Introduction to Hilbert Spaces with Applications

Bilinear functionals are often called sesquilinear. Note that a bilinear functional is linear with respect to the first variable and anti-linear with respect to the second variable. Clearly, bilinear functionals form a vector space. Example 4.3.1.

Inner product is a bilinear functional.

Example 4.3.2. Let A and B be operators on an inner product space E. Then ¢ 1(x,y)=(Ax,y), ¢ 2 (x,y)=(x,By), and ¢ 3 (x,y)=(Ax,By) are bilinear functionals on E. Example 4.3.3. Let f and g be linear functionals on a vector space E. Then ¢(x, y) = f(x)g(y) is a bilinear functional on E. Definition 4.3.2.

Let ¢ be a bilinear functional on E.

(a)¢ is called symmetric if ¢(x,y)=¢(y,x) for all x,yEE; (b) ¢is called positive if ¢(x, x)2:0 for every XE E; (c) ¢ is called strictly positive if it is positive and ¢ ( x, x) > 0 for all x (d) If E is a normed space, then ¢ is called bounded if

:;<'

0.

l¢(x, Y)l :o; KllxiiiiYII for some K > 0 and all x, y functional is defined by II¢ II=

E

E. The norm of a bounded bilinear

sup

l¢(x, Y)l.

llxiHIYII-1

Iff= gin Example 4.3.3, then ¢ 1 is symmetric and positive. Inner product is strictly positive. If operators A and B in Example 4.3.2 are bounded, then ¢ 1 , ¢ 2 , and ¢ 3 are bounded. Similarly, iff and g in Example 4.3.3 are bounded, then the defined bilinear functional is also bounded. Note that for a bounded bilinear functional

l¢(x, Y)l :o; II

for all x, y

E

E.

Definition 4.3.3 (Quadratic Form). Let ¢ be a bilinear functional on a vector space E. The functional

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Theory

is called the quadratic form associated with ¢. A quadratic form

0 such that for all x

E

E.

The norm of a bounded quadratic form is defined by 11<~>11 = sup I
Note that for a bounded quadratic form

Theorem 4.3.1 (Polarization Identity). Let

Proof.

E

(4.3.1)

E.

For any a, f3

E

C we have

( ax+ f3y) =¢(ax+ f3y, ax+ f3y) 2

2

=I a 1 <1>( x) + a,B¢ ( x, y) + iif3

Corollary 4.3.1. Let ¢ 1 and ¢ 2 be bilinear functionals on E. If
If ¢ 1 (x, x) = ¢ 2 (x, x) for all x

E

E, then the quadratic norms <1> 1 and 1 and ¢ 2 are equal.

<1> 2 are equal, and hence, by (4.3.1), the functionals ¢

The proof for operators is similar.

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Linear Operators on Hilbert Spaces

Theorem 4.3.2. A bilinear functional ¢ on E is symmetric if and only if the associated quadratic form

If ¢(x, y) = ¢(y, x) for all x, y E E, then
for every x E E, and thus

Then, for the associated quadratic form 'I' we have 'I'( x) = ¢(x, x) =
Thus, by Corollary 4.3.1, ¢(x, y) = t/J(x, y) for all x, y E E, i.e., ¢(x, y) = ¢(y, x). This completes the proof. Theorem 4.3.3. A bilinear functional ¢ on a normed space E is bounded if and only if the associated quadratic form

(4.3.2)

II
Since II
~x~=1

sup

l¢(x, y)l = II
~x~=~y~=1

if ¢ is bounded, then

2

2

Hence, by the parallelogram law,

Consequently sup

llxiHIYII=1

l¢(x, Y)l :S

sup

llx~=IIYII=1

2

II
2 )

=211<1>11·

Thus, if

Theory

146

Proof By Theorem 4.3.3, II II s 114> II· We need to show that the opposite inequality holds as well. Since 4> is symmetric, ¢ is real, by Theorem 4.3.2. Next, by the polarization identity, we obtain

Re 4> (x, y) = ~[(x + y)- (x- y)], and hence 5 ~II
IRe cf>(x, Y)l

2

=!II
llx- Yll

2

)

2 ),

by the parallelogram law. Let x andy be arbitrary fixed elements of E such that llxll = IIYII = 1, and let e be a complex number such that 1e1 = 1 and lcf>(x, y)l = ecp(x, y). Then

14> (x, y )I= ecp(x, y) =¢(ex, y) =IRe 4> (ex, y )Is !11<1>11( I exll 2 + IIY

n =II II,

and thus 114> II=

sup

llxii~IIYII=l

lcf>(x, Y)l

5

I II.

The proof is complete. Theorem 4.3.5. Let A be a bounded operator on a Hilbert space H Then the bilinear functional defined by 4> ( x, y) = ( x, Ay) is bounded and I A I = I 4> II.

Proof

For all x, y

E

H, by Schwarz's inequality, we have

lcf>(x, Y)l

5

l(x, Ay)l 5 llxll IIAyll

Thus 4> is bounded and I 4> II s 2

I A II.

11Axll = I(Ax, Ax)l

Therefore, for

Ax~

=

5

IIAII llxll IIYII·

On the other hand, we have

lcf>(Ax, x)l s 114>11 IIAxll llxll.

0, we have

IIAxll s 114>11 llxll.

Since the above inequality is trivially satisfied if Ax= 0, we obtain I All s I 4> II. The proof is complete. Theorem 4.3.6. Let 4> be a bounded bilinear functional on a Hilbert space H There exists a unique bounded operator A on H such that

cp(x, y) = (x, Ay)

for all x, y E H

Proof For a fixed y E H, cp(x, y) is a bounded linear functional on H Thus, by the Riesz Representation Theorem, there exists a unique element

Linear Operators on Hilbert Spaces

147

AyE H such that

YJ) + /3¢ (x, Y2)

(x, aAYJ+ f3AJ2),

and thus A( aYI + f3Y2) = aAYJ + f3Ay 2 •

Now we show that A is bounded. Since

= I
i(x, Ay)l

for some k > 0 and all

we obtain

Therefore, if Ay ~ 0, we get

which is also trivially satisfied if Ay = 0. This proves that A is bounded. To prove uniqueness notice that (x, Ay) = (x, By)

for all x, y

E

H

implies A = B. Definition 4.3.4 (Coercive Functionals). A bilinear functional ¢ on a norrued space E is called coercive if there exists a positive constant K such that

for all x E E. Coercive functionals are also called elliptic. Example 4.3.4. Let z be a continuous real valued function on [0, 1] such that min,E[O,Jl z(t)>O. The bilinear functional¢ defined on L\[0, 1]) by
is coercive. Indeed, we have

where K

= min,E[o, 11 z(t).

r

X(t)y(t)z(t) dt

Theory

148

The following theorem, proved by P. Lax and A N. Milgram in 1954, is an important generalization of the Riesz Representation Theorem. Applications of this theorem to boundary value problems are discussed in Chapter 6.

Theorem 4.3.7 (Lax-Milgram Theorem). Let

f(x) = ¢(x, x1 ) Proof.

for all x

H.

E

By Theorem 4.3.6, there exists a bounded operator A such that for all x, y

H.

E

Since ¢ is coercive, we have

K llxll 2 ~ ¢(x, x) = (x, Ax)~ II Axil llxll, and hence

IIAxll 2: Kllxll

for all x

E

H.

which implies x 1 = x 2 • Therefore A is one-to-one. Denote by 9l(A) the range of A. Let {x"} be a sequence of elements of H. If IIAxn- y II-i> 0, for some y E H, then

1

II Xn - Xm II ~ K II Axn - Axm II -i> 0, as m, n -i> oo. Therefore {x"} is a Cauchy sequence in H. Since His complete, there exists x E H such that llxn- x II-i> 0. Hence II Ax"- Axil -i> 0, because A is continuous. Consequently, Ax= y, and thus y E 9l(A). This proves that 9l(A) is a closed subspace of H. We will prove that 9l(A) =H. Suppose 9l (A) is a proper subspace of H. Then, by Theorem 3.10.4, there exists a non-zero x E H which is orthogonal to 9l (A), i.e.,

(x,Ay)=O

for ally

E

H.

In particular, we have 0 = i(x, Ax)i = l¢(x, x)l2: K llxll which contradicts the assumption x

~

0.

2

,

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Linear Operators on Hilbert Spaces

Iff is a bounded linear functional on H, then there exists a unique x 0 E H such that

f(x) = (x, x 0 )

for all xE H.

Since A is a one-to-one mapping and 9l(A) = H, there exists a unique x1 E H such that x 0 = Axh and hence for all x

E

H.

This proves the theorem.

4.4. Adjoint and Self-Adjoint Operators Let A be a bounded operator on a Hilbert space H. For every fixed x 0 E the functional f defined on H by f(x) =(Ax, x 0 ) is a bounded linear functional on H. Thus, by the Riesz Representation Theorem, there exists a unique y 0 E H such that f(x) = (x, y 0 ) for all x E H, or equivalently (Ax, x 0 ) = (x, y 0 ) for all x E H. If we denote by A* the operator which to every x 0 E H assigns that unique y 0 , then we have (Ax, y) = (x, A*y) for all x, y E H.

Definition 4.4.1 (Adjoint Operator). Let A be a bounded operator on a Hilbert space H. The operator A*: H...;. H defined by (Ax, y) = (x, A*y)

for all x, y

E

E

is called the adjoint operator of A. The following properties are direct consequences of the above definition:

(A+B)*=A*+B*, (aA)* = aA*, (A*)*= A, ,j* = .!P,

(AB)* = B* A*; for arbitrary operators A and B and scalar a. Theorem 4.4.1. The adjoint operator A* of a bounded operator A is bounded. 2 Moreover, we have I All= IIA*II and I A* All= IIAII •

150

Proof.

Theory

By Schwarz's inequality, for any x, y

E

H we have

ICA*x, Y)l = l(x, Ay)l :S llxiiiiAYII :S IIAIIIIxiiiiYII, and hence for y = A*x we obtain

IIA*xll 2 = (A*x, A*x) :S IIAIIIIxiiiiA*xll· Consequently,

IIA*II :S I All·

( 4.4.1)

This proves that the adjoint of a bounded operator is bounded. Therefore, inequality (4.4.1) can be used for A* instead of A, which gives

IIAII = II(A*)*II :S IIA*II· Thus IIAII = IIA*II· From the above we get

I A* All :S IIA*IIIIAII = IIAII 2 • On the other hand, for every x

E

H we have

11Axll 2 =(Ax, Ax)= (A* Ax, x):S I A* Axllllxll :S I A* Allllxll 2 • 2 This gives I A* All= IIAII , completing the proof. In general A and A* need not be equal. For instance, let H = C 2 and let A be defined by

A(zh z2 ) = (0, zJ. Then (A(x 1 , x2 ), (yh J2)) = X1Y2 and ( (xJ, for which A= A* are of special interest.

x2 ), A(y 1 , Y2)) = X 2YJ.

Operators

Definition 4.4.2 (Self-Adjoint Operator). If A= A*, i.e., (Ax, y) = (x, Ay) for all x, y E H, then A is called self-adjoint (or Hermitian). Example 4.4.1. Let H =eN and let { e!' ... ' eN} be the standard orthonormal base in H. Let A be an operator represented by matrix ( aiJ ), i.e., aii = (AeJ, ei), see Example 4.2.2. Then the adjoint operator A* is represented by the matrix bkJ = (A*eJ, ek). Consequently bkJ = (eJ, Aek) = (Aek> eJ) == aJk. Therefore the operator A is self-adjoint if aiJ = aJi. A matrix satisfying this condition is often called Hermitian.

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Linear Operators on Hilbert Spaces

Example 4.4.2. Let H be a separable infinite dimensional Hilbert space and let { e~> e2 , e 3 , ••• } be a complete orthonormal sequence in H. Let A be a bounded operator in H represented by an infinite matrix ( aiJ ), see Theorem 4.2.2. Like in the finite dimensional case, the adjoint operator A* is represented by the infinite matrix ( a1;). A is self-adjoint if a;1 = a1; for all i, j EN. Example 4.4.3.

r

LetT be a Fredholm operator on L 2 ([a, b]) defined by (Tx)(s)=

K(s, t)x(t) dt,

rr rr rr r r r r

where K is a function defined on [a, b] x [a, b] such that

IK (s, t)l 2 ds dt < 00.

Note that the condition is satisfied if K is continuous. We have ( Tx, y) =

K (s, t)x(t).Y"(S) ds dt

=

K(s, t)x(t)y(s) dsdt

x(t)

=

=(x,

This shows that

K (s, t)y(s) ds dt

K(s,t)y(s)ds).

(K*x)(s)=

K(t, s)x(t) dt.

Thus a Fredholm operator is self-adjoint if its kernel satisfies the equality K ( s, t) = K ( t, s).

Example 4.4.4.

Let A be the operator on L \[a, b]) defined by (Ax)( t) = tx( t).

Since (Ax, y) = A is self-adjoint.

r

tx(t)y(t) dt =

Lb x(t)ty(t) dt

=

(x, Ay),

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Theory

Consider the operator A defined on L\R) defined by

Example 4.4.5.

(Ax)(t) = e-ltlx(t).

This is also a bounded self-adjoint operator. Boundedness of A can be shown as in Example 4.2.5. Moreover, we have (Ax, y) =

L:

e- 1'1x(t)y(t) dt =

L:

x(t)[e-l'ly(t)] dt = (x, Ay).

Thus A is self-adjoint. Example 4.4.6. Let

(if¢ is symmetric),

and ¢(x, y)

=

(x, Ay) =(Ax, y) = (y, Ax)=
(if A is self-adjoint).

Theorem 4.4.2. Let A be a bounded operator on a Hilbert space H. Then operators T 1 =A* A and T 2 =A+ A* are self-adjoint. Proof.

We have (T1 x, y) =(A* Ax, y) =(Ax, Ay) = (x, A* Ay) = (x, T 1 y)

and ( T2 x, y) =((A+ A*)x, y) = (x, (A+ A*)*y) ='(x, (A+ A*)y) = (x, T2 y),

for all x, y

E

H.

Theorem 4.4.3. The product of two self-adjoint operators is selfadjoint if and only if the operators commute. Proof.

Let A and B be self-adjoint operators. Then (ABx, y) = (Bx, A*y) = (x, B* A*y) = (x, BAy).

Thus, if AB = BA, then AB is self-adjoint. Conversely, if AB is self-adjoint, then the above implies AB=(AB)*= BA,

which completes the proof.

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Linear Operators on Hilbert Spaces

Corollary 4.4.1.

If A is self-adjoint, then so is any polynomial of A anAn+· · ·+a 1 A+a 0

with real coefficients an, ... , a 0 • In the definition of the adjoint operator of A we have assumed that the domain of A is the entire space H. Then the existence and uniqueness of the adjoint operator A* was guaranteed by the Riesz Representation Theorem. In practice, we often deal with operators defined on a proper subspace of H, for instance the differential operator. In such a case the adjoint operator can be defined as follows. For an operator T, denote by ffi(T) the domain of T. We can always assume, without loss of generality, that ffi( T) is a vector space. Let A: ffi(A) -i> H and B: ffi(B) -i> H be operators, ffi (A), ffi(B) c H. Then B is called an adjoint operator of A if for every x

(Ax, y) = (x, By)

E

ffi(A) andy

E

ffi(B).

In general, an adjoint operator is not unique. However, it can be proved that, if ffi(A) is dense in H, then it is unique. Operators defined on a proper subspace of a Hilbert space will be discussed in more detail in Section 4.12.

Example 4.4.7. Consider the differential operator D in the space of all differentiable functions on R vanishing at infinity. Then

oo

d

(Dx, y) = f - x(t) y(t) dt = -00 dt =

L: x(t)[-

foo -00

d

x(t) -d y(t) dt t

:ry(t)] dt=(x,-Dy).

Thus - D is the adjoint of D.

Example 4.4.8. Consider operator T = id I dt in the space of all differentiable functions on R vanishing at infinity. We have (Tx,y)=

f

oo

d

-00

i

= -

_

i-x(t)y(t) dt dt

f

oo

-00

X(

dt) - y ( t) dt dt

00

=f -oo

x(t)[i_!}__y(t)J dt=(x, Ty). dt

Therefore, T is a self-adjoint operator.

Theory

154

Theorem 4.4.4. representation

Every bounded operator T on a Hilbert space H has a

T=A+iB, where A and B are self-adjoint. Moreover,

T* =A- iB. Proof.

Let T be a bounded operator on H. Define A=~(T+ T*)

and

B=tt(T- T*).

Then A and B have the desired properties. Moreover, for any x, y have

E

H we

(Tx, y) =((A+ iB)x, y) =(Ax, y) + i(Bx, y) =

(x, Ay) + i(x, By)

=

(x, (A- iB)y).

Hence T* =A- iB. In particular, if T is self-adjoint, then A= T and B = 0. Thus self-adjoint operators are like real numbers in C. The following property of self-adjoint operators is useful in the investigation of spectral properties of such operators, see Section 4.9. Theorem 4.4.5.

Let T be a self-adjoint operator in a Hilbert space H. Then ( 4.4.2)

II Til= sup I(Tx, x)l. llxll-1

Proof.

Let M=sup xn- 1 1(Tx,x)l. If llxll = 1, then 11

I(Tx, x)l :S II Txllllxll =II Txll :S II Tllllxll =II Til· Thus ( 4.4.3)

Let x

E

H and Tx

~

0. Define

/jfTxll - \/Txf

a-

and

Tx

z=-. a

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Linear Operators on Hilbert Spaces

Then

I Txll 2 = (T(ax), z) =~[(T(ax+z), ax+z)-(T(ax-z), ax-z)]

:S~M(II ax+ zll 2

2

+ I ax- zll 2 )

=~M(IIaxll + =~M[a llxll

2

2

llzll 2 ) 2 2 +(1/ a )11 Txii J= Mllxllll Txll·

Thus I Tx I :S M llx II· Since this inequality is trivially satisfied for x such that Tx = 0, we have I Til :SM. This, together with (4.4.3), proves the desired equality. Definition 4.4.3 (Anti-Hermitian Operator). Hermitian if A= -A*.

An operator A is called anti-

The operator in Example 4.4.7 is anti-Hermitian.

4.5. Invertible, Normal, Isometric, and Unitary Operators Let A be an operator defined on a subset ffi(A) of a vector space E. We will denote by 21l(A) the range of A, i.e., 21l(A) ={Ax: x E ffi(A)}. If A is an operator and ffi(A) is a vector subspace of E, then 9l(A) is a vector subspace of E. Definition 4.5.1 (Inverse Operator). Let A be an operator defined on a vector subspace of E. An operator B defined on 9l(A) is called the inverse of A if ABx = x for all x

E

9l(A)

and

BAx = x for all x

E

ffi(A).

An operator which has an inverse is called invertible. The inverse of A will be denoted by A- 1 • If an operator has an inverse, then it is unique. Indeed, suppose B 1 and B 2 are inverses of A Then B 1 = B 1 -P = B 1 AB 2 = .PB2 = B 2 •

Note also that

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Theory

Theorem 4.5.1.

Proof.

The inverse of an operator is linear.

For any x, y

E

9l(A) and a, f3

1

A - ( ax+ f3y)

E

C we have

=

A- ( aAA - 1 x + f3AA- 1 y)

=

A- 1 A(aA- 1 x+ f3A- 1 y)

1

=aA- 1 x+f3A- 1 y. Theorem 4.5.2.

An operator A is invertible if and only if Ax = 0 implies x

=

0.

Proof. If A is invertible and Ax=O, then x=A- 1 Ax=A- 10=0. Assume now that Ax= 0 implies x = 0. If Ax 1 = Ax2 , then A(x 1 - x 2 ) = 0 and thus x 1 - x 2 = 0. Consequently x 1 = x 2 , which proves that A is invertible. Theorem 4.5.3. If an operator A is invertible and vectors x 1 , ••• , Xn are linearly independent, then Ax 1 , ••• , Axn are linearly independent.

Proof. Suppose a 1 Ax 1 +· · ·+anAxn=O. Then A(a 1 x 1 +.· · ·+anXn)=O, and by Theorem 4.5.2, a 1 X1 + · · · + anxn = 0. Since x 1 , ••• , Xn are linearly independent, we get a 1 = · · · = a" = 0, and thus vectors Ax 1 , • •• , Ax" are linearly independent. If follows from the above theorem that, if E is a finite dimensional vector space and A is a linear invertible operator on E, then 9l (A)= E. As the following example shows, in infinite dimensional vector spaces this is not necessarily true.

Example 4.5.1.

Let E = 12 • Define an operator A on E by

A(x 1 , x 2 ,

•• • )

=

(0,

x~>

x2 ,

•• • ).

Clearly, this is a linear invertible operator on 12 whose range is a proper subspace of 12 • The next example shows that inverse of a bounded operator is not necessarily bounded.

Example 4.5.2.

Let E = 12 • Define an operator A on E by

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Linear Operators on Hilbert Spaces

A is bounded. Also is A invertible: A-\x~> x 2 ,

•• • )

= (x~> 2x 2 , 3x3 ,

••. ,

nxn, .. .).

However, A- 1 is not bounded. In fact, consider the sequence {en} of elements of 12 , where en is a sequence whose nth term is 1 and all the remaining terms are 0. Then I en I = 1 and

IIA-1enll =

n.

1

Therefore A- is unbounded. If E is finite dimensional, then the inverse of any invertible operator on E is bounded, because every operator on a finite dimensional space is bounded. Theorem 4.5.4. If operators A and B are invertible, then the operator AB is invertible and (AB)- 1 = B- 1 A- 1 • The easy proof is left as an exercise. Note that the sum of two invertible operators is not necessarily invertible: ji + (-Ji). Theorem 4.5.5. Let A be a bounded operator on a Hilbert space H such that 9l(A) =H. If A has a bounded inverse, then the adjoint A* is invertible and (A*)- 1 =(A - 1)*.

Proof.

It suffices to show that

(A- 1 )* A*x = A*(A- 1 )*x = x

for every x

E

H. Indeed, for any y

E

(4.5.1)

H we have

(y, (A - 1 )* A*x) = (A- 1 y, A*x) = (AA - 1 y, x) = (y, x) and

(y, A*(A- 1 )*x) = (Ay, (A- 1)*x) = (A- 1 Ay, x) = (y, x). Thus

(y, (A - 1 )* A*x) = (y, (A*(A- 1 )*x) = (y, x) which implies (4.5.1 ).

for all y

E

H,

( 4.5.2)

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158

In the above theorem we assume that iYt (A)= H. This assumption was used only to conclude that (4.5.2) implies (4.5.1). Thus it is enough to assume that iYt(A) is such that (y, x) = 0 for ally E iYt(A) implies x = 0. This is true if iYt(A) is dense in H. Corollary 4.5.1. If a bounded self-adjoint operator A has a bounded inverse A-\ then A-! is self-adjoint.

Definition 4.5.2 (Normal Operator). A bounded operator T is called a normal operator if it commutes with its adjoint, i.e., TT* = T* T. Obviously, every self-adjoint operator is normal. The following theorem will help us find examples of normal operators which are not self-adjoint. Theorem 4.5.6. for all x E H.

Proof.

A bounded operator Tis normal if and only if II Tx II= II T*x II

We have 2 (T*Tx, x) = (Tx, Tx) =II Txll •

If T is normal, then we also have 2 (T*Tx, x) = (TT*x, x) = (T*x, T*x) =II T*xll

for all x

E

H.

Thus IITxll = IIT*xll· Assume now II Txll =II T*xll· By the above argument we have

(TT*x, x)

=

(T*Tx, x)

for all x

E

H.

Therefore, by Corollary 4.3.1, TT* = T* T. Note that the condition II Txll =II T*xll is stronger than II Til= II T*ll .

.Example 4.5.3. Let H be a Hilbert space and let Tx = ix for all x E H. Since T* x =- ix =- Tx, T is not self-adjoint. On the other hand, II Tx II = II T*xll for all x E H, and thus Tis normal.

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Linear Operators on Hilbert Spaces

Theorem 4.5.7.

Proof.

If A is normal, then ( a!fi- A) is normal for any a

E

C.

Since (a!fi-A)*=(a!fi-A*), we have

(a!fi- A)(a!fi- A)*= lal 2 !fi- aA- aA* + AA* = ( a!fi- A)*( a!fi- A). This proves the theorem. Theorem 4.5.8. Let T be a bounded operator on a Hilbert space H and let A and B be self-adjoint operators on H such that T =A+ iB. Then Tis normal if and only if A and B commute.

Proof.

Assume T is normal. Since T* =A- iB, we have

TT* =(A+ iB)(A- iB) = A 2 + B 2 - i(AB- BA)

(4.5.3)

T*T= (A- iB)(A+ iB) = A 2 + B 2 + i(AB- BA).

(4.5.4)

and

Thus AB-BA= 0, which proves that A and B commute. On the other hand, if A and B commute, then by (4.5.3) and (4.5.4) we have

The proof is complete.

Definition 4.5.3 (Isometric Operator). A bounded operator T on a Hilbert space H is called an isometric operator if I Tx II= llx I for all x E H Example 4.5.4. Let {en}, n EN, be a complete orthonormal sequence in a Hilbert space H There exists a unique operator A such that Aen =en+! for all n EN. In fact, if X= ,L..,n=l anen, then 00

00

Ax=

L

anen+!·

n=l

Clearly, A is linear and 00

11Axll = L lanl 2 = llxll 2 • 2

n=l

Therefore A is an isometric operator. Operator A is called a one-sided shift operator.

Theory

160

Theorem 4.5.9. A bounded operator T defined on a Hilbert space H is isometric if and only if T* T = !} on H. Proof. hence

If Tis isometric, then for every x

E

H we have II Txll 2 = llxll 2 , and

for all x

(T*Tx,x)=(Tx, Tx)=(x,x)

This implies, by Corollary 4.3.1, T* T

= !fi.

E

H.

Similarly, if T* T

= !fi,

then

II Txll = .J( Tx, Tx) = .J(T*Tx, x) = .J(x, x) = llxll· Note that isometric operators "preserve inner product": (Tx, Ty) In particular, if x _i y, then Tx _i Ty. The operator in Example 4.5.1 is an isometric operator.

= (x, y).

Definition 4.5.4 (Unitary Operator). A bounded operator T on a Hilbert space H is called a unitary operator if T* T = TT* =!} on H. It is essential in the above definition that the domain arid the range of T

are the entire space H. Theorem 4.5.10. Proof.

Every unitary operator is normal.

It follows directly from the definition.

Note that the converse is not true. Every self-adjoint operator A such that II All~ 1 is a normal but not unitary operator. Theorem 4.5.11. T- 1 = T*.

An operator Tis unitary

if and only if it is invertible and

Proof. Assume that T is an invertible operator on a Hilbert space H such that T- 1 = T*. Then T* T = T- 1 T = !}

and TT* =

rr-

Therefore T is a unitary operator. The proof of the converse is similar.

1

= !fi.

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Linear Operators on Hilbert Spaces

Theorem 4.5.12.

Proof.

If Tis a unitary operator, then T- 1 and T* are unitary.

If T is unitary, then

(T- 1 )*T- 1 = T**T- 1 = TT- 1 =.P. Similarly, T-\T- 1 )* = .P, and thus T- 1 is unitary. Since T* Theorem 4.5.11, T* is also unitary.

= T-\

by

Example 4.5.5. Let H be the Hilbert space of all sequences of complex numbers x = ( ... , x_h x 0 , X1, •• . ) such that llxll = lxnl 2 < oo. The inner product is defined by

L.:oo

-oo Define an operator T by T(xn) = (xn_ 1 ). T is a unitary operator: T is invertible and 00

( Tx, y) =

00

L Xn-!Yn = L XnYn+! = (x, T-!y), -00

which implies T* = T-

Example 4.5.6.

1

-00

•

Let H = L 2 ([0, 1]). Define an operator Ton H by

(Tx)(t)=x(l-t). This operator is a one-to-one mapping of H onto H Moreover, we have T= T*= T- 1 • Tis a unitary operator.

4.6. Positive Operators Definition 4.6.1 (Positive Operator). An operator A is called positive if it is self-adjoint and (Ax, x);::: 0 for all x E H Example 4.6.1. Let ¢ be a non-negative continuous function on [a, b]. The multiplication operator A on L \[a, b]) defined by Ax=
r

is positive. In fact, for any x

(Ax,

x) =

E

r

L \[a, b]) we have

¢ ( t)x( t)x( t) dt =

¢ ( t)lx( t)l 2 dt;::: 0.

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Theory

Example 4.6.2. Let K be a positive continuous function defined on [a, b] x [a, b]. The integral operator Ton L 2 ([a, b]) defined by (Tx)(s)

rr

Lb K(s, t)x(t) dt

=

is positive. Indeed, we have

(Tx,x)=

K(s,t)x(t)x(t)dtds=

rr

K(s,t)ix(tWdtds~o

for all xEL 2 ([a,b]). Theorem 4.6.1. are positive.

Proof.

For any bounded operator A, the operators A* A and AA*

For any x

H we have

E

(A* Ax, x) =(Ax, Ax)= 11Axll 2 ~0 and

(AA*x, Theorem 4.6.2. is positive.

Proof.

If y

E

x) =

(A*x, A*x) = IIA*xll 2 ~0.

If A is an invertible positive operator, then its inverse A -J

ffi(A- 1 ), then y =Ax for some x

E

H, and then

(A- 1 y, y) = (A- 1 Ax, Ax)= (x, Ax) ~o. If A is a positive operator, we will write A~ 0. If the difference A- B of two self-adjoint operators is a positive operato1', i.e., A- B ~ 0, we will write A~ B. The defined relation has the following natural properties;

If A~B and

C~

If A~O and

a~O

If A~B and

B~

IfA~O

IIAII:::;l,

and

C~B+

D, then A+

(a ER), then

C, then

A~

D;

aA~O;

C;

thenA:::;.P;

If A~ 0, then there exists a> 0 (a

E

R) such that aA:::; .P.

Easy proofs are left as exercises.

Example 4.6.3. The product of two positive operators is not necessarily positive. Indeed, consider operators on R 2 defined by matrices

A=[~

~]

and

B = [:

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Linear Operators on Hilbert Spaces

It is easy to check that both A and B are positive operators, but the product AB is not.

Theorem 4.6.3. operator.

Proof.

The product of two commuting positive operators is a positive

Let A and B be commuting positive operators. Define for n = 1, 2, ....

Note that operators A" are self-adjoint and commuting. We will show, by induction, that (4.6.1) for all n EN. Clearly, (4.6.1) is satisfied for n = 1. Suppose now (4.6.1) holds for some kEN. Then

(Al(J5- Ak)x, x) = (Ak(J5- Ak)x, Akx) = ((J'i- Ak)Akx, Akx);::: 0 and

which means

Consequently

and

J5- Ak+! = (J'i- Ak) +Al ;:::0. This shows that (4.6.1) holds for k+1, and thus for all n EN, by induction. We have n

A!= Ai + A2 = Ai +A~+ A3

=••·= L k~!

and hence n

L Al =A!-An+!:S A

1•

k~!

Therefore n

L k~!

(Akx, Akx) :S (A 1 x, x).

Al + An+h

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Theory

This shows that the series L~~~ IIAnxll 2 converges and IIAnxll-i>O. Then

(f

Al )x = A 1 x- An+ 1x -i> A 1 x

as n -i> oo

1

or, equivalently,

n=l

Since B commutes with An for all n EN, we have 00

(ABx, x) = IIAII(BAJx, x) = IIAII

L

(BA~x, x)

n=l 00

= IIAII

L

(BAnx,Anx)~O.

n=l

This proves the theorem. The following theorem is an interesting analog of a property of real numbers.

a

Theorem 4.6.4. Let A 1 :S A 2 :S · · · :SAn ::; · · · be self-adjoint operators in Hilbert space H such that AnAm = AmAn for all m, n EN. If B is a self-adjoint operator on H such that AnB = BAn and An :S B for all n EN, then the sequence {An} converges strongly to a self-adjoint operator A and A 1 :SA :S B.

Proof.

Define en= B-An. Operators en commute with each other and

By Theorem 4.6.3, for n > m the operators

are positive. Hence

( e~x, x) ~ ( emenx, x) ~ ( e~x, x) for every x E H. Since, for an arbitrary fixed x E H, { ( elx, x)} is a nonincreasing sequence of non-negative numbers, it converges and thus lim (emenx,x)=lim (e~x,x).

m,n --?00

Hence

n --?00

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Linear Operators on Hilbert Spaces

as m, n-'? oo. Therefore { C,x} is a Cauchy sequence for every x E H. Consequently, {Cn}, and thus also {An}, are convergent. It is easy to check that the operator A defined by Ax= lim" ~oo A"x is self-adjoint and A 1 :SA :S B. Definition 4.6.2 (Square Root). A square root of a positive operator A is a self-adjoint operator B satisfying B 2 = A.

Theorem 4.6.5. Every positive operator A has a unique positive square root B. Moreover, B commutes with every operator commuting with A. Proof. and

Let A;::: 0 and let a> 0 (a

E

R) be such that a 2 A::; !fi. Define T0 = 0 (4.6.2)

for n = 0, 1, 2, .... Note that operators T" are self-adjoint (as polynomials of A with real coefficients) and positive. Moreover, they commute with every operator commuting with A. In particular, Tn Tm = Tm Tn for all m and n. For every n, we have (4.6.3) and Tn+!- T" =W!P- Tn-1)+(!fi- Tn))(T"- Tn-1).

(4.6.4)

In view of ( 4.6.3), we have T" :S!} for all n. Moreover

Indeed,

and if T"- Tn- 1;:::O, then Tn+ 1- T" ;:::O, by (4.6.4). By Theorem 4.6.4, the sequence { T"} converges to a positive self-adjoint operator T Letting n-'? oo in ( 4.6.2) yields T= T+~(a 2 A- T 2 ), i.e.,

Denote B = T I a. The operator B is obviously positive. Since, for each n EN, Tn commutes with every operator commuting with A, so do T and B.

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Theory

It remains to prove the uniqueness. Let C be a positive operator such that C 2 =A. Since C commutes with A, C commutes with B. Let x E H and let y 0 = (B- C)x. Then

(Byo, Yo)+ ( Cyo, Yo)= ((B + C)yo, Yo) =

((B

= ( (B

+ C)(B- C)x, Yo) 2

-

C 2 )x, y 0 ) = 0.

Since B and C are positive, we have (By 0 , y 0 ) = ( Cy0 , y 0 ) = 0. If D is a positive square root of B, then 2

2

II Dyoll = (D Yo, Yo)= (Byo, Yo)= 0.

Hence Dy0 = 0 and also By0 = D( Dy0 ) IIBx- Cxll

2

= 0. Similarly Cy0 = 0. Finally,

= ((B- C) 2 x, x) = ((B- C)y0 , x) = 0

for arbitrary x E H This proves B = C, completing the proof of the theorem.

Definition 4.6.3 (Positive Definite Operator). A self-adjoint operator is called strictly positive or positive definite if (Ax, x) > 0 for all x E H, x :;<' 0. If¢ in Example 4.6.1 is positive, then clearly A is strictly positive.

4.7. Projection Operators The projection theorem (Theorem 3.10.4) states that if Sis a closed subspace of a Hilbert space H, then for every x E H there exists a unique element yES such that x = y + z, where z E Sj_. Thus every closed subspace induces an operator on H which assigns to x that unique y.

Definition 4. 7.1 (Projection Operator). Let S be a closed subspace of a Hilbert space H The operator P on H defined by P(x) = y

for x=y+z,yES, and zESj_,

is called the projection operator onto S. The vector y is called the projection of x onto S. The projection operator onto a subspace S is often denoted by P5 .

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Linear Operators on Hilbert Spaces

From the uniqueness of the decomposition x = y + z, it follows that projection operators are linear. Since, by the Pythagorean formula, we have 2 2 2 llxll = IIYII + llzll , and hence 11Pxll

= IIYII 2 = llxll 2 -llzii 2 :S llxll 2 , bounded and liP I :S 1. The

2

zero operator is a projection operators are projection non-zero is P If projection operator on the zero subspace. operator, then liP II= 1, because, for every xES, we have Px = x. The identity operator !} is the projection operator onto the whole space H ( S =H).

Example 4.7.1. Let S be a closed subspace of a Hilbert space H, and let {e], e2 , •• • } be a complete orthonormal system inS (S as a closed subspace of a Hilbert space is a Hilbert space itself). Then the projection operator P onto S can be defined by

n=l

In particular, if S is of dimension 1 and v E S,

I vII= 1, then

Px = (x, v )v.

Example 4. 7.2. Let H = L \[ -71', 7T ]). Every x E H can be represented as x = y + z, where y is an even function and z is an odd function. The operator defined by Px=y is the projection operator onto the subspace of all even functions. This operator can be also defined as in Example 4. 7.1; namely we have

where ¢ 0 (t)=1, and
Example 4. 7.3.

Let h = L 2 ([ -71', 7T ]). Let P be an operator on H defined by (Px)(t)={

0 x( t)

if t :S 0, if t>O.

P is the projection operator onto the space of all functions which vanish for t :S 0. Definition 4. 7.2 (Idempotent Operator). if T 2 = T.

An operator Tis called idempotent

Theory

168

Theorem 4.7.1. A bounded operator is a projection (f and only !lit is idempotent and selj~adjoint. Proof. Let P be the projection operator onto a subspace S. Since P is the identity operator on S and Px E S for every x E H, we have P 2 x = P(Px) = Px

for all x

E

H

Moreover, if Px 1 = y 1 and Px2 = Y2, then (Px1, x2) = (YJ, X2) = (YJ, Y2) = (x1, Y2) = (x1, Px2).

Thus P is self-adjoint. Assume now that Tis an idempotent self-adjoint operator on H Define S={xEH: Tx=x}.

Since T is a bounded operator, S is a closed subspace of H If x

E

H, then

(X - Tx, z) = (X, z) - ( Tx, z) = (X, z) - (X, Tz) = ( x, z) - ( x, z) = 0 for all z E S,

because T is self-adjoint. Moreover, for all x E H,

T(Tx)= Tx

because T is idempotent. Consequently, for every x x=Tx+(x-Tx),

TxES

and

E

H, we have

x-TxES_j_.

This proves that T is the projection onto S. The proof is complete. Corollary 4.7.1. If Pis a projection operator onto a subspace of a Hilbert 2 space H, then ( Px, x) = I Px 1 for all x E H. Proof.

From Theorem 4.7.1, we have (Px, x) = (PPx, x) = (Px, P*x) = (Px, Px) =

11Pxll 2 •

Example 4. 7.4. If P is the projection operator onto a closed subspace S, then !} - P is the projection operator onto Sj_ and Sj_ = {x: Px = 0}. The operator !} - P is sometimes denoted by Pj_ and called the complementary projection operator.

In general, the sum of two projection operators is not a projection operator. For example, if P is a non-zero projection operator, then P + P = 2P is not a projection operator, because IIP+PII =2. On the other hand, P + Pj_ is a projection operator. We will give a necessary and sufficient condition for the sum of two projection operators to be a projection operator.

Linear Operators on Hilbert Spaces

169

Definition 4. 7.3 (Orthogonality of Projection Operators). operators P and Q are orthogonal if PQ = 0.

Two projection

Note that for any two projection operators P and Q we have PQ = P*Q* = ( QP)*. Thus PQ = 0 if and only if QP = 0. Theorem 4.7.2. Two projection operators PR: H...;. R and P5 : H...;. S are orthogonal if and only if R _i S. Proof.

Assume PRPs = 0. If x E R and y

E

S, then

Hence R_i S. Next, assume R _iS. If x E H, then P5 x E S, and thus P5 x _i R. Hence PR ( P5 x) = 0 for all x E H This means that PR and P5 are orthogonal. Theorem 4.7.3. The sum of two projection operators PR and P5 is a projection operator if and only if PRPs = 0. In this case PR + P5 = PRffiS. Proof.

If P

=

PR + P5 is a projection operator, then (PR +P5 ) 2 = PR + P 5

or PRPs + PsPR = 0. Multiplication of this equality by PR from the left yields PRPs + PRPsPR = 0. Multiplication of the last equality by PR from the right gives 2PRPsPR = 0. Hence PRPs = 0. Assume now PRPs = 0. Then also P 5 PR (PR

+ P5 ) 2 =

= 0, and PR + P 5 •

thus

Therefore, P is idempotent. Since P is also self-adjoint, as a sum of two self-adjoint operators, it is a projection, by Theorem 4.7.1. For any x E H, we have

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Theory

Moreover, if x

=

x 1 +x2 with x 1 E R, x 2 E S, then

so P is the identity on Rffi S. The proof is complete. Theorem 4.7.4. The product of two projection operators PR and P5 is a projection operator if and only if PR and P5 commute. In this case PRPs = PRns. Proof.

Assume that P

=

PRPs is a projection. Then P* = P and thus

PRPs = (PRPs)*

Conversely, if PRPs

=

PsPR, then

(PRPs)*

so P

=

= P~P~ = PsPR.

= P~P~ = PsPR = PRPs,

PRPs is self-adjoint. Moreover, P

2

= PRPsPRPs = P~P~ = PRPs = P,

so P is idempotent. Thus, P is a projection. For xEH, we have Px=PR(P5 x)=P5 (PRx), and hence PxERnS. Moreover, for x ERn S, we have Px = PR(P5 x) = PRx = x. We thus have P = PRns. This completes the proof. Theorem 4.7.5. Let RandS be two closed subspaces of a Hilbert space H and let PR and P5 be the respective projections. The following conditions are equivalent: (a) Rc S; (b) PsPR = PR; (c) PRPs = PR; (d) IIPRxll :S IIPsxll for all x

E

H

Proof. Assume R c S. Then PRx E S for all x PRx, and thus (a) implies (b).· If P5 PR = PR, then PR

so (b) implies (c). Assume now PRPs

=

E

H Consequently, P5 PRx =

= P~ = (PsPR)* = P~P~ = PRPs, PR. Then

IIPRxll = IIPRPsxll :S IIPRIIIIPsxll :S IIPsxll, for all xE H Thus (c) implies (d).

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Linear Operators on Hilbert Spaces

Finally, suppose (d) holds and (a) does not. Then there exists x E R x £ 8. Let X= y + z with y E 8 and z E: sj_. Hence

which contradicts (d). Thus R

c

S. The proof is now complete.

4.8. Compact Operators Compact operators constitute an important class of bounded operators. The concept originated from the theory of integral equations of the second kind. The compact operators also provide a natural generalization of finite dimensional operators, i.e., operators with finite dimensional range. Definition 4.8.1 (Compact Operator). An operator A in a Hilbert space H is called a compact operator (or completely continuous operator) if for every bounded sequence {xn} in H the sequence {Axn} contains a convergent subsequence.

We have the following theorem. Theorem 4.8.1.

Compact operators are bounded.

Proof. If an operator A is not bounded, then there exists a sequence {xn} such that llxn II= 1 (for all n EN) and IIAxn II-i> oo. Then {Axn} does not contain a convergent subsequence, which means that A is not compact. Example 4.8.1. Every operator on a finite dimensional Hilbert space is compact. Indeed, if A is an operator on eN, then it is bounded. Therefore, if {xn} is a bounded sequence, then {Axn} is a bounded sequence in eN. By the Bolzano-Weierstrass Theorem, {Axn} contains a convergent subsequence.

Not every bounded operator is compact. Example 4.8.2. The identity operator !} on an infinite dimensional Hilbert space H is not compact, although it is bounded. In fact, consider an orthonormal sequence {en} in H. Then the sequence !Pen = en does not contain a convergent subsequence.

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Theory

Example 4.8.3.

Let y and z be fixed elements of a Hilbert space H. Define Tx = (x, y)z.

Let {x"} be a bounded sequence, i.e., llx" II s; M for some M > 0 and all n EN. Since

the sequence {(xn, y)} contains a convergent subsequence {(xp"' y)}. Denote the limit of that subsequence by a. Then TxPn = (Xp"' y)z...;. az

as n...;. oo.

Therefore T is a compact operator. Example 4.8.4. Important examples of compact operators are integral operators T on L 2([ a, b]) defined by (Tx)(s) =

r

K(s, t)x(t) dt,

where a and b are finite and K is continuous. We will sketch the proof of compactness of such an operator. Let Xn E L 2 ([a, b]) and llxn II s; M for n = 1, 2, ... and some M > 0. Then IC Txn)(s)l s;

r r

IK (s, t)xn( t)l dt s; M max IK (s, t)lv'b- a'

and thus the sequence { Txn} is uniformly bounded. Moreover, for every s 1 , s 2E [a, b], we have I(Txn)(sl)-(Txn)(s2)1s;

s;)

IK(sl, t)-K(s2, t)llxn(t)l dt

r

IK(sl, t) -K(s2, tW d t )

r

lxn(tW dt

s; Mv'b- a max IK (s~> t)- K (s2, t)l. IE(a, b)

Since K is uniformly continuous, the last inequality implies that the sequence { Tx"} is equicontinuous. Therefore, by Arzela's theorem (see for example [A. Friedman, Foundations of Modern Analysis, Dover, 1982]), { Tx"} contains a uniformly convergent subsequence. This proves that Tis compact, because uniform convergence on [a, b] implies convergence in L\[a, b]).

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Linear Operators on Hilbert Spaces

Example 4.8.5. Let S be a finite dimensional subspace of a Hilbert space H. The projection operator Ps is a compact operator, as in Example 4.8.1.

Theorem 4.8.2. The collection of all compact operators on a Hilbert space H is a vector space. The easy proof is left as an exercise. Theorem 4.8.3. Let A be a compact operator on a Hilbert space H and let B be a bounded operator on H. Then AB and BA are compact. Proof. Let {x"} be a bounded sequence in H. Since B is bounded, the sequence {Bxn} is bounded. Next, since A is compact, the sequence {ABxn} contains a convergent subsequence, which means that the operator AB is compact. Similarly, since A is compact, the sequence {Ax"} contains a convergent subsequence {AxpJ· Now, since B is bounded (and thus continuous), the sequence {BAxn} converges. Therefore, the operator BA is compact.

The operator defined in Example 4.8.5 is a special case of a finite dimensional operator. Definition 4.8.2 (Finite Dimensional Operator). dimensional if its range is of finite dimension.

Theorem 4.8.4.

An operator is called finite

Finite dimensional bounded operators are compact.

Proof. Let A be a finite dimensional bounded operator and let {z 1 , be an orthonormal basis of the range of A. Define

••• ,

zd

an(X) =(Ax, Zn)

and for n = 1, ... , k. Since TnX =(Ax, Zn)Zn = (x, A*zn)Zn,

the operators Tn are compact as proved in Example 4.8.3. Since k

A=

L

Tn,

n=l

A is compact, by Theorem 4.8.2.

Theorem 4.8.5. If { Tn} is a sequence of compact operators on a Hilbert space Hand Tn...;. T, i.e., II Tn- Til-;. 0, then Tis also compact.

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Theory

Proof. Let {x"} be a bounded sequence, i.e., there exists a constant M such that llxn I :s; M for all n EN. Since T 1 is compacl, there exists a subsequence {x~)} of {xn} such that { Tx~)} is convergent. Similarly, the sequence {T2 x~)} contains a convergent subsequence {T2 x~2 )}. In general, for k 2::2, let {x~k)} be a subsequence of {x~k-J)} such that { Tkx~k)} is convergent. Consider the sequence {x~"l Since it is a subsequence of {xn} we can put xp, = x~"), where {pn} is an increasing sequence of positive integers. Obviously, the sequence { Tkxp,} converges for every kEN. We will show that the sequence { TxPJ converges too. Let c: > 0. Since I Tn - Tll-i> 0, there exists kEN such that I Tk- Til < c:/3M. Next, let k 1 EN be such that

for all n, m > k 1 • We have

I Txp,- TxPm I :s; I Txp,- Tkxp, II+ I Tkxp,- Tkxp, II+ I TkxPm- Txp'" I E

E

E

<-+-+-=c:

3 3 3

'

for sufficiently large n and m. Thus { TxPJ is a Cauchy sequence in H Completeness of H implies that { TxPJ converges. This proves the theorem. Corollary 4.8.1. The limit of a convergent sequence of finite dimensional operators is a compact operator.

Proof.

Finite dimensional operators are compact.

Theorem 4.8.6. Let T be a compact operator on a Hilbert space H Then the adjoint operator T* is compact.

Proof. Let {xn} be a bounded sequence, i.e., llxn I :s; M for all n EN. Define Yn = T*xn, n = 1, 2, .... Since T* is bounded, the sequence {Yn} is bounded. It thus contains a subsequence {ykJ such that the sequence { TykJ converges in H Now, for any m, n EN, we have

IIYk,"- Ydl 2 =II T*xk,,- T*xkJ 2 = ( T*(xkm- xk,), T*(xk"'- xkJ)

= ( IT* ( Xk, -

Xk, ) , ( Xk,, - Xk, ) )

=II TT*(xk,,- xkJ llllxk"'- xdj :s; 2M I Tykm - Tyk, II-i> 0,

Linear Operators on Hilbert Spaces

175

as m, n...;. oo. Therefore {yk,} is a Cauchy sequence in H, which implies that {Yk,} converges. This proves that T* is a compact operator. Theorem 4.8.7. Let T be a compact operator in a Hilbert space Hand let {cf>n} be an orthonormal sequence in H Then lim"-"" Tcf>n = 0.

Proof. Suppose the sequence does not converge to 0. Then there exists a subsequence { cf>PJ such that II T cf>PJ 2: E for some E > 0 and all n EN. Since { cf>PJ is bounded, the sequence { Tcf>PJ has a convergent subsequence { Tcf>q,J so that

Consequently,

On the other hand, since { cf>q,} is orthonormal, we have

This contradiction proves the theorem. Note that from the above theorem it follows that the inverse of a compact operator on an infinite dimensional Hilbert space, if it exists, is unbounded. Theorem 4.8.7 can be generalized in the following way: Theorem 4.8.8. Let T be a compact operator in a Hilbert space H and let Xn, x E H, n EN. If the sequence {xn} is bounded and Xn...;. x weakly, then Txn...;. Tx strongly, i.e., if (xn, y)...;. (x, y) for ally E H, then II Txn- Txll...;. 0.

Proof. Since the sequence {x"} is bounded, every subsequence of the sequence {Tx"} contains a strongly convergent subsequence {TxPJ. On the other hand, for every y E H we have (Txn,y)=(xn, T*y)...;.(x, T*y)=(Tx,y), so Tx"...;. Tx weakly. Thus, by Theorem 3.6.1, TxP"...;. Tx strongly. We have proved that every subsequence of { Tx"} contains a subsequence convergent strongly to Tx, which implies strong convergence of { T" x} (see 1.9 Exercises, (28)).

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Theory

Remarks. 1. In Theorem 4.8.8 the assumption that {xn} is bounded is superfluous. It can be proved that every weakly convergent sequence is bounded. (A proof can be found in [A. Friedman, Foundations of Modem Analysis, Dover, 1982]). Theorem 4.8.8 will be used in the proof of Theorem 4.10.1, where the sequence in question is bounded, so our weaker version is sufficient. 2. It has already been noted that compactness of operators is a stronger condition than boundednesss. For operators, boundedness is equivalent to continuity: bounded operators are exactly those operators which map strongly convergent sequences into strongly convergent sequences. It turns out that the condition in Theorem 4.8.8 is equivalent to compactness, i.e., compact operators can be characterized as those operators (defined on the entire space H) which map weakly convergent sequences into strongly convergent sequences. From this point of view, compactness is a stronger type of continuity. For this reason compact operators are called sometimes completely continuous operators. The above condition has been used by F. Riesz as the definition of compact operators. Hilbert has used still another (equivalent) definition of compact operators: an operator, A defined on a x weakly and Yn...;. y weakly implies Hilbert space H is compact if X (Axn, Yn) -i> (Ax, y). 11 ...;.

4.9. Eigenvalues and Eigenvectors Concepts discussed in this section play a central role in the theory of operators and their applications. First we need to introduce a number of new ideas. Definition 4.9.1 (Eigenvalue). Let A be an operator on a complex vector space E. A complex number A is called an eigenvalue of A if there is a non-zero vector u E E such that Au= Au.

(4.9.1)

Every vector u satisfying (4.9 .1) is called an eigenvector of A corresponding to the eigenvalue A. If E is a function space, eigenvectors are often called eigenfunctions. Example 4.9.1. Let S be a linear subspace of an inner product space E and let A be the projection onto S. The only eigenvalues of A are 0 and 1. Indeed, if for some A E C and 0 ~ u E E we have Au= Au, then Au= A 2 u,

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Linear Operators on Hilbert Spaces

because A 2 =A. Therefore A= 0 or A= 1. The eigenvectors corresponding to 0 are the vectors of E which are orthogonal to S. The eigenvectors corresponding to 1 are all elements of S. Note that, if A is not an eigenvalue of A, then the operator A- A.'fi is invertible, and conversely. If space E is finite dimensional and A is not an eigenvalue of A, then the operator (A- A.'fi) -J is bounded, because all operators in a finite dimensional space are bounded. The situation is more complicated for infinite dimensional spaces.

Definition 4.9.2 (Resolvent, Spectrum). space E. The operator

Let A be an operator on a normed

is called the resolvent of A. The values A for which AA is defined on the whole space E and is bounded are called regular points of A. The set of all As which are not regular is called the spectrum of A. The set of all eigenvalues, which is a subset of the spectrum, is called the point spectrum. The remaining part of the spectrum, i.e., the set of all As for which AA exists but is unbounded, is called the continuous spectrum.

Example 4.9.2. Let E be the space ~([a, b ]) of continuous functions on the interval [a, b ]. For a fixed u E ~([a, b ]), consider the operator A defined by (Ax)(t) = u(t)x(t).

Since (A-A!fi)- 1x(t) =

x(t)

,

u( t)- A

the spectrum of A consists of all A such that A-u( t) = 0 for some t E [a, b]. This means that the spectrum of A is exactly the range of u. If u( t) = c is a constant function, then A = c is an eigenvalue of A. On the other hand, if u is a strictly increasing function, then A has no eigenvalues. The spectrum of A in such a case is the interval (f( a), f( b)]. This is an example of operator with a purely continuous spectrum. The problem of finding eigenvalues and eigenvectors is called the eigenvalue problem. One of the main sources of eigenvalue problems in mechanics is the theory of oscillating systems. The state of a given system at a given timet may be represented by an element u( t) E H, where His an appropriate

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Theory

Hilbert space of functions. The equation of motion in the classical mechanics is (4.9.2) where A is an operator in H If the system oscillates, the time dependence of u is sinusoidal so that u( t) = v sin wt, where v is a fixed element of H If A is linear, then (4.9.2) becomes (4.9.3) This means that -w 2 is an eigenvalue of A. Physically, the eigenvalues of A correspond to possible frequencies of oscillations. In atomic systems, the frequencies of oscillations are visible as bright lines in the spectrum of light they emit. Thus the name spectrum arises from physical considerations. It is important to note that every eigenvector corresponds to exactly one eigenvalue, but there are always infinitely many eigenvectors corresponding to an eigenvalue. Indeed, every multiple of an eigenvector is an eigenvector. Moreover, several linearly independent vectors may correspond to the same eigenvalue. We have the following simple theorem: Theorem 4.9.1. The collection of all eigenvectors corresponding to one particular eigenvalue of an operator is a vector space.

The easy proof is left as an exercise. Definition 4.9.3 (Eigenvalue Space). The set of all eigenvectors corresponding to one particular eigenvalue A is called the eigenvalue space of A. The dimension of that space is called the multiplicity of A. An eigenvalue of multiplicity one is called simple or non-degenerate. An eigenvalue of multiplicity greater than one is called multiple or degenerate. In such a case the number of linearly independent eigenvectors is also called the degree of degeneracy. Example 4.9.3. Consider 2 L ([0, 21r ]) defined by

the

integral

operator

A: L \[0, 21T ]) ...;.

2,.

(Au)(t)=

f

0

cos(t-y)u(y) dy.

(4.9.4)

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Linear Operators on Hilbert Spaces

We will show that A has exactly one non-zero eigenvalue A= 'TT, and its eigenfunctions are u(t) =a cos t+b sin t

with arbitrary a and b. The eigenvalue equation is 2,-

(Au)(t) =

f

cos(t-y)u(y) dy =Au(t)

0

or cost

f

2,-

cosyu(y)dy+sint

f2,-

sinyu(y)dy=Au(t).

(4.9.5)

0

0

This means that, for A ~ 0, u is a linear combination of cosine and sine functions, i.e., u( t) =a cos t + b sin t,

(4.9.6)

where a, bE C. Substituting this into Equation (4.9.5) we obtain 7ra

= Aa

and

7Th

= Ab.

(4.9.7)

Hence A = 'TT, which means that A has exactly one non-zero eigenvalue and its eigenfunctions are given by (4.9.6). This is a two dimensional eigenspace, so the multiplicity of the eigenvalue is two. Equation ( 4.9.5) reveals that A = 0 is also an eigenvalue of A. The corresponding eigenfunctions are all the functions orthogonal to {cost} and {sin t}. Therefore, A= 0 is an eigenvalue of infinite multiplicity. The following theorems describe properties of eigenvalues and eigenvectors for some special classes of operators. Our main interest is in self-adjoint, unitary, and compact operators. Theorem 4.9.2. Let T be an invertible operator on a vector space E and let A be an operator on E. The operators A and TAT- 1 have the same eigenvalues. Proof. Let A be an eigenvalue of A. This means that there exists a non-zero vector u such that Au= Au. Since Tis invertible, Tu ~ 0 and TAr 1( Tu) =TAu= T(Au) = ATu.

Thus A is an eigenvalue of TAT- 1 •

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Theory

Assume now that A is an eigenvalue of TAT-\ i.e., TAT- 1 u=Au for some non-zero vector u = Tv E E. Then

which proves that A is an eigenvalue of A. Theorem 4.9.3. are real.

All eigenvalues of a self-adjoint operator on a Hilbert space

Proof. Let A be an eigenvalue of a self-adjoint operator A, and let u be an eigenvector of A, u ~ 0. Then A(u, u) =(Au, u) =(Au, u) = (u, Au)= (u, Au)= A(u, u). Since ( u, u) > 0, we conclude A =X. Theorem 4.9.4. All eigenvalues of a positive operator are non-negative. All eigenvalues of a strictly positive operator are positive. Proof. Let A be a positive operator, and let Ax= Ax for some x A is self-adjoint, we have O:S (Ax, x) = A(x, x) =A llxll

2

•

~

0. Since ( 4.9.8)

Thus A 2:0. The proof of the second part of the theorem is obtained by replacing"::;" by"<" in (4.9.8). Theorem 4.9.5. All eigenvalues of a unitary operator on a Hilbert space are complex numbers of modulus 1. Proof. Let A be an eigenvalue of a unitary operator A, and let u be an eigenvector of A, u ~ 0. Then

On the other hand, (Au, Au)= (u, A* Au)=
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Linear Operators on Hilbert Spaces

Proof. Let A be a self-adjoint operator and let u 1 and u2 be eigenvectors corresponding to distinct eigenvalues A1 and A2 , i.e., Au 1 = A1 u 1 and Au2 = A2 u2, A1 ~ A2. By Theorem 4.9.3, A1 and A2 are real. Then

A1( u~> u2 ) =(Au~> u 2 ) = ( u~> Au2 ) = ( u~> A2u2) = A2(u~> U2) = A2( u~> u2 ), and hence

Since A1 ~ A2 , we have ( u 1 , u 2 ) = 0, i.e., u 1 and u 2 are orthogonal. Suppose now A is a unitary operator. Then AA* =A* A=!} and I Au II= I ull· First, we will show that A1 ~ A2 implies A1 A2~ 1. Indeed, if A1A2= 1, then

A2 = AIA2A2 = AIIA21 2=AI' because

IA 2

1

= 1, by Theorem 4.9.5. Now A 1 A 2 (u~>

U2)

= (A 1 u~> A2u 2) =(Au~> Au2 ) =

(u 1 , u2 ).

Since A1A2 ~ 1, we get ( u~> u 2 ) = 0, which proves that the eigenvectors u 1 and u2 are orthogonal. The proof is complete. Theorem 4.9.7.

For every eigenvalue A of a bounded operator A we have

IAI:S IIAII· Proof. Let u be a non-zero eigenvector corresponding to A. Since Au= Au, we have

IIAull = IIAull, and thus

IAIIIull = IIAull :S IIAIIIIull· This implies

IAI :S I All.

Remark. If the eigenvalues are considered as points in the complex plane, the above result implies that all the eigenvalues of a bounded operator A lie inside the circle of radius I All. Corollary 4.9.1. the inequality

All eigenvalues of a bounded self-adjoint operator A satisfy

IAI :S

sup llxll"'l

{I(Ax, x)l}.

(4.9.9)

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Theory

Proof follows immediately from Theorem 4.4.5. It is natural to ask whether the absolute value of any eigenvalue actually attains the value IIAII· In general the answer is negative, but it is true for compact self-adjoint operators.

Theorem 4.9.8. If A is a non-zero compact self-adjoint operator on a Hilbert space H, then it has an eigenvalue A equal to either IIAII or -IIAII. Proof. Let {un} be a sequence of elements of H such that I Un II= 1 for all n EN and as n...;. oo.

(4.9.10)

Then IIA Un -IIAun I

Un 1 2 = (A 2 Un -IIAun I 2 Un, A 2 Un -II Au" I 2 Un) 2 2 2 4 2 = IIA unii -2IIAunii\A Un, Un)+IIAunll llunll 2 2 4 2 = IIA Un 1 -2IIAun 11\Aun, Aun) + IIAun 1 1 Un 1 4 2 2 = IIA Un 1 -IIAun 1 2 2 4 :S IIAII 11Aun 1 -IIAu" 1 = IIAun IICIIAII 2 -II Au" 1 2 ). I Au" I converges to I All, we obtain 2 2 as n...;. oo. (4.9.11) IIA Un -IIAun I Un II-i> 0 2

Since

2

The operator A 2 being the product of two compact operators is also compact. Hence there exists a subsequence {upJ of {u"} such that {A 2 uPJ converges. 2 Since I All~ 0, the limit can be written in the form IIAII v, v ~ 0. For every n EN we have

IIIIAII 2 v -IIAII 2 uPn I :S IIIIAII 2 v- A 2 uPn II+ IIA 2 uPn -IIAuPn ll 2 uPn I + IIIIAuPn ll 2 uPn -IIAII 2 uPn II· Thus, by (4.9.10) and (4.9.11), we have as n...;. oo or as n...;. oo.

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Linear Operators on Hilbert Spaces

This means that the sequence { urJ converges to v and therefore

The above equation can be written as

(A-IIAII)(A + IIAII)v = 0. If w =(A+ IIAII)v ~ 0, then (A -IIAII)w = 0, and thus I All is an eigenvalue of A. On the other hand, if w = 0, then -II All is an eigenvalue of A. This completes the proof. Corollary 4.9.2. If A is a non-zero compact self-adjoint operator on a Hilbert space H, then there is a vector w such that I wll = 1 and I(Aw,

w)l = sup {I( Ax, x)l}. llxll-51

Proof. Let w, I w I = 1, be an eigenvector corresponding to an eigenvalue A such that lA I= I All· Then 2 I(Aw, w)l = I(Aw, w)l = 1AIIIwll = IAI = IIAII = sup {I(Ax, x)l}, II xll"'1

by Theorem 4.4.5. Remarks. Theorem 4.9.8 guarantees the existence of at least one non-zero eigenvalue, but no more in general. The corollary gives a useful method for finding that eigenvalue by maximizing certain quadratic expression. The following result is another example of a theorem describing spectral properties of an operator. We will not prove this result. The interested reader can find a proof in [E. Kreyszig, Introductory Functional Analysis with Applications, Wiley, 1978, Theorem 9.2-3]. Theorem 4.9.9.

Let A be a self-adjoint operator in a Hilbert space H Define m = inf (Ax, x) llx11~1

and

M

=

sup (Ax, x). llxll~1

The spectrum of A lies in the closed interval [m, M]. Moreover, m and M belong to the spectrum. Theorem 4.9.10. The set ofdistinct non-zero eigenvalues {An} ofa self-adjoint compact operator A is either finite or limn_.oo An = 0.

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Theory

Proof. Suppose A has infinitely many distinct eigenvalues A,., n EN. Let un, for n EN, be an eigenvector corresponding to An such that II un II= 1. By Theorem 4.9.6, {un} is an orthonormal system. Moreover, by Theorem 4.8.7, we have 2

0 =lim IIAun 11 =lim (Aun, Aun) =lim (AnUn, Anun) n--;..oo

n--;..oo

n--;..oo

2

=lim A~llunll = lim A~. n--;..oo

n-;..oo

This proves the theorem. Example 4.9.4. We will find the eigenvalues and eigenfunctions of the operator A on L \[ 0, 21T]) defined by (Au)(x) = ["'" k(x- t)u(t) dt,

where k is a periodic function with period 21r, square integrable on [0, 21T ]. , As a trial solution we take

and note that (Aun)(x)=

f

2,-

k(x-t) ein'dt=einx

fx

0

k(s) einsds.

x-2~

Thus nEZ,

where

An=["'" k(s)

eins ds.

The set of functions {un}, n E Z, is a complete orthogonal system in L\[0,21T]). Note that A is self-adjoint if k(x)=k(-x) for all x, but the collection of eigenfunctions is complete even if A is not self-adjoint. Theorem 4.9.11. Let {Pn} be an orthogonal sequence of projection operators on a Hilbert space Hand let {An} be a sequence of numbers such that An...;. 0 as n...;. oo. Then (a) L:~J AnPn converges; (b) For each nEN, An is an eigenvalue of the operatorA=[:= 1 AnPn, and the only other possible eigenvalue of A is 0.

185

Linear Operators on Hilbert Spaces

(c) If all Ans are real, then A is self-adjoint. (d) If all projections P" are finite-dimensional, then A is compact. Proof.

(a) For every

mE

N, we have

Since the vectors A Pnx are orthogonal, 11

II

f.

AnPnx

r

n=k

=

f. IIAnPnxll f. IAni IIPnxll 2

2

=

n=k

2 •

n=k

Now, since A11 -i>O as n...;.oo, for every c>O we have

t A~~P~~xr$£ t IIP~~xii =E 1 t P~~xr$£ 1 t P~~rllxll , 2

II

n-k

2

2

n-k

2

n-k

2

n-k

(4.9.12) for all sufficiently large k and m. The sum L~~k P being a finite sum of projection operators, is a projection operator and its operator norm is 1. Thus (4.9.12) yields 11 ,

whenever k and m are sufficiently large. Thus the sequence of partial sums L~~k A11 P11 is a Cauchy sequence, and by Theorem 1.6.5, the series converges to a bounded operator on H (b) Denote the range of Pn by !Jt(Pn) and let n0 E N. If u E !Jt(Pno), then P nou = u and Pnu = 0 for all n ~ n0 , because the P" are orthogonaL Thus Au= A110 U, which shows that A110 is an eigenvalue of A To prove that there are no other non-zero eigenvalues, suppose u is an eigenvector corresponding to an eigenvalue A. Set V 11 = P11 u, n = 1, 2, ... , and let w = Qu, where Q is the projection on the orthogonal complement of !Jt(A). Then 00

u=

L. v"+w, n=l

with w _i iJt ( P") for all n EN. Clearly,

(4.9.13)

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Theory

since Pnw = 0 and A is continuous. Consequently, the eigenvalue equation has the form

or 00

L

(4.9.14)

(A-An)vn+Aw=O.

n=l

Since all vectors in (4.9.14) are orthogonal, the sum vanishes only if every term vanishes. Hence Aw = 0, and for every n EN either A =An or vn = 0. Finally, if u in (4.9.13) is a non-zero eigenvector, then either w ~ 0 or vk ~ 0 for some kEN. Therefore, A= 0 or A= Ak for some kEN, by ( 4.9.14). This proves the assertion. (c) Suppose all Ans are real. Since projections are self-adjoint operators, for any x, y E H we have 00

(Ax, y) =

L

00

(AnPnX, y) =

L

An(PnX, y)

n=l

n=l

00

=

L

00

An(x,Pny)=

n=l

L

(x,AnPny)=(x,Ay).

n=l

(d) A is the limit of a uniformly convergent sequence of compact operators, hence it is compact. Definition 4.9.4 (Approximate Eigenvalue). Let T be an operator on a Hilbert space H A scalar A is called an approximate eigenvalue of T if there exists a sequence of vectors {xn} such that llxn II= 1 for all n EN and II Txn - Axn II-i> 0 as n -i> oo. Obviously, every eigenvalue is an approximate eigenvalue. Example 4.9.5. Let {en} be a complete orthonormal sequence in a Hilbert space H Let An be a strictly decreasing sequence of scalars convergent to some A. Define an operator on H by 00

Tx =

L

An(x, en)en.

n=l

It is easy to see that every An is an eigenvalue of T, but A is not. On the other hand,

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Linear Operators on Hilbert Spaces

as n...;. oo. Thus A is an approximate eigenvalue of T. Note that the same is true if we just assume that An...;. A and An ~A for all n EN. For further properties of approximate eigenvalues see 4.13. Exercises at the end of this chapter.

4.10. Spectral Decomposition Let H be a finite-dimensional Hilbert space, say H = CN. It is known from linear algebra that eigenvectors of a self-adjoint operator on H form an orthogonal basis of H. The following theorems generalize this result to infinite dimensional spaces. Theorem 4.10.1 (Hilbert-Schmidt Theorem). For every self-adjoint compact operator A on an infinite dimensional Hilbert space H there exists an orthonormal system of eigenvectors {un} corresponding to non-zero eigenvalues {An} such that every element x E H has a unique representation in the form 00

X=

L

anun+v,

(4.10.1)

n=l

where an

E

C and v satisfies the equation Av = 0.

Proof. By Theorem 4.9.8 and Corollary 4.9.2 there exists an eigenvalue A1 of A such that IA 11= sup I(Ax, llxll"'l

x)l.

Let u 1 be a normalized eigenvector corresponding to A1 • We set

Q1 = {x E H: x _i u 1}, i.e., Q1 is the orthogonal complement of the set {u 1}. Thus Q1 is a closed linear subspace of H If x E Q1 , then (Ax, u 1 )

= (x, Au 1 ) = A 1 (x, u 1 ) = 0,

which means that x E Q 1 implies AxE Q 1 . Therefore A maps the Hilbert space Q1 into itself. We can again apply Theorem 4.9.8 and Corollary 4.9.2 with Q1 in place of H This gives an eigenvalue A2 such that IA 2 1= sup {I(Ax, llxllsl

x)l:

u E Q1}.

Let u2 be a normalized eigenvector of A2 • Clearly u 1 _i u 2 • Next we set

Q2 =

{X E

Q1; X

j_ U2},

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Theory

and repeat the above argument. Having eigenvalues A1 , ••• , An and the corresponding normalized eigenvectors u 1 , ••• , un, we define

and choose an eigenvalue An+J such that IAn+ll = sup {I(Ax, x)l:

U

(4.10.2)

E Qn}.

llxil"'i

For un+J we choose a normalized vector corresponding to An+J· This procedure can terminate after a finite number of steps. Namely, it can happen that there is a positive integer k such that (Ax, x) = 0 for all x E Qk. Then every element x of H has a unique representation

where Av = 0. Then

and the theorem is proved in this case. Now suppose that the described procedure yields an infinite sequence of eigenvalues {An} and eigenvectors {un}. Then {un}, as an orthonormal sequence, converges weakly to 0. Consequently, by Theorem 4.8.8, the sequence {Aun} converges strongly to 0. Hence

Denote by S the space spanned by vectors {un}. By the Projection Theorem (Theorem 3.10.4), every x E H has a unique decomposition x = u + v or 00

X=

L

anUn +v,

n=l

where v E Sj_. It remains to prove that Av = 0 for all v E Sj_. Let vES\ v:;60. Define w=v/llvll· Then (Av, v) = llvii\Aw, w).

Since

wE

Sj_ E Q,. for every n EN, by (4.10.2) we have 2

I(Av, v )I= II vii I(Aw, w )I:::::: II v 11

2

2

sup {I(Ax, x)l: u E Qn} =II vii 1An+ 1 1-i> 0.

llxll"'i

This implies (Av, v) = 0 for every v E Sj_. Therefore, by Theorem 4.4.5, the norm of A restricted to Sj_ is 0, and thus Av = 0 for all v E Sj_. This completes the proof.

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Linear Operators on Hilbert Spaces

Theorem 4.10.2 (Spectral Theorem for Self-Adjoint Compact Operators). Let A be a se{j~adjoint compact operator on an infinite dimensional Hilbert space H. Then there exists in H a complete orthonormal system (an orthonormal basis) {vn} consisting of eigenvectors of A. Moreover, for every x E H 00

Ax=

L

An( X, Vn)Vn,

(4.10.3)

n=l

where An is the eigenvalue corresponding to Vn· Proof. Most of this theorem is already contained in Theorem 4.10.1. To obtain a complete orthonormal system {vn} we need to add an arbitrary orthonormal basis of sj_ to the system {Un} defined in the proof of Theorem 4.1 0.1. The eigenvalues corresponding to those vectors from Sj_ all equal zero. Equality ( 4.1 0.3) follows from the continuity of A.

Theorem 4.10.3. For any two commuting self-adjoint compact operators A and B on a Hilbert space H, there exists a complete orthonormal system of common eigenvectors. Proof. Let A be an eigenvalue of A and let S be the corresponding eigenspace. For any x E S we have ABx = BAx = B(Ax) = ABx.

This means that Bx is an eigenvector of A corresponding to A, provided Bx ~ 0. In any case, Bx E S and hence B maps S into itself. Since B is a self-adjoint compact operator, by Theorem 4.1 0.2, S has an orthonormal basis consisting of eigenvalues of B, but these vectors are also eigenvectors of A, because they belong to S. If we repeat the same with every eigenspace of A, then the union of all these eigenvectors will be an orthonormal basis of H This proves the theorem. Theorem 4.10.4. Let A be a self-adjoint compact operator on a Hilbert space H with a complete orthonormal system of eigenvectors {vn} corresponding to eigenvalues {An}. Let Pn be the projection operator onto the space spanned by Vn. Then, for all X E H, (4.10.4) n=l

and (4.10.5) n=l

190

Proof.

Theory

From the Spectral Theorem (Theorem 4.10.2), we have (4.1 0.6) n=l

For every kEN, the projection operator Pk onto the one dimensional subspace Sk spanned by Vk is given by

Indeed, for every x E H we have X

= (X, Vk) Vk + L

(X,

Vn ) Vn ,

nT'k

where (x, vk)vk E sk and LnT'k (x, Vn)Vn j_ Sk. Thus (x, of x onto Sk. Now (4.10.6) can be written as

vk)vk

is the projection

n=l

and, by Theorem 4.10.2,

n=l

Hence, for all x

E

n=l

H,

which proves (4.10.5).

L

Note that the convergence of A"P" is guaranteed by Theorem 4.9.11 and is quite different from the convergence of A"P"x.

L

Remarks. 1. Theorem 4.1 0.4 can be considered as another version of the Spectral Theorem. This version is important in the sense that it can be extended to non-compact operators. It is also useful because it leads to an elegant expression for powers and more general functions of an operator. 2. It follows from Theorem 4.10.4 that a self-adjoint compact operator is an infinite sum of very simple operators. One dimensional projection operators are not only the simplest self-adjoint compact operators, but they are also the fundamental ones because any self-adjoint compact operator is a (possibly infinite) linear combination of them.

191

Linear Operators on Hilbert Spaces

Let A, A,., and P,. be as in Theorem 4.10.4. Then

because APnx = AnPnx for all x E H. Similarly, for any kEN, we get (4.10.7) n=l

More generally, for any polynomial p( t) = ant"+ . .. +a It, we have 00

p(A) =

L p(An)Pn. n=l

The constant term in p must be zero, because otherwise the sequence {p(An)} would not converge to zero. In order to deal with polynomials with a non-zero constant term a 0 , we have to add a 0!fi to the series. Note that in such a case p(A) is not a compact operator. The above result can be generalized in the following way. Definition 4.10.1 (Function ofan Operator). Letfbe a real valued function on R such that f( A)...;. 0 as A ...;. 0. For a self-adjoint compact operator A= L:~J AnPn we define 00

J(A) =

L J(An)Pn.

(4.10.8)

n=l

Theorem 4.9.11 ensures the convergence of the series in (4.1 0.8), and that f(A) is self-adjoint and compact.

Example 4.10.1. Let A= L.:~J A"P" be a self-adjoint compact operator such that all A" 's are non-negative. For any a> 0 we can define A"' by 00

A"'x =

L

A~Pnx.

n=l

Note that in the case a =!the above definition agrees with the Definition 4.6.2. Indeed, by (4.10.7), we have 00

(JAY=

L

00

(JA;;/Pn =

n=l

because all An's are non-negative.

L n=l

AnPn =A,

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Theory

Let A= L~~~J AnPn be a self-adjoint compact operator. of A by sine define We can

Example 4.10.2.

00

L

sin A=

(sin An)Pn.

n=l

Condition f( A)-'? 0 as A-'? 0 in Definition 4.1 0.1 can be replaced by AnPn boundedness off in a neighborhood of the origin. Indeed, if A= and Pnx =(x, vn)vn, then for any XE H we have

L.:=J

00

(f(A))x =

L f(An)(x, Vn)Vn, n=l

where convergence of the series is justified by Theorem 3.8.3, because lf(An)(x, vnW::o: Ml(x, vnW, 2 for some constant M, and hence {f(An)(x, An)} E 1 • Clearly, in this case we cannot expect f(A) to be a compact operator.

If eigenvectors {un} of a self-adjoint operator Ton a Hilbert space H form a complete orthonormal system in H and all eigenvalues are positive (or non-negative), then Tis strictly positive (or positive).

Theorem 4.10.5.

Proof. Suppose {un} is a complete orthonormal system of eigenvectors of T corresponding to eigenvalues {An}. For any non-zero vector u =

r.:=J

anun

E

H we have

=

L

00

00

00

an(u, Anun) =

n=l

L n=l

Anan(U, Un)

=L

Ananan

n=l

n=l

if all eigenvalues are non-negative. If all A,s are positive, then the last inequality becomes strict. This completes the proof.

4.11. The Fourier Transform In this section we introduce the Fourier transform in L \R) and discuss its basic properties. The definition of the transform in L \R) is not trivial. The

193

Linear Operators on Hilbert Spaces

integral _1_

../2ii

foo

e-ikxf(x) dx

-00

cannot be used as a definition of the Fourier transform in L\R) because not all functions in L\R) are integrable. It is however possible to extend the Fourier transform from L 1(R) n L\R) onto L 2 (R). In the first part of this section we discuss properties of the Fourier transform in L 1(R). Then we show that the extension onto L 2 (R) is possible and study properties of that extension. Let f be an integrable function on R. Consider the integral kE R.

( 4.11.1)

Since the function g(x) = e-ikx is continuous and bounded, the product e-ikxf(x) is a locally integrable function for any kE R (see Theorem 2.9.2). Moreover, since Ie-ikxl s 1 for all k, x E R, we have

and thus, by Theorem 2.9.3, the integral ( 4.11.1) exists for all k E R. Definition 4.11.1 (Fourier Transform in L 1(R)). j defined by 1 27T

~

A f(k) =

y

f

00

Let f

e-•"kx f(x) dx.

E

L 1(R). The function

(4.11.2)

-00

is called the Fourier transform of f. In some books the Fourier transform is defined without the factor 1j../2ii in the integral. Another variation is the definition without the "-" sign in the exponent, i.e.,

These details do not change the theory of Fourier transforms at all. Instead of"]" the notation ".'JP{f(x)}" can also be used. The latter is especially convenient if instead of a letter "f" or "g" we want to use an 2 expression describing a function, for example .'JP{ e-x }. We will use freely both symbols.

194

Theory

Example 4.11.1. (a) Let a >0. Then gi{e-a
../2ii

-00

=-1-fo ex(a-ik)dx+-1-foo e-x(a+ik)dx

../2ii

../2ii

-00

0

(b)

1 -k2/4 =--e

vT7i

f

oo

1 -k2!4. e -u2 d u=-e

-/2

-00

The following theorem is an immediate consequence of Definition 4.11.1. Theorem 4.11.1.

Let f, g E L 1(R) and a

gi(f +g)= gi(f) + gi(g)

Theorem 4.11.2. tinuous function. Proof.

E

C. Then

and

gi( a f)= agi(f).

The Fourier transform of an integrable function is a con-

Let fE L 1(R). For any k, hER, we have i}(k+h)-}(k)i=

::S

lvb L:

e-ikx(eihx_1)f(x)

1 foo leihx_1llf(x)l dx. ~ v27T -oo

Since Ieihx -1llf(x)l ::S 2lf(x)l

and lim le'hx -11 = 0 h~o

for all x

E

R,

we conclude lim h~o

~foo

v 27T _00

leihx_1llf(x)ldx=O,

dxl (4.11.3)

195

Linear Operators on Hilbert Spaces

J

by the Dominated Convergence Theorem. This proves continuity of (Actually, since (4.11.3) is independent of k, we have proved that is uniformly continuous.)

J

Recall from Section 2.6 that the integral S:oo lf(x)l dx defines a norm in 1 L (R). This norm will be denoted by II· ll1, i.e., 11!11 1 =

L:

lf(x)l dx

for

/E L 1(R).

Theorem 4.11.3. If / 1,/2 , ••• EL 1(R) and ll!nll 1-i>O as n-i>OO, then the sequence {jn} converges to uniformly on R.

J

Proof.

First notice that l](k)l :S

vh L:

vh L:

le-ikxf(x)l dx :S

lf(x)l dx.

Thus

f

00

" " sup lfn(k)f(k)l R

1 :S ~ v2~

-oo

lfn(x)- f(x)l dx,

proving the theorem.

IffE L 1(R), then

Theorem 4.11.4 (Riemann-Lebesgue Lemma). lim l](k)l = 0. lkl-;.oo Since e-ikx = -e-ikx-,-' we have

Proof.

](k) = __1_ foo e-ik(x-,-fk)f(x) dx = __1_

v2iT - oo

v2iT

foo -oo

e-ikxf(x-7!.) dx. k

Hence

f v2ii

f v2ii

00

00

f(k) "

=1[-

2

1

e-·h f(x) d x -1-

-00

1 -1=

foo

2vlii -oo

-00

e-•"kx f ( x-~) dx ] . k

~)) dx. e- ikx( f(x)- f ( x-k

We thus have

1 2vlii

l](k)I:S- foo -00

Jt(x)-J(x-7!.)1 dx. k

Since limJkl_.ooS:oolf(x)-f(x-~/k)l dx=O, by Theorem 2.4.2, we have limlkl-;.oo lf(k)l = 0.

196

Theory

Note that the space ~0 (R) of all continuous functions on R which vanish at infinity (i.e., limlxl~oof(x) = 0), is a normed space with respect to the norm defined by

IIIII =sup lf(x)l. XER

Theorems 4.11.1-4.11.4 show that the Fourier transform is a continuous linear operator from L 1(R) into ~0 (R).

Theorem 4.11.5.

Let f

E

L 1(R). Then

(a) @P{f(x)}=@P{f(-x)}. (b) @P{f(x- y)} = @P{f(x)} e-iky. (c) @P{f(ax)} = (1ja )@P{f(xj a)}, a> 0. The proof follows easily from Definition 4.11.1.

Theorem 4.11.6. If f is a continuous piecewise differentiable function, f,f' E L 1(R), and limlxl~oof(x) = 0, then @P{f'} = ik@P{f}.

Proof.

Simple integration by parts gives

_1_ foo f'(x) e-ikx dx =-1-[f(x) ~ -oo ~ =

e-ikx]~oo+~ foo

f(x) e-ikx dx

..[2;i -oo

ik](k).

Corollary 4.11.1. Iff is a continuous piecewise n-times differentiable function, f, f', ... , f(n) E L 1(R), and limlxl~oo f(k)(x) = 0 fork= 0, ... , n -1, then @P{f(")} = (i)"k"@P{f}. Because of our definition of the Fourier transform it is better to redefine the convolution of two functions, f, g E L 1(R) as follows:

1 (f * g)(x) = ~

foo

v27T -oo

f(x- u)g(u) du.

The main reason is the simplicity of the formula in the next theorem.

Theorem 4.11.7 (Convolution Theorem). @P{f} @P{g}.

Let f, g E L 1(R). Then @P{f * g} =

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Linear Operators on Hilbert Spaces

Proof. Let f, g E L 1(R) and h = and we have " h(k)=

1 27T

f7C: y

1 27T

=-

=-127T =

foo

.

h(x)e-'kxdx=

foo ·g(u) foo -oo

foo

1 27T

foo

.

e-•kx

-00

1 27T

f7C: y

foo

f(x-u)g(u)dudx

-00

.

e-•kxf(x-u) dxdu

-oo

g(u)

-00

~ foo

v27T

*g. Then hE L 1 (R), by Theorem 2.15.1,

f7C: y

-00

f

foo

e-ik(x+u)f(x) dx du

-00

g(u) e-iku du

-oo

~ foo

v27T

e-ikxf(x) dx = g(k)}(k).

-oo

We will now discuss the extension of the Fourier transform onto L\R).

In the following theorem, and in the remaining part of this section, denotes the norm in L 2 (R), i.e.,

llfll2 =

11·11 2

~ L: lfCxW dx

Theorem 4.11.8. Let f be a continuous function on R vanishing outside a A 2 bounded interval. Then f E L (R) and

Proof. Suppose first that f vanishes outside the interval [ -7r, 7T ]. Using Parseval's formula for the orthonormal sequence of functions on [ -7r, 7T] cf>n(X ) =

1

v2iT e

-inx

,

n=0,±1,±2,... ,

we get

Since the above inequality holds also for g(x) = e-i$xf(x) instead of f(x), we obtain 00

IIIII~=

L licn+gW, n=-oo

198

Theory

since IIIII~= I gil~. Integration of both sides with respect tog from 0 to 1 yields

Iff does not vanish outside [ -7r, 7T ], then we take a positive number A for which the function g(x) =/(Ax) vanishes outside [ -7r, 7T ]. Then

and thus

The proof is complete. The space of all continuous functions on R with compact support is dense in L\R). Theorem 4.11.8 shows that the Fourier transform is a continuous mapping from that space into L\R). Since the mapping is linear, it has a unique extension to a linear mapping from L\R) into itself. This extension will be called the Fourier transform on L\R). Definition 4.11.2 (Fourier Transform in e(R)). Let/E L\R) and let {cf>n} be a sequence of continuous functions with compact support convergent to fin L\R), i.e., II!- cf>n 1 2 -i> 0. The Fourier transform off is defined by (4.11.4) where the limit is with respect to the norm in L\R). Theorem 4.11.8 guarantees that the limit exists and is independent of a particular sequence approximating f. It is important to remember that the convergence in L\R) does not imply pointwise convergence and therefore the Fourier transform of a square integrable function is not defined at a point, unlike the Fourier transform of an integrable function. We can say that the Fourier transform of a square integrable function is defined almost everywhere. For this reason we cannot say that, iff E L 1(R) n L \R), then the Fourier transform defined by (4.11.2) and the one defined by (4.11.4) are equal. To be precise, we should say that the function defined by ( 4.11.2) belongs to the equivalence class of square integrable functions defined by

199

Linear Operators on Hilbert Spaces

( 4.11.4). In spite of this difference, we will use the same symbol to denote both transforms. It will not cause any misunderstanding. The following theorem is an immediate consequence of Definition 4.11.2 and Theorem 4.11.8. IffE L 2(R), then

Theorem 4.11.9 (Parseval's Relation).

Remark. In physical problems, the quantity II! 11 2 is a measure of energy, and llill 2 represents the power spectrum off

Theorem 4.11.10.

Let fE L 2 (R). Then

" f(k) =lim

1

r;::;n-.coy27T

fn

e-·"k xf(x) dx,

-n

( 4.11.5)

where the convergence is with respect to the norm in L 2 (R). Proof.

For n = 1, 2, 3, ... , define if lxl< n, if lxl ::2: n.

Then II/- fn ll2 -i> 0, and thus IIi- in 11 2-i> 0 as n -i> oo.

Theorem 4.11.11.

If f. gEL 2 (R), then

L:!(x)g(x) dx= Proof.

f_cocoi(x)g(x) dx.

For n = 1, 2, 3, ... , define if lxl < n, if lxl ::2: n,

and if lxl< n, iflx1:2:n. Since

( 4.11.6)

200

Theory

we have

The function e-iX$gn(x)fm(g) 2

is integrable over R and thus the Fubini Theorem can be applied. Consequently

and

L: Jm(x)gn(X) dx = L""""/m(Hgn(g) df Since

llg- g" llr'-' 0 and II§- in 1 2

-i>

0, by letting n -i> oo we obtain

L:Jm(x)g(x) dx= L:fm(x)g(x) dx, by the continuity of inner product. For the same reason, by letting m -i> oo, we get

L: }(x)g(x) dx =

J:

f(x)g(x) dx,

completing the proof. The following technical lemma will be useful in the proof of the important inversion theorem for the Fourier transform in L 2 (R). Lemma 4.11.1.

Proof.

Let f

E

2

L (R) and let g

= fX

A

Then f =g.

From Theorems 4.11.9 and 4.11.11, and the equality g =

Cf. g)= cJ, g)= cJ,}) =IIi II~= IIIII ~-

X

f we obtain (4.11.7)

Hence also (f,

g)= llfll~.

(4.11.8)

Finally, by Parseval's equality,

llill~= llgll~= 11111~= 11!11~-

(4.11.9)

201

Linear Operators on Hilbert Spaces

Using (4.11. 7 -4.11.9) we get

This

II!- ill~= U- i.f- g)= 11111~- C!. §)- C!. §)+ 11§11~= o. shows that f = l

Theorem 4.11.12 (Inversion of Fourier Transforms in L 2 (R)). Then

f(x)

1 f" =lim ~ n-;.co v27T -n

Letf E L 2 (R).

e'kx" f(k) dk,

where the convergence is with respect to the norm in L 2 (R).

Proof.

Let f

E

L 2 (R). If g =], then, by Lemma 4.11.1, f(x)

1 f" e-•kx g(k) dk =~ g(x) =lim ~ n-;.coy27T -n 1 f" e'·~ex= n-;.coy27T lim ~ g(k) dk -n

Corollary 4.11.2.

IffEL\R)nL 2 (R), then the equality 1 /(x) = ~ v27T

f""

e'"kx" f(k) dk

(4.11.10)

-co

holds almost everywhere in R. The transform defined by (4.11.10) is called the inverse Fourier transform. One of the main reasons for introducing the factor 1/ v27i in the definition of the Fourier transform is the symmetry of the transform and its inverse: 1 .'JP{f(x)} = ~ y

27T

f""

.

e-'kxf(x) dx,

-a)

Theorem 4.11.13 (General Parseval's Relation).

If f. g E L\R), then

L:f(x)g(x) dx= f_""""j(k)g(k) dk.

202

Theory

Proof.

The polarization identity C!. g)= ~elf+ gl -If- gl + ilf + igl 2

2

2

-

ilf- igl

2

)

implies that every isometry preserves inner product. Since the Fourier 2 transform is an isometry on L (R), we have (/,g)= (i, g). The following theorem summarizes the results of this section. It is known as the Plancherel Theorem.

For every fE L\R) there exists

Theorem 4.11.14 (Plancherel Theorem). }EL 2 (R) such that:

(a) Iff E L 1 (R) n L 2 (R), then }(k) = (1/..f27i) f::oo e-ik:x f(x) dx. (b) ll}(k)- (1/..f27i) e-ikxf(x) dxll 2 -i> 0 and II/( x)- (1/ ..f27i) eikx}( k) dkll 2 -i> 0 as n -i> oo. (c) II! II~= 11111~. (d) The mapping f

Cn Cn

-i>

Jis a Hilbert space isomorphism of L \R) onto L (R). 2

Proof. The only part of this theorem which remains to be proved is the fact that the Fourier transform is "onto". Let fE L\R) and define h=

J

,..

and

g =h.

Then, by Lemma 4.11.1,

!=h=i and hence

f=g. This shows that every square integrable function is the Fourier transform of a square integrable function. Theorem 4.11.15.

Proof.

The Fourier transform is an unitary operator on L 2 (R).

First note that

.'JP{g}(k)=

1

171:::

v2~

·~ex-

foo

e-' g(x)dx=

-oo

1

171:::

v2~

foo

·~ex

e' g(x)dx=.'JP- 1{g}(k).

-oo

Now, using Theorem 4.11.11, we obtain 1 (.'JP{f}, g)= f2= YL~

1 = ~ v2~

foo -oo

foo

_ 1 .'JP{f}(x)g(x) dx = ~ v2~

foo

/(x).'JP{g}(x) dx

-oo

/(x).'JP- 1{g}(x) dx = (/, g;:- 1 {g}).

-oo

This shows that g;:-J = .'JP*, and thus .'1P is unitary.

203

Linear Operators on Hilbert Spaces

The Fourier transform can be defined for functions in L 1 (RN) by ](k) =

\N/2 JRN r e-ik·xt(x) dX,

(217"

where k = ( k 1 , ••• , kN ), x = ( X~o ... , xN) and k · x = k 1 x 1 + · · · + kNxN. The theory of the Fourier transform in L\RN) is similar to the one dimensional case. Moreover, the extension to L 2 (RN) is possible and it has similar properties, including the Inversion Theorem and the Plancherel Theorem.

4.12. Unbounded Operators Boundedness of an operator was an essential assumption in almost every theorem proved in this chapter. Methods used were developed with boundedness or continuity in mind. However, in the most important applications of the theory of Hilbert spaces we often have to deal with operators which are not bounded. In this section we will briefly discuss some basic problems, concepts and methods in the theory of unbounded operators. An operator A defined in a Hilbert space H, i.e., iYt(A) c H, is called unbounded if it is not bounded. Therefore, to show that an operator A is unbounded it suffices to find a sequence of elements x" E H such that llxnllsM (for some M and all nEN) and IIAxnll-i>OO. Since for linear operators boundedness is equivalent to continuity, unboundedness is equivalent to discontinuity (at every point). Consequently, we can show that an operator A is unbounded by finding a sequence {x"} convergent to 0 such that the sequence {Axn} does not converge to 0. One of the most important unbounded operators is the differential operator, see Example 4.2.3. Other important unbounded operators arise from the quantum mechanics and will be discussed in Chapter 7. In physical applications it is natural to assume that all eigenvalues are real. For this reason self-adjoint operators are of special interest. It will be convenient to adopt the following convention: when we say "A is an operator on a Hilbert space H" we mean that the domain of A is the whole space H, and when we say "A is an operator in a Hilbert space H" we mean that the domain of A is a subset of H. If the domain of a bounded operator A is a proper subspace of a Hilbert space H, then A can be extended to a bounded operator defined on the entire space H. More precisely, there exists a bounded operator B defined on H, ffi(B) = H, such that Ax= Bx for every x E ffi(A). Moreover, we can always find B such that liB II= IIAII, see 4.13. Exercises, (2). We may thus

204

Theory

always assume that the domain of a bounded operator is the whole of H. In the case of unbounded operators the above is impossible. For instance the domain of the differential operator cannot be extended onto H. On the other hand, it may be still possible to extend the domain of an unbounded operator in such a way that, although the domain of the extension is not the whole space, it has better properties. Extension of unbounded operators is one of the important problems of the theory.

Definition 4.12.1 (Extension of Operators). a vector space E. If

Let A and B be operators on

ffi(A)c ffi(B)

and

Ax=Bx

for every x

E

ffi(A),

then B is called an extension of A, and we write A c B. When performing typical operations on unbounded operators, we have to remember about the domains. For instance, the operator A+ B is defined for all x E ffi(A) n ffi(B), i.e., ffi(A +B)= ffi(A) n ffi(B). It may happen that ffi(A) n ffi(B) = {0} and then the sum A+ B does not make sense. Similarly,

ffi(AB) = {x E ffi(B): Bx E ffi(A)}. The usual properties need not hold. Although, we have the equality (A+ B) C = AC + BC, in general, the inclusion AB + AC c A(B +C) cannot be replaced by equality.

Definition 4.12.2 (Densely Defined Operator). An operator A defined in a normed space E is called densely defined if its domain is a dense subset of E, i.e., cl ffi(A) =E. The differential operator D = dj dx is densely defined in L 2 (R), because the space of differentiable square integrable functions is dense in L 2 (R).

Definition 4.12.3 (Adjoint of a Densely Defined Operator). Let A be a densely defined operator in a Hilbert space H. Denote by ffi(A*) the set of ally E H for which (Ax, y) is a continuous functional on ffi(A). The adjoint A* of A is the operator defined by (Ax, y) = (x, A*y)

for all xEffi(A) and yEffi(A*).

In the above definition A has to be densely defined in order to ensure the uniqueness of the adjoint A*.

205

Linear Operators on Hilbert Spaces

Theorem 4.12.1. space H

Let A and B be densely defined operators in a Hilbert

(a) If A c B, then B* c A*. (b) Ifffi(B*) is dense in H, then B c B**. Proof.

First note that A c B implies (Ax, y) = (x, B*y)

for all xEffi(A) and all yEffi(B*).

(4.12.1)

for all xEffi(A) and all yEffi(A*).

(4.12.2)

On the other hand, we have (Ax, y) = (x, A*y)

Comparing (4.12.1) and (4.12.2), we conclude that ffi(B*) c ffi(A*) and A*(y) = B*(y) for ally E ffi(B*). This proves (a). To prove (b) observe that the condition (Bx, y) = (x, B*y)

for all xE ffi(B) and ally E ffi(B*),

can be rewritten as (B*y, x) = (y, Bx)

for all yEffi(B*) and all xEffi(B).

(4.12.3)

Therefore, since ffi(B*) is dense in H, B** exists and we have (B*y, x) = (y, B**x)

for all y E ffi(B*) and all x E ffi(B**).

(4.12.4)

From (4.12.3) and (4.12.4) it follows that ffi(B) c ffi(B**) and B(x) = B**(x) for any x E ffi(B). The proof is complete. Theorem 4.12.2. If A is a one-to-one densely defined operator in a Hilbert space H such that its inverse A -J is densely defined, then A* is also one-to-one and (4.12.5) Proof. Let yEffi(A*). Then for every xEffi(A- 1) we have A- 1 xEffi(A) and hence (A- 1 x,A*y)=(AA- 1x,y)=(x,y). This means that A*yE ffi((A- 1)*) and

(4.12.6) 1

Next, take an arbitrary yE ffi(A- )*. Then, for each xE ffi(A), we have AxE ffi(A - 1 ). Hence (Ax, (A- 1 )*y) =(A -I Ax, y) = (x, y).

This shows that (A- 1 )*y E ffi(A*) and A*(A - 1 )*y =(A -I A)*y = y.

Equality (4.12.5) follows from (4.12.6) and (4.12.7).

(4.12.7)

206

Theorem 4.12.3. B*A*c (AB)*.

Theory

If A, B, and AB are densely defined operators in H, then

Proof. Suppose xE ffi(AB) and yEffi(B*A*). Since xEffi(B) and A*yE f0 ( B*), it follows that (Bx, A*y) = (x, B* A*y). On the other hand, since BxEffi(A) and yEffi(A*), we have

(A(Bx), y) = (Bx, A*y). Hence

(A(Bx), y) = (x, B*(A*y)). Since this holds for all x (AB)y.

E

ffi(AB), we have y E ffi((AB)*) and (B* A*)y =

Self-adjoint operators have been already discussed in Section 4.4. In that section however, we limited our discussion to bounded operators. Without the boundedness condition the matter is more delicate.

Definition 4.12.4 (Self-Adjoint Operator). Let A be a densely defined operator in a Hilbert space H. A is called self-adjoint if A= A*. Remember that A= A* means that ffi(A*) = ffi(A) and A(x) = A*(x) for all x E ffi(A). If A is a bounded densely defined operator in H, then A has a unique extension to a bounded operator on H, and then its domain as well as the domain of its adjoint is the whole space H. In the case of unbounded operators the situation is much more complicated. It is possible that a densely defined operator A has an adjoint A* such that A(x) = A*(x) whenever xEffi(A)nffi(A*), but ffi(A*):;<'ffi(A), and thus A is not selfadjoint. The following relaxation of the conditions in Definition 4.12.4 seems reasonable in the case of unbounded operators.

Definition 4.12.5 (Symmetric Operator). A densely defined operator A in a Hilbert space H is said to be symmetric if (Ax, y) = ( x, Ay) for every X, y E ffi(A). For bounded operators on a Hilbert space symmetry is equivalent with self-adjointness. Generally, every self-adjoint operator is symmetric. In the first of the following examples we have an unbounded self-adjoint operator. Example 4.12.2 shows that a symmetric operator need not be self-adjoint.

207

Linear Operators on Hilbert Spaces

Example 4.12.1.

Consider H

=

F and an operator in H defined by

This is a one-to-one and self-adjoint operator. The subspace iY't(A) = ffi(A - 1) is dense in H, and it consists of all sequences (Yn) E 12 such that

The inverse A- 1 is defined by

Clearly, A- 1 is an unbounded operator. Indeed, if ek = (8kn), k= 1, 2, ... , then II ek II = 1 and

By Theorem 4.12.2,

Thus A -J is self-adjoint. Example 4.12.2.

Consider the operator A= i dj dt with the domain 2

ffi(A) = {fE L ([a, b]): f' is continuous and /(a)= f(b)

Then

(A!, g)=

r r

if'(t)g(t) dt

= if(b)g(b)- if(a)g(a)=

f(t)ig'(t) dt

=

0}.

r

if(t)g'(t) dt

= (/, Ag).

This shows that A is symmetric. On the other hand, 9Ll (A*) is not the same as ffi(A), because for every fE ffi(A) the equality (A/, g)=(/, Ag) holds for functions g which do not satisfy the condition g(a)=g(b)=O and therefore are not in the domain of A Consequently, A is not self-adjoint. The following theorem gives a nice characterization of symmetric operators.

208

Theory

Theorem 4.12.4. A densely defined operator A in a Hilbert space H is symmetric if and only if A c A*.

Proof.

Suppose A c A*. Since (Ax, y) = (x, A*y)

for all x

E

ffi (A) and all y E ffi (A*),

(4.12.8)

we have

(Ax, y) = (x, Ay)

for all x, y

E

ffi(A).

(4.12.9)

Thus A is symmetric. If A is symmetric, then (4.12.8) and (4.12.9) hold, which implies Ac A*. Now we are going to look at the concept of a closed operator. In some sense it is a weaker version of boundedness. Recall that by the graph of an operator A defined on ffi(A) c E 1 with the range iYt(A)c E 2 , where E 1 and E 2 are vector spaces, we mean 'tl(A)= {(x,Ax):xEffi(A)}. The graph of A:ffi(A)-i>iY't(A) is thus a subset of E 1 x E 2 • Note that, if A c B, then 'tl(A) c 'tl( B).

Definition 4.12.6 (Closed Operator). An operator A from a normed space E 1 into a normed space E 2 is called closed if its graph 'tl(A) is a closed subspace of E 1 x E 2 , i.e., Xn E

ffi(A),

Xn

-i>

X

and Ax" -i> y implies

X E

ffi(A) and Ax= y.

Observe that the domain ffi(A) of a closed operator A does not have to be closed. It can be proved that every closed operator from a Banach space into a Banach space is bounded. (This is kriown as the Closed Graph Theorem.) Thus the domain of an unbounded operator in a Hilbert space H cannot be closed; in particular it cannot be the whole of H. Theorem 4.12.5.

Proof.

The inverse of a closed operator is closed.

If 'tl(A) = {(x, Ax): x

E

ffi(A)} is closed, then obviously

'ti(A -J) = {(Ax, x): x

E

ffi ( x)}

is closed. Theorem 4.12.6. If A is a densely defined operator in a Hilbert space H, then A* is a closed operator.

209

Linear Operators on Hilbert Spaces

Proof.

If Yn E ffi(A *), Yn...;. y, and A *yn...;. z, then for any x E ffi( A) we have (Ax, y) =lim (Ax, Yn) =lim (x, A*yn) = (x, z). n--;..oo

n--;..oo

Hence y E ffi(A*) and A*y = z. Theorem 4.12.7. space H.

Let A be a densely defined closed operator in a Hilbert

(a) For any u, v E H there exist a unique x E ffi(A) andy E ffi(A*) such that Ax+ y = u and x- A* y = v. (b) For any v E H there exists a unique x E ffi(A* A) such that A*(Ax) + x = v. Proof. (a) Consider the Hilbert space H 1 = H x H. Since A is closed, 'tl(A)is a closed subspace of H 1 • Thus, by Theorem 3.10.4, HI= 'ti(A)EB 'tl(A)J_,

with 'tl(A) n 'tl(A)J_ = {0}. Now, [z,y]E'ti(A)J_ if and only if ([x,Ax],[z,y])=O for all xEffi(A) or, equivalently, (X, Z) + (Ax, y) = 0

for all x E ffi(A).

Thus [z,y]E'ti(A)J_

ifandonlyif (Ax,y)=(x,-z)

forallxEffi(A),

that is, if and only if yEffi(A*)

and

z=-A*y.

Consequently, if [v,u]EHxH, then there exist unique xEffi(A) and y E ffi(A*) such that [v, u] = [x, Ax]+ [ -A*x, y],

which completes the proof of (a). (b) If u=O in (a), it follows that there are unique xEffi(A) and yEffi(A*) such that Ax+y=O and x-A*y=v, i.e., x-A*(-Ax)=v for A*(Ax) + x = v. The property of being a closed operator is a desirable one. If a given operator A is not closed, is it possible to extend A to a closed operator? The following important theorem shows that if A is symmetric then it can be always extended to a closed operator.

210

Theory

Theorem 4.12.8.

If A is a symmetric densely defined operator in a Hilbert space H, then there exists a closed symmetric operator B such that A c B.

Proof. We start by defining the domain of B: ffi(B) will the set of all x E H for which there exist X 1 , x 2 , • •• E ffi (A) and y E H such that x"...;. x and A(xn)...;. y. It is easy to check that ffi(B) is a vector space and ffi(A) c ffi(B). The operator B is defined as follows: B(x) =lim A(xn),

(4.12.10)

n~oo

where xl>x 2 , . . . Effi(A) are such that Xn-i>X and the limit limn~ooA(xn) exists. First of all, we have to show that (4.12.10) defines an operator, that is, we have to show that, for every x E ffi(B), the value B(x) is independent of a particular sequence {xn}. Suppose, A(xn)-i> y, A(zn)-i>

W.

By the symmetry of A, we have

for every u E ffi (A). Since the inner product is continuous, (4.12.11) implies ( u, y - w) = (Au, x - x) = 0.

In other words, y- w _i ffi(A). Since ffi(A) is dense in H, we conclude y = z. Therefore B is a well-defined operator in H. Let xEffi(A). If, in (4.12.10), we put xn=x for all nEN, we obtain A(x) = B(x). Thus B is an extension of A. Let x, y E ffi(B). Then there exist sequences {x"} and {y"} of elements of ffi(A) such that

Axn...;. Bx, Yn

-i>

y,

Since A is symmetric, we have

By letting n...;. oo, we get (Bx, y) = (x, By).

Therefore, B is symmetric.

211

Linear Operators on Hilbert Spaces

Finally, to prove that B is closed, we take an arbitrary sequence {x.,} in qj}(B) such that

(4.12.12) for some x, y E H, and prove that x E qj}(B) and Bx = y. By the definition of qj}(B), for every mEN there exists Ym E qj}(B) such that 1

llxm- Ym II<m

and

Therefore, by (4.12.12), we have Ym

But this means that x

E

-i> X

and

Aym

-i>

y.

qz; (B) and Bx = y. The proof is complete.

Note that the extension constructed in the above proof is the minimal extension of A to a closed operator. Spectral decomposition of operators is one of the most important tools in the theory of operators and its applications. We have discussed some spectral properties of bounded operators in Section 4.10. The problem is much more difficult in the case of unbounded operators. The following theorem is an example of a spectral theorem for unbounded operators. Theorem 4.12.9. Let A be an operator on a Hilbert space H If there exists a complex scalar A such that the resolvent (A!fi- A)- 1 is a normal and compact operator, then A can be represented as a weighted sum (possibly infinite) of projections. Sketch of the Proof. Since (A.'fi- A)- 1 is normal and compact it can be represented as a weighted sum of projections, i.e., 00

L

(A.'fi- A)-I=

AnPn.

n=l

Hence 00

(A.'fi- A)=

L

(An)-! Pn

n=l

and 00

A=

L n=l

1

(A- A;; )Pn.

212

Theory

An operator A is said to have a compact-normal resolvent if there exists a scalar A such that (A.'fi- A)- 1 is a compact and normal operator. To apply the above theorem we need to determine whether a given operator A has a compact-normal resolvent. We close this section with the following, rather interesting, remark. Let A be a closed operator in a Hilbert space H. We know that this does not imply boundedness of A. On the other hand, it is always possible to redefine the inner product on ffi(A) such that ffi(A) becomes a Hilbert space and A becomes a bounded operator on ffi(A). In fact, for x, y E ffi(A) define ( x, y) 1 = ( x, y) + (Ax, Ay),

where ( ·, ·) denotes the inner product in H. The proof of completeness of ffi(A) with respect to the norm

llxll1 = llxll 2 + 11Axll 2 , and the boundedness of A in this new Hilbert space is left as an exercise.

4.13. Exercises

(1) If A is an operator on H such that Ax _i x for every x E H, show that A=O.

(2) Let A be a bounded operator defined on a proper subspace of a Hilbert space H. (a) Define an operator A 1 on the closure ffi(A) of the domain of A by A 1x =lim Ax"' n->OO

where

Xn E

ffi(A) and

Xn...;.

x.

Show that A 1 is well defined, i.e., A 1 x does not depend on a particular choice of the sequence {x"}. Show that A 1 is a linear and bounded operator defined on ffi(A). (b) Define an operator B on H by where x 1 is the projection of x onto ffi(A). Show that B is a bounded operator on H. (c) Show that IIAII =liB II· Since A= Bon ffi(A), B is an extension of A.

(3) Let