Chapter 8 Hermitian Operators in Hilbert Space

CHAPTER 8

Hermitian Operators in Hilbert Space

Given the Hilbert space H (over the complex numbers, finite dimensional or infinite dimensional; the elements are denoted by x, y, . . and the inner product of x and y is denoted by ( x , y)), the set 2 of all bounded Hermitian operators in H i s a real vector space, partially ordered by defining that A 5 B holds for A and B in 2 whenever (Ax, x ) S (Bx, x ) holds for all x E H . The set A? is not a Riesz space, except in the trivial case that the Hilbert space H i s one-dimensional. We will prove that A? has a property analogous to Dedekind completeness, namely, if (A, : z E {T}) is an upwards directed subset of A? which is bounded from above, then A = sup A, exists in A?. We denote the positive cone of 2 by 2'. It is well-known that every A E 2' has a unique square root in 2 +i.e., , there exists a unique B E A?' satisfying B 2 = A . Denoting, for any A E 2,the square root of A' by IAI, we set

.

A+

=

3(A+IA/),

and we will prove that, among all B E 2 commuting with A and satisfying B L A and B 2 0 (where 0 is the null element of @), the element A + is the smallest. We believe the proof is a little simpler than usually presented. Given the subset 9of 2 such that any two members of 9commute, the commutant W ( 9 ) of 9 is by definition the set of all B E 2 commuting with every A E 9, and the second commutant W'(9) is the commutant of %'(9).We will prove that V'(9 )is a Dedekind complete Riesz space, super Dedekind complete if the Hilbert space H i s separable. The classical spectral theorem for a bounded Hermitian operator A is now a fairly immediate consequence of Freudenthal's spectral theorem in the Riesz space %?"({A}), and also the spectrum of A in the classical sense is exactly the spectrum of A considered as an element of the Riesz space. The spectral theorem for a bounded normal operator follows easily. In order to further clarify the structure of the second commutant, it will be shown that if W is a subset of 2 such that W is a (real) linear subspace of 369

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HERMITIAN OPERATORS I N HILBERT SPACE

[CH.8 , 8 53

8 as well as a ring (in the algebraic sense), and if 92 is closed under the operation of taking strong limits, then W is a Riesz space with respect to the partial ordering inherited from 2.In addition, if H is separable and W contains the identity operator, then U"(92) = W . Furthermore, it will be shown that any second commutant W ' ( 9 )is an abstract L, space, i.e., the operator norm satisfies IlSUP (1-41, 1Bl)Il

=

max (IIAIL 11B11)

for all A and B in W'( 9). The ordered vector space L is called an anti-lattice whenever, for any two elementsf, g E L ,the element inf (f,g ) exists in L if and only iff and g are comparable (i.e.,fz g or f 5 9 ) . We will prove R. V. Kadison's theorem (1951) that 8 is an anti-lattice. 53. The ordered vector space of Hermitian operators in a Hilbert space

Let H be a Hilbert space (over the complex numbers); the elements of H will be denoted by x, y , z, . and the inner product of x and y by ( x , y ) . As well-known, the bounded linear operator A in H (i.e., from H into H ) is called Hermitian (or self-adjoint) if

..

(Ax, Y ) = (x, AY)

holds for all x and y in H. The set of all bounded Hermitian operators in H will be denoted by H,arbitrary elements of H by A , B y . .., the identity operator and null operator by E and 8 respectively. The following lemma is well-known. Lemma 53.1. (i) If A and B are elements of 8, then the product A B (in the usual operator sense) is an element of Af i f and only if A and B commute (i.e., ifand only i f A B = BA). (ii) If A , E 8for n = 1 , 2 , . . . and A,, converges weakly to the bounded linear operator A (i.e., (A,x, y ) + (Ax, y ) as n -+ 00 for all x , y E H ) , then A E 8. Hence, if A,, E &for n = 1,2, . . and A , converges strongly to the bounded linear operator A (i.e., IIAx-A,xll + 0 as n -+ co for all x E H ) , then A E H.

.

Proof. (i) We have A B E 2 if and only if (ABx, y ) = ( x , ABy) holds for all x , y E H . Also, on account of A E Af and B E 8, we have (x, ABy) = (Ax, BY) = @AX, Y )

CH. 8, 531

for all x , y

HERMITIAN OPERATORS IN HILBERT SPACE

E H.

371

It follows that A B E 2 f if and only if (ABX,Y) = ( B A x , y )

holds for all x , y E H , and this is equivalent to A B = BA. (ii) Let A , E 2 for n = 1,2, . . ., and let A be a bounded linear operator such that ( A , x , y ) --f ( A x , y ) as n + 03 for all x , y E H . Then soA~2f.

(Ax,y )

=

lim ( A , x , y )

=

lim ( x , A n y ) = ( x , A y ) ,

The set 2 f is a real vector space under the usual algebraic operations. Introducing in 2 f a partial ordering by defining that A 2 B holds whenever ( A x , x ) 2 ( B x , x ) holds for all x E H, the set 2 f becomes an ordered vector space (i.e., A 2 B implies A + C 2 B + C for all C E2P and aA 2 aB for all real a 2 0). Note that it is not completely trivial to prove that A 2 B and B 2 A together imply that A = B . We briefly recall the proof. If A 2 B and B 2 A hold simultaneously, then ( A x , x ) = (Bx, x ) holds for all x E H . Replacing x by x +y in this inequality, we find that ( A x , r)+ (AY, x ) = (Bx, v)+ (BY, x ) holds for all x , y E H . Observing now that the number (Ay, x ) = ( y , A x ) is the complex conjugate of ( A x , y), we derive from the last formula that the real parts of ( A x , y ) and (Bx, y ) are equal for all x, y E H . Replacing y by iy, it follows that the imaginary parts of ( A x , y ) and ( B x , ~ are ) also equal. Hence, we have ( A x , y ) = ( B x , ~for ) all x , y E H. This implies A = B. The positive cone 2f' of 2 f consists of all A E 2 satisfying ( A x , x ) 2 0 for all x E H,the elements of 2' are usually called positive (semi-) definite Hermitian operators. If the Hilbert space H is one-dimensional, every element of 2 is a real multiple of the identity operator, and so every two elements of %' commute. Hence, 2 f is now a real commutative algebra (i.e., simultaneously a ring and a real linear space), isomorphic to the algebra of all real numbers and such that the isomorphism is order preserving in both directions. As soon as the Hilbert space H i s at least two-dimensional, 3? is no longer an algebra, as shown by the following simple lemma.

Lemma 53.2. (i) If A E Z , and A commutes with every member of 2fy then A is a real multiple of the identity operator E. (ii) rfthe Hilbert space H i s at least two-dimensional, then 2 f is no algebra. Proof. (i) If H is one-dimensional, the statement is true. Hence, assume

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HERMITIAN OPERATORS I N HILBERT SPACE

[CH. 8 , 5 53

that H i s at least two-dimensional, and let A commute with evzry element of Z . Then A commutes in particular with every orthogonal projection P in the Hilbert space H, and so every x in the range of P satisfies A x = APx = PAX, which shows that A x is also in the range of P. Every closed linear subspace of H is the range of an appropriate orthogonal projection, so it follows that A leaves every closed linear subspace of H invariant, in particular A leaves every one-dimensional subspace invariant. This implies that for every X E H there exists a real 1, such that A x = I,x. It remains to prove that Ax = AY for x and y linearly independent. To this end, given linearly independent x and y , let (z, t } be an orthonormal basis of the subspace spanned by x and y , and let B be the Hermitian operator satisfying Bz = t and Bt = z on this subspace, while B is identicallyzero on the orthogonal subspace. Then

1 , t = At = ABz

=

BAz

=

1,Bz = 1,t,

and so 1, = 1,. It follows easily that 1, = 5. (ii) If H is at least two-dimensional, the set Z contains an element A that is no real multiple of the identity E, and so by part (i) there exists an element B E 2 such that A and B do not commute. But then, by part (i) of the preceding lemma, A B is no element of Z .This shows that Z is no algebra. We recall the generalized Schwarz's inequality, according to which

](AX,Y)12 5 (Ax7 x)(Ay,V ) holds for all A E 2' and all x , y E H. The proof is based upon the inequality

0

5 ( A ( x - ~ Y )x, - 1 ~ )= ( A X ,~ ) - 1 ( A y x)-X(AX, , y)+AX(Ay, y ) ,

holding for all complex 1.If (Ay, y ) # 0, we choose

I = (x7 AY)/(AY7 Y)7 and the inequality to be proved follows immediately. If (Ay, y ) = 0, we have to prove that (Ax, y ) = 0. Assuming that (Ay, y ) = 0, but (Ay, x ) = reiq with r > 0 and cp real, we choose 1 = Rediq with R positive and so large that Rr > (Ax, x ) . This immediately yields a contradiction in the above inequality, and so (Ax, y ) must be zero. We also observe that A E 2' implies A" E #' for n = 2,3, . Indeed, for n even, say n = 2k, we have

..

( A Z k x , x )= llAkxl122 0

CH. 8,s 531

HERMITIAN OPERATORS IN HILBERT SPACE

373

for every x E H, and for n odd, say n = 2k+ 1, we have

(P+ 'x, x)

= (AA'x, Akx)

20

for every x E H. It follows that any polynomial in A with non-negative coefficients is an element of S+, and since the product of any two polynomials '.Furtherof this kind is again of the same kind, the product is again in S more, if A E &?+ is given and the number y A is defined by YA

= sup ((Ax,

x)/11x112

:x #

o},

then y A = 11A11. Indeed, it follows from (Ax, x)

that that

I IIAxll * llxll 5 IlAll

*

llX1l2

5 11A11. Conversely, we have by the generalized Schwarz's inequality llAX1l2 = (Ax, Ax)

5 ('4x9 X)+(AAX, Ax)+ 5 Y ~ l l ~ l l Y ~ l l ~ ~ l l ~

and so llAxll 5 ~ ~ l l xiflAx l # 0. The same inequality holds if A x = 0, and hence we have llAll 5 y A . Combining the results, we obtain y A = 11A11. In particular it follows from 6 5 A 5 E that j l A 5 1, so IlAll 6 1. But then we have 11A"11 5 1 for n = 1,2,. ., and so 6 5 A" 6 E in view of the just established results.

.

Lemma 53.3 (J. P. Vigier [l], 1946). such that 6 I A -B 5 E, then IIAx-Bx114

5

If A andB are Hermitian operator3

{(Ax, x)-(Bx,

x))11x112

holdsfor every x E H.

Proof. Writing A - B = C, we have C

2 6 (i.e., C E &+), and so

I Icx114= (cx,cx)2 I (cx,x ) ( ~ 2 x , cX) by the generalized Schwarz's inequality. In view of the remarks above, it follows from 6 5 C E that llCll 5 1, and so we have

llCX1l4 6 (CX, x)11x112, which is the desired result. We will prove finally in this section that 3" has a property analogous to Dedekind completeness of a Riesz space.

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HERMITIAN OPERATORS I N HILBERT SPACE

[CH. 8,§ 53

Theorem 53.4. (i) If (A, : t E {z}) is an upwards direcfedsubset of 8 such that the subset is boundedfrom above, then A = sup A, exists in 2,and for in ( t }such that every E > 0 and every x E H there exists an index IIAx-A,xll < E holdsfor all A, 1 (ii) If (A, : z E { 7 } ) is the same as above and the Hilbert space H is separable, there exists an increasing sequence(A, :n = 1, 2, . . .) in (A, : z E {z}) such that A,,x + A x holds for all x E H simultaneously (i.e., the sequence of operators ATnconverges strongly f o the operator A). Proof. (i) We choose z0 E {z}. The sets (A, : z E {z}) and (A, : z E {z}, A, 2 A,,) have the same upper bounds, so we may just as well assume that A, >= A,, holds for all z E {z}. But then, subtracting A,, and multiplying by an appropriate positive constant, we may restrict ourselves to the case that 8 5 A, 5 E holds for all T. Under this assumption, if A,. 1 A,, we have by Vigier's lemma that

II&x-A,xI14

I { ( A z , ~. ,) - ( A x , x)}IIxII'

holds for every X E H . Since, fixing x, the upwards directed set of real numbers (A,x, x ) has a finite supremum, it follows that for every e > 0 there exists an index t8,,in {z} such that IIA,.x-A,xlj

In particular, far n = 1,2,


for all A,, A,.

2

. . ., there exists an index zX," in {z]

IIA,x- A,, ,,XII < n-l

for all A,

such that

2 A,*,-,

and we may assume that ATx,n+l

holds for all n. Then

2

*rx,n

I I A , x . , ~ - A , x , n ~ I
1 n, and so the sequence

.

(A,,,, x : n = 1 , 2 , . .)

in H converges to an element in H , which we denote by A x . Note that A x is uniquely determined in the sense that if (A,.*,,, : n = 1,2,

. . .)

is another increasing sequence such that

IIA~x-&,,,,xII

< a-1

for all A,

1 A,;,",

CH.

8,s 531

HERMITIAN OPERATORS IN HILBERT SPACE

375

then A,.x,,,x converges also to Ax. The proof follows by observing that there exists an operator A, satisfying A, 2 A,=, and A , 2 A,,x,,, simultaneously, and so I I A T X . n ~ - A r * x , n ~< I I 2n-I. It is easily verified now that the thus defined operator A in H is linear. Indeed, for the proof that A is additive, let x and y in H be given, set z = x + y , and observe that

1

llAx-A,xll < 2n-' llAy-ATyll < 2n-1 IIAz-A,zll < 2n-'

for all A, 2 A,*, n , for all A, 5 ATY,, for all A, 2 A,=, .

(1)

,,, ATY, and A,=.,, simultaneHence, for all A, greater than or equal to ATX, ously, the three inequalities in (1) hold simultaneously, and so in view of A,z = A,x+A,y we have IIAz-(Ax+Ay)lI < 6n-I.

..

This holds for n = 1, 2, .; it follows that Az = A x + A y . It is evident that A is a bounded operator. Indeed, for llxll = 1 we have

.

l l ~ x l sl IIA,,,,xII+~~-'

s 1+2n-'

for n = 1,2,. ., so A is bounded with IlAll 5 1. Furthermore, we have A E X , i.e., A is Hermitian. Indeed, given x and y in H, there exist increasing sequences Arx,,,and ATY, ,, such that Arx,,,x and ATY,y converge to Ax and Ay respectively, and hence (cf. the inequalities in (1)) there exists an increasing sequence (ATn:n = 1,2, .) such that

..

A,x + A x and ATny+ A y hold simultaneously. It follows that (A,,x, y ) = (x, A , J ) converges to ( A x , y ) as well as to (x, Ay), and so (Ax, y ) = (x, Ay), i.e., A is Hermitian. Finally, it follows from

AX, x ) - ( A , x ,

x) IS IIAx-A,xlI

*

llxll

5 2n-'IIxll

for all A, 2 A,*,, that (Ax, x) differs an arbitrarily small amount from the number sup, (A,x, x), and hence

). =

x,

holds for all x E H. In other words, A = sup A, holds in X .

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[CH.8 , s 53

HERMITIAN OPERATORS I N HILBERT SPACE

(ii) We assume now that His separable, so there exists a countable dense set (x, : n = 1,2, . . .) in H. For each x, there exists a sequence (Ank : k = 1, 2 , . . .)

C

( A , : T E {TI)

such that Ankincreases as k increases and Ankx, -P Ax, as k + 00. More precisely, Ankcan be chosen such that

IIAx,-A,x,l( < k-‘ holds for all A, 2 Ank.We will prove the existence of an increasing sequence (& :k = 1,2, . . .) in (A, : z E {T)) such that Bkxn-+ Ax, as k + 00 holds for all x, simultaneously. Since it may be assumed by part (i) that 0 5 A - A , - E holds for all z, it is sufficient by Vigier’s lemma to show that (Ax,, xn)I (Bkxn,x,) tends to zero as k -+ 00 for every x, . The proof is, naturally, a variant upon the diagonal procedure. We choose B , = A,,, and then B2 E (A, : z E {t}) such that B, 2 A 1 2 , B2 2 A 2 2and B2 2 B,; generally, if B,, . . .,Bk- have been chosen already, we choose BkE (A, : z E {T}) such that Bk is greater than or equal to A l k , A Z k , . . ., A k k , B k - 1 . NCW, if n iS given, let k 2 n. Then Bk 2 Ank, and SO

,

0

5

( A x , , xn)-(Bkxn,

xn)

5 (Ax,, X n ) - ( A n k X n , xn) 5 IIAxn-Ankxnll

’

Ilxnll

5 k-lIIxnll*

As observed above, this shows that IIAX,-&x,,ll tends to zero a s k + 00 for every x,. Note also that on account of 0 5 Bk 5 E we have llBkll 6 1 for all k. Given now the arbitrary element x in H a n d the number > 0, choose x, from the dense subset such that ] ~ x - x , ,< ~ ~+E and then determine ko such that IIBkx,-Ax,ll < 3& holds for all k 2 ko.It follows then from

!IAx-

Bkxl

I6

IIAx-Axnll

+I IAxn-

I f II B k x n - Bkxll 6 IIx-xnll + I I A x n - B k x n l l

BkxnI

+

l~xn-xll

that IIAx-Bkxll c & for all k 2 k o . In other words, Bkx converges to Ax as k + co for every x E H, i.e., Bk converges strongly to A. Exercise 53.5. Let H be the Hilbert space consisting of all complexvalued functions x ( t ) on the real line such that x ( t ) # 0 for at most countably many t, and with Ix(t)l’ < 00. The inner product in H is

ct

t

where j j denotes the complex conjugate of y.

CH. 8,0541

THE SQUARE ROOT

377

OF A POSITIVE HERMITIAN OPERATOR

Show the existence in 2'P of an upwards directed set (A, :z E {z}) with supremum A such that for no increasing sequence (ATn: n = 1,2, . . .) in (A, : z E (z}) we have A,x + A x for all x E H simultaneously.

54. The square root of a positive Hermitian operator It is a well-known theorem that every A E 2' has a unique square root A* in &+,i.e., there exists a unique element A%in 2'P' such that (A*)2 = A . For the proof we may obviously restrict ourselves to the case that 8 5 A E holds, and we present C. Visser's construction of the square root ([l], 1937). We set A , = E and

s

for n = 0 , 1 , 2 , . . .,

A,,+1 = A,,+$(A-Ai)

and we will prove the existence of an operator A , in S ' satisfying 4,J A , and A t = A. For the proof, let B = E - A and B,, = E- A,, for n = 0, 1,2, . . . Then 8 5 B E as well as B, = 8 and B,,,, = +(B+B:)

for all n,

as follows from a simple calculation. We show first that, for every n, the operators B,, and B,,, -B,, are polynomials in B with non-negative coefficients. For n = 0 this holds in view of B, = 8 and B, = +B. Assume now that, for some value of n, we know already that B,,-, and B,,-B,,-l are polynomials in B with non-negative coefficients. Then B,,, and hence B,, +B,,- 1, are also such polynomials, and so

,

Bn + 1 -B, = +(B,Z -B.'-

1)

= HBn +Bn - 1)(Bn-Bn

,

- 1)

is also such a polynomial (note that we use here that B,,- and B,, commute). The induction proof is now complete; it follows that we have

e = B, s B, 5 B~ 6 ...t.

Since 0 = B, 6 E, and since 8 5 B,, 5 E implies 8 B,,,, = $(B+B:) 5 E, we have then that

e = B,

I;B,

6 B,

s ...t

5 B."

E, and so

E,

whence on account of Theorem 53.4 there exists an operator B, E A?' such that B,, 1B, and B,,x -P B,x for every x E H. Returning to the operators A,, = E-B,,, we conclude, therefore, that A , = E-B, satisfies

E = A , Z A , Z A ~ B. . . ~ ~ , z e

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HERMITIAN OPERATORS IN HILBERT SPACE

[CH.

8,854

and A,x + A , x for every x E H . This implies A i x + A k x for every x E H. Indeed, using that IIA,I 1 5 1 holds for all n and writing y = A , x, we have

IIA,ZX-A:X~I

5

IIA,~X-A,A,X~~+IIA,A,X-A:X~~

= IIAn(Anx-Amx)Il+ ll(An-Am)AmxII

5

l l ~ n ~ - ~ m ~ l l + l l A n Y - ~ m Y l0 l +

as n + co. Taking limits now (n + a)in the formula A , + ~ x= A,x++(Ax-A,ZX),

we find that A: x = A x holds for every x

E H,

so A: = A .

As a first consequence we note that if soae Hermitian C commutes with A , then C commutes with every An,and henct. C commutes with A , . Secondly, it follows then that if A and C are commuting elements of &+,then A C = C A is again in 8 ' (note that A C = C A E & follows already from Lemma 53.1 (i)). For the proof, let A , be the just obtained square root of A which, therefore, commutes with C, and observe that (CAX,X) = ( C A , A,x,

X) =

( A , CA,x,

X) =

(CA,x, A , x ) 2 0

holds for every x E If. Uniqueness of the square root is not ntccssary for this proof. More generally, we have now that if A 2 B holds in 2,and C is an element of 2"' commuting with A and By then C A 2 CB. Indeed, since C and A - B are commuting elements in %+,we have C ( A - B ) E 2"+ by the result just established, so CA 2 CB. Note, as a particular case, that if A 2 B 2 8 and A and B commute, then A 2 2 A B 3 B 2 . In the above construction of the square root of A , where 8 have used A . = E and

..

An+1

5 A 5 E, we

= An++(A-A,Z)

as n = 0, 1,2, . for the approximating sequence. We introduce another sequence (A; : n = 0, 1,2, .) by defining A; = 8 and

..

= A;++{A-(A:)2)

for n = 0 , 1 , 2 , .

.. .

Similarly as above, writing again B = E - A and BL = E - A : , we have BA = E and % + l = 3(B+(B;)'),

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THE SQUARE ROOT OF A POSITIVE HERMITIAN OPERATOR

319

and so it follows by induction (and by means of the just established result that the product of commuting elements in A?’ is again in 2’)from

B;+, -B; = +(B;+B,’-,)(B;-B;-,)

that (B; : n = 0,1,2, . . .) is a decreasing sequence in A?’,possming therefore an infimum B&. In other words, writing A‘, = E- Bk , we have

8 = A; 5 A ; 5 A; 5 ...I A& 5 E and AAx + A’, x for every x E H. Exactly as above it follows that (A’,)’ = A , i.e., A‘, is also a square root of A . Collecting the thus obtained results, we have

. .. t A ; 2 ...1 A ,

8 = A; 5 A ; 5 A; 5

with (A&)’ = A,

E = A. 2 A , 2 A’

with A: = A .

Writing again B,, = E- A,, and B; = E- A: for all n, we hake Bo I;B ; , and so it follows by induction from B,,+l = +(B+Bz) and B:+l = +{B+(B:)’}

that B, 5 BA holds for all n, i.e., A: 5 A, holds for all n. But then (since A: is increasing and A, is decreasing as n increases) we have A; 5 A, for all m, n = 0,1,2, . . ., and so A’, 5 A , . It follows that A,

=

A’,+(A,-A’,)

with A , - & 2 8. Furthermore, since all A: are polynomials in A, and since A , commutes with any polynomial in A , it follows that A , and A’, commute; hence A: = (A:)’+2A&(Aw -A’,)+(A, -A’,)’.

But A: = (A’,)’ = A , and the other t e r n are elements of A?+.It follows that (A, -A’,)’ = 8, and so A , = A’, . This shows already that the square roots A , and A’, of A are equal. For the uniqueness proof, assume now that C E 2“ is another positive square root of A, so C 2 8 and Cz = A . Then C and A commute in view of C A = A C = C 3 ,and this implies that any polynomial in C and A commutes with all A , and all A:. Also, it follows from C’ = A 2 E that llCX11’ =

(CX,

for all x E H, so llCll

Cx) = (CZx, x) = (Ax, x)

5

1, and hence 8

llx11’

5 C 5 E. The sequence

3 80

(A: : n = 0,1,2,

HERMITIAN OPERATORS IN HILBERT SPACE

. . .) is now defined by A:

=

= A Y + + { P ~ - ( A : ' ) ~ ) for

C and

n = 0,1,2,.

[CH.

8,s 54

.. .

Introducing B:' = E- A:' for all n, we have

B ; , and so it is easily derived by induction that B,, 5 BA' and Bo 5 B t 5 BA holds for all n, i.e., A; 5 A: 5 A,, holds for all n. In the proof we use that BA', as a polynomial in A and C, commutes with all B,, and BA, so B,, 5 Bi' 5 Bi implies B.' S (B;')' 5 (B;)', and hence B,,+l 5 Bikl 5 Bi+ 1. Observing now that A:' = C holds for all n (in view of A: = C and C 2 = A), we find that A; S C 5 A, holds for all n, and so A& 5 C 5 A,. But A& = A,, so A& = C = A,, i.e., the positive square root of A is unique. One last remark for later purposes. All A; are polynomials in A without constant term (i.e., without a term which is a nonzero multiple of E), so the positive square root of A is the strong limit of a sequence of polynomials in A without constant term. There exist other proofs, even shorter ones, for the uniqueness of the square root. The present proof, however, shows also that the sequence (A, : n = 0,1,2, . . .) with An+1

= A~++(A-A,Z)

always converges to the square root of A no matter how A . is chosen, provided only that 8 5 A . 5 E holds and A , commutes with A (cf. Exercise 54.3 at the end of the present section). We consider now an arbitrary Hermitian operator A. The positive square root of A' will be denoted by IAI, and we set

+

A" = *(A IAI). This notation might cause confusion; we do not wish to suggest tbat A + = sup (A, 8) holds in A?. In the next section it will be proved, however, that this formula holds in an appropriate subspace of 2. We prove here that A+ is always an upper bound of the set consisting of A and 8.

Theorem 54.1. Given A E 2, we have A + ly, IA,1 2 A iutd IAI 2 - A .

2 8 and A + 2 A or, equivalent-

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THE SQUARE ROOT OF A POSITIVE HERMITIAN OPERATOR

381

Proof. It is evident that the statements are equivalent. Hence, since A’ and (-A)’ have the same square root [ A ! ,it is sufficient to prove that IAI 2 A holds. Multiplying, if necessary, by a positive constant, we may assume that -E A S E, and so O 5 A’ S E. We set B = E - A 2 and we define (B,, : n = 0 , 1 , 2 , . . .) by Bo = Oand

B , + ~= +(B+B,Z) As shown above, we have O = Bo

for n = 0,1,2,.

.. .

5 BI B B2 5...r B , 5 E,

and E-B, is the square root IAI. For the proof that IAI 2 A holds we have to show, therefore, that B, = E - ] A ] S E - A , and on account of B, T B, it will be sufficient to show that B,, 5 A - A holds for all n. For n = 0 this is evident. Assume that B,, 5 E - A has been proved for some value of n, and observe that B, and A commute. Then we have B: (E- A)’, which can be rewritten as

B,Z 5 E - 2 A + A Z = 2E-2A-(E-A’) and so

2B,,+, = B+B,Z

= 2E-2A-B,

5 2(E-A),

which is the desired result for n+ 1. This completes the proof. We will prove next that, among all Hermitian B, commuting with a given Hermitian A and satisfying B 2 A as well as B 5 8, the operator A’ is the smallest. Theorem 54.2. If A and B are commuting Hermitian operators such that B 2 A as well as B 2 8, then B 2 A’. Equivalently, if A and C are commuting Hermitian operators such that C 2 A as well as C 2 - A , then C 2 IAI. Proof. Let A and B be commuting Hermitian operators such that B 2 A as well as B 2 8. Setting C = 2B-A, we have C 2 A as well as C 2 - A , and C commutes with A . Hence, if we prove that C 2 IAI, it will follow that 2 B - A 2 IAI, i.e., B 2 A’. Assume, therefore, that C 2 A as well as C 2 - A , and C commutes with A . Then C- A and C + A are in Z?’ and they commute, so ( C + A ) ( C - A ) 2 8, i.e., C 2 2 A’. It follows then from the construction of the square root that (C”* 2 (A2)* = IAI.

Since C E Z?’ (because C obtain C 2 IAI.

2 A and C 5 - A implies 2C 2 A - A

= O), we

382

HERMITIAN OPERATORS IN HILBERT SPACE

[CH.8,s 55

Exercise 54.3. Let A and C be commuting Hermitian operators between 8 and E, and set

for n = 0,1,2,. . . . .A: = C, A,*,, = A,*++{A-(A:)'} Show that A,* converges strongly, although not necessarily monotonely, to the positive square root of A.

55. The second commutant of a commuting set of Hermitian operators Let 9 be a subset of i%? such that all elements of 9 commute mutually. The set @')( 9)of all B in 2 commuting with every element of 9 is called the commutant of B, and for n = 2,3, . . . the set Y ( " ) ( 9of ) all B in X commuting with every element of Y("-')(B) is called the n-th commutant of 9.Evidently, every V(")(9) is a real linear subspace of 2,and every real polynomial in the elements of 9, with a constant term consisting of a real multiple of E included, is contained in every V(")(9). Also, if the se9 )and Ak converges weakly quence (Ak :k = 1,2, . . .) is contained in e(")( to the bounded linear operator A (cf. Lemma 53.1 (ii)), then A E V(")(g). The sets %(')(a), V(')(9) and V(3)(9)will be denoted briefly by V', V" and V'".

Lemma 55.1. Wehave%?" c V ' a n d V ( " + 2 ) ( 9= ) V("l(L2)for everyn, i.e.,

. . ., 6= ) . .. .

fg' = C p ' S -- Y ( 9 = g t f

=~

( 4= ) ~ (

Furthermore, V" is a commutative ring, i.e., Y" is a ring in the algebraic sense the elements of which commute mutually.

Proof. Note first that, generally, g1c 5Bz implies u'(L21)13 V'(9'). Hence, 9 c W implies V' 13 W'. Similarly, 9 c W' implies F ' 3 V'". On the other hand, every element of %' commutes with every element of Y,', and so v' c v'".It follows that v'" = V', and this implies that V("") = V(") holds for every n = 1, 2, . In order to show that the elements of V" commute mutually, assume that A and B are in v". Then A E Y,' and B E V" c V', so A and B commute since every element of V" commutes with every element of v'. It follows, in view of this commutativity property, that the product of two operators from Y,' is Hermitian in the first place, and once this has been established it follows then also that the product of operators from Y" is again an element of v",so Y" is a ring.

. ..

CH.8 , s 551

THE SECOND COMMUTANT

383

By way of example, let 9 consist only of the identity operator E. Then the first commutant U' satisfies v' = JZ' and the second cornmutant Y" consists of all real multiples of E (cf. Lemma 53.2). This shows that the first commutant " ( 9 )is not necessarily a ring. Note, furthermore, that B1 c implies Y'(91) 3 W ( 9 2 ) ,and so v''(gl)c %?"(92). Adding the does not change identity operator E to 9,if E is not already contained in 9, v' and W'. There is enough material available now to show that, for any subset 9of mutually commuting elements of S, the second commutant of 9 is a Riesz space. It was already observed that W ' ( 9 ) is a real linear subspace of X , and with respect to the ordering inherited from S the space v''(9)is evidently an ordered vector space. Theorem 55.2. For any subset 9of mutually commuting elements of A?, the second commutant 9 )is a Dedekind complete Riesz Apace. If the Hilbert space H is heparable, the Riesz space U''(9 )is super Dedekind complete. %"I(

Proof. In order to show that U " ( 9 ) is a Riesz space, it is sufficient to show that sup (A, 8) exists in W'( 9)for every A E W'( 9).We will prove that the element A + = +(A+IAI) satisfies this condition. I n the first place, since [A1 is the strong limit of a sequence of polynomials in A', and since all these polynomials are elements of v"(9),we have IAl E W ' ( 9 ) , and so A + = -)(A+IAI) ~ v " ( 9 ) . Secondly, it is evident that v''(9)c v ' ( { A } ) , and so, if B E U ' ' ( 9 ) is an upper bound of the set consisting of A and 8, then B 2 A + by Theorem 54.2. This shows that A + = sup (A, 0) holds in B).As observed above, it follows that v''(9)is a Riesz space. If (A, :z E {z}) is an upwards directed subset of ""(9) such that the set is bounded from above, then A = sup A, exists in JZ' by Theorem 53.4. In order to show that %"(29) is Dedekind complete, it is sufficient to show that A EW'(B),i.e., it is sufficient to show that A commutes with every B E 'X'( 9). Tc this end, let x E H and B E U'(9)be given, and set y = Bx. As shown in the proof of Theorem 53.4, there exists an increasing sequence (ATn: n = 1,2, . . .) in (A, : z E {z}) such that A,,x + Ax and Army+ A y hold simultaneously. This implies that %'I(

ABx = Ay

=

lim Arny = lim A,,Bx = lim BA,,x = BAx.

Thus ABx = BAx holds for every x

E

H, which is the desired result.

384

HERMITIAN OPERATORS IN HILBERT SPACE

[CH.8,§ 55

If the Hilbert space H is separable, there exists an increasing sequence (Az, : n = 1,2, . . .) in (A, : z E {z}) such that A,x + Ax holds for all x E H simultaneously (cf. Theorem 53.4 (ii)). By the Dedekind completeness of V'(53) the element B = sup A,, exists in %"(9 )and A,, converges strongly to B (once again by Theorem 53.4). It follows that B = A, so A = sup A,,. Hence, the given set (A, : z E {z}) has a countable subset with the same supremum. This shows that W ' ( 9 )is super Dedekind complete. This theorem has been known for a long time, but it seems that until around 1952 no simple proof was available (i.e., a proof without making use of the spectral theorem for Hermitian operators). Then, for the case that 53 consists of one Hermitian operator, the theorem was proved in an elementary manner by V. I. Sobolev ([l], 1952); the proof was extended to the general case by V. D. Lyubovin ([l], [2], 1954). For any Hemitian operator A , let N ( A ) and R(A) denote the null space of A and the closure of the range of A respectively. As well-known, N ( A ) and R ( A )are mutually orthogonal closed linear subspaces of the Hilbert space H, and H i s the direct sum of N ( A ) and R(A). In the following series of lemmas we collect some simple properties of the Riesz space W ' ( 9 ) , where 9 is again a set of mutually commuting elements of %.

Lemma 55.3. (i) For A and B in al'(9)we have A IB if and only if AB = 8, and also if and only if A' 1B 2 . Also, A IB holds if and only if the closed linear subspaces R(A) and R(B) are orthogonal in H. (ii) For A , B and C in W ' ( 9 ) it follows from A IB that AC IBC. Proof. (i) For the proof that A IB holds if and only if AB = 8, we assume first that A 2 8 and B 2 8, and we write C = inf (A, B). Note that A, B and C commute mutually, since all elements of W ' ( 9 ) commute mutually. Assuming first that AB = 8, we have C 2 5 AB = 8, so C 2 = 8, which implies C = 8, i.e., A IB. For the proof of the converse it may be assumed without loss of generality that A IB and 8 5 A 5 E as well as 8 5 B S E. Then 8 5 AB 6 A and 8 AB 5 B, so

8 5 AB 5 inf (A, B ) = 8, i.e., AB = 8. Now assume that A and B are arbitrary elements of W'(52) satisfying A IB. This means that IAI 1IBI, and so IAI * IBI = 8 in view of what was proved already. Since IAI 1B1 and lABl are both equal to the positive square

CH. 8, S 551

3 85

THE SECOND COMMUTANT

root of A2B2,we have, therefore, that I A B= ~ I A ~I .B I= 8,

which implies that AB = 8. Conversely, if A and B are elements of %"( 9) satisfying A B = 8, then lABl = 8, so IAI IBI = 8, which shows that IAl 1 .IBI, i.e., A 1B. On account of these results we have now that

-

AIBoAB=eo(AB)Z=eoAzBZ= eoAzIBZ. Finally, lei again A and B be elements of U"(L3) satisfying A _L B, so AB = 8. Furthermore, let x and y be elements of the Hilbert space H such that xis in the range of A and y is in the range of Byi.e., x = Ax' and y = By' for appropriate x', y' E H. Then (x, y ) = (Ax', By') = (x', ABy')

= 0,

so the ranges of A and B are orthogonal in H. Hence, the closures R ( A ) and R(B) of these ranges are also orthogonal. Conversely,if A and Bare elements of W ' ( 9 )such that R(A) IR(B),then (Ax, B y ) = 0 for all x, y E H, i.e., ( x , ABy) = 0 for all x , y E H. This implies that AB = 8, so A J- B. (ii) If A, B and C are elements of U"(9) and if A IBy then AB = 8, so ABCZ = 8, i.e., (AC)(BC) = 8, which implies that AC IBC.

Lemma 55.4. (i) Given A E %'"(9), we have A + J- A - in the Riesz space sense, which implies that the closed linear subspaces R ( A + ) and R ( A - ) are orthogonal. It follows that

ll~x1I2= IIIAlxllZ = ll~+x112+11~-x11z holdsfor every x E H. Hence llAll = 11 IAI 11, and we have Ax = 0 ifandonly i f

[ A ~= x 0. (ii) FotAandBinV"(L3)wehave IAl 5 1B1 ifandonlyifIIAxJI 4 IlBxll holds for every x E H. (iii) I f A, ( n = 1,2, .) and A are elements of Y"(L3) such that A,xO + Ax, holds for some x, E H (not necessarily A,x + Ax for all x E H ) , then A:x, --r A+X0.

..

Proof. (i) By the preceding lemma R( 4 + ) and R ( A - ) are orthogonal subspaces of H , and so A + x IA-x holds for every x E H. Since A = A+-A-

and IAI = A + + A - ,

386

HERMITIAN OPERATORS I N HILBERT SPACE

it follows that

llAx1I2 = IIA+xl12 +llA-x1(2=

[CH.8,§ 55

11 IAIx112.

(ii) If A and Bare elements of U ” ( 9 )satisfying IAl 6 lBl, then A 2 5 B2, and so l l A ~ 1 1=~ (A’x, X ) 5 (B’x, X) = (IBx(I2 holds for every x E H. Conversely, if A and B are elements of U “ ( 9 ) such that IIAxll 5 IIBxll holds for every x E H, then A’ 5 B2, and so IAl 5 IBI. (iii) Since IA+-ATI 6 IA-A,I, we have by (ii) that

As observed earlier, the identity operator E is an element of %”( B), and it is obvious that the ideal in W ’ ( 9 ) generated by E is the whole space 9).

%’I(

Lemma 55.5. (i) I f P is a component of E in 9),then P is the orthogonal projection on some closed linear subspace of the Hilbert space H. (ii) Given A in %”( 9),let P be the component of E in the principal band generated in C“(9)by A . Then P is the orthogonal projection on R(A). In other words, we have N ( P ) = N ( A ) and R(P) = R(A). It follows that i f A and B are elements of %”( 9 )generating the same principal band in W’( 9), then N(A) = N(B) and R ( A ) = R(B). Conversely, i f A and B are elements of %”( 9 )such that R ( A ) = R(B) or, equiualently, N ( A ) = N(B), then A and B have the same disjoint complement by Lemma 55.3, and hence A and B generate the same band in W’( 9). (iii) I f P , and P, are components of E such that P , and P, are elemenrs of %”(9), then P , IP , holds in %“(9) if and only if R ( P , ) and R(P2) are orthogonal subspaces of H. %If(

Proof. (i) If P is a component of E in W ‘ ( 9 ) , then we have P I( E - P ) by definition, and so P ( E - P ) = 8 by Lemma 55.3. It follows that P 2 = P. Hence, P is a bounded Hermitian operator satisfyingP 2 = P. As well-known, this implies that P is the orthogonal projection on some closed linear subspace of the Hilbert space H. (ii) Under the given hypotheses, we need only prove that N ( P ) = N(A). Since A is disjoint from E - P , we have A ( E - P ) = 0, so A = AP. This shows that x E N(P) implies x E N(A). For the converse, observe that since P is an element of the band generated by A there exists an upwards directed set (B, : z E {z}) in the ideal generated by A such that 8 5 B, P. Hence, if

CH. 8 , § 551

387

THE SECOND COMMUTANT

x E N ( A ) , then A x = 0 and lAlx = 0, so (B, x , x ) = 0 for all t E {t}, and it

follows that

(Px, X) = SUP (B,x,

X)

= 0.

This implies Px = 0, so x E N(P). (iii) Follows from Lemma 55.3. In several parts of the lemmas in this section it was assumed that A and B are members of W ' ( 9 ) for some set 9 of mutually commuting Hermitian operators, and then several properties of A and B were proved depending upon various additional conditions imposed upon A and B. It may be asked whether these results depend upon the choice of 9. Fortunately this is not the case. In order to see this, note first that the assumption that A and B are members of W ' ( 9 )for some 9 of the kind referred to is equivalent to the assumption that A and Bare commuting Hermitian operators. Tndeed, given the commuting Hermitian operators A and B, we can take for 9the set conand 92 are subsets of sisting of A and B. Secondly, as observed earlier, if 9, 22' of mutually commuting operators such that g1c B2, then W'(Bl) c %"( 9,).Both of these are Dedekind complete Riesz spaces, so if Y is any subset of W'(91) which is bounded from above, then Y has a supremum in V'(9,)as well as in W'(9,). These suprema are the same. Indeed, if Y consists of two elements, A and B say, then sup ( A , B) is equal to A + (B-A)', which does not depend on any elements in U"(9,) but not in W ' ( 9 , ) . Hence, if Y is a finite set, the supremum depends only on the members of %"'(g1). Now, let Y = ( A , : T E {t}) be an infinite subset of %"'(g1), let B, be its supremum in W'( g1) and B, its supremum in%"'( 9,). In view of the preceding remarks we may assume that Y is directed upwards. Then, as shown in the proof of Theorem 53.4, we have (B1x, X) = SUP (A,x,

X)

= (B,x,

X)

for every x E H . Hence B, = B 2 . To mention one more example. in Lemma 55.4 (i), where we have one member A of W ' ( 9 ) , we can just as well state that we are given one Hermitian operator A , and the formula A + 1A - can now be understood to hold in the Riesz space V ' ( { A } ) , i.e., in the second commutant of the set consisting of A alone. As observed, A + 1A - holds just as well, however, in %"'(g) for any set 9 of mutually commuting Hermitian operators having A as one of its members. A similar remark holds for Lemma55.5 (ii). It has been proved in Lemma 55.1 of the present section that W'(9) is a commutative ring with respect to the ordinary multiplication of operators

388

HERMITIAN OPERATORS I N HILBERT SPACE

[CH. 8 , s 56

on the Hilbert space H . The identity operator E is the unit element with respect to this multiplication. It has been proved also (in Theorem 55.2) that W ’ ( 9 )is a Dedekind complete Riesz space and hence, according to the results in section 45, W ’ ( 9 ) is Riesz isomorphic to the Riesz space “{V’(9)} of all real continuous functions on a certain compact HausdorlT space, such that E corresponds in this isomorphism to the function identically equal to one. In “{W’(9)} we have the ordinary pointwise multiplication of functions. Transferring this back to W’(93), we obtain another commutative multiplication in W’(9), also with E as unit element. It is a reasonable conjecture that these two multiplications in V ‘ ( 9 )are the same. This is indeed the case, but it seems easier to postpone the proof until after the spectral theorem in the next section.

Exercise 55.6. If it is given that A and B are members of 2’ such that AB = 6, it follows easily that BA = 6, and so we derive from Lemma 55.3 (i) that A IB holds in the Riesz space $?’({A, B } ) , i.e., A + B = [ A-BI. It is interesting to observe that, without assuming in advance that A and B commute, it follows for Hermitian A and B from A + B = IA-BI that AB = BA = 6. In order to prove this, show first that A and B are members of &+; next, show that AB+BA = 6, which will imply then that AB = BA. Hint:Observe first that A+B-IA-BI

6 A+B-(A-B)

= 28,

and similarly A + B - I A - B [ 6 24. Hence it follows from A + B = [A-BI that A and B are members of 2’. Next, note that (A+B)’ = IA-BI’ = (A-B)’,

so AB+BA = 6. It follows readily from AB = -BA that A’B’ = B2A2, so B2 commutes with A’. But then B’ commutes with the positive square root of A’, i.e., B’ commutes with A, so AB’ = B’A. This implies, similarly, that AB = BA. It follows that AB = BA = 6. This proof, as presented in a paper by D. M. Topping ([l], 1965), is due to H. Leptin. 56. The spectral theorem for Hermitian operators and normal operators As in the preceding section, let 9 be a subset of 2 such that all members of 59 commute mutually, and let W’(9) be the second commutant of 9. We recall that 8 and E are the notations for the null operator and the identity operator respectively. It will be investigated first what the notion of E-uni-

CH. 8,6561

THE SPECTRAL THEOREM

389

form convergence of a sequence in %"'(9 )amounts to in terms of the usual Hilbert space terminology. We recall that the sequence (A, :n = 1, 2, . . .) in V ' ( 9 )is said to converge E-uniformly to A E V'(9 )whenever there exists for every E > 0 a natural number n, such that IA -A,I 5 EEholds for all n 5 n,. Lemma 56.1. If A , (n = 1,2, .. .) and A are members o f Q " ( 9 ) , then A, converges E-uniformly to A if and only if I IA -A,[ I -+ 0 as n -+ 00.

Proof. In the space V ' ( 9 )the sequence A, (n = I , 2, . . .) converges Euniformly to A if and only if, for every E > 0, there exists a natural number n, such that IA-A,I 5 ~ E f o all r n 2 n,, i.e., (IA -A&,

X)

S ~ ( xx) , for all x E H and all n 2 n,.

This is equivalent to

5 E for all n 2 n,, i.e., equivalent to IIIA-A,III 5 E for all n 2 n,. Since IA-A,I and A-A, sup {(IA-A,lx, x ) : llxll

5

1)

have the same operator norm (cf. Lemma 55.4 (i)), the last inequality is equivalent to (IA-.4,11 5 E for all n 2 n,. This shows that A, converges E-uniformly to A if and only if A, converges to A in the operator norm. Given the operator A in W ' ( 9 ) ,we will consider now the spectral system (Pa : - co < x < 03) of A with respect to E. We recall that Pa is the component of E in the band B, generated by (uE-A)'; the order projection on the band B, (in the Dedekind complete Rieszspace W'(9)) is denoted by 9,. Hence, we have Pa = 9, E for every a. We will write A, = 9, A for every u. It follows from Lemma 55.5 that Pa is an orthogonal projection in the Hilbert space Hsuch that Pa and (uE- A)' have the same null space. The range of Pa,which is the orthogonal complement of the null space of Pa,is then the closure of the range of (uE-A)+. Since Pa increases as u increases (i.e., c1 < fl implies Pa Ps in V'(9),so (Pax,x ) 5 (Psx,x ) for all x E H), it follows immediately that the null space N(P,) decreases and the range R(P,) increases as CL increases. Furthermore, there exist real numbers a and b such that aE 6 A 5 bE holds, and we have Pa = 8 for u 5 a and Pa = E for u > b, i.e., R(P,) consists only of the null element of H for u 6 a, and R(P,) is the whole space H for u > b. Before proceeding with the investigation of the projections Pa in the spectral system of A, we recall the general theorem that if Pl and P2are commuting orthogonal projections in the Hilbert space H, then P = P1P2is also

390

HERMITIAN OPERATORS IN HILBERT SPACE

[CH. 8,s 56

an orthogonal projection with range equal to the intersection of the ranges of PI and P2. We include a brief proof. It is trivial to see that P is Hermitian and P 2 = P, so P is an orthogonal projection. It is also evident that R(P,) n R(P,) is included in R(P). For the converse, assume that x E R(P), so x = P , P z x . Then P, x = P: P 2 x = P , P z x = x , so x E R(P,). Similarly, we have x E R(P2), so x E R(P,) n R(P2). Returning to the spectral system of A , we observe that, for - 00 < ci S p < 00, the element P,-P, is also a component of E (more precisely, P,-Pa is the component of E in the band B, n B:). It is evident from Pa S P, that Pa is an element of the band B,, so Pa I(E-P,), i.e., P,(E-P,)

=

(E-P,)P,

=

e,

and so Pa = PUP, = P,P,. It follows that PB-P, = P,(E-P,)

=

(E-P,)P,,

which implies (in view of what was observed above about commuting orthogonal projections) that the range of the orthogonal projection P, -P, in the Hilbert space H is the intersection of the ranges of P, and E-Pa, i.e., R(P,-P,)

= R(P,) n N(P,),

which is the same as the orthogonal complement of R(P,) with respect to R(P,). Finally, if -00 < a, S p, S ci2 5 p2 < 00, we have so the ranges of PB,-Pa, and PB2-Pa, are mutually orthogonal subspaces of H.

Lemma 56.2. (i) The operator A, A,

=

=

B,A satisfies

P,A = AP,,

and SO we have A,x = A x for all x E R(P,) and A,x = 0 for all x E N(P,). Furthermore, A leaves R(P,) and N(P,) invariant, i.e., x E R(P,) implies A x E R(P,) and x E N(P,) implies A x E N(P,). (ii) We have (Ax, x ) S a(x, x ) for all x E R(P,) and ( A x , x ) 2 a(x,x ) for all x E N(P,). Hence, for - 00 < a S p < a,we have a(x, x ) S ( A x , x) S B(x, x )

for all x E R(P,) n N(P,) = R(P, -Pa). Furthermore, A leaves R(P,-P,) invariant.

CH. 8,s 561

Proof. (i) Since A ,

I(E-Pa), we have A,(E-P,)

so A ,

=

A,P,

391

THE SPECTRAL THEOREM

=

=

(E-P,)A, =

e,

IPa, we have = P,(A - A , ) = e,

P,A,. Since ( A - A,)

( A -A,)P,

so AP, = A,Pa and P,A = P,A,. It follows that A , = P,A

=

AP,.

In order to show that A leaves R(P,) invariant, assume that x E R(P,). Then A x = AP,x = P,Ax, so A x E R(P,). In order to show that A leaves N(P,) invariant, assume that x E N(P,), i.e., Pax = 0. Then P,Ax = A P g = 0, so A x E N(P,). Note that since N(P,) and R(P,) are orthogonal complements in H , we can also derive the invariance of one from the invariance of the other one by means of the theorem that if a Hermitian operator leaves a linear subspace invariant, then the operator also leaves the orthogonal complement invariant. (ii) By Theorem 38.4 (i) we have 9 , A S aP, for all a, i.e., A , S aP,. It follows that (A,x, x ) 5 a(P,x, x ) for all x E H, i.e., (AP,x, x ) S a(P,x, x ) for all x E H.

In particular, for x E R(P,), it follows that (Ax, x ) By Theorem 38.4 (ii) we have A-Y,A

5 a(x, x).

2 a(E-Pa) for all a,

i.e., A - A , 2 a(E-Pa). It follows that ( A x , x ) - (AP,x, x ) 2 a(x, x ) - .(Pax, x ) for all x E H. In particular, for x E N(P,), it follows that (Ax, x ) 2 a(x, x). Since A leaves R(PB)and N(P,) invariant, A also leaves the intersection R ( P B )n N(P,) = R(PB-P,) invariant. Given now the real numbers a and b and the positive number E such that U E 5 A 5 (b-&)E,

we assign to any partition n = n(ao, a t , . . ., a,) of [a, b ] the lower sum s(n; A ) and the upper sum t(n;A), given by

n

392

[a. 8,B 56

HERMITIAN OPERATORS I N HILBERT SPACE

The sums s(n; A) and t(n;A) are Hermitian operators of the form B k Q k , where the operators Qk (k = 1,. . ., n) are orthogonal projections in the Hilbert space H with mutually orthogonal ranges such that the direct sum of these ranges is the whole Hilbert space H , and the B k (k = 1, .. ., n ) are real numbers. Hence, the numbers Pk are the eigenvalues of & B k Q k , and the ranges R(Qk) are the corresponding eigenspaces. Furthermore, we have s(n; A) 5 A 5 r(n; A), and e 6 t ( n ;A ) - + ; A ) 5 I ~ I E , where In1 is the maximal interval length in the partition

e 6 A - s ( ~ ;A ) 5 e 5 t(n; A ) - A 5

R.

Hence

lnl~, I~IE,

which implies that the operator norms IIA-s(n; A)ll and Ilt(n;A)- All are less than or at most equal to 1x1. The spectral theorem for the Hermitian operator A follows now from Freudenthal's spectral theorem in Riesz spaces (Theorem 40.2), as follows.

Theorem 56.3 (Spectral theorem for a Hermitian operator). If A is a Hermitian operator in the Hilbert space H with spectral system (Pa ; - co < u < co), and the real numbers a, b are such that aE

5 A =< (b-E)E

for some E > 0, and if (n,,:n = 1,2, . . .) is a sequence of partitions of [a, b], each of which is a rejinement of its predecessor and such that In1 . 10, then s(n,,;A ) 4 A 6 t(n,,;A ) holdsfor every n, and s(n,,;A ) as well as t(n,,;A ) converge to A in the operator norm, i.e., IIA-s(nn; A)II + 0, llA-t(nn; A)ll

+

0

as n 4 ax Furthermore, in the Riesz space %'"({A)) or, for that matter, even in the ordered vector space 2,the increasing sequence ~(n,,; A ) has A as its supremum, and the decreasing sequence t(x,,; A ) has A as its infimum. Proof. Follows by applying Freudenthal's spectral theorem in the Riesz

CH. 8,s 561

393

THE SPECTRAL THEOREM

space %“({A}) and by observing that, on account of Lemma 56.1, E-uniform convergence is the same as convergence in the operator norm (usually called uniform convergence of operators). As already observed in the discussion of Freudenthal’s spectral theorem, the spectral theorem is often expressed by writing A

J-mu dP, . a?

=

We shall make no use of this notation. Note, however, that in Exercises 56.11, 56.12 and 56.13 it is indicated how the spectral system (Pg: - 00 < u < 0 0 ) generates a projection valued “measure” in the real line; the integral udP, can be interpreted as a Stieltjes-Lebesguetype integral with respect to the thus generated measure. The uniqueness result in Theorem 40.8, when applied to the Hermitian operator A in the last theorem, yields the following result.

Theorem 56.4. Let (Q, : - co < u < co) be a set of orthogonalprojections in the Hilbert space H , all Q, members of the second commutant W‘({A}),

and such that (i) Q, is increasing as u increases, (ii) Q,TQ,asuTP, Q , = e f o r a S a a n d Q , = E f o r u > b , (iii) for every sequence of partitions K , , ( U , , ~ , a n l , . . ., a,,,,,,) of [a, b ], each n,+ a refinement of K,, and such that ln,,l 1 0, the sequence m,

=

C1

k=

un,k-l(Q a* -Q am,k-a )

satisfies s; 7 A, i.e., (stx, x ) (Ax, x ) for every x E H (since, in any case, si increases as n increases, this condition is satisfied in particular whenever si converges to A in the operator norm). Then (Q,: - 00 < u < co) is the spectral system of A. Generally, for bounded linear operators, if IIA-A,Il 0 and IIB-BnII + 0 as n + co,then (IAB-A,,B,II 4 0 as n 4 00. Applying this to the case that A is the same Hermitian operator as before and B is a power of A, it follows easily by induction that -+

llAk-?(nn; A)ll

-+

0 as n

+ 00

for any natural number k, where (s(K,; A) :n = 1,2, . . .) is the same sequence of lower sums as before. The k-th power of s(n,; A) is easily determin-

394

HERMITIAN OPERATORS IN HILBERT SPACE

[CH. 8,s 56

ed. Indeed, since

with mutually disjoint terms, we have

Similarly, where

-,o

I I ~ ~ - t ~ ( i qA, ,) ;I ~

as n

--f

00,

There are several further remarks. In the first place, given the set 9 of mutually commuting elements of X , we consider the Yosida representation " { W ' ( 9 ) } of the Dedekind complete Riesz space W ' ( 9 ) . The elements of "{%'"( are 9) the } real continuous functions on the compact topological space $, the points of which are the maximal ideals in W ' ( 9 ) . The vector space structure and the lattice structure in "{W'( g ) }are the usual ones, so " { V f ( 9 )is}a Riesz space. The operator A E W ' ( 9 ) is mapped onto the function "u E " { W f ( 9 ) } where , the value "u(J) of "u at the point J E $ is the (unique) real number L satisfying A -AE E J. This mapping is a Riesz isomorphism of %"( 9 )onto "{V'(9)}. We return to the problem (mentioned in the preceding section) whether the operator multiplication in W ' ( 9 ) is the same as the multiplication inherited from the pointwise multiplication of the functions in the Yosida representation " { W ' ( 9 ) } .First we compare operator multiplication of components P, and P2 of E (these components are members of %"(9), of course) and the pointwise multiplication of the corresponding functions "p, and "p2. The components PIand P2 are commuting orthogonal projections in the Hilbert space H,let R(P,)and R(P2) be their ranges. Then P = P,P2 = P2P, is the orthogonal projection with range R ( P l ) n R(P2),and obviously P is a lower bound of P, and P2. Any other member A of W f ( 9 ) ,satisfying 8 5 A 5 PI and 8 6 A 5 P2,must vanish on each of the null spaces N ( P l ) and N ( P 2 ) ,so the range of A is included in R(P) = R(P,)n R(P2),which implies that A vanishes on the orthogonal complement of R(P). From the hypotheses it follows that 0 5 (Ax, x) 5 (x, x) holds for all x E R(P), so in view of (Px, x) = (x, x) holding for all these x, it follows now immediately that A 5 P.This shows

CH.8,§ 561

THE SPECTRAL THEOREM

395

that P is the greatest lower bound of P, and P,, i.e.,

P,P, = P2Pl

=

inf (P,,P,),

where inf (P,,Pz)denoles the infimum of P, and P2 in the Riesz space W'(9). The corresponding functions "pl and "p2 can assume only the values zero and one (since "pl and "p, are components of the function identically equal to one), so A

PI "P2

= ^P2 "P1 = inf ("PI 9

"PZ), where inf ("p,, "p2)denotes the infimum of "p, and "p2 in the Riesz space "{W'( 9)}Since . inf (P,,Pz)and inf ("pl, "p2)are corresponding elements in the Riesz isomorphism, it follows that P,P2 and "pi "p2are corresponding. But then A B and " d b correspond for A and B finite linear combinations of components of E in W"(9). Observe, finally, that for arbitrary members A and B of ""(9) there exist sequences A,, and B,,(n = 1, 2, . . .) in W"(9), each A,, and each B,, a finite linear combination of components of E, such that llA- A,,I I and IIB- B,,I I tend to zero as n + 00. It follows that IIAB- A , B.11 tends to zero, and the functions "a,,"b,,, corresponding to A,,B,,, converge uniformly in "{W'(L2)} to " d b . This shows that A B and "a"b are corresponding elements under the Riesz isomorphism. Hence, operator multiplication in W ' ( 9 ) is the same as the multiplication inherited from the pointwise multiplication of functions in "{W'( 9)). Secondly, it will be proved that the definition of the spectrum of an element in an Archimedean Riesz space with a strong unit, as presented in section 44 (cf. also Theorem 45.1), is equivalent to the familiar definition of the spectrum of a Hermitian operator A when we apply the general definition to the Riesz space %"({A}) or, more generally, to any Riesz space U " ( 9 ) of which the given operator A is a member. We recall the general definition. Given the Archimedean Riesz space L with strong unit e, we consider the Yosida representation "L. The spectrum o f f € L (with respect to the unit e ) is, by definition, the range of the function "E' "L corresponding tof, i.e., the real number c1 is in the spectrum off whenever there exists a maximal ideal J i n L such that u e - f E J. Assuming now that the Riesz space L is Dedekind complete (or even only Dedekind a-complete), the representation " L consists of all real continuous functions on the compact topological space f of all maximal ideals, and so it follows easily that for any given g E L the number zero is not in the spectrum of g if and only if there exists a function "h in "L such that "g"h = "h"g = "e, i.e., "g(J)"h(J) = 1 for all points J E$. The element h in L , corresponding to "h, is then the inverse of g with respect to the multiplication in L inherited from the point-

396

HERMITIAN OPERATORS IN HILBERT SPACE

[CH.

8,s 56

wise multiplication of the functions in " L , i.e., h = g-' in the usual notation. Hence, givenf E L and the real number ao, we have that a. is not in the spectrum off if and only if aoe-fhas an inverse in "L with respect to the multiplication in L inherited from "L. We consider now the Dedekind complete Riesz space W'(9), where 9 is again a collection of mutually commuting Hermitian operators in the Hilbert space H , and we observe first that the identity operator E is a strong unit in C'(9).Applying the remarks made above to the present situation, we obtain the result that, for any A in C"(9) and any real number a. ,the number a. is not in the spectrum of A if and only if a. E-A has an inverse in W'(9) with respect to the multiplication in W'(9) inherited from the pointwise multiplication in A{W(9)}. This multiplication, however, is the ordinary operator multiplication, as proved in the preceding remark. Hence, a. fails to be in the spectrum of A if and only if (aoE-A)-' exists as a member of W'( 9).This is almost the familiar definition. Indeed, according to the usual definition in Hilbert space theory we say that a. fails to be in the spectrum of A whenever ( a o E - A ) - l exists as a bounded linear operator. In order to show that the two definitions are equivalent, it remains to show that if A is a member of C"( 9 )and A - ' exists as a bounded linear operator, then A-' is also a member of W ' ( 9 ) . It is trivial to prove that A-' is Hermitian, and so all we have to show is that A-' commutes with every member of W ( 9 ) . For this purpose, let B in C ' ( 9 )and x and y in H be given, and set z = A-'y, so y = Az. Then ( A - ' B x , y ) = (A-'Bx, Az) = (Bx, Z ) = (x, Bz) = (A-'x, ABz) = (A-Ix, BAz) = ( B A - l x , v). This holds for all x , y E H , so A - ' B = BA-'. The proof is thus complete. We will now extendthe spectral theorem to bounded normaloperators in the Hilbert space H. For this purpose we recall some elementary facts. For any bounded linear operator T (from H into H ) there exists a unique bounded linear operator T* (the adjoint of T) such that (Tx,y) = (x, T*y) holds for all x and y in H. We have T* = T if and only if T is Hermitian (Le., if and only if T is self-adjoint), and T is called normal whenever T and T* commute, i.e., whenever TT*= T*T. For any bounded linear operator T, setting

A = +(T+T*)and B = -$(T-T*), the operators A and B are Hermitian, and T = A + i B and T* = A-iB.

CH.8,o 561

397

THE SPECTRAL THEOREM

This is the decomposition of an arbitrary T into Hermitian components. The decomposition is unique (Le., if A +iB = A'+iB' with A , B, A', B' Hermitian, then A = A' and B = B'). Furthermore, if we have T = A + i B with A and B Hermitian, then T is normal if and only if A and B commute. Assume now that N i s a normal bounded linear operator, and let N = A + iB be the decomposition into Hermitian components. We consider the Riesz space %"({A, B ) ) or, more generally, any Riesz space W'( 9),where 9 is a set of commuting Hermitian operators of which A and B are members. Let (P,: - 00 < a < a)and (Pi : - 00 < B < m) be the spectral systems of A and B respectively. All members of these spectral systems are members of %'"(9), and so they commute mutually. Now,for brevity, we shall denote by [a1 , a,; B1, B z ) the half open rectangle in the a/?-plane, given by ( ( a y B ) :a1

5

a

< a t , Bi

sB<

B2).

The corresponding operator

Q c ~uz: .

= (p=~-pui)(Pi~-ph1)

~ 1 , ~ 2 )

is an orthogonal projection in H with range equal to the intersection of the ranges of the projections Pal -Pu,and Pil -Pil. It follows immediately that Q ~ u i . u z ; ~ i~, z )

and

Q[a;a4:B3.P4)

have mutually orthogonal ranges (i.e., their product equals 0) as soon as the rectangles [a1, a,; pl, /I2)and [a3, a4; B 3 , B4) in the aB-plane are disjoint. Choose the real numbers a, b and the positive number E such that aE 5 A I; (b-E)E and aE 6 B 5 (b-e)E hold, let n(aO, al,. ., an) and n'(po, PI , ...,B,) be partitions of [a, b], and let

.

m

be the corresponding lower sums. According to the spectral theorem for Hermitian operators, s(n; A ) and s(n'; B ) converge in the operator norm to A and B respectively as 1111 0 and ln'l '+0 ; hence,

+

s(n, n'; N ) = s(n; A ) i s(n'; B )

converges in the operator norm to N = A + i B as

1% n'l

= max (bl, In'l)

398

HERMITIAN OPERATORS

[CH. 8,g 56

IN HILBERT SPACE

tends to zero. In order to investigate the sum s(x, A'; N ) more closely, we set Qkj

= ~ p a k - P a k - ] j ( P i j - ~ ~ , - ,fjo r k = 1 , .

. ., n

and j = 1 , . . ., u i .

Observing that m

we find

s(n, n'; N ) = s ( x ; A ) + i s ( n ' ; B ) n

n

=

m

m

C C (ak-l k=l j=1

+iBj-I)Qkj.

The sum s(z, n'; N ) might appropriately be called, therefore, a kind of lower sum corresponding to the partition (n,z') of the square [a, b ; a, b ] .The partition (z, d)is determined by the partitions a and d along the axes. As observed above, we have IIN-s(n, d ;N)II

+0

as I(n, A')I

-,0.

Similarly, we have where

(IN-r(n, n'; N)II

4

n

t(n, X ' ; N ) = C k=l

0 as

((71,

.)'I

+ 0,

m

C (ffk+iSj)Qkj. j=1

More generally, the following holds. Theorem 56.5 (Spectral theorem for a normal operator). With the notations introduced above, we have n

m

where C k j is any complex number in the rectangle times this is written as N = fJabC

dQr.

[ f f k - 1, f f k ;

Pi-

1,

Pi]. Some-

CH. 8 , § 561

399

THE SPECTRAL THEOREM

Once more, let 93 bs a set of mutually commuting (bounded) Hermitian operators in the Hilbert space H , and let U " ( 9 ) be the second commutant of 9.Any operator N = A + iB, with A and B members of V'(Q), is normal. The set of all normal operators of this kind will be denoted by M . Evidently, N is a complex linear subspace of the complex linear space of all bounded linear operators on H . The members of N commute mutually, and N is a ring in the algebraic sense (i.e., if N , and N , are members of N ,then N1N2 = N2N , is a member of N). For A , B, . . . members of C"( 9),let a, "b, . . . be the corresponding members of the Yosida representation "{V'(9)}. For N = A + i B e M , we will say that the complex function n = "a+i"b represents N . Evidently, "n is a continuous complex function on the compact Hausdorff space $ of all maximal ideals J in u"(9). Conversely, given any continuous complex function on $, the real and imaginary parts "a and "b of "n correspond to members A and B of %''(.9), and so "n corresponds to the member N = A + iB of N.It follows immediately that JV is ring isomorphic to the ring of all continuous complex functions on $. For any N E N ,we define the spectrum of N to consist of all complex numbers in the range of the function "n which corresponds to N . It is evident from the definition that the number zero is not in the spectrum of N if and only if "n has an inverse, i.e., if and only if there exists a continuous complex function "m on $ such that A

A

"n(J)"m(J) = "m(J)"n(J) = 1

holds for all J E f . This is equivalent, therefore, to saying that zero is not in the spectrum of N if and only if there exists a member M of N such that M N = NM = E holds. If this is the case, then M is the inverse operator N-' of N . Conversely, if N is a member of N and the inverse N-' exists as a bounded linear operator, then N-' is also a member o f N , and so zero is not in the spectrum of N.The proof that N-' is a member of N is not completely trivial. The details follow. First of all, we infer from the wellknown formula "-')* = (N*)-' that

N-'(N-')* = N-'(N*)-'

=

(N*N)-' =

("*)-I

- (N*)-"-'

=

(N-')*N-',

so N-' is normal. Let N-' = P + i Q be the decomposition of N-' into Hermitian components, and note that ( N - ' ) * = P - i Q . Since Nis a member of N ,any Hermitian A in the first commutant W ( 9 )commutes with N , i.e.,

400

[CH.8,5 56

HERMITIAN OPERATORS IN HILBERT SPACE

A N = NA. Multiplying on the left and the right by N - ' , we obtain N - ' A = A N - ' , so A commutes with N-' = P + iQ. Similarly,anyA in%?'( G) commutes with ( N - ' ) * = P - i Q . Hence A commutes with P and Q separately. It follows that P and Q are members of W ' ( 9 ) and so N-' is a member of Jlr. Returning to the spectrum of N , we have proved thus that the number zero is not in the spectrum of N if and only if N-' exists as a bounded linear operator. It is an easy consequence that any given complex number c( is not in the spectrum of N if and only if (aE-N)-' exists as a bounded linear operator. Hence, our definition of the spectrum of a normal operator is equivalent to the familiar definition. It follows also that if N = A iB is the decomposition into Hermitian components of a normal operator Nand G is any set of commuting Hermitian operators of which A and B are members, then the spectrum of N is independent of the particular choice of 9. For any N = A + iB E N the , spectrum of N is the range of the continuous function "n = "a+i"b on the compact Hausdorff space $, and so the spectrum of N is a compact subset of the complex plane. The adjoint operator N* = A - i B corresponds to the complex conjugate function "n = "a-i"b, so the spectrum of N* is obtained from the spectrum of N by reflection with respect to the real axis. Since N is a ring, the operator NN* = N*N corresponds to the function " n 5 = l"n12. Hence, the positive Hermitian operator NN* has its spectrum equal to the range of the function I %I2. Now, and this holds for any Hermitian operator A , there is an obvious relation between the operator norm IlAll and the values of the function "a corresponding to A , as follows. Writing IlAll = m for brevity, we have -mE 5 A 5 mE, so - m 5 ^a(J) 5 m for all J E $, it., I^a(J)I 5 m for all J E $. Applying this to the Hermitian operator NN* = N*N, and observing that

+

we obtain the result that l"n12(J) IIN1I2 holds for all J, i.e., I"nI(J) 5 IlNIl for all J. This shows that the range of "n is contained in the subset (z : lzl 6 IlNll) of the complex plane. In other words, the spectrum of N is a compact subset of the set (z : Izl 5 IlNll). It was observed above that the spectrum of NN* = N*Nis the range of the function IAnlz. I n particular, if Nis a unitary operator, i.e., if NN* = N*N = E, then I "n1'(J) = 1 for every J E $, so the spectrum of N is a subset of the boundary of the unit circle in the complex plane.

CH. 8,s 561

THE SPECTRAL THEOREM

40 1

Exercise 56.6. (i) Show that if A E W'(9) corresponds to the function "a, then the uniform norm of "a is exactly equal to IlAll. Hence, one at least of the numbers llAll and -1IAll is in the spectrum of A. (ii) Show that if the normal operator N E JV corresponds to the function "n, then the uniform norm of "n is exactly equal to 11N11. Hence, the spectrum of N has at least one number of absolute value exactly equal to I lNlI. Hint: We may assume that A is positive. For brevity, write IlAll = m. If "a(J) < m for all J, then the function m- "a has an inverse, so mE- A has an inverse. It follows that, for some positive constantp, we have Il(mE-A)xll B pllxll for all x , so ((mE-A)x, x ) 2 p ( x , x ) for all x. This implies m- (Ax, x ) 5 p for all x with llxll = 1. Contradiction. Exercise 56.7. Let Pa(- co < a < co) be the spectral system of the Hermitian operator A. Show that a, is not in the spectrum of A if and only if there is a neighborhood of a, such that Pa is constant in this neighborhood, i.e., there exist real numbers a1 and a, such that a1 < a, < a, and PoIz-Palis the null operator. Hint: Combine Theorem 45.1 and Lemma 55.5 (ii). Exercise 56.8. Let N = A + i B be the decomposition into Hermitian components of the normal bounded linear operator N , and let (Pa : - co < a < a)and (Pi : - co < jl < co) be the spectral systems of A and B respectively. Also, let [xl,x,; y l , y z ]be a rectangle in the complex plane, completely outside the circle (z : IzI 5 IlNlI), and let Qo = (pxz-pxl)(p~2-p~,)

be the corresponding orthogonal projection. Show that Q , is the null operator. Hint: With the notations of Theorem 56.5 we have

so holds for every x E H. Choose E > 0 such that the given rectangle is contained in (z : IzI > IINII + E ) . Assume now that Q , is not the null operator. Then there exists x,, E H such that Ilxol I = 1 and Q , x, = xo , so (Q,x, ,x,) = 1. With the notations of Theorem 56.5, choose partitions 'II and n' such that the given rectangle is inside the square [a, b; a, b] and such that xl, xz and y , ,y , are among the endpoints of the subintervals of n and n' respectively.

402

HERMITIAN OPERATORS I N HILBERT SPACE

[CH.

8,s 56

For ( x , x ' ) we have c = C C lckj12(Qkjxo, =

XO)

=

(CC )'1ckj12(Qkjxo,xo),

C C ICkjI'(QkjQox0,

XO)

where the last summation is exclusively over the combinations (k,j ) for which the corresponding rectangle lies inside [ x , , x,; yl ,y 2 ] .Hence

On the other hand, we have

Inequalities (2) and (3) together contradict (1). Exercise 56.9. Let N = A + iB be the same as in the preceding exercise, and also let Paand Pi be the same. For any rectangle p = [xl ,x,; yl ,y z ] in the complex plane, we shall denote the corresponding orthogonal projection

briefly by Q ( p ) . Assume now that the origin of the complex plane is not in the spectrum of N. Show that there exists a rectangle p around the origin such that Q ( p ) is the null operator. Hint: There exists a number C > 0 such that IINxlI 2 Cllxll for all x E H , i.e., (N*Nx, x ) 2 C211x112for all x E H. Take a rectangle p around the origin and contained in (z : IzI 5 *C), and show (similarly as in the preceding exercise) that Q ( p ) is the null operator. Exercise 56.10. For the same operator N as in the prueding exercise, assume that the complex number zo = xo+ iyo is not in the spectrum of N . Show that there exists a rectangle p around zo such that Q @ ) is the null operator. Hint: Observe that the origin is not in the spectrum of N-zoE, and apply the result of the preceding exercise. Exercise 56.11. Let Pa(- m < a < m) be the spectral system of the Hermitian operator A in the Hilbert space H . For any half open interval D = [a1 , a,) and any x E H , write P ( D ) = Pa,-Pa,and p x ( D ) = (P(D)x,x ) .

CH. 8, 6 561

THE SPECTRAL THEOREM

403

Since the real function g,(a) = (Pax,x ) is monotonely increasing in a and left continuous at every point ao(on account of P, t Pa,,as a 1cr,), it follows along familiar lines that on the semi-ring of all intervals D = [al , crz) the set function p, is a non-negative and countably additive measure. The measure p, is extended by means of the Carathbodory extension procedure. Every Borel subset of the real line is then px-measurable. This is done for every x E H , and hence every p x is a Stieltjes-Lebesgue measure on the a-field (a-algebra) of all Borel sets. Since p,(D) = (P(D)x,x ) 5 (x, x ) holds for every D = [a1, a2), it follows easily that p J E ) 5 Ilxll’ holds for every Borel set E, in particular for E equal to the real line itself. Now, keep the Borel set E fixed, and show that the square root of px(E),as a function of x, is a seminorm in the Hilbert space H , obeying the parallelogram law and majorized by the norm llxll in H, i.e., pc,,(E) = I c 1 2 p x ( ~for ) a11 complex c, {PX+Y(E)l3 P x +Y@)

PJE)

s {Px(E)13+ {PY(E)I*Y

+P.x - y(E) = 2P,(E) +2PY(E)Y

(4)

5 IlXllZ.

Hence, by a well-known theorem of J. von Neumann and P. Jordan (for a proof we refer to Wilansky [I], section 8.3), the functional qE(x, Y ) = i { ~ x + y ( E )-A - y ( E l + i ~ xd+E ) - i ~ x- iy(E)>

is a semi-inner product (i.e., it has all the properties of an inner product except that q E ( x ,x ) m a y be zero without x being the null element). Also, rp,(x, x ) = p,(E) for all x , and so I ~ E ( X U)l2 Y

5 p ~ ( E ) p y ( ~ )llxl12 ’ IlYl12

holds for all x , y in H. Hint: The formulas (4) hold for E = D = [al,a2), and by familiar measure theoretic arguments can then be shown to hold for every Borel set E.

Exercise 56.12. This is a continuation of the preceding exercise. For E and y fixed, q E ( x , y )is a bounded linear functional on H, so there exists y1 E H such that q E ( x ,y ) = ( x , yl) holds for all x. Evidently, y1 depends on y and E. Writingy, = T(E)y,it follows immediately that T ( E ) is a bounded linear operator satisfying IIT(E)II 5 1. Hence, q E ( x , y )= ( x , T (E )y )holds for all x , y in H, and IIT(E)I I 5 1. Also, since qE(x,y ) and q E ( y ,x ) are complex conjugate numbers, it follows that T ( E ) is Hermitian, so q E ( x , y )=

404

[CH. 8,§ 56

HERMITIAN OPERATORS IN HILBERT SPACE

(T(E)x,y ) holds for all x , y in H. For all x

E Hand

(T(D)x,x ) = CpD(X, x) = A ( D ) =

all D = [ a , , r 2 )we have

( W ) X ,

x),

so (since T ( D ) and P ( D ) are Hermitian) we may conclude that T ( D ) coincides with the orthogonal projectionP(D) = Pa,-Pa,. Show that if E is afinite union of half open intervals (open on the right), then T ( E ) is also an orthogonal projection, and note that the collection of all such finite unions is a ring. Finally, by observing that the smallest monotone family containing a given ring is the same as the smallest a-ring containing the same ring (cf. the measure theory book by P. Halmos [l], section 6), show that T ( E )is an orthogonal projection for every Borel set E. Hint: Observe that a finite union of half open intervals can be written as a disjoint finite union of such intervals. For the last statement, fix x E H and prove that the collection of all Borel sets F satisfying

(T(F)x, T(F)x) = (T(F)x,x ) is a monotone family, which, therefore, must coincide with the collection of all Borel sets.

Exercise 56.13. This is a continuation of the preceding exercise. For any Borel set E, write P(E) instead of T ( E) . Thus the set function P,defined for the Borel sets, is now an extension of the original set function P, defined

only for the half open intervals. The set function P is a projection valued measure on the a-field of the Borel sets (i.e., P(+) is the null operator, P ( E ) is an orthogonal projection for every Borel set E, and P is a-additive, which means that P ( E ) = P(E,)+P(E,) for E = El u E2 with El and E2 disjoint and P(En)t P( E) for En t E). Furthermore, if the real numbers a, b have the property that the given Hermitian operator A is between a times and b times the identity operator, then P(E) is equal to the identity operator for any Econtaining [a, b] in its interior. Show that, for any Borel sets El and E 2 , we have

P(El u &)+f‘(E1 n E2) = P(E1)+P(E2)7

P(E1 n E2)

= P(El)P&),

(5)

so in particular P(E,)P(E,) is the null operator for El and E2 disjoint. Given a partition n = (. . ., LL,, a,,, a , , . . .) of the real line, A is approximated in norm by

CH.8 , § 571

so

405

THE STRUCTURE OF THE SECOND COMMUTANT

( A x , x ) = lim

1a,(P(D,)x, x) = lim C a,$,(&)

Inl-10

Inl-tO

The sum on the right is exactly the Stielljes-Lebesgue integral (with respect to the measure p x ) of the step function equal to on Dk for each k. Hence, we obtain

s

( A x , x ) = adp,. Since p, is generated by the monotone function g,(cx) be written as

=

(Pax,x), this can

( A x , x) = / a d g x ( a ) = J a d ( P a x , x). Show, similarly, that for x and y fixed the set function p* ( E ) = (P(E)x,y ) is a complex measure on the a-field of the Bore1 sets, and show that

( A x , y) = s a d p - = s a d ( P a x ,y). Hint: In order to prove the first formula in (5), add P(El n E,) to both sides of P(E, u E,) = P(E1)+P(E, - € 1 ) .

From P(El n E,)

5 P(E,) S P(E, u

E,) it follows that

P(E,)P(E, n E,) = P(E, n E,) and P(El)P(E1u E,) = P(El). Hence, multiplying the first formula in ( 5 ) by P(E,), we obtain P(El)+P(El n Ed = P(El)+P(E,)P(E,)Y which proves the second formula in (5).

57. The structure of the second commutant In this section we will prove some additional theorems about the second commutant %”( g ) ,where 9 is once more a set of mutually commuting Hermitian operators in the Hilbert space H . For each A E 9 the null space N ( A ) of A is a closed linear subspace of H, and so H’ = n ( N ( A ) : A

E

9)

is also a closed linear subspace of H. Let H” be the orthogonal complement of

406

HERMITIAN OPERATORS IN HILBERT SPACE

[CH.8,s 57

H', and let P' and P" = E-P' be the orthogonal projections in H having as their ranges the subspaces H' and H" respectively. It is a reasonable conjecture that the behavior of an element of %"( 9 )on the subspace H cannot be of great importance; we will prove that this is indeed true. We first recall that if the Hermitian operator B leaves a closed linear subspace H 1 invariant, then B leaves the orthogonal complement of H , invariant. Indeed, given that Bx E H , holds for all x E H I , any y IH , satisfies (By, x ) = ( y , B x ) = 0 for all x E H , ,so By IH I . Furthermore, if A and B are commuting Hermitian operators, then B leaves the null space N ( A ) and its orthogonal complement R ( A ) invariant (we recall that, by our definitions, R ( A ) is the closure of the range of A). Indeed, x E N ( A ) implies ABx = BAx = 0, so Bx E N(A). This shows that B leaves N ( A ) invariant, so that, by the preceding remark, B leaves R ( A ) invariant as well.

Theorem 57.1. As above, let 9 be a set of mutually commuting Hermitian operators in the Hilbert space H , let H' = n ( N ( A ) : A E 9 )and let H" be the orthogonal complement of H . Finally, let P' and P" be the orthogonalprojections on H and H' respectively. Then the following holds. (i) Each B E V(B) leaves H and H' invariant, and among the elements of W( 9 )are all Hermitian B which vanish on H" (i.e., Bx = 0for all x E H I ) . These are exactly the Hermitian B satisfying A B = BA = 8for all A E 9. (ii) Each B E "'(23) leaves H' and H" invariant, atid the restriction of B on H' is a real multiple of the identity operator on H'. In other words, there exists a real number a such that B - aP' is the null operator on H' (and, of course' B - UP'is equal to B on H ' ) . In the converse direction, any real multiple of P' is an element of W ' ( 9 ) . (iii) The principal band d generated by P' in W ' ( 9 ) consists of all real multiples of P'; the band L3 generated in V ' ( 9 )by the elements of 9 is the disjoint complement of d ,and we have

at @ 9l

=

W'(9).

Proof. (i) Since H and H" are orthogonal complements, it is sufficient to show that each B E V ( 9 )leaves H invariant. Given B E %'(9 we ), have A B = BA for every A E 9, and so B leaves N ( A ) invariant for every A E 9. But then B leaves H' = n ( N ( A ) :A E 9 ) invariant. Now, let B be Hermitian such that B x = 0 for all x E H'.This is equivalent to saying that H' is included in N(B), i.e., R(B) c H . Observing that

CH.8,s 571

THE STRUCTURE OF THE SECOND COMMUTANT

407

H c N(A) or, equivalently, R(A) c H'holds for all A E 9,we derive from R(B) c H and R(A) c H' for all A E 9 that R(B) is orthogonal to all R(A), and so AB = BA = 8 for all A E 9. It follows in particular that B commutes with every A E 9, and so B E W ( 9 ) . Conversely, if B is Hermitian such that AB = BA = 8 for all A E 9, then R ( B ) c N(A) for all A E 9, so R(B) c n (N(A) :A

E

9)= H'.

This implies that HI' c N(B), i.e., B vanishes on H'. (ii) The operator P' is Hermitian, and P'x = 0 holds for every x E H". It follows from part (i) of the present proof that P' is a member of W( 9). Each B E W ' ( 9 ) commutes therefore with P', and so B leaves the null space H'of P'invariant. But then B leaves the orthogonal complement H' invariant as well. The operator B in C"( 9)commutes with every C in W( 9), and so the restriction of B on H commutes with the restriction of C on H . But on account of what was proved in part (i), if we let C run through V ( 9 ) ,the restriction of C o n H runs through the set of all Hermitian operators in the Hilbert space H'. Hence, by Lemma 53.2 (i), the restriction of B on H must be a real multiple of the identity operator on H . In other words, there exists a real number a such that B-UP' is equal to the null operator on H'. Of course, B-UP' is equal to B on H ' . In the converse direction, every real multiple of P' is evidently an element of W ' ( 9 ) . (iii) If B is a member of the ideal generated in W ' ( 9 ) by P', then B is between two multiples of P', and so B vanishes on H ' . Furthermore, by part (ii), B is on H a real multiple of the identity operator. These facts together show that B is a real multiple of P' on the whole space H. It follows now that if B is a member of the band d generated in u"(9)by P', then B is also a real multiple of P'. Since AP' = P'A = 8 holds for every A E 9, we have P' IA for every A E 9, and so P' IB holds for every B in the band 3? generated in %"(9 ) by the elements of 9. This shows that the bands CQZ and 3? are disjoint. Since "'(9)is Dedekind complete, the direct sum d'0 3? is a band in %"'(9 and ), not merely an ideal. Let CEC ' ( 9 )be disjoint to d 0 3?, so C I P' and C I A for all A E 9. Then CA = AC = 8 for all A E 9, i.e., R(C) is contained in N(A) for every A E 9.It follows that R ( C ) is contained in the intersection of all N(A), i.e., R(C)c H . But C IP' implies that CP'= P'C = 8, so R(C)is contained in the null space H ' of P'. But then R ( C ) is = g"(9). contained in H' n H", so C = 8. It follows that d 0

408

HERMITIAN OPERATORS IN HILBERT SPACE

[CH.8 , s 57

As observed earlier, every polynomial in the elements of 9 and with real coefficients, a constant term consisting of a real multiple of E included, is a member of %"(9), and it may be asked in a rather vague sense how dense the set of these polynomials is lying in Evidently, the band genersimply because the ideal ated by these polynomials is the whole of %'"(9), generated by E alone is already the whole of W ' ( 9 ) . The band generated by the polynomials without constant term is the band &? introduced in part (iii) of the last theorem (note that this makes a difference only if E is no member of 9).The following result, for the case that the Hilbert space H is separable, is more precise.

%''(a).

Theorem 57.2. If H is separable, then for any member A of %"( 9 )there exists a sequence (A,, : n = 1,2, .) in the Riesz subspace generated by die polynomials in the elements of 9 such that A,, converges strongly to A , i.e., IIAx-A,,xll --+ 0 for every X E H .

..

Proof. Polynomials in the elements of 9, with real coefficients, will be denoted by q( 9), q 1 ( 9 ) ,q2(B),. . .. The proof is divided into several parts. (i) Given x E H, let H [ x ]be the closed linear subspace of H spanned by all elements of the form q ( 9 ) x . Then we have H [ y ]IH [ x ] if y IH [ x ] . For the proof it is sufficient to show that (zl , z 2 ) = 0 for z1 = q 1 ( 9 ) y and z2 = q2(9 ) x . This, however, is evident on account of

(z1,z 2 ) = (41(=wY> 4 2 ( 9 ) x ) = ( A 9 1 P ) q 2 ( 9 ) x ) = (YY q 3 ( 9 ) x )

= 0.

(ii) The space H i s the direct sum of at most countably many mutually orthogonal subspaces H[x,,](n= 1,2, .). Indeed, since H is separable, it is evident in the first place that the number of mutually orthogonal H [ x ] , with each x # 0, is at most countable. That H is indeed the direct sum of such subspaces follows then from Zorn's lemma. The use of Zorn's lemma can be avoided by starting with a countable dense set (y,, : n = 1,2, .) in H, and defining x1 = y l , then x2 the component of y 2 orthogonal to H [ x , ] , then x3 the component of y 3 orthogonal to H [ x , ] 0 H [ x 2 ] ,and so on. In order to show that H i s then the direct sum of the H[x,,](for those values of n for which x,, # 0), it is sufficient to show that z l H[x,,]for all n implies that z = 0. This is easy; if z 1H[x,,]for all n, then z Iy . for all n, and so z = 0. (iii) Any A E W ' ( 9 ) leaves every H [ x ] invariant, i.e., y E H [ x ] implies that A y E H [ x ] . For the proof, let x be given, and denote the orthogonal projection on H [ x ] by P. Every B E 52 leaves H [ x ] invariant (in view of the definition of H [ x ] ) ,so PBP = BP, which shows that BPis Hermitian (simply

..

..

CH.8, $571

409

THE STRUCTURJ? OF THE SECOND COMMUTANT

becausePBPis Hermitian). But then Pand Bcommute by Lemma 53.1 (i), i.e.,

P E C ( 9 ) .It follows, on account of A E W'(g), that A and P commute.

Hence, y E H [ x ] implies that Ay = APy = PAY, so Ay E H [ x ] . (iv) Now, let A E W ( 9 )be given. According to part (ii), H is the direct sum of the mutually orthogonal subspaces H[x,,](n= 1,2, . . .). We may assume that Ilx,,ll = 1 for all n. We set x, = n-lx,,, which implies Ax, = n-'Ax,,. Observe that, in view of what was proved in part (iii), the components of Ax, in the mutually orthogonal subspacesH[x,,]are exactly the elements n-'Ax,. Once more by part (iii) we have Ax, E H [ x , ] ; hence, by the definition of H [ x , ] , there exists a sequence of polynomials ( q k ( 9 ) : k = 1,2, . . .) such that q k ( 9 ) x o + Ax, as k --t a.

x:

zy

The component of qk(9)xo in H[x,,]is n-'q,(g)x,, and, as observed above, the component of Ax, is n-'Ax,,. Hence i.e.,

n-'q,(Q)x,,

+ n-'Ax,,

q k ( 9 ) x , , + Ax,, as k

--t

as k

co for n

+ 00,

= 1,2,.

..

But then, for n fixed, q k ( 9 ) x + Ax holds for every x of the form x = q( 9)x,,, i.e., the formula q J 9 ) x + Ax holds for a set of x dense in H[x,,]. Since this is true for n = 1,2, . . ., the formula holds for a set of x dense in H. Let K be this dense set. Corresponding to the given operator A E W ( 9 ) ,there exists a number m > 0 such that -mE 6 A 5 mE. The operators R, = sup ( q k ( 9 ) , -mE) are members of the Riesz subspace of W"( 9)generated by the polynomials q@), and RkX = (sup ( q k , -mE))x

+ (SUP

( A , -mE))x = AX

holds for all x E K by Lemma 55.4 (iii). Similarly, the operators Ak = inf (R,, M E )are members of the same Riesz subspace, and A,x + Ax holds for all x E K.Inaddition, thenorms llAkll satisfy llAkll 5 m for k = 1, 2, . . .. Now, since K is dense in H, since A,x + Ax holds for all x E K and llAkll 5 m for all k,it follows easily that Akx + Ax holds for all x E H. This completes the proof. The above theorem and the proof of it are related to a theorem of J. von Neumann, F. Riesz, Y. Mimura and B. Sz. -Nagy (cf. the book by F. Riesz and B. Sz.-Nagy [l], section 129). In the terminology of the theorem referred

410

HERMITIAN OPERATORS I N HILBERT SPACE

[CH.

8,s 57

to, we have proved that every A in W'(B) is a "function" of the elements of B. Given the set 9 of mutually commuting elements of 2,it was shown in Lemma 55.1 that the second commutant W'( 9), besides being a linear subspace of 2,is also a commutative ring. In addition, it is easy to see that if the sequence ( A , :n = 1,2, .) is containec in W'(9 )and A, converges strongly (or even only weakly) to the boundedlinear operator A , then A is in V'(9). These conditions are, therefore, necessary in order that a subset of 2 be the second commutant of some other subset of mutually commuting elements of 3".It may be asked whether these conditions are also sufficient. We will show now that, in a separable Hilbert space, this is indeed the case.

..

Theorem 57.3. Let W be a subset of 8 satisfying the following conditions. (i) W is a (real) linear subspace of X . (ii) W is a ring in the algebraic sense, i.e., A , B E W impIies that A B E 9 (and hence all members of W are mutuaIIy commuting). (iii) If A , E W f o r n = 1,2, . . . and the sequence (A, : n = 1,2, . . .) converges strongly to the bounded linear operator A , then A E 9. Then W is a Riesz space with respect to the ordering inherited from 2. In addition, if the Hilbert space H is separable and if the identity operator E is a member of 9, then %"'(W) = 9.

Proof. For the proof that W is a Riesz space with respect to the ordering inherited from X , it is sufficient to prove that for every A E 92 we have A + E W and A + = sup ( A , 8 ) with respect to the ordering referred to. As before, A + is here defined by A" = + ( A IAI). For the purpose of this part of the proof, let A E W be given. Then A' E 92,and so any polynomial in A' without constant term is also a member of W . Since IAl is the strong limit of a sequence of polynomials of this kind, it follows that IAl E 9, and so A" = + ( A+ IAI) E W.Hence, since A + 2 A and A" 2 8 hold in 2, the This shows already that A + is an upper bound same inequalities hold in 9. of A and 8 in W . In addition, we have by Theorem 54.2 that A + is the smallest among all Hermitian operators that commute with A and are upper bounds of A and 8. Since any member of W commutes with A , it follows that A" is the smallest among all members of W that are upper bounds of A and 0. In other words, A + = sup ( A , 0) holds in 9. This completes the proof that W is a Riesz space. Note that W is a linear subspace of W'(W),and that for every A E W

+

CH.8,s 571

THE STRUCTURE OF THE SECOND COMMUTANT

41 1

the equality A + = sup ( A , 8) does hold not only in W but also in W'(W). This shows that 9is a Riesz subspace of W'(W). Assume now that the Hilbert space H i s separable, and E E 9. Then every polynomial in the elements of W is again in W and so, in view of W being a Riesz subspace of W'(W), it is evident that the Riesz subspace of %"(W) generated by the polynomials in the elements of W is the space 9 itself. By the preceding theorem every member of %'"(a) is the strong limit of a sequence of elements in this Riesz subspace, i.e., every member of W ' ( W ) is the stronglimit of a sequence of members of B.By hypothesis the strong limit of a sequence of members of W is a member of W ; it follows that c 9. It is obvious that W c W'(W), and hence = W. For the first part of this theorem, cf. also B.Z. Vulikh ([l], 1957).

%"(a)

%''(a)

Any Riesz space L which is at the same time a normed linear space is called a normed Riesz space whenever norm and order in L are compatible, i.e., whenever llfll 5 llgll holds for every pairf, g E L satisfying If1 5 Igl. Note that this is equivalent to requiring that 11 If1 11 = l l f l l holds for all f~ L and IIf1 I 6 1 Igl I holds for every pair f, g E L satisfying 0 6 f 5 g. Normed Riesz spaces will be investigated in later chapters. For the present we restrict ourselves to observing that, for any subset 9 of mutually commuting elements of 2,the Riesz space W ' ( 9 ) is a normed Riesz space with respect to the ordinary operator norm. Indeed, it was proved in Lemma 55.4 (i) that II IAl II = llAll holds for every A E W ' ( 9 ) and it follows from Lemma 55.4 (ii) that 8 5 A 5 B implies IlAll 5 11B11. Indeed, it was proved in Lemma 55.4 (ii) that, for A , B E %"( 9),we have IAl 5 IBI if and only if 11A.xIl 5 IlBxll holds for every x E H. The normed Riesz space L is said to be an abstract L, space whenever

holds for every pair f, g E L ' . It is evident that the name is derived from the fact that the usual norm in any L , space satisfies this condition. It will be shown now that W'(9) is an abstract L, space. The proof could be based on the Yosida representation "{W'( g ) }by showing that the operator norm in %"( 9 )corresponds with the uniform norm in the space "{W'( 9))of continuous functions (cf. Exercise 56.1). We prefer to present an elementary proof, based on the spectral theorem.

Lemma 57.4. Let PI,. . ., P,,be orthogonalprojections in the Hilbert space

412

HERMITIAN OPERATORS I N HILBERT SPACE

[CH.

8,g 57

H such that P i p j = 8 for i # j , and let S = x a i P i and i=l

..

T = ZziPi, I= 1

.

where a1, ., a, and z1 , . .,7, are non-negative numbers. It i s evident that S and T are Hermitian and commuting, and so SUP

holds in W'({Pl ,

( S , T ) = *(S+T+IS-TI)

...,P,}).The norms satisfy

IlSUP (S7 Till = max (IlSll, IITll).

Proof. It follows from

(S- T), =

that IS-TI =

Hence, observing that m a

we obtain

(Ci, T i )

c (ai-zt)2Pi n

t= 1

x loi-rJPt. n

i= 1

= t.(ot+zi+lai-ziI),

sup ( S , T ) =

c max n

i= 1

(ai,

zI)Pt.

Without loss of generality we may assume now that P i # 6 for i = 1, . . .,n. It is evident then from the definition of S that llSll = max (ai : 1 6 i

5 n),

and similar formulas hold for IIT1 I and for I lsup (S, T)I I. Hence IlSUP

(S T)lI

= max {max (Qi, 1Sisn

Ti)}

= max (IlSll, IlTll).

.

Lemma 57.5. If A , S1, S , ,. . are Hermitian and mutually commuting, and i f llA-S,,Il + 0, then 11 lAl-IS,,l 11 + 0. Proof. The second commutant W ' (9)of 9 = {A, S,, S2, . . .} is a Riesz space containing A and lAl, all S,, and all IS,,l. Hence

IlAl-lSnIl 6 IA-Snl,

CH.8,§ 571

THE STRUCTURE OF THE SECOND COMMUTANT

413

and so, keeping in mind that 11 IBI 11 = llBl1 holds for any Hermitian B, we obtain II IAl-ISnl I I 5 IIA-SnII 0. +

.

Corollary 57.6. If A, B, S, , S, , . . and TI, T 2 ,. . .are Hermitian and mutually commuting, and if I IA - S,,l I + 0 and I I B - T,I I + 0, then

IISUP

(A, B)-~uP ( S n y G)II

+

0.

-

Proof. The sequence S,,- T,,converges in norm to A B. Hence, by the above lemma, IS,,-T,,I converges in norm to IA -BI. It follows immediately that sup (S,,, T,,) converges in norm to sup (A, B). Theorem 57.7. If 9 is a set of mutually commuting Hermitian operators, and if A and B are members of W ( 9 )such that A 2 8 and B 2 8, then lbUP (A, B)II = max (IIAIL IlBll). Hence, W’( 9 )is an abstract L, space.

Proof. We may assume without loss of generality that 8 5 A 0 5 B 5 E. The spectral systems of A and B will be denoted by

5 E and

(P,(A) : -a < 1 < a)and (PJB) : -a < ,u < a) respectively. Any PA@)and any PJB) is a member of W ‘ ( 9 ) , and so they commute, i.e., PA(A)P,(B) = ~/I(B)~A(A). Consider now two partitions of the interval [0,2], defined by 0 = A0 < A1 <

... < 1, = 2,

0 = / l o < p1 < ... < & = 2,

and let

.

Qij

= { ~ A , + , ( A ) - P A , ( A ) } { i(B)-pp,(B)I P,+

.

for i = 0, 1, . .,n- 1 and j = 0, 1, . .,m - 1. Every Q i j is an orthogonal projection in H , and Q,Q,, = 8 for ( i , j ) # (k,1). Hence, if aij and 7 i j are non-negative numbers, and if ij

we have by Lemma 57.4 that

ij

414

HERMITIAN OPERATORS I N HILBERT SPACE

This holds in particular if a i j = l i and z i j = P j for all i and j , so

i.e., S and Tare the lower sums of A and B respectively with respect to the partitions introduced above. Consider now a sequence of such partitions, each one a refinement of its predecessor, and let S,, T, (n = 1,2, . . .) be the corresponding lower sums of A and B. By the spectral theorem S, and T, converge in norm to A and B respectively and, as elements of V’(9),these operators all commute among each other. Hence, it follows from the last corollary that sup (S,,, T,) converges in norm to sup ( A , B). But then ~ ~ S llTnll , , ~ ~and , llsup (S,, T,)II converge to 11A11, IlBll and llsup ( A , B)II respectively, sc from

IISUP

(Sn, Tn)II = max (IISnIIy IITnII),

which holds for n = 1,2,

. . ., we derive finally that

llsup (A, B)II = max (IIAIL IIBII).

Exercise 57.8. It was shown in Theorem 57.2 that if the Hilbert space H is separable and 9 is a set of mutually commuting members of 2,then any member of V ’ ( 9is ) the strong limit of some sequence in the Riesz subspace generated in V ’ ( 9 by ) the polynomials in the elements of 9. In order to show that this does not necessarily hold if H is non-separable, let H be the same Hilbert space as in Exercise 53.5, i.e., H consists of all complex functions x ( t ) on the real line satisfying x ( t ) # 0 for at most countably many t, and such that xtlx(t)lz < co. The inner product is ( x , y ) = C t x ( f ) J ( t ) .Given the real number t o ythe one-dimensional linear subspace H(to) of H is defined by H(t,,) = ( x :x

E H,

~ ( t= ) 0 for all t #

to),

and P(to) is the orthogonal projection on H(to).Let 9 = ( P ( t ) : --co < t < a).

The set 9 is a subset of 2 (i.e., all P ( t ) are Hermitian) and P(s)P(t) = P(t)P(s) = 8 for s # t. Show that any A E V ( 9 )is of the form ( W t )= 4 M t )

CH.8 , 5 581

KADISON'S ANTI-LATTICE THEOREM

415

for some real bounded function a ( t ) and that, conversely, any a(t) of this kind thus determines a member of W ( 9 ) . At the next step, show that U"(9) = W ( 9 ) . Show also that if A, B E W ' ( 9 ) correspond thus to a(t) and P(t) respectively, then sup ( A , B ) corresponds to the function max (a(t),p(t)). Finally, show that if An(. = 1, 2, . . .) and A are members of C"(9),and A , converges strongly to A , then a,(f) converges pointwise to u(t). Show now that the Riesz subspace of W ' ( 9 ) generated by 9 consists of all members of U"(9) corresponding to real a ( t ) such that act) # 0 for at most finitely many t. Hence, the Riesz subspace generated by the polynomials in the elements of 9 (i.e., the Riesz subspace generated by 9 and the identity operator E ) consists of all members of W ' ( 9 ) corresponding to real a(t) of the form a ( t ) = ae(t)+a,(t), where a is a real number, e(t) = 1 for all t, and a,(t) # 0 for at most finitely many t. It follows that if the member A of W'(9 )is the strong limit of some sequence in this Riesz subspace, then the function a(t) corresponding to A is of the form a ( t ) = ae(t)+a,(t), with a2(t) # 0 for at most countably many t. This shows that not every member of W ' ( 9 ) is such a strong limit. 58. Kadison's anti-lattice theorem Once more, let &' be the ordered vector space of all bounded Hermitian operators in the Hilbert space H . It may be asked whether i@ is a Riesz space with respect to the ordering. Following R. V. Kadison ([I], 1951) we will prove that i@ is no Riesz space unless the Hilbert space H is onedimensional. We will prove, in fact, that &' is what is called an anti-lattice. The definition follows. Definition 58.1. The ordered vector space L is called an anti-lattice whenever, for every pairf, g E L, the element inf (f,g ) exists i f and only i f f and g are comparable elements (i.e.,f 2 g or f 4 g ) . Note that if L is an ordered vector space with a strong unit, then L is an anti-lattice whenever, for every pair u, v in the positive cone L ', the element inf (u, v) exists if and only if u and u are comparable. We turn to the proof that 2 is an anti-lattice.

416

HERMITIAN OPERATORS IN HILBERT SPACE

[CH. 8,s 58

Lemma 58.2. Let P I and P , be orthogonal projections in the Hilbert space H such that inf ( P , , P,) = 8 holds in 2. Then P , and P2 have mutually orthogonal ranges, i.e., P , P2 = P2P I = 8. Proof. Let the ranges of P , and P , be R ( P , ) and R(P,) respectively, and denote the orthogonal projection on the closure of the algebraic sum R(P,) R(P2) by P , . It is evident that P , 5 P 3 and P , 5 P 3 . Write D = P 3 - P 2 . It follows from D 2 8 that P , - D 6 P , , and it follows from P , 5 P 3 that

+

Hence

so

Pi-D

-

PI D PI

5 P3-D

= P2.

5 inf (Pi ,P 2 ) = 8, 5D

= P3-PZ.

This shows that the range of P I is included in the range of the orthogonal projection P 3 - P 2 . Since the ranges of P,-P, and P p are mutually orthogonal subspaces, it follows that the ranges of P I and P2 are mutually orthogonal. In the next theorem we present a much stronger result. Theorem 58.3. If P , and P , are orthogonal projections in the Hilbert dpace H such that inf ( P , , P 2 ) = 8 holds in 2,then P , = 8 or P2 = 6 .

Proof. Assume that P , # 8 and Pz # 8. Then there exist elements x # 0 in R(P,) and y # 0 in R(P,). We have x Iy by the preceding lemma, and we may assume that llxll = llyll = 1. It follows that P , x = x , P l y = 0 and P 2 x = 0, P 2 y = y. Let M be the subspace of H spanned by x and y , and let P be the orthogonal projection on M . The projections P , and P , leave M invariant, and the restrictions of P 1 and P2 on M have the matrices

with respect to { x , y } as an orthonormal basis. Now, note that the matrix

with real coefficients and with respect to the same basis, corresponds to a

CH. 8,s 581

KADISON’S ANTI-LATTICE THEOREM

417

positive definite operator on M if and only if a 2 0, c 2 0 and b2 5 ac. It follows that the operator A on M , corresponding to the matrix

is not negative definite on M . In other words, the member AP = PAP of 2 does not satisfy AP 6 8. On the other hand, since P , - A corresponds on M with the matrix

(-$

-f5),

it follows that P,- A is positive definite on M , and this implies that AP S P I . Indeed, given z E H , we set z = x + y with x E M and y 1M , and so

(APz, Z) = (AX,x + y )

=

( A X ,X )

5 (PI x , x ) 5 (P,X , x ) + (Ply, u) = (PI z, z), where it is used in the last equality that P I leaves M and the orthogonal complement of M invariant. It follows now that AP 5 P i , and similarly it is proved that AP S P 2 . But then we have AP

5 inf ( P , , P 2 ) = 8,

contradicting the earlier result that AP 5 8 does not hold. Thus we arrive at a contradiction, and so P , = 8 or P2 = 8 must hold.

Theorem 58.4. 8 is an anti-lattice. Proof. It is evident that if A and B are members of 8 such that A 5 B or A 2 B holds, then inf ( A , B ) exists in 2. Conversely, let A‘, B‘ E 8 be given and assume that inf (A’, B’) = C exists in H.Writing A’- C = A and B’- C = B, we have A , B E 8 and inf ( A , B ) = 8. The proof will be complete if we show that one at least of A and B is equal to 8. To this end, assume that A # 8 and B # 8. Then. by the spectral theorem, there exist orthogonal projections P,,P2 and positive numbers a,, a2 such that 8 < 6,P,

5 A and 8 < B2P2 6 B

(for 6, P , we can take one of the nonzero terms in an appropriate lower sum for A , and similarly for b2P2).It follows from inf ( A , B ) = 8 that inf ( 6 , P , , g 2 P 2 ) = 0, and hence inf ( P , , P2)= 8. But then, in view of the

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HERMITIAN OPERATORS I N HILBERT SPACE

[CH. 8 , s 58

preceding theorem, we have PI = 8 or P, = 8. Contradiction. Hence, one at least of A and B must be equal to 8. As visible from the proof of Theorem 58.3, the proof that 2 is an antilattice in tbe general case is reduced to the case that the Hilbert space H is two-dimensional. It is reasonable to conjecture that if the Hilbert space H is real and two-dimensional (i.e., H is the ordinary real plane), then the fact that 3 is an anti-lattice has a simple geometrical meaning. This is indeed so. For any A # 8 in 2, consider the set of all x E H satisfying (Ax, x ) 5 1, and call this the indicatrix of A . For A 8 the indicatrix 'I of A is an elliptic disk with the origin as center, where the area between two different parallel lines is also to be regarded as an elliptic disk. If 8 < B 5 A , the indicatrix I A of A is contained in the indicatrix IB of B, i.e., if a member of 2+becomes larger, then its indicatrix becomes smaller. The anti-lattice theorem states that if A and B are incomparable in 2, then inf (A, B ) does not exist or, equivalently, sup (A, B ) does not exist. Since we can always add a sufficiently large multiple of the identity operator E to A and By this is the same as stating that if A 2 E, B 2 E and A , B are incomparable, then sup (A, B ) does not exist. The indicatrix of E is the unit circle in the plane, so A 2 E implies that the indicatrix IA of A is an elliptic disk with the origin as center and contained in the unit circle. Similarly for the indicatrix IB of B. Since A and B are incomparable, neither of IA and I B is contained in the other, i.e., the boundary of IA n Is is no ellipse. Assume now that C = sup (A, B ) exists. This means that C 2 A and C 2 By and any other upper bound C' of A and B satisfies C' 2 C. Hence, the indicatrix Ic of C is an elliptic disk (with the origin as center) contained in IA n I B , and any other elliptic disk with the origin as center and contained in' I n IB is also contained in Ic. This, however, is evidently false. Hence, sup (A, B ) does not exist in 8.

=-

Exercise 58.5. Find the geometrical meaning (in a real two-dimensional Hilbert space) of Theorem 58.3. Hint: The indicatrix of the orthogonal projection on a one-dimensional subspace is the area between two different parallel lines (equidistant from the origin). Also, if A and 8 are incomparable, the boundary of the indicatrix of A is a hyperbola.