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(yo) =£ 0. For any vector x e H, we can then write C. Let x € V. Show that 9x{ip) = V>(x),
«*>-£U»w(^»>)0(s/o) This implies that
( * (2/o) - & ) which shows that x yoeiV ^(2/0)
Hence 0=
(x~ -zH\yo
Wo)
It follows that 4>{x) = (j>{y0) {x, j/o) = (x,4>(y0)y0)
.
Put z = 0(yo) 2/0- Then we have 0(x) = (x, z) for all x € H. To prove uniqueness, assume that we have two representations
•
Weak convergence
Definition 3.6.1. A sequence (xn) in a Hilbert space H is said to be weakly convergent with weak limit x, if for all y G H, the sequence of complex numbers
72
Functional Analysis: Entering Hilbert Space
({xn, y)) has the limit ( x , y), in other words, if ( x n , y) —> ( x , y) for n —> oo, for all y £ H. If a sequence (xn) in if converges weakly to x £ H, we write x n —*• x for n —> oo. In this context, ordinary convergence xn —> x for n —> oo is often called strong convergence. Remark 3.6.2. Using Riesz' Representation Theorem, xn —*• x for n —• oo if and only if (j){xn) —> (x) for n —> oo for every bounded linear functional <> / on if. In this formulation, the notion of weak convergence can also be introduced in the more general setting of Banach spaces. Lemma 3.6.3. If xn -+ x for n —> oo, then the weak limit x £ H is unique. Proof. Suppose xn —>• x and xn —*• z for n —• oo. Then (xn,y) —• ( x , y) and (xn,y) —* (z,y) for n —> oo, for all y £ if. Since limits are unique in C, it follows that (x,y) = (z,y) for all y £ if, and hence x — z £ H1- = {0}. • Example 3.6.4. Let (e„) be an orthonormal sequence in the Hilbert space H. For y £ H, we have by Bessel's inequality, Theorem 3.4.5, that the series X^Li \(y> en)\2 is convergent and hence (en, y) —> 0 = (0, y) for n —> oo, for all y £ H. This shows that e n —*• 0 for n —> oo. On the other hand, cP
n
_ pc
||2 _ | U ||2 i lip ||2 _ o m || — || c n 11 T 11 c m 11 — ^ 5
for all n j^ m, so that (e n ) is not a Cauchy sequence, in particular, it is not strongly convergent. This justifies the name weak convergence. Proposition 3.6.5. Let (xn) be a sequence in the Hilbert space H. If (xn) converges strongly to x forn —• oo, then (xn) converges weakly to x forn —> oo, in other words, if xn —•> x for n —» oo, then xn —>• x for n —> oo. Proof. If xn —> x for n —> oo, we have 11 x„ — a; 11 —> 0 for n —> oo. From the Cauchy-Schwarz inequality follows | ( z n , y ) - ( a ; , y ) | = |(:En-:E,2/)|<||:En-:E||||y||
for all y £ if .
This implies that ( i „ , y) —• (a;, y) for n —* oo, for all y £ H, proving that xn —*• a; for n —* oo. D The difference between strong convergence and weak convergence is a genuine infinite dimensional phenomenon; if the Hilbert space is finite dimensional, weak convergence will imply strong convergence.
73
Chapter 3: Theory of Hilbert Spaces
Theorem 3.6.6. Let H be a finite dimensional Hilbert space. Then every weakly convergent sequence (xn) in H is strongly convergent. Proof. Let {ei,...,efc} be an orthonormal basis for H. Then any vector z £ H admits a unique decomposition z = (z,ei)ei H
\-{z, ek)ek
.
Weak convergence of the sequence (xn) to x £ H implies (xn, e$) —> (x, e,) for n —+ oo, for all i = 1 , . . . , k. Hence k
\\x - xn\\2 = ^ | ( x , e,) - (xn, e j ) | 2 —» 0 for n —> oo . i=l
This proves that xn —> a; for n —> oo and hence completes the proof.
•
Theorem 3.6.7. Every weakly convergent sequence in a Hilbert space H is bounded. Proof. Let (xn) be a weakly convergent sequence in the Hilbert space H with weak limit x € H. We have to prove that the sequence (xn) is bounded, in other words, that the set of elements {xn} is a bounded subset of H. We shall argue by contradiction. Hence we assume that the sequence (xn) is unbounded. Clearly, the sequence (xn — x) is then also unbounded. Let K € R be a real number K > 1, which we shall specify later. We can then extract a subsequence (yn) from the unbounded sequence (xn — x) such that \\yn|| >Kn for each n £ N. Now define the sequence (zn) in H by a recursive procedure. For n = 1 put zi =2/1, and then for n > 2, Tf-n+1
( Z n - l i3M) j/n l(2n-l,!M)|||y„|
A
if
(z„-i,2/„) 7^0
if
On-i,2/n) = 0
= Zn-l+ <
#-n+l
IIJMII
Clearly, ||z n — Zn-i II = K~n+1 for each n > 2. By the sum formula for a geometric series, it then follows that K-n
\\zm-zn\\
<
1_K-1
for all m > n > 2. This proves that (zn) is a Cauchy sequence in the Hilbert space H. Consequently, (zn) is convergent, say with limit z € H. First we consider the inner products {zn,yn) for n > 2.
Functional Analysis: Entering Hilbert Space
74
If on t h e one h a n d (zn-i, \lz
V )\~\(z
yn) ^ 0, we have
1 4- K-n+l
J^n-UUn) \{zn-i,yn)\\\yn\\
V_n_
,i '
\{zn-i,yn)\\\yn\ Z
2
> K-n+l\( n-l
, yn)\\\yn\\
=
Kn~^
|(*n-l,yn)|||i/n|| If on the other hand {zn-\,
\\jjn '
yn) = 0, trivially
l(«n,2/n)| = - ^ r f
foreachn>2.
Next we consider the inner products (z, yn) for n > 2. I ( 2 , 2/n)| = |(^rx + 2 - 2 n , y n ) | = \(zn, > \(zn,Vn)\
~
> \{zn,yn)\
-
1
K"-
||z-zn||||yn||
1
1 - K-
l _K(l-K-{1-K)-l\\y1n)\\ 1
> K{1 - K- )
(z-zn,yn)\
\{z~Zn,yn)\
K~n £»•
11?/ II -> 11 Un 11
yn) +
|| "
||
2/n
"
n
K -1 = 1
if we p u t K = 3.
Now recall t h a t (yn) is a subsequence of (xn — x). Hence the above argument proves t h a t there is a subsequence of ( ( z , xn — x)) t h a t does not converge t o 0, contradicting t h a t the sequence (xn) converges weakly t o x. This completes the proof of the theorem. •
Chapter 4
Operators on Hilbert Spaces
Applications of functional analysis often involve operations by continuous linear mappings on the elements in a normed vector space, in particular, a Hilbert space. The terminology used in this connection for such an operation is mostly that of a bounded linear operator rather than a continuous linear mapping. For a normed vector space V, we denote by B(V)
the space of bounded linear operators
T : V —> V ,
in other words, the space of continuous linear mappings T : V —> V.
4.1
The adjoint of a bounded linear operator
We open right away with the definition of the adjoint operator of a bounded linear operator on a Hilbert space and its operator norm properties. Theorem 4.1.1. Let H be a Hilbert space and T e B(H) a bounded linear operator on H. Then there is a unique bounded linear operator T* S B(H) on H, called the adjoint operator of T, satisfying (Tx,y)
= (x,T*y)
for all
x,y € H .
For the operator norms of T and T*, it holds that ||T*|| = | m i
and
||r*T|| = ||T||2.
Furthermore, it holds that (T*)* =T
and
(ST)* = T*S* ,
for all bounded linear operators S,T G B(H). 75
76
Functional Analysis: Entering Hilbert Space
Proof.
For every vector y G H, we can define a bounded linear functional
by
The functional cpy is indeed linear, since T is linear and the inner product is linear in the first argument. That (fry is bounded follows by the computation, using Cauchy-Schwarz' inequality, \
=
(\\T\\\\y\\)\\x\\.
By Riesz' Representation Theorem, there is a unique z G H such that
for all x G H .
To every y G H, there is in other words a uniquely determined z G H for which (Tx,y)
= (x,z)
for all x G H .
We can therefore define a map T* : H —> H by letting T*(y) = z. The map T* is characterized by the equation (Tx,y)
= (x,T*y)
for all x,y e # .
Assertion 4.1.2. T* is a linear operator. Proof. For arbitrary vectors 2/1,2/2 € i? and scalars a i , 02 G C, we have the following computation for every x G H, (x,T*(aiyi
+a 2 2/ 2 )) = {Tx, axyi
+a2y2)
= W[{Tx, y1) + a2~{Tx, y2) = cn{x,T*y1)
+
cx2-{x,T*y2)
= {x,a1T*y1+a2T*y2)
.
Since this holds for all x G H, we conclude that T*(ai2/i + a 2 j/ 2 ) = aiT*j/i + a2T*2/2 , completing the proof that T* is linear. Assertion 4.1.3. T* is bounded. Proof.
For y G H, we have the computation ||T*j/|| 2 = (T*2/,T*2/) = (T(T*2/),2/) <||T(T*y)||||2/||<||r||||T*2/||||2/||.
•
Chapter 4: Operators on Hilbert Spaces
77
Prom this follows that ||T*y||<||T||||y||
for all y 6 H .
[Trivial, if \\T*y\\ = 0; by division, if ||T*y|| ^ 0.] This proves that T* is bounded, and furthermore, that 11T* \ \ < \ \ T11.
D
Assertion 4.1.4. ||T*|| = ||T|| and \\T*T\\ = \\T\\2. Proof. Interchanging the roles of T and T* in the argument showing that ||T*|| < ||T||, we get \\Tx\\2 = (Tx,Tx)
=
(x,T*(Tx))
<\\x\\\\T*(Tx)\\<\\x\\\\T*\\\\Tx\\, from which follows, ||Ta:||<||r*||||s||
forallxeff,
proving that | | r | | < ||T*||. Altogether, we conclude that | | T | | = ||T*||. For 11 a; 11 = 1, we have the computation ||Tx|| 2 = (Tx,Tx)
= (x,T*Tx)
< \\x\\ \\T*T\\ \\x\\ = \\T*T\\ ,
proving that | | r | | 2 < | | T * r | | . Since | | T T | | < ||T*||||T|| = | | T | | | | r | | = ||T||2 , we get altogether ||T*T|| = j | y 112, completing the proof of the assertion.
•
Assertion 4.1.5. (T*)* = T and (ST)* = T*S* for any pair of bounded linear operators S,T € B(H). Proof.
For all pairs of vectors x, y € H, we have the computation (Tx, y) = (x, T*y) = (T*y,x)
= (y,(T*)*x)
from which follows that Tx = (T*)*x for all x£H, The computation (x, (ST)*y) = (STx,y)
= ((!*)*x,
y) ,
proving that (T*)* = T.
= (Tx, S*y) = (x, T*S*y) ,
valid for all x, y G H, proves that (ST)*y = T*S*y or, in other words, that (ST)* = T*S*.
for all y G H , O
Functional Analysis: Entering Hilbert Space
78
This completes the proof of Theorem 4.1.1.
•
Using Theorem 3.5.9 we can prove the following theorem. Theorem 4.1.6. Every bounded linear operator T G B(H) on a Hilbert space H induces the orthogonal splitting H = ker(r*) e T(H) , where ker(T*) denotes the kernel ofT* and T(H) is the closure of the image T(H) for T. Proof.
By the Projection Theorem,
H = T(HJ± ®T(HJ , valid for every closed linear subspace in H, in particular T{H). Assertion 4.1.7. TjH)x
= T ( F ) - 1 = ker(T*).
Proof. Let y G T(H)-1. Then (x,T*y)
= (Tx,y)=0
for all x G H ,
proving that T*y = 0, i.e. y G ker(T*). On the other hand, let y G ker(T*). Then {Tx, y) = (x,T*y)
= (x, 0) = 0 for all x G H ,
proving that y G T(H)X. Altogether, T(H)X = ker(T*). It is easy to prove that T{H)X = T{H)X. This completes the proof of the assertion and hence of the theorem. • Example 4.1.8. A bounded linear operator T : Ck —> Ck is represented by a k x k - matrix [7V,-] with respect to the canonical basis { e i , . . . , efc} in Cfc. In terms of the inner product in Cfc, the matrix elements are determined by ^ ij
=
(•* ej)
e
i) •
Let [T*j] be the matrix representing the adjoint operator T* : Cfc —> Ck. Then
Conclusion: T/ie matrix representation for the adjoint linear operator T* is the conjugate transpose of the matrix representation for T.
Chapter 4: Operators on Hilbert Spaces
79
In the case of real vector spaces, the adjoint operator T* : Rfe —> Rk for a bounded linear operator T : ]Rfc —> Rk is represented by the transposed matrix to the matrix representing T. Definition 4.1.9. Let H be a Hilbert space. A bounded linear operator T on H is called self-adjoint if T = T*. Lemma 4.1.10. Let T be a bounded, self-adjoint linear operator on a Hilbert space H. Then (Tx, x) is a real number for all x € H. Proof.
For every x e H, we have the computation (Tx,x)
= (x,T*x)
= (x,Tx)
= (Tx,x)
,
proving that (Tx, x) is a real number.
•
For self-adjoint operators, there is an alternative method for determining the operator norm. Theorem 4.1.11. Let T be a bounded, self-adjoint linear operator on a Hilbert space H. Then | | T | | = s u p { | (To;, a:) 11 a; € # , ||a;|| = 1} • Proof. For each vector x e H with \\x\\ = 1, we get by the Cauchy-Schwarz inequality, |(Ta;,;r:)|<||Tx||||x||<||T||||a;||2 = ||T||. This proves that the number M T = s u p { | ( T x , 2 ) | | x e # , | | x | | = l} exists, and that MT < \\T\\. A simple calculation shows that (T(x + y),x + y)-(T(x-y),x-y)=2(Tx,y)
+ 2(Ty,x)
,
where each term on the left hand side is a real number since T is self-adjoint. From the definition of MT follows that {T{x + y),x + y) < M T | | x + y|| 2 , -{T(x
- y), x - y) < M T | | x - j / | | 2 .
To prove these inequalities, write x + y = \\x + y\\u and x — y = \\x — y\\v for unit vectors u,v £ H.
Functional Analysis: Entering Hilbert Space
80
Using the parallelogram law, Proposition 3.5.1, for the second inequality we get, (Tx,y)
+ (Ty,x)
= ^((T(x
+ y),x +
<±MT(\\x
+ y\\2 +
<MT()\x\\2
y)-(T(x-y),x-y)) \\x-y\\2)
\\y\\2).
+
Assertion 4.1.12. 11 Tx | | < MT11 x 11 for all x G H. Proof.
The assertion is trivial for Tx = 0. So assume that Tx ^ 0. Put
\\Tx\\ Then we have the computations, (Tx, y) + (Ty, x) = - ^
(Tx,Tx)
+J t t
= \\x\\\\Tx\\
+
= \\x\\\\Tx\\
+ ^-(Tx,Tx)
(T(Tx),x)
ML(Tx,T*x) (T*=T)
= 211^1111^1 MT(lk||2 + ||?/||2)=MT(||a;||2
|2
+
I
J x^ | | T a ; | | 2 ) = 2 M T | | i C | | 2 .
By the inequality preceding the assertion we conclude that \\Tx\\ <MT\\x\\ as asserted.
for all x
€H, •
From the assertion we infer that | | T | | < Mr- Together with the already established inequality MT < \\T\\ this proves that | | T | | = Mr, and the proof of the theorem is complete. • In finite dimensions we know that a linear operator T : Cfe —•
Chapter 4: Operators on Hilbert Spaces
81
Theorem 4.1.13. Let T be a bounded, self-adjoint linear operator on a Hilbert space H. IfT(H) is dense in H, then T has an inverse operator defined on the image T(H) ofT. Proof. By Theorem 4.1.6 we get the splitting H = ker(T*) © T(H) from which follows that ker(T*) = {0}, since T(H) is dense in H. But then ker(T) = {0}, since T* = T, proving that T is injective. Altogether, we have a well defined inverse linear operator T*1 : T(H) —> H as stated in the theorem. • Remark 4.1.14. The inverse linear operator T _ 1 in Theorem 4.1.13 is not necessarily bounded; in fact, most often in applications it is not! If T _ 1 is unbounded its domain of definition cannot be extended to all of H, since it is equivalent for T _ 1 to be bounded and to admit an extension from the dense subspace T(H) to all of H. That these statements are equivalent follows since the inverse mapping for a bijective, bounded linear mapping between Banach spaces is always bounded. The latter fact is a corollary to the famous Open Mapping Theorem (precise statement on p. 104). Theorem 4.1.15. Let P be a bounded, self-adjoint linear operator on the Hilbert space H, and assume that P is an idempotent operator, i.e. P2 = P. Then P is the orthogonal projection onto the closed linear subspace M = {z € H\Pz = z}
(fixed space for P).
Proof. Let I : H —> H denote the identity operator on H, i.e. I(z) = z for all z£H. Then M = { z e H | Pz = z} = { z e HI (P - I)z = 0} = ker(P - / ) . Since P — I is a continuous linear operator on H, it follows from the description of M as the kernel of P — I that M is a closed linear subspace in H. By Theorem 3.5.9, we then have the orthogonal splitting H = M 8 M± . Hence every vector z G H admits a unique decomposition, z = x + y, x e M,y G M1- . From the decomposition z = x + y with x G M and y G ML, we get Pz = Px + Py = x + Py. Since P(Py) = P2y = Py, we have Py G M. But Py also belongs to M1-. This follows from the computation ( x , Py) = (Px, y) = (x, y) = 0, valid for all x G M. Since M n M 1 = {0}, we conclude that Py = 0.
Functional Analysis: Entering Hilbert Space
82
For z = x + y, x G M, y E M^, it follows altogether that Pz = x . In other words, P is exactly the orthogonal projection onto the fixed space M for P, and the proof is complete. • For obvious reasons, a bounded, self-adjoint, idempotent linear operator P on a Hilbert space H is called a projection operator. As we have seen in Theorem 4.1.15, a projection operator is exactly the orthogonal projection onto the corresponding fixed space. Proposition 4.1.16. For a projection operator P on a Hilbert space H, the operator norm satisfies \\P\\ < 1 • If P ^0, the operator norm \\P\\ = 1 • Proof.
For all x G H we have the computation \\Px\\2 = {Px, P x ) = ( P 2 x , x) = {Px,x)<
||Px||||x|| ,
proving that ||Pa;|| < 1 for all x G H with | | x | | = 1. It follows that | | P | | < 1. If P ^ 0, the fixed space M = {z € H\Pz = z] for P contains nonzero vectors [with z = Pu, we have Pz = P(Pu) = P2u = Pu = z\. For a unit vector x G M, we have ||Pa;|| = | | x | | = 1, and consequently, the operator norm must be | | P | | = 1. •
4.2
Compact operators
In a metric space, in particular in a normed vector space, the notions of compactness (defined by the covering property) and sequential compactness (every sequence contains a convergent subsequence) are equivalent notions. When the notion of compactness is used in normed vector spaces, we are therefore free to use the definition most suitable for our purposes. In functional analysis it is often convenient to use the definition: A subset K in a normed vector space V is compact, if every sequence (xn) of elements in K contains a convergent subsequence (xnk) with limit point x G K. In a finite dimensional normed vector space V we know by the Heine-Borel Theorem that a subset K C V is compact if and only if it is a closed and bounded subset. It is always true that a compact subset K C V of a normed vector space V is closed and bounded. In general, the converse is only true in finite dimensions. Example 4.2.1. Let H be a separable Hilbert space. Consider the unit ball C = {xeH\\\x\\
< 1} .
Chapter 4: Operators on Hilbert Spaces
83
Clearly, C is a closed and bounded subset of H. But C is not compact if H is infinite dimensional. This follows since an arbitrary orthonormal sequence (xn) in H is contained in C and has no convergent subsequence (xnk); a subsequence is not even a Cauchy sequence, since \\xni — xn. || 2 = 2 for all i =fi j . A linear operator between normed vector spaces is by definition a bounded, or, continuous, operator if it maps bounded subsets in the domain vector space into bounded subsets in the image vector space. A compact, or, completely continuous, operator has a stronger property. Definition 4.2.2. Let V and W be normed vector spaces. A bounded linear operator T : V —> W is said to be a compact, or completely continuous, operator if it maps every bounded subset A in V into a subset T{A) in W with compact closure T(A) in W. By Example 4.2.1, the unit ball in a Hilbert space H is not compact in infinite dimensions. Hence the identity operator I on H is not compact. Somehow, compactness of an operator is related to finite dimensionality of its image. Definition 4.2.3. A bounded linear operator T : V —> W between normed vector spaces V and W is said to have finite rank if dimT(V) is finite, where dimT(V) is the dimension of the linear subspace in W formed by the image for T in W. Proposition 4.2.4. A bounded linear operator of finite rank is a compact operator. Proof. Let T : V —> W be a bounded linear operator between normed vector spaces V and W for which dimT(V) is finite. Since T is bounded, it maps a bounded subset A C V into a bounded subset T(A) C T{V) C W. Also the closure T(A) is contained in T(V), since T(V) is finite dimensional and hence a closed linear subspace in W. Consequently, T(A) is bounded and closed in a finite dimensional vector space and therefore compact. • In finite dimensions, weak convergence coincides with convergence in norm and hence it is to be expected that compact linear operators behave well with respect to weak convergence. This is indeed true, and the following result motivates why compact linear operators are also called completely continuous operators. Theorem 4.2.5. Let H be a Hilbert space and let (xn) be a weakly convergent sequence with weak limit x. Then for every compact linear operator T on H, the image sequence (Txn) converges strongly (in norm) to Tx. In other words:
Functional Analysis: Entering Hilbert Space
84
For a compact linear operator T G B{H) it holds that xn —*• x for n —> oo => Txn —> Tx for n —> oo . Proof.
If x„ —^ x for n —> oo we have for all y G H, (Txn,y) = (x„,T*y)->(x,T*y) = (Tx,y)
for
n - • oo .
This proves that Txn —*• Tx for n —> oo. Since weak limits are unique and strong convergence implies weak convergence, the only possible strong limit for (Txn) is Tx. Assertion 4.2.6. Txn —» Tx for n —> oo. Proof. The proof is indirect. Suppose (Txn) does not converge in norm to Tx. Then we can extract a subsequence (Txnk) of (Txn) such that \\Txnk - Tx\\ > S for all k G N and some 5 > 0 . Since (xn) is bounded by Theorem 3.6.7 and T is a compact linear operator, we can extract a subsequence (Txntc ) of {Txnk) for which Txnk —> y € H for Z —> oo. Since Ta; nt —^ Tx for 2 —> oo, we must have y = Tx, which is not possible by the construction of Txnk. Thereby we have obtained a contradiction and hence Txn —• Tx for n —> oo. • This completes the proof of the theorem.
•
Example 4.2.7. Let (e„) be an orthonormal sequence in the Hilbert space H. Prom Example 3.6.4 we know that en —*• 0 for n —> oo. Hence by Theorem 4.2.5 we conclude that Ten —> 0 for n —> oo for every compact linear operator T G B(iJ). From the example we infer that the inverse operator (if it exists) of a compact linear operator on a separable Hilbert space is always unbounded. Proposition 4.2.8. Let H be an infinite dimensional, separable Hilbert space. Suppose T G B{H) is a compact linear operator on H, which has an inverse T _ 1 on T(H). Then T _ 1 is always an unbounded linear operator. Proof. Let (e„) be an orthonormal sequence in H; for example an orthonormal basis in H. Then Ten —> 0 for n —> oo, but on the other hand, ||T~ 1 (Te„) || = ||e„ || = 1 for all n G N, and hence T~l cannot be continuous. In other words, the linear operator T _ 1 is unbounded. •
Chapter 4-' Operators on Hilbert Spaces
85
The set of compact linear operators on a Hilbert space H is a closed subset in the Banach space (operator norm) of bounded linear operators B(H) on H. This is a consequence of the following theorem. Theorem 4.2.9. Let H be a Hilbert space and suppose that (Tn) is a sequence of compact linear operators in B(H) converging (in operator norm) to the bounded linear operator T G B(H). Then T is a compact linear operator. Proof. It is sufficient to prove that if (xn) is a bounded sequence in H, then it is possible to extract a subsequence (xnk) such that (Txnk) is convergent. Let (x„) be a bounded sequence in H. Since T\ is compact, (x„) has a subsequence (x^) such that (Tix^) is convergent. Since T2 is compact, (x^) has a subsequence (x^) such that (T2X^) is convergent. Continuing this way we get for every k G N, a subsequence (x*) of (x n ) such that (Tfcx^) is convergent. Now consider the diagonal sequence (x™). It is obvious that (IfcX™) is convergent for every k G N. Assertion 4.2.10. (Tx™) is a Cauchy sequence in H. Proof. Given e > 0. Since the original sequence (x n ) is bounded, there exists a positive constant C such that ||x™|| < C for all n g N. Since Tn —> T for n —> 00 (in operator norm), we can choose a number k G N, which we then keep fixed, such that \\T-Tk\\<^. Since (T^x™) is convergent, there exists a number no G N, such that \\Tkxnn-TkxZ\\
for all
n,m>n0.
For n,m > no we now have the computation \\TxZ-TxZ\\
+ \\TkxZ-Tkx™\\
+
l\Tkx™-Tx™\\
< | | T - T f c | | | | < | | + ! + ||Tfc-T||||x™||
This proves that (Tx™) is indeed a Cauchy sequence in H.
•
By completeness of H, the Cauchy sequence (Tx™) is convergent in H. Since (re") is a subsequence of (x n ), this proves that T is a compact linear operator. •
Functional Analysis: Entering Hilbert Space
86
The following theorem provides among others a method for constructing compact linear operators. Theorem 4.2.11. Let (en) be an orthonormal basis for the infinite dimensional, separable Hilbert space H and let (An) be an arbitrary sequence of complex numbers. For every x £ H, consider the infinite series oo
i x = y ^ An {x, cn )en . «=i
(1) The infinite series is convergent and the sum defines a linear operator T on H if the sequence (Xn) is bounded. (2) T exists and is bounded if and only if the sequence (Xn) is bounded. (3) T exists and is a compact linear operator if and only if Xn —> 0 for n —y oo. Proof. The series Ylri=i(x> en)^n is convergent for all x € H by Theorem 3.4.5 (Bessel's inequality). Then by the comparison test, the infinite series Tx — "Y^n=\ ^n{x, en)en is also convergent for all x £ H if the sequence (A„) is bounded, in particular, if A„ —> 0 for n —> oo. In case of existence, clearly T is a linear operator. This proves (1). Since oo
oo
HTzH^XilAnnOcen)!2
and
||x||2 = £
| (x, en) | 2 ,
and since 11Ten \\ — \Xn\ for all n £ N, it follows immediately that T exists and is bounded if and only if the sequence (A„) is bounded. This proves (2). In order to prove (3), we define the sequence of operators (T^) by k
Tkx = J2 Xn (x,
en) en
for all x £ H .
n=l
Since Tk has finite rank, all the operators T^ are compact linear operators by Proposition 4.2.4. When the sequence (A„) is bounded, we can define the bounds Kk=sup{\Xn\2}
for all k £ N .
n>k
Then for all k E N, we get the estimate oo
\\Tx-Tkx\\2=
]T n=k+l
|An|2|(x,en)|2
Chapter 4: Operators on Hilbert Spaces
87
Prom this estimate follows that 11T — T^ \ \ < \[Kk and hence that ||T-Tfc||->0
forfc^oo,
if
Xn -> 0
for n -+ 00 .
This proves that Tk —> T for k —» oo, and hence by Theorem compact linear operator, if An —> 0 for n —> oo. Now assume that the sequence (An) does not converge to Then there exist an e > 0 and a subsequence (A n J of (An) such for all k £ N. Consider the corresponding subsequence (enfc) of orthonormal sequence, (enfc) is weakly convergent to 0. Since
4.2.9 that T is a 0 for n —> oo. that | Xnk \ > e (e n ). Being an
1 1 T ^ - Tenj || 2 = || A ni e„ 4 - A n .e„. || 2 = | \ni | 2 + | An. | 2 > 2e2 , for all i y^= j , the sequence (Ten/c) sequences, and hence in particular, we conclude by Theorem 4.2.5 that This completes the proof of the
can have no subsequences that are Cauchy (Tenk) does not converge to 0. From this T is not a compact linear operator. theorem. •
Example 4.2.12. Let (e„) be an orthonormal basis in the infinite dimensional, separable Hilbert space H. Define oo
..
Tx — z7 —(x,en)en —' n
for all x E H .
n—l
Then T is a compact linear operator on H by Theorem 4.2.11. The following computation shows that the operator T is self-adjoint, {Tx,y)
= y~]-(x,
en)(y,
e„) = (x,Ty)
for all x,y € H .
ra=l
Since T(fcefc) = efc for all k € N, it follows that T(.H") is dense in H. Hence T has an inverse operator T _ 1 : T(iJ) —> iJ by Theorem 4.1.13. Since T _1 (e fc ) = fcefe for all fceN, clearly T _ 1 is unbounded.
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Chapter 5
Spectral Theory
A linear map T : Ck —> Ck is represented by a complex k x fc-matrix with respect to the canonical basis in Cfc. If the linear map T is self-adjoint, we know from Example 4.1.8 that it is represented by a conjugate symmetric matrix. From basic linear algebra, it is then well known that there exists an orthonormal basis {ej}^ = 1 for Cfe in which the matrix for T can be diagonalized, or, which amounts to the same thing, that the linear map itself is given by fc Tx = ^2\j{x,
ej)ej,
for all x G Ck .
i=i
Here (x,ej) is the j t h coordinate of x in an orthonormal basis {ej}k=1 of eigenvectors for T corresponding to the eigenvalues {Xj}k=1 for T. The eigenvalues for T are the complex numbers X, for which the equation Tx = Xx has nontrivial solutions x e Cfc, or, by an equivalent formulation, for which the linear operator T — XI is not injective, where I denotes the identity operator on Cfc. In infinite dimensions, the situation is much more complicated, but for selfadjoint, compact linear operators a corresponding theory can be developed. It culminates in the famous spectral theorem - the main goal of this chapter.
5.1
The spectrum and the resolvent
Let H he a Hilbert space and let T : H —> H he a linear operator. Often in the following, a linear operator T may not be defined on all of H but only on a linear subspace of H called the domain of definition and denoted by D(T). Let T : H —> H be a linear operator with domain of definition D(T) C H. 89
90
Functional Analysis: Entering Hilbert Space
For A G C, we define the linear operator TX=T-XI
with D{TX) = D{T) ,
where / denotes the identity operator on H. Remark 5.1.1. Integral equations of the form {T-XI)u
=f
for
u,f£L2([a,b]),
AGC ,
defined in the Hilbert space H = L2([a,b]), are known under the name of Fredholm integral equations of the second kind. They occur frequently in applications in the physical sciences. In the general setting presented above, we consider the equation T\x = y
for each y £ H .
If Tx is injective, we can solve the equation to get x = T^y _1
for all y G Tx(H) = D{T^)
.
X
If TA is also bounded, and TX(H) = D{TX ) is dense in H, then T^1 admits a unique extension to all of H, such that the equation Txx = y can be solved for all y € H. This is the ideal situation from the point of view of the equation. However, it will not always be the case. The operator Rx(T) = Tx-l = (T-XI)-1
,
defined for those A G C where Tx is injective, is called the resolvent of T. We are particularly interested in the situation where D(Rx(T)) = Tx(H) is dense in H and R\(T) is a bounded linear operator. Definition 5.1.2. The resolvent set for T, denoted by p(T), is the set of A G C for which R\(T) exists as a densely defined and bounded linear operator on H. The complement a(T) —C\ p(T) is called the spectrum for T. For A G cr(T) various things can happen. First of all, maybe the linear map Tx is not injective so that the resolvent R\(T) is not defined. This is the case if and only if the equation (T — XI) x = 0 has a nontrivial solution
x GH ,
or equivalently, Tx = Xx
for a nontrivial vector
x GH .
Chapter 5: Spectral
Theory
91
If this is the case, we say that A is an eigenvalue of T, with corresponding eigenvector x e H. The equation Tx = Ax is always satisfied for x = 0 and hence we emphasize that in order for A to be an eigenvalue for T, the equation shall be satisfied for an eigenvector x ^ 0. Furthermore, we notice that the collection of all eigenvectors, belonging to an eigenvalue A, form a linear subspace H\ of H, called the eigenspace corresponding to A. The linear subspace H\ can be of finite or infinite dimension depending on the case in question. Definition 5.1.3. The subset crP{T) of the spectrum a{T) for T consisting of the eigenvalues for T is called the point spectrum for T. The subset crc(T) of the spectrum a(T) for T consisting of the complex numbers A £ C for which R\(T) exists as a densely defined but unbounded operator on H, is called the continuous spectrum for T. Finally, the subset ay(T) of the spectrum cr(T) for T, where R\(T) exists but is not densely defined, is called the residual spectrum, for T. All possibilities for A £ cr(T) are covered by Definition 5.1.3, and hence the spectrum a{T) for T is the disjoint union of the point spectrum, the continuous spectrum and the residual spectrum for T. In concrete cases, the resolvent set p(T) and/or parts of the spectrum for T may be empty. Example 5.1.4. Let T : Ck —> Cfc be a linear map with corresponding matrix T = [Tij]. Then it is well known that T has k eigenvalues A i , . . . , Afc (counted with multiplicity) and that the eigenvalues are the roots of the characteristic polynomial P(X) = det(T - AI) for T. Since RX(T) £ B{Ck), when A is not an eigenvalue, it follows that the spectrum o~{T) is a pure point spectrum, namely cr(T) = {Aj}j =1 , and that the resolvent set is the complex plane except finitely many points, namely p(T) = C \ {Xj}k=1. Example 5.1.5. Let d = d/dx be the differentiation operator with domain of definition D(d) = C 1 ^ , 1]) in the Hilbert space H = L 2 ([0,1]). Then (d - XI)eXx = 0 for all A £ C . It follows that the spectrum a(d) is a pure point spectrum, namely a{&) = C, and consequently, that the resolvent set p{d) = C \ o~{d) = 0. The unbounded linear operator d in Example 5.1.5 has an empty resolvent set. On the other hand, this is never the case for a bounded linear operator. This will be an immediate consequence of interesting topological properties of the spectrum for a bounded linear operator. For the proof of these properties, we need a general result on existence of bounded inverse operators for small
92
Functional Analysis: Entering Hilbert Space
perturbations of an identity operator proved by the German mathematician Carl Gottfried Neumann (1832-1925). Lemma 5.1.6 (Neumann's Lemma). Let V be an arbitrary Banach space, and let T G B(V) be a bounded linear operator with operator norm \\T\\ < 1. Then the operator I — T has a bounded inverse operator (I — T ) _ 1 on V. Proof.
Consider the infinite series, oo n=l
in the Banach space B(V). The series is convergent in B(V) by the comparison test, since | | T n | | < \\T\\n f o r a l l n e N , and the geometric series X^^Lo ll^ll™ *s convergent, since | | T | | < 1. Elementary computations show that S(I-T)=I=(I-
T)S .
This proves that I — T is invertible with inverse operator (/ - T)""1 = S.
•
Then we are ready to study the topological properties of the resolvent set p(T) and the spectrum o~(T) for a bounded linear operator T on a Hilbert space H. Theorem 5.1.7. Let H be a Hilbert space and let T € B(H) be a bounded linear operator on H. Then the resolvent set p{T) for T is an open set in C and the spectrum a(T) for T is a closed set in C. Proof. If p(T) = 0, the theorem is trivially true. Assume therefore that p(T) =£ 0. Let fi e p(T) be an arbitrary point in p(T). Then R^T) = (T —/x/) _ 1 is a densely defined, bounded linear operator. For A G C, we have T - AJ = (T - /*/) - (A - p)I = {T- »I)(I - (A -
riR^T))
.
By Neumann's Lemma, this rewriting shows that T—XI is an invertible, densely defined (namely on T^(H)) operator with bounded inverse, when \\(\-fi)Rll(T)\\
=
\\-v\\\Rll(T)\\
in other words, for | A — /x j < ||i? M (T)|| _ 1 . This describes a circle in C with center //, thereby proving that p{T) is an open set in C.
Chapter 5: Spectral
Theory
93
Since a(T) — C \ p(T), the spectrum a(T) is a closed set in C.
•
Theorem 5.1.8. Let H be a Hilbert space and let T e B(H) be a bounded linear operator on H. Then the spectrum o(T) for T is a compact set in C, contained in the circle with radius \\T\\ and center 0. Proof.
Assume that | A | > 11T11. Then
T-XI = -X(I-\T) has a bounded inverse, since || (1/A)T|| < 1, and hence A e p{T). Consequently, A e o(T) implies that |A| < | | T | | , proving that a(T) is a bounded set in C. But cr(T) is also a closed set in C, and hence cr{T) is a compact set in C • Remark 5.1.9. For a bounded linear operator T G B(H) it follows in particular that A 6 p(T) for |A| > | | T | | . Hence p(T) =£ 0 for a bounded linear operator. 5.2
Spectral theorem for compact self-adjoint operators
We now turn our attention to bounded, self-adjoint linear operators T on a Hilbert space H. For such an operator T, we have the alternative formula for the operator norm | | T | | proved in Theorem 4.1.11, \[T\\ = B\xp{\{Tx, x)\\x
e H,\\x\\ = 1} .
Hence we get the following corollary from Theorem 5.1.8. Corollary 5.2.1. LetT be a bounded, self-adjoint linear operator on a Hilbert space H. Then for all A G o-(T) in the spectrum for T it holds that |A| < s u p { | ( T a ; , a ; ) | | a ; e f l r ) | | a ; | | = l } . Example 5.2.2. Let (e„) be an orthonormal basis for the infinite dimensional, separable Hilbert space H. Then the one-sided shift operator T on H with respect to this basis is defined by oo
oo
T(x) - T(^2(x,
en)en)
n=l
= ^{x,
e„)e„_i
for all x e H .
n=2
We shall determine the spectrum a(T) for T. For that purpose, suppose oo
x = 2_\ixi n=l
e
n)en
is an eigenvector for T ,
Functional Analysis: Entering Hilbert Space
94
with corresponding eigenvalue A. Then we get recursively, (x,e2) = A(x,ei) (x, e 3 ) = A(x, e 2 ) = A 2 (x, ei) (x, e 4 ) = X(x, e 3 ) = A 3 (x, ei)
(x,en)
=
Xn-1(x,e1)
Now further suppose that (x, ei) = 1. Then we get {x, en) — A™-1 for all n € N, and hence x = J2^=i(xy e n ) e « is an eigenvector corresponding to A 6 C if and only if the sum defines a vector x € H, which by Proposition 3.4.4 amounts to X)^Li |^™ _1 1 2 < °°- The latter is certainly the case if and only if | A | < 1. Hence the open unit disc is contained in the spectrum a(T). On the other hand, for all x 6 H, we have, oo
oo
||Tx|| 2 = ^ | ( x , e „ ) | 2 < ^ | ( x ra=2
! e n
)|
2
= ||x||2,
ri=l
and since ||Te„|| = 1 for n > 2, we get | | T | | = 1. By Theorem 5.1.8, it follows that the spectrum a(T) for the shift operator T must be contained in the closed unit disc. Since criT) is a closed set in C containing the open unit disc, the spectrum
Since T is self-adjoint (not necessarily bounded) we have, {Tx,y)
= (x,Ty)
for
all x,y € H .
If Tx = Xx for A £ C and a vector x =f= 0, we get A(x, x) = (Ax, x) = (Tx, x) = (x, Tx) = (x, Ax) = A(x, x) . Since (x, x) = | | x | | 2 ^ 0, we conclude that A = A, proving that A is real.
Chapter 5: Spectral
Theory
95
If now also Ty = \xy for fi 6 C and a vector y ^ 0, we get \(x,y)
= (\x,y)
= (Tx,y)
= (x,Ty)
= (x,ny)
= JL{x,y) = fi(x,y)
From this we conclude that ( x , y ) = 0 i f A ^ / i . This completes the proof of the theorem.
. D
We are now heading for a proof that a self-adjoint, compact linear operator on a Hilbert space actually has an eigenvalue. We begin with the following result showing that at least asymptotically there is an eigenvalue. Theorem 5.2.4. Let T be a bounded, self-adjoint linear operator on a Hilbert space H. Then there exist a real number A and a sequence (xn) of unit vectors in H (11 xn 11 = 1 for all n € N), such that (T — \I)xn
—» 0 for
n —> oo .
Furthermore, either A = | | T | | or A = — | | T | | has this property. Proof.
Since \\T\\=8ap{\{Tx,x)\\x£H,\\x\\
= l},
there is a sequence (yn) with ]\yn || = 1 in H such that |(Ty„,2/n)|^||T||
for
n - oo .
Now, since (Tyn, yn) is a real number, there is a subsequence (xn) of (yn) for which either (Txn, xn) —> ||T||, or, (Txn, xn) —> —||T||, for n —> oo. Choose A = ± | | T11 such that ( T x n , xn) —• A for n —> oo. Then we have
II (r - A/)xn ||2 = [TXn - \xn, rx n - \xn) = ||T : C n || 2 + A 2 | | a ; „ | | 2 - 2 A ( T a : „ , x n ) <||r||2+A2-2A(Txn,:En) = 2A2 - 2X(Txn,xn)
- • 2A2 - 2A2 = 0 ,
for n —* oo. This proves that || (T — A/)x n || —> 0 for n —* oo, and hence that (T - A i > „ -^ 0 for n -^ oo. • Theorem 5.2.4 states that a bounded, self-adjoint linear operator T on a Hilbert space H has either ||!T|| or — | | T | | as an approximate eigenvalue. In the limit, the sequence (x n ) functions as an eigenvector. The only problem is that the sequence (xn) does not necessarily converge. When the operator T
96
Functional Analysis: Entering Hilbert Space
is compact and self-adjoint, the sequence does converge and we get a proper eigenvalue as stated in the following theorem. Theorem 5.2.5. Let T be a compact, self-adjoint linear operator on a Hilbert space H. Then at least one of the numbers \\T\\ and — | | T | | is an eigenvalue forT. Proof. The case T = 0 is trivial, and hence we assume that T / 0 . Consider a sequence (x„) in H chosen in accordance with Theorem 5.2.4 and related to A = ±11T \ \. Since T is compact, the sequence (x n ) has a subsequence ( i „ J such that (Txnk) is convergent. Prom the way (xn) was chosen, it follows that Txnk — Xxnk —+ 0 for k —• oo. Since Ax„fc = Txnk — (Txnk — Xxnk) for all k £ N, also the sequence (^Xxnk) must be convergent, and hence xnk —> XQ for k —> co for some xo € H. By continuity of the norm 11 • 11, it follows that ||xo|] = 1, since \\xnk || = 1 for each k 6 N. By the construction of XQ £ H, it follows that TXQ = XXQ- Since Xo ^ 0, this proves that A = ± | | T | | is an eigenvalue for T. D During the proof of Theorem 5.2.5, we have produced a unit vector XQ G H, such that Txo = Xxo, with A = ± | | T | | . Prom this we get, \\Tx0\\
= \\Xx0\\
= \X\\\xo\\
= \\T\\
= s u p { | ( T x , x ) | | a ; e i ? , | | x | | = 1} , showing that the supremum is attained for x = xo and hence is actually a maximum value. We get therefore the following corollary to Theorem 5.2.5. Corollary 5.2.6. Let T be a compact, self-adjoint linear operator on a Hilbert space H. Then the operator norm ofT can be determined as a maximum value, | | T | | = m a x { | ( T x , x ) | | x e # , | | x | | = l} . Moreover, the maximum value is attained for a nonzero eigenvector corresponding to the eigenvalue \\T\\ or — \\T\\. Proposition 5.2.7. Let T be a compact linear operator on the Hilbert space H, and assume that T has a nonzero eigenvalue X. Then the corresponding eigenspace H\ = {x £ H | Tx = Ax}
is finite dimensional.
Proof. We proceed by an indirect proof. Assume therefore that H\ is infinite dimensional. Then we can choose an orthonormal sequence (x n ) in H\, such
Chapter 5: Spectral Theory
97
that ||Txn-ro;TO||2 = |A|2||a;n-xm||2 = 2|A|2>0,
for n ^ m ,
proving that (Txn) contains no convergent subsequences. Hence T cannot be compact and we have a contradiction. This proves that H\ is finite dimensional. O Then we are ready for the main theorem in this chapter. Theorem 5.2.8 (Spectral Theorem - Compact Operators). Let T be a compact, self-adjoint linear operator on a Hilbert space H of finite dimension, or of separable, infinite dimension. Then H admits an orthonormal basis (e n ) consisting of eigenvectors for T. In the finite dimensional case, the numbering of the finite sequence of basis vectors ( e i , . . . , efc) can be chosen such that the corresponding finite sequence of eigenvalues (X\,..., Afc) decreases numerically, |Ai|>|A2|>...|Afc| .
In the basis of eigenvectors, the operator T is described by, k
Tx = ^
k
\n(x,
en)en
for x = ] P ( x , e n )e„ .
n=l
n=l
In the separable, infinite dimensional case, possible eigenvalues Xn = 0 must be admitted anywhere in the sequence. With this tacitly understood, the numbering of the infinite sequence of basis vectors (e„) can be chosen such that the sequence of corresponding eigenvalues (An) decreases numerically, I Ai | > [ A2 I > . . . I A„ I > . . .
and
X„ —> 0 for n —* 00 .
In the basis of eigenvectors, the operator T is described by, 00
Tx=
] P A „ ( X , en)en ra=l
00
for
x - ^ ( s , en)en
.
n=l
Proof. By Corollary 5.2.6, the compact, self-adjoint linear operator T admits at least one eigenvalue, namely Ai = ± m a x { | ( T a ; , x ) | | x e H, \\x\\ = 1} . Choose a corresponding normalized eigenvector e\ G H. Let Qi = {ei}-1- be the orthogonal complement to the 1-dimensional subspace spanned by the vector e\. Being an orthogonal complement, Q% is a
98
Functional Analysis: Entering Hilbert Space
closed subspace of the Hilbert space II and hence Qi is itself a Hilbert space. For x € Qi, we have the following computation (Tx, ei) = (a;, Tex) = A ^ x , e x ) = 0 , showing that Q\ is invariant under T in the sense that Tx G Q\ if a; 6 Q\. Hence T can be considered as a compact, self-adjoint linear operator on the Hilbert space Q\. Working inside Q\, we then get a second eigenvalue for T, A2 = ± m a x { | ( T x , a ; ) | | x G <5i,||a;|| = 1} , and a normalized eigenvector e 2 G Qi for A2. Clearly, | Ai | > | A2 | and ei J. e 2 . Next consider the orthogonal complement Q 2 to the 2-dimensional subspace spanned by the vectors e\ and e 2 . Working inside Q 2 , we find a third eigenvalue A3 for T corresponding to a normalized eigenvector e^ G Q 2 . Proceeding in this manner, we get an orthonormal sequence of eigenvectors (e„) and a decreasing sequence, |A1|>|A2|>...|A„|>... , defined by eigenvalues for T and associated with a sequence of subspaces • • • C Qn C Q n _i C • • • C Qi c H, such that each eigenvalue satisfies, |A„| = max{|(Ta:, x ) | | x G Q „ _ i , | | x | | = 1} . If H is finite dimensional, say of dimension k, then this procedure terminates after k steps, and we have an orthonormal basis consisting of the k eigenvectors for T. If H is infinite dimensional, the orthonormal sequence (e n ) converges weakly to 0, and hence (Ten) converges to 0 in norm since T is compact, thereby proving that | A„ | = 11 Ten 11 —• 0 for n —• 00. In the infinite dimensional case, let M be the linear subspace of if consisting of all convergent series J2n°=i anen- Then (e„) is an orthonormal basis for M and via this basis, M can be identified with the complete space I2. Hence M is a complete subspace, and therefore a closed subspace, in H. By the Projection Theorem 3.5.10, we can then decompose H as the orthogonal sum H = M © M1-. Hence every vector x € H can be written uniquely in the form x — z + y, for vectors z e M and y G M1-. Since every element y G M1- is orthogonal to all the eigenvectors e n , the subspace M1- of i J is contained in all the subspaces Qn of H.
Chapter 5: Spectral Theory Assertion 5.2.9. Restricted to M1, Ty = 0 for ally G M x .
99
the operator T is the zero operator, i.e.
Proof. The assertion is trivially true for y = 0. For an arbitrary nonzero vector y G M1- (if such a vector exists), we write y in the form y = \\y\\yi, whereyi = y/\\y\\. Then we get (Ty,y) = \\y\\2 {Tyu Vl) . Since?/! G Qn for all n G N, it follows that | (Ty\, f/i) | < | An | for all n G N, and consequently, that |(Ty,2/)|<||y||2|A„|^0
for
n -> oo ,
since | An | —• 0 for n —> oo. We conclude that (Ty, y) = 0 for all y G M- 1 , from which follows (Exercise 56) that T is the zero operator on M- 1 . • Now choose an orthonormal basis for M1- supplementing the basis in M to an orthonomal basis for H = M © M1-. Since the restriction of T to M1- is the zero operator, each of the basis vectors from M1- is an eigenvector for T corresponding to the eigenvalue 0. Finally, we assign numbers to the eigenvectors in the combined basis for H, for example by sticking the eigenvectors from M x in between the sequence of eigenvectors from M on the even numbers as long as needed to include all eigenvectors from M1-. In this way we obtain an orthonormal basis of eigenvectors for T with the properties in the theorem. In particular, we still have that An —> 0 for n —> oo, since all the new eigenvalues added to the sequence of eigenvalues from M are 0. In the finite dimensional case, every vector x G H admits the (unique) expansion in the orthonormal basis of eigenvectors, k
x — y j \ x, en) e n . n=l
Then we have k
Tx = T\^^2(x, n=l
k
en)en)
k x
— Y2( >
e
re
™) ™ = ^
n=l
A n (a;, e n ) e n .
n=l
In the separable, infinite dimensional case, every vector x G H admits the (unique) expansion in the orthonormal basis of eigenvectors, oo
x = > ( x, Gnjen . 71 = 1
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Functional Analysis: Entering Hilbert Space
By continuity of T we then have k
k
Tx = T[ lim V " 0 , en)en) \ k—too ^—' n=l
= lim T( V V x , /
fc—too
k =
V. *—' n=l
en)en) /
oo
lim V V x , e n ) T e n = V ^ A n ( a ; , e „ ) e „ . fc—too
*—^
z
—'
This completes the proof of the Spectral Theorem.
•
Exercises
Basic prerequisites for the exercises from topology, including completeness of metric spaces (Banach fixed point theorem), and the theory of normed vector spaces, including the operator norm of a bounded linear mapping, can all be found in the general reference V.L. Hansen: Fundamental Concepts in Modern Analysis, World Scientific, 1999. A short summary of the necessary results from metric topology is given in Chapter 1.
Chapter 1 Exercise 1. Consider the metric space (M,d) where M = [l,oo) and d is the usual distance in M. Define the mapping T : M — - > M by Tx=
x 1 - +2 x
for x e M .
Show that T is a contraction and determine the minimal contraction factor A. Determine the fixed point for T. Exercise 2. A mapping T of a metric space (M, d) into itself is called a weak contraction if d(Tx, Ty) < d(x, y)
for all x, y G M, x =/= y .
1) Show that T has at most one fixed point. 2) Show that T does not necessarily have a fixed point. Hint: Consider the function Tx = x + \ for x > 1. Exercise 3. Let T be a mapping from a complete metric space (M, d) into itself, and assume that there is a natural number m such that Tm is a contrac101
102
Functional Analysis: Entering Hilbert Space
tion. Show that T has one and only one fixed point. Exercise 4. In mathematics one often considers iteration schemes of the form xn =
g{xn-i)
for a real-valued, differentiable function g of class C 1 . Show that the sequence (xn) is convergent for any choice of XQ if there is a real number a in the interval 0 < a < 1, such that \g'{x)\
for all x e l .
Exercise 5. A general method to approximate the solution to an equation is to try to bring the equation into the form x = g{x), and then choose an XQ, and use the iteration scheme xn = g(xn-\). Assume that g is a C1-function in the interval [XQ — 5, XQ + 5], and that \g'{x) \ < a < 1 for x £ [XQ — 5, x$ + 6], and moreover \g{x0) -x0\
< (1 -a)6
.
Show that there is one and only one solution x £ [XQ — S, XQ + 6} to the equation, and that xn —> x. Exercise 6. Solve by iteration the equation f{x) = 0 for / € Cx{[a, b]), where f(a) < 0 < f(b) and / ' is bounded and strictly positive in [a, 6]. Hint: Take g(x) = x — Xf(x) for a suitable choice of A. Exercise 7. Show that the equation f(x) = x3 + x — 1 = 0 can be solved by the iteration xn = g{xn^i)
= 1/(1 + xl_x)
.
Find xi,£2,£3 for XQ = 1, and find an estimate for \x — xn\. Exercise 8. Show that any finite dimensional linear subspace U in a normed vector space V is a closed set in V. Exercise 9. Let / : X —• Y be an arbitrary mapping between topological spaces X and Y. In the product space X x Y equipped with the product topology, consider the graph G(f) of / , i.e. the subset
G(f) = {(xJ(x))\x€X}
.
103
Exercises
1) Suppose that Y is a Hausdorff space. Prove that the graph G(f) of / is a closed subset in X x Y if / : X —• Y is a continuous mapping. (A topological space Y is called a Hausdorff space if, for every pair of points 2/i; 2/2 ^Y with 2/1 ^ 2/2, there exists a corresponding pair of mutually disjoint, open sets U\ and U2 in Y, such that y\ 6 C/i and 2/2 £ C^2-) 2) Is it necessary that Y is a Hausdorff space for 1) to hold? Hint: Consider the graph of the identity mapping lx : X —> X. Exercise 10. Show that a closed linear subspace of a Banach space is itself a Banach space. Exercise 11. Let (V, ||-||y) and (W, \\-\\w) be normed vector spaces. Define the product V ® W of normed vector spaces by giving the set V ®W = {(x,y)\x
&V,y eW}
,
the obvious entrywise defined vector space structure and the norm \\(x,y)\\1
= \\x\\v + \\y\\w
for
x G V, yeW
.
Now assume that (V, || • ||y) and (W, || • \\w) are Banach spaces. 1) Prove that (V ® W, \\ • \\i) is a Banach space. 2) Prove that the graph G(f) of a continuous, linear mapping / :V - • W is a closed linear subspace of (V"
1- anxn + y ,
,an S C and a unique vector y &V with f(y) = 0.
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Functional Analysis: Entering Hilbert Space
2) Suppose t h a t W is finite dimensional and t h a t / : V -> W is a surjective, bounded linear mapping. Prove t h a t for every open ball Br(x0) in V there exists an open ball Bs{f{x0)) in W such t h a t Bs(f(x0)) C f(Br(x0)). Hint: Suitably exploit t h a t the image of a ball in V under / is t h e image of a ball in a finite dimensional subspace X of V and t h a t there exists a radius t > 0 such t h a t \\f(x) - f(xQ)\\ > i for s e 1 with \\x - x0\\ = 1. 3) Prove the Open Mapping Theorem for W finite dimensional. 4) Try to prove the general Open Mapping Theorem. It needs other ideas and possibly you have to consult references in the Bibliography for inspiration. E x e r c i s e 1 3 . Prove t h e following corollary t o t h e Open Mapping Theorem. T h e o r e m (Banach's Theorem). Let (V, \\-\\v) and {W,\\-\\w) be spaces. Then the inverse linear mapping f~l : W —> V to a bijective, linear mapping f : V —> W is also bounded.
Banach bounded
E x e r c i s e 14. Prove the following corollary to Banach's Theorem. T h e o r e m (Closed G r a p h Theorem). A linear mapping / : V —» W between Banach spaces (V, || • \\v) and (W, || • 11vv) is continuous if and only if the graph G(f) of / is a closed linear subspace in (V ® W, || • | | i ) . Hint: Consider the linear mappings t h a t project the graph G{f) of / in V ® W onto V and W and write / as a suitable composition of linear mappings. D e f i n i t i o n . Let T : D{T) c V —> W be a linear operator between normed vector spaces (V, || • ||y) and (W, || • \\w) denned in the linear subspace D(T) of V. T h e n T is called a closed linear operator if the graph G{T) of T is a closed linear subspace of the product space (V" © W, || • | | i ) ; cf. Exercise 14. E x e r c i s e 1 5 . Prove t h a t a linear operator T : D(T) C V —> W between normed vector spaces (V, | | - | | y ) and (W, \\-\\w) is a closed linear operator if and only if for each sequence (xn) in the domain of definition D{T) of T t h a t converges to a point x eV, and for which the image sequence (Txn) converges to a point y eW, the limit point x actually belongs to D(T) and Tx = y. E x e r c i s e 16. Prove t h a t a linear operator T : V —> W between Banach spaces (V, || • | | v ) and (W, \\-\\w) is bounded if and only if it is closed. E x e r c i s e 17. Let T : D{T) C V —> W be a closed linear operator between two normed vector spaces defined in the linear subspace D(T) of V. Show t h a t ker(T) is a closed subspace of V•
105
Exercises
Exercise 18. Let V be a Banach space and let T be a closed linear operator T : D(T) cV -^V defined in the linear subspace D(T) of V. Prove that A+T and TA are closed linear operators for any bounded linear operator A : V —> V.
Chapter 2 Exercise 19. Let p b e a real-valued continuous function defined for a; > 0, and assume that lmxj^oo ip(x) exists (and is finite). Show that for every e > 0 there exists an integer n € N and real constants a,k,k = 0,1, ...,n, such that n
\ip(x) - V^a fe e~ fcx | < e for all
x>0.
fc=0
Exercise 20. Prove the inequalities of Holder, Cauchy-Schwarz and Minkowski for the spaces of absolute pth-power Riemann integrable functions with compact support 7?.Q(M). Exercise 21. Prove that if (/„) is a Cauchy sequence in (TZ%(R), || • || p ), then there exists a Cauchy sequence (gn) in (C0(1R), 11 • | | p ) such that 11 / „ — gn \ \p —* 0 for n —> oo. Exercise 22. For an arbitrary real number h € R, define the operator 77, on L2(M) by Thf{x) = f{x - h)
for / € L 2 (E) .
Show that Th is bounded. Exercise 23. Let c denote the set of convergent complex sequences (xn). Show that c is a Banach space when equipped with the norm II (Zn) I loo = SUp|£ n | . Hint: Show that (c, || • ||oo) is a closed linear subspace in the Banach space of bounded complex sequences (l°°, j| -1|oo)Exercise 24. In the Banach space of bounded complex sequences (Z°°, || - ]|oo), we consider the subset CQ consisting of the sequences converging to 0 and the
106
Functional Analysis: Entering Hilbert Space
subset coo consisting of the sequences with only a finite number of elements different from 0. 1) Show that Co and coo are linear subspaces of l°°. 2) Investigate if Co and/or coo are Banach spaces. Exercise 25. Consider the Volterra integral equation: x(t) — fi / k(t, s)x(s)ds = v(t),
t G [a, b] ,
Ja
where v G C([a,b]), k <= C([a,b}2) and fx G C. Show that the equation has a unique solution x G C([a, b}) for any \x G C. Hint: Write the Volterra integral equation in the form x = Tx where
Tx = v(t) + fi / k(t, s)x(s)ds . Ja Take XQ G C([a,b\) and define by iteration xn+\ that
\Tmx(t)-Tmy(t)\<
= Txn. Show by induction
M^^^-dooOc.y) ,
where c = max|fc| and d00(x,y) = sup t € [ a 6 ] \x(t) — y(t) \. Then show (by looking at d00(Tmx,Tmy)) that Tm is a contraction for some m and argue that T must then have a unique fixed point in the metric space (C([a, b]), doo). Exercise 26. Let / G L J (R). 1) Can we conclude that f(x) —> 0 for | x \ —> oo ? 2) Can we find a,b e l such that | /(a;) | < 6 for | x \ > a ? Exercise 27. In the vector space C([a, b}) of real-valued, continuous functions in the interval [a,b], consider the functions eo(t),ei(t),...,en(t), where ej(t) is a polynomial of degree j , for each j = 0,1,..., n. Show that eo, ei,..., e„ are linearly independent in C([a, &]). Exercise 28. Let U\ and £/2 be subspaces of the vector space V. Show that U\C\Ui is a subspace in V. Is U\UU2 always a subspace? If not, state conditions such that U\ U U2 is a subspace in V. Exercise 29. Let V denote the set of all real n x n-matrices. Show that V is a vector space, when equipped with the obvious entrywise defined operations.
Exercises
107
Is the set of all regular n x n-matrices a subspace of VI Is the set of all symmetric n x n-matrices a subspace of VI Exercise 30. In the space C([a, b]) of real-valued, continuous functions in the interval [a, b], consider the sets Ui= the set of polynomials defined on [a, 6]. • U2— the set of polynomials denned on [a, b] of degree < n. U%= the set of polynomials defined on [a, b] of degree = n. UA= the set of all / G C{[a, b]) with f(a) = f(b) = 0. U5=C\[a,b\). Which ones of the sets Ui, % = 1,2,..., 5 are subspaces of C([a, 6])? Exercise 31. In C([— 1,1]) we consider the sets U\ and Ui consisting of the odd and the even functions in C([—1,1]), respectively. Show that U\ and Ui are subspaces and that U\ n Ui = {0}. Show that every / G C([—1,1]) can be written in the form / = f\ + fi, where /1 G C/i and fi G Ui, and that this decomposition is unique. Exercise 32. Consider the space C 1 ([a, 6]) of real-valued, differentiate functions of class C 1 in the interval [a, b] with the uniform norm ||/||oc=
SUP | / ( i ) | . te[a,6]
1) Prove that ll/IU=
SUp \f(t)\
.
te(a,b)
2) Show that ((^([a, &]), || • ||oo) is not a Banach space. 3) Show that 11/11;,= sup | / ( t ) | + sup | / ' ( t ) | te[a,b]
is also a norm in C1^^}) norm.
t€[a,b]
and that C^fo, b]) is a Banach space with this
Exercise 33. Let / G C([a, b}) and consider the p-norms
\\f\\p = ([ Ja
\f(t)\pdt)K
P>1,
Functional Analysis: Entering Hilbert Space
108
and the uniform norm ||/||oo=
SUP t€{a,b]
\f(t)\.
Show that | | / | | p -> ll/Hoo forp—> oo. Exercise 34. Let V be a normed vector space and let x\,..., Xk be A; linearly independent vectors from V. Show that there exists a positive constant m such that for all scalars ai € C, i = 1 , . . . , k, we have Hccixi H
\-akxk\\
> m(\ai\-\
\- \ak\).
Exercise 35. Let T be a linear operator from a finite dimensional vector space into itself. Prove that T is injective if and only if T is surjective. Exercise 36. Let T : C°°(M.) - • C°°(K) be the linear mapping defined by
Tf = /'. 1) Show that T is surjective. 2) Is T injective? Exercise 37. (Riesz' Lemma) Let V be a normed vector space and let U be a closed subspace of V with U / V. Let a be a real number in the interval 0 < a < 1. Show that there exists a vector v EV such that 11 u 11 = 1 and
11 v — u 11 > a
for all u e C/ .
Exercise 38. In Z°°, the vector space of bounded sequences, we consider the sets U\ and C/2, where U\ denotes the set of sequences with only a finite number of elements different from 0, and U2 the set of sequences with all but the N first elements different from 0. Are Ui and/or U2 closed subspaces in l°°l Are U\ and/or U2 finite dimensional? Exercise 39. Let T : V —> W be a linear operator between normed vector spaces V and W. Show that the image T(V) is a subspace of W. Show that the kernel (or nullspace) ker(T) is a subspace of V. If T is bounded, is it true that T(V) and/or ker(T) are closed? Exercise 40. In the Banach space lp, for 1 < p < 00, we are given a sequence
Exercises
109
(xn) converging to an element x, where xn = (xni,xn2,...)
and
x = (xx,x2,
• • •) .
Show that if xn —> x in Zp, then xnk —• #& for all fc € N. If :r„fc —• Xfc for all k & N, is it true that x„ —> x in / p ? Exercise 41. Prove that the vector space of finitely non-zero sequences is isomorphic to the vector space of all polynomial functions defined in a closed and bounded interval [a, b] with a < b. Exercise 42. Let T be a linear mapping from R m to E™, both equipped with the 2-norm. Let (ay-) denote a real n x m matrix corresponding to T. Show that T is a bounded linear operator with | | T | | 2 < YnLi 5Z™=i a!jExercise 43. Let [a, b] be an arbitrary closed and bounded interval. Show that L*([a,b})cLH[a,b}).
Chapter 3 Exercise 44. Let T : V —> W be an injective linear mapping of the vector space V into the vector space W. Let (•, • )\y be an inner product in W. 1) Show that we define an inner product (•, -)y in V by the definition {x,y)v
= {Tx,Ty)w
forx,yeV.
2) Show that T is a bounded linear operator, when V is equipped with the inner product (•, -)y and W with the inner product (•, -)w Determine the operator norm 11T11 of T. Exercise 45. Let H be a Hilbert space of infinite dimension. Show that H is separable by a basis if and only if it contains a countable dense subset. Does part of this result hold in greater generality than for Hilbert spaces? Exercise 46. This exercise examines separability of the sequence spaces. 1) Prove that the sequence spaces lp, p > 1, are separable. 2) Prove that the space Z°° of bounded complex sequences is not separable. Hint: First prove that the subset B of sequences (xn) with all elements xn either 0 or 1 is uncountable. Next prove (indirectly) that l°° does not contain a countable dense subset by exploiting that any two mutually different sequences
Functional Analysis: Entering Hilbert Space
110
in B have norm distance 1 and hence cannot both be contained in an open ball of radius 1/3 in l°°. Exercise 47. Let J = [a, b] be a bounded interval and consider the linear mapping T from C([a, b]) into itself, given by
Tf{t) = f f(s)ds. Ja
We assume that C([a, b]) is equipped with the uniform norm. Show that T is bounded and find 11T | |. Show that T is injective and find T " 1 : T(C([a,b])) -> C([a,b]). Is T~l bounded? Exercise 48. Let T be a bounded linear operator from a normed vector space V into a normed vector space W, and assume that T is surjective. Assume that there is a c > 0 such that ||Tx|| > c | | x | |
forallzeV.
Show that T _ 1 exists and that T _ 1 is bounded. Exercise 49. Prove that in a real vector space with inner product we have
(x>v) = ^(Wx + vW2 - \ \ x - y \ \ 2 ) , and in a complex vector space with inner product we have (z, V) = i {\\x + yf
- )\x - yf
+ i\\x + iyf
- i\\x -
iyf).
These are the so-called Polarization identities. They show that, in a Hilbert space, the inner product is determined by the norm. Exercise 50. Let V be a real normed vector space, and assume that the norm satisfies
lk + 2/|| 2 +IN-y|| 2 = 2(||x||2+||yi|2) for all x, y £ V. Show that (x,y) = \(\\x +
y\\2-\\x-y\\2)
Exercises
111
defines an inner product in V and that the norm is induced by this inner product. Exercise 51. Prove that the uniform norm on the space of continuous functions C{[a,b\) is not induced by an inner product. Exercise 52. Prove that in a real vector space with inner product we have that (x + y, x — y) = 0 if 11 x \ | = 11 y | |. In the case V = M2, this corresponds to a well-known geometric statement. Which statement? Exercise 53. Let Vi, i = 1 , . . . , k, be vector spaces equipped with inner products (•, • )i, respectively. We define the product space 0 i = 1 Vi, in analogy with Exercise 11. Show that we can define an inner product in ®™=1 Vi by k
((xi,x2,...,xk),
(yi,V2, • • • ,yk)) =
Yl{xi,yi)i,
i=i
and that (^)™=1 Vi with this inner product is a Hilbert space if all the inner product spaces Vj, i = 1 , . . . , k, are Hilbert spaces. Exercise 54. Let x and y be vectors in a vector space with an inner product. Show that (x, y) = 0 if and only if 11 x + ay 11 = 11 a; - ay 11 for all scalars a. Moreover, show that (x, y) = 0 if and only if \\x +
ay\\>\\x\\
for all scalars a. Exercise 55. Let x and y be vectors in a complex vector space with an inner product, and assume that
||z + 2/||2 = l k l l 2 + IMI 2 Does this imply that (x, y) = 0? Exercise 56. Let V be a complex vector space with an inner product and assume that T € B{V).
112
Functional Analysis: Entering Hilbert Space
Show that (Tx, y) = 0 for all x, y £ V if and only if T is the zero operator. Show next that (Tx,x) = 0 for all x G V if and only if T is the zero operator. Are these results true if the vector space is a real vector space? Exercise 57. Let T be a linear operator T : L 2 (K) —> L2(R) satisfying that / > 0 implies that Tf > 0. Show that
lim/l)ll>l|r/ll for all / G L2(R). Show that T is bounded. Exercise 58. Consider the Hilbert space H = L2([—K,TT}). Using classical Fourier theory we can define an orthonormal basis (e„) in H by en{t) = -^=eint,
neZ.
[This follows, since it is known that (e n ) is an orthonormal basis in the space of smooth functions in [—n, n], which is dense in H.) For / G if we define the Fourier transform f by f(x} =
2iJ s/>, / ( ' ) e
*•
1) Show that / exists for all i £ l . 2) Use the function / to express the Fourier expansion of / G H in terms of the orthogonal basis {^/2nen). 3) Let 7 € R and / G H be given and define the function g by
9(t) = f(t)e
-i"ft
Find <7(x). 4) Show that for any 7 G M and / G H we have ^
2
1
E l/(" + 7)| = ^ll/llt n— — 00
5) Take / = 1 and 7 = £, 9 0 {pvr | p G Z} and show that 1
2
sin (0)
°°
f^
1
(n7r + ^) 2 '
Exercises
113
Exercise 59. Let H = L2([0,1]), and consider the operator
Tf(x) = V3xf(x3) . 1) Show that T G B(H) and find | | T | | . 2) Show that T " 1 exists and that T~l g&H, and find \\T~l\\.
e B(H). Determine T~lg(y)
for
Exercise 60. Let M be a subset of a Hilbert space H. Show that M x is a closed subspace of H. Show that M c {M^)L, and show that [ML)L is the smallest closed subspace containing M. Exercise 6 1 . Let (xn) be an orthogonal sequence in a Hilbert space H, satisfying that oo
^||a;„||2
Show that the series X ^ l i xn is convergent in H. Is this still true if we drop the orthogonality assumption? Exercise 62. Let H be an infinite dimensional, separable Hilbert space. Show that there is a sequence of vectors (xn) such that 11 xn \ \ = 1 for all n, and such that (xn, x) —> 0 for all x G H. Exercise 63. Let H be a Hilbert space. Show that | | * - z | | = ||a;-y|| + ||i/-z|| if and only if y = ax + (1 — a)z for some a € [0,1]. Exercise 64. Let (e„) be an orthonormal basis for L2([0,1]). Construct an orthonormal basis for L2([a, 6]), where [a, b] is an arbitrary closed and bounded interval. Exercise 65. Let (e„) be an orthonormal sequence in the space of square integrable functions L2([a,b]) in a closed and bounded interval [a, b], with the property that for any continuous function / € L2([a, b}) and any e > 0 we can
114
Functional Analysis: Entering Hilbert Space
find an integer N G N and constants a\, a2,...,
ajv such that
N
||/-^afcefc||2 < e . fc=i
Show that (e n ) is an orthonormal basis for L2([a,b}). Exercise 66. Construct the sequence of Haar-functions (hn) in L 2 ([0,1]) by defining hi(t) = 1, for t £ [0,1], and hn for n > 2 by \/2™
for
k—1 2k — 1 — — < f < -—-r2A; — 1 k
/i2-+fc(i) = -V2™
for
——r
h2m+k{t) =
< t < —-
h2™+k(t) = 0 else, for /c = 1, 2 , . . . , 2 m and m = 0,1, 2 , . . . . 1) Sketch the graphs of h\, h2, • • •, h&. 2) Show that (hn) is an orthonormal sequence in L2([0,1]). 3) Show that (hn) is an orthonormal basis in L 2 ([0,1]). Exercise 67. Let H be a Hilbert space and let P and Q denote the orthogonal projections onto the closed subspaces M and N, respectively. Show that if M -L N, then P + Q is the orthogonal projection onto M ®JV. Exercise 68. Let P and Q denote orthogonal projections in a Hilbert space, and assume that PQ = QP. Show that P + Q-PQ is an orthogonal projection and find the image of P + Q — PQ. Exercise 69. Let
Exercises
115
5, T G B[V) satisfies the canonical commutator relation: [S, T}=ST-TS
=I .
Hint: Show by induction that STn - TnS = nTn~l, n G N, and use this to estimate ||5|| and ||T||. Exercise 73. Let (e n ) denote an orthonormal basis in a Hilbert space H, and define the operator T by oo
oo
k=l
k=l
Show that T G B{H) and find ||T||. Show that T is injective and find T"1. Exercise 74. In the Hilbert space of square summable sequences I2, define the operator T{{xn)) = {yn) , by setting yi—x\
and
yn = — ^ (x\ + x2 H
\-xn)
for n > 2 .
Show that the operator T is bounded and not surjective. Exercise 75. Let (efc) be an orthonormal basis in a Hilbert space H, and let T G B(H). Define for j , k G N the numbers tjk = (Tej,ek)
•
Show that oo
fc=l 2
and that J^fcLi l*jfc| < oo for j G N. The matrix (tjk) is called the matrix form, for T with respect to the orthonormal basis (efc). Let A, B G -B(if) have the forms ( a ^ ) and (bjk), respectively. Find the forms for A + B and AB.
Functional Analysis: Entering Hilbert Space
116
Chapter 4 Exercise 76. Let V denote a normed vector space with norm || • ||. The vector space of bounded linear functionals ip : V —> C on V is called the dual space of V and is denoted by V*. Introduce the dual norm 11 • 11 * in V* by taking 11 tp 11 * to be the operator norm of the bounded linear functional
¥> € V* ,
defines an element gx G V** in the double dual V**. Show that the mapping x —> gx is a linear and injective mapping from V toV**, and that 115*11** = ||a;||. If x —> gx is also surjective, 1/ is said to be reflexive. Show that a Hilbert space is reflexive. Exercise 77. We consider the space of sequences lp, where p > 1. Let y e lq where l/p+ 1/q = 1. (If p = 1 then y £ l°°, the space of bounded sequences.) Show that oo
x->^2 xiyi i=i
defines an element y* G (F)* with dual norm \\y* ||* = ||y|| g . Exercise 78. Let cp denote a bounded linear functional on a Hilbert space H, and assume that the domain D{
= h(xi,x)
+ h(x2,x)
,
h(x, x\ + x2) = h(x, x\) + h(x, x2) , h(axi,x2)
= ah(xi,x2)
,
h(xi,ax2)
= ah(xi,x2)
•
We say that h is bounded if there is a constant c > 0 such that \h(xi,x2)\
< c||a;i ||||a:2 ||
for all X\,x2 G H. The norm \\h\\ is defined as the smallest possible c.
Exercises
117
Show that there is a S G B(H) such that h(xi,x2)
= (Sxi,x2)
,
and that this representation is unique. Show also that \\h\\ = \\S\\. A sesquilinear form is called Hermitian if h{x,y) = h(y,x) for all x,y £ H. If, moreover, h(x,x) > 0, the form is called positive semidefinite. Show that in this case we have Schwarz' inequality: \h(x,y)\2
<
h{x,x)h(y,y)
for all x,y € H. Exercise 80. In the Hilbert space I2, we define an operator T : D(T) —• I2 by T
{(xn))
= (anxn) ,
where (a n ) is a complex sequence. Find the maximal possible D(T) and show that T is linear. Show that D(T) is dense in I2. Show that if (o„) is bounded, then D{T) = I2 and T is bounded. Exercise 81. Consider in L2(M.) the operator Q defined by Q/(x) = xf{x)
,
with D(Q) = {fe
L2(R) | Qf e L2(R)} .
Show that Q is linear but not bounded. Show that D(Q) is dense in L2(IR). In quantum mechanics, Q is called the position operator. Exercise 82. Consider in L2(R) the operator P defined by
"'--4ox with D(P) = {fe L2(R) | Pf € L2(R)} . Show that P is linear but not bounded. Show that D(P) is dense in
L2(R).
Functional Analysis: Entering Hilbert Space
118
In quantum mechanics, P is called the momentum operatorExercise 83. Let V be a normed space and assume that T e B(V) is bijective. Show that if T _ 1 is bounded, then
IIT - 1 !!
> HTir 1 .
Exercise 84. Let (e^) be an orthonormal basis in a Hilbert space iJ, and let T : D(T) —> K be a linear operator from H into the Hilbert space K. Show that if ek e D{T) for all fcgN, then D(T) is dense in H. Exercise 85. Let T : X —» Y be a bounded linear operator between two normed spaces, and let A C X be compact. Show that T(A) is closed. Exercise 86. Let T be a self-adjoint operator in a Hilbert space H. Show that if D{T) = H, then T is bounded. Exercise 87. Let T be a bounded linear operator on a Hilbert space H, and assume that M and N are closed subspaces of H. Show that T{M) C N if and only if T*{N^) c M- 1 . Show, moreover, that ker(T)=T*(H)- L and
ker(r) x =T*(H). Exercise 88. Let T be a bounded linear operator on a Hilbert space H with 11T11 = 1, and assume that we can find XQ £ H such that TXQ — XQ. Show that also T*XQ = xo. Exercise 89. Let (e n ) denote an orthonormal basis in a Hilbert space H, and
Exercises
119
consider the operator oo
oo
T(^akek)
= ^akek+1
.
fc=i fe=i
Find the adjoint T* and show that T* is an extension of T~x. Exercise 90. Let (e„) denote an orthonormal basis in a Hilbert space H, and consider the operator oo
oo
T{^akek)
= ^2\/k-
lafcefe_i .
Show that T is a densely defined, unbounded operator, and find T*. Exercise 91. Let T G B(H). Show that we can write T as
T = A + iB where A and B are uniquely determined, bounded, self-adjoint operators. Exercise 92. Show that T e B(H) is self-adjoint if and only if one of the following conditions is satisfied: (Tx,x) = (x,Tx)
for all
x£H
or (Tx, x) e R
for all
x G H.
Exercise 93. Let S and T be bounded, self-adjoint operators on a Hilbert space. Show that ST + TS and i(ST - TS) are self-adjoint. Exercise 94. Let T be a bounded self-adjoint operator. Define the numbers m = inf{(Ta;,:r) | \\x\\ = 1 } and M = sup{(T:z,;r) | ||a;|| = 1}. Show that cr(T) C [m, M], and show that both m and M belong to a(T). Show that ||T|| = max{|m|, \M\}.
120
Functional Analysis: Entering Hilbert Space
Exercise 95. Consider in L2(R) the operator Q defined by
Qf(x) = xf(x) , with D(Q) = {f£
L2(R) | Qf e L2(R)} .
Show that Q is self-adjoint. Exercise 96. Show that the set of self-adjoint operators is closed in B(H). Exercise 97. Let T £ B(H). An operator is isometric if \\Tx\\ = \\x\\ for all x G H• Show that the following conditions are equivalent for T G B{H): a) T is isometric. b) T*T = I. c) (Tx,Ty) = (x,y) for all x,y £ H. Exercise 98. Let T € B(H) be an isometric operator. Show that T(H) is a closed subspace. Show that T(H) = H if H is finite dimensional. Give an example of an isometric operator with T(H) ^ H. Exercise 99. Let T G B{H) be an isometric operator, and let M and TV denote closed subspaces of the Hilbert space H. Show that T(M) = TV => TiM2-) C N1-. Show that T is isometric if and only if, for any orthonormal basis (e^), the sequence (Te*,) is an orthonormal sequence. Exercise 100. Let M be a closed linear subspace in the Hilbert space H. Suppose that T : H —> H is a mapping satisfying the following conditions: (i) {Tx, y) = {x,Ty) for all x,y G H; (ii) Tx = x when a; G M; and (hi) Ty = 0 when y G M - 1 . Prove that T is the projection operator with fixed space M. Exercise 101. Let T £ B(H) be an isometric operator. Show that TT* is a projection operator and determine the range. Exercise 102. Consider the Hilbert space L 2 ([0, oo)). Let h > 0 and define
Exercises
121
the operator T by Tf(x)
= 0
for
Tf(x)
= f{x -h)
0< x
h<x .
Show that T is isometric and determine T*. Find TT* and T*T. Exercise 103. An operator T £ B(H) is called unitary if it is isometric and surjective. Show that the following conditions are equivalent for an operator T e B(H): a) T is unitary. b) T is bijective and T " 1 = T*. c) T*T = TT* = I. d) T and T* are isometric. e) T is isometric and T* is injective. f) T* is unitary. Exercise 104. Let (e^) denote an orthonormal basis in a Hilbert space H, and let T e £ ( F ) be given by OO
OO
fc=i fe=i Show that T is unitary if and only if |Afc| = 1 for all k. Exercise 105. An operator T e B(H) is normal if TT Show that T is normal if and only if ||T*x|| = ||Tx|| for all
x&H.
Exercise 106. An operator T 6 B(H) is called positive if (Tx, x) > 0 for all x € H, and we write T > 0. Prove the following: a) T > 0 implies that T is self-adjoint. b) If S, T > 0, a > 0, then S + aT > 0. c) If T > 0 and S e £ ( # ) , then 5*T5 > 0. d) If T G £ ( # ) , then T*T > 0. e) If T is an orthogonal projection, then T > 0. Exercise 107. Let PM and Pjv denote the orthogonal projections onto the closed subspaces M and TV of a Hilbert space H. Show that M c N implies
Functional Analysis: Entering Hilbert Space
122
that the operator PJV — PM is positive, and we write PM < PNExercise 108. An operator T € B(H) is called a contraction if \\Tx\\ < \\x\\ for all
x€H
.
Show that the following conditions are equivalent for an operator T 6 B(H): a) T is a contraction.
b) imi < i. c) T*T
Chapter 5 Exercise 112. Consider in L2(M) the operator Q defined by
Qf(x) = xf(x) , with D(Q) = {fe
L2(R) | Qf e L2(R)} .
1) Determine the set of eigenvalues for Q. 2) Determine the resolvent set p(Q) for Q. Exercise 113. Let (e n ) denote an orthonormal basis in a Hilbert space H,
Exercises
123
and consider the operator oo
oo
akek+\ • fe=i fc=i Determine ||T|| and a(T). Exercise 114. Let (e n ) denote an orthonormal basis in a Hilbert space H, and consider the operator oo
oo
T{^2akek)
= y~]Vkakek~i
.
fc=l fc=2
Determine the spectrum o~(T), and find for each eigenvalue the corresponding eigenvectors. Exercise 115. Let (e„) denote an orthonormal basis in a Hilbert space H. We define the sequence {fk)kez by /o = ei , fk = e2k+i fk = e_2fc In this way (/fc)kez shift operator S by
IS a n
for
k > 0,
for
k <0.
orthonormal basis. We define the double-sided
oo
oo
k=—oo
fc=—oo
Show that 5 is a bounded operator, and show that S has no eigenvalues. Exercise 116. For a real number h / 0, define the operator TH on L 2 (R) by r h /(a;) = f(x - h)
for i £ l .
1) Show that Th has no eigenvalues. 2) Show that all complex numbers A with | A | ^ 1 belong to the resolvent set p(Q). (It is, in fact, true that a(rh) = {A e C | |A| = 1}.) Exercise 117. Let T € B(H) where H is a Hilbert (or just Banach) space. Show that \\R\{T)\\ -> 0 for |A| - • oo.
124
Functional Analysis: Entering Hilbert Space
Exercise 118. Let (en) denote an orthonormal basis in a Hilbert space H, and let (rk) be all the rational numbers in ]0,1[ arranged as a sequence. Consider the operator oo
oo
T(^2akek)
= ^rfcofeefc .
k=l
k=l
Show that T is self-adjoint and that | | T | | = 1. Find p(T) and determine the point spectrum and the continuous spectrum for T. Exercise 119. Let T € B(H) be unitary. Show that
| * | = 1} •
Exercise 120. Let T € B(H) be normal. Show that
\\(T-\i)x\\ = \\cr-\i)x\\ for all x € H. Show that oy(T) is empty. Exercise 121. Let (e^) denote an orthonormal basis in a Hilbert space H, and define the operator T by OO
OO
T(^akek)
-.
= Y^ Take-k-\ •
k=l
fc=2
Show that T is compact and find T*. Find ap(T) and ap{T*). Exercise 122. Let (e^) denote an orthonormal basis in a Hilbert space H, and assume that the operator T has the matrix representation (tjk) with respect to the basis (ek). Show that oo
oo
3 = 1 fe=l
implies that T is compact. Let (fk) denote another orthonormal basis in H, and let Sjk = (Tfj,fk) so that (sjk) is the matrix representation of T with respect to the basis (fk)Show that oo
oo
oo
oo
2
EEM = EEM2j= l
fc=l
j = l fc=l
Exercises
125
An operator satisfying J2^Li Z^feli \tjk\2 < oo is called a general HilbertSchmidt operator. Exercise 123. For a general Hilbert-Schmidt operator, we define the HilbertSchmidt norm || • \\HS by oo
oo
j=i fc=i
Show that this is a norm, and show that
imi
-1
-e-l*-*l/(t)dt .
/
Show that Kf £ £ 2 (R), and that K is linear and bounded with norm < 1. Show that the function |e~l x ~'l does not belong to L 2 (R 2 ), so that K is not a Hilbert-Schmidt operator. Exercise 125. Let K denote the Hilbert-Schmidt operator with kernel k(x, t) = sin(a;) cos(t)
for
0 < x, t < 2ir .
Show that the only eigenvalue for K is 0. Find an orthonormal basis for ker(X). Exercise 126. Let K denote the Hilbert-Schmidt operator with continuous kernel k on L2 (I) where 7 is a closed and bounded interval. Show that all the iterated kernels kn are continuous on I2, and show that
IIM2 < ||*H5 • Show that if|A|||fc||2 < 1, then the series 00 /
j •*
n=l
is convergent in
L2(I).
k
n
Functional Analysis: Entering Hilbert Space
126
Exercise 127. Let K and L denote the Hilbert-Schmidt operators with continuous kernels k and I on L2(I) where / is a closed and bounded interval. We define the trace of K, tr(K) by tr(K) = / k(x,x)dx
,
and similarly for L. Show that \tr{KL)\
<
\\K\\HS\\L\\HS
and \tr(Kn)\
< \\K\\nHS,
n>2.
Moreover, if (Kn), (Ln) denote sequences of Hilbert-Schmidt operators like above, where \\Kn ~ K\\HS - 0 and
\\Ln - L\\HS - • 0,
then tr{KnLn)
- • tr(KL)
.
Exercise 128. Let K denote the Hilbert-Schmidt operator on L 2 ([0,1]) with kernel k(x,t) = x + t . Find all eigenvalues and eigenfunctions for K. Solve the equation Ku = fiu + f, f € L2([0,1]) when /j, is not in the spectrum for K. Exercise 129. Let K denote the Hilbert-Schmidt operator on L2([— | , | ] ) with kernel k(x,t) = cos(a; — t) • Find all eigenvalues and eigenfunctions for K. Solve the equation Ku = fiu + / , / G I / 2 ( [ - | , | ] ) when fi is not in the spectrum for K.
Exercises
127
Exercise 130. Let K denote the Hilbert-Schmidt operator on L 2 ([-7r, TT]) with kernel k(x,t) = (cos(x) + cos(i)) 2 . Find all eigenvalues and eigenfunctions for K, and find an orthonormal basis for ker(K). Exercise 131. Let K denote a self-adjoint Hilbert-Schmidt operator on L2(I) with kernel k. Show that 11if || = ||A;||2 if and only if the spectrum for K consists of only two points.
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Bibliography
Main
sources:
V.L. Hansen: Fundamental Concepts in Modern Analysis, World Scientific, 1999. M. Pedersen: Functional Analysis in Applied Mathematics and Engineering, Chapman & Hall/CRC, 2000. Other
sources:
S.K. Berberian: Introduction to Hilbert Space, Oxford University Press, 1961. Y. Choquet-Bruhat, C. de Witt-Morette, M. Dillard-Bleick: Analysis, Manifolds and Physics, North-Holland Publishing Company, 1977. S. Hartman, J. Mikusinski: The Theory of Lebesgue Measure and Integration, Pergamon Press Inc., 1961. L. Hormander: The Analysis of Linear Partial Differential Operators III, Springer-Verlag, 1985. W. Rudin: Functional Analysis, McGraw-Hill Inc., 1973. M.E. Taylor: Partial Differential Equations II, Springer-Verlag, 1996. E. Zeidler: Applied Functional Analysis: Applications to Mathematical Physics, Springer-Verlag, 1995. E. Zeidler: Applied Functional Analysis: Main Principles and Their Applications, Springer-Verlag, 1995.
129
130
Suggestions
Functional Analysis: Entering Hilbert Space
for further
reading:
J.B. Conway: A Course in Functional Analysis, Springer-Verlag, 1990 (21 ed.). P.D. Lax: Functional Analysis, John Wiley & Sons Inc., 2002. G.K. Pedersen: Analysis Now, Springer-Verlag, 1989. W. Rudin: Real and Complex Analysis, McGraw-Hill Book Co., 1966.
List of Symbols
Symbol
Explanation
T*
adjoint operator
75
B(V)
the space of bounded linear operators on V
75
[S,T]
commutator of operators
C([a,b])
the space of continuous functions in [a, b]
23
Co(R)
the space of continuous functions with compact support
32
CQ° (R)
the space of smooth functions with compact support
42
D{T)
domain of definition for linear operator
89
V*
dual space
116
||-||*
dual norm
116
H\
eigenspace for eigenvalue A
91
l(xn)]
equivalence class of Cauchy sequence
27
~
equivalence relation
27
I
identity operator
81
inner product
49
inverse operator
80
ker(^>)
kernel of linear functional
71
H(A)
Lebesgue measure of set A
39
(•, •) T
_1
p
L
p
l
1
M-
th
space of p
Page
power Lebesgue integrable functions
115
36
the space of absolute p-summable sequences
45
the orthogonal complement to subset M
68
131
132
Functional Analysis: Entering Hilbert Space
Symbol
Explanation
V®W
product vector space
WWv
p-norm
32
P
projection operator
82
P(T)
resolvent set for linear operator T
90
fto(R)
the space of Riemann integrable functions
43
a(T)
spectrum for linear operator T
90
°c(T)
continuous spectrum for linear operator T
91
aP(T)
point spectrum for linear operator T
91
Gr{T)
residual spectrum for linear operator T
91
sum of series
55
support (/)
support of function
32
ll-lloo
uniform norm
23
weak convergence of sequence (xn) to x
71
XJI
X
Page 103
Index
adjoint operator, 75
Closed Graph Theorem, 104 closed operator, 104 set, 10 closure of set, 11 commutator relation, 115 compactness, 13, 82 compact operator, 83 compact support, 32 comparison test, 56 completely continuous operator, 83 complete metric space, 12 Completion Theorem, 26 contact point, 11 continuity of mapping, 9, 10 continuous spectrum, 91 contraction factor, 13 in metric space, 13 operator, 122 weak, 101 convergence of sequence, 11 coordinates of vector, 16
Banach, Stefan, vii Banach space, 18, 23 Banach Fixed Point Theorem, 14 Banach's Theorem, 104 basis in vector space finite dimension, 56 infinite dimension, 56 Bernstein polynomial, 30 Bessel, Friedrich Wilhelm, 60 Bessel's equation, 60 inequality, 60 Best Approximation Theorem, 61 bounded operator, 19, 75 set, 13 bump function, 42 Cantor, Georg, 41 Cantor set, 41 Cauchy, Augustin-Louis, 11 Cauchy condition for sequences, 12 for series, 56 Cauchy-Schwarz inequality inRn, 6 for continuous functions, 34 for £ 2 -functions, 38 for I2 sequences, 46 in inner product spaces, 52 Cauchy sequence, 12, 23
dense subspace, 11, 26 discrete metric, 9 distance, 5 divergent series, 56 domain of definition, 89 dual space, 116 dual norm, 116 eigenvalue, 91
133
134
Functional Analysis: Entering Hilbert Space
eigenvector, 91 eigenspace, 91 equivalence class, 3, 27 equivalence relation, 3, 27 essentially bounded, 37 Euclidean space, 7 finite rank operator, 83 fixed point, 14 space, 81 Fourier expansion 59 Fourier transform, 112 Frechet, Maurice, 6 Fredholm, Erik Ivar, vii Fredholm integral equation, 90 function algebra, 37 fundamental sequence, 12 Gram-Schmidt procedure, 59 greatest lower bound, 4, 8 Haar-functions, 114 Hausdorff space, 103 Hermitian symmetry, 49 Hilbert, David, vii Hilbert space, 54 Holder, Otto, 34 Holder's inequality for continuous functions, 34 in L p -spaces, 38 in Zp-spaces, 46 idempotent operator, 81 identity operator, 81 image, 2 indicator function, 39 induced metric, 9 norm, 52 infimum, 4, 8 inner product, 6, 49 integral (Riemann), 43 integral equation, 90, 106 interior of set, 11 point, 11 inverse operator, 80
isometric mapping, 26 operator, 120 iteration of mapping, 13 kernel of linear functional, 71 of linear operator, 78 least upper bound, 4, 7 Lebesgue, Henri, 36 Lebesgue measurable set, 39 Lebesgue measure, 39 Levy, Paul, vii L p -spaces, 36 P-spaces, 45 limit point, 11 maximal element, 3 metric, 6 Euclidean, 7 metric space, 6 minimal element, 3 Minkowski, Hermann, 34 Minkowski inequality for continuous functions, 34 for L p -functions, 38 for lv sequences, 47 momentum operator, 118 Neumann, Carl Gottfried, 92 Neumann, John von, vii Neumann's Lemma, 92 norm, 16 Euclidean, 6 operator, 79 supremum, 23 uniform, 23 normal operator, 121 normed vector space, 16 open ball, 9 set, 9 Open Mapping Theorem, 103 operator, 18 closed, 104 compact, 83 completely continuous, 83
135
Index finite rank, 83 Hilbert-Schmidt, 125 trace, 126 idempotent, 81 isometric, 120 normal, 121 positive, 121 projection, 82 self-adjoint, 79 unitary, 121 operator norm bounded operator, 20 self-adjoint operator, 79 ordered set, 3 ordering relation, 3 orthogonal complement, 68 projection, 66, 68 set, 58 subsets, 68 sum, 69 vectors, 54 orthonormal basis, 58 sequence, 58 set, 58 Parallelogram Law, 66 Parceval's equation, 64 partial sum, 55 point spectrum, 91 polarization identities, 110 position operator, 117 positive definite, 5, 16, 49 positive operator, 121 preimage, 2 product vector space, 103 projection operator, 82 Projection Theorem, 70 Pythagoras' theorem, 54 reflexive space, 116 relation, 2 residual spectrum, 91 resolvent, 90 Riemann integral, 43 Riesz, Frigyes, 70
Riesz' Lemma, 108 Riesz-Fischer Theorem, 65 Riesz' Representation Theorem, 71 Schauder basis, 56 self-adjoint operator, 79 separability, 56 by a basis, 57 topological space, 57 sequentially compact, 13, 82 series, 55 sesquilinear mapping, 116 shift operator, 93 double-sided, 123 space Banach, 23 complete metric, 12, 26 dual, 116 Euclidean, 7 Hilbert, 54 inner product, 49 Lp, 36 P, 45 reflexive, 116 Spectral Theorem, 97 spectrum, 90 strong convergence, 72 sum of series, 55 support of function, 32 supremum, 4, 7 supremum norm, 23 symmetry, 5 term in series, 55 total set, 64 topology, 10 triangle inequality, 5, 16, 33 uniform norm, 23 uniform scaling, 16 unit ball in Hilbert space, 82 in normed vector space, 19 vector space finite dimension, 16, 56 infinite dimension, 56 Volterra, Vito, vii
136
Functional Analysis: Entering Hilbert Space
Volterra integral equation, 106 weak convergence, 71 weak limit, 71 Weierstrass Approximation Theorem, 28 Young's inequality, 33
his book presents basic
elements
of
the theory of Hilbert spaces and operators on Hilbert spaces, culminating in a proof of the spectral theorem
for
compact,
self-adjoint
operators on separable Hilbert spaces. II exhibits a construction of the space of p"' power Lebesgue integrable functions by a completion procedure with respect to a suitable norm in a spate of continuous
functions,
including proofs of the basic
inequalities
of
Holder and Minkowski.
The r-spaces thereby emerges
in
N
C
IO
A k I A I A
IN A
N
A
L
\ / O l O
L Y U I O
Entering Hilbert Space
direct
analogy with a construction ol the real numbers Irom the rational numbers. This allows grasping the main ideas more rapidly. Other important Banach spaces arising from function spaces and sequence spaces are also treated.
ISBN 981-256-563-9
YEARS 01 PCM ISHINC, I
9 "789812 565631"
www.worldscientitic.com
