 isenough.

1.12. Brief on the Analytic Functional Calculus.

Exercise 1.12.1: Let C be an oriented curve in C, f a continuous functionfrom C to a Banach space E, and P the set of finite ordered partitions ofC.(1) Show that for every ε > 0 there is a δ > 0 with the property that

for every pair of oriented partitions P1,P2 satisfying ‖Pk‖ ≤ δ fork = 1, 2, one has ‖R(f,P1)−R(f,P2)‖ ≤ ε.

(2) Verify the estimate∥∥∥∥∫C

f(λ) dλ

∥∥∥∥ ≤ ∫C

‖f(λ)‖ d|λ| ≤ supλ∈C‖f(λ)‖`(C).

Solution:(1) Since f is a continuous function with compact domain C, it is uni-

formly continuous. Thus, given ε > 0, we may choose δ > 0 such that|z − w| < δ implies |f(z)− f(w)| < ε

2`(C) for any z, w ∈ C.

Suppose P1 and P2 are two partitions with ‖P1‖ < δ and ‖P2‖ < δ.Let P3 = P1 ∪ P2 be the common refinement. Let γ0, . . . , γN denotethe points in P1.Let k ∈ 0, . . . , N − 1. Suppose P3 contains the additional pointsη1, . . . , ηm between γk and γk+1. (It is possible that m = 0, i.e. thatthere are no additional points.) We consider the partial Riemann sums

Rk(f,P1) := f(γk+1)(γk+1 − γk)

Rk(f,P3) := f(η1)(η1 − γk) +

m−1∑j=1

f(ηj+1)(ηj+1 − ηj) + f(γk+1)(γk+1 − ηm)


for the portion Ck of C between γk and γk+1. (In the case m = 0 wehave instead Rk(f,P3) = Rk(f,P1).)Note that we can expand

Rk(f,P1) = f(γk+1)

((η1 − γk) +

m−1∑j=1

(ηj+1 − ηj) + (γk+1 − ηm)

).

Then

|Rk(f,P3)−Rk(f,P1)|=∣∣∣∣(f(η1)− f(γk)

)(η1 − γk)+

m−1∑j=1

(f(ηj+1)− f(γk)

)(ηj+1 − ηj

+(f(γk+1)− f(γk)

)(γk+1 − ηm)

∣∣∣∣≤|f(η1)− f(γk)||η1 − γk| +

m−1∑j=1

|f(ηj+1)− f(γk)||ηj+1 − ηj |

+|f(γk+1)− f(γk)||γk+1 − ηm|

≤ ε

2`(C)

(|η1 − γk|+

m−1∑j=1

|ηj+1 − ηj |+ |γk+1 − γk|)

≤ ε

2`(C)`(Ck).

Summing over k, we have

|R(f,P3)−R(f,P1)| =

∣∣∣∣∣∑k

Rk(f,P3)−∑k

Rk(f,P1)

∣∣∣∣∣≤∑k

|Rk(f,P3)−Rk(f,P1)|

≤ ε

2`(C)

∑k

`(Ck) =ε

2.

The same calculation holds with P2 in place of P1. By the TriangleInequality, it follows that

|R(f,P1)−R(f,P2)| ≤ ε.

(2) For every partition P one has by the Triangle Inequality

‖R(f,P)‖ ≤∑k

‖f(γk+1)‖|γk+1 − γk| = R(‖f‖,P)

where the right-hand side denotes the Riemann sum of the function‖f‖ : C → R with respect to d|λ|. Taking the limit of both sides as‖P‖ → 0, we have ‖

∫Cf(λ) dλ‖ ≤

∫C‖f(λ)‖d|λ|. For the second half,

if M = supλ∈C ‖f(λ)‖, then ‖f(λ)‖ ≤M at each point λ, so that∫C

‖f(λ)‖d|λ| ≤∫C

Md|λ| = M`(C).

Notation: For exercise 2-4, let E be a Banach space and T ∈ B(E).


Exercise 1.12.2: Let D = z ∈ C : |z| < R be an open disc containingσ(T ). Let f : D → C be an analytic function defined on D with powerseries

f(z) =

∞∑n=0

cnzn, z ∈ D.

Show that the infinite series of operators∞∑n=0

cnTn

converges absolutely in the sense that∑n |cn|‖Tn‖ <∞.

Solution: Choose some R with r(T ) < R < R. The spectral radius formula

implies that ‖Tn‖ < Rn for sufficiently large n, say n ≥ N , so∑n≥N

|cn|‖Tn‖ <∑n≥N

|cn|Rn

which converges because power series converge absolutely on proper sub-discs.

Exercise 1.12.3: Give a definition of sinT and cosT using power series.Solution: We define

sinT =

∞∑k=0

(−1)k

(2k + 1)!T 2k+1, cosT =

∞∑k=0

(−1)k

(2k)!T 2k.

By the previous exercise, these series are absolutely convergent for any T ,since the complex-valued functions sin and cos are entire.

Exercise 1.12.4: Use your definitions in the preceding exercise to show that

(sinT )2 + (cosT )2 = 1.

Solution: This is immediate from the holomorphic functional calculus, whichsends the complex function sin2 z + cos2 z − 1 to 0. Alternatively, one canprove it directly by manipulating the above series: Since

sin2 T =

∞∑j,k=0

(−1)j+k

(2k + 1)!(2j + 1)!T 2(j+k+1) =

∞∑`=1

(−1)`−1T `∑s odd

1

s!(2`− s)!

and

cos2 T =

∞∑j,k=0

(−1)j+k

(2k)!(2j)!T 2(j+k) = 1 +

∞∑`=1

(−1)`T 2`∑s even

1

s!(2`− s)!,

we have

sin2 T + cos2 T = 1 +

∞∑`=1

(−1)`T 2`

( ∑s even

1

s!(2`− s)!−∑s odd

1

s!(2`− s)!

)

= 1 +

∞∑`=1

(−1)`T 2`2∑s=0

(−1)s

s!(2`− s)!.

But, by the Binomial Theorem,

2∑s=0

(−1)s

s!(2`− s)!=

1

(2`)!

2∑s=0

(−1)s(

2`

s

)=

1

(2`)!(1− 1)2` = 0,


so that sin2 T + cos2 T = 1.

2. Operators on Hilbert Space

2.1. Operators and Their C∗-Algebras.

Exercise 2.1.1: Let [·, ·] : H × H → C be a sesquilinear form defined on aHilbert space H. Show that [·, ·] satisfies the polarization formula

4[ξ, η] =

3∑k=0

ik[ξ + ikη, ξ + ikη

].

Solution: We expand the right-hand side using sesquilinearity:

3∑k=0

ik[ξ + ikη, ξ + ikη

]=

3∑k=0

ik(

[ξ, ξ] + ik[η, ξ] + (−i)k[ξ, η] + [η, η])

=

(3∑k=0

ik

)[ξ, ξ] +

(3∑k=0

(−1)k

)[η, ξ] +

3∑k=0

[ξ, η] +

(3∑k=0

ik

)[η, η]

= 4[ξ, η]

Exercise 2.1.2: Let A ∈ B(H) be a Hilbert space operator. The quadraticform of A is the function qA : H → C defined by qA(ξ) = 〈Aξ, ξ〉. Thenumerical range and numerical radius of A are defined, respectively,by

W (A) = qA(ξ) : ‖ξ‖ = 1 ⊆ C,w(A) = sup|qA(ξ)| : ‖ξ‖ = 1.

(a) Show that A is self-adjoint iff qA is real-valued.(b) Show that w(A) ≤ ‖A‖ ≤ 2w(A) and deduce that qA = qB only when

A = B.Solution:

(a) We have

∀ξ : 〈Aξ, ξ〉 ∈ R ⇔ ∀ξ : 〈Aξ, ξ〉 = 〈Aξ, ξ〉 ⇔ ∀ξ : 〈Aξ, ξ〉 = 〈ξ, Aξ〉⇔ ∀ξ : 〈Aξ, ξ〉 = 〈A∗ξ, ξ〉 ⇔ ∀ξ, η : 〈Aξ, η〉 = 〈A∗ξ, η〉⇔ A = A∗

The penultimate equivalence uses the polarization identity from theprevious exercise, which implies that the values of the sesquilinearform (ξ, η) 7→ 〈Aξ, η〉 are determined by the values of the quadraticform ξ 7→ 〈Aξ, ξ〉.

(b) By Cauchy-Schwarz, |qA(ξ)| ≤ ‖Aξ‖‖ξ‖ ≤ ‖A‖‖ξ‖2 for all ξ, so thatw(A) ≤ ‖A‖. For the other direction, the polarization formula impliesthat

‖A‖ = sup‖η‖≤1‖ξ‖≤1

|〈Aξ, η〉| ≤ sup‖η‖≤1‖ξ‖≤1

1

4

3∑k=0

|qA(ξ + ikη)| ≤ sup‖χ‖≤

√2

|qA(χ)| = 2w(A).

Finally, observe that qA+B = qA + qB . Then

‖A−B‖ = 0⇔ w(A−B) = 0⇔ qA−B = 0⇔ qA − qB = 0⇔ qA = qB .


Exercise 2.1.3: Show that the adjoint operator A 7→ A∗ in B(H) is weaklycontinuous but not strongly continuous.

Solution: If Tν → 0 in WOT, then for any ξ, η ∈ H,

〈T ∗ν ξ, η〉 = 〈ξ, Tνη〉 = 〈Tνη, ξ〉 → 0,

so that T ∗ν → 0 in WOT. Hence adjunction is WOT-continuous.However, if S is the right shift on `2, then S∗n → 0 in SOT, whereas

Sn 6→ 0 in SOT, so that adjunction is not SOT-continuous.Exercise 2.1.4: Show that the only operators that commute with all opera-

tors in B(H) are the scalar multiples of the identity.Solution: For each ξ, η ∈ H let ξ ⊗ η∗ denote the rank-one operator given

by (ξ ⊗ η∗)(y) = 〈y, η〉ξ. If an operator T commutes with all rank-oneoperators, then for any ξ and η,

‖η‖2(Tξ) = T (ξ ⊗ η∗)(η) = [(ξ ⊗ η∗) T ](η) = 〈Tη, η〉ξ.

Note that this implies that every ξ is an eigenvector of T . Let cξ denote thecorresponding eigenvalue. Returning to the above equation with Tξ = cξξand Tη = cηη, we have

‖η‖2cξξ = cη‖η‖2ξ,

so that cξ = cη. That is, all vectors have the same eigenvalue. Thus T is ascalar times the identity.

Exercise 2.1.5: Let C be the closure in the strong operator topology of theset of all unitary operators in B(H). Show that C consists of isometries.

Solution: Suppose Uν are unitaries and S an operator such that Uν → S inSOT. Then for any ξ ∈ H,

‖Sξ‖ = ‖ limνUνξ‖ = lim

ν‖Uνξ‖ = ‖ξ‖

since ‖Uνξ‖ = ‖ξ‖ for all ν. Thus S is an isometry.Note that we never used the fact that Uν are actually unitaries, not

isometries; this proof shows that the set of isometries is SOT-closed.Exercise 2.1.6: Show that the unilateral shift S belongs to C be exhibiting a

sequence of unitary operators U1, U2, . . . that converges to S in the strongoperator topology.

Solution: Define unitaries Un on the standard ONB by

Unek =

ek+1 1 ≤ k ≤ n− 1

e1 k = n

ek k > n.

This is unitary because it maps an ONB to an ONB (it permutes thestandard basis). Fix x ∈ `2, and for each n, let xn be the projection of xonto the span of (e1, . . . , en−1). Then Unxn = Sxn, so that

‖Unx− Sx‖ = ‖Un[xn + (x− xn)]− S[xn + (x− xn)]‖ = ‖Unxn − Sxn + (Un − S)(x− xn)‖= ‖(Un − S)(x− xn)‖ ≤ ‖Un − S‖‖x− xn‖ ≤ 2‖x− xn‖ → 0.

Hence Un → S in SOT.


Exercise 2.1.7: Let (X,µ) be a σ-finite measure space and let f : X → C bea bounded complex-valued Borel function. Show that the essential rangeof f can be characterized as the intersection⋂

g(X) : g ∼ f

of all the closed ranges of all bounded Borel functions g : X → C that agreewith f almost everywhere (dµ).

Solution: Suppose λ is in the essential range of f , and g = f a.e. Thenfor every n ∈ N, g−1(B(λ, 1/n)) has positive measure; in particular, it isnonempty, so that g(X) contains points in B(λ, 1/n) for every n. Thus λis in the closure of g(X).

Conversely, suppose λ is not in the essential range of f . Let r > 0 suchthat f−1(B(λ, r)) has measure zero. Let µ ∈ C\B(λ, r). Define g : X → Cby

g(x) =

f(x) f(x) /∈ B(λ, r)

µ else.

Then g agrees with f a.e., and λ /∈ g(X).

2.2. Commutative C∗-Algebras.

Exercise 2.2.1: LetA be a unital Banach algebra and a0, a1, . . . and b0, b1, . . .be two sequences of elements of A such that

∑‖an‖ <∞ and

∑‖bn‖ <∞,

and let x =∑an, y =

∑bn. Prove that the product xy is given by the

series xy =∑cn, where

cn = a0bn + a1bn−1 + · · ·+ anb0,

the series∑cn being absolutely convergent in the sense that

∑‖cn‖ <∞.

Solution: Let us introduce the notation

xn = a0 + · · ·+ an

yn = b0 + · · ·+ bn

Xn = ‖a0‖+ · · ·+ ‖an‖Yn = ‖b0‖+ · · ·+ ‖bn‖

Sn =

n∑k=0

ck

An =∑

j+k≤n

‖aj‖‖bk‖ =

n∑k=0

‖ck‖.

Then An ≤ XnYn, so that An is bounded above and hence converges, i.e.∑ck converges absolutely. Now

‖xnyn − Sn‖ =

∥∥∥∥∥∥∥∥∑j,k≤nj+k>n

ajbk

∥∥∥∥∥∥∥∥ ≤ A2n −An


which tends to zero since An is a convergent sequence. Thus,∞∑k=0

ck = limn→∞

Sn = limn→∞

xnyn = xy.

We note that the key estimate above, using A2n−An as an upper bound,can be pictured as a relationship between squares and right triangles on aquarter-infinite lattice. If Sn indicates an n × n square and Tn an n × nright triangle, then Tn ⊆ Sn ⊆ T2n whereas Sn ⊆ T2n ⊆ S2n. Here thepoints of the lattice represent pairs (j, k) over which one sums ajbk.

Exercise 2.2.2: Let A be a C∗-algebra.(a) Show that the involution in A satisfies ‖x∗‖ = ‖x‖.(b) Show that if A contains a unit 1, then ‖1‖ = 1.

Solution:(a) Obviously this is true if x = 0; for nonzero x, divide the inequality‖x‖2 = ‖x∗x‖ ≤ ‖x∗‖‖x‖ by ‖x‖ to obtain ‖x‖ ≤ ‖x∗‖. Applyingthe same result to x∗ yields ‖x∗‖ ≤ ‖x‖, from which it follows that‖x∗‖ = ‖x‖.

(b) For any x ∈ A,

1∗x∗ = (x1)∗ = x∗ = (1x)∗ = x∗1∗

so that 1∗ is a multiplicative identity; hence 1∗ = 1. Then ‖1‖ =‖1∗1‖ = ‖1‖2 and, since 1 6= 0, we can divide both sides by ‖1‖ toobtain ‖1‖ = 1.

Notation: In the following exercises, X and Y denote compact Hausdorffspaces, and θ : C(X)→ C(Y ) denotes an isomorphism of complex algebras.We do not assume continuity of θ.

Exercise 2.2.3: Let p ∈ Y . Show that there is a unique point q ∈ X suchthat (θf)(p) = f(q) for all f ∈ C(X).

Solution: Let I = f ∈ C(Y ) : f(q) = 0, which is a maximal ideal in C(Y )corresponding to the MLF of evaluation at q. Since θ is an isomorphism,J = θ−1I is a maximal ideal in C(X). As proved in section 1.10, everymaximal ideal in C(X) is the zero set of a unique point p ∈ X.• Given f ∈ C(Y ), let λ = f(q). Then f−λ1 ∈ I, so that θ−1(f)−λ1 ∈J . Hence (θ−1f)(p)−λ = 0, so (θ−1f)(p) = λ. Thus p has the desiredproperty.

• If x ∈ X with x 6= p, then there is some g ∈ J with g(x) 6= 0, by theuniqueness result from chapter 1.10 (or by Urysohn’s lemma). Since(θg)(q) = 0, this x does not have the desired property. Hence p isunique.

Exercise 2.2.4: Show that there is a homeomorphism φ : Y → X such thatθf = f φ.

Solution: The previous exercise gives us a bijection φ : Y → X with thisproperty, so we need only show φ is continuous. (This will imply it’s ahomeomorphism, since it’s a continuous bijection from a compact space toa Hausdorff space.) Suppose not; then there is some closed F ⊆ X suchthat G = φ−1(F ) is not closed in Y . Let y ∈ G \ G and let x = φ(y).By Urysohn’s lemma, there exists f ∈ C(X) such that f(x) = 1 and fvanishes on F . But then θf = 0 on G, hence on G, while also (θf)(y) = 1,a contradiction. Hence φ is continuous.


Exercise 2.2.5: Conclude that θ is necessarily a self-adjoint linear map inthe sense that θ(f∗) = θ(f)∗, f ∈ C(X).

Solution: This is immediate since (f φ)∗ = f∗ φ.Exercise 2.2.6: Formulate and prove a theorem that characterizes unital

algebra homomorphisms θ : C(X) → C(Y ) in terms of certain mapsφ : Y → X. Which maps φ give rise to isomorphisms?

Solution: Theorem: For every unital algebra homomorphism θ : C(X) →C(Y ) there exists a unique continuous function Θ : Y → X such thatθ = · Θ (that is, for all φ ∈ C(X), θ(φ) = φΘ). Moreover, θ is (injective,surjective) iff Θ is (surjective, injective). In particular, θ is an isomorphismiff Θ is a homeomorphism.

Proof: Let ΨX : X → Σ(C(X)) be the canonical homeomorphism x 7→(φ 7→ φ(x)), and similarly with ΨY . Define θ : Σ(C(Y )) → Σ(C(X)) by

θ = · θ, that is, for all ω ∈ Σ(C(Y )) we have θ(ω) = ω θ, which in turn

means that for all φ ∈ C(X) we have θ(ω)(φ) = ω(θ(φ)). Define Θ : Y → X

by Θ = Ψ−1X θ ΨY .

Let φ ∈ C(X). By definition of Θ, we have φ Θ = φ Ψ−1X θ ΨY . In

case it’s not immediately obvious that this equals θ(φ), let’s slowly unwindthe definitions:• Let y ∈ Y .• ΨY (y) is the multiplicative linear functional on C(Y ) given by ψ 7→ψ(y).

• θ(ΨY (y)) is the multiplicative linear functional on C(X) given by ψ 7→θ(ψ)(y).

• Ψ−1X (θ(ΨY (y))) is the unique point x ∈ X such that, for all ψ ∈ C(X),

ψ(x) = [θ(ΨY (y))](ψ).

• φ(Ψ−1X (θ(ΨY (y)))) is, therefore, equal to [θ(ΨY (y))](φ).

• By the definition of θ(ΨY (y)), this is θ(φ)(y).Thus, θ(φ) = φ Θ as claimed.

This proves the existence of Θ; for uniqueness, suppose there were adifferent map Ξ with the same properties. Let y ∈ Y such that Θ(y) 6= Ξ(y).By Urysohn’s lemma, there exists φ ∈ C(X) with φ(Θ(y)) 6= φ(Ξ(y)).Hence Θ is unique.

The last assertions are a consequence of the fact that surjectivity andinjectivity correspond to left and right cancellations, and that the corre-spondence between θ and Θ is given by a contravariant functor. Thus, ifZ is any other compact Hausdorff space and λ1, λ2 : C(Z)→ C(X) unitalhomomorphisms, corresponding to continuous maps Λ1,Λ2 : X → Z viaλi = · Λi, then (λi θ) = · (Θ Λi). Therefore, if Θ is right-cancellative,in that Θ Λ1 = Θ Λ2 implies Λ1 = Λ2, then θ is left-cancellative, mean-ing that λ1 θ = λ2 θ implies λ1 = λ2. For the other direction (left-cancellativity of Θ corresponds to right-cancellativity of θ) we note that Θ

has the same cancellativity properties as θ, and repeat the above argument.In more concrete terms, however, suppose Θ is not surjective. Let x be

in the complement of the range, which is open; by Urysohn, there existsφ ∈ C(X) which is zero on the range of Θ and such that φ(x) = 1. Thenθ(φ) = φ Θ = 0, proving that θ is not injective. Conversely, if θ is not


injective, let φ ∈ C(X) be nonzero such that φ Θ = 0; then φ is zero onthe range of Θ, but is not zero everywhere, so that Θ is not surjective.

Similarly, if Θ is not injective, let y1 6= y2 such that Θ(y1) = Θ(y2).Choose ψ ∈ C(Y ) with ψ(y1) = 1 and ψ(y2) = 0. If θ were injective, therewould exist φ ∈ C(X) with θ(φ) = ψ. But then

φ(Θ(y1)) = θ(φ)(y1) = ψ(y1) = 1 6= 0 = ψ(y2) = θ(φ)(y2) = φ(Θ(y2))

which is impossible since Θ(y1) = Θ(y2). Hence θ is not surjective. Con-versely, suppose θ is injective. Then it is a homeomorphism onto its rangeX, so that θ factors as ι χ where χ : Y → X is a homeomorphism andι : X → X is the inclusion map. Correspondingly, we get the factorizationθ = ηθ, where θ : C(X)→ C(X) is the map θ = ·ι, and η : C(X)→ C(Y )

is the isomorphism η = · χ. Thus θ will be surjective iff θ is; but this isthe restriction map, which is surjective by Urysohn’s lemma again (or, ifyou like, by the Tietze extension theorem).

Notation: In the remaining exercises, let H be a Hilbert space and let T ∈B(H)−1 be an invertible operator. Define θ : B(H)→ B(H) by

Θ(A) = TAT−1, A ∈ B(H).

Exercise 2.2.7: Show that θ is an automorphism of the Banach algebra struc-ture of B(H).

Solution: Clearly Θ is an algebraic automorphism; it is bounded because

‖Θ(A)‖ ≤ ‖T‖‖A‖‖T‖−1‖ ⇒ ‖Θ‖ ≤ ‖T‖‖T‖−1,

and its inverse A 7→ T−1AT is bounded in the same way.Exercise 2.2.8: Show that θ(A∗) = θ(A)∗ for all A ∈ B(H) iff T is a scalar

multiple of the identity operator.Solution: The “if” direction is clear. Conversely, suppose Θ(A∗) = Θ(A)∗

for all A. That is,

TA∗T−1 = (TAT−1)∗ = T−∗A∗T ∗ for all A ∈ B(H).

This implies

T ∗TA∗ = A∗T ∗T for all A ∈ B(H),

i.e. that T ∗T ∈ B(H)′ = CI. Let r > 0 such that T ∗T = r2I. Theequation Θ(T ∗) = Θ(T )∗ implies T ∗T = TT ∗, so that T is normal. ThenU = 1

rT satisfies

U∗U =1

r2T ∗T = I =

1

r2TT ∗ = UU∗

and hence is unitary.

2.3. Continuous Functions of Normal Operators.

Exercise 2.3.1: Show that the spectrum of a normal operator T ∈ B(H) isconnected if and only if the C∗-algebra generated by T and 1 contains noprojections other than 0 and 1.

Solution: By the spectral theorem, C∗(T ) contains a nontrivial projection iffC(σ(T )) does. But a nontrivial projection in C(σ(T )) is the characteristicfunction of a nontrivial union of connected components, which can onlyexist of σ(T ) is disconnected.


Notation: Consider the algebra C of all continuous functions f : C → C.There is no natural norm on C, but for every compact subset X ⊆ C thereis a seminorm

‖f‖X = supz∈X|f(z)|.

Exercise 2.3.2: Given a normal operator T ∈ B(H), show that there is anatural extension of the functional calculus to a *-homomorphism f ∈ C →f(T ) ∈ B(H) that satisfies ‖f(T )‖ = ‖f‖σ(T ).

Solution: Given T , we obtain a restriction homomorphism ρT : C → C(σ(T ))which satisfies ρT (f) = ‖f‖σ(T ). Composing this with the functional cal-culus yields the result.

Exercise 2.3.3 (Continuity of the functional calculus): Fix a functionf ∈ C and let T1, T2, . . . be a sequence of normal operators that convergesin norm to an operator T . Show that f(Tn) converges in norm to f(T ).

Solution: First note that T is normal, since norm convergence implies T ∗n →T ∗ and therefore TT ∗ − TT ∗ = limTnT

∗n − T ∗nTn = 0.

Let K ⊆ C be a compact set containing all the σ(Tn). Since one even-tually has r(Tn) = ‖Tn‖ ≤ ‖T‖+ 1, this is possible.

Let Pk be a sequence of polynomials converging uniformly to f in C(K).Given ε > 0, choose k such that ‖Pk − f‖K < ε. Since Pk(Tn)→ Pk(T )

in norm (by norm-continuity of products and sums), there is N ∈ N suchthat ‖Pk(Tn)− Pk(T )‖ < ε for n > N . Then for n > N one has

‖f(Tn)− f(T )‖ = ‖f(Tn)− Pk(Tn) + Pk(Tn)− Pk(T ) + Pk(T )− f(T )‖≤ ‖f(Tn)− Pk(Tn)‖+ ‖Pk(Tn)− Pk(T )‖+ ‖Pk(T )− f(T )‖ < 3ε.

Thus f(Tn)→ f(T ).

2.4. The Spectral Theorem and Diagonalizations.

Exercise 2.4.1: Let X be a Borel space, f a bounded complex-valued Borelfunction on X, and µ and ν two σ-finite measures on X. The multiplicationoperator Mf defines bounded operators A on L2(X,µ) and B on L2(X, ν).Assuming that µ and ν are mutually absolutely continuous, show that thereis a unitary operator W : L2(X,µ)→ L2(X, ν) such that WA = BW .

Solution: By Radon-Nikodym, dν = hdµ for some nonnegative function h.(Nonnegative because L2 spaces are defined with respect to positive mea-sures.) Define W : L2(X,µ) → L2(X, ν) by (Wφ)(x) = 1√

h(x)φ(x). Then

WA and BW are both Mf/√h, i.e.

(WAφ)(x) =f(x)√h(x)

φ(x) = (BWφ)(x)

for every φ ∈ L2(X,µ).Exercise 2.4.2: Show that every diagonalizable operator on a separable Hilbert

space is unitarily equivalent to a multiplication operator Mf acting onL2(X,µ) where (X,µ) is a probability space, that is, a measure space forwhich µ(X) = 1.

Solution: By definition, if T ∈ B(H) is diagonalizable, there is a σ-finitemeasure space (X, ν), a unitary U : H → L2(X, ν), and a function φ ∈L∞(X, ν) such that such that UT = MφU . Let X =

⋃Xi where ν(Xi)


is finite and the Xi are disjoint. Define a measure µ on X by µ(E ∩Xi) = ν(E∩Xi)

2iν(Xi). Then µ is a probability on X which is mutually absolutely

continuous with respect to ν. By the previous exercise, Mφ is unitarilyequivalent to a multiplication operator on L2(X,µ), so T is as well.

Notation: The following exercises concern the self-adjoint operator A definedon the Hilbert space `2(Z) by

Aξn = ξn+1 + ξn−1.

Exercise 2.4.3: Show that A is diagonalizable by exhibiting an explicit uni-tary operator W : L2(T, dθ/2π) → H for which WMf = AW , wheref : T→ R is the function f(eiθ) = 2 cos θ. Deduce that the spectrum of Ais the interval [−2, 2] and that the point spectrum of A is empty.

Solution: The unitary is just the Plancherel transform, i.e. the operatorgiven by

(Wg)(n) = n =

∫Tf(ζ)ζ−ndζ =

∫ 1

0

f(e2πiθ)e−2πinθ dθ.

For any g ∈ L2(T) one then has Wg =∑n g(n)ξn where ξn is the stan-

dard basis for `2(Z). Hence

AWg =∑n

g(n)[ξn+1 + ξn−1] =∑m

[g(m− 1) + g(m+ 1)]ξm

whereas

WMfg =∑n

fg(n)ξn =∑n

[g(n− 1) + g(n+ 1)]ξn.

Exercise 2.4.4: Let U be the operator defined on L2(T, dθ/2π) by

(Uf)(eiθ) = f(e−θ).

Show that U is a unitary operator that satisfies U2 = 1 and which commuteswith Mf . [Typo in the book: says “commutes with A”.]

Solution: Clearly U2 = 1. It’s also clear that ‖Uφ‖ = ‖φ‖, so that U is anisometry; since it’s also invertible, it’s unitary. Finally, for any φ ∈ L2(T),

(UMfφ)(eiθ) = (Mfφ)(e−iθ) = 2 cos(−θ)φ(e−iθ) = 2 cos(θ)φ(e−iθ) = (MfUφ)(eiθ)

so that U commutes with Mf .Exercise 2.4.5: Let B be the set of all operators on L2(T, dθ/2π) that have

the form Mf + MgU where f, g ∈ L∞(T, dθ/2π) and U is the unitaryoperator of the preceding exercise. Show that B is *-isomorphic to the C∗-algebra of all 2 × 2 matrices of functions M2(B0), where B0 is the abelianC∗-algebra L∞(X,µ), X being the upper half of the unit circle X = z ∈T : im z ≥ 0 and µ being the restriction of the measure dσ = dθ/2π to X.

Solution: First, we note that one can recover f and g from the operatorM(f) +M(g)U ; that is, the linear map Φ : L∞(T)⊕ L∞(T)→ B given byΦ(f⊕g) = M(f)+M(g)U is injective. This is clear from a consideration ofthis operator applied to the characteristic functions of the upper and lowersemicircles.

Next, we introduce notation for some of the relevant maps betweenspaces:


• M : L∞(T)→ B(L2(T)) is the *-algebra homomorphism that takes fto the multiplication operator by f

• Pu : L∞(T)→ L∞(X) and P` : L∞(T)→ L∞(Y ) are the *-homomorphismsof projection onto the upper and lower semicircles, which we denoteby X and Y ; we view L∞(X) and L∞(Y ) as subalgebras of L∞(T)

• F : L∞(T) → L∞(T) is the *-homomorphism (Ff)(eiθ) = f(e−iθ),which satisfies F2 = id. Note that F(L∞(X)) = L∞(Y ) and con-versely.

• Ψ : B →M2(B0) is the map

Ψ(M(f) +M(g)U) =

[Pu(f) Pu(g)

Pu(F(g)) Pu(F(f))

],

which we will prove to be a *-algebra isomorphism.Here are some elementary relations between these maps, all easy to ver-

ify:• UM(f) = M(F(f))U• FP` = PuF and FPu = P`FIt is clear that the map Ψ is bijective and linear. It is self-adjoint because

Ψ(

[M(f) +M(g)U ]∗)

= Ψ(M(f)∗ + UM(g)∗

)= Ψ

(M(f) +M(F(g))U

)=

[Pu(f) Pu(F(g))

Pu(F(F(g))) Pu(F(f))

]=

[Pu(f) Pu(g)

Pu(F(g)) Pu(F(f))

]∗and is multiplicative because

Ψ(

[M(f) +M(g)U ][M(h) +M(k)U ])

= Ψ(M(f)M(h) +M(f)M(k)U +M(g)UM(h) +M(g)UM(k)U

)= Ψ

(M(f)M(h) +M(f)M(k)U +M(g)M(F(h))U +M(g)M(F(k))

)= Ψ

(M(fh+ gF(k)) +M(fk + gF(h))U

)=

[Pu(fh+ gF(k)) Pu(fk + gF(h))

Pu(F(fk + gF(h))) Pu(F(fh+ gF(k)))

]=

[Pu(f)Pu(h) + Pu(g)Pu(F(k)) Pu(f)Pu(k) + Pu(g)Pu(F(h))

Pu(F(f))Pu(F(k)) + Pu(F(g))Pu(h) Pu(F(f))Pu(F(h)) + Pu(F(g))Pu(k)

]=

[Pu(f) Pu(g)

Pu(F(g)) Pu(F(f))

] [Pu(h) Pu(k)

Pu(F(k)) Pu(F(h))

]= Ψ(M(f) +M(g)U)Ψ(M(h) +M(k)U).

Notation: The following exercises ask you to compare the operator A to arelated operator B that acts on the Hilbert space L2([−2, 2], ν) where νis Lebesgue measure on the interval [−2, 2]. The operator B is defined by(Bf)(x) = xf(x).


Exercise 2.4.6: Show that B has spectrum [−2, 2] and that it has no pointspectrum. Deduce that for every f ∈ C([−2, 2]) we have ‖f(A)‖ = ‖f(B)‖.

Solution: The spectrum of a multiplication operator is its essential range,which in this case is [−2, 2]. Since A and B are both normal (in fact, self-adjoint) and have the same spectrum, the functional calculi for the two ofthem together imply that ‖f(A)‖ = ‖f(B)‖ for all f ∈ C([−2, 2]).

I’m not sure why he mentioned the point spectrum, as it’s unnecessaryfor the above proof.

Exercise 2.4.7: Show that A and B are not unitarily equivalent.Solution: First, we show that B′ = B′′, where ′ denotes the commutant.

• Clearly every multiplication operator commutes with B.• Conversely, suppose T commutes with B = Mx.• Let g = T1. We will show that g ∈ L∞([−2, 2]) and that B = Mg.• For any polynomial p, p(Mx) = Mp(x), so that T commutes with Mp(x)

as well.• It follows that T commutes with Mf for any f ∈ C([−2, 2]). Indeed,

since polynomials are dense in C([−2, 2]) by the Weierstrass approx-imation theorem, given any f ∈ C([−2, 2]) one can find polynomialspn → f uniformly; then ‖Mpn−Mf‖ → 0, because Mpn−Mf = Mpn−fand the norm of a multiplication operator is the essential supremumof its symbol. Then

‖TMf −MfT‖ = ‖T (Mf −Mp) + TMp −MpT + (Mp −Mf )T‖ = ‖T (Mf −Mp) + (Mp −Mf )T‖≤ ‖T‖‖Mf −Mp‖+ ‖Mp −Mf‖‖T‖ → 0.

• If we knew that T commuted with Mf for any f ∈ L∞([−2, 2]), wecould quickly infer that g ∈ L∞([−2, 2]) and that T = Mg as follows:For any h ∈ L∞([−2, 2]),

Th = TMh1 = MhT1 = Mhg = Mgh

so that T andMg agree on the dense subspace L∞([−2, 2]) of L2([−2, 2]).If we know that g ∈ L∞([−2, 2]), so that Mg is bounded, we can con-clude that T = Mg. If g /∈ L∞, let hn be the characteristic function ofx : |g(x)| > n, and it would follow that

‖Thn‖ = ‖ghn‖ > n‖hn‖

so that T would not be bounded.Unfortunately I don’t know a good way to show that T commutes withMg for all g ∈ L∞, since C([−2, 2]) is not dense in L∞([−2, 2]).• However, we can still prove that g ∈ L∞ and that T = Mg as follows:

Let φ ∈ L2([−2, 2]). Let fn ∈ C([−2, 2]) such that fn → φ in L2. Letfnj be a subsequence such that fnj → φ a.e. (see for instance Theorem3.12 in ([Rud91]) or Exercise 6.9 in ([Fol99]) for the existence of such).

Then gfnj = TfnjL2

→ Tφ. Let fnjk be a sub-subsequence such that

gfnjka.e.→ Tφ. Then

Tφ = lima.e.

Tfnjk = lima.e.

gfnjk = gφ,

so that Tφ = Mgφ for all φ ∈ L2. Since T is bounded, one must haveg ∈ L∞([−2, 2]) as well.


• I have a feeling that there’s a simpler way to prove that B′ consistsonly of multiplication operators, but I don’t know what it is. Sugges-tions welcome!

• If M denotes the algebra of multiplication operators, the fact thatB ⊆ M implies thatM′ ⊆ B′ =M. On the other hand, sinceMis commutative, we also have M ⊆ M′. Hence B′′ = M′ = M =B′.

On the other hand, it is not true that A′ = A′′. If it were, it would

follow that A′ = A′′, where A := M2 cos θ is the operator shown to be

unitarily equivalent to A in Exercise 3. However, U ∈ A′ as previously

noted, so that A′′ ⊆ U′. But the latter does not contain multiplication

operators of non-even functions, so it is strictly smaller than A′, whichcontains all multiplication operators.

Thus, we have found a property which is true of B and false of A, andwhich is a unitary invariant. It follows that B is not unitarily equivalentto A.

Exercise 2.4.8: Show that A is unitarily equivalent to B ⊕B.Solution: It is clear that A is unitarily equivalent to the direct sum of M2 cos θ

on L2([0, π], dθ/2π) and M2 cos θ on L2([−π, 0], dθ/2π), and that these areunitarily equivalent to each other. Hence it suffices to show that A ∼ Twhere T is M2 cos θ on L2([0, π], dθ/2π).

Define a linear map Φ : L2([−2, 2]) → L2([0, π], dθ/2π) by (Φf)(θ) =

2√πf(2 cos θ)

√sin θ. Then

‖Φf‖2 =1

2π

∫ π

0

|(Φf)(θ)|2 dθ

= 2

∫ π

0

|f(2 cos θ)|2 sin θ dθ

=1

4π

∫ 2

−2

|f(u)|2 du = ‖f‖2

so that Φ is an isometry. Moreover, Φ is bijective, with

(Φ−1g)(x) =1

2√π(4− x2)

g(cos−1(x/2)),

so that Φ is unitary. Finally, TΦ = ΦB, proving that T and B are unitarilyequivalent.

2.5. Representations of Banach *-Algebras.

Exercise 2.5.1: Let A = A∗ ⊆ B(H) be a self-adjoint algebra of operatorson a Hilbert space, and let

N = ξ ∈ H : Aξ = 0

be the null space of A. Show that the orthogonal complement of N is theclosed linear span of AH = Tξ : T ∈ A, ξ ∈ H and that both N and[AH] are A-invariant subspaces.


Solution: Let (AH) denote the (non-closed) linear span of AH, which hasclosure [AH]. Then (AH)⊥ = [AH]⊥. But show (AH)⊥ = N , because

h ∈ (AH)⊥ ⇔ ∀ξ1, . . . , ξn ∈ H : ∀a1, . . . , an ∈ A :⟨h,∑

aiξi

⟩= 0

⇔ ∀ξ1, . . . , ξn ∈ H : ∀a1, . . . , an ∈ A :∑〈h, aiξi〉 = 0

⇔ ∀ξ1, . . . , ξn ∈ H : ∀a1, . . . , an ∈ A :∑〈a∗i h, ξi〉 = 0

⇔ ∀a ∈ A : ah = 0

⇔ h ∈ N.

Exercise 2.5.2: Let A be a Banach *-algebra with unit 1, and let π ∈rep(A,H) be a representation of A. Show that π is nondegenerate iffπ(1) = 1H .

Solution: Because π(1) = π(1∗) = π(1)∗ and π(1) = π(12) = π(1)2, we seethat π(1) must be a projection P ∈ B(H). Then for any h ∈ H and a ∈ A,

π(a)h = π(1a)h = π(1)π(a)h = Pπ(a)h

so that π(a)h ∈ PH; thus, PH is the essential subspace for π. By definition,π is nondegenerate iff this is all of H, which happens iff P = 1H .

Exercise 2.5.3: LetA be a Banach *-algebra. A representation π ∈ rep(A,H)is said to be cyclic if there is a vector ξ ∈ H with the property that the setof vectors π(A)ξ is dense in H. Show that a representation π ∈ rep(A,H)is nondegenerate iff it can be decomposed into a direct sum of cyclic sub-representations in the following sense: There is a family Hi ⊆ H, i ∈ Iof nonzero subspaces of H that are mutually orthogonal, π(A)-invariant,that sum to H, and such that for each i ∈ I there is a vector ξi ∈ Hi withπ(A)ξi = Hi.

Solution: If π is a direct sum of cyclic subrepresentations, then it must bedegenerate, as π(A)H contains each Hi and therefore all of H.

Conversely, suppose π is nondegenerate. Let Z be the family of allsubrepresentations of π which are direct sums of cyclic subrepresentations,with the partial order that µ ≤ ν if µ is a subrepresentation of ν. ClearlyZ is nonempty, as any vector ξ ∈ H gives rise to a cyclic subrepresentationπξ on the subspace π(A)ξ. Moreover, every chain in Z has an upper boundgiven by direct summation over its members. By Zorn’s lemma, there is amaximal representation ψ ∈ Z. Suppose ψ is a proper subrepresentationof π, on a proper subspace H ′ ⊆ H. Let ξ ∈ (H ′)⊥ be any nonzero vector;then ψ ⊕ πξ is an element of Z strictly greater than ψ, a contradiction.Hence ψ = π and π is a direct sum of cyclic subrepresentations.

Exercise 2.5.4: LetA be a Banach *-algebra. A representation π ∈ rep(A,H)is said to be irreducible if the only closed π(A)-invariant subspaces of Hare the trivial ones 0 and H. Show that π is irreducible iff the commutantof π(A) consists of scalar multiples of the identity operator.

Solution: A closed subspace of H is π(A)-invariant iff its projection com-mutes with π(A). Hence π is irreducible iff π(A)′ contains no nontrivialprojections. This in turn is equivalent to π(A)′ being one-dimensional be-cause π(A)′ is a von Neumann algebra, and therefore is generated by itsprojections. (This fact was mentioned on page 42 of the text.)


Exercise 2.5.5: Let X be a compact Hausdorff space and let π be an irre-ducible representation of C(X) on the Hilbert space H. Show that H is one-dimensional and that there is a unique point p ∈ X such that π(f) = f(p)1for all f ∈ C(X).

Solution: By the previous exercise, π(C(X))′ contains only the scalar op-erators. But π(C(X)) is commutative, so that π(C(X)) ⊆ π(C(X))′.Hence π(C(X)) contains only scalar operators, so that π has the formπ(f) = ω(f)1, where ω : C(X) → C is a nonzero homomorphism, aka amultiplicative linear functional. As previously established (chapter 1.10),the only MLF’s on C(X) are point evaluations. Finally, any closed sub-space of H will be invariant under scalar operators, so the irreducibility ofπ implies that H is one-dimensional.

2.6. Borel Functions of Normal Operators.

Exercise 2.6.1: Show that for every f ∈ B(X) and every ε > 0 there is afinite linear combination of characteristic functions in B(X) (i.e. a simplefunction)

g = c1χE1+ · · ·+ cnχEn

such that ‖f − g‖ ≤ ε.Solution: Let C be tiled by squares with diameter ε; a finite collectionS1, . . . , Sn of them cover f(X). Let Ei = f−1(Si) ⊆ X, and choose anarbitrary point ci ∈ Si for each i. Then g =

∑ciχEi satisfies ‖f − g‖ ≤ ε.

Exercise 2.6.2: Let (X,B) be a Borel space. For every σ-finite measure µon X let πµ be the representation of B(X) on L2(X,µ) defined by

πµ(f)ξ(p) = f(p)ξ(p), ξ ∈ L2(X,µ).

(a) Show that πµ is a σ-representation of B(X) on L2(X,µ). (Notice thatthe definition of σ-representation makes sense in this more generalcontext.)

(b) Given two σ-finite measures µ, ν on (X,B), show that πµ and πν areunitarily equivalent iff µ and ν are mutually absolutely continuous.

(c) Deduce that a multiplication operator acting on the L2 space of aσ-finite measure is unitarily equivalent to a multiplication operatoracting on the L2 space of a finite measure space.

Solution:(a) Let fn be uniformly bounded functions in B(X) (say ‖fn‖ ≤M for

all n) such that fn(p) → 0 for all p ∈ X. For any ξ ∈ L2(X,µ), thefunctions ξ2f2

n are dominated by M2ξ2, so

lim ‖πµ(fn)ξ‖2 = lim

∫|fn(p)|2|ξ(p)|2 dµ =

∫lim |fn(p)|2|ξ(p)|2 dµ = 0

by the dominated convergence theorem.(b) Suppose µ ∼ ν. By Radon-Nikodym, there is a measurable function

h such that dν = h dµ, with h nonzero a.e. Define U : L2(X, ν) →L2(X,µ) by (Uf)(x) =

√|h(x)|f(x); then

‖Uf‖2 =

∫X

|h(x)||f(x)|2 dµ =

∫X

|f(x)|2 dν = ‖f‖2


so that U is an isometry, and since f 7→ 1√|h(x)|

is an inverse for U ,

it is unitary. Finally, since πµU = Uπν , we see that πµ and πν areunitarily equivalent.Conversely, suppose µ and ν are not mutually absolutely continuous.Then there exists a Borel subset E ⊆ X such that either µ(E) = 0while ν(E) 6= 0, or ν(E) = 0 while µ(E) 6= 0; WLOG suppose theformer. Then πµ takes 1

Eto the zero operator, whereas πν takes 1

E

to a nonzero operator; hence πµ and πν cannot be unitarily equivalent.(c) If (X,µ) is a σ-finite measure space, let X1, X2, . . . be disjoint mea-

surable subsets of X with each µ(Xi) finite. Define a measure ν by

ν(E ∩Xi) =µ(E)

2iµ(Xi).

Then ν ∼ µ, and ν(X) = 1.Note that this exercise is very similar to 2.4.1.

2.7. Spectral Measures.

Exercise 2.7.1: Let N1 ∈ B(H1) and N2 ∈ B(H2) be two normal operatorsacting on finite-dimensional Hilbert spaces H1, H2. Show that there is aunitary operator W : H1 → H2 such that WN1W

−1 = N2 iff N1 and N2

have the same spectrum and the same multiplicity function.Solution: Clearly, unitary equivalence preserves both spectrum and multi-

plicity. For the converse, suppose N1 and N2 have the same spectrumand multiplicity. For each λ ∈ σ(N1) = σ(N2), let Eλ,1 and Eλ,2 be thecorresponding eigenspaces for N1 and N2. By hypothesis, these have thesame dimension, so that there is a unitary Wλ : Eλ,1 → Eλ2 . Because theoperators are normal, they’re diagonalizable, so that

Hj =⊕

λ∈σ(Nj)

Eλ,j for j = 1, 2.

We can therefore define a unitary W : H1 → H2 by W = ⊕Wλ. One hasWPλ,1 = Pλ,2W , where Pλ,j is the projection onto Eλ,j , so that

W∑λ

λPλ,1 =∑λ

λPλ,2W ⇒ WN1 = N2W.

Exercise 2.7.2: Calculate the spectral measure of the multiplication opera-tor X defined on L2([0, 1]) by (Xξ)(t) = tξ(t), 0 ≤ t ≤ 1.

Solution: The spectral measure is given by P (S) = M(1S) for each Borel

subset S ⊆ [0, 1], where M(f) denotes the multiplication operator inducedby f .

More generally, we will prove the following: Let X be a locally compactHausdorff space, µ a positive σ-finite Borel measure on X, and φ : X → Ca bounded Borel function with essential range K ⊆ C. Then the spectralmeasure corresponding to the multiplication operator M(φ) on L2(X,µ) isgiven by P (S) = M(1

φ−1(S)) for Borel subsets S ⊆ K.

The proof amounts to the following observations:(1) In general, if P is the spectral measure for some normal operator T ,

then P (S) is the operator 1S(T ) defined via the Borel functional cal-

culus. This observation was made at the end of section 2.7.


(2) The Borel functional calculus is applied to multiplication operatorsby composition: ψ(M(φ)) = M(ψ φ) for a bounded Borel functionψ defined on the essential range of φ. One proves this formula instages, by “promoting” ψ through the different levels of the functionalcalculus. If ψ is a polynomial, the formula is immediate. If ψ ∈ C(K),choose polynomials pn → ψ in C(K); then pnφ→ ψφ in B∞(X), sothat M(pn φ)→M(ψ φ) in B(L2(X,µ)). Since ψ(M(φ)) is definedby the continuous functional calculus to be the limit of M(pn φ), thisshows that ψ(M(φ)) = M(ψ φ). Finally, to extend to ψ ∈ B∞(K),note that ψ 7→M(ψφ) is a σ-representation of B∞(K) which extendsthe continuous functional calculus, and since we know ψ 7→ ψ(M(φ))is the only such, it follows that ψ(M(φ)) = M(ψ φ).

(3) Applying our second remark to the function ψ = 1S, we have 1

S(M(φ)) =

M(1Sφ). Applying the first remark, this becomes P (S) = M(1

Sφ).

Finally, we note that 1S φ = 1

φ−1(S).

Exercise 2.7.3: A resolution of the identity is a function λ : R → Pλ ∈B(H) from R to the projections on a Hilbert space with the followingproperties:• λ ≤ µ⇒ Pλ ≤ Pµ• Relative to the strong operator topology,

limλ→−∞

Pλ = 0, limλ→+∞

Pλ = 1.

• Right continuity: For every λ ∈ R,

limµ→λ+

Pµ = Pλ.

Early formulations of the spectral theorem made extensive use of resolutionsof the identity. It was gradually realized that these objects are equivalentto spectral measures, in much the same way that Stieltjes integrals areequivalent to integrals with respect to a measure. This exercise is related tothe bijective correspondence that exists between resolutions of the identityand spectral measures on the real line.(a) Consider the Borel space (R,B) of the real line. Given a spectral

measure E : B → B(H), show that the function Pλ = E((−∞, λ]),λ ∈ R, is a resolution of the identity.

(b) Given two spectral measures E,F : B → B(H) that give rise to thesame resolution of the identity, show that E = F .

Solution:(a)

(1) If λ ≤ µ, then (−∞, λ] ⊆ (−∞, µ], so

Pλ = E((−∞, λ]) ≤ E((−∞, µ]) = Pµ.

(2) By the previous property, it suffices to prove this with λ replacedby an integer argument m. Now

1H

= E(R) = E

(∐n∈Z

(n, n+ 1]

)=∑n∈Z

E((n, n+ 1]).


We use the fact that the tails of a convergent sum tend to zero.A negative tail of the above sum has the form∑n≤m

E((n, n+ 1]) = E((−∞, n+ 1]) = Pm+1,

from which we see that Pm → 0 as m → −∞. Similarly, apositive tail is of the form

E((m,∞)) = 1H− E((−∞,m]) = 1

H− Pm,

from which we see that Pm → 1H

as m→∞.(3) We write

E((λ, λ+ 1]) = E

( ∞∐n=1

(λ+

1

n+ 1, λ+

1

n

])=

∞∑n=1

E

((λ+

1

n+ 1, λ+

1

n

])and again use the fact that the tails of a convergent sum tendto zero. A tail of this sum has the form

∞∑n=N

E

((λ+

1

n+ 1, λ+

1

n

])= E

((λ, λ+

1

N

])= Pλ+1/N − Pλ.

Thus, Pλ+1/N → Pλ as N →∞; appealing to the first property

again, this implies that Pµ → Pλ as µ→ λ+.(b) Let

A = S ∈ B : E(S) = F (S).We will show that A is a σ-algebra which contains the half-open in-tervals, which implies B ⊆ A and hence that B = A.

(1) E(∅) = 0 = F (∅), so that ∅ ∈ A.(2) If S ∈ A, then E(S′) = 1 − E(S) = 1 − F (S) = F (S′), so that

S′ ∈ A. Thus A is closed under complementation.(3) If S1, S2 ∈ A, then E(S1 ∩ S2) = E(S1)E(S2) = F (S1)F (S2) =

F (S1 ∩ S2), so that S1 ∩ S2 ∈ A. Thus A is closed under finiteintersections.

(4) If T1, T2, · · · ∈ A, let S1 = T1 and

Sj+1 = Tj+1 \j⋃i=1

Ti = Tj+1 ∩j⋂i=1

T ′i ,

so that Si is a disjoint family with the property that⋃ni=1 Si =⋃n

i=1 Ti for each n. Note also that Si ∈ A for each i, by previ-ously established properties of A. Then

E

( ∞⋃i=1

Ti

)= E

( ∞⋃i=1

Si

)=

∞∑i=1

E(Si) =

∞∑i=1

F (Si) = F

( ∞⋃i=1

Si

)= F

( ∞⋃i=1

Ti

)so that

⋃∞i=1 Ti is in A. Thus A is closed under countable unions,

and hence is a σ-algebra. (So far we’ve retraced the standardexercise that a collection of sets is a σ-algebra iff it contains theempty set and is closed under complementation, finite intersec-tions, and countable disjoint unions.)


(5) Finally, given any half-open interval (a, b]), we have

E((a, b]) = Pb − Pa = F ((a, b])

so that (a, b] ∈ A.

2.8. Compact Operators.

Exercise 2.8.1: Let A be a compact operator on a Hilbert space H. Showthat for every sequence of mutually orthogonal unit vectors ξ1, ξ2, · · · ∈ Hwe have

limn→∞

‖Aξn‖ = 0.

Solution: If not, we may assume, by taking a subsequence and rescalingA if needed, that ‖Aξn‖ ≥ 1 for all n. Taking another subsequence ifneeded and using the compactness of A, we have Aξn → η for some η(which automatically satisfies ‖η‖ ≥ 1). This implies Re 〈Aξn, η〉 ≥ 1

2 forsufficiently large n; WLOG, by taking a tail if needed, for all n. Now letxN = ξ1 + ξ2 + · · ·+ ξN . We have ‖xN‖ =

√N , but

Re 〈AxN , η〉 ≥N

2⇒ |〈AxN , η〉| ≥

N

2⇒ ‖AxN‖‖η‖ ≥

N

2

⇒ ‖A‖‖xN‖‖η‖ ≥N

2⇒ ‖A‖ ≥

√N

2‖η‖

for all N , contradicting the boundedness of A.Exercise 2.8.2: Let e1, e2, . . . be an orthonormal basis for a Hilbert space H

and let A ∈ B(H). Show that A is compact iff

limn→∞

‖(1− En)A(1− En)‖ = 0,

where En denotes the projection onto spane1, . . . , en.Solution: If A has this property, it is the norm limit of finite-rank operators,

hence is compact. Conversely, if A is compact, the preceding exercise im-plies that for any ε > 0 there exists N such that ‖Aen‖ < ε for n ≥ N ; thisimplies that ‖A(1− En)‖ < ε, so that ‖(1− En)A(1− En)‖ < ε.

Exercise 2.8.3: Verify the polarization formula for bounded operators on aHilbert space H:

4B∗A =

3∑k=0

ik(A+ ikB)∗(A+ ikB).

Solution: It’s tempting to cite the fact that (A,B) 7→ B∗A is an inner prod-uct, but since this is B(H)-valued rather than C-valued, we’d have to de-velop the theory of Hilbert C∗-modules to substantiate this. So we can


make the direct calculation:

3∑k=0

ik(A+ ikB)∗(A+ ikB) =

3∑k=0

ik(A∗ − ikB∗)(A+ ikB)

=

3∑k=0

ik(A∗A− ikB∗A+ ikA∗B +B∗B)

=

(3∑k=0

ik

)A∗A+

(3∑k=0

1

)B∗A+

(3∑k=0

(−1)k

)A∗B +

(3∑k=0

ik

)B∗B

= 4B∗A.

Exercise 2.8.4: Let ‖A‖2 = 〈A,A〉1/22 for every Hilbert-Schmidt operator A.(a) Let A1, A2, . . . be a sequence in L2 that satisfies

limm,n→∞

‖Am −An‖2 = 0.

Show that there is an operator A ∈ B(H) such that ‖An −A‖ → 0 asn→∞.

(b) Show that L2 is a Hilbert space relative to the inner product 〈A,B〉2 =trace B∗A.

Solution:(a) This will follow from the more general fact that ‖A‖ ≤ ‖A‖2. For this,

let x ∈ H; then

‖A∗x‖2 =∑n

|〈A∗x, en〉|2 =∑n

|〈x,Aen〉|2 ≤∑n

‖Aen‖2‖x‖2

by Parseval’s identity and the Cauchy-Schwarz inequality. It followsthat ‖A‖2 = ‖A∗‖2 ≤ ‖A‖22.

(b) Let An be Cauchy in L2. Let A be the norm limit from part (a).Let ε > 0, and choose N ∈ N such that ‖Aj − Ak‖2 < ε for j, k ≥ N .Let k ≥ N .We seek to prove that ‖A − Ak‖2 ≤ ε. It is tempting to use thecomputation

‖A−Ak‖2 = ‖ limjAj −Ak‖2 = lim

j‖Aj −Ak‖2 ≤ ε.

However, the notation limj here denotes the norm limit, which can-not be moved through the ‖ · ‖2 symbol because the Hilbert-Schmidtnorm is not continuous in the operator-norm topology if H is infinite-dimensional. (For instance, consider a projection of rank n; this hasoperator norm 1 but Hilbert-Schmidt norm

√n.) Thus, another proof

approach must be found. We will use the fact that an infinite sum ofnonnegative terms is the supremum over finite sums.For any fixed m ∈ N,

(A−Ak)em = (limjAj −Ak)em = lim

j

((Aj −Ak)em

),


so that for any M ∈ N,

M∑m=1

‖(A−Ak)em‖2 =

M∑m=1

limj‖(Aj −Ak)em‖2 ≤

M∑m=1

supj≥N‖(Aj −Ak)em‖2

= supj≥N

M∑m=1

‖(Aj −Ak)em‖2 ≤ supj≥N‖Aj −Ak‖22 ≤ ε.

Thus,

‖A−Ak‖22 = supM∈N

M∑m=1

‖(A−Ak)em‖2 ≤ ε.

It follows that ‖A−Ak‖2 → 0, so that L2 is complete.Exercise 2.8.5: Show that a multiplication operator Mf is self-adjoint and

has nonnegative spectrum iff 〈Mfξ, ξ〉 ≥ 0 for every ξ ∈ L2(X,µ).Solution: Suppose Mf is self-adjoint and has nonnegative spectrum. Since

the spectrum is the essential range, it follows that f is nonnegative a.e.Then for any ξ ∈ L2(X,µ),

〈Mfξ, ξ〉 =

∫X

(Mfξ)(x)ξ(x) dµ(x) =

∫X

f(x)ξ(x)ξ(x) dµ(x) ≥ 0

because the integrand is nonnegative a.e.Conversely, suppose f does not have nonnegative spectrum. Let λ ∈

σ(Mf ) \ [0,∞), and let N be a neighborhood of λ such that one of theconditions(1) Re z < 0(2) Im z < 0(3) Im z > 0

holds throughout N ; for convenience, we assume the first. Let V = f−1(N),which has positive measure because λ is in the essential range of f . Then

〈Mf1V ,1V 〉 =

∫V

f(x) dµ(x)

which has strictly negative real part, because the integrand does a.e. Hencethere exists ξ ∈ L2 such that 〈Mfξ, ξ〉 is not nonnegative.

2.9. Adjoining a Unit to a C∗-Algebra.

Exercise 2.9.1: Let A be a nonunital C∗-algebra and let π : A→ Ae be thenatural map of A into its unitalization. Considering Ae as a C∗-algebra,suppose that there is an isometric *-homomorphism σ : A → B of A intoanother unital C∗-algebra B such that σ(A) is an ideal of codimension 1 inB. Show that there is a unique isometric *-isomorphism θ : Ae → B suchthat θ π = σ.

Solution: The map θ : Ae → B defined by θ(a + λ1) = σ(a) + λ1 is a *-homomorphism with the property θ π = σ. It is surjective because itsrange includes σ(A) and 1. It is injective because if θ(a + λ1) = 0, thenσ(a) = −λ1, and since σ(A) ∩ C1 = 0 it follows that σ(a) = λ = 0 andhence, by the injectivity of σ, that a = λ = 0.


Exercise 2.9.2: Let K be the C∗-algebra of compact operators on a Hilbertspace H. Show that the space of operators λ1 + K : λ ∈ C,K ∈ K is aC∗-algebra *-isomorphic to Ke.

Solution: It is easy to see by direct calculation that the space of operators inquestion is a unital C∗-algebra in which K is an ideal of codimension one.The previous exercise implies that it is isomorphic to Ke.

Exercise 2.9.3: Let X be a compact Hausdorff space and let F be a properclosed subset of X. Let A be the ideal of all functions f ∈ C(X) that vanishthroughout F . Note that A is a C∗-algebra in its own right.(a) Show that A has a unit if and only if F is both closed and open.(b) Assuming that F is not open, identify the unitalization of A in concrete

terms by exhibiting a compact Hausdorff space Y such that Ae 'C(Y ), describing the precise relationship of Y to X and F .

Solution:(a) If F is clopen, the function 1− 1

Fis continuous, and is a unit for A.

Conversely, suppose A has a unit φ. Given any point x ∈ X \F , thereexists f ∈ C(X) with f(x) = 1 and f = 0 on F ; since f ∈ A, φf = f ,so that φ(x) = 1. Hence φ = 1 − 1

F. Since this is continuous, F is

clopen.(b) If F is not open, X \ F is locally compact but not compact. The

algebra A is isomorphic to C0(X \ F ), so its unitization is isomorphic

to C(X \ F ), where ∼ denotes the one-point compactification.

2.10. Quotients of C∗-Algebras.

Exercise 2.10.1: Let e1, e2, . . . be an orthonormal basis for a separableHilbert space H, and let En be the projection on the span of e1, . . . , en.Show that an operator T ∈ B(H) is compact iff

limn→∞

‖T − TEn‖ = 0,

and deduce that En : n ∈ N is an approximate unit for the C∗-algebraK.

Solution: The “if” direction is true because a norm-limit of compact opera-tors is compact; this is true even on Banach spaces (Exercise 1.4.2). The“only if” direction is a consequence of the spectral theorem for compactnormal operators (Theorem 2.8.2).

Exercise 2.10.2: Let U be a unitary operator on a Hilbert space H. Thenσ(U) ⊆ T, and hence there is a unique representation ρ ∈ rep(C(T), H)satisfying ρ(f) = f(U) for f ∈ C(T). Identify ker ρ as an ideal in C(T),identify the quotient C(T)/ ker ρ in concrete terms as a commutative C∗-algebra, and similarly describe the natural factorization ρ = ρ π, where

π : C(T)→ C(T)/ ker ρ

is the natural map onto the quotient C∗-algebra.Solution: The existence and uniqueness of ρ deserves a little explanation,

since ρ is not just the continuous functional calculus for U if σ(U) is aproper subset of T. It is Moreover, even the symbol f(U) needs defining

in this case. We define it to mean f(U) where f is the restriction of f toσ(U). Then ρ is unique because it is determined on the set of polynomials,which are dense in C(T); for existence, note that the composition F r


has the desired properties, where F : C(σ(U)) → B(H) is the continuousfunctional calculus and r : C(T)→ C(σ(U)) is restriction.

Now on to the exercise. Because F is injective,

ker ρ = ker r = f ∈ C(T) : f = 0 on σ(U).

Now the restriction map r is surjective (by Urysohn’s lemma), so r = rπwhere π : C(T) → C(T)/ ker r is the quotient map (the same as the πmentioned in the problem, since ker ρ = ker r) and r is an isomorphism.Hence C(T)/ ker ρ = C(T)/ ker r ' C(σ(U)).

Finally, ρ = F r = F r π, so that ρ = F r is the composition of thecontinuous functional calculus with the identification between C(T)/ ker ρand C(σ(U)).

Exercise 2.10.3: Show that there is a compact Hausdorff space βR and anisometric *-isomorphic of Cb(R) onto C(βR).

Solution: Since Cb(R) is a unital C∗-algebra, the Gelfand spectrum Σ(Cb(R))fits the bill.

Exercise 2.10.4: For every t ∈ R, show that there is a (naturally defined)point t ∈ βR, and that the map t 7→ t is a homeomorphism of R onto adense subspace of βR.

Solution: We define t to be evaluation at t, that is, t(f) = f(t) for f ∈ Cb(R).• t 7→ t is continuous because if tα → t in R, then ∀f ∈ Cb(R) : f(tα)→f(t); that is, ∀f ∈ Cb(R) : tα(f)→ t(f) in C. By the definition of theGelfand topology, this is the same as tα → t in βR.

• t 7→ t is injective because if t1 6= t2, we can use Urysohn’s lemma to finda function g ∈ Cb(R) with g(t1) = 0 and g(t2) = 1; then t1(g) 6= t2(g),so t1 6= t2. Note that in a metric space, you don’t need Urysohn; youcan use the explicit formula

g(x) = max

(0, 1− 2d(x, t2)

d(t2, t1)

).

• t 7→ t is open because if tα 6→ t in R, there exists (by taking a subnetif necessary) a neighborhood N of t such that tα /∈ N for all α. Thenby Urysohn’s lemma, there exists g ∈ Cb(R) with g(t) = 1 and with gsupported in N . Then

0 = tα(g) 6→ t(g) = 1.

• Suppose the range R is not dense. Let φ ∈ βR be a point and U aneighborhood of φ disjoint from R. Using the basis for the weak-*topology, there exist an ε > 0, an n ∈ N, and function f1, . . . , fn ∈Cb(R) such that |φ(fk)−ω(fk)| < ε for all ω ∈ U and all k = 1, . . . , n.Let

f(x) =

n∑k=1

|fk(x)− φ(fk)|2.

Since no x is in U , we must have f(x) ≥ ε2 for every x ∈ R. Thus, f

is invertible, which implies φ(f) is nonzero, a contradiction. Hence Ris dense in βR.This density argument is taken from chapter 6.5 of ([Zhu93]).


Exercise 2.10.5: Identifying R with its image in βR, the subspace βR \R iscalled the corona of R. Show that the corona is closed (and hence, R is anopen subset of βR).

Solution: As suggested in the hint, we will show that the corona is preciselythe closed subset

ω ∈ βR : ∀f ∈ C0(R) : ω(f) = 0.

Clearly this subset is contained in the corona, since if t ∈ R then thereexists f ∈ C0(R) (in fact, f can be compactly supported) with f(t) 6= 0and hence with t(f) 6= 0. On the other hand, suppose ω ∈ βR \ R. Letf ∈ C0(R). Let ε > 0 and choose compact K ⊆ R such that |f | < ε on

R \K. Because R is dense in βR, there is a net tα ⊆ R such that tα → ωin βR. Then every subnet converges to ω as well; this implies that onlyfinitely many points of the net lie in K, as otherwise a subnet of tα wouldconverge to a point t ∈ K, and therefore the corresponding subnet of tαwould converge to t rather than ω. Then |tα(f)| < ε for sufficiently largeα, so that

|ω(f)| = | limαtα(f)| ≤ ε.

This is true for arbitrary ε > 0, so ω(f) = 0.Exercise 2.10.6: Deduce that the quotient C∗-algebra Cb(R)/C0(R) is iso-

metrically isomorphic to C(βR \ R).Solution: Define ρ : Cb(R) → C(βR \ R) by ρ = r Γ where Γ : Cb(R) →C(βR) is the Gelfand transform and r : C(βR) → C(βR \ R) is restric-tion. By Urysohn’s lemma, r is surjective, and since Γ is an isomor-phism, ρ is surjective as well. Moreover, because Γ is an isomorphism,ker ρ = Γ−1(ker r), which equals C0(R) by the previous exercise. HenceCb(R)/C0(R) ' C(βR \ R) by the First Isomorphism Theorem.

Exercise 2.10.7: A compactification of R is a pair (φ, Y ) where Y is acompact Hausdorff space and φ : R→ Y is a continuous map such that φ(R)is dense in Y . Show that (t 7→ t, βR) is a universal compactification of R inthe following sense: If (φ, Y ) is any compactification of R, then there is a

unique extension of φ : R→ Y to a continuous surjection φ : βR→ Y . Hint:The map φ induces a *-isomorphism of C(Y ) onto a unital C∗-subalgebraof Cb(R).

Solution: Let Ψ : R→ βR denote the map t 7→ t. Define Φ : C(Y )→ Cb(R)by Φ = · φ; that is, for any f ∈ C(Y ), Φ(f) = f φ. To see that Φ isinjective, suppose Φ(f) = 0, so f φ = 0. Thus, f is zero on the range ofφ; but this is dense in Y , so that f must be zero.

Let Φ : C(Y ) → C(βR) be the map Φ = Γ Φ where Γ : Cb(R) →C(βR) is the Gelfand transform; since the latter is an isomorphism, Φ isalso injective. By Exercise 2.2.6, there exists a unique continuous functionu : βR → Y with Φ = · u; moreover, the injectivity of Φ implies thesurjectivity of u. Finally, we have Φ = Γ−1Φ and taking the correspondingmaps on topological spaces as in Exercise 2.2.6, we get u Ψ = φ.

3. Asymptotics: Compact Perturbations and Fredholm Theory

3.1. The Calkin Algebra.


3.2. Riesz Theory of Compact Operators.

3.3. Fredholm Operators.

3.4. The Fredholm Index.

4. Methods and Applications

4.1. Maximal Abelian von Neumann Algebras.

4.2. Toeplitz Matrices and Toeplitz Operators.

4.3. The Toeplitz C∗-Algebra.

4.4. Index Theorem for Continuous Symbols.

4.5. Some H2 Function Theory.

4.6. Spectra of Toeplitz Operators with Continuous Symbol.

4.7. States and the GNS Construction.

4.8. Existence of States: The Gelfand-Naimark Theorem.

References

[Con78] John B. Conway, Functions of one complex variable, second ed., Graduate Texts inMathematics, vol. 11, Springer-Verlag, New York, 1978. MR 503901 (80c:30003)

[Fol99] Gerald B. Folland, Real analysis, second ed., Pure and Applied Mathematics (New York),

John Wiley & Sons Inc., New York, 1999, Modern techniques and their applications, AWiley-Interscience Publication. MR 1681462 (2000c:00001)

[Rud87] Walter Rudin, Real and complex analysis, third ed., McGraw-Hill Book Co., New York,

1987. MR 924157 (88k:00002)[Rud91] , Functional analysis, second ed., International Series in Pure and Applied Math-

ematics, McGraw-Hill Inc., New York, 1991. MR 1157815 (92k:46001)[SS05] Elias M. Stein and Rami Shakarchi, Real analysis, Princeton Lectures in Analysis, III,

Princeton University Press, Princeton, NJ, 2005, Measure theory, integration, and Hilbert

spaces. MR 2129625 (2005k:28024)[Zhu93] Ke He Zhu, An introduction to operator algebras, Studies in Advanced Mathematics,

CRC Press, Boca Raton, FL, 1993. MR 1257406 (94k:46124)

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