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1. Introduction 22. Bounded operators in normed and Banach spaces 43. Hilbert Spaces 84. Inverse and ajoint operators 145. Unitary Equivalence 176. Sufficient Conditions for an infinite matrix to determine a bounded
operator 207. Integral Operators and their Hilbert-Schmidt Condition an Schur Test 238. The Spectrum 269. Complex Methods and the Spectrum 3110. The Continuous Functional Calculus 3511. Compact Operators 39
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1. Introduction
In classical analysis, properties such as continuity, differentiability etc are studiedat the level of individual functions. Functional analysis shifts the focus to classes(usually vector spaces) of functions and (usually linear) transformations or operatorsbetween them. In addition, the abstract structures involved are also studied, evenwhen they are not necessarily spaces of functions.
In the main we study linear operators between Hilbert spaces. This theory hasimportant applications, particularly in quantum mechanics. To illustrate the maintheme, consider the set of n×n complex matrices, Mn(C). This set is closed underaddition, scalar multiplication and matrix product, so if p(z) =
∑kj=0 cjz
k is anypolynomial we may define a new matrix by
k∑j=0
cjAk
and call it p(A). In the general case it is not clear how to define more generalfunctions of a matrix (although power series provide one possibility, for suitablefunctions and matrices). However, in the special case of a diagonal matrix, it’seasy: we set
f(diag(λ1, . . . , λn)) = diag(f(λ1), . . . , f(λn))and the definition is justified by its properties: it agrees with the previous definitionif f is a polynomial and we also have the identities
(λf + µg)(A) = λf(A) + µg(A)
(fg)(A) = f(A)g(A)
(f g)(A) = f(g(A))
for any diagonal A.Now any Hermitian matrix A = A† can be unitarially diagonalised: there is a
unitary matrix U so that UAU† is diagonal. If we now define
f(A) = U†f(UAU†)U
one may check that the previous functional calculus is thereby extended to Hermit-ian matrices.
The main aim of this course is to develop the theory of operators in Hilbert spaceso as the understand the extend to which the above can be extended. Diagonal,Hermitian, and unitary matrices have analogues in the general theory, and there isa general diagonalisation theory: unfortunately this is an ε too advanced for thepresent course. However we will develop a general functional calculus, and see howdiagonalisation can be proved for some special classes of self-adjoint operators.
Although our main focus will be Hilbert space operators, we will start by devel-oping some theory in the more general context of normed spaces and Banach spaces,introduced in Topics in Analysis B (TAB). As in that course, K means “either R orC”.
Defintion 1.1. Suppose X is a vector space over K, e.g. Kn or C([a, b],K). A normon X is a mapping ‖·‖ : X → R such that:
(1) if x ∈ X then ‖x‖ ≥ 0 and ‖x‖ = 0 if and only if x = 0;(2) if x ∈ X and t ∈ K then ‖tx‖ = |t| ‖x‖;(3) if x, y ∈ X then ‖x+ y‖ ≤ ‖x‖ + ‖y‖ (triangle inequality).
A vector space with a norm defined on it is called a normed space.
Recall that any norm gives rise to a metric.
Theorem 1.2. If X is a normed space and we define d(x, y) = ‖x− y‖, then d isa metric on X, known as the metrix induced by the norm.
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Proof. This comes immediately from the definition:(1) since ‖z‖ ≥ 0 for all z, d(x, y) = ‖x− y‖ ≥ 0 for all x and y, and
d(x, y) = 0 ⇐⇒ ‖x− y‖ = 0 ⇐⇒ x− y = 0 ⇐⇒ x = y
(2) d(x, y) = ‖x− y‖ = ‖(−1)(y − x)‖ = |−1| ‖y − x‖ = ‖y − x‖ = d(y, x);(3) d(x, y) = ‖x− y‖ = ‖(x− z) + (z − y)‖ ≤ ‖x− z‖ + ‖z − y‖ = d(x, z) +
d(z, y).
Continuity, convergence, compactness, completeness, etc. are defined on normedspaces using the metric induced by the norm.
Defintion 1.3. A normed space which, as a metric space, is complete (i.e., everyCauchy sequence converges) is called a Banach space or B-space.
Example 1.4. Various norms can be placed on Kn, e.g.
‖x‖1 =N∑j=1
|xj |
‖x‖∞ =n
maxj=1|xj |
‖x‖ =
∞∑j=1
|xj |21/2
and any one of these norms makes KN into a B-space. The third of these, theEuclidian norm, was discussed in Metric Spaces; the other two were exercies inTAB.
A useful alternative characterisation of Banach spaces was also given in TAB.
Defintion 1.5. If (xn)n∈N is a sequence of vectors in a normed space X, thenthe serives
∑∞n=1 xn is called absolutely convergent if the real series
∑∞n=1 ‖xn‖ is
convergent.
Lemma 1.6 (Lemma 1.21 in TAB). Let X be a normed space. Then the followingare equivalent:
(1) X is a Banach space;(2) every absolutely convergent series in X is convergent.
An elementary, but useful, result is the following:
Theorem 1.7. Suppose that (xn)n∈N is a sequence in a normed space, and xn → xas n→∞. Then ‖xn‖ → ‖x‖. If (yn)n∈N is another sequence converging to y ands, t ∈ K, then sxn + tyn → sx+ ty as n→∞.
Proof. Using the triangle inequality, we have
‖xn‖ ≤ ‖xn − x‖ + ‖x‖and also
‖x‖ ≤ ‖x− xn‖ + ‖xn‖from which we get
‖x‖ − ‖x− xn‖ ≤ ‖xn‖ ≤ ‖xn − x‖ + ‖x‖and so the ‘sandwitch theorem’ gives the required result because ‖xn − x‖ → 0 asn→∞.
The rest of the result is a simple exercise in the triangle inequality.
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2. Bounded operators in normed and Banach spaces
In normed spaces there is a nice reformulation of continuity for linear maps.
Theorem 2.1. Let T be a linear operator between normed spaces E and F . ThenTFAE
(1) T is continuous(2) there exists an M ≥ 0 such that ‖Tx‖ ≤M ‖x‖ for all x ∈ E.
Proof. Suppose (2) holds. Then ‖Tx− Ty‖ = ‖T (x− y)‖ ≤ M ‖x− y‖. Continu-ity is immediate (take δ = ε
M ), and hence (2) =⇒ (1).Now suppose T is continuous. Then it is continuous at 0 ∈ E, so there exists a
δ > 0 such that ‖x‖ < δ =⇒ ‖Tx− T0‖ < 1; i.e. ‖Tx‖ < 1 for ‖x‖ < δ. For
any x 6= 0,∥∥∥ δ
2‖x‖x∥∥∥ = δ
2 < δ, so∥∥∥T δ
2‖x‖x∥∥∥ < 1; i.e. ‖Tx‖ ≤ 2
δ ‖x‖. Since this also
holds for x = 0, (2) then holds with M = 2δ .
Defintion 2.2. An operator T between normed spaces E and F is said to bebounded if there exists an M > 0 such that
‖Tx‖ ≤M ‖X‖for all x ∈ E.
The operator norm of T is the infimum of all such constants M , and is denoted‖T‖:
‖T‖ = infM ≥ 0 : ‖Tx‖ ≤M ‖x‖ ∀x ∈ E.The set of bounded operators between E and F is denoted B(E,F ). We usually
abreviate B(E,E) as B(E).
Remark 2.3. Theorem 2.1 shows that boundedness and continuity are eqivalent forlinear maps. Typically it is easier to check boundedness. An immediate consequenceof the definitions is that
‖Tx‖ ≤ ‖T‖ ‖x‖which is used heavily.
Note: in this course, the term “operator” always refers to linear maps.
Remark 2.4. To check that ‖T‖ ≤ M we must show that ‖Tx‖ ≤ M ‖x‖ for allx. To check that ‖T‖ > M we need only find one x such that ‖Tx‖ > M ‖x‖. Toshow that T is unbounded we need a sequence (xn) in E such that
‖xn‖ = 1∀n but ‖Txn‖ → ∞ as n→∞.Usually we only get (and indeed, only require) an estimate for ‖T‖.Example 2.5. Every normed space E has an identity operator I, given by Ix = xfor all x ∈ E. Obviously, ‖I‖ = 1.
Example 2.6 (Differentiation). Consider the space X of smooth (C∞) functionson R which are 2π-periodic and integrable on (−π, π), with the norm defined as‖f‖ =
∫ π−π |f |. Then differentiaion is a linear map D : X → X.
Let xn(t) = eint
2π . Then ‖xn‖ = 1 for all n.
(Dxn)(t) =ineint
2π= inxn(t)
So ‖Dxn‖ = n, which diverges as n→∞. Thus D is unbounded.
Theorem 2.7. If T is a linear map between normed spaces E and F , then thefollowing expressions all give ‖T‖ if T is bounded or ∞ if it is not:
supx∈Ex 6=0
‖Tx‖‖x‖
supx∈E‖x‖=1
‖Tx‖ supx∈E‖x‖≤1
‖Tx‖ .
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Proof. Exercise.
Theorem 2.8. If E and F are normed spaces over a field K then B(E,F ) is anormed space over K with respect to the operator norm.
Proof. The linear structure is as usual: (λT + µS)(x) = λTx+ µSx.Check that this definition makes B(E,F ) a vector space, and that the operator
norm is indeed a norm. (Exercise)
Theorem 2.9. If E,F,G are normed spaces over K and T ∈ B(E,F ), S ∈ B(F,G),then ST ∈ B(E,G) and ‖ST‖ ≤ ‖S‖ ‖T‖.
Proof. Exercise, using the definitions.
Remark 2.10. In particualar this shows that B(E) is a normed algebra (closed underproducts as well as linear combinations).
Theorem 2.8 also means that we have a natural notion of convergence in B(E,F ):
Tn → T iff ‖Tn − T‖ → 0
as n→∞. This is called convergence in norm or uniform convergence.Warning: there are other notions of convergence of operators, namely strong and
weak convergence. These are different from uniform convergence, and care shouldbe taken not to confuse these.
Theorem 2.11. Let E be a normed space and F be a Banach space. Then B(E,F )is a Banach space with respect to the operator norm.
Proof. By Theorem 2.8, B(E,F ) is a normed space. We need only check for com-pleteness. Accordingly, let (Tn)n∈N be a Cauchy sequence in B(E,F ). For anyx 6= 0 and any ε > 0 then there exists an N such that for all n,m > N
‖Tn − Tm‖ ≤ε
‖x‖and hence
‖Tnx− Tmx‖ = ‖(Tn − Tm)(x)‖ ≤ ‖Tn − Tm‖ ‖x‖ ≤ ε.
Then (Tnx) is Cauchy in F and hence converges as n → ∞. Define Tx =limn→∞ Tnx for x 6= 0 and T0 = 0. It is then easy to check that T is linear(Exercise). So we need to show that T is bounded and ‖Tn − T‖ → 0 as n → ∞.To do this, go back to the fact that Tn is Cauchy. For any ε > 0, there exists a Nsuch that
n,m > N =⇒ ‖Tn − Tm‖ < ε
and so‖Tnx− Tmx‖ ≤ ε ‖x‖ for all x ∈ E.
Letting m→∞ we have (by Theorem 2.7)
‖Tnx− Tx‖ ≤ ε ‖x‖ ∀x ∈ E.
Then (1) ‖Tx‖ = ‖Tnx+ Tx− Tnx‖ ≤ (‖Tn‖ + ε) ‖x‖ for all x ∈ E. Therefore Tis bounded. (2)
‖(Tn − T )x‖‖x‖
=‖Tnx− Tx‖‖x‖
≤ ε ∀x ∈ E, x 6= 0
so ‖Tn − T‖ ≤ ε and hence for all ε > 0 there exists an N such that ‖Tn − T‖ ≤ εfor all n > N ; i.e. Tn → T as n→∞.
The final result is a key technical result that often allows to aviod awkward issues.
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Theorem 2.12 (Bounded Linear Tranformation (BLT) theorem). Suppose (E1, ‖.‖1)and (E2, ‖.‖2) are Banach spaces and that F is a dense linear subspace of E1. (Then(F, ‖.‖1) is a normed space, but not necessarily a Banach space).
If T is a bounded operator between F and E2 then there exists a unique boundedoperator T between E1 and E2 that
(a) extends T (i.e T x = Tx for all x ∈ F )(b) has
∥∥∥T∥∥∥ = ‖T‖.
Proof. First note that if (yn) is a sequence in F with yn → 0 then Tyn → 0 as T isbounded and therefore continuous (Theorem 2.1). Next choose any x ∈ E1. Thenthere exists a sequence (xn) in F with xn → x. This is Cauchy, so given any ε > 0there exists an N such that for any n,m > N , ‖xn − xm‖ < ε and hence
‖Txn − Txm‖ = ‖T (xn − xm)‖ ≤ ‖T‖ ‖xn − xm‖ ≤ ε ‖T‖ .
Thus (Txn) is a Cauchy sequence in E2 and therefore converges.Note that if f(xn) is any other sequence in F with xn → x then yn = xn−xn → 0,
hence Tyn → 0 and limn→∞ Txn = limn→∞ T xn.Hence we may define
T x = limn→∞
Txn
without any ambiguity.It is easy to check that T is linear. (Exercise)To prove a, choose xn = x for all n, and hence T x = Tx.To prove b, observe that∥∥∥T x∥∥∥ =
∥∥∥ limn→∞
Txn
∥∥∥ = limn→∞
‖Txn‖ ≤ limn→∞
‖T‖ ‖xn‖ = ‖T‖ ‖x‖
for all x ∈ E1, so that∥∥∥T∥∥∥ ≤ ‖T‖.
On the other hand,∥∥∥T∥∥∥ = supx∈E1‖x‖≤1
∥∥∥T x∥∥∥ ≥ supx∈F‖x‖≤1
∥∥∥T x∥∥∥ = supx∈F‖x‖≤1
‖Tx‖ = ‖T‖
and so∥∥∥T∥∥∥ = ‖T‖.
The proof of uniquness is left as an exercise (consider the case T = 0 first).
Example 2.13. Let E1 = l2(Z), E2 = L2(−π, π) and F be the subspace of termi-nating sequence. Define the linear map T : F → E2 by
(Tc)(t) =∑n∈Z
cn√2π
eint
for any c = (cn)n∈Z ∈ F . Note that since c terminates the sum trivially converges.We calculate
‖Tc‖2 =∫ π
−π|(Tc)(t)|2 dt
=∫ π
−π
∑n∈Z
∑m∈Z
cncm2π
e−i(n−m)t dt
=∑n∈Z
∑m∈Z
cncm2π
∫ π
−πe−i(n−m)t dt
and, noting that the integral evaluates to 2πδmn, conclude that
‖Tc‖2 =∑n∈Z|cn|2 = ‖cn‖2 .
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All manipulations are obviously valid as all terms involve only finitely manyterms of a sequence. Thus we can see that T : F → E2 is bounded with ‖T‖ = 1.
As F is dense in E1, we infer from the BLT Theorem the existance of a uniqueT : E1 → E2 extending T and with
∥∥∥T∥∥∥ = 1.
To see what T does take any c ∈ l2(Z). Then c = limn→∞ fn where fn ∈ F suchthat
(fn)m =
cm |m| < n
0 otherwise.
Thus T c = limn→∞ Tfn, which we may write as
(T c)(t) = limN→∞
∑n∈Z|n|<N
cn√2π
eint
noting that the limit is in L2 (not meaning pointwise).Finally, observe that, for all c ∈ l2(Z)∥∥∥T c∥∥∥2
= limn→∞
∥∥∥T fn∥∥∥2
= limn→∞
‖Tfn‖2 = limn→∞
‖fn‖2 = ‖c‖2 .
The interchanges of sums and integrals were all taken care of by the BLT Theo-rem.
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3. Hilbert Spaces
We begin by recalling some definitions and basic results from Topics in AnalysisB.
Defintion 3.1. An inner product space is a vector space X over the field K withan inner product defined on it; that is, a function 〈·, ·〉 : X ×X → K such that forx, y, z ∈ X and s, t ∈ K,
(1) 〈sx+ ty, z〉 = s 〈x, z〉+ t 〈y, z〉 (linearity on the left);(2) 〈x, y〉 = 〈y, x〉 (conjugate symmetry);(3) 〈x, x〉 ≥ 0 and 〈x, x〉 = 0 if and only if x = 0 (positive definite property).
The norm of a vector x is defined by ‖x‖ = 〈x, x〉12 , taking the non-negative square
root.Vectors x and y are said to be orthogonal , written x ⊥ y, if 〈x, y〉 = 0.
An immediate consequence of the definition is that the inner product is conjugate-linear on the right:
〈x, sy + tz〉 = s 〈x, y〉+ t 〈x, z〉 .Note that in Mathematical Physics it is conventional to use an inner product 〈· | ·〉which is conjugate-linear on the right and linear on the left!
Example 3.2. The standard inner product on KN is
〈x, y〉 =N∑j=1
xjyj =⇒ ‖x‖ =
N∑j=1
|xj |21/2
.
In R2 and R3 this reduces to the dot product
〈x, y〉 = x · y = x1y1 + x2y2 + x3y3
which is the geometrical motivation behind the definition of orthogonality. OnC([a, b],K) we can define
〈f, g〉 =∫ b
a
fg =⇒ ‖f‖ =
(∫ b
a
|f |2)1/2
.
If w : [a, b]→ R is continuous and strictly positive, we can define a weighted version:
〈f, g〉 =∫ b
a
wfg.
Theorem 3.3 (Cauchy-Schwarz and Triangle Inequalities). If X is an inner productspace over K, then the norm induced by the inner product is a norm in the senseof Definition 1.1, and we have the important Cauchy-Schwartz inequality : for anyx, y ∈ X, we have
|〈x, y〉| ≤ ‖x‖ ‖y‖with equality if and only if y is a scalar multiple of x.
Theorem 3.4. Suppose X is an inner product space normed space over K, (xn)n∈Nand (yn)n∈N are sequences converging to x and y respectively. Then 〈xn, yn〉 →〈x, y〉.Theorem 3.5 (Pythagoras’ Theorem and the Parallelogram Identity). If X is aninner product space and x, y ∈ X then
‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2)
(parallelogram identity). If, in addition, x and y are orthogonal then
‖x+ y‖2 = ‖x‖2 + ‖y‖2
(Pythagoras’ Theorem).
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Defintion 3.6 (Hilbert space). A Hilbert space is a complete inner product space;that is, an inner product space which is coplete in the metric induced by the innerproduct.
In particular, any Hilbert space is a Banach space. In fact, a Hilbert space maybe characterised as a Banach space in which the norm obeys the parallelogramidentity.
Example 3.7. The following are the most common examples of Hilbert spaces.
(1) l2 is the set of all sequence (xj)j∈N in K such that∑∞j=1 |xj |
2 converges.The inner product is defined by 〈x, y〉 =
∑∞j=1 xj |yj | and the norm by
‖x‖2 =∑∞j=1 |xj |
2. If we want to be clear on the field we write l2R
or l2C
.(2) It is sometimes convenient to do the same with other countable index sets,
especially Z. If A is countable, we consider the set l2K(A) of all functions
from A to K such that∑α∈A |xα|
2 converges and define the inner product〈x, y〉 =
∑α∈A xαyα.
(3) If (wj) is a sequence of positive real numbers we can define the weightedHilbert space l2
K(N;w) to be the set of all sequences (xj)j∈N such that∑∞
k=1 wj |xj |2 converges, with inner product 〈x, y〉 =
∑∞j=1 wjxjyj .
(4) We can combine the previous two examples in the obvious way to defineweighted l2 spaces over arbitrary index sets.
(5) If S is a non-null measureable subset of RN we can define L2K(S) to be the
space of all measureable functions f : S → K such that |f |2 is integrable overS, quotiented by the null functions. The inner product is 〈f, g〉 =
∫Sfg.
(6) Finally, if w is an æpositive measureable function on S, we can define aweighted L2 space L2
K(S;w) to be those measureable functions f : S → K
such that w |f |2 is integrable over S, again quotiented by null functions,with the inner product 〈f, g〉 =
∫Swfg.
Defintion 3.8. If S is a subset of a Hilbert space H, its orthogonal complement isdefined by
S⊥ = x ∈ H : for all y ∈ S, 〈x, y〉 = 0.
Theorem 3.9 (Orthogonal decompositions). Suppose H is a Hilbert space. IfS ⊆ H then S⊥ is a closed subspace of H.
If U is a closed subspace of H then for any H there exist unique vectors u ∈ Uand v ∈ U⊥ such that x = u+ v.
If S ⊆ H then (S⊥)⊥ is the closed span of S, i.e. the smallest closed subspacecontaining S.
Because every Hilbert space is also a Banach space, Theorem 2.11 applies. Notethat B(E,F ) is a Banach space if E and F are Hilbert spaces.
We augment Theorem 2.7 by
Theorem 3.10. If T is a linear map between Hilbert spaces H and K then
supx∈Hy∈K
‖x‖=‖y‖=1
|〈Tx, y〉|
is equal to ‖T‖ if T is bounded, and infinite otherwise.
Proof. From Theorem 2.7,
‖T‖ = supx∈H‖x‖=1
‖Tx‖
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if T is bounded. But by the ‘extremal’ case of Cauchy-Schwarz,
‖Tx‖ = supy∈K‖y‖=1
|〈Tx, y〉| .
The result follows.
Defintion 3.11. If T ∈ B(H) for a Hilbert space H, the map x 7→ 〈Tx, x〉 is thequadratic form of T . Inner products of the form 〈Tx, y〉 are called matrix elementsof T .
Theorem 3.12. If T ∈ B(H) for H a complex Hilbert space, the matrix elementsof T may be reconstructed according to
(1) 〈Tx, y〉 = 14 [〈T (x+ y), x+ y〉 − 〈T (x− y), x− y〉+
i 〈T (x+ iy), x+ iy〉 − i 〈T (x− iy), x− iy〉].
Proof. Verify directly by expanding out the RHS.
This reconstruction is known as polarisation and the identity
〈x, y〉 = 14 [〈x+ y, x+ y〉 − 〈x− y, x− y〉+ i 〈x+ iy, x+ iy〉 − i 〈x− iy, x− iy〉] .
obtained by putting T = I is called the polarisation identity for Hilbert spaces overC.
There is a simpler polarisation identity for real Hilbert spaces:
〈x, y〉 = 14 (‖x+ y‖2 − ‖x− y‖2)
However, there is no exact analogue of Theorem 3.12. (Exercise: see what goeswrong.)
In a similar vein, knowledge of the matrix elements of T ∈ B(H,K) for all yspecifies Tx uniquely (for either K = R or C). For suppose there exists z ∈ Ksuch that 〈z, y〉 = 〈Tx, y〉 for all y ∈ K, set y = Tx − z and rearrange to find〈Tx− z, Tx− z〉 = 0, which gives Tx = z. This proves
Theorem 3.13. If T ∈ B(H,K) for Hilbert spaces H and K, then T is determineduniquely by its matrix elements.
Corollary 3.14. If T ∈ B(H) for H a complex Hilbert space, T is determineduniquely by its quadratic form.
We can go further, using the Riesz-Fischer theorem (Topics in Analysis B).
Theorem 3.15 (Riesz-Fischer). If (eα)α∈A is an orthonormal system in a Hilbertspace H indexed by a countable set A then the following are equivalent:
(1) (eα) is complete (〈x, eα〉 = 0 ∀α ∈ A =⇒ x = 0)(2) (eα) is closed (∀x ∈ H,x =
∑α∈A 〈x, eα〉 eα)
(3) the linear span of the (eα) is dense in H(4) Parseval: ∀x, y ∈ H, 〈x, y〉 =
∑α∈A 〈x, eα〉 〈eα, y〉.
Moreover, if c : A → K is such that∑α∈A |cα|
2 converges, then∑α∈A cαeα con-
verges in H.
Theorem 3.16. Let T ∈ B(H,K) for Hilbert spaces H and K with orthonormalsystems (fβ)β∈B and (eα)α∈H respectively, with A and B countable. Then, for allx ∈ H,
(∗) Tx =∑α∈A
∑β∈B
〈x, fβ〉 〈Tfβ , eα〉 eα.
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Proof. Apply part (2) of Riesz-Fischer to Tx and x.
(∗∗) Tx =∑α∈A〈Tx, eα〉 eα
andx =
∑β∈B
〈x, fβ〉 fβ .
As T is bounded and therefore continuous, this gives
Tx =∑β∈B
〈x, fβ〉Tfβ
so
〈Tx, eα〉 =
⟨∑β∈B
〈x, fβ〉Tfβ , eα
⟩=∑β∈B
〈x, fβ〉 〈Tfβ , eα〉 .
Finally, substitute in (∗∗) to get (∗).
The coefficients aαβ = 〈Tfβ , eα〉 form an infinite matrix (aαβ)α∈A,β∈B called thematrix of T with respect to (fβ) and (eα).
If K = H we call aαβ = 〈Tfβ , fα〉 the matrix with respect to (fβ)β∈B .
Example 3.17. Multiplication operators on l2 Let (mj)j∈N be a sequence in K.For any x ∈ l2 we may define a new sequence
(Mx)j = mjxj
It is not clear if this defines a map from l2 → l2, but it does define a linear mapfrom F to l2, where F ⊆ l2 consists of terminating sequences.
Note that
(†) ‖M‖ = supx∈F‖x‖=1
‖Mx‖ ≥ supj∈N
∥∥∥Me(j)∥∥∥ = sup
j∈N
∥∥∥mje(j)∥∥∥ = sup
j∈N|mj |
where e(j) is the j-th standard basis vector in l2 - i.e. the sequence (e(j))i = δij .Accordingly, M is unbounded if (mj) is unbounded.On the other hand, suppose (mj) is bounded. Then
‖Mx‖ =√∑j∈N|mjxj |2 ≤
√sup |mj |2
∑j ∈ N |xj |2 = sup
j∈N|mj | ‖x‖
for all x ∈ F . Thus M is bounded, ,and ‖M‖ ≤ supj∈N |mj |. Combining with (†),‖M‖ = supj∈N |mj |.
Thus M ∈ B(F, l2) so by BLT there is a unique extension M ∈ B(l2) with∥∥∥M∥∥∥ = supj∈N |mj |.Now let x ∈ l2 be arbitrary, and let x(n) → x with x(n) ∈ F . Then
(Mx)j =⟨Mx, e(j)
⟩= limn→∞
⟨Mx(n), e(j)
⟩= limn→∞
⟨Mx(n), e(j)
⟩= limn→∞
mj
⟨x(n), e(j)
⟩= mj
⟨x, e(j)
⟩= mjxj
so M acts exactly as we would have hoped.
12
The matrix M with respect to the standard bases (e(j)) is
ajk =⟨Me(k), e(j)
⟩=⟨mke
(k), e(j)⟩
= mkδjk
so multiplication operators generalise the notion of a diagonal matrix. Clearly wecan find a functional calculus by setting
(f(M)x)j = f(mj)xj .
Example 3.18 (Multiplication operators on L2(a, b)). Recall that each f ∈ L2(a, b)is an equivlence class of functions agreeing almost everywhere, which adds a smallcompliction here. Suppose m is a measurable function on (a, b) with |m(t)| ≤ C fora.a. t ∈ (a, b); such a C is called a essential upper bound for |m|. If we define
(Mf)(t) = m(t)f(t)
it is again easy to see that M is linear. We can estimate∫ b
a
|(Mf)(t)|2 dt =∫ b
a
|m(t)|2 |f(t)|2 dt
≤ C2
∫ b
a
|f(t)|2 dt
which shows that M is bounded with ‖M‖ ≤ C. This is true for any essential upperbound, so we have
(∗) ‖M‖ ≤ infC ≥ 0 : |m(t)| ≤ C for a.a. t ∈ (a, b)
The right-hand side is called the essential supremum of |m|, denoted ess-sup |m|.(As a point of fine-detail, we also need to observe that multiplication by m mapsfunctions that agree a.e. to functions that also agree a.e., and is therefore well de-fined on L2(a, b) whose elements are actually equivalence classes of square integrablefunctions agreeing a.e..)
But now choose any ε > 0. By definition of the essential supremum, the set of t ∈(a, b) for which |m(t)| > ess-sup |m| − ε has positive measure µ so its characteristicfunction χ is a measurable square-integrable function with ‖χ‖ =
õ. Noting that
‖Mχ‖2 =∫ b
a
|m(t)|2 |χ(t)|2 ≥ (ess-sup |m| − ε)2µ = (ess-sup |m| − ε)2 ‖χ‖2
we therefore have‖M‖ ≥ ess-sup |m| − ε
for any ε > 0. Together with our earlier observation we have ‖M‖ = ess-sup |m|,which is analogous to the previous example l2. We can work out the matrix of Mwith respect to the complete orthonormal system (ej)j∈Z given by ej(t) = eijt by
〈Mek, ej〉 =1
2π
∫ π
−πm(t)ek(t)ej(t) dt
=1
2π
∫ π
−πm(t)eikte−ijt dt
=1
2π
∫ π
−πm(t)ei(k−j)t dt
=1
2π
∫ π
−πm(t)ei(j−k)t dt
= 〈m, ej−k〉
so the (jk)th entry of the matrix is the (j − k)th Fourier coefficient of m.
13
Remark 3.19. We have seen that every bounded operator between Hilbert spacescan be represented by an infinite matrix. There is no general criterion on infinitemattrices that characterises those arrising from bounded operators.
We will study some sufficient conditions later.
14
4. Inverse and ajoint operators
Let H and K be Hilbert spaces with identity operators IH and IK .
Defintion 4.1. An operator T ∈ B(H,K) is invertible if there exists an operatorT−1 ∈ B(K,H) such that T−1T = IH and TT−1 = IK .
Note that T−1 is required to be bounded.
Theorem 4.2 (Properties of the inverse). Let H,K,L be Hilbert spaces. Then
(1) If T ∈ B(H,K) is invertable then∥∥T−1
∥∥ ≥ ‖T‖−1
(2) If T ∈ B(H,K) and S ∈ B(K,L) are invertable then ST ∈ B(H,L) isinvertable, and (ST )−1 = T−1S−1
(3) The invertable operators of B(H) form a group under operator productswith IH as identity.
Proof. For (1) note that 1 = ‖IH‖ =∥∥T−1T
∥∥ ≤ ∥∥T−1∥∥ ‖T‖ so
∥∥T−1∥∥ ≥ ‖T‖−1.
The rest is easy.
Later we will also show that the invertable operators in B(H) form an opensubset.
Example 4.3. Multiplication on l2. Define M : l2 → l2 by (Mx)j = mjxj for somebounded sequence (mj)j∈N. The only candidate for M−1 would be (M−1x)j =mj−1xj and this is bounded iff (mj
−1)j∈N is bounded, or infj |mj | > 0.∥∥M−1∥∥ = sup
j
∣∣mj−1∣∣ = (inf
j|mj |)−1
We see that ∥∥M−1∥∥ ≥ ‖M‖−1
holds because inf |mj | ≤ sup |mj | and so∥∥M−1
∥∥ = ‖M‖−1 holds iff inf |mj | =sup |mj |; i.e. |mj | is constant.
Example 4.4. Multiplication on L2(a, b). Define M as in 3.18. The inverse mapis
(M−1f)(t) = m(t)−1f(t)
which is bounded iff∣∣∣m(t)−1
∣∣∣ is essentially bounded from above. i.e. ess-inf |m(t)| >0, where ess-inf = supC : |m(t)| ≥ C for a.a. t.
Defintion 4.5. Let H and K be Hilbert spaces and T ∈ B(H,K). The adjoint T ∗
of T is a bounded operator T ∗ : K → H such that
〈Tx, y〉 = 〈x, T ∗y〉
for all x ∈ H, y ∈ K.An operator T ∈ B(H) is self-adjoint (sometimes ‘Hermition’ or ‘symmetric’) if
T = T ∗; i.e.〈Tx, y〉 = 〈x, Ty〉 ∀x, y ∈ H.
Obviously IH is always self-adjoint.The existance of adjoints depends on the Riesz-Frechet theorem (TAB).
Theorem 4.6 (Riesz-Frechet). If H is a Hilbert space over K and φ is a continuouslinear map from H to K then there is a unique u ∈ H such that
φ(y) = 〈y, u〉 ∀y ∈ H.
Theorem 4.7 (Existance of Adjoints). Suppose T ∈ B(H,K), H,K Hilbert spaces.Then there is a unique operator T ∗ ∈ B(K,H) such that 〈Tx, y〉 = 〈x, T ∗y〉 for allx ∈ H, y ∈ K. Moreover, ‖T ∗‖ = ‖T‖.
15
Proof. For any y ∈ K define φy(x) = 〈Tx, y〉. φy is continuous and linear becauseT is continuous and linear. Riesz-Frechet implies there exists a unique z ∈ H suchthat φy(x) = 〈x, z〉 for all x ∈ H, and we define T ∗(y) = z. Check that T ∗ is linear:
〈x, T ∗(λw + µz)〉 = 〈Tx, λw + µz〉= λ 〈Tx, w〉+ µ 〈Tx, z〉= 〈x, λT ∗w〉+ 〈x, µT ∗z〉= 〈x, λT ∗w + µT ∗z〉
for all x. By the uniqueness part of Riesz-Frechet, T ∗(λw + µz) = λT ∗w + µT ∗z,so T ∗ is linear.
‖T ∗‖ = supy∈Kx∈H
‖x‖=‖y‖=1
|〈T ∗y, x〉| = supy∈Kx∈H
‖x‖=‖y‖=1
|〈x, T ∗y〉| = supy∈Kx∈H
‖x‖=‖y‖=1
|〈Tx, y〉| = ‖T‖
Theorem 4.8 (Properties of Adjoints). Suppose H, K and L are Hilbert spaces.Then
(1) If T ∈ B(H,K) then (T ∗)∗ = T(2) If S, T ∈ B(H,K) and λ, µ ∈ K then (λS + µT )∗ = λS∗ + µT ∗
(3) If S ∈ B(H,K) and T ∈ B(K,L) then (TS)∗ = S∗T ∗
(4) T ∈ B(H,K) is invertable iff T ∗ is invertable and (T ∗)−1 = (T−1)∗
(5) If T ∈ B(H,K) then ‖T ∗T‖ = ‖TT ∗‖ = ‖T‖2
Proof. Routine except for (5). ‖T ∗T‖ ≤ ‖T ∗‖ ‖T‖ = ‖T‖2. Also for any x ∈ H‖Tx‖2 = 〈Tx, Tx〉 = 〈x, T ∗Tx〉 ≤ ‖x‖ ‖T ∗Tx‖ ≤ ‖T ∗T‖ ‖x‖2 which implies that‖T‖ ≤
√‖T ∗T‖.
The proof for ‖TT ∗‖ is similar and follows by replacing T by T ∗.
Remark 4.9. We have previously shown that B(H) is a complete normed algebrawith ‖ST‖ ≤ ‖S‖ ‖T‖ (a Banach algebra). As it is closed under taking adjoints,and the adjoint map is involutive (1), antilinear (2), and preserves norms, B(H) isa Banach ∗-algebra, or a B∗-algebra. Property (5) makes B(H) a C∗-algebra andis sometimes called the C∗-property.
Example 4.10 (Adjoints of multiplication operators). Suppose (mj)j∈N is a boundedsequence in K, so (Mx)j = mjxj defines a bounded operator on l2
K. Then for all
x, y ∈ l2K
〈x, M∗y〉 = 〈Mx, y〉 =∞∑j=1
(Mx)jyj =∞∑j=1
mjxjyj =∞∑j=1
xjmjyj = 〈x, M ′y〉
where (M ′y)j = mjyj . Thus M∗ and M ′ have identical matrix elements, so M∗ =M ′ (Theorem 3.13).
Similarly, if m is measureable and essentially bounded on (a, b) the operator Mon L2(a, b) (Mf)(t) = m(t)f(t) has adjoint
(M∗f)(t) = m(t)f(t).
Thus is either case M is self-adjoint if mj = mj for all j, or m(t) = m(t) for a.a.t.
Theorem 4.11 (Matrix of the Adjoint). Suppose H and K are Hilbert spaceswith countable orthonormal systems (fβ)β∈B and (eα)α∈A respectively. Then T ∈B(H,K) has matrix
aαβ = 〈Tfβ , eα〉
16
with respect to (fβ) and (eα).Similarly, T ∗ has matrix
(a∗)βα = 〈T ∗eα, fβ〉 = aαβ
i.e. the matrix of T ∗ is the conjugate transpose of the matrix of T (as for the finitedimensional case).
Proof. (a∗)βα = 〈T ∗eα, fβ〉 = 〈fβ , T ∗eα〉 = 〈Tfβ , eα〉 = aαβ
Note therefore that T ∈ B(H) is self-adjoint iff its matrix wrt a single basis isconjugate symmetric.
Defintion 4.12. Let H and K be Hilbert spaces. The range and null space of anoperator T : H → K are defined by
RT = Tx : x ∈ H (range)
NT = x ∈ H : Tx = 0 (null space).
The rank and nullity of T are defined as
rank(T ) = dimRT and nullity(T ) = dimNT .Note that the null space has an alternate name - the kernel. However, a different
use of ‘kernel’ will appear later, in the context of integral operators.
Theorem 4.13. LetH,K be Hilbert spaces and T ∈ B(H,K). ThenNT = (RT∗)⊥
and RT = (NT∗)⊥ (where · denotes closure).
Proof.
x ∈ NT ⇐⇒ Tx = 0
⇐⇒ 〈Tx, y〉 = 0 ∀y ∈ K⇐⇒ 〈x, T ∗y〉 = 0 ∀y ∈ K
⇐⇒ x ∈ (RT∗)⊥.
Next, note that we have NT∗ = (RT )⊥, so (NT∗)⊥ = (RT )⊥⊥ = RT .
17
5. Unitary Equivalence
Unitary operators determine Hilbert space isomorphisms - they preserve linearityand inner products.
Defintion 5.1. Let H and K be Hilbert spaces. An operator U ∈ B(H,K) isunitary if U∗U = IH and UU∗ = IK . If such a U exists, H and K are said to beunitarialy equivalent , or isomorphic.
Theorem 5.2. If U ∈ B(H,K) is unitary then(1) 〈Ux, Uy〉 = 〈x, y〉 for all x,∈ H;(2) ‖U‖ = 1.
Proof. (1) 〈Ux, Uy〉 = 〈x, U∗Uy〉 = 〈x, IHy〉 = 〈x, y〉.(2)
‖U‖ = supx∈H‖x‖=1
‖Ux‖ = supx∈H‖x‖=1
〈Ux, Ux〉
= supx∈H‖x‖=1
〈x, x〉 = 1.
Theorem 5.3. Let H and K be Hilbert spaces and U ∈ B(H,K). Then TFAEU is unitaryU∗ is unitaryU is invertable and U−1 = U∗
U is surjective and an isometry, i.e. ‖Ux‖ = ‖x‖ for all x ∈ H.
Proof. 1), 2, 3 are all equivalent by the definitions.1 =⇒ 4 because U is invertable, and is therefore surjective and U is isometric
by Theorem 5.2.4 =⇒ 1 Note that by the polarisation identity
〈x, y〉 =3∑k=0
ik∥∥x+ iky
∥∥2
(if K = C), so
〈Ux, Uy〉 =3∑k=0
ik∥∥Ux+ ikUy
∥∥2
=3∑k=0
∥∥U(x+ iky)∥∥
=3∑k=0
ik∥∥x+ iky
∥∥2as U is an isometry
= 〈x, y〉 .
Thus 〈U∗Ux, y〉 = 〈x, y〉 for all x, y ∈ H, so U∗U = IH as they have the samematrix elements (Theorem 3.13). As U is surjective, for any z ∈ K there exists anx ∈ H such that z = Ux. Then UU∗z = UU∗Ux = Ux = z, so UU∗ = IK .
Invertability also requires injectivity as well as surjectivity, so unitraries are bi-jective linear isometries.
Example 5.4. Let (wj)j∈N be a sequence in (0,∞). Then trivially∞∑j=1
wj |xj |2 =∞∑j=1
∣∣∣∣w 12j xj
∣∣∣∣2 .
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So the map U : l2(N;w)→ l2(N) given by
(Ux)j =√wjxj
is an isometry. Can check that
(U∗x)j =1√wjxj .
(N.B. l2(N;w) consists of seqeuences (xj) such that∑∞j=1 wj |xj |
2< ∞. So U is
well defined.)
Proof.
〈y, U∗x〉 = 〈Uy, x〉
=∞∑j=1
(Uy)jxj
=∞∑j=1
yj√wjxj
=∞∑j=1
wjyjxj√wj
= 〈y, V x〉
where (V x)j = xj√wj
. Therefore U∗x = V x.So U∗U = Il2(N;w); UU∗ = Il2(N) and hence U is unitary.
The same idea works in L2 spaces.Also, suppose Φ : [a, b]→ R is differentiable with Φ′(t) > 0 for all t.∫ Φ(b)
Φ(a)
|f(t)|2 dt =∫ b
a
|f(Φ(s))|2 Φ′(s) ds
soU : L2(Φ(a),Φ(b))→ L2((a, b); Φ′)(Uf)(s) = f(Φ(s))
is an isometry, and also surjective as L2((a, b); Φ′) consists of (equivalence classesof) measureable functions g on (a, b) such that
∫ ba|g(s)|2 Φ′(s) ds < ∞, so Uf = g
for f = g Φ−1.
Can find many unitaries from the Riesz-Fischer Theorem.
Example 5.5. Let H be a Hilbert space with countable orthonormal system(eα)α∈A (we say H is separable). We define a map U : H → l2(A) by (Uf)α =〈f, eα〉.U is obviously linear. To find the ajoint, note that
〈x, U∗y〉 = 〈Ux, y〉
=∑α∈A
(Ux)αyα
=∑α∈A〈x, eα〉 yα
=∑α∈A〈x, eα〉 〈eα, z〉
where z =∑α∈A yαeα (which exists, by R-F and has 〈z, eα〉 = yα). So 〈x, U∗y〉 =
〈x, z〉 by R-F (Parseval relation).As this holds for all x, z = U∗y.Check using R-F that U∗U = IH and UU∗ = Il2(A) (Ex.). Thus U is unitary.
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Theorem 5.6. Any separable infinite-dimensional Hilbert space is unitarially equiv-alent to l2(N) (Just label A by N).
So any theorem which deals solely with the properties of a separable, infinite-diemensional Hilbert space can be proved by considering the single example l2(N).
Example 5.7 (Fourier series). In H = L2(−π, π), the functions en(t) = 1√2π
eint
for n ∈ Z form a complete orthonormal system. We thereform a unitary operatorU : L2(−π, π)→ l2(Z)
(Uf)n = 〈f, en〉 =1√2π
∫ π
−πf(t)e−int dt
whose adjoint (and hence inverse) is
(U∗c)(t) =∑n∈Z
cnen(t) =1√2π
∑n∈Z
cneint.
(This appears also in Example 2.13 where it was shown to be an isometry.) Recallthat convergence of the series is in L2, not pointwise.
Defintion 5.8. Operators T ∈ B(H) and S ∈ B(K) are said to be unitarilyequivavlent if there is a unitary U : H → K such that UTU∗ = S.
Example 5.9 (Convolutions and Fourier series). If f and g are integrable 2π-periodic functions, they have a convolution
(f ∗ g)(t) =∫ π
−πf(t− s)g(s) ds
which is also integrable and 2π-periodic, and obeys f ∗ g = g ∗ f .Calculate
(f ∗ en)(t) = (en ∗ f)(t) =1√2π
∫ π
−πein(t−s)f(s) ds = fnen(t)
where fn =∫ π−π f(s)e−ins ds.
Now let F be the subspace of L2(−π, π) consisting of finite linear combinationsof the en, i.e. trigonometric polynomials of the form
g(t) =∑n∈Z
gnen(t) =1√2π
∑n∈Z
gneint
with only finitely many nonzero gn’s.By linearity, (f ∗ g)(t) =
∑n∈Z fngnen(t) and belongs to L2(−π, π). Thus g 7→
f ∗ g defines a linear operator T : F → L2(−π, π). By Example 5.7,
(UTg)n = (U f ∗ g)n = fngn = fn(Ug)n = (SUg)nwhere S is the operator of multiplication by (fn)n∈Z on l2(Z), which is bounded
|fn| =∣∣∣∣∫ π
−πf(t)e−int dt
∣∣∣∣ ≤ ∫ π
−π|f(t)| dt = ‖f‖L1
(so (fn) is a bounded sequence).Thus UTg = SUg for all g ∈ F , or equivalently, Tg = U∗SUg for g ∈ F (as U
is unitary). Observing that ‖T‖ ≤ ‖U∗‖ ‖S‖ ‖U‖ = ‖S‖ ≤ ‖f‖L1 , we use 2.12 toshow that there exists a unique bounded extension T : L2(−π, π)→ L2(−π, π) (F isdense in L2(−π, π). But U∗SU is such an extension, so T = U∗SU , or UTU∗ = S.
Thus T is unitarially equivalent to the simple multiplication operator S. Laterwe will see that (T g) = f ∗ g for all g ∈ L2, not just g ∈ F .
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6. Sufficient Conditions for an infinite matrix to determine abounded operator
If (ajk)j,k∈N is an infinte matrix, then to each x ∈ l2 we my define a sequenceTx = ((Tx)j)j∈N by
(∗) (Tx)j =∞∑k=1
ajkxk
and ask whether x 7→ Tx defines a bounded operator T on l2. We give two sufficientconditions.
Defintion 6.1. An infinite matrix (ajk)j,k∈N is Hilbert-Schmidt if∞∑j=1
∞∑k=1
|ajk|2
converges.
Theorem 6.2 (Hilbert-Schmidt operators). Any H-S matric (ajk)j,k∈N defines abounded operator T on l2 by (∗) with ‖T‖ ≤ ‖T2‖ where
‖T‖2 =
√√√√ ∞∑j=1
∞∑k=1
∣∣∣a2jk
∣∣∣T is called a H-S operator and ‖T‖2 is it’s H-S norm.
Proof. First note thta for each j ∈ N, the sequence (ajk)k∈N ∈ l2 as∞∑k=1
|ajk|2 ≤∞∑j=1
∞∑k=1
|ajk|2 <∞.
Taking x ∈ l2 and calculate∞∑j=1
|(Tx)j |2 =∞∑j=1
∣∣∣∣∣∞∑k=1
ajkxk
∣∣∣∣∣2
but this is an l2 inner-product, so
≤∞∑j=1
(∞∑k=1
|ajk|2)(∞∑l=1
|xl|2) Cauchy-Schwartz
= ‖T‖22 ‖x‖2.
Thus Tx ∈ l2 and ‖Tx‖ ≤ ‖T‖2 ‖x‖. Linearity of T is clear.
There is a more general definito of H-S operators and teh H-S norm betweengeneral Hilbert spaces. It reduces to the above for operators on l2.
Theorem 6.3 (Schur test). Suppose (ajk)j,k∈N is a infinte matrix with∞∑k=1
|ajk| ≤ α for all j
and∞∑j=1
|ajk| ≤ β for all k.
Then T (defined by (∗)) is a bounded operator on l2 with
‖T‖ ≤√αβ.
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Proof. As in Theorem 6.2 we need only estimate∑∞j=1 |(Tx)j |2 =
∑∞j=1 |
∑∞k=1 ajkxk|
2.
Trick: write ajkxj as (θjk |ajk|12 )(|ajk|
12 |xk|) where |θjk| = 1 Then, for each j ∈ N,
(θjk |ajk|12 )k∈N ∈ l2 as
∑∞k=1
∣∣∣∣θjk |ajk| 12 ∣∣∣∣2 =∑∞k=1 |ajk| ≤ α.
Also, (|ajk|12 |xk|)k∈N ∈ l2 because, noting that |xk|2 ≤ ‖x‖2,
∞∑k=1
∣∣∣∣|ajk| 12 xk∣∣∣∣2 ≤ ‖x‖2 ∞∑k=1
|ajk| ≤ α ‖x‖2 .
Thus, by Cauchy-Schwartz:∣∣∣∣∣∞∑k=1
ajkxk
∣∣∣∣∣2
≤ (∞∑k=1
∣∣∣∣θjk |ajk| 12 ∣∣∣∣2)(∞∑k=1
(|ajk|12 |xk|)2) ≤ α
∞∑k=1
|ajk| |xk|2 .
Thus∞∑j=1
|(Tx)j |2 ≤ α∞∑j=1
∞∑k=1
|ajk| |xk|2 = α
∞∑k=1
∞∑j=1
|ajk| |xk|2 ≤ αβ∞∑k=1
|xk|2 = αβ ‖x‖2
The interchange of summation is justified as the resulting expression is absolutelyconvergent (compare Tonelli & Fubini). To show that x 7→ Tx is a linear map ofl2 → l2 and
‖T‖ ≤√αβ.
Remark. Both Schur and H-S tests generalise in an obvious wat to maps betweenl2(A) and l2(B) for countable (perhaps finite) A and B.
Example 6.4. Consider ajk = 1jk , i.e. (Tx)j =
∑∞k=1
xk
jk
(ajk) passes the H-S test:∞∑j=1
∞∑k=1
|ajk|2 =∞∑j=1
∞∑k=1
1j2k2
= (∞∑k=1
1k2
)2 = (π2
6)2.
Thus T is bouned and ‖T‖ ≤ ‖T‖2 = π2
6 .But Schur’s test fails:
∞∑k=1
|ajk| =∞∑k=1
1jk
which diverges!
Example 6.5. Consider ajk = 1(j−k)2+1 so (Tx)j =
∑∞k=1
xk
(j−k)4+1 . H-S fails,because
∞∑j=1
∞∑k=1
|ajk|2 ≥∞∑j=1
|ajj |2 =∞∑j=1
1
divergent.However,
∞∑k=1
|ajk| =∞∑k=1
1(j − k)4 + 1
≤∞∑
k=−∞
1(j − k)4 + 1
=∞∑
n=−∞
1n4 + 1
<∞.
Can estimate∞∑
n=−∞
1n4 + 1
= 1 + 2∞∑n=1
1n4 + 1
≤ 1 + 2∞∑n=1
1n4
= 1 +π4
45
using the standard result1∑∞n=1
1n4 = π4
90 .
1cf Blue-Peter Annual
22
Clearly∑∞j=1 |ajk| obeys the same bound, so we may take α = β = 1 + π2
45 and
deduce that T is bounded with ‖T‖ ≤ 1 + π4
45 .
Remark. Adding the matrices of Examples 6.4 and 6.5 together we obtain a matrixwhich fails both tests, but nonetheless defines a bounded operator with norm ≤π2
6 + π4
45 + 1.
Example 6.5 is a special case of a general result:
Theorem 6.6. If∑∞j=−∞ |cj | is finite then the convolution operator
(Tx)j =∞∑
k=−∞
cj−kxk
is bounded on l2(Z) with ‖T‖ ≤∑∞k=−∞ |ck|.
Proof. Use Schur:∑∞k=−∞ |cj−k| =
∑∞k=−∞ |ck| =
∑∞j=−∞ |cj−k| <∞.
We can do better, using unitary equivalence.
Theorem 6.7. Suppose∑∞j=−∞ |cj | < ∞ and define the convolution operator T
as in Theorem 6.6. Then T is unitarially equivalent to multiplication by
m(t) =∞∑
j=−∞cjeijt
on L2(−π, π) and‖T‖ = max
t∈[−π,pi]|m(t)| .
Proof. The sum defining m(t) is uniformly convergent so m(t) is continuous (Weier-strauss M-test). So defining (Mf)(t) = m(t)f(t) for f ∈ L2(−π, pi), we have
‖M‖ = supt∈[−pi,pi]
|m(t)| = maxt∈[−π,pi]
|m(t)|
(cts fun on compact interval attains bounds) (Example 3.18).To establish unitary equivalence let en(t) = 1√
2πeint and let U : L2(−π, pi) →
l2(Z) be (Uf)n = 〈f, en〉 which is unitary (Example 5.7). (U∗x)(t) =∑∞n=−∞ xnen(t)
(UMen)k = 〈Men, ek〉 =⟨∑∞
j=−∞ cjej+n, ek
⟩= ck−n Hence on F , the subspace
of terminating sequences
(UMU∗x)k =∞∑
n=−∞ck−nxn = (Tx)k for all k
Thus UMU∗ and T all agree on F .But F is dense and UMU∗ is bounded and must therefore be the unique BLT
extension on T .Thus T = UMU∗ so ‖T‖ = ‖M‖.
23
7. Integral Operators and their Hilbert-Schmidt Condition an SchurTest
Defintion 7.1. An integral operator on L2(a, b) is an operator of the form
(∗) (Tf)(t) =∫ b
a
k(t, s)f(s) ds
(here a can be −∞ and b can be +∞). The (measureable) function k is called thekernel of the ooperator.
Integral operators are much like infinite matrices: a given kernel might or mightnot give rise to a bounded operator, and it is in general impossible to tell. TheHilbert-Schmidt and Schur tests carry over with a little modification.
Defintion 7.2 (Hilbert-Schmidt kernels). Suppose k is a measureable function on(c, d)× (a, b) such that ∫ d
c
∫ b
a
|k(t, s)|2 dsdt
is finite. Then k is called a Hilbert-Schmidt kernel .
Theorem 7.3. If k is a Hilbert-Schmidt kernel on (c, d)× (a, b) then (∗) defines abounded integral operator T from L2(a, b) to L2(c, d) with norm less than or equalto the Hilbert-Schmidt norm
‖T‖2 =
(∫ d
c
∫ b
a
|k(t, s)|2 dsdt
) 12
.
Theorem 7.4 (Schur’s Test for Integral Operators). Suppose k is a measureablefunction on (c, d)× (a, b) and there exist constants α and β such that for almost allt ∈ (c, d) ∫ b
a
|k(t, s)| ds ≤ α
and for almost all s ∈ (a, b) ∫ d
c
|k(t, s)| dt ≤ β.
Then (∗) defines a bounded operator T from L2(a, b) to L2(c, d) with norm less
than or equal to (αβ)12 .
Remark 7.5. Implicit in both theorems is the fact that the integral (∗) exists anddefines an element of L2(c, d) for every f ∈ L2(a, b). Just as in the matrix case, thetests are independent and there exist bounded operators defined by kernels whichfail both tests. Both tests works on any subsets of RN with positive measure as wellas on intervals.
Theorem 7.6 (Convolution on [−π, π]). Suppose k is a 2π-periodic function thatin integrable over [−π, π]. Define T : L2(−π, π)→ L2(−π, π) by
(*) (Tf)(t) =∫ π
−πk(t− s)f(s) ds
Then T is bounded and ‖T‖ ≤∫ π−π |k|; in fact ‖T‖ = supj∈Z
∣∣∣∫ π−π k(t)e−ijt dt∣∣∣.
Proof. Schur test:∫ π
−π|k(t− s)| ds =
∫ π
−π|k(s)| ds for all t (by periodicity of k)
=∫ π
−π|k(t− s)| dt
24
so T is bounded with ‖T‖ ≤∫ π−π |k|.
Previously (Example 5.9) we studied T on the dense subspace of trigonometricpolynomials and showed that it has a unique bounded extension to L2(−π, π).Schur’s test shows that (*) defines a bounded operator on all of L2(−π, π) – this istherefore the extension.
Then the result of Example 5.9 shows that T is unitarially equivalent to mul-tiplication by (mj)j∈Z mj =
∫ π−π k(t)e−ijt dt on l2(Z) which has operator norm
supj |mj |.
NB Riemann-Lebesgue lemma implies mj → 0 as |j| → ∞ so (Mx)j = mjxj isnot invertible (Example 3.17). Hence T is not invertable.
Theorem 7.7 (Convolution on R). If k ∈ L1(R) define
(Tf)(t) =∫ ∞−∞
k(t− s)f(s) ds.
Then T is a bounded operator on L2(R) with ‖T‖ ≤∫∞−∞ |k| = ‖k‖L1 .
Proof. Schur.
Example 7.8 (Green functions). Consider the D.E.
−f ′′(t) + f(t) = g(t)
on [−a, a] with b.c.’s f(±a) = 0.The solution is
f(t) =∫ a
−aG(t, s)g(s) ds
where the Green function is
G(t, s) =
A sinh(a+ t) sinh(a− s) t ≤ sA sinh(a− t) sinh(a+ s) t > s
and A = 1sinh 2a .
By the Hilbert-Schmidt condition, G defines a bounded operator
(Tg)(t) =∫G(t, s)g(s) ds
on L2(−a, a) which is Hilbert-Schmidt.
Example 7.9. By analogy with infinite matrices, it is reasonable to expect that ifT is an integrable operator with kernel k(t, s) then its adjoint should be the integraloperator with the conjugate transpose kernel. Formally, this seems to work: if Tmaps L2(a, b) to L2(c, d) by
(Tf)(t) =∫ b
a
k(t, s)f(s) ds
then
〈Tf, g〉 =∫ d
c
∫ b
a
k(t, s)f(s) dsg(t) dt
=∫ d
c
∫ b
a
k(t, s)f(s)g(t) dsdt
=∫ b
a
∫ d
c
k(t, s)f(s)g(t) dtds
=∫ b
a
f(s)∫ d
c
k(t, s)g(t) dtds
= 〈f, T ′g〉
25
where
(T ′g)(s) =∫ d
c
k(t, s)g(t) dt
so we conclude T ∗ = T ′. However, some care is needed in making the chage of orderin the integral. To use Tonelli’s Theorem we need one of∫ d
c
∫ b
a
|k(t, s)| |f(s)| ds |g(t)| dt∫ b
a
|f(s)|∫ d
c
|k(t, s)| |g(t)| dtds
to exist. Since f ∈ L2(a, b) the first integral will exist if the other factor is in L2,i.e. if the integral operator with kernel |k| maps L2(a, b) to L2(c, d). Alternatively,the second integral will exist if the transpose of |k| defines an integral operator fromL2(c, d) to L2(a, b). In either case, the calculation above is justified.
Any kernel which satisfies the Hilbert-Schmidt condition (Theorem 7.3) or Schur’stest (Theorem 7.4) automatically satisfies the first condition, but there are kernelsfor which the change of order is simply not valid: it turns out that there adjointsare not integral operators at all. An example [see §7.2 of P.R. Halmos & V.S. Sun-der, Bounded Integral Operators on L2 spaces, (Springer-Verlag, Berlin, 1978)] isk(t, s) = e−2πi
Rts on R × [0, 1], where
∫t is the greatest integer less than or equal
to t.Assuming all this works, in the case (a, b) = (c, d), T is self-adjoint providied
k(s, t) = k(t, s) for a.a. t and s. The converse is also true (we omit the measure-theoretic details).
As usual, any subset of RN with positive Lebesgue measure can be used in placeof the intervals (a, b) and (c, d).
26
8. The Spectrum
The spectrum generalises the set of eigenvalues familiar in matrix theory.
Defintion 8.1. If H is a Hilbert space over K and T ∈ B(H) then the spectrumσ(T ) is defined by
σ(T ) = λ ∈ K : λI − T is not invertable.
The resolvent set of T is
ρ(T ) = λ ∈ K : λI − T is invertable = K \ σ(T ).
The spectral radiusr(T ) = sup|λ| : λ ∈ σ(T ).
If λ ∈ ρ(T ) thenRλ(T ) = (λI − T )−1
is called the resolvent of T .2
An eigenvalue of T is λ ∈ K such that ∃x ∈ H \ 0 Tx = λx.Clearly every eigenvalue of T belongs to σ(T ), as (λI − T )x = 0, so λI − T is
not invertable.The eigenspace associated with λ is NλI−T (the null space).
Theorem 8.2 (Neumann Series). If H is a Hilbert space over K, T ∈ B(H) andλ ∈ K, then consider the Neumann series
∞∑n=0
Tn
λn+1
where T 0 = I. Then(1) If |λ| > ‖T‖ then the series is absolutely convergent.(2) If the series converges then λ ∈ ρ(T ) and
S =∞∑n=0
Tn
λn+1= Rλ(T ).
In particular, if |λ| ≥ ‖T‖ we have λ ∈ ρ(T ). Hence r(T ) ≤ ‖T‖. figure1
Proof.Note that ‖Tn‖ ≤ ‖T‖n. Thus
∑∞n=0
Tn
λn+1 converges absolutely for λ > ‖T‖.Suppose series converges.
(λI − T )S =∞∑n=0
Tn
λn−∞∑n=0
Tn+1
λn+1= I.
Similarly, S(λI−T ) = I, so λI−T is inverible (so λ ∈ ρ(T )) and S = (λI − T )−1 =Rλ(T ).
Remark 8.3. If |λ| > ‖T‖ there are useful Neumann series estimates
‖Rλ(T )‖ ≤∞∑n=0
‖Tn‖|λ|n+1 ≤
1|λ|
∞∑n=0
(‖T‖|λ|
)n=
1|λ| − ‖T‖∥∥∥∥Rλ(T )− 1
λI
∥∥∥∥ ≤ 1|λ|
∞∑n=1
(‖T‖|λ|
)n=
‖T‖|λ| (|λ| − ‖T‖
Theorem 8.4. If H is a Hilbert space, the set of invertable operators G(H) is anopen subset of B(H) and S 7→ S−1 is a homeomorphism of G(H) onto itself.
2Beware that some authors write Rλ(T ) = (T − λI)−1.
27
Proof. First consider operators ‘near’ I. If ‖I − S‖ < 1 then by Theorem 8.21 ∈ ρ(I − S) and S = I − (I − S) is invertible with inverse
S−1 = R1(I − S).
So I has an open neighbourhood in G(H). Moreover, the Neumann series estimatesgive ∥∥S−1 − I
∥∥ = ‖R1(I − S)− I‖ ≤‖I − S‖
1− ‖I − S‖so S → I =⇒ S−1 → I; i.e. S 7→ S−1 is continuous at I.
If T is any element of G(H) ans ‖S − T‖ <∥∥T−1
∥∥−1 then
(∗)∥∥ST−1 − I
∥∥ =∥∥(S − T )T−1
∥∥ ≤ ‖S − T‖ ∥∥T−1∥∥ < 1
so S ∈ G(H) (so T has an open neighbourhood of invertable operators) Hence G(H)is open.
Moreover, (∗) also shows that S → T then ST−1 → I, so TS−1 = (ST−1)−1 → I(continuity of inversion at I) and hence S−1 = T−1TS−1 → T−1. Thus inversionis continuous at T .
As inversion is self-inverse, it follows that it is a homeomorphism of G(H) ontoitself.
Corollary 8.5. If T ∈ B(H) (H a Hilbert space) then ρ(T ) is open and σ(T ) iscompact (i.e. closed and bounded).
Proof. Define F : K→ B(H) by F (λ) = λI − T (clearly continuous). Then ρ(T ) =F−1(G(H)) and is therefore open, as G(H) is open.
Thus σ(T ) = K \ ρ(T ) is closed. Moreover, Theorem 8.2 showed that
σ(T ) ⊆ λ ∈ K : |λ| ≤ ‖T‖,
so σ(T ) is also bounded.
Remark 8.6. If H is a complex Hilbert space then σ(T ) is nonempty for any T ∈B(H) (see Section 9). However, this need not be true if H is a real Hilbert space.
Example. H = R2; T =(
0 −11 0
). Then λI − T =
(λ 1−1 λ
)∈ G(H) for all
λ ∈ R, so ρ(T ) = R, σ(T ) = ∅.(If H = C2, we’d have σ(T ) = ±i.)
The spectrum is preserved by unitary equivalence.
Theorem 8.7. Let H,K be Hilbert spaces and U ∈ B(H,K) be unitary. Then
σ(T ) = σ(UTU∗)
for any T ∈ B(H).
Proof. Note that S is invertiable iff USU∗ is, with inverse US−1U∗. Thus
λ ∈ ρ(T ) ⇐⇒ λIH − T ∈ G(H)
⇐⇒ U(λIH − T )U∗ ∈ G(K)
⇐⇒ λIK − UTU∗ ∈ G(K)
⇐⇒ λ ∈ ρ(UTU∗).
Note the importance of the preservation of identities, UIHU∗ = IK . S 7→ USV ∗
for U, V unitaries preserves invertability but not identities (unless V = U) and sodoes not generally preserve the spectrum.
28
Example 8.8 (Multiplication operators). Suppose (mj)j∈N is bounded and defineM ∈ B(l2
C(N)) by
(Mx)j = mjxj
for all x ∈ l2.Then
((λI −M)x)j = λxj −mjxj = (λ−mj)xji.e. another multiplication operator. So λI −M is invertible iff
infj|λ−mj | > 0.
Thus (ex.) σ(M) = mj : j ∈ N (closure in C).Note that each mj is an eigenvalue, Me(j)mje
(j) (where e(j) is the jth standardbasis vector), but the spectrum may contain additional points; e.g. if mj = 1
j then
σ(m) = 1j
: j ∈ N = 0 ∪ 1j
: j ∈ N,
or, if j 7→ mj is a surjection of N onto Q ∩ [0, 1], then
σ(T ) = Q ∩ [0, 1] = [0, 1]
with eigenvalues precisely at the rational points.Multiplication of L2(a, b)→ C is measurable and essentially bounded, so
(Mf)(t) = m(t)f(t)
defines M ∈ B(L2(a, b)). Then ((λI −M)f)(t) = (λ −m(t))f(t), so λ ∈ ρ(M) iff∃ε > 0 s.t. |λ−m(t)| ≥ ε for at ∈ (a, b). Hence λ ∈ σ(M) iff
t ∈ (a, b) : |λ−m(t)| < εhas nonzero measure for all ε > 0; i.e. λ belongs to the essential range of m. Soσ(M) = ess-ranm.
If m is continuous, we may see directly that
σ(M) = m(t) : t ∈ (a, b)and the only points in σ(M) not in Rm come from limits at a and b. So if a and bare finite and m is continuous on [a, b],
σ(M) = m(t) : t ∈ [a, b].
In general, no eigenvalues m(t)f(t) = λf(t) a.a. t m(t) = ambda at ∈ t : f(t) 6=0.
Example 8.9 (Convolutions). Example 5.9 and Theorem 6.7 shows that convo-lution ops are unitarially equivalent to multiplication ops Thus e.g. if (cj)j∈Z,∑∞j=−∞ |cj | <∞ then T defined on l2(Z) by (Tx)j =
∑∞k=−∞ cj−kxk, is unitarially
equivalent to multiplication by
m(t) =∞∑
j=−∞cjeijt
on L2(−π, π). Therefore σ(T ) = ∑∞j=−∞ cjeijt : t ∈ [−π, π] as m is cts on [−π, π]
(M-test)Similarly if K is 2π periodic integrable on [−π, π] then T , defined on L2[−π, π]
by
(Tf)(t) =∫ π
−πk(t− s)f(s) ds
is unitarialy equivalent to multiplication by a sequence in l2, where
cj =∫ π
−πk(t)e−ijt dt.
29
Thus
σ(T ) =∫ π
−πk(t)e−ijt dt : j ∈ Z
cj → 0 as j →∞ by Riemann-Lebesgue lemma. Ths means
σ(T ) = 0 ∪ ∫ π
−πk(t)e−ijt dt : j ∈ Z.
The relationship between spectrum of T and its adjoint
Theorem 8.10. If H is a Hilbert space and T ∈ B(H) then
σ(T ∗) = λ : λ ∈ σ(T ).
Proof. S ∈ G(H) ⇐⇒ S∗ ∈ G(H), so λ ∈ ρ(T ) ⇐⇒ λI − T ∈ G(H) ⇐⇒(λI − T )∗ ∈ G(H) (λI − T )∗ = λI − T ∗, so λ ∈ ρ(T ) ⇐⇒ λ ∈ ρ(T ∗) thusλ ∈ σ(T ) ⇐⇒ λ ∈ σ(T ∗).
Useful e.g. Q7 Sheet 4
Theorem 8.10 shows that if T = T ∗ then σ(T ) is invariant under comple conju-gation but we want to show that σ(T ) ⊆ R as it is for Hermition matrices. (Thesimple argument used for matrices inly tells us about eigenvalues of general boundedoperators.)
Theorem 8.11. Let H be a complete Hilbert space and T ∈ B(H). Then σ(T ) ⊂R.
Proof. Let λ = α+ iβ, α, β ∈ R, β 6= 0. One may calculate
‖(λI − T )x‖2 = ‖(αI − T )x‖2 + |β|2 ‖x‖2 ≥ |β|2 ‖x‖2
then NλI−T = 0 since β 6= 0. So λI − T is injective and therefore invertible onits range. Thus there is alinear operators S : RλI−T → H such that
S(λI − T )x = x∀x ∈ H(λI − T )Sy = y∀y ∈ RλI−TSet x = Sy in (8.11) gives
‖Sy‖ ≤ |β|−1 ‖(λI − T )Sy‖ = |β|−1 ‖y‖
for all y ∈ RλI−T , so S is bounded. If RλI−T is dense in H, S has a unique boundedextension S ∈ B(H). Now S is a left-inverse to λI − T , because S is; to see that Sis a right-inverse, take any y ∈ H and yn → y with yn ∈ RλI−T , and note that
(λI − T )Sy = limn→∞
(λI − T )Syn = limn→∞
(λI − T )Syn = limn→∞
yn = y
Thus λI − T ∈ G(H), and so λ /∈ σ(T ). Hence σ(T ) ⊆ R, and σ(T ) 6= R as σ(T ) iscompact.
To show that RλI−T is dense, argue as before to show∥∥(λI − T ∗)x∥∥2 ≥ |β|2 ‖x‖2
Then, from Theorem 4.13,
RλI−T = N⊥λI−T∗ = 0⊥ = H.
The spectrum of polynomial functions of bounded operators are also easily de-termined.
If p(z) =∑nr=0 crz
r, cr ∈ K, then p(T ) =∑nr=0 crT
r (where, by convention,T 0 = I). Check:
(λp+ µq)(T ) = λp(T ) + µq(T )
(pq)(T ) = p(T )q(T ) = q(T )p(T ).
30
Theorem 8.12. Let H be a complex Hilbert space, T ∈ B(H), p a polynomialover C. Then
σ(p(T )) = p(σ(T ))def= p(z) : z ∈ σ(T ).
Proof. Suppose µ ∈ σ(p(T )) and set q(z) = p(z)−µ. Factorising q(z) = α∏ni=1(z−
λi) for α, λ1, . . . , λn ∈ C. So
p(T )− µI = q(T ) = α
n∏i=1
(T − λiI)
The LHS is not invertible; therefore at least one factor in the RHS is not invertible(recall G(H) is a group). Thus λi ∈ σ(T ) for some i. As q(λi) = 0 we havep(λi) = µ; i.e. µ ∈ p(σ(T )).
On the other hand, suppose for some λ that p(λ) /∈ σ(p(T )). Consider q(z) =p(z)− p(λ). As q(λ) = 0, we may write q(z) = (z − λ)r(z) for some polynomial r.As before,
p(T )− p(λ)I = q(T ) = (T − λI)r(T ) = r(T )(T − λI).
Now LHS is invertible so
I = (T − λI)r(T )(p(T )− p(λ)I)−1
AlsoI = (p(T )− p(λ)I)−1
r(T )(T − λI)Thus T − λI has left and right bounded inverses, therefore T − λI is invertable,so λ /∈ σ(T ). Taking contrapositive, λ ∈ σ(T ) =⇒ p(λ) ∈ σ(p(T )). Henceσ(p(T )) = p(σ(T )).
Remark 8.13. Depends critically on the fact that C is algebraically closed: poly-
nomials factorise down to linear factors. Note: for H = R2, T =(
0 −11 0
), then
σ(T ) = ∅, but T 2 = −I so σ(T 2) = −1.
31
9. Complex Methods and the Spectrum
We have already seen that the polynomial spectral mapping theorem only holds incomplex Hilbert spaces. In this section we will illustrate further how techniques fromcomplex analysis may be used to find important infromation about the spectrum ofbounded operators on complex Hilert spaces.
The facts we will use are these:• A function f : C→ C is complex differentiable at w ∈ C if
f(z)− f(w)z − w
has a finite limit, which we denote f ′(w), as z → w If f is complex differ-entiable at every point of an open set U , f is said to be holomorphic in U .If a function is holomorphic on all of C, it is said to be entire. If there is a(possibly empty) subset A of U such that (i) A has no limit point in U , (ii)f is holomorphic on U \ A, (iii) f has a pole at each point of A, then f issaid to be meromorphic on U .
• If f is holomorphic in an open set U and w ∈ U , then f may be representedby its Taylor series at w in any disk centred on w and contained in U :
f(z) =∞∑n=0
f (n)(w)(z − w)n
n!.
• If f is meromorphic in a simply connected subset U of C and C is a closed,non-self-intersecting curve of finite length in U that does not meet any polesof f , then
12πi
∮C
f(z) dz = R
where C is traversed in the anticlockwise direction, R is the sum of theresidues of poles of f contained within C and the residue of f at a pole atw ∈ U is limz→w(z − w)f(z).
Special cases of the residue theorem give us Cauchy’s formulae
f (n)(w) =n!2πi
∮C
f(z)(z − w)n+1
dz
where C is any contour about w neither enclosing nor meeting any poles of f , andthe consequent Cauchy estimates∣∣∣f (n)(w)
∣∣∣ ≤ n!Rn
supz:|z−w|=R
|f(z)|
obtained by taking C to be the circular contour of radius R about w (assuming thisneither encloses nor meets any poles of f . We also have
Theorem 9.1 (Liouville’s theorem). Every bounded entire function is constant.
Proof. Use Cauchy’s estimates with w = 0 for circles of arbitrarily large R to deducethat f (n)(0) = 0 for n ≥ 1. Then by the Taylor series representation f(z) = f(0)for all z.
Theorem 9.2. Let T ∈ B(H). Then λ 7→ Rλ(T ) = (λI − T )−1 is a B(H)-valuedcomplex differentiable function on ρ(T ).
Proof. Note that λ 7→ λI−T is continuous from C→ B(H). Choose µ ∈ ρ(T ); thenλI − T is invertible for all λ in a neihbourhood of µ (ρ(T ) is open; Corollary 8.5)and as inversion is continuous (Theorem 8.4), we have Rλ(T ) → Rµ(T ) as λ → µ.Next, with µ ∈ ρ(T ), use the resolvent identity (Q1, Sheet 4)
Rλ(T )−Rµ(T ) = −(λ− µ)Rλ(T )Rµ(T )
32
soRλ(T )−Rµ(T )
λ− µ= −Rλ(T )Rµ(T )→ −Rµ(T )2
as λ→ µ.
Corollary 9.3. For any T ∈ B(H) and x, y ∈ H, λ→ 〈Rλ(T )x, y〉 is holomorphicon ρ(T ).
Theorem 9.4. For any T ∈ B(H), σ(T ) 6= ∅.
Proof. Suppose σ(T ) = ∅. Then ρ(T ) = C, so for any x, y ∈ H, 〈Rλ(T )x, y〉 isentire. Moreover, λ 7→ ‖Rλ(T )‖ is bounded: (i) for |λ| ≥ ‖T‖ we have
‖Rλ(T )‖ ≤ 1|λ| − ‖T‖
(Neumann series estimate; Remark 8.3) so ‖Rλ(T )‖ < 1‖T‖ for |λ| > 2 ‖T‖; ii) inside
the compact disc |λ| ≤ 2 ‖T‖, λ 7→ ‖Rλ(T )‖ is continuous and therefore bounded.Therefore λ 7→ 〈Rλ(T )x, y〉 is a bounded entire function, and therefore constant
by Liouville (Theorem 9.1).As |〈Rλ(T )x, y〉| ≤ ‖x‖ ‖y‖ ‖Rλ(T )‖ → 0 as |λ| → ∞ we have 〈Rλ(T )x, y〉 = 0
for all x, y ∈ H. Thus Rλ(T ) = 0 for all λ ∈ C. But Rλ(T ) is invertible (inverseλI − T ). Contradiction!
The following links the spectral radius and the norm.
Theorem 9.5. Let T ∈ B(H). Then the spectral radius is
r(T ) = limn→∞
‖Tn‖1n = inf
n≥1‖Tn‖
1n .
Proof. We have r(T ) ≤ ‖T‖ (Theorem 8.2). Also r(Tn) ≤ ‖Tn‖. By the polynomialspectral mapping formulas
σ(Tn) = λn : λ ∈ σ(T )
so r(Tn) = r(T )n. Thus r(T ) = r(Tn)1n ≤ ‖Tn‖
1n for all n = 1, 2, 3, . . . . Thus
r(T ) ≤ infn≥1 ‖Tn‖1n .
Now note that I − zT is invertible on |z| < r(T )−1 and set S(z) = (I − zT )−1;we have S(z) = z−1Rz−1(T ) for 0 < |z| < 1
r(T ) , and S(0) = I. repeat fig 1 fig 2 Onthe possibly smaller disk |z| = 1
‖T‖ , we have
S(z) =∞∑n=0
znTn
by the Neumann series (Theorem 8.2). For arbitrary x, y ∈ H define Sx,y(z) =〈S(z)x, y〉, which is holomorphic in |z| < 1
r(T ) by Corollary 9.3. Therefore it has aTaylor series about 0, but we already know
(∗) Sx,y(z) =∞∑n=0
zn 〈Tnx, y〉
in |z| < 1‖T‖ . By uniquness of Taylor coeffs, (∗) holds in all of |z| < 1
r(T ) .Next, we use Cauchy’s formula to extract the coeffs
(†) 〈Tnx, y〉 =1
2πi
∮C
Sx,y(z)zn+1
dz
33
where C is any contour surrounding 0 within the disk |z| < 1r(T ) . In particular, we
have Cauchy estimates
|〈Tnx, y〉| ≤ Rn sup|z|=R−1
|Sx,y(z)| ≤ Rn ‖x‖ ‖y‖ sup|z|=R−1
‖S(z)‖︸ ︷︷ ︸MR
for C the circle of radius 1R about 0, R > r(T ). MR is finite because z 7→ S(z) is
continuous and C is compact.Now from Theorem 3.10,
‖Tn‖ = supx,y∈H
‖x‖=‖y‖=1
|〈Tnx, y〉| ≤ RnMR
i.e. ‖Tn‖1n ≤ RM
1n
R for all n ≥ 1.
Given any ε > 0 choose R = r(T ) + ε2 and N large enough that M
1n
R < 1 + ε2R
for all n > N (possible as M1n
R → 1 as n→∞. Then
‖Tn‖1n ≤ R(1 +
ε
2R= R+
ε
2= r(T ) + ε
for all n > N . Thus we see both that ‖Tn‖1n → r(T ) and r(T ) = infn≥1 ‖Tn‖
1n .
Corollary 9.6. If T ∈ B(H) is self-adjoint, then r(T ) = ‖T‖.
Proof. The C∗-property of the norm gives
‖T‖2 = ‖T ∗T‖ =∥∥T 2
∥∥ .Iterating: ‖T‖2
n
=∥∥T 2n∥∥ for all n ≥ 1 so∥∥∥T 2n
∥∥∥ 12n
= ‖T‖ .
But∥∥T 2n∥∥ 1
2n is a subsequence of a convergent sequence (Theorem 9.5), so tends tothe same limit. But this is constant, so we have the result.
Remark 9.7. Equation (†) may be rewritten
〈Tnx, y〉 =1
2πi
∮|z|= 1
R
⟨(I − zT )−1
x, y⟩
zn+1dz =
12πi
∮|w|=R
wn 〈Rw(T )x, y〉 dw
(ex: paramaterize. z = R−1eiθ, θ ∈ [−π, π], change variables to φ = −θ and writein terms of w = Reiφ.
Thus for any polynomial p,
〈p(T )x, y〉 =1
2πi
∮Γ
p(w) 〈Rw(T )x, y〉 dw
for Γ any contour surrounding σ(T ) in ρ(T ).
This may be used as the basis of a holomorphic functional calculus for B(H).
Theorem 9.8. Let Ω be an open subset of C containing σ(T ) for some T ∈ B(H).Then for every holomorphic function f on Ω there is an operator f(T ) ∈ B(H) withmatrix elements
〈f(T )x, y〉 =1
2πi
∮Γ
f(w) 〈Rw(t)x, y〉 dw
for Γ any contour surrounding σ(T ) in Ω.The following hold:• If f is polynomial f(T ) reproduces earlier def.
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• If f, g are holomorphic on Ω and λ, µ ∈ C then
(λf + µg)(T ) = λf(T ) + µg(T )
(fg)(T ) = f(T )g(T ).
• There is a general spectral mapping formula
σ(f(T )) = f(σ(T )) = f(λ) : λ ∈ σ(T ).
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10. The Continuous Functional Calculus
We consider a complex Hilbert space H. If T ∈ B(H) is self-adjoint, σ(T ) is acompact subset of R contained in [−‖T‖ , ‖T‖]. The continuous functions on σ(T )form a Banach space C(σ(T )) with respect to
‖f‖∞ = supt∈σ(T )
|f(t)|
We establish a functional calculus for continuous functions of T .
Theorem 10.1 (Continuous functional calculus). Let T ∈ B(H) be self-adjoint.There is a unique map φ : C(σ(T ))→ B(H) with the following properties:
(1) Writing 1 for the continuous function 1(t) = 1 and t for t(t) = t,
φ(1) = I φ(t) = T.
(2) φ is a ∗-algebrahomomorphism: for f, g ∈ C(σ(T )) and λ, µ ∈ Cφ(λf + µg) = λφ(f) + µφ(g) φ(fg) = φ(f)φ(g) φ(f) = φ(f)∗
(where f(t) = f(t)). Together with 1 this means that φ(p) = p(T ) for anypolynomial p and also that φ(f)φ(g) = φ(g)φ(f).
(3) φ is continuous, and ‖φ(f)‖B(H) = ‖f‖∞.(4) If Tx = λx then φ(f)x = f(λ)x.(5) Spectral mapping property σ(φ(f)) = f(σ(T )).
We begin with useful facts about polynomial functions of T .
Lemma 10.2. If p and q are polynomils with complex coefficients and S ∈ B(H)then
(1) (pq)(S) = p(S)q(S)(2) p(S)∗ = p(S∗)(3) if S = S∗ then ‖p(S)‖ = ‖p‖∞ = sups∈σ(S) |p(s)|
Proof. 1 and 2 - exercies.For (3), the C∗-property gives
‖p(t)‖2 =∥∥p(T )∗p(T )
∥∥ = ‖p(T ∗)p(T )‖ = ‖p(T )p(T )‖ = ‖(pp)(T )‖
but [(pp)(T )]∗ = (pp)(T ∗) = pp(T ), so ‖(pp)(T )‖ = r(pp(T )) (Corollary 9.6) there-fore ‖(pp)(T )‖ = sup|λ| : λ ∈ σ(pp(T )) = sup|pp(t)| : t ∈ σ(T ) by the spectralmapping for polynomials. Thus ‖p(T )‖ = sup|p(t)| : t ∈ σ(T ) = ‖p‖∞.
Recall Corollary 5.5 in Topics in Analysis B:
Theorem 10.3 (Weierstrauss Approximation Theorem). Suppose f ∈ C([a, b]).Then for any ε > 0 there exists a polynomial p such that
supt∈[a,b]
|f(t)− p(t)| ≤ ε.
The same is true if [a, b] is replaced by any other compact subset of R, e.g. σ(T ).Given f ∈ C(σ(T )) extend to a function f ∈ C([−‖T‖ , ‖T‖]). Given ε > 0, applyTheorem 10.3 to f , obtaining polynomial p and then note supt∈σ(T ) |f(t)− p(t)| ≤supt∈[−‖T‖,‖T‖]
∣∣∣f(t)− p(t)∣∣∣ ≤ ε.
Thus the polynomials are dense in C(σ(T )) with respect to ‖·‖∞.
Proof of Theorem 10.1. Begin by defining φ(p) = p(T ) for any polynomial p. Part3 of Lemma 10.2 shows that this is a continuous map to B(H). As polynomials aredense in C(σ(T )), the BLT Theorem shows that φ extends uniquely to a continuousmap φ : C(σ(T )) → B(H) with ‖φ(f)‖ = ‖f‖∞. This constructs φ and proves 1,3, and the linearity part of 2.
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To prove the rest of 2, introduce sequences pn and qn of polynomials, pn → fand qn → g in C(σ(T )). Then pnqn → fg, so
φ(fg) = limn→∞
φ(pnqn) = limn→∞
(pnqn)(T ) = limn→∞
pn(T )qn(T ) =(
limn→∞
pn(T ))(
limn→∞
qn(T ))
= φ(f)φ(g).
Similarly
φ(f)∗ = ( limn→∞
φ(pn))∗ = limn→∞
φ(pn)∗ = limn→∞
φ(pn) = φ(f)
where we use part 2 of Lemma 10.2. So 2 is proved. Also, for 4
φ(f)ψ = limn→∞
φ(pn)ψ = limn→∞
pn(T )ψ = limn→∞
pn(λ)ψ = f(λ)ψ.
To prove 5, suppose first that λ is not in the range of f . Then t 7→ g(t) =(λ− f(t))−1 is continuous on σ(T ) and by the homomorphism property, φ(g) is aninverse to λI − φ(f) (h(t) = λ− f(t) and note h(t)g(t) =, so φ(h)φ(g) = φ(1) = I)Thus λ /∈ σ(φ(f)), so
σ(φ(f)) ⊆ f(t) : t ∈ σ(T ).Now suppose λ = f(t) for some t ∈ σ(T ) and take a sequence of polynomials
pn → f . By the polynomial spectral mapping theorem, pn(T ) ∈ σ(pn(T )) for all n;i.e. Sn = pn(t)I−φ(pn) is not invertible. But Sn → f(t)I−φ(f) = λI−φf . As thenon-invertable operators form a closed subset of B(H) (Theorem 8.4) we concludethat λ ∈ σ(φ(f)). So 5 is proved.
Now obtain some consequences, writing f(T ) for φ(f).
Theorem 10.4. Suppose T ∈ B(H) is self-adjoint and f ∈ C(σ(T )).If f is real-valued then f(T ) is self-adjoint.If |f(t)| = 1 for all t, f(T ) is unitary.
Proof.f(T )∗ = f(T ) = f(T ).f(T )∗f(T ) = f(T )f(T ) = (ff)(T ) = 1(T ) = I. Similarly, f(T )f(T )∗ = I.
Theorem 10.5. Let T ∈ B(H) be self-adjoint, g ∈ C(σ(T )) be real-valued andf ∈ C(g(σ(T ))). Then (f g)(T ) = f(g(T )).
Proof. As g(T ) is self-adjoint we may define f(g(T )) using the functional calculus.If (pn) and (qn) are polynomial sequences with pn → f , qn → g as n → ∞, thenpn qn → f g in C(σ(T )) (exercise) and as one may check that pn(qn(T )) =pn qn(T ) the result follows.
Example 10.6. Suppose T ∈ B(H) is positive, i.e. T = T ∗ nd 〈Tx, x〉 ≥ 0 for allx ∈ H. If λ < 0 we have
‖(T − λI)x‖ ‖x‖ ≥ |〈T − λx, x〉| = 〈Tx, x〉 − λ ‖x‖2 ≥ −λ ‖x‖2 ≥ 0.
So ‖(T − λI)x‖ ≥ −λ ‖x‖. The proof of Theorem 8.11 shows that λ /∈ σ(T ).
Example 10.7. Let T ∈ B(H) be self-adjoint with σ(T ) ⊂ [0,∞). For any n ∈ Ndefine continuous functions f(t) = tn, g(t) = t1/n on σ(T ). By Theorem 10.5,g(T )n = f(g(T )) = (f g)(T ) = t(T ) = T . So g(T ) is a n’th root of T and g(T ) =g(T )∗ (as g is real-valued) and σ(g(T )) = g(σ(T )) = λ1/n : λ ∈ σ(T ) ⊆ [0.∞).
Moreover, if S any other self-adjoint operator with Sn = T and spectrumS ⊆[0,∞) then S = (g f)(S) = g(f(S)) = g(Sn) = g(T ). Thus, g(T ) is the uniqueself-adjoint n’th root with spectrum in [0,∞) and we write T 1/n for g(T ).
Theorem 10.8. T ∈ B(H) is positive iff T = T ∗ and σ(T ) ⊆ [0,∞).
Proof. =⇒ Example 10.6 ⇐= IF T = T ∗ and σ(T ) ⊆ [0,∞) then T = S2
for S = T 1/2 and S = S∗ so T = S∗S and hence 〈Tx, x〉 = ‖Sx‖2 ≥ 0 for allx ∈ H.
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Example 10.9. Let T ∈ B(H). Then T ∗T is positive, so has a postivei squareroot |T | = (T ∗T )1/2. [If T = T ∗, |T | = f(T ) with f(t) = |t|.]
Now ‖|T |x‖2 =⟨|T |2 x, x
⟩= 〈T ∗tx, x〉 = ‖Tx‖2 so |T | y = |T |x ⇐⇒ Ty =
Tx. Thus there is a linear (Exercise) map U : R|T | → RT such that U |T |x =Tx which is well-defined and obeys ‖Uy‖ = ‖y‖ for all y ∈ R|T |. Extend U by(BLT) continuity to an isometry U : R|T | → RT . Define U to vanish on R⊥|T | =
N|T | we have defined U : R|T | ⊕ |T |⊥︸ ︷︷ ︸[H] → H. U is a partial isometry and we
have T = U |T | (NB: N|T | = NT .) This is called polar decomposition (cf withmodulus/argument for reiθ in C). σ(U) ⊆ z ∈ C : |z| = 1, σ(|T |) ⊆ [0,∞),σ(T ) ⊆ C.
Example 10.10. Let T ∈ B(H) we self-adjoint with ‖T‖ ≤ 1. Define f : [−1, 1]→C, f(t) = t + i
√1− t2, f ∈ C(σ(T )). As |f(t)|2 = 1 for all t, f(T ) is unitary. But
also 12 (f(t)+f(t)) = t, so 1
2f(T )+f(T )∗) = t(T ) = T . i.e. T is a linear combinationof two unitary operators.
In general, any T ∈ B(H) is a linear combination of at most four unitaries(exercise).
We conclude with two indications of where the theory goes next.1. An operator is normal if it commutes with its adjoint, i.e. , TT ∗ = T ∗T .
Examples include self-adjoint and unitary operators. If T is normal then, givenany polynomial p(z, z), we may define an operator p(T, T ∗) in the obvious way; itis clear that we have the multiplicative property (pq)(T, T ∗) = p(T, T ∗)q(T, T ∗) aswell as linearity. One may also show that
‖p(T, T ∗)‖ = supz∈σ(T )
|p(z, z)|
and this permits us to define a continuous functional calculus for normal operatorsusong the generalisation of the Weierstrass theorem to functions of two variables.The main difference is that the spectrum of a normal operator need not be in thereal line and so we work with continuous function on σ(T ) ⊂ C.
2. Fixing any x ∈ H and self-adjoint T ∈ B(H), the map f 7→ 〈f(T )x, x〉 isa continuous linear map from C(σ(T )) to C which is positive in the sense that iff(t) ≥ 0 for all t ∈ σ(T ), then 〈f(T )x, x〉 ≥ 0. A general result on intergrationtheory (the Riesz-Markov theorem) implies that we may write
(∗) 〈f(T )x, x〉 =∫σ(T )
f(t) dµx(t)
where µx is a so-called spectral measure on the σ(T ); unlike the Lebesgue measureencountered in Topics in Analysis A, µx may assign nonzero measure to single-pointsets [and indeed generally does so at eigenvalues of T ]. Armed with more detailedintergration theorey we may use (∗) to define f(T ) [via its quadratic form] for amuch wider class of functions f (the bounded Borel functions on R) while preservingmost of the properties of the continuous functional calculus. If x has the propertythat finite linear combinations of the vectors (Tnx)∞n=0 are dense in H, we say thatx is cyclic, then there is a unitary map V : L2(σ(T ),dµx) → H defined so thatV f = φ(f)x, which is in fact unitary, and
(V ∗TV f)(t) = tf(t).
So bounded self-adjoint operators with cyclic vectors are unitarially equivalent tomultiplication operators on a suitable L2 space. A general bounded self-adjointoperator can be expressed as a direct sum of such multiplication operators. This isthe eventual analohue of the unitary diagonalisaion of Hermition matrices.
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The main reason that general spectral theory is more involved than for finitedimensional matrices is that not all spectral points are eigenvalues. The last topicin the course converns the class of compact operators for which every spectral point,apart from perhaps 0, represents a (finite dimensional) eigenspace. Here we will beable to obtain a general spectral theorem without high-powered integration theory.
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11. Compact Operators
Every N × N compelx matrix has at least one eigenvalue, and at most N . Incontrast, the Hilbert space situation is much more complicated: an operator canhave no eigenvalues, or finitely many, countably many, even uncountably many.There is a class of operator which are genuinely infinite-dimensional in characterbut share many of the similar properties of finite matrices, called compact operators.
Recall from Metric Spaces that a subset S of a metric space is compact if everybounded sequence in S has a convergent subsequence. The Heine-Borel theorementails that every closed bounded subset of a finite-dimensional Hilbert space iscompact (as every such space is isometric to RN for some N). However this is nottrue in infinite dimensions.
Example 11.1. The closed unit ball in an infinite-dimensional Hilbert space is notcompact. To see this, take an orthonormal sequenece (en)n∈N which lies in the ball,but has no convergent subsequence, because ‖en − em‖ =
√2 for all m 6= n.
Defintion 11.2. If H and K are Hilbert space then a linear operator T : H → K iscalled compact if whenever C is a bounded subset of H, T (C) has compact closurein K.
Equivalently, T is compact if whenever (xn) is a bounded sequence in H, (Txn)has a convergent subsequence.
Example 11.3. Every finite rank operator T ∈ B(H,K) is compact, because,for any bounded sequence (xn) in H, (Txn) is a bounded sequence in the finite-dimensional space RT , and therefore has a convergent subsequence.
Theorem 11.4. If (Tn)n∈N is a norm-convergent sequence of compact operatorsthen T = limn→∞ Tn is compact.
Proof. Take any bounded sequence (xn)n∈N. Then for each n, we may find a sub-sequence (x(n)
m )m∈N of (xn) such that (1) x(n+1) is a subsequence of x(n) for each nand (2) (Tnx
(n)m )m∈N is convergent. We consider the diagonal sequence ym = x
(m)m
which is a subsequence of (xn). One may show that (Tym)m∈N is convergent [exer-cise sheet].
Example 11.5. In l2, multiplication by any sequence (mj)j∈N with mj → 0 iscompact, as M is a limit of finite rank operators [exercise sheet]. The same is truein L2(Z) if mj → 0 as |j| → ∞.
Theorem 11.6. Let H,K and L be Hilbert spaces and T ∈ B(H,K) be compact.Then ST is compact whenever S ∈ B(K,L) and TS is compact whenever S ∈B(L,H).
Proof. Exercise.
As the set Com(H,K) of compact operators between H and K is a vector space[exercise] the above results prove that Com(H) = Com(H,H) is a closed two-sidedideal in B(H).
Example 11.8. Consider a Hilbert-Schmidt operator on l2
(Tx)j =∞∑k=1
ajkxk
‖T‖2 =
√√√√ ∞∑j=1
∞∑k=1
∣∣∣a2jk
∣∣∣ <∞.
40
We know T ∈ B(l2) and ‖T‖ ≤ ‖T‖2. We show that R ∈ Com(l2). Define TN withmatrices
a(N)jk =
ajk j < N
0 j > N.
Each TN has rank ≤ N . Moreover, ‖T − TN‖22 =∑∞j=N+1
∑∞k=1 |ajk|
2 → 0 asN → ∞. Thus ‖T − TN‖ → 0 as N → ∞, so T is the limit of sequence of finiterank operators, and is hence compact.
Hilbert-Schmidt intergral operators are also compact (exercise). (There is anoverarching theory of H-S ops.)
Our main goal is to prve the spectral theorem for self-adjoint compact operators.
Lemma 11.9. Let T be a bounded self-adjoint operator and λ ∈ R. If there isan ε > 0 such that ‖(λI − T )x‖ ≥ ε ‖x‖ for all x then λ ∈ ρ(T ). Equivalently, ifλ ∈ σ(T ) then for every ε > 0 there exists x ∈ H with ‖(λI − T )x‖ < ε ‖x‖ andhence a sequence (xn) with ‖xn‖ = 1 for all n and (λI − T )xn → 0.
Proof. Adapt the proof of Theorem 8.11.
Theorem 11.10. If T ∈ Com)H) is self-adjoint and nonzero then one or both of±‖T‖ are eigenvalues of T .
Proof. If H is complex, we know that r(T ) = ‖T‖ and σ(T ) ⊂ R. Thus we maychoose λ ∈ σ(T ) with |λ| = ‖T‖; λ 6= 0 as T 6= 0. As λ ∈ σ(T ) there must exista sequence xn with ‖xn‖ = 1 and (λI − T )xn → 0 (by Lemma 11.9). Choosing asubsequence (ym) of (xn) so that Tym → w, this implies that λym also convergesto w and
λw − Tw = limm
(λym − Tym) = 0.
But ‖w‖ = |λ| limm ‖ym‖ = |λ| > 0 so λ is an eigenvalue of T . One may also givea proof that works in real Hilbert spaces as well but we omit this.
Lemma 11.11. Let T ∈ Com(H) be self-adjoint and µ > 0. If Vµ is a subspace ofH with an orthonormal basis of eigenvectors of T with eigenvalues exceeding µ inmagnitude, then Vµ has finite dimension. In particular, any eigenspace of nonzeroeigenvalue is finite dimensional.
Proof. Otherwise there would be an infinite orthonormal sequence (un) in Vµ ofeigenvectors with eigenvalues λn, |λn| > µ. But as
‖Tun − Tum‖2 = |λn|2 + |λm|2 > 2µ2
for n 6= m, (Tun) can have no Cauchy subsequence, contradicting compactness ofT .
Corollary 11.12. If T ∈ Com(H) be self-adjoint then either T has finitely manynonzero eigenvalues or T has a countable sequence of eigenvalues tending to 0.
Proof. As unit eigenvectors of distinct eigenvalue are orthonormal, there can onlybe finitely many eigenvalues exceeding any µ > 0 in magnitude by Lemma 11.11. IfΛ is the set of nonzero eigenvalues of T we may write
Λ =⋃n∈N
Λ ∩ (R \ [−1/n, 1/n])
which shows that Λ is a countable union of finite sets and therefore countable. IfΛ is not finite, arrange its elements in decreasing order of magnitude (positive onesfirst, in the case of equal and opposite pairs). Given any ε > 0 all but finitely manyof its elements are less than ε in magnitude so there is an N for which n > N implies|λn| < ε. Thus λn → 0.
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Lemma 11.13. Let T ∈ Com(H) be self-adjoint and let V be the subspace of Hconsisting of vectors that are orthogonal to every eigenvector of nonzero eigenvalue.Then V = NT .
Proof. Take any x ∈ V and observe that 〈Tx, un〉 = 〈x, Tun〉 = λn 〈x, un〉 = 0 forall n. Thus Tx ∈ V . As an intersection of closed sets, V is closed, and hence can beregarded as a Hilbert space in its own right. The restriction T |V of T to V is clearlycompact but can have no nonzero eigenvalues. Hence T |V = 0 by Theorem 11.10,i.e., V ⊆ NT . On the other hand, any element of NT is an eigenvector of T withzero eigenvalue and hence orthogonal to every eigenvector of nonzero eigenvalue.Thus V = NT .
Theorem 11.14. Let T ∈ Com(H) be self-adjoint. Then there is a completeorthonormal sequence (un)rankT
n=1 forRT so that Tun = λnun and λn → 0. Moreover,
Tx =rankT∑n=1
λn 〈x, un〉un.
Proof. Choose orthonormal bases for each nonzero eigenvalue of T , and assemblethese to give the (un). The closed span of the un has NT as its orthogonal com-plement Lemma 11.13 and is therefore equal to (NT )⊥ = RT . Thus the (un) area complete orthonormal system in RT . Thus the (un) are a complete orthonormalsystem in RT . The formula for Tx is left as an exercise.
Corollary 11.15. Let H be separable, of dimension N ∈ N ∪ ∞. If T ∈Com(H) is self-adjoint then T is unitarily equivalent to a multiplication operatorin l2(1, . . . , N) (understanding this as l2(N) if N =∞).
Proof. Extend (if necessary) the basis constructed in the theorem to a countableorthonormal basis for H, with the additional vectors spanning the nullspace of T .Denote this basis as (vn)Nn=1; we have Tvn = µnvn for all n. The map U : H →l2(1, . . . , N) defined by (Ux)n = 〈x, vn〉 is unitary and had vn = U∗en, where enis the n’th standard basis vector of l2(1, . . . , N) (see Example 5.5). Calculating(UTU∗en = UTvn = µnUvn = µnen, we see that T is unitarily equivelent tomultiplication by (µn) on l2(1, . . . , N) as required.
Thus the spectral theorem tells us that for any compact self-adjoint operatorthere is some kind of transform, analogous to mapping a function to its classicalFourier series, which converts the operator into diagonal form.
Even non-self-adjoint compact operators have a canonical form.
Theorem 11.16 (Schmidt decomposition). Let T ∈ Com(H). Then there areorthonormal sequences (un)rankT
n=1 and (vn)rankTn=1 and positive real numbers λn so
that
Tx =rankT∑n=1
µn 〈x, un〉 vn
for all x ∈ H, and µn → 0 as n → ∞ if rankT is infinite. The (vn) are completefor RT , and the (un) are complete for RT .
Proof. (Sketch) First show that |T | is compact (exercises), apply the spectral theo-rem to yield the (un), and use the polar decomposition to write T = U |T | for somepartical isometry U . The required result follows with vn = Uun.
This is also known as the singular value decomposition of T : the µn’s are calledthe singular values, and are equal to positive square roots of eigenvalues of T ∗T .As a consequence we see that every T ∈ Com(H) is a limit of finite rank operators,so the compacts are precisely the norm closure of the finite rank operators.
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Example 11.17. We have already seen (Example 5.9) that the convolution oper-ator
(Tf)(t) =∫ π
−πk(t− s)f(s) ds
where k is 2π-periodic can be unitarily diagonalised by means of Fourier series.This may be though of as an example of Theorem 11.16, with un(t) = eint/
√2π,
vn = ωnun and the scalars ωn ∈ C and µn ∈ R fixed so that
µnωn =∫ π
−πe−inuk(u) du,
with µn ≥ 0 and |ωn| = 1.If k(−u) = k(u) a.e., then T is self-adjoint and ωn = ±1; this then becomes an
example of Theorem 11.14, with λn = ωnµn.
Example 11.18. We know that the equation −f ′′ = λf on [0, a] (finite a) subjectto f(0) = f(a) = 0 has a discrete spectrum of eigenvalues λn = (nπ/a)2 (n ∈ N)with eigenfunctions
un(x) =
√2a
sin(nπx
a
)forming a complete basis for L2(0, a). (Devotees of Quantum Mechanics will recog-nise the ‘particle in a box’.)
This may be understood in terms of the bounded integral operator
(T0f)(x) =∫ a
0
G(x, y)f(y) dy
where
G(x, y) =
x(y − a)/a x < y
y(x− a)/a x > y
is the Green function for the differential operator L0f = −f ′′. Eigenfunctions ofT0 are eigenfunctions of L, and because T0 is Hilbert-Schmidt and self-adjoint weknow that there is a complete basis of such eigenfunctions.
Of more interest is the modified equation (Lf)(x)def= −f ′′(x) + V (x)f(x) =
λf(x), where V is (for simplicity) a continuous function on [0, a] with sup |V | <‖T0‖−1. One may argue that there is a solution operator T for this equation, givenby
T = (I + T0M)−1T0
where (Mf)(x) = V (x)f(x). The inverse exists because ‖T0M‖ < 1 (cf the proof ofTheorem 8.4) and T is compact because T0 is (Theorem 11.6). If V is real-valued,T turns out to be self-adjoint, and so there is again a complete basis for L2(0, a)consisting of eigenfunctions of L.
