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					                  7. Operator Theory on Hilbert spaces

    In this section we take a closer look at linear continuous maps between Hilbert
spaces. These are often called bounded operators, and the branch of Functional
Analysis that studies these objects is called “Operator Theory.” The standard
notations 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 space
L(H, H) is simply denoted by B(H).
     Given T ∈ B(H1 , H2 ) and S ∈ B(H2 , H3 ), their composition S◦T ∈ B(H1 , H3 )
will be simply denoted by ST .
     We know that B(H) is a unital Banach algebra. This Banach algebra will
be our main tool used for investigating bounded operators. The general theme of
this section is to study the interplay between the operator theoretic properties of
a bounded linear operator, say T : H → H, and the algebraic properties of T as
an element of the Banach algebra B(H). Some of the various concepts associated
with bounded linear operators are actually defined using this Banach algebra. For
example, 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 product
topology. 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)           min C ≥ 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
                                                                            1
with xn , yn ≤ 1, but such that |φ(xn , yn )| ≥ n2 . If we take vn = n xn and
       1                                                    1
wn = n yn , then on the one hand we have vn , wn ≤ n , ∀ n ≥ 1, which forces
limn→∞ (vn , wn ) = (0, 0) in X × Y, so by (iii) we have limn→∞ φ(vn , wn ) = 0. On
the other hand, we also have
                                         |φ(xn , yn )|
                       |φ(vn , wn )| =                 ≥ 1, ∀ n ≥ 1,
                                             n2
                                             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 can
                                                                   1                1
assume both x and y are non-zero. Consider the vectors v = x x and w = 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. Let
C ≥ 0 is as in (iv). Let (xn )n→∞ ⊂ X and (yn )n→∞ ⊂ Y be convergent sequences
with 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 above
inequality, 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 been
obtained when we proved the implication (iii) ⇒ (iv).
    Notation. With the notations above, the number defined in (iii), which is
also 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) φ −→ 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 be
a sesquilinear map. The following are equivalent (equip H1 × H2 with the product
topology):
      (i) φ is continuous;
     (ii) there exists T ∈ B(H1 , H2 ), such that
                      φ(ξ1 , ξ2 ) = (T ξ1 |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        ζ −→ 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 unique
       ˜
vector 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 + (˜|ζ)H2 = φ(ξ, ζ) + φ(η, ζ) = φ(ξ + η, ζ) = φξ+η (ζ), ∀ ζ ∈ H2 ,
                  η
                                                                        ˜ ˜
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         ˜
                                               ξ −→ ξ ∈ H2 ,
with
                        φ(ξ, ζ) = (T ξ|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 is
obvious.
    (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 the
Cauchy-Bunyakovski-Schwartz inequality we have
|φ(ξ1 , ξ2 )| = |(T ξ1 |ξ2 )H2 | ≤ T ξ1 · 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 .
                            §7. Operator Theory on Hilbert spaces                    303


Moreover, S ∈ B(H2 , H1 ) is unique, and satisfies S = T . This follows from
the 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       (ξ1 , ξ2 ) −→ (T ξ1 |ξ2 )H2 ∈ C,
then there exists a unique operator S ∈ B(H2 , H1 ), such that
(5)                 (T ξ1 |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 . By
the above Remark, for any two vectors ξ1 ∈ H1 , ξ2 ∈ H2 , we have the identities
                                 (T ξ1 |ξ2 )H2 = (ξ1 |T ξ2 )H1 ,
                                 (ξ2 |T ξ1 )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
(6)                          T    = T , ∀ T ∈ B(H1 , H2 );
(7)                        (T ) = T, ∀ T ∈ B(H1 , H2 );
(8)                    (S + T ) = S + T ∀ S, T ∈ B(H1 , H2 );
(9)                               ¯
                          (λT ) = λT ∀ T ∈ B(H1 , H2 ), λ ∈ C.
        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. We
denote by X the operator whose adjoint is the left hand side, we denote by Y
the operator in the right hand side, so we must show X = Y , and we prove this
equality 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 pretty
obvious that
            (Xξ|η) = (Sξ + T ξ|η) = (Sξ|η) + (T ξ|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 gives
the fact that B(H) is a unital involutive Banach algebra. (See Section 5 for the
terminology.) This will be crucial for the development of the theory.
       Another important set of results deals with the kernel and the range.
304                CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


    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) Ker T = (Ran T )⊥ ;
     (ii) Ran T = (Ker T )⊥ .
    Proof. (i). If we start with some vector η ∈ Ker T , then for every ξ ∈ H1 ,
we have
                            (η|T ξ)H2 = (T η|ξ)H1 = 0,
thus proving that η ⊥ T ξ, ∀ ξ ∈ H1 , i.e. η ∈ (Ran T )⊥ . This proves the inclusion
                                       Ker T = (Ran T )⊥ .
To prove the inclusion in the other direction, we start with some vector η ∈
Ker T = (Ran T )⊥ , and we prove that T η = 0. This is however pretty clear
since we have η ⊥ (T T η), i.e.
                       0 = (η|T T 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 :
                                                       ⊥
                        Ran T = [Ran T ]⊥                  = (Ker T )⊥ .
   Example 7.1. Let n ≥ 1 be some integer, and consider the Hilbert space Cn ,
whose inner product is the standard one:
                       n
             (ξ|η) =         ¯
                             ξk ηk , ∀ ξ = (ξ1 , . . . , ξn ), η = (η1 , . . . , ηn ) ∈ Cn .
                       k=1
The Banach algebra B(Cn ) is obviously identified with the algebra Matn×n (C) of
n × n matrices with complex coefficients. The adjoint operation then corresponds
to a (familiar) operation in linear algebra. For A ∈ Matn×n (C), say A = [ajk ]n
                                                                               j,k=1 ,
one takes A = [bjk ]n  j,k=1  to be the conjugate transpose of A, i.e. the b’s are
defined as bjk = akj . A similar identification works with B(Cm , Cn ), identified
                   ¯
with 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 = T T .
      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
                                     (T ξ|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 )∞ of normal operators, which converges
                                         n=1
(in norm) to some T ∈ B(H), then (Tn )∞ converges to T , and since the multi-
                                         n=1
plication map
                       B(H) × B(H) (X, Y ) −→ XY ∈ B(H)
is continuous, have T T = limn→∞ Tn Tn and T T = limn→∞ Tn Tn , so we imme-
diately get T T = T T .
     C. For T ∈ B(H), the following are equivalent (see Remark 3.1):
        • T is self-adjoint;
                           §7. Operator Theory on Hilbert spaces                       305


         • the sesquilinear map
                         φT : H × H         (ξ, η) −→ (T ξ|η)H ∈ C
          is sesqui-symmetric, i.e. (T ξ|η) = (T η|ξ), ∀ ξ, η ∈ H;
       • (T ξ|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 observation
that, if (Tn )∞ converges to some T , then we have
              n=1

                           (T ξ|ξ) = lim (Tn ξ|η), ∀ ξ ∈ H.
                                       n→∞

So if for example all Tn ’s are self-adjoint, then this proves that (T ξ|ξ) ∈ R, ∀ ξ, ∈ H,
so T is selfadjoint. Likewise, if all Tn ’s are positive, then (T ξ|ξ) ≥ 0, ∀ ξ ∈ H, so T
is positive.
     E. Given Hilbert spaces H1 and H2 , and an operator T ∈ B(H1 , H2 ), it follows
that the operators T T ∈ B(H1 ) and T T ∈ B(H2 ) are positive. This is quite
obvious, since
                     (T T ξ|ξ) = (T ξ|T ξ) = T ξ      2
                                                          ≥ 0, ∀ ξ ∈ H1 ,
                   (T T 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) ≥ (T ξ|ξ), ∀ ξ ∈ H.
The transitivity and reflexivity properties are clear. For the antisymmetry, one
must show that if T ≥ S and S ≥ T , then S = T . This is however clear, because
the difference X = S − T is self-adjoint, and satisfies
(11)                              (Xζ|ζ) = 0, ∀ ζ ∈ H.
Using polarization (see Section 3), we have
                                       3
                                  1
                       (Xξ|η) =             i−k X(ξ + ik η) ξ + ik η ,
                                  4
                                      k=0

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.
306                CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


      Exercise 2 ♦ . Prove that for an operator T ∈ B(H) the following are equivalent:
        • T is normal;
        • T ξ = T ξ , ∀ ξ ∈ H.
    Exercise 3*. For a self-adjoint operator, Theorem 7.1 has a slightly improved
version. Prove that if T ∈ B(H) be self-adjoint, then one has the equality
                          T = sup |(T ξ|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 first
that
(∗)                               |(T ζ|ζ) ≤ M ζ   2
                                                       , ∀ ζ ∈ H.
Use Theorem 7.1 which gives
                          T = sup |(T ξ|η)| : ξ, η ∈ H,         ξ , η ≤1 .
This reduces the problem to proving that, whenever ξ , η ≤ 1, one has the inequality
                                         |(T ξ|η)| ≤ M.

Prove this inequality, under the extra assumption that (T ξ|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 linear
subspace. Then the orthogonal projection PX : H → H is a positive bounded
operator. Indeed, if one starts with some ξ ∈ H, using the fact that ξ − PX ξ ∈ X⊥ ,
it follows that (ξ − PX ξ) ⊥ PX ξ, so we have
                       0 = (PX ξ|ξ − PX ξ) = (P ξ|ξ) − (PX ξ|PX ξ),
which proves that
                     (PX ξ|ξ) = (PX ξ|PX ξ) = PX ξ          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 orthogonal
           projection 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 = Ran Q. First of all,
we must show that X is closed. This is pretty obvious, since the equality Q2 = Q
gives the equality X = Ker(I − Q). To prove that Q = PX , we must prove two
things:
     (a) Qξ = ξ, ∀ ξ ∈ X;
      (b) Qξ = 0, , ∀ ξ ∈ X⊥ .
                                 §7. Operator Theory on Hilbert spaces                                  307


The first property is clear, since X = Ker(I − Q), To prove the second property, we
use Proposition 7.2 to get
                             X⊥ = (Ran Q)⊥ = Ker Q = Ker Q.
    Convention. An operator Q ∈ B(H) with Q2 = Q = Q will be simply called
a 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 normal
and self-adjoint are already described in algebraic terms. Besides these, there are
other 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 T ξ = 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) T T = IH2 .
     C. For an operator U ∈ B(H1 , H2 ), the following are equivalent:
         (i) U is unitary;
        (ii) U U = IH1 and U U = IH2 .
    Proof. A. (i) ⇒ (ii). Using polarization, applied to the sesquilinear form
                             φ : H1 × H1        (ξ, η) −→ (T T ξ|η) ∈ C,
it follows that, for every ξ, η ∈ H, one has the equalities
                    3                                         3
               1                                      1
   φ(ξ, η) =             i−k φ(ξ + ik , ξ + ik η) =               i−k T T (ξ + ik η) ξ + ik η =
               4                                      4
                   k=0                                    k=0
                    3                                                       3
               1                                                      1
           =             i−k T (ξ + ik η) T (ξ + ik η) =                         i−k T (ξ + ik η) 2 .
               4                                                      4
                   k=0                                                     k=0

Using the fact that T is an isometry, and polarization again (for the inner product),
the above computation continues with
                                  3                                    3
                             1                                    1
             φ(ξ, η) =                 i−k T (ξ + ik η)   2
                                                              =             i−k ξ + ik η    2
                                                                                                =
                             4                                    4
                                 k=0                                  k=0
                                  3
                             1
                         =             i−k ξ + ik η ξ + ik η = (ξ|η).
                             4
                                 k=0
308                  CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


Since we now have
                                (T T ξ|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
                         = (T ξ|T ξ) = (T T ξ|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 above
equality actually forces U −1 = U , so we also get U U = U U −1 = IH2 .
    (ii) ⇒ (i). Assume U U = IH1 and U U = IH2 , and let us prove that U is a
unitary. On the one hand, these two equalities prove that U is both left and right
invertible, so U is bijective. On the other hand, by part A, it follows that U is an
isometry, so U is indeed unitary.

      In the study of bounded linear operators, positivity is an essential tool. This
is 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 T T ≥ αIH2 .
    Proof. (i) ⇒ (ii). Assume T is invertible. Then T : H2 → H1 is also
invertible, with inverse (T )−1 = (T −1 ) . Define the number α = T −1 −2 . Using
Proposition 7.1, we also have α = (T )−1 −2 . Remark now that if one starts with
some ξ ∈ H1 , then the norm inequality
                                ξ = T −1 (T ξ) ≤ T −1 · T ξ
gives T ξ ≥ T −1         −1
                              · ξ , so taking squares we get
            (T T ξ|ξ)H1 = (T ξ|T ξ)H2 = T ξ        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 T T − αIH2 ∈ B(H2 ) is
proven the exact same way.
    (ii) ⇒ (i). Assume there exists α > 0 such that T T ≥ αIH1 and T T ≥ αIH2 ,
and let us show that T is invertible. First of all, using the inequality T T ≥ αIH1 ,
we have
                   T ξ 2 = (T ξ|T ξ) = (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 Proposition
2.1, this also proves that Ran T is a Banach space, when equipped with the norm
coming from H2 , i.e. Ran T is closed in H2 . In particular, by Proposition 7.2 we
have
(12)                             Ran T = Ran T = (Ker T )⊥ .
                          §7. Operator Theory on Hilbert spaces                 309


Notice that exactly as above, we also have
                                     √
                            T η ≥ α η , ∀ η ∈ H2 ,
so T is injective as well, which means that Ker T = {0}, so the equality (12) gives
Ran T = H2 , so T is also surjective.

    Remark 7.3. With the notations above, the condition that T is invertible is
also equivalent to the condition
     (iii) both operators T T ∈ B(H1 ) and T T ∈ B(H2 ) are invertible.
Indeed, if T is invertible, then so is T , so the products T T and T T are also
invertible. Conversely, if both T T and T T are invertible, and if we put X =
(T T )−1 and Y = (T T )−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 every
complex number λ ∈ C R, the operator X = λI −T is invertible. Write λ = a+ib,
with a, b ∈ R with b = 0. We are going to apply Lemma 7.2, so we need to consider
the operators X X and XX . It turns out that
                      X X = XX = |λ|2 I − 2(Re λ)T + T 2 ,
so all we need is the existence of a constant α > 0, such that X X ≥ αI. But this
is clear, since
                 X X = (a2 + b2 )I − 2aT + T 2 = b2 I + (aI − T )2 ,
and the positivity of (aI −T )2 = (aI −T ) (aI −T ) (see Remark 7.2.E) immediately
gives X X ≥ b2 I.
    (ii). By part (i) we only need to prove that, for every number a ∈ (−∞, 0), the
operator X = aI − T is invertible. As before, we have
                          X X = XX = a2 I − 2aT + T 2 ,
and then the positivity of −2aT and of T 2 = T T (see Remark 7.2.F), forces
X X ≥ a2 I. Since a = 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 generality
we can assume that T = 1. Since radH (T ) ≤ T = 1 (see Section 5), all we have
to prove is the fact that SpecH (T ) contains one of the numbers ±1. Equivalently,
310                CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


we must prove that either −I − T or I − T is non-invertible. Consider (see Remark
7.2.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 fact
by contradiction. Assume that X − I is invertible, so by Lemma 7.2 there exists
some 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 4 ξ|ξ) = T 2 ξ    2
                                      ≤ ( T · T ξ )2 ≤ T ξ        2
                                                                      = (T 2 ξ|ξ), ∀ ξ ∈ H,
which reads:
                                    X ≥ X 2 ≥ 0.
In particular this gives (I − X) − (X − I)2 = X − X 2 ≥ 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 2 ξ|ξ) = T ξ 2 , ∀ ξ ∈ H,
which gives                  √
                      T ξ ≤ 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 the
development of the theory.
   Proposition 7.7. Let H1 and H2 be Hilbert spaces. For every operator T ∈
B(H1 , H2 ), one has the identity
                                                      2
(14)                                      T T = T         .
       Proof. Fix T ∈ B(H1 , H2 ). Consider the sesquilinear map
                           φ : H1 × H1     (ξ, η) −→ (T T ξ|η)H1 ∈ C.
By Theorem 7.1, we know that T T                 =   φ . Notice however that, for every
ξ ∈ H1 with ξ ≤ 1, one has
                      φ ≥ |φ(ξ, ξ)| = |(T T ξ|ξ)| = |(T ξ|T ξ)| = T ξ 2 ,
so we get
                           φ ≥ sup       T ξ : ξ ∈ H1 , ξ ≤ 1 = T ,
                                                       2
thus proving the inequality T T = φ ≥ T                       . 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 Banach
algebra. Then every -homomorphism Φ : A → B(H) is contractive, in the sense
that one has the inequality
(15)                                  Φ(a) ≤ a , ∀ a ∈ A.
                         §7. Operator Theory on Hilbert spaces                     311


     Proof. Fix a -homomorphism Φ : A → B(H). We can assume that A is
                                                                     ˜
unital, and Φ(1) = I. (If not, we work with the unitized algebra A, which is
again an involutive Banach algebra, and with the map Φ : A → B(H) defined
                                                          ˜   ˜
    ˜
by φ(a, α) = Φ(a) + αI, a ∈ A, α ∈ C, which clearly defines a -homomorphism
           ˜
satisfying Φ(1) = I.)
     To prove (15) we start with an arbitrary element a ∈ A, and we consider the
element b = a a. On the one hand, the operator Φ(b) = Φ(a) Φ(a) ∈ B(H) is
obviously 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
                                            2
(18)                                 Φ(a)       ≤ 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 result
suggests 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 develop
an entire machinery, which enables one to do “algebraic” Operator Theory. At the
core of this development is the following concept.
     Definition. An involutive Banach algebra A is called a C -algebra, if its norm
satisfies the C -norm property, i.e. one has
                               a a = a 2 , ∀ a ∈ A.
Remark that if A is unital, with unit 1 = 0, then 1 = 1, and the above condition
forces 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 fact
that, 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 equipped
with the natural involution f = f .¯
312                CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


     Remarks 7.4. A. The types normal and self-adjoint make sense for elements
in 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 Proposition
7.4 as definitions.
     D. The characterization of invertible elements given in Remark 7.3 is valid if
B(H) is replaced with an arbitrary unital C -algebra. More explicitly, if A is a
unital 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 this
is true even when A is only a unital -algebra.)
     The following result analyzes the Gelfand correspondence discussed in Section
5, in an important particular case.
    Theorem 7.2. Let H be a Hilbert space, and let T ∈ B(H) be a normal
operator. 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) γ −→ 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 following
notation. If P (s, t) is a polynomial in two variables, say
                                             N
                               P (s, t) =           αmn sm tn ,
                                            m,n=0

with complex coefficients, we define the operator
                                            N
(19)                        P (T , T ) =           αmn (T )m T n ,
                                           m,n=0

with the convention that (T )0 = T 0 = I. We denote by C[s, t] the algebra of
polynomials in two variables. With these notations we have the following.
      Claim 1: The correspondence
(20)                     Π : C[s, t]   P −→ P (T , T ) ∈ B(H)
            gives rise to a surjective unital algebra homomorphism Π : C[s, t] → T0 .
                          §7. Operator Theory on Hilbert spaces                     313


It is obvious that Π is linear, and satisfies Π(1) = I, where 1 denotes the con-
                                                   N          m n
stant polynomial 1. Remark that, if P (s, t) =     m,n=0 αmn s t and Q(s, t) =
  K
     βk βk sk t , are polynomials in two variables, then using the fact that T
  k, =0
commutes with T (i.e. T T = T T ), we get
                                      N    K
            P (T , T )Q(T , T ) =                 αmn βk (T )m T n (T )k T =
                                    m,n=0 k, =0
                                      N    K
                               =                  αmn βk (T )m+k T n+ ,
                                    m,n=0 k, =0

so if we consider the polynomial P Q, we have
                         P (T , T )Q(T , T ) = (P Q)(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 Ran Π ⊂ T0 . On the other hand, since Π is
an algebra homomorphism, it follows that Ran Π is a subalgebra of T0 . Finally,
since Ran Π ⊃ {I, T, T }, it follows that we have in fact the equality Ran Π =
alg({I, T, T }) = T0 .
     Having proven Claim 1, let us observe that this already gives the fact that T0 is
commutative (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] also
carries an involution defined as follows. If P ∈ C[s, t] is a polynomial, say P (s, t) =
   N            m n                                       #          N            n m
   m,n=0 αmn s t , then one defines the polynomial P (s, t) =                ¯
                                                                     m,n=0 αmn s t .
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 = Ran Π is a -subalgebra of B(H). Taking the closure, it follows that
T is a C -subalgebra of B(H).
     B-E. In order to prove these statements, we are going to consider first a larger
C -subalgebra, which will have properties A-E, and we will eventually prove that
this larger algebra in fact coincides with T. Consider the set
                 M = Q ∈ C[s, t] : P (T , T ) invertible in B(H) ,
and define
               A0 = P (T , T )[Q(T , T )]−1 : P ∈ C[s, t], Q ∈ M .
Remark that, if Q1 , Q2 ∈ M, then Q1 Q2 ∈ M. One also has the implication Q ∈
M ⇒ Q# ∈ M. A simple computation shows that if X1 , X2 ∈ A0 are represented
as Xk = Pk (T , T )[Qk (T , T )]−1 with Pk ∈ C[s, t] and Qk ∈ M, k = 1, 2, then one
has the equalities
                  X1 X2 = (P1 P2 )(T , T )[(Q1 Q2 )(T , T )]−1 ,
               X1 + X2 = (P1 Q2 + P2 Q1 )(T , T )[(Q1 Q2 )(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).
314                CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


     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 the
inclusion 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 contains
I, we always have the inclusion
                         SpecH (X) ⊂ SpecA (X), ∀ X ∈ A.
Secondly we observe that, if we start with some operator X ∈ A0 , written as
X = P (T , T )[Q(T , T )]−1 , with P ∈ C[s, t] and Q ∈ M, then A contains all
operators of the form (λI − X)−1 , with λ ∈ C SpecH (X). This is due to the fact
that for every λ ∈ C we can write
             λI − X = Pλ (T , T )[Q(T , T )]−1 = [Q(T , T )]−1 Pλ (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 by
construction. In particular this means that λI − X is invertible in A, for every
λ ∈ C SpecH (X), so we have the inclusion
                          C    SpecH (X) ⊂ C      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 the
equality
(22)                          γ(X ) = γ(X), ∀ X ∈ A.
Using continuity, it suffices to prove that (22) holds only for X ∈ A0 . Start with
some arbitrary operator X ∈ A0 , and consider the operators X1 = 1 (X + X ) and
                                                                  2
X2 = 2i (X − X ). Since A0 is a -subalgebra of B(H(, we know that both X1 and
      1

X2 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, combined
with 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 that
γ(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.
                         §7. Operator Theory on Hilbert spaces                  315


Recall that the Gelfand correspondence is defined by
                                       ˆ
                             ΓA (X) = X, ∀ X ∈ A,
      ˆ
where X : Char(A) → C is the Gelfand transform of X, defined by
                          ˆ
                          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)                            ˆ
                               X = X , ∀ X ∈ A0 .
Fix X ∈ A0 . Since A is a C -algebra, we have the equality X 2 = X X . Notice
that the element Y = X X still belongs to A0 . On the one hand, by Claim 3 we
have 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.
                                                     ˆ
By Corollary 5.4 we know however that radA (Y ) = Y , so we now have
(25)                                  X   2     ˆ
                                              = Y .
                                           ˆ    ˆ ˆ
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      ˆ ˆ    ˆ
                                     = (X) X = X        2
                                                            ,
and then (23) is clear.
    Let us show now that A has property E. We consider the Gelfand transform
      ˆ
GT = T of T , which defines a function GT ∈ C Char(A) , given by
                         GT : Char(A)         γ −→ 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 Ran GT = SpecA (T ), so the
only other thing to prove is the fact that GT is injective. (We use here the fact
that Char(A) is compact.) What we have to prove is the fact that if two characters
γ1 , γ2 ∈ Char(A) have the property that γ1 (T ) = γ2 (T ), then γ1 = γ2 . By Claim
3, 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 )   γ2 P (T , T )
          γ1 P (T , T )[Q(T , T )]−1 =                  =               =
                                          γ1 Q(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 .
316                CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


    We now proceed with the proof of the fact that the “old” C -algebra T has
properties B-E. As mentioned before this will be done by comparing it with A. We
know 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) by
A. On the one hand, since ΓA is an isometric -homomorphism, it follows that A
is a C -subalgebra of C(K). On the other hand, it is clear that A contains the
constant function 1 = ΓA (I). Finally, A separates the points of K, which means
that whenever γ1 , γ2 ∈ K are such that γ1 = γ2 , then there exists f ∈ A with
f (γ1 ) = f (γ2 ). This is however clear, because if we take GT = ΓA (T ) ∈ A, then
by Claim 4, we know that we must have GT (γ1 ) = GT (γ2 ). We now use the
Stone-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. Now
we are done, since A has properties B-E.

    Theorem 7.2 has a multitude of applications. The first one is a generalization
of Proposition 7.6.
    Corollary 7.2 (Spectral Radius Formula for normal operators). Let H be a
Hilbert 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 know
that 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, so
we 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 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 operator
X ∈ 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 the
operators T1 and T2 is non-invertible in A. One way to rewrite this is:
(26)                          0 ∈ SpecA (T1 ) ∪ SpecA (T2 ).
                           §7. Operator Theory on Hilbert spaces                 317


Since T1 and T2 are self-adjoint, they are normal. In particular, if one considers
the C -algebras Tk = alg({I, Tk , Tk }), k = 1, 2, then by Theorem 7.2 we have the
equalities
(27)                      SpecTk (Tk ) = SpecH (Tk ), k = 1, 2.
Notice that, since we have the inclusions Tk ⊂ A ⊂ B(H), again by the results in
Section 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 be
non-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 contains
at least one operator X = 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 following
     Particular 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 the
following.
       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) is
equivalent 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 equality
SpecA (X) = SpecH (X), and everything follows from Propoistion 7.5.(i).
       Claim 2: The space A = Ran ΓA 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 one
starts with two characters γ1 = γ2 , then there exists X ∈ A with γ1 (X) = γ2 (X),
                  ˆ
so if we put f = X = ΓA (X) ∈ A, we have
                           f (γ1 ) = γ1 (X) = γ2 (X) = f (γ2 ).
    Using Claims 1 and 2, we see now that the proof will be finished once we
show that ΓA is isometric. (Among other things this will force A to be closed in
318              CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


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 ∈ A
and 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 not
contain 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 D
is 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 that
its Gelfand correspondence ΓD : D → C Char(D) is an isometric -isomorphism.
Notice that, since I ∈ A, the correspondence
                                ˜
                           Θ : A (X, λ) −→ X + λI ∈ D
is 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 character
φ(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 , the
homeomorphism φ : T → Char(D) gives rise to an isometric -isomorphism
                      Φ : C Char(D)      f −→ 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 dim D ≥ 2, which
among other things forces dim C(T ) ≥ 2. In particular the space T = Char(A)∪{0}
cannot 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 (X) = γ(X), ∀ γ ∈ Char(A),
                                   γ     ˜
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
                         §7. Operator Theory on Hilbert spaces                    319


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 ∈ D
with Θ(Z) = f . Of course, when we write Z = X + αI, with X ∈ A and α ∈ C, by
Claim 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 to
C0 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 homeomorphism
described in Theorem 7.2. This gives rise to an isometric -isomorpism
                ΘT : C SpecH (T )      f −→ f ◦ FT ∈ C Char(T) .
When we compose this -isomorphism with the inverse of the Gelfand correspon-
dence Γ−1 : C Char(T) → T, we get an isometric -isomorphism
       T

                       ΦT = Γ−1 ◦ ΘT : C SpecH (T ) → T.
                             T

The map ΦT is called the Continuous Functional Calculus correspondence associated
with T . The following result summarizes the properties of this construction.
    Theorem 7.4 (Properties of Continuous Functional Calculus). Let H be a
Hilbert space, and let T ∈ B(H) be a normal operator.
      A. The continuous functional calcululus correspondence ΦT is the unique map
         Φ : 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(f g) = 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) Ran ΦT = 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), we
use the definition. First of all we know that the function h = ΘT (ι) : Char(T) → C
is defined as ΘT (ι) = ι ◦ FT , where FT is the homeomorphism
                         FT : Char(T)      γ −→ γ(T ) ∈ K.
It then follows that we have
                           h(γ) = γ(T ), ∀ γ ∈ Char(T).
320              CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


                                    ˆ
This is, of course, the same as h = T = ΓT (T ), the Gelfand transform of T , so we
immediately get
                                ΦT (ι) = Γ−1 (h) = T.
                                          T

    Next we prove that ΦT is the unique map with properties (i)-(vi). Start with
an 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 follows
that A is a closed in C(K), in the norm topology. On the other hand, since both Φ
and ΦT satisfy (i)-(vi) it follows that in fact A is a C -subalgebra of C(K), which
contains 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 it
follows 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 function
f : 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, for
every continuous function f : SpecH (T ) → C, one has:
(30)                   f (T ) = max |f (λ)| : λ ∈ SpecH (T ) ,
(31)                       SpecH f (T ) = f SpecH (T ) .

The equality (31) is referred to as the Spectral Mapping Formula. By Theorem
7.4.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 of
functional calculus, the same operator can be defined in the following equivalent
way. 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]}
                           §7. Operator Theory on Hilbert spaces                          321


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 )∞ ⊂ C[s, t], such that
                                                         n=1
limn→∞ fPn − f = 0, i.e.
               lim             ¯
                      max{|Pn (λ, λ) − f (λ)| : λ ∈ SpecH (T )} = 0.
               n→∞

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 ) = lim Pn (T , T ).
                                          n→∞

    C. The continuous functional calculus is an extension of the entire functional
calcululs defined in Section 5. If F : C → C is an entire function, i.e. F is defined
by a power series
                                          ∞
                              F (ζ) =         αn ζ n , ∀ ζ ∈ C,
                                        n=0
with infinite radius of converges, then the operator
                                                 ∞
                                     F (T ) =         αn T n
                                                n=0

coincides with f (T ), where f = F Spec (T ) . This is a consequence of the fact
                                        H
that, if one defines the sequence of finite sums, as polynomials (in one variable)
           n
Pn (t) = k=0 αk tk , then we know that
                            lim     max |F (λ) − Pn (λ)| = 0,
                           n→∞      λ∈K

for every compact subset K ⊂ C. In particular if we consider K = SpecH (T ), we
have limn→∞ f − fPn = 0, so we get
                                                n                ∞
           f (T ) = lim fPn (T ) = lim                αk T k =         αn T n = F (T ).
                     n→∞               n→∞
                                                k=0              n=0

    D. Consider the locally compact space SpecH (T ) {0}, and the commutative
C -algebra C0 SpecH (T ) {0} . This algebra is identified with a C -subalgebra
in C SpecH (T ) , according to the two cases below, as follows:
    If 0 ∈ SpecH (T ), then C0 SpecH (T ) {0} = C SpecH (T ) .
    If 0 ∈ SpecH (T ), then C0 SpecH (T ) {0} = f ∈ C SpecH (T ) : f (0) = 0 .
    Using the Stone-Weierstrass Theorem (for locally compact spaces), it follows
that the algebra
                        P0 = {fP : P ∈ C[s, t], P (0, 0) = 0}
is dense in C0 SpecH (T ) {0} . This means that the condition that f belongs
to C0 SpecH (T ) {0} is equivalnet to the existence of a sequence of polynomials
(Pn )∞ ⊂ P0 (i.e. without constant term), such that
     n=1

               lim             ¯
                      max{|Pn (λ, λ) − f (λ)| : λ ∈ SpecH (T )} = 0,
               n→∞

and in this case one has f (T ) = limn→∞ Pn (T, T ). This argument proves that the
C -subalgebra C0 SpecH (T ) {0} ⊂ C0 SpecH (T ) is equivalently described as

                        C0 SpecH (T )           {0} = alg({ι, ¯}),
                                                              ι
322                CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS


where ι : SpecH (T ) → C is the inclusion map. Consequently one has the equality
                 alg({T, T }) = f (T ) : f ∈ C0 SpecH (T )      {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 commutative
C -algebra alg({T, T }).
     (ii). Remark that, since SpecH (X) = SpecH (f (T )) = Ran f , the composition
g ◦ f makes sense for every g ∈ C SpecH (X) . Consider the map
                    Ψ : C SpecH (X)        g −→ (g ◦ f )(T ) ∈ B(H).
It is trivial that Ψ has all properties (i)-(vi) from Theorem 7.4.A, so one has the
equality Ψ = ΦX , where ΦX is the continuous functional calculus correspondence
for X.

   Comment. The results of Proposition 7.5.(i) and Proposition 7.6 are true if
B(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 with
the 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 (Tλ ξ)λ∈Λ ⊂ H is convergent.
     (iii) The map T : H → H defined by
                               T ξ = lim Tλ ξ, ∀ ξ ∈ H,
                                     λ∈Λ

           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 )⊥ : (Ker T )⊥ → H2 is an
isometry.
     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.
                          §7. Operator Theory on Hilbert spaces                 323


    Proof. C. Consider the closed subspaces X1 = (Ker V )⊥ ⊂ H1 and X2 =
(Ker V )⊥ ⊂ H2 . By Proposition 7.2 we know that X1 = Ran V and X2 = Ran V .
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 by
examining the operator W = V X1 : X1 → H2 . On the one hand, we clearly have
the inclusion Ran W ⊂ Ran W . On the other hand, we also know that, since for
every vector ξ ∈ H1 we have
                                                ⊥
                  ξ − P1 ξ ∈ X⊥ = (Ker V )⊥
                              1                     = Ker V = Ker V,
we have
               V ξ = V (ξ − P1 ξ) + V P1 ξ = V P1 ξ = W P1 ξ, ∀ ξ ∈ H1 ,
so in particular we have Ran V ⊂ Ran W . This gives the equality
(33)                               Ran W = Ran V.
Finally, we know that W is an isometry. In particular, it follows that Ran W is
closed, and then the equality (33) gives
                   Ran W = Ran W = Ran V = (Ker V )⊥ = X2 .
This proves that, when regarded as an operator W : X1 → X2 , W is unitary.

				
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