Ar Ve Son Spectral Solutions

BY

BY WILLIAWILLIAM M ARARVESONVESON

Contents Contents

1.

1. SpectrSpectral Thal Theory eory and Band Banacanach Algh Algebrasebras 22 1.1

1.1. . OriOrigingins of Ss of Spectpectral Tral Theoheoryry 22 1.2

1.2. . The SThe Spectpectrum rum of an of an OperOperatoatorr 44 1.3.

1.3. BanacBanach h AlgebAlgebras: Exras: Exampleampless 55 1.4.

1.4. The The RegulRegular ar RepreRepresensentatiotationn 88 1.5

1.5. . The GThe Geneeneral Lral Lineinear Grar Group oup of of AA 99 1.6.

1.6. SpectrSpectrum of aum of an Elemn Element ent of a Banof a Banach ach AlgebrAlgebraa 1111 1.7

1.7. . SpecSpectratral l RadRadiusius 1212

1.8

1.8. . IdeIdeals als and and QuoQuotietientntss 1414 1.9.

1.9. CommCommutatutative ive BanacBanach Algh Algebrasebras 1515 1.1

1.10. 0. ExaExamplmples:es: C C ((X X ) and the Wiener Algebra) and the Wiener Algebra 1616 1.11.

1.11. SpectraSpectral Pl Permaermanence nence TheorTheoremem 1717 1.12.

1.12. Brief oBrief on the Ann the Analytic alytic FFunctiunctional Caonal Calculuslculus 1818 2.

2. OperOperatoators ors on Hin Hilberlbert Spt Spaceace 2121 2.1

2.1. . OperOperatoators rs and and TheTheirir C C 

∗∗

_{-Algebras-Algebras 2121}

2.2

2.2. . ComCommumutattativivee C C 

∗∗

_{-Algebras-Algebras 2323}

2.3.

2.3. ConContinutinuous Fous Functiounctions of Normns of Normal Operatal Operatorsors 2626 2.4.

2.4. The SThe Spectral pectral TheorTheorem anem and Diagd Diagonalizonalizationationss 2727 2.5.

2.5. RepreRepresensentatiotations of ns of BanacBanach *-Alh *-Algebragebrass 3131 2.6.

2.6. Borel Borel FFunctiounctions of ns of NormaNormal Operatl Operatorsors 3333 2.7

2.7. . SpecSpectratral l MeaMeasursureses 3434 2.8

2.8. . ComCompacpact t OperOperatoatorsrs 3737 2.9

2.9. . AdjAdjoinoining ing a Una Unit tit to ao a C C 

∗∗

_{-Algebra-Algebra 3939}

2.1

2.10. 0. QuoQuotietientnts s of of C C 

∗∗

_{-Algebras-Algebras 4040}

3.

3. AsympAsymptotictotics: s: CompaCompact Perct Perturbturbations anations and Frd Fredholm Tedholm Theoryheory 4242 3.1

3.1. . The The CalCalkin kin AlgAlgebrebraa 4242 3.2.

3.2. Riesz Riesz TheorTheory of y of CompCompact act OperatOperatorsors 4343 3.3.

3.3. FFredhoredholm lm OperatoOperatorsrs 4343 3.4

3.4. . The The FFredredholholm Im Indendexx 4343 4.

4. MetMethods hods and and AppApplicalicatiotionsns 4343 4.1.

4.1. MaximMaximal Abelial Abelian van von Neon Neumann umann AlgebrAlgebrasas 4343 4.2.

4.2. TToeplitz oeplitz MatriMatrices ances and Td Toeplitz Ooeplitz Operatoperatorsrs 4343 4.3

4.3. . The The TToeploeplitzitz C C 

∗∗

-Algebra-Algebra 4343 4.4.

4.4. IndeIndex Theox Theorem frem for Conor Contintinuous Suous Symboymbolsls 4343 4.

4.5. 5. SoSomeme H H 22 Function TheoryFunction Theory 4343 4.6.

4.6. SpectrSpectra of Ta of Toeplitz Operoeplitz Operatorators with Cons with Continutinuous Symous Symbolbol 4343 4.7.

4.7. StatStates aes and tnd the Ghe GNS CoNS Construnstructionction 4343 4.8.

4.8. ExistExistence oence of Statf States: The Gelfandes: The Gelfand-Naim-Naimark Theark Theoremorem 4343 References

References 4343

1 1

1.

1. SpectrSpectral al TheorTheory y and and BanacBanach h AlgebrasAlgebras

1.1.

1.1. Origins of Spectral Theory.Origins of Spectral Theory. (1)

(1) Fix a sequeFix a sequencence

_{{

aann

}}

of numbers with 0of numbers with 0 < < 

≤

≤

aann

≤

≤

M M , and let, and let AA :: 22

→

→

22 be the operator determined by (

be the operator determined by (AxAx))nn == aannxxnn.Show that.Show that AA is a boundedis a bounded operator on

operator on 22, and exhibit a bounded operator, and exhibit a bounded operator BB onon 22 such thatsuch that ABAB == BA

BA == 11..

Solution.

Solution. AA is bounded becauseis bounded because



AxAx

_

22 ==





n n

((AxAx))nn((AxAx))nn ==





n n a annxxnnaannxxnn ==





n n

||

a ann

||

22

||

xxnn

||

22

≤

≤





n n M  M 22

_||

xxnn

||

22 == M M 22



xx



22 The inverse operator

The inverse operator BB is determined by (is determined by (BxBx))nn == _aa11_nnxxnn; this operator is; this operator is bounded by 1

bounded by 1//, by the same argument., by the same argument. 

(2) Let

(2) Let

_{{

aann

}}

be a bounded increasing sequence of positive numbers and letbe a bounded increasing sequence of positive numbers and let D

Dnn == aa11aa22. . . a. . . ann. Show that the sequence. Show that the sequence DDnn converges to a nonzero limitconverges to a nonzero limit D

D((AA) ) iff iff 

∞

∞





n n=1=1 (1 (1

₋

₋

aann)) <<

∞

∞

.. Solution.

Solution. To clarify, when we sayTo clarify, when we say



(1(1

₋

₋

aann)) <<

∞

∞

, we mean to say that the, we mean to say that the sum converges to a finite number. In particular, the case where it converges sum converges to a finite number. In particular, the case where it converges to

to

_−∞

_−∞

(for example, if all the(for example, if all the aann are equal to 2) is disallowed.are equal to 2) is disallowed.

It is clear that either of the conditions whose equivalence we seek to It is clear that either of the conditions whose equivalence we seek to establish implies that

establish implies that aann

↑↑

1, so we will assume this to be the case.1, so we will assume this to be the case. The condition that

The condition that DDnnapproaches a nonzero limit approaches a nonzero limit is equivalent tois equivalent to



jj((

−

−

lnln aajj)) <<

∞

∞

. We want to show this is equivalent to. We want to show this is equivalent to



_jj(1(1

₋

₋

aajj)) <<

∞

∞

. One direction. One direction is

is trivitrivial, bal, becausecausee

₋

₋

lnln xx

_≥

_≥

11

₋

₋

xx for all positivefor all positive xx (just compare the deriva-(just compare the deriva-tives to the right and left of the intersection point

tives to the right and left of the intersection point xx = 1). = 1). FFor the otheor the otherr direction, we note that

direction, we note that

₋

₋

lnln x <x < 2(12(1

₋

₋

xx) in a neighborhood of ) in a neighborhood of  xx = 1 (again,= 1 (again, this follows by comparing the derivatives at

this follows by comparing the derivatives at xx = 1); since= 1); since aann

→

→

1, one has1, one has

−

−

lnln aajj << 2(12(1

−

−

aajj) for sufficiently large jj, so that the result follows by the) for sufficiently large , so that the result follows by the comparison test.

comparison test.

We note that this argument is standard in complex analysis in the We note that this argument is standard in complex analysis in the con-text of

text of the Weiethe Weierstrarstrass ss factfactorizaorization theorem; tion theorem; see for see for instainstance ([nce ([ Rud87Rud87])]) Theorem 15.5 or ([

Theorem 15.5 or ([Con78Con78]) Proposition 5.4.]) Proposition 5.4. 

(3) Let

(3) Let kk((x,x, yy) be a continuous function on) be a continuous function on

_{{

((x,x, yy) ) : : 00

_≤

_≤

yy

_≤

_≤

xx

_≤

_≤

11

_}}

and letand let f 

f 

_∈∈

C C ([0([0,, 1]). Show that the function1]). Show that the function gg : : [0[0,, 1]1]

_→

_→

CC defined bydefined by gg((xx) =) =

 

 

x x 0 0 k k((x,x, yy))f f ((yy)) dydy is

is concontintinuouuous, s, and that and that the linear mapthe linear map K K  :: f f 

_→

_→

gg defidefines nes a a bounboundedded operato

operator r onon C C ([0([0,, 1]).1]).

Solution.

Solution. LetLet M M  ==

_

f f 

_

supsup andand LL = max= max

||

kk((x,x, yy))

||

. Fix any. Fix any xx

∈∈

[0[0,, 1].1]. Let

Let  > > 0. Since0. Since kk is uniformly continuous (its domain is compact), oneis uniformly continuous (its domain is compact), one can choose

((xx22,, yy22))



< δ < δ 11. Let. Let δ δ = min(= min(δ,/Lδ,/L). For 0). For 0

≤

≤

h < h < δ δ  one hasone has

||

gg((xx ++ hh))

₋

₋

gg((xx))

_||

==





 

 

x x++hh 0 0 k k((xx ++ h,h, yy))f f ((yy)) dydy

₋

₋

 

 

x x 0 0 k k((x,x, yy))f f ((yy)) dydy





= =





 

 

x x 0 0

[[kk((xx ++ h,h, yy))

₋

₋

kk((x,x, yy)])]f f ((yy)) dydy ++

 

 

x x++hh x x k k((xx ++ h,h, yy))f f ((yy)) dydy





≤

≤

 

 

x x 0 0

||

k

k((xx ++ h,h, yy))

₋

₋

kk((x,x, yy))

_||||

f f ((yy))

_||

dydy ++

 

 

x x++hh x x

||

k k((xx ++ h,h, yy))

_||

f f ((yy))

_||

dydy

≤

≤

M M 

 

 

x x 0 0

||

k k((xx ++ h,h, yy))

₋

₋

kk((x,x, yy))

_||

dydy ++ M M 

 

 

x x++hh x x

||

k k((xx ++ h,h, yy))

_||

dydy

≤

≤

MM ++ MLh <MLh < 22MM.. A similar calculation holds for

A similar calculation holds for

₋

₋

δ δ < < hh

_≤

_≤

0 (the second integral is reversed).0 (the second integral is reversed). Thus

Thus gg((xx ++ hh))

_→

_→

gg((xx) ) asas hh

_→

_→

0, so that gg is continuous at0, so that is continuous at xx.. For the boundedness of 

For the boundedness of  K K , note that for each, note that for each xx

_∈∈

[0[0,, 1],1],

||

gg((xx))

_||

==





 

 

x x 0 0 k k((x,x, yy))f f ((yy)) dydy





≤

≤

 

 

x x 0 0

||

k k((x,x, yy))

_||||

f f ((yy))

_||

dydy

≤

≤

M M 

 

 

x x 0 0

||

k k((x,x, yy))

_||

dydy

≤

≤

MM LL == LL

_

f f 

_

supsup Taking the supremum over

Taking the supremum over xx

_∈∈

[0[0,, 1], we have1], we have

_

gg

_

supsup

≤

≤

LL



f f 



supsup. . ThThusus



K K 

_

opop

≤

≤

LL.. 

(4)

(4) FFor the or the kerkernelnel kk((x,x, yy) = 1 for 0) = 1 for 0

_≤

_≤

yy

_≤

_≤

xx

_≤

_≤

1 consider the corresponding1 consider the corresponding Volterra operator

Volterra operator V V  :: C C ([0([0,, 1])1])

_→

_→

C C ([0([0,, 1]), i.e.1]), i.e. ((VV f f )()(xx) =) =

 

 

x x 0 0 f  f ((yy)) ddyy, , f  f  

_∈∈

C C ([0([0,, 1])1]).. Give

Given n a a functfunctionion gg

_∈∈

C C ([0([0,, 1])1]), , shoshow w thathat t the equatthe equationion VV f f  == gg has ahas a solution

solution f f 

_∈∈

C C ([0([0,, 1]) iff 1]) iff  gg is continuously differentiable andis continuously differentiable and gg(0) = 0.(0) = 0.

Solution.

Solution. Clearly every elementClearly every element VV f f  of the range is continuously differen-of the range is continuously differen-tiable and has (

tiable and has (VV f f )(0) = 0. Conversely, suppose)(0) = 0. Conversely, suppose gg satisfies those hypothe-satisfies those hypothe-ses; let

ses; let f f  == gg



_{. Then (. Then (VV f f ))}



_{= f =f  == gg}



_{, so that, so that VV f f  andand gg differ by an additivediffer by an additive}

constant; but (

constant; but (VV f f )(0) = 0 =)(0) = 0 = gg(0), so that(0), so that VV f f  == gg.. 

(5) Let

(5) Let kk((x,x, yy), ), 00

_≤

_≤

x,x, yy

_≤

_≤

1 be a continuous function on the unit square, and1 be a continuous function on the unit square, and consid

consider er the the boundebounded d operatoperatoror K K  defined ondefined on C C ([0([0,, 1]) by1]) by ((Kf Kf )()(xx) =) =

 

 

1 1 0 0 k

k((x,x, yy))f f ((yy)) dy,dy, 00

_≤

_≤

xx

_≤

_≤

11.. Let

Let BB11 ==

{{

f f 

∈∈

C C ([0([0,, 1]) :1]) :



f f 



 ≤

≤

11

}}

be the closed unit ball inbe the closed unit ball in C C ([0([0,, 1]).1]). Show that

Show that K K  is is aa compact compact  operator in the sense that the norm closure of operator in the sense that the norm closure of  the image

Solution.

Solution. ClearlyClearly KBKB11 is a pointwise bounded family of functions (sinceis a pointwise bounded family of functions (since K K  is

is a a bounded operatorbounded operator, , by a by a straistraightghtforwforward generalizard generalization of ation of exerexercise 3).cise 3). Given

Given   >> 0, use the uniform continuity of 0, use the uniform continuity of  kk to chooseto choose δ δ >> 0 such that0 such that

||

kk((xx11,, yy11))

−

−

kk((xx22,, yy22))

||

< <  wheneverwhenever



((xx11,, yy11))

−

−

((xx22,, yy22))



< δ < δ . Then for any. Then for any f 

f 

_∈∈

BB11 and anyand any xx11,, xx22

∈∈

[0[0,, 1] satisfying1] satisfying

||

xx11

−

−

xx22

||

< δ < δ ,,

||

((Kf Kf )()(xx11))

−

−

((Kf Kf )()(xx22))

||

==





 

 

1 1 0 0

[[kk((xx11,, yy))

−

−

k((xk x22,, yy)])]f f ((yy)) dydy





≤

≤

 

 

1 1 0 0

||

k

k((xx11,, yy))

−

−

kk((xx22,, yy))

||||

f f ((yy))

||

dydy

≤

≤

 

 

1 1 0 0 

_··

11 dydy == .. Thus,

Thus, KBKB11 is uniformly equicontinuous; by the Arzela-Ascoli theorem, itis uniformly equicontinuous; by the Arzela-Ascoli theorem, it has compact closure.

has compact closure. Not

Note: e: The hinThe hint is t is mismislealeadinding, asg, as KBKB11 is is not not uniuniforformly mly LipLipscschithitz z inin gen

generaeral; l; indindeedeed, , it it mamay y concontaitain n funfunctictions whicons which h are not are not LipLipscschithitz. z. FForor examp

example, le, supposesuppose kk((x,x, yy) ) ==

√ 

√ 

xx andand f f  is the constant functionis the constant function f f ((yy) = 1;) = 1; then (

then (Kf Kf )()(xx) =) =

√ 

√ 

xx is not Lipschitis not Lipschitz. z. One needs to use uniform ratheOne needs to use uniform rather thanr than Lipschitz continuity.

Lipschitz continuity.

 

1.2.

1.2. The Spectrum of an Operator.The Spectrum of an Operator. (1)

(1) Give explicit Give explicit examples of examples of bounded operatorsbounded operators A,A, BB onon 22_{((NN) such that) such that ABAB ==} 1

1 andand BABA is the projection onto a closed infinite-dimensional subspace of is the projection onto a closed infinite-dimensional subspace of  infinite codimension.

infinite codimension.

Solution.

Solution. We defineWe define AA andand BB on the standard basis vectors byon the standard basis vectors by BeBekk == ee22kk and and Ae Aekk ==



eek/k/22 kk eveneven 0 0 eellssee.. Then

Then ABAB == 11 andand BABA is the projection onto spanis the projection onto spanee22,, ee44,, ee66, . . ., . . ...  (2) Let

(2) Let AA andand BB be the operators defined onbe the operators defined on 22_{((NN) ) bebe} A

A((xx11,, xx22, . . ., . . . ) = (0) = (0,, xx11,, xx22, . . ., . . . )) B

B((xx22,, xx22, . . ., . . . ) = ) = ((xx22,, xx33,, xx44, . . ., . . . )) Show that

Show that

_

AA

_

==

_

BB

_

= 1, and compute both= 1, and compute both BABA andand ABAB. . DedDeduce thuce thatat A

A is injective but not surjective,is injective but not surjective, BB is surjective but not injective, and thatis surjective but not injective, and that σ

σ((ABAB))

_

== σσ((BABA).).

Solution.

Solution. ClearlyClearly AA is isometric andis isometric and BB is contractive; sinceis contractive; since

_

BB(0(0,, 11,, 00,, 00, . . ., . . . ))

_

==



(0(0,, 11,, 00,, 00, . . ., . . . ))

_

we see thatwe see that

_

BB

_

= 1.= 1. BA

BA == 11,, whereaswhereas ABAB is projection onto the complement of the span of is projection onto the complement of the span of  ee11. . The formeThe former of r of these impthese implies the injectilies the injectivity of vity of  AA and surjectivity of and surjectivity of  BB;; if 

if  AA were surjective orwere surjective or BB injective, the compositioninjective, the composition ABAB would have to bewould have to be as well, but it’s not.

as well, but it’s not. The non-injectivity of 

The non-injectivity of  ABAB implies that its spectrum includes 0, whichimplies that its spectrum includes 0, which σ

(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 + BC A. Then

D(1

₋

BA) = 1+BC A

₋

BA

₋

BCABA = 1+B(C 

₋

CAB )A

₋

BA = 1+BA

₋

BA = 1 and

(1

₋

BA)D = 1

₋

BA+BC A

₋

BABCA = 1

₋

BA+B(C 

₋

ABC )A = 1

₋

BA+BA = 1 so that 1

₋

BA is invertible. Interchanging A and B yields the other direction of implication.

We note that norm properties enter nowhere into this calculation; the

result holds in any unital algebra. 

(4) Use the result of the preceding exercise to show that for any two bounded operators A, B acting on a Banach space, σ(AB) and σ(BA) agree except perhaps for 0: σ(AB)

_{\ {}

0

_}

= σ(BA)

_{\ {}

0

_}

.

Solution. For λ

_

= 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 for every sequence of elements xn

∈

X  satisfying



_n



xn



<

∞

, there is an element y

_∈

X  such that

lim

n

_→∞



y

−

(x1+

· · ·

+ xn)



= 0.

Solution. Suppose E  is Banach. Let

_{

xn

}

be an absolutely summable se-quence. Let yn =



ni=1xi 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, it converges to a limit y. Thus, every absolutely summable sequence in E  is summable.

Conversely, suppose E  is a normed space in which every absolutely summable sequence is summable. Let

_{

xn

}

be a Cauchy sequence in E . Let

_{

xnk

}

be a subsequence such that



xnk+1

−

xnk



< 2

−

k_{. Let y} k =

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 side must as well. Hence the sequence xnk converges. But then

{

xn

}

is a Cauchy

sequence 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 f n = 1[0,1/n]. Then

1 = f n(0) = (f n

∗

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 convolution of  L1 _{functions with L}

_∞

_{functions is continuous ([Fol99] Proposition 8.8).} But then the convolution of  g with a characteristic function would have to be a continuous characteristic function, which is impossible. 

(3) For every n = 1, 2, . . . let φn be a nonnegative function in L1(R) such that φn vanishes outside the interval [

−

1/n, 1/n] and

 

∞

−∞

φn(t) dt = 1.

Show that φ1, φ2, . . . is an approximate identity for the convolution algebra L1_{(R) in the sense that}

lim

n

_→∞



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





 

R f (x

₋

y)φn(y) dy

−

f (x)

 

R φn(y) dy





dx

≤

 

_R

 

1/n

−

1/n

|

f (x

₋

y)

₋

f (x)

_||

φn(y)

|

dydx =

 

1/n

−

1/n φn(y)

 

R

|

f y(x)

−

f (x)

|

dxdy =

 

1/n

−

1/n φn(y)



f y

−

f 



1dy.

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:} ˆ

f (ξ ) =

 

∞

−∞

eitξf (t) dt, ξ  

_∈

R.

Show that ˆf  belongs to the algebra C 

_∞

(R) of all continuous functions on

R _{that vanish at}

_∞

_.

Solution. Fix ξ 

_∈

R. Let  > 0 be given. Let K 

_⊆

R be a compact subset such that

 

_R

\

K 

|

f 

|

< . Now for any η

∈

R,

|

f (ξ )ˆ

₋

f (η)ˆ

_|

=





 

R eiξtf (t) dt

₋

 

R eiηtf (t) dt





=





 

R



eiξt

₋

eiηt



f (t) dt





≤

 

_R



eiξt

₋

eiηt



_|

f (t)

_|

dt

≤

2 +

 

K 



eiξt

₋

eiηt



_|

f (t)

_|

dt. As η

_→

ξ , eiηt

→

eiξt _{uniformly on K , so that the above expression is less} than 3 for η sufficiently close to ξ . This proves that ˆf  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 identification of the Fourier transform with the Gelfand transform; my presentation in Jorgensen’s class in spring 2011 goes into this. A more elementary proof, however (thank to [SS05]), is as follows: For any ξ 

_

= 0, the change of  variables x



_{= x +} π ξ yields ˆ f (ξ ) =

₋

 

R f 



x



₋

π ξ 



e ix_ξ dx



.

Averaging this with the original expression for ˆf (ξ ) yields ˆ f (ξ ) = 1 2

 

R



_f (x)

₋

_f 



_x

₋

π ξ 



e ixξ_dx

from which

_|

f (ξ )ˆ

_{| ≤}

1₂

_

f 

₋

f π/ξ



. The result follows by the L1 continuity

of translation. 

(5) Show that the Fourier transform is a homomorphism of the convolution algebra L1_{(R) onto a subalgebra}

A

of C 

_∞

(R) which is closed under complex conjugation and separates points of R.

Solution. Linearity is obvious. Multiplicativity is a well-known consequence of Fubini’s theorem. For self-adjointness, transform the function f 

∗

_{(x) =}

f (

₋

x). For separation of points, suppose

_|

ξ 

_|

<

_|

η

_|

and consider the Fourier transform of the characteristic function of [0 , 2π

1.4. The Regular Representation.

(1) Let E  and F  be normed linear spaces with E 

_

=

_{

0

_}

. Show that B(E, F ) is a Banach space iff  F  is a Banach space.

Solution. Suppose F  is Banach. Let

_{

T n

}

be a Cauchy sequence in B(E, F ). For each e

_∈

E  and each m, n

_∈

N,



T me

−

T ne



=



(T m

−

T n)e

 ≤ 

T m

−

T n



e



so that

_{

T ne

}

is a Cauchy sequence in F . Define the function T  : E 

→

F  by T e = lim T ne. Clearly this is linear; it is bounded because



T e

_

=

_

lim T ne



= lim



T ne

 ≤

(lim



T n



)



e



so that

_

T 

_{ ≤}

lim

_

T n



(which limit exists because

{

T n

}

is a Cauchy sequence in R). Finally, we must prove that T n

→

T  in norm, not just pointwise. Given  > 0, choose N 

_∈

N such that

_

T m

−

T n



<  for m, n

_≥

N . Then for n

_≥

N  and e

_∈

E ,



T e

₋

T ne



=



lim

m T me

−

T ne



= limm



T me

−

T ne

≤





e



since

_

T me

−

T ne



< 



e



for each m. Hence



T 

−

T n

 →

0.

Conversely, suppose B(E, F ) is Banach. Let

_{

f n

}

be a Cauchy sequence in F . Let e

_∈

E  be some nonzero vector. Define linear maps T n : E 

→

F  by T ne = f n and T n = 0 on the (algebraic) complement of the span of  e. Clearly T n is linear with norm



_

f _en

_



. Moreover, since

(T m

−

T n)x =



α(f m

−

f n) x = αe

0 else,

we see that

_

T m

−

T n



=



f m

_

−

_e

_

f n



, so that

{

T n

}

is Cauchy in B(E, F ). Let T  be its limit, and let f  = T e. Then



f 

₋

f n



=



T e

−

T ne

 ≤ 

T 

−

T n



e

 →

0.

Thus F  is Banach. 

(2) Let A

_∈

B(E ) be an operator with the property that there is a sequence

{

Am

}

of finite-rank operators such that



A

−

An

 →

0. Show that A is compact.

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



=



lim

n Akyn

−

limn Ajyn



= limn



(Ak

−

Aj)yn

 ≤ 

Ak

−

Aj



since

_

(Ak

−

Aj)yn

 ≤ 

Ak

−

Aj



for each n. This proves that

{

zk

}

is Cauchy; let its limit be z. Then for each k,

Since all three terms on the right approach 0 as k

_{→ ∞}

, we see that Ayn

→

z. Thus, we have found a subsequence

{

yn

}

such that

{

Ayn

}

converges. 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 cluster point of  A; see 11.5 of my putative functional HW. It’s also used to prove

Arzela-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. Let Anbe the truncated multiplication operator which 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 is compact.

Conversely, suppose an

→

0. Let  > 0 and ank a subsequence with

|

ank

|

>  for each k. Then

{

enk

}

is a sequence of unit vectors such that

{

Aenk

}

has no convergent subsequence (indeed,



Aenk

−

Aenj



> 

√ 

_{2), so}

that 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 fact

compact. 

(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 with}

kernel k



_{, then}

_

_A

₋

_A



_{ ≤ }

_k

₋

_k



_

_{sup. Now by Stone-Weierstrass, there}

exists a sequence kn of kernels of the form kn(x, y) =



N _j=1n  p(n)_j (x)q _j(n)(y), with p(n)_j and q _j(n) 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 function f  : [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 functions

t

_→

1

₋

tx with y

_→

y

−

1. 

(2) Show that for every element x

_∈

G there is an  > 0 with the following property: For every element y

_∈

G satisfying

_

y

₋

x

_

<  there is an arc in G connecting y to x.

Solution. Let  =

_

x

−

1



−

1_{. If  h}

∈

A is such that x + h

_∈

G and

_

h

_

< , then let f  be a path from 1 to 1 + x

−

1_{h as in the previous problem; then} t

_→

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

−

x or (1

₋

x)

−

1_{, where x}

∈

A satisfies

_

x

_

< 1. Show that G0 is the connected component of  1 in G.

Solution. Let

_I 

denote the connected component of the identity. Note that this is the same as the path component of the identity, because G is locally path 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, then g(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 is again in G0.

•

G0 is clopen, because open subgroups of topological groups are always clopen: 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

_→

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 topology on G/G0 makes it into a discrete group.

Solution. The connected component of the identity is a subgroup, as shown above; it is normal because, if  f  is a path from 1 to x and y

_∈

G, then t

_→

yf (t)y

−

1 _{is a path from 1 to yxy}

₋

1_{, so that G}

0 is closed under conjugation by elements of  G.

The cosets of G0in G are precisely the connected components of  G; to see this, let Gx denote the component of some x

∈

G, and note that xG0

⊆

Gx and x

−

1_G

x

⊆

G0. As a result, the preimage of any subset of  G/G0 will be a 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 isomorphic to the algebra of complex numbers.

Solution. Such an algebra would have to have the same linear structure but a different multiplication. The simplest way to do this is to define all products to be zero.

Note that it is impossible to give a unital counterexample, because the multiplicative structure of a unital one-dimensional algebra is determined

by (λ1)(µ1) = λµ1. 

(2) Let X  be a compact Hausdorff space and let A = C (X ) be the Banach algebra of all complex-valued continuous functions on X . Show that for every f 

_∈

C (X ), σ(f ) = f (X ).

Solution. The invertible functions in C (X ) are precisely the functions which are 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 (T f )(x) = xf (x). What is the spectrum of  T ? Does T  have point spectrum?

Solution. T  has no point spectrum, because if  f  is a function such that xf (x) = f (x) then f (x) = 0 for x

_

= 1, so that f  = 0 in L2.

The spectrum is [0, 1]. It is at most this, because for λ /

_∈

[0, 1], the inverse of  T 

₋

λ1 is the operator (Sf )(x) = _x

₋

1_λf (x). On the other hand, if  λ

_∈

[0, 1] then such an inverse S  cannot exist; when f (x) is the constant function f (x) = 1, we must have (Sf )(x) = _x1

−

λ a.e., but this is not square integrable.

We recall that, more generally, the spectrum of a multiplication operator

M f on L2 is the essential range of  f . 

For the remaining exercises, let

_{

an

}

be a bounded sequence of complex numbers 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 aunilateral 

weighted shift .

Solution. This operator is just the composition of the multiplication oper-ator A



determined by

_{

an

}

with the unilateral shift S . These are both

bounded by previous work. 

(5) Let A

_∈

B(H ) be a weighted shift as above. Show that for every complex number λ with

_|

λ

_|

= 1 there is a unitary operator U λ

∈

B(H ) such that U AU 

−

1 = λA.

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

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

Solution. For any µ, λ

_∈

C with

_|

λ

_|

= 1, an operator T  is the inverse for A

₋

µ1 iff U T U 

−

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, because of the factorization A = SA



_{mentioned above.}

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



An

_

= sup k



Anek



. Now Anek =





k+n

₋

1



j=k aj





ek+n so that



Anek



1/n = sup k





k+n

₋

1



j=k aj





1/n .

This approaches 0. Explicitly, given  > 0, choose N  such that

_|

an

|

< /2 for n

_≥

N ; let M  be the product of all a1, . . . aN with absolute value greater than 1, if there are any; choose N 2 such that M (/2)N 2

−

N < ; then the geometric mean of any N 2 consecutive terms of the sequence

{

an

}

is less than .

For amusement, one could apply AM-GM and do the above epsilontics with arithmetic rather than geometric means.



1.7. Spectral Radius.

(1) Let

_{

an

}

be a sequence with an

→

0. Show that the associated weighted shift 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 _n!1 .

First Solution. We have the explicit expression

|

∆n

|

=

 

1 0

 

xn−1 0

 

xn−2 0 . . .

 

x2 0 1dx1dx2. . . d xn.

One can see inductively that performing the first k integrals yields the integrand xkk+1

k! , so that one ends up integrating xn−1

n

(n

₋

1)! from 0 to 1, yielding

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

∆n,r =

{

(x1, . . . , xn)

∈

[0, r]n : x1

≤ ·· · ≤

xn

≤

r

}

for each r

_≥

0. By homothety,

_|

∆n,r

|

= rn

|

∆n

|

.

Now the base case

_|

∆1

|

= _1!1 is clear; for the induction, we note that

|

∆n+1

|

=

 

∆n+1 1dV n+1 =

 

1 0

 

∆_n,xn+1 1dV ndxn+1 =

 

1 0 xn_n+1

_|

∆n

|

dxn+1 =

|

∆n

|

n + 1. 

Third Solution. We note that the permutation group S n acts isometrically on the cube [0, 1]n_{. Also,}



π

_∈

S n

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



= id. Referring to the cycle decom-position of π, if (a1a2. . . ) is a cycle in standard form (i.e. smallest element first), then

∆n

∩

π(∆n)

⊆ {

(x1, . . . , xn)

∈

[0, 1]n : xa1 = xa2

}

which has volume zero.

Thus, [0, 1]n _{is the union of  n! almost-disjoint regions of equal volume,}

so each must have volume _n!1 . 

(3) Let k(x, y) be a Volterra kernel as in Example 1.1.4, and let K  be its corresponding integral operator on the Banach space C ([0, 1]). Estimate the norms

_

K n

_

by showing that there is a positive constant M  such that for every f 

_∈

C ([0, 1]) and every n = 1, 2, . . . ,



K nf 

_{ ≤}

M  n n!



f 



. Solution. By induction, (K nf )(t) =

 

∆n,t k(t, xn)k(xn, xn+1) . . . k(x2, x1)f (x1)dx1. . . d xn so that

|

(K nf )(t)

_{| ≤}

 

∆n,t

|

k(t, xn)

|

. . .

|

k(x2, x1)

||

f (x1)

|

dx1. . . d xn

≤

 

∆n,t M n

_

f 

_

dx1. . . d xn = M n



f 

_

tn n!

≤

M n



f 

_

n! where M  =

_

k

_

C ([0,1]2₎. Hence,

_

K n

_{ ≤}

M _n!n. 

(4) Let K  be a Volterra operator as in the preceding exercise. Show that for every complex number λ

_

= 0 and every g

_∈

C ([0, 1]), the Volterra equation of 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 get} this result, or use the fact that _(n!)n1/n

→

e), the previous exercise implies

that the spectral radius of  K  is 0, so that the spectrum is

_{

0

_}

. Hence

1.8. Ideals and Quotients.

(1) Let V  and W  be finite-dimensional vector spaces over Cand let T  : V 

_→

W  be a linear map satisfying T V  = W , and having kernel K  =

_{

x

_∈

V  : T x = 0

_}

. Then we have a short exact sequence of vector spaces

0

_−→

K 

_−→

V 

_−→

W 

_−→

0. Show that dim V  = dim K + dim W .

Solution. Let

_{

k1, . . . , kn

}

be a basis for K  and

{

w1, . . . , wm

}

a basis for W . For each j = 1, . . . , m let vj

∈

V  be a vector such that T vj = wj. We will show that B =

_{

k1, . . . , kn, v1, . . . , vm

}

is a basis for V . It is linearly independent 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 have a1k1 +

· · ·

+ ankn = 0, which by the independence of  k1, . . . , kn implies a1 =

· · ·

= an = 0.

Furthermore, B spans V , because if  v

_∈

V , there exist scalars c1, . . . , cm such that T v = c1w1+

· · ·

+ cmwm; but then v

−

(c1v1+

· · ·

+cmvm)

∈

K , so that v

₋

(c1v1+

· · ·

+ cmvm) = d1k1+

· · ·

+ dnkn for some scalars d1, . . . , dn. We thus have

v = d1k1+

· · ·

+ dnkn+ c1v1+

· · ·

+ cmvm.



(2) For n = 1, 2, . . . let V 1, V 2, . . . , V  n be finite-dimensional vector spaces and set V 0 = V n+1 = 0. Suppose that for k = 0, . . . , n we have a linear map T k : V k

→

V k+1 such that 0 T 0

→

V 1 T 1

→

V 2 T 2

→

. . .T n−1

→

V n T n

→

0 is exact. Show that



n_k=1(

₋

1)k_{dim V} 

k = 0.

Solution. By exactness,

_R

(T k

₋

1) = ker T k, so that dim

R

(T k

₋

1) = dim ker T k. By the Rank Theorem, dim ker T k+dim

R

(T k) = dim V k. Substituting the former equation into the latter,

dim V k = dim

R

(T k

₋

1) + dim

R

(T k). Thus n



k=1 (

₋

1)kdim V k = n



k=1

(

₋

1)k[dim

_R

(T k

₋

1)+dim

R

(T k)] =

−

dim

R

(T 0)+(

−

1)ndim

R

(T n) since the sum telescopes. But both T 0 and T n 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 a discontinuous linear functional f  : E 

_→

C.

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

If 

_{

eα

}

is a unit basis of the infinite-dimensional normed space E , let

{

en

}

be a countable subset. Define f (en) = n and f (eα) = 0 for all other

α. Then f  is discontinuous. 

(4) Let A be a complex algebra and let I  be a proper ideal of  A. Show that I  is 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  is the canonical quotient map, so that I  is not maximal.

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

simple. 

(5) Let A be a unital Banach algebra, let n be a positive integer, and let ω : A

_→

M n be a homomorphism of complex algebras such that ω(A) = M n. Show that ω is continuous. Deduce that every multiplicative linear functional f  : A

_→

C is continuous.

Solution. Since M n is simple, the previous problem implies that the kernel of 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 followed by an isomorphism of finite-dimensional algebras), so ω is continuous. The

final comment is simply the case n = 1. 

1.9. Commutative Banach Algebras.

(1) Show that if A is nontrivial in the sense that A

_

=

_{

0

_}

(equivalently, 1

_

= 0), one has sp(A)

_

=

_{

0

_}

.

Solution. Apply Theorem 1.9.5 to the element 1. 

(2) Show that ω

_→

ker ω is a bijection of the Gelfand spectrum onto the set of  maximal ideals in A.

Solution. This maps into the set of maximal ideals as claimed, because ker ω has codimension one. It is injective because if ker ω1 = ker ω2 then ker(ω1

−

ω2) includes both ker ω1 and 1 and hence is all of  A. Finally, it is surjective, because if  I  is a maximal ideal, then A/I  is a division algebra and hence isomorphic to C by Gelfand-Mazur; hence the quotient map A

_→

A/I  has the form a

_→

ω(a) + I  for some complex homomorphism ω,

from which it follows that I  = ker ω. 

(3) Show that the Gelfand map is an isometry iff 

_

x2



=

_

x

_

2 _{for every x}

∈

A.

Solution. One always has

_

xˆ2



=

_

xˆ

_

2_{; hence, if the Gelfand map is an} isometry, 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) =

_

xˆ

_

as well, so that

_

x

_

=

_

xˆ

_

. 

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



x

_∈

A : lim

n

_→∞



x n



1/n = 0



.

Show that rad(A) is a closed ideal in A with the property that A/rad(A) has no nonzero quasinilpotents (such a commutative Banach algebra is called

semisimple ).

Solution. The quasinilpotent elements are precisely those with spectrum

{

0

_}

, 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 be

zero. 

(5) Let A and B be commutative unital Banach algebras and let θ : A

_→

B be a unital algebra homomorphism.

(a) Show that θ induces a continuous map ˆθ : sp(B)

_→

sp(A) by ˆθ(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 ˆθ 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

ˆ

θ(ων )(a) = ων (θ(a))

→

ω(θ(a)) = ˆθ(ω)(a) for every a

_∈

A, so that ˆθ(ων )

→

θ(ω) in sp(A).ˆ

(b) Suppose (aν , θ(aν ))

→

(a, b). Then for every ω

∈

sp(B), ω(θ(aν ))

→

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

ω(θ(aν )) = ˆθ(ω)(aν )

→

θ(ω)(a) = ω(θ(a)).ˆ

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. By semisimplicity, 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 convergent power series f (z) =

∞



n=0 anzn with



_|

an

|

<

∞

.

Prove the following analogue of Wiener’s theorem: If  f 

_{∈ B} 

satisfies f (z)

_

= 0 for every z

_∈

D¯, then 1_f 

_{∈ B} 

.

Solution. First, we show that the maximal ideal space of 

_B 

(also known as the disk algebra A(D)) is homeomorphic to D. Clearly point evaluations are 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 with evaluation at z0 on all polynomials; but the latter are dense in

B 

, so ω is evaluation 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 right shift 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 space of  1(Z+) is homeomorphic to ¯D, with the Gelfand transform being the Fourier transform a

_→



anzn. By the previous exercise, f  has no zeros in 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 consists of a (convergent infinite) sum of translates of  a. Hence the span of such translates includes the delta function; but, being translation-invariant, it therefore 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 unbounded component of C

_\

_{σA(x). Show that for every λ}

_∈

_Ω

∞

there is a sequence

of polynomials p1, p2, . . . such that lim

n

_→∞



(x

−

λ1)

−

1

−

 pn(x)



= 0.

Solution. Let B be the unital subalgebra generated by x. Then σB(x) has the same unbounded component as σA(x); in particular, λ /

∈

σB(x), so that x

₋

λ1 has an inverse in B; but this inverse must be a norm limit of 

polynomials, because everything in B is. 

(2) Let A be a unital Banach algebra that is generated by

_{

1, x

_}

for some x

_∈

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 complecomple-ment. We wish to show that K  = σA(x). By Theorem 1.9.5, it suffices to show that for each ζ 

∈

K  there 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 multiplicative linear functional on the closure of the set of all polynomials in x; but this

is all of A. 

(3) Deduce the following theorem of Runge: Let X 

_⊆

C be a compact set whose complement is connected. Show that if  f (z) = p(z)/q (z) is a rational function with q (z)

_

= 0 for z

_∈

X , then there is a sequence of polynomials f 1, f 2, . . . such that

sup

z

_∈

X

|

f (z)

−

f n(z)

| →

0.

Solution. Let

_A

be the unital subalgebra of C (K ) generated by the identity function—that is, the algebra of norm limits of polynomials. We wish to show 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 σ

_A

(id) = K  as well. By Spectral Mapping, σ

_A

(q ) = σ

_A

(q (id)) = q (K ), which by hypothesis does not contain 0. Hence q  is invertible in

_A

.

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

enough. 

1.12. Brief on the Analytic Functional Calculus.

Exercise 1.12.1: Let C  be an oriented curve in C, f  a continuous function from C  to a Banach space E , and

_P 

the set of finite ordered partitions of  C .

(1) Show that for every  > 0 there is a δ > 0 with the property that for every pair of oriented partitions

_P 

1,

P 

2 satisfying

P 

k

 ≤

δ  for k = 1, 2, one has

_

R(f,

_P 

1)

−

R(f,

P 

2)

 ≤

.

(2) Verify the estimate





 

C  f (λ) dλ





≤

 

C 



f (λ)

_

d

_|

λ

_{| ≤}

sup λ

_∈

C 



f (λ)

_

(C ). Solution:

(1) Since f  is a continuous function with compact domain C , it is uni-formly continuous. Thus, given  > 0, we may choose δ > 0 such that

|

z

₋

w

_|

< δ  implies

_|

f (z)

₋

f (w)

_|

< _2(C ) for any z, w

_∈

C .

Suppose

_P 

1 and

P 

2 are two partitions with

P 

1



< δ  and

P 

2



< δ . Let

_P 

3 =

P 

1

∪ P 

2 be the common refinement. Let γ 0, . . . , γ  N  denote the points in

_P 

1.

Let k

_{∈ {}

0, . . . , N  

₋

1

_}

. Suppose

_P 

3 contains the additional points η1, . . . , ηm between γ k and γ k+1. (It is possible that m = 0, i.e. that there are no additional points.) We consider the partial Riemann sums Rk(f,

P 

1) := f (γ k+1)(γ k+1

−

γ k) Rk(f,

P 

3) := f (η1)(η1

−

γ k) + m

₋

1



j=1 f (ηj+1)(ηj+1

−

ηj) + f (γ k+1)(γ k+1

−

ηm)

for the portion C k of  C  between γ k and γ k+1. (In the case m = 0 we have instead Rk(f,

P 

3) = Rk(f,

P 

1).)

Note that we can expand Rk(f,

P 

1) = f (γ k+1)



(η1

−

γ k) + m

₋

1



j=1 (ηj+1

−

ηj) + (γ k+1

−

ηm)



. Then

|

Rk(f,

P 

3)

−

Rk(f,

P 

1)

|

=







f (η1)

−

f (γ k)



(η1

−

γ k)+ m

₋

1



j=1



f (ηj+1)

−

f (γ k)



(ηj+1

−

ηj +



f (γ k+1)

−

f (γ k)



(γ k+1

−

ηm)





≤|

f (η1)

−

f (γ k)

||

η1

−

γ k

|

+ m

₋

1



j=1

|

f (ηj+1)

−

f (γ k)

||

ηj+1

−

ηj

|

+

_|

f (γ k+1)

−

f (γ k)

||

γ k+1

−

ηm

|

≤

_2(C )



|

η1

−

γ k

|

+ m

₋

1



j=1

|

ηj+1

−

ηj

|

+

|

γ k+1

−

γ k

|



≤

_2(C ) (C k). Summing over k, we have

|

R(f,

_P 

3)

−

R(f,

P 

1)

|

=







k Rk(f,

P 

3)

−



k Rk(f,

P 

1)





≤



k

|

Rk(f,

P 

3)

−

Rk(f,

P 

1)

|

≤

_2(C )



k (C k) =  2.

The same calculation holds with

_P 

2 in place of 

P 

1. By the Triangle Inequality, it follows that

|

R(f,

_P 

1)

−

R(f,

P 

2)

| ≤

.

(2) For every partition

_P 

one has by the Triangle Inequality



R(f,

_P 

)

_{ ≤}



k



f (γ k+1)

|

γ k+1

−

γ k

|

= 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

_

 

_C f (λ) dλ

_{ ≤}

 

_C 

_

f (λ)

_

d

_|

λ

_|

. For the second half, if  M  = sup_λ

_∈

_C 

_

f (λ)

_

, then

_

f (λ)

_{ ≤}

M  at each point λ, so that

 

C 



f (λ)

_

d

_|

λ

_{| ≤}

 

C 

M d

_|

λ

_|

= M (C ).

Exercise 1.12.2: Let D =

_{

z

_∈

C _:

_|

_z

_|

_{< R}

_}

_{be an open disc containing}

σ(T ). Let f  : D

_→

C be an analytic function defined on D with power series f (z) =

∞



n=0 cnzn, z

∈

D. Show that the infinite series of operators

∞



n=0 cnT n

converges absolutely in the sense that



_n

_|

cn

|

T n



<

∞

.

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

_

T n



< ˜Rn _{for sufficiently large n, say n}

≥

N , so



n

_≥

N 

|

cn

|

T n



<



n

_≥

N 

|

cn

|

R˜n

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

Exercise 1.12.3: Give a definition of sin T  and cos T  using power series. Solution: We define sin T  =



∞

k=0 (

₋

1)k (2k + 1)!T  2k+1_{, cos T  =}



∞

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 (sin T )2+ (cos T )2 = 1.

Solution: This is immediate from the holomorphic functional calculus, which sends the complex function sin2z + cos2z

₋

1 to 0. Alternatively, one can prove it directly by manipulating the above series: Since

sin2T  =



∞

j,k=0 (

₋

1)j+k (2k + 1)!(2 j + 1)!T  2(j+k+1)₌



∞

=1 (

₋

1)

−

1T 



s odd 1 s!(2

₋

s)! and cos2T  =



∞

j,k=0 (

₋

1)j+k (2k)!(2 j)!T  2(j+k) _{= 1 +}



∞

=1 (

₋

1)T 2



s even 1 s!(2

₋

s)!, we have sin2T  + cos2T  = 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,}