7. Operator Theory on Hilbert spaces

7. Operator Theory on Hilbert spaces

In this section we take a closer look at linear continuous maps between Hilbertspaces. These are often called bounded operators, and the branch of FunctionalAnalysis that studies these objects is called “Operator Theory.” The standardnotations in Operator Theory are as follows.

Notations. If H1 and H2 are Hilbert spaces, the Banach space

L(H1,H2) = T : H1 → H2 : T linear continuous will be denoted by B(H1,H2). In the case of one Hilbert space H, the spaceL(H,H) is simply denoted by B(H).

Given T ∈ B(H1,H2) and S ∈ B(H2,H3), their composition ST ∈ B(H1,H3)will be simply denoted by ST .

We know that B(H) is a unital Banach algebra. This Banach algebra willbe our main tool used for investigating bounded operators. The general theme ofthis section is to study the interplay between the operator theoretic properties ofa bounded linear operator, say T : H → H, and the algebraic properties of T asan element of the Banach algebra B(H). Some of the various concepts associatedwith bounded linear operators are actually defined using this Banach algebra. Forexample, for T ∈ B(H), we denote its spectrum in B(H) by SpecH(T ), that is,

SpecH(T ) = λ ∈ C : T − λI : H → H not invertible,and we define its spectral radius

radH(T ) = max|λ| : λ ∈ SpecH(T )

.

We start with an important technical result, for which we need the following.

Lemma 7.1. Let X and Y be normed vector spaces. Equip X×Y with the producttopology. For a sesquilinear map φ : X× Y → C, the following are equivalent:

(i) φ is continuous;(ii) φ is continuous at (0, 0);(iii) sup

|φ(x, y)| : (x, y) ∈ X× Y, ‖x‖, ‖y‖ ≤ 1

<∞;

(iv) there exists some constant C ≥ 0, such that

|φ(x, y)| ≤ C · ‖x‖ · ‖y‖, ∀ (x, y) ∈ X× Y.

Moreover, the number in (iii) is equal to

(1) minC ≥ 0 : |φ(x, y)| ≤ C · ‖x‖ · ‖y‖, ∀ (x, y) ∈ X× Y

.

Proof. The implication (i) ⇒ (ii) is trivial.(ii) ⇒ (iii). Assume φ is continuous at (0, 0). We prove (iii) by contra-

diction. Assume, for each integer n ≥ 1 there are vectors xn ∈ X and yn ∈ Y

with ‖xn‖, ‖yn‖ ≤ 1, but such that |φ(xn, yn)| ≥ n2. If we take vn = 1nxn and

wn = 1nyn, then on the one hand we have ‖vn‖, ‖wn‖ ≤ 1

n , ∀n ≥ 1, which forceslimn→∞(vn, wn) = (0, 0) in X × Y, so by (iii) we have limn→∞ φ(vn, wn) = 0. Onthe other hand, we also have

|φ(vn, wn)| = |φ(xn, yn)|n2

≥ 1, ∀n ≥ 1,

300

§7. Operator Theory on Hilbert spaces 301

which is impossible.(iii) ⇒ (iv). Assume φ has property (iii), and denote the number

sup|φ(x, y)| : (x, y) ∈ X× Y, ‖x‖, ‖y‖ ≤ 1

simply by M . In order to prove (iv) we are going to prove the inequality

(2) |φ(x, y)| ≤M · ‖x‖ · ‖y‖, ∀(x, y) ∈ X× Y.

Fix (x, y) ∈ X×Y. If either x = 0 or y = 0, the above inequality is trivial, so we canassume both x and y are non-zero. Consider the vectors v = 1

‖x‖x and w = 1‖y‖y.

We clearly have

|φ(x, y)| =∣∣φ(‖x‖v, ‖y‖w)

∣∣ = ‖x‖ · ‖y‖ · |φ(v, w)|.Since ‖v‖ = ‖w‖ = 1, we have |φ(v, w)| ≤M , so the above inequality gives (2).

(iv) ⇒ (i). Assume φ has property (iv) and let us show that φ is continuous. LetC ≥ 0 is as in (iv). Let (xn)n→∞ ⊂ X and (yn)n→∞ ⊂ Y be convergent sequenceswith limn→∞ xn = x and limn→∞ yn = y, and let us prove that limn→∞ φ(xn, yn) =φ(x, y). Using (iv) we have

|φ(xn, yn)− φ(x, y)| ≤ |φ(xn, yn)− φ(xn, y)|+ |φ(xn, y)− φ(x, y)| == |φ(xn, yn − y)|+ |φ(xn − x, y)| ≤≤ C · ‖xn‖ · ‖yn − y‖+ C · ‖xn − x‖ · ‖y‖, ∀n ≥ 1,

which clearly forces limn→∞ |φ(xn, yn)− φ(x, y)| = 0, and we are done.To prove the last assertion we observe first that every C ≥ 0 with

|φ(x, y)| ≤ C · ‖x‖ · ‖y‖, ∀ (x, y) ∈ X× Y,

automatically satisfies the inequality C ≥ M . This is a consequence of the aboveinequality, restricted to those (x, y) ∈ X×Y, with ‖x‖, ‖y‖ ≤ 1. To finish the proof,all we have to prove is the fact that C = M satisfies (iv). But this has already beenobtained when we proved the implication (iii) ⇒ (iv).

Notation. With the notations above, the number defined in (iii), which isalso equal to the quantity (1), is denoted by ‖φ‖. This is justified by the following.

Exercise 1. Let X and Y be normed vector spaces over C. Prove that the space

S(X,Y) =φ : X× Y → C : φ sesquilinear continuous

is a vector space, when equipped with pointwise addition and scalar multiplication.Prove that the map

S(X,Y) 3 φ 7−→ ‖φ‖ ∈ [0,∞)defines a norm.

With this terminology, we have the following technical result.

Theorem 7.1. Let H1 and H2 be Hilbert spaces, and let φ : H1 ×H2 → C bea sesquilinear map. The following are equivalent (equip H1 ×H2 with the producttopology):

(i) φ is continuous;(ii) there exists T ∈ B(H1,H2), such that

φ(ξ1, ξ2) = (Tξ1|ξ2)H2 , ∀ (ξ1, ξ2) ∈ H1 ×H2,

where ( . | . )H2 denotes the inner product on H2.

302 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Moreover, the operator T ∈ B(H1,H2) is unique, and has norm ‖T‖ = ‖φ‖.

Proof. (i) ⇒ (ii). Assume φ is continuous, so by Lemma 7.1 we have

(3) |φ(ξ, ζ)| ≤ ‖φ‖ · ‖ξ‖ · ‖ζ‖, ∀ ξ ∈ H1, ζ ∈ H2.

Fix for the moment ξ ∈ H1, and consider the map

φξ : H2 3 ζ 7−→ φ(ξ, ζ) ∈ C.

Using (3), it is clear that φξ : H2 → C is linear continuous, and has norm ‖φξ‖ ≤‖φ‖·‖ξ‖. Using Riesz’ Theorem (see Section 3), it follows that there exists a uniquevector ξ ∈ H2, such that

φξ(ζ) = (ξ|ζ)H2 , ∀ ζ ∈ H2.

Moreover, one has the equality

(4) ‖ξ‖H2 = ‖φξ‖ ≤ ‖φ‖ · ‖ξ‖H1 .

Remark that, if we start with two vectors ξ, η ∈ H1, then we have

(ξ|ζ)H2 + (η|ζ)H2 = φ(ξ, ζ) + φ(η, ζ) = φ(ξ + η, ζ) = φξ+η(ζ), ∀ ζ ∈ H2,

so by the uniqueness part in Riesz’ Theorem we get the equality ξ + η = ξ + η.Likewise, if ξ ∈ H1, and λ ∈ C, we have

(λξ|ζ)H2 = λ(ξ|ζ)H2 = λφ(ξ, ζ) = φ(λξ, ζ) = φλξ(ζ), ∀ ζ ∈ H2,

which forces λξ = λξ. This way we have defined a linear map

T : H1 3 ξ 7−→ ξ ∈ H2,

withφ(ξ, ζ) = (Tξ|ζ)H2 , ∀ (ξ, ζ) ∈ H1 ×H2.

Using (4) we also have

‖Tξ‖H2 ≤ ‖φ‖ · ‖ξ‖H1 , ∀x ∈∈ H1,

so T is indeed continuous, and it has norm ‖T‖ ≤ ‖φ‖. The uniqueness of T isobvious.

(ii) ⇒ (i). Assume φ has property (ii), and let us prove that φ is continuous.This is pretty clear, because if we take T ∈ B(H1,H2) as in (ii), then using theCauchy-Bunyakovski-Schwartz inequality we have

|φ(ξ1, ξ2)| = |(Tξ1|ξ2)H2 | ≤ ‖Tξ1‖ · ‖ξ2‖ ≤ ‖T‖ · ‖ξ1‖ · ‖ξ2‖, ∀ (ξ1, ξ2) ∈ H1 ×H2,

so we can apply Lemma 7.1. Notice that this also proves the inequality ‖φ‖ ≤ ‖T‖.Since by the proof of the implication (i) ⇒ (ii) we already know that ‖T‖ ≤ ‖φ‖,it follows that in fact we have equality ‖T‖ = ‖φ‖.

Remark 7.1. For a sesquilinear map φ : H1×H2 → C, the conditions (i), (ii)are also equivalent to

(ii’) there exists S ∈ B(H2,H1), such that

φ(ξ1, ξ2) = (ξ1|Sξ2)H1 , ∀ (ξ1, ξ2) ∈ H1 ×H2.


Moreover, S ∈ B(H2,H1) is unique, and satisfies ‖S‖ = ‖T‖. This follows fromthe Theorem, applied to the sesquilinear map ψ : H2 ×H1 → C, defined by

ψ(ξ2, ξ1) = φ(ξ1, ξ2), ∀ (ξ2, ξ1) ∈ H2 ×H1.

Notice that one clearly has ‖φ‖ = ‖ψ‖, so we also get ‖T‖ = ‖S‖.In particular, if we start with an operator T ∈ B(H1,H2) and we consider the

continuous sesquilinear map

H1 ×H2 3 (ξ1, ξ2) 7−→ (Tξ1|ξ2)H2 ∈ C,then there exists a unique operator S ∈ B(H2,H1), such that

(5) (Tξ1|ξ2)H2 = (ξ1|Sξ2)H1 , ∀ (ξ1, ξ2) ∈ H1 ×H2.

This operator will satisfy ‖S‖ = ‖T‖.

Definition. Given an operator T ∈ B(H1,H2), the unique operator S ∈B(H2,H1) that satisfies (5) is called the adjoint of T , and is denoted by T ?. Bythe above Remark, for any two vectors ξ1 ∈ H1, ξ2 ∈ H2, we have the identities

(Tξ1|ξ2)H2 = (ξ1|T ?ξ2)H1 ,

(ξ2|Tξ1)H2 = (T ?ξ2|ξ1)H1 .

(The second identity follows from the first one by taking complex conjugates.)

The following result collects a first set of properties of the adjoint operation.

Proposition 7.1. A. For two Hilbert spaces H1, H2, one has

‖T ?‖ = ‖T‖, ∀T ∈ B(H1,H2);(6)

(T ?)? = T, ∀T ∈ B(H1,H2);(7)

(S + T )? = S? + T ? ∀S, T ∈ B(H1,H2);(8)

(λT )? = λT ? ∀T ∈ B(H1,H2), λ ∈ C.(9)

B. Given three Hilbert spaces H1, H2, and H3, one has

(10) (ST )? = T ?S? ∀T ∈ B(H1,H2), S ∈ B(H2,H3)

Proof. The equality (6) has already been discussed in Remark 7.1 The identity(7) is obvious. To prove the other identities we employ the following strategy. Wedenote by X the operator whose adjoint is the left hand side, we denote by Ythe operator in the right hand side, so we must show X? = Y , and we prove thisequality by proving the equality

(Xξ|η) = (ξ|Y η), ∀ ξ, η.For example, to prove (8) we put X = S + T and Y = S? + T ?, and it is prettyobvious that

(Xξ|η) = (Sξ + Tξ|η) = (Sξ|η) + (Tξ|η) = (ξ|S?η) + (ξ|T ?η) =

= (ξ|S?η + T ?η) = (ξ|Y η), ∀ ξ ∈ H1, η ∈ H2.

The other identities are proven the exact same way.

Comment. If we work with one Hilbert space H, then the above result givesthe fact that B(H) is a unital involutive Banach algebra. (See Section 5 for theterminology.) This will be crucial for the development of the theory.

Another important set of results deals with the kernel and the range.


Proposition 7.2 (Kernel-Range Identities). Let H1 and H2 be Hilbert spaces.For any operator T ∈ B(H1,H2), one has the equalities

(i) KerT ? = (RanT )⊥;(ii) RanT ? = (KerT )⊥.

Proof. (i). If we start with some vector η ∈ KerT ?, then for every ξ ∈ H1,we have

(η|Tξ)H2 = (T ?η|ξ)H1 = 0,thus proving that η ⊥ Tξ, ∀ ξ ∈ H1, i.e. η ∈ (RanT )⊥. This proves the inclusion

KerT ? = (RanT )⊥.

To prove the inclusion in the other direction, we start with some vector η ∈KerT ? = (RanT )⊥, and we prove that T ?η = 0. This is however pretty clearsince we have η ⊥ (TT ?η), i.e.

0 = (η|TT ?η)H2 = (T ?η|T ?η)H1 = ‖T ?η‖2,which forces T ?η = 0.

(ii). This follows immediately from part (i) applied to T ?:

RanT ? =([RanT ?]⊥

)⊥ = (KerT )⊥.

Example 7.1. Let n ≥ 1 be some integer, and consider the Hilbert space Cn,whose inner product is the standard one:

(ξ|η) =n∑

k=1

ξkηk, ∀ ξ = (ξ1, . . . , ξn), η = (η1, . . . , ηn) ∈ Cn.

The Banach algebra B(Cn) is obviously identified with the algebra Matn×n(C) ofn × n matrices with complex coefficients. The adjoint operation then correspondsto a (familiar) operation in linear algebra. For A ∈ Matn×n(C), say A = [ajk]nj,k=1,one takes A? = [bjk]nj,k=1 to be the conjugate transpose of A, i.e. the b’s aredefined as bjk = akj . A similar identification works with B(Cm,Cn), identifiedwith Matn×m(C).

The adjoint operation is used for defining certain types of operators.

Definitions. Let H be a Hilbert space.A. We say that an operator T ∈ B(H) is normal, if T ?T = TT ?.B. We say that an operator T ∈ B(H) is self-adjoint, if T ? = T .C. We say that an operator T ∈ B(H) is positive, if

(Tξ|ξ)H ≥ 0, ∀ ξ ∈ H.

Remarks 7.2. A. Every self-adjoint operator T ∈ B(H) is normal.B. The set T ∈ B(H) : T normal is closed in B(H), in the norm topology.

Indeed, if we start with a sequence (Tn)∞n=1 of normal operators, which converges(in norm) to some T ∈ B(H), then (T ?

n)∞n=1 converges to T ?, and since the multi-plication map

B(H)×B(H) 3 (X,Y ) 7−→ XY ∈ B(H)is continuous, have T ?T = limn→∞ T ?

nTn and TT ? = limn→∞ TnT?n , so we imme-

diately get T ?T = TT ?.C. For T ∈ B(H), the following are equivalent (see Remark 3.1):• T is self-adjoint;


• the sesquilinear map

φT : H ×H 3 (ξ, η) 7−→ (Tξ|η)H ∈ C

is sesqui-symmetric, i.e. (Tξ|η) = (Tη|ξ), ∀ ξ, η ∈ H;• (Tξ|ξ) ∈ R, ∀ ξ ∈ H.

In particular, we see that every positive operator T is self-adjoint.Using the terminology from Section 3, the condition that T is positive is equiv-

alent to the condition that φT is positive definite.D. The sets:

B(H)sa = T ∈ B(H) : T ? = T,B(H)+ = T ∈ B(H) : T positive

are also closed in B(H), in the norm topology. This follows from the observationthat, if (Tn)∞n=1 converges to some T , then we have

(Tξ|ξ) = limn→∞

(Tnξ|η), ∀ ξ ∈ H.

So if for example all Tn’s are self-adjoint, then this proves that (Tξ|ξ) ∈ R, ∀ ξ,∈ H,so T is selfadjoint. Likewise, if all Tn’s are positive, then (Tξ|ξ) ≥ 0, ∀ ξ ∈ H, so Tis positive.

E. Given Hilbert spaces H1 and H2, and an operator T ∈ B(H1,H2), it followsthat the operators T ?T ∈ B(H1) and TT ? ∈ B(H2) are positive. This is quiteobvious, since

(T ?Tξ|ξ) = (Tξ|Tξ) = ‖Tξ‖2 ≥ 0, ∀ ξ ∈ H1,

(TT ?η|η) = (T ?η|T ?η) = ‖T ?η‖2 ≥ 0, ∀ η ∈ H2.

F. The space B(H)sa is a real linear subspace of B(H).G. The space B(H)+ is a convex cone in B(H)sa, in the sense that• if S, T ∈ B(H)+, then S + T ∈ B(H)+;• if S ∈ B(H)+ and α ∈ [0,∞), then αS ∈ B(H)+.

H. Using G, one can define a order relation on the real vector space B(H)sa by

S ≥ T ⇐⇒ S − T ∈ B(H)+.

This is equivalent to the inequality

(Sξ|ξ) ≥ (Tξ|ξ), ∀ ξ ∈ H.

The transitivity and reflexivity properties are clear. For the antisymmetry, onemust show that if T ≥ S and S ≥ T , then S = T . This is however clear, becausethe difference X = S − T is self-adjoint, and satisfies

(11) (Xζ|ζ) = 0, ∀ ζ ∈ H.

Using polarization (see Section 3), we have

(Xξ|η) =14

3∑k=0

i−k(X(ξ + ikη)

∣∣ξ + ikη),

and then (11) forces(Xξ|η) = 0 ∀ ξ, η ∈ H,

we must have X = 0.From now on, we are going to write T ≥ 0 to mean that T is positive.


Exercise 2♦. Prove that for an operator T ∈ B(H) the following are equivalent:• T is normal;• ‖Tξ‖ = ‖T ?ξ‖, ∀ ξ ∈ H.

Exercise 3*. For a self-adjoint operator, Theorem 7.1 has a slightly improvedversion. Prove that if T ∈ B(H) be self-adjoint, then one has the equality

‖T‖ = sup|(Tξ|ξ)| : ξ ∈ H, ‖ξ‖ ≤ 1

.

Hints: Denote the right hand side of the above equality by M . Use the Cauchy-Bunyakovski-Schwartz inequality, to get the inequality ‖T‖ ≤ M . To prove the inequality ‖T‖ ≤ M , show firstthat

(∗) |(Tζ|ζ) ≤ M‖ζ‖2, ∀ ζ ∈ H.

Use Theorem 7.1 which gives

‖T‖ = sup|(Tξ|η)| : ξ, η ∈ H, ‖ξ‖, ‖η‖ ≤ 1

.

This reduces the problem to proving that, whenever ‖ξ‖, ‖η‖ ≤ 1, one has the inequality

|(Tξ|η)| ≤ M.

Prove this inequality, under the extra assumption that (Tξ|η) ∈ R, using (∗), polarization, and

the Parallelogram Law.

Exercise 4. A. Prove that, if T ∈ B(H) is self-adjoint, then one has the in-equalities −‖T‖I ≤ T ≤ ‖T‖I.

B. Prove that, if S, T ∈ B(H) are such that S ≥ T ≥ 0, then ‖S‖ ≥ ‖T‖.

Example 7.2. Let H be a Hilbert space, and let X ⊂ H be a closed linearsubspace. Then the orthogonal projection PX : H → H is a positive boundedoperator. Indeed, if one starts with some ξ ∈ H, using the fact that ξ−PXξ ∈ X⊥,it follows that (ξ − PXξ) ⊥ PXξ, so we have

0 = (PXξ|ξ − PXξ) = (Pξ|ξ)− (PXξ|PXξ),

which proves that

(PXξ|ξ) = (PXξ|PXξ) = ‖PXξ‖2 ≥ 0, ∀ ξ ∈ H.

It turns out that orthogonal projections can be completely characterized alge-braically. More explicitly one has the following.

Proposition 7.3. Let H be a Hilbert space. For a bounded operator Q ∈ B(H),the following are equivalent:

(i) there exists a closed subspace X ⊂ H, such that Q = PX - the orthogonalprojection onto X;

(ii) Q = Q? = Q2.

Proof. The implication (i) ⇒ (ii) is trivial.(ii) ⇒ (i). Assume Q = Q? = Q2, and let us prove that Q is the orthogonal

projection onto some closed subspace X ⊂ H. We define X = RanQ. First of all,we must show that X is closed. This is pretty obvious, since the equality Q2 = Qgives the equality X = Ker(I − Q). To prove that Q = PX, we must prove twothings:

(a) Qξ = ξ, ∀ ξ ∈ X;(b) Qξ = 0, , ∀ ξ ∈ X⊥.


The first property is clear, since X = Ker(I −Q), To prove the second property, weuse Proposition 7.2 to get

X⊥ = (RanQ)⊥ = KerQ? = KerQ.

Convention. An operator Q ∈ B(H) with Q2 = Q = Q? will be simply calleda projection.

The above example illustrates the basic idea that spatial (i.e. operator theo-retic) properties of bounded operators can often be translated into algebraic prop-erties, stated in terms of the involutive Banach algebra B(H). The types normaland self-adjoint are already described in algebraic terms. Besides these, there areother types of operators whose spatial properties can be characterized algebraically(see Proposition 7.4 below), and these are introduced below.

Definitions. Let H1 and H2 be Hilbert spaces.A. An operator T ∈ B(H1,H2) is called an isometry, if ‖Tξ‖ = ‖ξ‖, ∀ ξ ∈ H1.B. An operator T ∈ B(H1,H2) is said to be a coisometry, if its adjoint T ? ∈

B(H2,H1) is an isometry.C. An operator U ∈ B(H1,H2) is called a unitary, if U is a bijective isometry.

The algebraic characterizations for these types of operators are as follows.

Proposition 7.4. Let H1 and H2 be Hilbert spaces.A. For an operator T ∈ B(H1,H2), the following are equivalent:

(i) T is an isometry;(ii) T ?T = IH1 .

B. For an operator T ∈ B(H1,H2), the following are equivalent:(i) T is a coisometry;(ii) TT ? = IH2 .

C. For an operator U ∈ B(H1,H2), the following are equivalent:(i) U is unitary;(ii) U?U = IH1 and UU? = IH2 .

Proof. A. (i) ⇒ (ii). Using polarization, applied to the sesquilinear form

φ : H1 ×H1 3 (ξ, η) 7−→ (T ?Tξ|η) ∈ C,it follows that, for every ξ, η ∈ H, one has the equalities

φ(ξ, η) =14

3∑k=0

i−kφ(ξ + ik, ξ + ikη) =14

3∑k=0

i−k(T ?T (ξ + ikη)

∣∣ ξ + ikη)

=

=14

3∑k=0

i−k(T (ξ + ikη)

∣∣T (ξ + ikη))

=14

3∑k=0

i−k‖T (ξ + ikη)‖2.

Using the fact that T is an isometry, and polarization again (for the inner product),the above computation continues with

φ(ξ, η) =14

3∑k=0

i−k‖T (ξ + ikη)‖2 =14

3∑k=0

i−k‖ξ + ikη‖2 =

=14

3∑k=0

i−k(ξ + ikη

∣∣ ξ + ikη)

= (ξ|η).


Since we now have(T ?Tξ|η) = (ξ|η), ∀ ξ, η ∈ H1,

by Lemma 7.1 (the uniqueness part) we get T ?T = IH1 .The implication (ii) ⇒ (i) is trivial, since the equality T ?T = IH1 gives

‖Tξ‖2 = (Tξ|Tξ) = (T ?Tξ|ξ) = (ξ|ξ) = ‖ξ‖2, ∀xi ∈ H1.

B. This is immediate, by applying part A to T ?.C. (i) ⇒ (ii). Assume U is unitary. On the one hand, since U is an isometry,

by part A we get U?U = IH1 . On the other hand, since U is bijective, the aboveequality actually forces U−1 = U?, so we also get UU? = UU−1 = IH2 .

(ii) ⇒ (i). Assume U?U = IH1 and UU? = IH2 , and let us prove that U is aunitary. On the one hand, these two equalities prove that U is both left and rightinvertible, so U is bijective. On the other hand, by part A, it follows that U is anisometry, so U is indeed unitary.

In the study of bounded linear operators, positivity is an essential tool. Thisis illustrated by the following technical result.

Lemma 7.2. Let H1 and H2 be Hilbert spaces. For an operator T ∈ B(H1,H2),the following are equivalent:

(i) T is invertible;(ii) there exists a constant α > 0, such that T ?T ≥ αIH1 and TT ? ≥ αIH2 .

Proof. (i) ⇒ (ii). Assume T is invertible. Then T ? : H2 → H1 is alsoinvertible, with inverse (T ?)−1 = (T−1)?. Define the number α = ‖T−1‖−2. UsingProposition 7.1, we also have α = ‖(T ?)−1‖−2. Remark now that if one starts withsome ξ ∈ H1, then the norm inequality

‖ξ‖ = ‖T−1(Tξ)‖ ≤ ‖T−1‖ · ‖Tξ‖

gives ‖Tξ‖ ≥ ‖T−1‖−1 · ‖ξ‖, so taking squares we get

(T ?Tξ|ξ)H1 = (Tξ|Tξ)H2 = ‖Tξ‖2 ≥ ‖T−1‖−2‖ξ‖2 = α(ξ|ξ)H1 .

This proves precisely that((T ?T − αIH1)ξ

∣∣ ξ )H1

≥ 0, ∀ ξ ∈ H1,

so T ?T − αIH1 ∈ B(H1) is positive. The positivity of TT ? − αIH2 ∈ B(H2) isproven the exact same way.

(ii) ⇒ (i). Assume there exists α > 0 such that T ?T ≥ αIH1 and TT ? ≥ αIH2 ,and let us show that T is invertible. First of all, using the inequality T ?T ≥ αIH1 ,we have

‖Tξ‖2 = (Tξ|Tξ) = (T ?ξ|ξ) ≥ α(ξ|ξ) = α‖ξ‖2,so we get

‖Tξ‖ ≥√α‖ξ‖, ∀ ξ ∈ H1.

On the one hand, this shows that T is injective. On the other hand, by Proposition2.1, this also proves that RanT is a Banach space, when equipped with the normcoming from H2, i.e. RanT is closed in H2. In particular, by Proposition 7.2 wehave

(12) RanT = RanT = (KerT ?)⊥.


Notice that exactly as above, we also have

‖T ?η‖ ≥√α‖η‖, ∀ η ∈ H2,

so T ? is injective as well, which means that KerT ? = 0, so the equality (12) givesRanT = H2, so T is also surjective.

Remark 7.3. With the notations above, the condition that T is invertible isalso equivalent to the condition

(iii) both operators T ?T ∈ B(H1) and TT ? ∈ B(H2) are invertible.Indeed, if T is invertible, then so is T ?, so the products T ?T and TT ? are alsoinvertible. Conversely, if both T ?T and TT ? are invertible, and if we put X =(T ?T )−1 and Y = (TT ?)−1, then we have the equalities

(XT ?)T = IH1 and T (T ?Y ) = IH2 ,

so T is both left and right invertible.

Proposition 7.5. Let H be a Hilbert space.(i) Every self-adjoint operator T ∈ B(H) has real spectrum, i.e. one has the

inclusion SpecH(T ) ⊂ R.(ii) Every positive operator T ∈ B(H) has non-negative spectrum, i.e. one

has the inclusion SpecH(T ) ⊂ [0,∞).

Proof. (i). Let T ∈ B(H) be self-adjoint. We wish to prove that for everycomplex number λ ∈ CrR, the operator X = λI−T is invertible. Write λ = a+ib,with a, b ∈ R with b 6= 0. We are going to apply Lemma 7.2, so we need to considerthe operators X?X and XX?. It turns out that

X?X = XX? = |λ|2I − 2(Reλ)T + T 2,

so all we need is the existence of a constant α > 0, such that X?X ≥ αI. But thisis clear, since

X?X = (a2 + b2)I − 2aT + T 2 = b2I + (aI − T )2,

and the positivity of (aI−T )2 = (aI−T )?(aI−T ) (see Remark 7.2.E) immediatelygives X?X ≥ b2I.

(ii). By part (i) we only need to prove that, for every number a ∈ (−∞, 0), theoperator X = aI − T is invertible. As before, we have

X?X = XX? = a2I − 2aT + T 2,

and then the positivity of −2aT and of T 2 = T ?T (see Remark 7.2.F), forcesX?X ≥ a2I. Since a 6= 0, by Lemma 7.2, it follows that X is indeed invertible.

The above result can be nicely complemented with the one below.

Proposition 7.6 (Spectral Radius Formula for self-adjoint operators). Let H

be a Hilbert space. For every self-adjoint operator T ∈ B(H), one has the equality

radH(T ) = ‖T‖.

Proof. It T = 0, there is nothing to prove, so without any loss of generalitywe can assume that ‖T‖ = 1. Since radH(T ) ≤ ‖T‖ = 1 (see Section 5), all we haveto prove is the fact that SpecH(T ) contains one of the numbers ±1. Equivalently,


we must prove that either −I −T or I −T is non-invertible. Consider (see Remark7.2.D) the positive operator X = T 2, so that we have

X − I = (−I − T )(I − T ) = (I − T )(−I − T ),

which means that we must prove that X − I is non-invertible. We prove this factby contradiction. Assume that X − I is invertible, so by Lemma 7.2 there existssome constant α ∈ (0, 1) such that

(13) αI ≤ (X − I)?(X − I) = (X − I)2.

Remark that, since ‖T‖ = 1, we have the inequality

0 ≤ (T 4ξ|ξ) = ‖T 2ξ‖2 ≤ (‖T‖ · ‖Tξ‖)2 ≤ ‖Tξ‖2 = (T 2ξ|ξ), ∀ ξ ∈ H,

which reads:X ≥ X2 ≥ 0.

In particular this gives (I −X)− (X − I)2 = X −X2 ≥ 0, so we also have

(I −X) ≥ (X − I)2.

Using (13) this forces the inequality I −X ≥ αI, which can be re-written as

(1− α)I ≥ X.

In other words, we have

(1− α)‖ξ‖2 = (1− α)(ξ|ξ) ≥ (Xξ|ξ) = (T 2ξ|ξ) = ‖Tξ‖2, ∀ ξ ∈ H,

which gives‖Tξ‖ ≤

√1− α · ‖ξ‖, ∀ ξ ∈ H.

This forces ‖T‖ ≤√

1− α, which contradicts the assumption that ‖T‖ = 1.

Although the following result may look quite “innocent,” it is crucial for thedevelopment of the theory.

Proposition 7.7. Let H1 and H2 be Hilbert spaces. For every operator T ∈B(H1,H2), one has the identity

(14) ‖T ?T‖ = ‖T‖2.

Proof. Fix T ∈ B(H1,H2). Consider the sesquilinear map

φ : H1 ×H1 3 (ξ, η) 7−→ (T ?Tξ|η)H1 ∈ C.

By Theorem 7.1, we know that ‖T ?T‖ = ‖φ‖. Notice however that, for everyξ ∈ H1 with ‖ξ‖ ≤ 1, one has

‖φ‖ ≥ |φ(ξ, ξ)| = |(T ?Tξ|ξ)| = |(Tξ|Tξ)| = ‖Tξ‖2,so we get √

‖φ‖ ≥ sup‖Tξ‖ : ξ ∈ H1, ‖ξ‖ ≤ 1

= ‖T‖,

thus proving the inequality ‖T ?T‖ = ‖φ‖ ≥ ‖T‖2. The other inequality is immedi-ate, since ‖T ?T‖ ≤ ‖T ?‖ · ‖T‖ = ‖T‖2.

Corollary 7.1. Let H be a Hilbert space, and let A be an involutive Banachalgebra. Then every ?-homomorphism Φ : A → B(H) is contractive, in the sensethat one has the inequality

(15) ‖Φ(a)‖ ≤ ‖a‖, ∀ a ∈ A.


Proof. Fix a ?-homomorphism Φ : A → B(H). We can assume that A isunital, and Φ(1) = I. (If not, we work with the unitized algebra A, which isagain an involutive Banach algebra, and with the map Φ : A → B(H) definedby φ(a, α) = Φ(a) + αI, a ∈ A, α ∈ C, which clearly defines a ?-homomorphismsatisfying Φ(1) = I.)

To prove (15) we start with an arbitrary element a ∈ A, and we consider theelement b = a?a. On the one hand, the operator Φ(b) = Φ(a)?Φ(a) ∈ B(H) isobviously self-adjoint, so by Proposition 7.6, we know that

(16) ‖Φ(b)‖ = radH

(Φ(b)

).

Since Φ is an algebra homomorphism with Φ(1) = 1, we have the inclusion

SpecH

(Φ(b)

)⊂ SpecA(b),

which then gives the inequality

radH

(Φ(b)

)≤ radA(b).

Using the inequality radA(b) ≤ ‖b‖, the above inequality, combined with (16), yields

(17) ‖Φ(b)‖ ≤ ‖b‖.

On the other hand, using Proposition 7.7, we know that

‖Φ(b)‖ = ‖Φ(a)?Φ(a)‖ = ‖Φ(a)‖2,

so (17) reads

(18) ‖Φ(a)‖2 ≤ ‖b‖.

Finally, since A is an involutive Banch algebra, we have

‖b‖ = ‖a?a‖ ≤ ‖a?‖ · ‖a‖ = ‖a‖2,

and then (18) clearly gives (15).

The identity (14) is referred to as the C?-norm condition. The above resultsuggests that this property has interesting applications. As shall see a little later,this condition is at the heart of the entire theory. It turns out that one can developan entire machinery, which enables one to do “algebraic” Operator Theory. At thecore of this development is the following concept.

Definition. An involutive Banach algebra A is called a C?-algebra, if its normsatisfies the C?-norm property, i.e. one has

‖a?a‖ = ‖a‖2, ∀ a ∈ A.

Remark that if A is unital, with unit 1 6= 0, then 1? = 1, and the above conditionforces ‖1‖ = 1.

Remark that, given a C?-algebra A, then any closed subalgebra B ⊂ A, with

b? ∈ B, ∀ b ∈ B,

is a C?-algebra. In this situation, we say that B is a C?-subalgebra of A.

Examples 7.3. A. Using the above terminology, Proposition 7.7 gives the factthat, if H is a Hilbert space, then B(H) is a unital C?-algebra.

B. If Ω is a locally compact space, then C0(Ω) is a C?-algebra, when equippedwith the natural involution f? = f .


Remarks 7.4. A. The types normal and self-adjoint make sense for elementsin a C?-algebra.

B. The type projection also makes sense for an element in a C?-algebra. Pro-jections are defined as those elements satisying condition (ii) from Proposition 7.3.

C. The types (co)isometry and unitary make sense for elements in a unital C?-algebra. For these types one uses conditions A.(ii), B.(ii), C.(ii) from Proposition7.4 as definitions.

D. The characterization of invertible elements given in Remark 7.3 is valid ifB(H) is replaced with an arbitrary unital C?-algebra. More explicitly, if A is aunital C?-algebra, then for an element a ∈ A, the conditions

• a is invertible in A,• both a?a and aa? are invertible in A,

are equivalent. The proof is identical to the one given in Remark 7.3. (In fact thisis true even when A is only a unital ?-algebra.)

The following result analyzes the Gelfand correspondence discussed in Section5, in an important particular case.

Theorem 7.2. Let H be a Hilbert space, and let T ∈ B(H) be a normaloperator. Consider the closed subalgebra T = alg(I, T, T ?).

A. T is a commutative C?-subalgebra of B(H).B. One has the equality SpecT(T ) = SpecH(T ).C. Every character of T is involutive, i.e. one has the equality

Char(T) = Char?(T).

D. The Gelfand correspondence ΓT : T → C(Char(T)

)is an isometric ?-

isomorphism.E. The map FT : Char(T) 3 γ 7−→ γ(T ) ∈ C establishes a homeomorphism

FT : Char(T) ∼−→ SpecT(T ).

Proof. A. We begin with an explicit description of the algebra

T0 = alg(I, T, T ?),

which by construction is dense in T. For this puprose we introduce the followingnotation. If P (s, t) is a polynomial in two variables, say

P (s, t) =N∑

m,n=0

αmnsmtn,

with complex coefficients, we define the operator

(19) P (T ?, T ) =N∑

m,n=0

αmn(T ?)mTn,

with the convention that (T ?)0 = T 0 = I. We denote by C[s, t] the algebra ofpolynomials in two variables. With these notations we have the following.

Claim 1: The correspondence

(20) Π : C[s, t] 3 P 7−→ P (T ?, T ) ∈ B(H)

gives rise to a surjective unital algebra homomorphism Π : C[s, t] → T0.


It is obvious that Π is linear, and satisfies Π(1) = I, where 1 denotes the con-stant polynomial 1. Remark that, if P (s, t) =

∑Nm,n=0 αmns

mtn and Q(s, t) =∑Kk,`=0 βk`βk`s

kt`, are polynomials in two variables, then using the fact that Tcommutes with T ? (i.e. TT ? = T ?T ), we get

P (T ?, T )Q(T ?, T ) =N∑

m,n=0

K∑k,`=0

αmnβk`(T ?)mTn(T ?)kT ` =

=N∑

m,n=0

K∑k,`=0

αmnβk`(T ?)m+kTn+`,

so if we consider the polynomial PQ, we have

P (T ?, T )Q(T ?, T ) = (PQ)(T ?, T ).

This proves that Π : C[s, t] → B(H) is indeed a unital algebra homomorphism.On the one hand, it is obvious that RanΠ ⊂ T0. On the other hand, since Π isan algebra homomorphism, it follows that RanΠ is a subalgebra of T0. Finally,since RanΠ ⊃ I, T, T ?, it follows that we have in fact the equality RanΠ =alg(I, T, T ?) = T0.

Having proven Claim 1, let us observe that this already gives the fact that T0 iscommutative (being a quotient of the commutative algebra C[s, t]). In particular,the closure T of T0 is also commutative. Remark also that the algebra C[s, t] alsocarries an involution defined as follows. If P ∈ C[s, t] is a polynomial, say P (s, t) =∑N

m,n=0 αmnsmtn, then one defines the polynomial P#(s, t) =

∑Nm,n=0 αmns

ntm.It is clear that the homomorphism (20) is in fact a ?-homomorphism, i.e. it satisfies

(P#)(T ?, T ) = [P (T ?, T )]?, ∀P ∈ C[s, t],

so in fact T0 = RanΠ is a ?-subalgebra of B(H). Taking the closure, it follows thatT is a C?-subalgebra of B(H).

B-E. In order to prove these statements, we are going to consider first a largerC?-subalgebra, which will have properties A-E, and we will eventually prove thatthis larger algebra in fact coincides with T. Consider the set

M =Q ∈ C[s, t] : P (T ?, T ) invertible in B(H)

,

and defineA0 =

P (T ?, T )[Q(T ?, T )]−1 : P ∈ C[s, t], Q ∈ M

.

Remark that, if Q1, Q2 ∈ M, then Q1Q2 ∈ M. One also has the implication Q ∈M ⇒ Q# ∈ M. A simple computation shows that if X1, X2 ∈ A0 are representedas Xk = Pk(T ?, T )[Qk(T ?, T )]−1 with Pk ∈ C[s, t] and Qk ∈ M, k = 1, 2, then onehas the equalities

X1X2 = (P1P2)(T ?, T )[(Q1Q2)(T ?, T )]−1,

X1 +X2 = (P1Q2 + P2Q1)(T ?, T )[(Q1Q2)(T ?, T )]−1,

so it follows that A0 is a subalgebra of B(H). If X ∈ A0 is represented as X =P (T ?, T )[Q(T ?, T )]−1 with P ∈ C[s, t] and Q ∈ M, then one also has the equality

X? = (P#)(T ?, T )[(Q#)(T ?, T )]−1,

so in fact A0 is a ?-subalgebra of B(H).


We take A = A0 (the closure of A0 in the norm topology). By construction,it follows that A is a C?-subalgebra of B(H). Notice that one obviously has theinclusion A ⊃ T.

Claim 2: One has the equality

(21) SpecH(X) = SpecA(X), ∀X ∈ A0.

First of all (see Section 5), since A is a closed subalgebra of B(H), which containsI, we always have the inclusion

SpecH(X) ⊂ SpecA(X), ∀X ∈ A.

Secondly we observe that, if we start with some operator X ∈ A0, written asX = P (T ?, T )[Q(T ?, T )]−1, with P ∈ C[s, t] and Q ∈ M, then A contains alloperators of the form (λI −X)−1, with λ ∈ C r SpecH(X). This is due to the factthat for every λ ∈ C we can write

λI −X = Pλ(T ?, T )[Q(T ?, T )]−1 = [Q(T ?, T )]−1Pλ(T ?, T ),

where Pλ(s, t) = λQ(s, t)−P (s, t), so if λI−X is invertible, it follows that Pλ ∈ M

and consequently (λI − X)−1 = Q(T ?, T )[Pλ(T ?, T )]−1 belongs to A0 ⊂ A byconstruction. In particular this means that λI − X is invertible in A, for everyλ ∈ C r SpecH(X), so we have the inclusion

C r SpecH(X) ⊂ C r SpecA(X),

which is equivalent to the inclusion

SpecH(X) ⊃ SpecA(X).

In particular, using Claim 2, we immediately see that A has property B, i.e.one has the equality

SpecA(T ) = SpecH(T ).Claim 3: Every character of A is involutive, i.e. one has the equality

Char(A) = Char?(A).

Fix a character γ ∈ Char(A). What we need to prove is the fact that we have theequality

(22) γ(X?) = γ(X), ∀X ∈ A.Using continuity, it suffices to prove that (22) holds only for X ∈ A0. Start withsome arbitrary operator X ∈ A0, and consider the operators X1 = 1

2 (X +X?) andX2 = 1

2i (X −X?). Since A0 is a ?-subalgebra of B(H(, we know that both X1 andX2 belong to A0. On the one hand, by Claim 2, we have the equalities

SpecA(Xk) = SpecH(Xk), k = 1, 2.

On the other hand, X1 and X2 are seff-adjoint, so by Proposition 7.5, combinedwith Corollary 3.5, the above equality gives

γ(Xk) ∈ SpecA(Xk) = SpecH(Xk) ⊂ R, k = 1, 2.

Finally, since we clearly have X = X1 + iX2 and X? = X1 − iX2, the fact thatγ(X1) and γ(X2) are real gives

γ(X?) = γ(X1)− iγ(X2) = γ(X1) + iγ(X2) = γ(X).

Let us consider now the Gelfand correspondence ΓA : A → C(Char(A)

).

Claim 4: ΓA is an isometric ?-homorphism.


Recall that the Gelfand correspondence is defined by

ΓA(X) = X, ∀X ∈ A,

where X : Char(A) → C is the Gelfand transform of X, defined by

X(γ) = γ(X), ∀ γ ∈ Char(A).

By Claim 3, it follows that ΓA is a ?-homomorphism. To prove that ΓA is isometric,by density it suffices to check the equality

(23) ‖X‖ = ‖X‖, ∀X ∈ A0.

Fix X ∈ A0. Since A is a C?-algebra, we have the equality ‖X‖2 = ‖X?X‖. Noticethat the element Y = X?X still belongs to A0. On the one hand, by Claim 3 wehave the equality

(24) SpecA(Y ) = SpecH(Y ).

On the other hand, since Y ? = Y , by Proposition 7.6, we have

‖Y ‖ = radH(Y ) = max|λ| : λ ∈ SpecH(Y )

,

so using (24) we get the equality ‖X‖2 = radA(Y ) - the spectral radius of Y in A.By Corollary 5.4 we know however that radA(Y ) = ‖Y ‖, so we now have

(25) ‖X‖2 = ‖Y ‖.

Since ΓA is a ?-homomorphism, we have Y = X?X = (X)?X. Since C(Char(A)

)is a C?-algebra, the equality (25) finally gives

‖X‖2 = ‖(X)?X‖ = ‖X‖2,and then (23) is clear.

Let us show now that A has property E. We consider the Gelfand transformGT = T of T , which defines a function GT ∈ C

(Char(A)

), given by

GT : Char(A) 3 γ 7−→ γ(T ) ∈ C.

Claim 5: The map GT is a homeomorphism GT : Char(A) ∼−→ SpecA(T ).We already know that GT is continuous, and we have RanGT = SpecA(T ), so theonly other thing to prove is the fact that GT is injective. (We use here the factthat Char(A) is compact.) What we have to prove is the fact that if two charactersγ1, γ2 ∈ Char(A) have the property that γ1(T ) = γ2(T ), then γ1 = γ2. By Claim3, we know that

γ1(T ?) = γ1(T ) = γ2(T ) = γ2(T ?),so we also get

γ1

(P (T ?, T )

)= γ2

(P (T ?, T )

), ∀P ∈ C[s, t].

Using the above equality we also get

γ1

(P (T ?, T )[Q(T ?, T )]−1

)=γ1

(P (T ?, T )

)γ1

(Q(T ?, T )

) =γ2

(P (T ?, T )

)γ2

(Q(T ?, T )

) =

= γ2

(P (T ?, T )[Q(T ?, T )]−1

), ∀P ∈ C[s, t], Q ∈ M.

In other words we have the equality

γ1(X) = γ2(X), ∀X ∈ A0.

Since A0 is dense in A, and γ1 and γ2 are continuous, this finally forces γ1 = γ2.


We now proceed with the proof of the fact that the “old” C?-algebra T hasproperties B-E. As mentioned before this will be done by comparing it with A. Weknow that T is a C?-subalgebra of A.

Claim 6: One has the equality ΓA(T) = C(Char(A)

).

Denote for simplicity the compact space Char(A) by K, and the space ΓA(T) byA. On the one hand, since ΓA is an isometric ?-homomorphism, it follows that Ais a C?-subalgebra of C(K). On the other hand, it is clear that A contains theconstant function 1 = ΓA(I). Finally, A separates the points of K, which meansthat whenever γ1, γ2 ∈ K are such that γ1 6= γ2, then there exists f ∈ A withf(γ1) 6= f(γ2). This is however clear, because if we take GT = ΓA(T ) ∈ A, thenby Claim 4, we know that we must have GT (γ1) 6= GT (γ2). We now use theStone-Weierstrass Theorem (see Section 6) to conclude that A = C(K).

Using Claim 6 we now see that the equality ΓA(T) = C(Char(A)

)forces on the

the one hand the equality ΓA(A) = C(Char(A)

). In particular, by Claim 5, we see

that ΓA : A → C(Char(A)

)is an isometric ?-isomorphism. On the other hand the

injectivity of ΓA, combined with the equality ΓA(T) = ΓA(A) forces T = A. Nowwe are done, since A has properties B-E.

Theorem 7.2 has a multitude of applications. The first one is a generalizationof Proposition 7.6.

Corollary 7.2 (Spectral Radius Formula for normal operators). Let H be aHilbert space, and let T ∈ B(H) be a normal operator. Then one has the equality

radH(T ) = ‖T‖.

Proof. Put T = alg(I, T, T ?). On the one hand, by Theorem 7.2, we knowthat SpecH(T ) = SpecT(T ). In particular we also have the equality

radH(T ) = radT(T ).

On the other hand, from the properties of the Gelfand transform (see Section 5),it follows that radT(T ) = ‖ΓT(T )‖. Now we are done, because ΓT is isometric, sowe have ‖ΓT(T )‖ = ‖T‖.

A particularly interesting consequence of Theorem 7.2 is the following.

Corollary 7.3 (Spectral Permanence Property). Let H be a Hilbert space,and let A be a C?-aubalgebra of B(H) with A 3 I. Then one has the equality

SpecA(X) = SpecH(X), ∀X ∈ A.

Proof. We already know from Section 5 that one always has the inclusion

SpecA(X) ⊃ SpecH(X), ∀X ∈ A.

To prove the inclusion in the other direction, we fix for the moment an operatorX ∈ A and some number λ ∈ SpecA(X), and we prove that λ ∈ SpecH(X).Consider the operators T1 = (λI − X)?(λI − X) and T2 = (λI − X)(λI − X)?.Obviosuly both T1 and T2 belong to A. By Remark 7.4.D it follows that one of theoperators T1 and T2 is non-invertible in A. One way to rewrite this is:

(26) 0 ∈ SpecA(T1) ∪ SpecA(T2).


Since T1 and T2 are self-adjoint, they are normal. In particular, if one considersthe C?-algebras Tk = alg(I, Tk, T ?

k ), k = 1, 2, then by Theorem 7.2 we have theequalities

(27) SpecTk(Tk) = SpecH(Tk), k = 1, 2.

Notice that, since we have the inclusions Tk ⊂ A ⊂ B(H), again by the results inSection 5, we have the inclusions

SpecTk(Tk) ⊃ SpecA(Tk) ⊃ SpecH(Tk), k = 1, 2,

and then the equalities (27) will force the equalities

SpecA(Tk) = SpecH(Tk), k = 1, 2.

Using these equalities, combined with (26), we now have

0 ∈ SpecH(T1) ∪ SpecA(T2),

that is, one of the operators T1 = (λI −X)?(λI −X) or T2 = (λI −X)(λI −X)?

is non-invertible in B(H). By Remark 7.3 this forces the operator λI − X to benon-invertible in B(H), so λ indeed belongs to SpecH(X).

The following result is a generalization of Theorem 7.2.D.

Theorem 7.3 (Gelfand-Naimark). Let H be a Hilbert space, and let A ⊂ B(H)be a commutative C?-subalgebra, which is non-trivial, in the sense that it containsat least one operator X 6= 0.

(i) The character space Char(A) is non-empty;(ii) The Gelfand correspondence ΓA : A → C0

(Char(A)

)is an isometric ?-

isomorphism.

Proof. We start with the followingParticular Case: Assume A contains the identity operator I.

As in the proof of Theorem 7.2, one key step in the proof is contained in thefollowing.

Claim 1: Every character of A is involutive, i.e. one has the equality

(28) Char(A) = Char?(A).

To prove this fact we use Proposition 5.4, which states that the equality (28) isequivalent to the fact that every self-adjoint element X ∈ A has real spectrum, i.e.SpecA(X) ⊂ R. But this is clear, because by Corollary 7.3 we have the equalitySpecA(X) = SpecH(X), and everything follows from Propoistion 7.5.(i).

Claim 2: The space A = RanΓA is dense in C(Char(A)

).

We already know that, since ΓA is a unital algebra homomorphism, A is a subalge-bra of C

(Char(A)

), which contains the unit 1. By Claim 1, we know that in fact A

is a ?-subalgebra of C(Char(A)

), so using the Stone-Weierstrass Theorem, we only

need to show that A separates the points of Char(A). But this is obvious, for if onestarts with two characters γ1 6= γ2, then there exists X ∈ A with γ1(X) 6= γ2(X),so if we put f = X = ΓA(X) ∈ A, we have

f(γ1) = γ1(X) 6= γ2(X) = f(γ2).

Using Claims 1 and 2, we see now that the proof will be finished once weshow that ΓA is isometric. (Among other things this will force A to be closed in


C(Char(A)

), so by Claim 2, this will force the equality C

(Char(A)

)= A.) But

this is obvious, since by Corollary 7.3 we have

‖ΓA(X)‖ = radA(X) = radH(X), ∀X ∈ Aand every X ∈ A is normal, so by Corollary 7.2 the above equalities yield

‖ΓA(X)‖ = ‖X‖, ∀X ∈ A.Having proven the Theorem in the above particular case, we now proceed with

the general case. By excluding the particular case, we assume here that A does notcontain the indentity operator I. Consider then the space

D = A+ CI = X + λI : X ∈ A, λ ∈ C.Since A ⊂ B(H) is a closed linear subspace, it follows (see Proposition 2.1) that Dis also closed. In fact, D is also a ?-subalgebra, so we get the fact that D is a C?-subalgebra of B(H). Since D contains I, by the above particular case it follows thatits Gelfand correspondence ΓD : D → C

(Char(D)

)is an isometric ?-isomorphism.

Notice that, since I 6∈ A, the correspondence

Θ : A 3 (X,λ) 7−→ X + λI ∈ Dis a ?-isomorphism. On the one hand, this ?-isomorphism gives rise to a homeo-morphism Char(D) ' Char(A). On the other hand, we also know (see Section 5)that Char(A) is identified with Char(A) ∪ 0 (this space regarded as a subset in(A∗)1), so we now have a homeomorphism

φ : Char(A) ∪ 0 → Char(D).

Explicitly, the map φ is given as follows. One starts with some γ ∈ Char(A) ∪ 0(so either γ : A → C is a character, or the zero map), and one defines the characterφ(γ) = γ : D → C by

γ(X + αI) = γ(X) + α, ∀X ∈ A, α ∈ C.Furthermore, if we denote the compact space Char(A) ∪ 0 simply by T , thehomeomorphism φ : T → Char(D) gives rise to an isometric ?-isomorphism

Φ : C(Char(D)

)3 f 7−→ f φ ∈ C(T ).

Note now that the composition

(29) Θ = Φ ΓD : D → C(T )

is an isometric ?-isomorphism. Since A is non-trivial, we have dimD ≥ 2, whichamong other things forces dimC(T ) ≥ 2. In particular the space T = Char(A)∪0cannot be a singleton, so Char(A) is indeed non-empty.

Claim 3: If we identify C0

(Char(A)

)=

f ∈ C(T ) : f(0) = 0

, as a closed

sub-algebra of C(T ), then we have the equality

ΓA = Θ∣∣A : A → C0

(Char(A)

).

Indeed, if we start with some element X ∈ A, and we take f = ΓD ∈ C(Char(D)

),

then the function h = Θ(X) = f φ ∈ C0

(Char(A)

)⊂ C(T ) is defined by

h(γ) = (f φ)(γ) = f(γ) = γ(X) = γ(X), ∀ γ ∈ Char(A),

i.e. h = ΓA(X).Since Θ is an isometric ?-isomorphism, by Claim 3, it follows that ΓA is an

isometric ?-homomorphism. To finish the proof, all we need to prove now is the


fact that ΓA is surjective. Start with some f ∈ C0

(Char(A)

). When we view f as

an element in C(T ), using the isomorphism (29), there exists some element Z ∈ Dwith Θ(Z) = f . Of course, when we write Z = X +αI, with X ∈ A and α ∈ C, byClaim 3 we have

f = Θ(X) + αΘ(I) = ΓA(X) + α1,

where 1 ∈ C(T ) is the constant function 1. Since both f and ΓA(X) belong toC0

(Char(A)

), i.e. f(0) = 0 = [ΓA(X)](0), the above equality forces α = 0, so we

actually get f = ΓA(X), and we are done.

Definition. Let H be a Hilbert space, and let T ∈ B(H) be a normal operator.Let T = alg(I, T, T ?) and let FT : Char(T) → SpecH(T ) be the homeomorphismdescribed in Theorem 7.2. This gives rise to an isometric ?-isomorpism

ΘT : C(SpecH(T )

)3 f 7−→ f FT ∈ C

(Char(T)

).

When we compose this ?-isomorphism with the inverse of the Gelfand correspon-dence Γ−1

T : C(Char(T)

)→ T, we get an isometric ?-isomorphism

ΦT = Γ−1T ΘT : C

(SpecH(T )

)→ T.

The map ΦT is called the Continuous Functional Calculus correspondence associatedwith T . The following result summarizes the properties of this construction.

Theorem 7.4 (Properties of Continuous Functional Calculus). Let H be aHilbert space, and let T ∈ B(H) be a normal operator.

A. The continuous functional calcululus correspondence ΦT is the unique mapΦ : C

(SpecH(T )

)→ B(H) with the following properties.

(i) Φ(f + g) = Φ(f)Φ(g), ∀ f, g ∈ C(SpecH(T )

);

(ii) Φ(αf) = αΦ(f), ∀ f ∈ C(SpecH(T )

), α ∈ C;

(iii) Φ(fg) = Φ(f)Φ(g), ∀ f, g ∈ C(SpecH(T )

);

(iv) Φ(f) = Φ(f)?, ∀ f ∈ C(SpecH(T )

);

(v) Φ(1) = I, where 1 ∈ C(SpecH(T )

)denotes the constant function 1;

(vi) Φ(ι) = T , where ι : SpecH(T ) → C is the inclusion map.B. If we denote the C?-algebra alg(I, T, T ?) by T, then he continuous func-

tional correspondence ΦT also satisfies(i) RanΦT = T;(ii) ‖ΦT (f)‖ = ‖f‖, ∀ f ∈ C

(SpecH(T )

).

(iii) SpecH

(ΦT (f)

)= SpecT

(ΦT (f)

)= Ran f , ∀ f ∈ C

(SpecH(T )

).

Proof. Denote for simplicity the compact set SpecH(T ) by K.A. First we show that ΦT has properties (i)-(vi). The fact that ΦT has proper-

ties (i)-(v) is clear. (What these conditions say is simply the fact that ΦT : C(K) →B(H) is a unital ?-homomorphism.) To prove that ΦT also has property (vi), weuse the definition. First of all we know that the function h = ΘT (ι) : Char(T) → Cis defined as ΘT (ι) = ι FT , where FT is the homeomorphism

FT : Char(T) 3 γ 7−→ γ(T ) ∈ K.

It then follows that we have

h(γ) = γ(T ), ∀ γ ∈ Char(T).


This is, of course, the same as h = T = ΓT(T ), the Gelfand transform of T , so weimmediately get

ΦT (ι) = Γ−1T (h) = T.

Next we prove that ΦT is the unique map with properties (i)-(vi). Start withan arbitrary map Φ : C(K) → B(H) with properties (i)-(vi). Consider the space

A =f ∈ C(K) : Φ(f) = ΦT (f)

.

On the one hand, since both Φ and ΦT are continuous (by Corollary 7.1), it followsthat A is a closed in C(K), in the norm topology. On the other hand, since both Φand ΦT satisfy (i)-(vi) it follows that in fact A is a C?-subalgebra of C(K), whichcontains the constant function 1. By (vi) it follows that A also contains ι : K → C,so in particular A separates the points of K. By the Stone-Weierstrass Theorem itfollows that A = C(K), so we indeed have Φ = ΦT .

B. Properties (i) and (ii) are obvious, since by construction ΦT : C(K) → T

is an isometric ?-isomorphism. The first equality in (iii) is clear, by Corollary 7.2.The second equality is a consequence of the fact that, since ΦT is an isomorphism,one has the equality

SpecT

(ΦT (f)

)= SpecC(K)(f), ∀ f ∈ C(K).

Then the desired equality follows from the well known equality

SpecC(K)(f) = Ran f, ∀ f ∈ C(K).

Notation. Given a normal operator T ∈ B(H), and a continuous functionf : SpecH(T ) → C, the operator ΦT (f) ∈ T will be denoted simply by f(T ).

Using this notation, properties (ii) and (iii) from Theorem 7.4.B state that, forevery continuous function f : SpecH(T ) → C, one has:

‖f(T )‖ = max|f(λ)| : λ ∈ SpecH(T )

,(30)

SpecH

(f(T )

)= f

(SpecH(T )

).(31)

The equality (31) is referred to as the Spectral Mapping Formula. By Theorem7.4.B.(i) we also know that

(32) alg(I, T, T ?) =f(T ) : f ∈ C

(SpecH(T )

).

Remarks 7.5. Let T ∈ B(H) be a normal operator.A. During the proof of Theorem 7.2, we defined, for every polynomial in two

variables P ∈ C[s, t], an operator denoted by P (T ?, T ). Using the properties offunctional calculus, the same operator can be defined in the following equivalentway. One considers the continuous function fP : SpecH(T ) → C, defined by

fP (λ) = P (λ, λ), ∀λ ∈ SpecH(T ),

and then one has the equality P (T ?, T ) = fP (T ).B. Conversely, the operators of the form P (T ?, T ), defined algebraically, as in

(19), can be used for constructing the operators of the form f(T ), f ∈ C(SpecH(T )

).

Indeed, by the Stone-Weierstrass Theorem, it follows that the algebra

P = fP : P ∈ C[s, t]


is dense in C(SpecH(T )

). So if one starts with an arbitrary continuous function

f : SpecH(T ) → C, then there exists a sequence (Pn)∞n=1 ⊂ C[s, t], such thatlimn→∞ ‖fPn

− f‖ = 0, i.e.

limn→∞

[max|Pn(λ, λ)− f(λ)| : λ ∈ SpecH(T )

]= 0.

Using the properties of functional caluclus, one has limn→∞ ‖fPn(T )− f(T )‖ = 0,

so in fact we can define f(T ) by

f(T ) = limn→∞

Pn(T ?, T ).

C. The continuous functional calculus is an extension of the entire functionalcalcululs defined in Section 5. If F : C → C is an entire function, i.e. F is definedby a power series

F (ζ) =∞∑

n=0

αnζn, ∀ ζ ∈ C,

with infinite radius of converges, then the operator

F (T ) =∞∑

n=0

αnTn

coincides with f(T ), where f = F∣∣SpecH(T )

. This is a consequence of the factthat, if one defines the sequence of finite sums, as polynomials (in one variable)Pn(t) =

∑nk=0 αkt

k, then we know that

limn→∞

[maxλ∈K

|F (λ)− Pn(λ)|]

= 0,

for every compact subset K ⊂ C. In particular if we consider K = SpecH(T ), wehave limn→∞ ‖f − fPn

‖ = 0, so we get

f(T ) = limn→∞

fPn(T ) = lim

n→∞

n∑k=0

αkTk =

∞∑n=0

αnTn = F (T ).

D. Consider the locally compact space SpecH(T ) r 0, and the commutativeC?-algebra C0

(SpecH(T ) r 0

). This algebra is identified with a C?-subalgebra

in C(SpecH(T )

), according to the two cases below, as follows:

If 0 6∈ SpecH(T ), then C0

(SpecH(T ) r 0

)= C

(SpecH(T )

).

If 0 ∈ SpecH(T ), then C0

(SpecH(T )r0

)=

f ∈ C

(SpecH(T )

): f(0) = 0

.

Using the Stone-Weierstrass Theorem (for locally compact spaces), it followsthat the algebra

P0 = fP : P ∈ C[s, t], P (0, 0) = 0is dense in C0

(SpecH(T ) r 0

). This means that the condition that f belongs

to C0

(SpecH(T ) r 0

)is equivalnet to the existence of a sequence of polynomials

(Pn)∞n=1 ⊂ P0 (i.e. without constant term), such that

limn→∞

[max|Pn(λ, λ)− f(λ)| : λ ∈ SpecH(T )

]= 0,

and in this case one has f(T ) = limn→∞ Pn(T, T ?). This argument proves that theC?-subalgebra C0

(SpecH(T ) r 0

)⊂ C0

(SpecH(T )

)is equivalently described as

C0

(SpecH(T ) r 0

)= alg(ι, ι),


where ι : SpecH(T ) → C is the inclusion map. Consequently one has the equality

alg(T, T ?) =f(T ) : f ∈ C0

(SpecH(T ) r 0

).

which can be regarded as a non-unital version of (32)

The continuous functional calculus is functorial in the following sense.

Proposition 7.8. Let T ∈ B(H) be a normal operator, and let f ∈ C(SpecH(T )

).

(i) The operator X = f(T ) ∈ B(H) is normal.(ii) For every g ∈ C

(SpecH(X)

), one has the equality g(X) = (g f)(T ).

Proof. (i). This is obvious, since f(T ) is an element of the commutativeC?-algebra alg(T, T ?).

(ii). Remark that, since SpecH(X) = SpecH(f(T )) = Ran f , the compositiong f makes sense for every g ∈ C

(SpecH(X)

). Consider the map

Ψ : C(SpecH(X)

)3 g 7−→ (g f)(T ) ∈ B(H).

It is trivial that Ψ has all properties (i)-(vi) from Theorem 7.4.A, so one has theequality Ψ = ΦX , where ΦX is the continuous functional calculus correspondencefor X.

Comment. The results of Proposition 7.5.(i) and Proposition 7.6 are true ifB(H) is replaced with a unital C?-algebra.

The following technical result is extremely useful.

Lemma 7.3. Let H be a Hilbert space, and let (Tλ)λ∈Λ ⊂ B(H) be a net withthe following properties:

• Tλ is positive, for each λ ∈ Λ;• for any λ, µ ∈ Λ with λ µ, the operator Tλ − Tµ is positive;• supλ∈Λ ‖Tλ‖ <∞.

Then the net (Tλ)λ∈Λ also has the following properties.(i) the net (‖Tλ‖)λ∈Λ ⊂ [0,∞) is increasing, in the sense that, whenever

λ µ, one has ‖Tλ‖ ‖Tµ‖.(ii) For every ξ ∈ H, the net (Tλξ)λ∈Λ ⊂ H is convergent.(iii) The map T : H → H defined by

Tξ = limλ∈Λ

Tλξ, ∀ ξ ∈ H,

is a positive bounded operator, with ‖T‖ = limλ∈Λ ‖Tλ‖.

Definition. Let H1 and H2 be Hilbert spaces. An operator T ∈ B(H1,H2)is said to be a partial isometry, if its restriction T

∣∣(Ker T )⊥

: (KerT )⊥ → H2 is anisometry.

Proposition 7.9. For an operator V ∈ B(H1,H2), the following are equiva-lent:

(i) V is a partial isometry;(i?) V ? is a partial isometry;(ii) V ?V ∈ B(H1) is a projection;

(ii?) V V ? ∈ B(H2) is a projection.


Proof. C. Consider the closed subspaces X1 = (KerV )⊥ ⊂ H1 and X2 =(KerV ?)⊥ ⊂ H2. By Proposition 7.2 we know that X1 = RanV ? and X2 = RanV .Denote by Pk ∈ B(Hk) the orthogonal projection on Xk, k = 1, 2. (i) ⇒ (i?).Assume V is a partial isometry, and let us prove that V ? is also a partial isometry.We must prove that V ?

∣∣X2

: X2 → H1 is an isometry. This is shown indirectly byexamining the operator W = V

∣∣X1

: X1 → H2. On the one hand, we clearly havethe inclusion RanW ⊂ RanW . On the other hand, we also know that, since forevery vector ξ ∈ H1 we have

ξ − P1ξ ∈ X⊥1 =[(KerV )⊥

]⊥= KerV = KerV,

we haveV ξ = V (ξ − P1ξ) + V P1ξ = V P1ξ = WP1ξ, ∀ ξ ∈ H1,

so in particular we have RanV ⊂ RanW . This gives the equality

(33) RanW = RanV.

Finally, we know that W is an isometry. In particular, it follows that RanW isclosed, and then the equality (33) gives

RanW = RanW = RanV = (KerV ?)⊥ = X2.

This proves that, when regarded as an operator W : X1 → X2, W is unitary.

7. Operator Theory on Hilbert spaces

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Transcript of 7. Operator Theory on Hilbert spaces