| T x, x |. It is enough to prove that

Proof. Let ± = sup x ¤1

| T x, y | ¤ ± x y

for all x and y. We can obviously assume that x and y are nonzero. Moreover, we may

multiply y by a complex number of modulus one, so we can assume that T x, y ≥ 0. Then

T (x + y), x + y ’ T (x ’ y), x ’ y = 4 Re T x, y = 4| T x, y |.

so

± ±

2

+ x ’ y 2) = 2

+ y 2 ).

| T x, y | ¤ ( x+y (x

4 2

Now apply this result with x replaced by y / x x and y replaced by x / y y.

Proof of spectral theorem for compact self-adjoint operators. We ¬rst show that T has a

nonzero eigenvector. If T = 0, this is obvious, so we assume that T = 0. Choose a sequence

xn ∈ X with xn = 1 so that | T xn , xn | ’ T . Since T is self-adjoint, T xn , xn ∈ R,

so we may pass to a subsequence (still denoted xn ), for which T xn , xn ’ » = ± T .

27

Since T is compact we may pass to a further subsequence and assume that T xn ’ y ∈ X.

Note that y ≥ |»| > 0.

Using the fact that T is self-adjoint and » is real, we get

2 2

’ 2» T xn , xn + »2 xn 2

T xn ’ »xn = T xn

2 2

’ 2»2 = 0.

¤2 T ’ 2» T xn , xn ’ 2 T

Since T xn ’ y we infer that »xn ’ y as well, or xn ’ y/» = 0. Applying T we have

T y/» = y, so » is indeed a nonzero eigenvalue.

To complete the proof, consider the set of all orthonormal subsets of X consisting of

eigenvectors of T . By Zorn™s lemma, it has a maximal element S. Let W be the closure

of the span of S. Clearly T W ‚ W , and it follows directly (since T is self-adjoint), that

T W ⊥ ‚ W ⊥ . Therefore T restricts to a self-adjoint operator on W ⊥ and thus, unless

W ⊥ = 0, T has an eigenvector in W ⊥ . But this clearly contradicts the maximality of S

(since we can adjoin this element to S to get a larger orthonormal set of eigenvectors).

Thus W ⊥ = 0, and S is an orthonormal basis.

The following structure result on the set of eigenvalues is generally considered part of

the spectral theorem as well.

Theorem. If T is a compact self-adjoint operator on a Hilbert space, then the set of

nonzero eigenvalues of T is either a ¬nite set or a sequence approaching 0 and the corre-

sponding eigenspaces are all ¬nite dimensional.

Remark. 0 may or may not be an eigenvalue, and its eigenspace may or may not be ¬nite.

Proof. Let ei be an orthonormal basis of eigenvectors, with T ei = »i ei . Here i ranges over

some index set I. It su¬ces to show that S = { i ∈ I | |»i | ≥ } is ¬nite for all > 0. Then

if i, j ∈ I

T ei ’ T ej 2 = »i ej ’ »j ej 2 = |»i |2 + |»j |2 ,

so if i, j ∈ S, then T ei ’ T ej 2 ≥ 2 2 . If S were in¬nite, we could then choose a sequence

of unit elements in X whose image under T has no convergent subsequence, which violates

the compactness of T .

Suppose, for concreteness, that X is an in¬nite dimensional separable Hilbert space and

that { en }n∈N is an orthonormal basis adapted to a compact self-adjoint operator T on X.

Then the map U : X ’ l2 given by

U( cn en ) = (c0 , c1 , . . . ),

n

is an isometric isomorphism. Moreover, when we use this map to transfer the action of T

to l2 , i.e., when we consider the operator U T U ’1 on l2 , we see that this operator is simply

multiplication by the bounded sequence (»0 , »1 , . . . ) ∈ l∞ . Thus the spectral theorem says

that every compact self-adjoint T is unitarily equivalent to a multiplication operator on l2 .

(An isometric isomorphism of Hilbert spaces is also called a unitary operator. Note that

it is characterized by the property U — = U ’1 .)

A useful extension is the spectral theorem for commuting self-adjoint compact operators.

28

Theorem. If T and S are self-adjoint compact operators in a Hilbert space H and T S =

ST , then there is an orthonormal basis of X whose elements are eigenvectors for both S

and T .

Proof. For an eigenvalue » of T , let X» denote the corresponding eigenspace of T . If

x ∈ X» , then T Sx = ST x = »Sx, so Sx ∈ X» . Thus S restricts to a self-adjoint

operator on X» , and so there is an orthonormal basis of S“eigenvectors for X» . These are

T “eigenvectors as well. Taking the union over all the eigenvalues » of T completes the

construction.

Let T1 and T2 be any two self-adjoint operators and set T = T1 + iT2 . Then T1 =

(T + T — )/2 and T2 = (T ’ T — )/(2i). Conversely, if T is any element of B(X), then we

can de¬ne two self-adjoint operators from these formulas and have T = T1 + iT2 . Now

suppose that T is compact and also normal, i.e., that T and T — commute. Then T1 and

T2 are compact and commute, and hence we have an orthonormal basis whose elements

are eigenvectors for both T1 and T2 , and hence for T . Since the real and imaginary parts

of the eigenvalues are the eigenvalues of T1 and T2 , we again see that the eigenvalues form

a sequence tending to zero and all have ¬nite dimensional eigenspaces.

We have thus shown that a compact normal operator admits an orthonormal basis

of eigenvectors. Conversely, if { ei } is an orthonormal basis of eigenvectors of T , then

T — ei , ej = 0 if i = j, which implies that each ei is also an eigenvector for T — . Thus

T — T ei = T T — ei for all i, and it follows easily that T is normal. We have thus shown:

Spectral Theorem for compact normal operators. Let T be a compact operator on a

Hilbert space X. Then there exists an orthonormal basis for X consisting of eigenvectors of

T if and only if T is normal. In this case, the set of nonzero eigenvalues form a ¬nite set or

a sequence tending to zero and the eigenspaces corresponding to the nonzero eigenvalues are

¬nite dimensional. The eigenvalues are all real if and only if the operator is self-adjoint.

The spectrum of a general compact operator. In this section we derive the structure

of the spectrum of a compact operator (not necessarily self-adjoint or normal) on a complex

Banach space X.

For any operator T on a complex Banach space, the resolvent set of T , ρ(T ) consists

of those » ∈ C such that T ’ »1 is invertible, and the spectrum σ(T ) is the complement.

If » ∈ σ(T ), then T ’ »1 may fail to be invertible in several ways. (1) It may be that

N (T ’ »1) = 0, i.e., that » is an eigenvalue of T . In this case we say that » belongs to the

point spectrum of T , denoted σp (T ). (2) If T ’ »1 is injective, it may be that its range is

dense but not closed in X. In this case we say that » belongs to the continuous spectrum

of T , σc (T ). Or (3) it may be that T ’ »1 is injective but that its range is not even dense

in X. This is the residual spectrum, σr (T ). Clearly we have a decomposition of C into the

disjoint sets ρ(T ), σp (T ), σc (T ), and σr (T ). As an example of the continuous spectrum,

consider the operator T en = »n en where the en form an orthonormal basis of a Hilbert

space and the »n form a positive sequence tending to 0. Then 0 ∈ σc (T ). If T en = »n en+1 ,

0 ∈ σr (T ).

29

Now if T is compact and X is in¬nite dimensional, then 0 ∈ σ(T ) (since if T were

invertible, the image of the unit ball would contain an open set, and so couldn™t be pre-

compact). From the examples just given, we see that 0 may belong to the point spectrum,

the continuous spectrum, or the residual spectrum. However, we shall show that all other

elements of the spectrum are eigenvalues, i.e., that σ(T ) = σp (T ) ∪ {0}, and that, as in the

normal case, the point spectrum consists of a ¬nite set or a sequence approaching zero.

The structure of the spectrum of a compact operator will be deduced from two lemmas.

The ¬rst is purely algebraic. To state it we need some terminology: consider a linear

operator T from a vector space X to itself, and consider the chains of subspaces

0 = N (1) ‚ N (T ) ‚ N (T 2 ) ‚ N (T 3 ) ‚ · · · .

Either this chain is strictly increasing forever, or there is a least n ≥ 0 such that N (T n ) =

N (T n+1 ), in which case only the ¬rst n spaces are distinct and all the others equal the

nth one. In the latter case we say that the kernel chain for T stabilizes at n. In particular,

the kernel chain stabilizes at 0 i¬ T is injective. Similarly we may consider the chain

X = R(1) ⊃ R(T ) ⊃ R(T 2 ) ⊃ R(T 3 ) ⊃ · · · ,

and de¬ne what it means for the range chain to stabilize at n > 0. (So the range stabilizes

at 0 i¬ T is surjective.) It could happen that neither or only one of these chains stabilizes.

However:

Lemma. Let T be a linear operator from a vector space X to itself. If the kernel chain

stabilizes at m and the range chain stabilizes at n, then m = n and X decomposes as the

direct sum of N (T n ) and R(T n ).

Proof. Suppose m were less than n. Since the range chain stabilizes at n, there exists x with

T n’1 x ∈ R(T n ), and then there exists y such that T n+1 y = T n x. Thus x ’ T y ∈ N (T n ),

/

and, since kernel chain stabilizes at m < n, N (T n ) = N (T n’1 ). Thus T n’1 x = T n y, a

contradiction. Thus m ≥ n. A similar argument, left to the reader, establishes the reverse

inequality.

Now if T n x ∈ N (T n ), then T 2n x = 0, whence T n x = 0. Thus N (T n ) © R(T n ) = 0.

Given x, let T 2n y = T n x, so x decomposes as T n y ∈ R(T n ) and x ’ T n y ∈ N (T n ).

The second lemma brings in the topology of compact operators.

Lemma. Let T : X ’ X be a compact operator on a Banach space and »1 , »2 , . . . a

sequence of complex numbers with inf |»n | > 0. Then the following is impossible: There

exists a strictly increasing chain of closed subspaces S1 ‚ S2 ‚ · · · with (»n 1 ’ T )Sn ‚

Sn’1 for all n.

Proof. Suppose such a chain exists. Note that each T Sn ‚ Sn for each n. Since Sn /Sn’1

contains an element of norm 1, we may choose yn ∈ Sn with yn ¤ 2, dist(yn , Sn’1 ) = 1.

If m < n, then

T ym ’ (»n 1 ’ T )yn

∈ Sn’1 ,

z :=

»n

30

and

T ym ’ T yn = |»n | yn ’ zn ≥ |»n |.

This implies that the sequence (T yn ) has no Cauchy subsequence, which contradicts the

compactness of T .

We are now ready to prove the result quoted at the beginning of the subsection.

Theorem. Let T be a compact operator on a Banach space X. Then any nonzero ele-

ment of the spectrum of T is an eigenvalue. Moreover σ(T ) is either ¬nite or a sequence

approaching zero.

Proof. Consider the subspace chains N [(»1 ’ T )n ] and R[(»1 ’ T )n ] (these are closed

subspaces by a previous result). Clearly »1 ’ T maps N [(»1 ’ T )n ] into N [(»1 ’ T )n’1 ],

so the previous lemma implies that the kernel chain stabilizes, say at n. Now R[(»1’T )n ] =

N [(»1 ’ T — )n ] (since the range is closed), and since these last stabilize, the range chain

a

stabilizes as well.

Thus we have X = N [(»1 ’ T )n ] • R[(»1 ’ T )n ]. Thus

R(»1 ’ T ) = X =’ R(»1 ’ T )n = X =’ N (»1 ’ T )n = 0 =’ N (»1 ’ T ) = 0.

In other words » ∈ σ(T ) =’ » ∈ σp (T ).

Finally we prove the last statement. If it were false we could ¬nd a sequence of

eigenvalues »n with inf |»n | > 0. Let x1 , x2 , . . . be corresponding nonzero eigenvectors

and set Sn = span[x1 , . . . , xn ]. These form a strictly increasing chain of subspaces (re-

call that eigenvectors corresponding to distinct eigenvalues are linearly independent) and

(»n 1 ’ T )Sn ‚ Sn’1 , which contradicts the lemma.

The above reasoning also gives us the Fredholm alternative:

Theorem. Let T be a compact operator on a Banach space X and » a nonzero complex

number. Then either (1) »1 ’ T is an isomorphism, or (2) it is neither injective nor

surjective.

Proof. Since the kernel chain and range chain for S = »1 ’ T stabilize, either they both

stabilize at 0, in which case S is injective and surjective, or neither does, in which case it

is neither.

We close this section with a result which is fundamental to the study of Fredholm

operators.

Theorem. Let T be a compact operator on a Banach space X and » a nonzero complex

number. Then

dim N (»1 ’ T ) = dim N (»1 ’ T — ) = codim R(»1 ’ T ) = codim R(»1 ’ T — ).

31

Proof. Let S = »1 ’ T . Since R(S) is closed

[X/ R(S)]— ∼ R(S)a = N (S — ).

=

Thus [X/ R(S)]— is ¬nite dimensional, so X/ R(S) is ¬nite dimensional, and these two

spaces are of the same dimension. Thus codim R(S) = dim N (S — ).

For a general operator S we only have R(S — ) ‚ N (S)a , but, as we now show, when

R(S) is closed, R(S — ) = N (S)a . Indeed, S induces an isomorphism of X/ N (S) onto

R(S), and for any f ∈ N (S)a , f induces a map X/ N (S) to R. It follows that f = gS

for some bounded linear operator g on R(S), which can be extended to an element of X —

by Hahn-Banach. But f = gS simply means that f = S — g, showing that N (S)a ‚ R(S — )

(and so equality holds) as claimed.

Thus

N (S)— ∼ X — / N (S)a = X — / R(S — ),

=

so codim R(S — ) = dim N (S)— = dim N (S).

We complete the theorem by showing that dim N (S) ¤ codim R(S) and dim N (S — ) ¤

codim R(S — ). Indeed, since R(S) is closed with ¬nite codimension, it is complemented by a

¬nite dimensional space M (with dim M = codim R(S). Since N (S) is ¬nite dimensional,

it is complemented by a space N . Let P denote the projection of X onto N (S) which is

a bounded map which to the identity on N (S) and to zero on N . Now if codim R(S) <

dim N (S), then there is a linear map of N (S) onto M which is not injective. But then T ’

f P is a compact operator and »1 ’ T + f P is easily seen to be surjective. By the Fredholm

alternative, it is injective as well. This implies that f is injective, a contradiction. We

have thus shown that dim N (S) ¤ codim R(S). Since T — is compact, the same argument

shows that dim N (S — ) ¤ codim R(S — ). This completes the proof.

VI. Introduction to General Spectral Theory

In this section we skim the surface of the spectral theory for a general (not necessarily

compact) operator on a Banach space, before encountering a version of the Spectral The-

orem for a bounded self-adjoint operator in Hilbert space. Our ¬rst results don™t require

the full structure of in the algebra of operators on a Banach space, but just an arbitrary

Banach algebra structure, and so we start there.

The spectrum and resolvent in a Banach algebra. Let X be a Banach algebra with

an identity element denoted 1. We assume that the norm in X has been normalized so that

1 = 1. The two main examples to bear in mind are (1) B(X), where X is some Banach

space; and (2) C(G) endowed with the sup norm, where G is some compact topological

space, the multiplication is just pointwise multiplication of functions, and 1 is the constant

function 1.

In this set up the resolvent set and spectrum may be de¬ned as before: ρ(x) = { » ∈

C | x ’ »1 is invertible }, σ(x) = C \ ρ(x). The spectral radius is de¬ned to be r(x) =

sup |σ(x)|. For » ∈ ρ(x), the resolvent is de¬ned as Rx (») = (x ’ »1)’1 .

32

Lemma. If x, y ∈ X with x invertible and x’1 y < 1, then x ’ y is invertible,

∞

’1

(x’1 y)n x’1 ,

(x ’ y) =

n=0

and (x ’ y)’1 ¤ x’1 /(1 ’ x’1 y ).

Proof.

(x’1 y)n x’1 ¤ x’1 x’1 y ¤ x’1 /(1 ’ x’1 y ),

n

so the sum converges absolutely and the norm bound holds. Also

∞ ∞ ∞

’1 n ’1 ’1

(x’1 y)n+1 = 1,

n

(x ’ y) = y) ’

(x y) x (x

n=0 n=0 n=0

and similarly for the product in the reverse order.

As a corollary, we see that if |»| > x , then »1 ’ x is invertible, i.e., » ∈ ρ(x). In other

words:

Proposition. r(x) ¤ x .

We also see from the lemma that lim»’∞ Rx (») = 0. Another corollary is that if

» ∈ ρ(x) and |µ| < Rx (») ’1 , then » ’ µ ∈ ρ(x) and

∞

Rx (»)n+1 µn .

Rx (» ’ µ) =

n=0

Theorem. The resolvent ρ(x) is always open and contains a neighborhood of ∞ in C and

the spectrum is always non-empty and compact.

Proof. The above considerations show that the resolvent is open, and so the spectrum is

closed. It is also bounded, so it is compact.

To see that the spectrum is non-empty, let f ∈ X — be arbitrary and de¬ne φ(») =

f [Rx (»)]. Then φ maps ρ(x) into C, and it is easy to see that it is holomorphic (since we

have the power series expansion

∞

f [R(»)n+1 ]µn

φ(» ’ µ) =

n=0

if µ is su¬ciently small). If σ(x) = …, then φ is entire. It is also bounded (since it tends to 0

at in¬nity), so Liouville™s theorem implies that it is identically zero. Thus for any f ∈ X — ,

f [(»1 ’ x)’1 ] = 0 . This implies that (»1 ’ x)’1 = 0, which is clearly impossible.

33

Corollary (Gelfand“Mazur). If X is a complex Banach division algebra, then X is

isometrically isomorphic to C.

Proof. For each 0 = x ∈ X, let » ∈ σ(x). Then x ’ »1 is not invertible, and since X is a

division algebra, this means that x = »1. Thus X = C1.

Now we turn to a bit of “functional calculus.” Let x ∈ X and let f be a complex

function of a complex variable which is holomorphic on the closed disk of radius x about

the origin. Then we make two claims: (1) plugging x into the power series expansion of f

de¬nes an element f (x) ∈ X; and (2) the complex function f maps the spectrum of x into

the spectrum of f (x). (In fact onto, as we shall show later in the case f is polynomial.)

To prove these claims, note that, by assumption, the radius of convergence of the power

∞ n

series for f about the origin exceeds x , so we can expand f (z) = n=0 an z where

|an | x n < ∞. Thus the series an xn is absolutely convergent in the Banach space

X; we call its limit f (x). (This is the de¬nition of f (x). It is a suggestive abuse of

notation to use f to denote the this function, which maps a subset of X into X, as well as

the original complex-valued function of a complex variable.) Now suppose that » ∈ σ(x).

Then

∞ ∞ ∞

n n

f (»)1 ’ f (x) = an (» 1 ’ x ) = (»1 ’ x) an Pn (»1 ’ x),

an Pn =

n=1 n=1 n=1

where

n’1

»k xn’k’1 .

Pn =

k=0

∞

n’1

Note that Pn ¤ n x an Pn converges to some y ∈ X. Thus

, so n=1

f (»)1 ’ f (x) = (»1 ’ x)y = y(»1 ’ x).

Now f (»)1 ’ f (x) can™t be invertible, because these formulas would then imply that »1 ’ x

would be invertible as well, but » ∈ σ(x). Thus we have veri¬ed that f (») ∈ σ f (x) for

all » ∈ σ(x).

Theorem (Spectral Radius Formula). r(x) = limn’∞ xn 1/n

= inf n xn 1/n

.

Proof. If » ∈ σ(x), then »n ∈ σ(xn ) (which is also evident algebraically), so |»n | ¤ xn .

This shows that r(x) ¤ inf n xn 1/n .

Now take f ∈ X — , and consider

∞

’1

»’n’1 f (xn ).

φ(») = f [(»I ’ x) ]=

n=0

Then φ is clearly holomorphic for » > x , but we know it extends holomorphically to

» > r(x) and tends to 0 as » tends to in¬nity. Let ψ(») = φ(1/»). Then ψ extends

34

analytically to zero with value zero and de¬nes an analytic function on the open ball of

radius 1/r(x) about zero, as does, therefore,

∞

f (»n xn ).

ψ(»)/» =

n=0

This shows that for each |»| < 1/r(x) and each f ∈ X — , f (»n xn ) is bounded. By the

uniform boundedness principle, the set of elements »n xn are bounded in X, say by K.

Thus xn 1/n ¤ K 1/n /|»| ’ 1/|»|. This is true for all |»| < 1/r(x), so lim sup xn 1/n ¤

r(x).

Corollary. If H is a Hilbert space and T ∈ B(H) a normal operator, then r(T ) = T .

Proof.

= sup T x, T x = sup T — T x, x = T — T ,

2

T

x ¤1 x ¤1

since T — T is self-adjoint. Using the normality of T we also get

T —T = sup T — T x, T — T x = sup T T — T x, T x = sup T — T 2 x, T x

2

x ¤1 x ¤1 x ¤1

= sup T 2 x, T 2 x = T 2 2 .

x ¤1

Thus T 2 = T 2 . Replacing T with T 2 gives, T 4 = T 4 , and similarly for all powers

of 2. The result thus follows from the spectral radius formula.

As mentioned, we can now show that p maps σ(x) onto σ p(x) if p is a polynomial.

Spectral Mapping Theorem. Let X be a complex Banach algebra with identity, x ∈ X,

and let p a polynomial in one variable with complex coe¬cients. Then p σ(x) = σ p(x) .

Proof. We have already shown that p σ(x) ‚ σ p(x) . Now suppose that » ∈ σ p(x) .

By the Fundamental Theorem of Algebra we can factor p ’ », so

p(x) ’ »1 = aΠn (x ’ »i 1),

i=1

for some nonzero a ∈ C and some roots »i ∈ C. Since p(x) ’ »1 is not invertible, it

follows that x ’ »i 1 is not invertible for at least one i. In orther words, »i ∈ σ(x), so

» = p(»i ) ∈ p σ(x) .

35

Spectral Theorem for bounded self-adjoint operators in Hilbert space. We now

restrict to self-adjoint operators on Hilbert space and close with a version of the Spectral

Theorem for this class of operator. We follow Halmos™s article “What does the Spec-

tral Theorem Say?” (American Mathematical Monthly 70, 1963) both in the relatively

elementary statement of the theorem and the outline of the proof.

First we note that self-adjoint operators have real spectra (not just real eigenvalues).

Proposition. If H is a Hilbert space and T ∈ B(H) is self-adjoint, then σ(T ) ‚ R.

Proof.

| (»1 ’ T )x, x | ≥ | Im (»1 ’ T )x, x | = | Im »| x 2 ,

so if Im » = 0, »1 ’ T is injective with closed range. The same reasoning shows that

¯

(»1 ’ T )— = »1 ’ T is injective, so R(»1 ’ T ) is dense. Thus » ∈ ρ(T ).

Spectral Theorem for self-adjoint operators in Hilbert space. If H is a complex

Hilbert space and T ∈ B(H) is self-adjoint, then there exists a measure space „¦ with

measure µ, a bounded measurable function φ : „¦ ’ R, and an isometric isomorphism

U : L2 ’ H such that

U ’1 T U = Mφ

where Mφ : L2 ’ L2 is the operation of multiplication by φ. (Here L2 means L2 („¦, µ; C),

the space of complex-valued functions on „¦ which are square integrable with respect to the

measure µ.)

Sketch of proof. Let x be a nonzero element of H, and consider the smallest closed sub-

space M of H containing T n x for n = 0, 1, . . . , i.e., M = { p(T )x | p ∈ PC }. Here PC is the

space of polynomials in one variable with complex coe¬cients. Both M and its orthogonal

complement are invariant under T (this uses the self-adjointness of T ). By a straightfor-

ward application of Zorn™s lemma we see that H can be written as a Hilbert space direct

sum of T invariant spaces of the form of M . If we can prove the theorem for each of these

subspaces, we can take direct products to get the result for all of H. Therefore we may

assume from the start that H = { p(T )x | p ∈ PC } for some x. (In other terminology, that

T has a cyclic vector x.)

Now set „¦ = σ(T ), which is a compact subset of the real line, and consider the space

C = C(„¦, R), the space of all continuous real-valued functions on „¦. The subspace of

real-valued polynomial functions is dense in C (since any continuous function on „¦ can

be extended to the interval [’r(T ), r(T )] thanks to Tietze™s extension theorem and then

approximated arbitrarily closely by a polynomial thanks to the Weierstrass approximation

theorem). For such a polynomial function, p, de¬ne Lp = p(T )x, x ∈ R. Clearly L is

linear and

|Lp| ¤ p(T ) x 2 = r p(T ) x 2 ,

by the special form of the spectral radius formula for self-adjoint operators in Hilbert space.

Since σ p(T ) = p σ(T ) , we have r p(T ) = p L∞ („¦) = p C , and thus,

2

|Lp| ¤ x p C.

36

This shows that L is a bounded linear functional on a dense subspace of C and so extends

uniquely to de¬ne a bounded linear functional on C.

Next we show that L is positive in the sense that Lf ≥ 0 for all non-negative functions

f ∈ C. Indeed, if f = p2 for some polynomial, then

Lf = p(T )2 x, x = p(T )x, p(T )x ≥ 0.

√

For an arbitrary non-negative f , we can approximate f uniformly by polynomials pn , so

f = lim p2 and Lf = lim Lp2 ≥ 0.

n n

We now apply the Riesz Representation Theorem for the representation of the linear

functional L on C. It state that there exists a ¬nite measure on „¦ such that Lf = f dµ

for f ∈ C (it is a positive measure since L is positive). In particular, p(T )x, x = p dµ

for all p ∈ PR .

We now turn to the space L2 of complex-valued functions on „¦ which are square inte-

grable with respect to the measure µ. The subspace of complex-valued polynomial func-

tions is dense in L2 (since the measure is ¬nite, the L2 norm is dominated by the supremum

norm). For such a polynomial function, q, de¬ne U q = q(T )x. Then

2 2

|q|2 dµ = q 2

Uq = q(T )x = q(T )x, q(T )x = q (T )q(T )x, x =

¯ L2 .

Thus U is an isometry of a dense subspace of L2 into H and so extends to an isometry of

L2 onto a closed subspace of H. In fact, U is onto H itself, since, by the assumption that

x is a cyclic vector for T , the range of U is dense.

Finally, de¬ne φ : „¦ ’ R by φ(») = ». If q is a complex polynomial, then (Mφ q)(») =

»q(»), which is also a polynomial. Thus

U ’1 T U q = U ’1 T q(T )x = U ’1 (Mφ q)(T ) x = Mφ q.

Thus the bounded operators U ’1 T U and Mφ coincide on a dense subset of L2 , and hence

they are equal.

For a more precise description of the measure space and the extension to normal oper-

ators, see Zimmer.