198

14 Hilbert Spaces Basics

14 Hilbert Spaces Basics

Fig. 14.1. The picture behind the proof of the Schwarz inequality.

(BRUCE: Perhaps this should be move to between Chapters 7 & 8?) Definition 14.1. Let K be a complex vector space. An inner product on K is a function, h·|·i : K × K $ C> such that 1. hd{ + e||}i = dh{|}i + eh||}i i.e. { $ h{|}i is linear. 2. h{||i = h||{i. 3. k{k2 := h{|{i  0 with equality k{k2 = 0 i { = 0= Notice that combining properties (1) and (2) that { $ h}|{i is anti-linear for fixed } 5 K> i.e.

The following identity will be used frequently in the sequel without further mention,

= k{k2 + k|k2 + 2Reh{||i=

from which it follows that 0  k|k2 k{k2  |h{||i|2 with equality i } = 0 or equivalently i { = k|k2 h{||i|= p Corollary 14.3. Let (K> h·|·i) be an inner product space and k{k := h{|{i= Then the Hilbertian norm, k·k> is a norm on K= Moreover h·|·i is continuous on K × K> where K is viewed as the normed space (K> k·k)= Proof. If {> | 5 K> then, using the Schwarz’s inequality,

h}|d{ + e|i = d ¯h}|{i + ¯eh}||i=

k{ + |k2 = h{ + ||{ + |i = k{k2 + k|k2 + h{||i + h||{i

° °2 2 ° h{||i ° ° = k{k2 + |h{||i| k|k2  2Reh{| h{||i |i 0  k}k2 = ° {  | ° ° 2 4 k|k k|k k|k2 2 |h{||i| = k{k2  k|k2

(14.1)

Theorem 14.2 (Schwarz Inequality). Let (K> h·|·i) be an inner product space, then for all {> | 5 K |h{||i|  k{kk|k and equality holds i { and | are linearly dependent. Proof. If | = 0> the result holds trivially. So assume that | 6= 0 and 2 observe; if { = | for some  5 C> then h{||i =  k|k and hence 2

|h{||i| = || k|k = k{kk|k= Now suppose that { 5 K is arbitrary, let } := {  k|k2 h{||i|= (So } is the “orthogonal projection” of { onto |> see Figure 14.1.) Then

k{ + |k2 = k{k2 + k|k2 + 2Reh{||i

 k{k2 + k|k2 + 2k{kk|k = (k{k + k|k)2 =

Taking the square root of this inequality shows k·k satisfies the triangle inequality. Checking that k·k satisfies the remaining axioms of a norm is not routine and will be left to the reader. If {> {0 |> | 0 5 K> then |h{||i  h{0 || 0 i| = |h{  {0 ||i + h{0 ||  | 0 i|  k|kk{  {0 k + k{0 kk|  | 0 k  k|kk{  {0 k + (k{k + k{  {0 k) k|  | 0 k = k|kk{  {0 k + k{kk|  | 0 k + k{  {0 kk|  | 0 k from which it follows that h·|·i is continuous. Definition 14.4. Let (K> h·|·i) be an inner product space, we say {> | 5 K are orthogonal and write { B | i h{||i = 0= More generally if D  K is a set, { 5 K is orthogonal to D (write { B D) i h{||i = 0 for all | 5 D= Let DB = {{ 5 K : { B D} be the set of vectors orthogonal to D= A subset V  K is an orthogonal set if { B | for all distinct elements {> | 5 V= If V further satisfies, k{k = 1 for all { 5 V> then V is said to be orthonormal set.

14 Hilbert Spaces Basics

199

Proposition 14.5. Let (K> h·|·i) be an inner product space then (14.2)

for all {> | 5 K= 2. (Pythagorean Theorem) If V  K is a finite orthogonal set, then ° ° ° X °2 X ° ° {° = k{k2 = (14.3) ° ° ° {5V

B

3. If D  K is a set, then D

14 Hilbert Spaces Basics

More example of Hilbert spaces will be given later after we develop the Lebesgue integral, see Example 23.1 below.

1. (Parallelogram Law) k{ + |k2 + k{  |k2 = 2k{k2 + 2k|k2

200

{5V

is a closed linear subspace of K=

Remark 14.6. See Proposition 14.54 for the “converse” of the parallelogram law. Proof. I will assume that K is a complex Hilbert space, the real case being easier. Items 1. and 2. are proved by the following elementary computations;

Definition 14.9. A subset F of a vector space [ is said to be convex if for all {> | 5 F the line segment [{> |] := {w{ + (1  w)| : 0  w  1} joining { to | is contained in F as well. (Notice that any vector subspace of [ is convex.) Theorem 14.10. Suppose that K is a Hilbert space and P  K is a closed convex subset of K= Then for any { 5 K there exists a unique | 5 P such that k{  |k = g({> P ) = inf k{  }k= }5P

Moreover, if P is a vector subspace of K> then the point | may also be characterized as the unique point in P such that ({  |) B P= Proof. Uniqueness. By replacing P by P  { := {p  { : p 5 P } we may assume { = 0= Let  := g(0> P ) = inf p5P kpk and |> } 5 P> see Figure 14.2.

k{ + |k2 + k{  |k2

= k{k2 + k|k2 + 2Reh{||i + k{k2 + k|k2  2Reh{||i

= 2k{k2 + 2k|k2 > and

° ° ° X °2 X X X ° ° { {| |i = h{||i ° =h ° ° ° {5V {5V |5V {>|5V X X h{|{i = k{k2 = = {5V

{5V

Fig. 14.2. The geometry of convex sets.

Item 3. is a consequence of the continuity of h·|·i and the fact that DB = _{5D Nul(h·|{i) where Nul(h·|{i) = {| 5 K : h||{i = 0} — a closed subspace of K= Definition 14.7. A Hilbert space is an inner product space (K> h·|·i) such that the induced Hilbertian norm is complete. Example 14.8. Suppose [ is a set and  : [ $ (0> 4) > then K := c2 () is a Hilbert space when equipped with the inner product, X hi |ji := i ({) j¯ ({)  ({) = {5[

In Exercise 14.6 you will show every Hilbert space K is “equivalent” to a Hilbert space of this form with   1=

By the parallelogram law and the convexity of P> 2k|k2 + 2k}k2 = k| + }k2 + k|  }k2 |+} 2 = 4k || + k|  }k2  4 2 + k|  }k2 = 2

(14.4)

Hence if k|k = k}k = > then 2 2 + 2 2  4 2 + k|  }k2 > so that k|  }k2 = 0= Therefore, if a minimizer for g(0> ·)|P exists, it is unique. Existence. Let |q 5 P be chosen such that k|q k = q $   g(0> P )= Taking | = |p and } = |q in Eq. (14.4) shows 2 2p + 2q2  4 2 + k|q  |p k2 =

Passing to the limit p> q $ 4 in this equation implies,

14 Hilbert Spaces Basics

201

2 2 + 2 2  4 2 + lim sup k|q  |p k2 > p>q$4

4

i.e. lim supp>q$4 k|q  |p k2 = 0= Therefore, by completeness of K> {|q }q=1 is convergent. Because P is closed, | := lim |q 5 P and because the norm q$4 is continuous, k|k = lim k|q k =  = g(0> P )= q$4

So | is the desired point in P which is closest to 0= Now suppose P is a closed subspace of K and { 5 K= Let | 5 P be the closest point in P to {= Then for z 5 P> the function j(w) := k{  (| + wz)k2 = k{  |k2  2wReh{  ||zi + w2 kzk2

202

14 Hilbert Spaces Basics

2 2. Obviously Ran(SP ) = P and SP { = { for all { 5 P . Therefore SP = SP . 3. Let {> | 5 K> then since ({  SP {) and (|  SP |) are in P B >

hSP {||i = hSP {|SP | + |  SP |i = hSP {|SP |i = hSP { + ({  SP {)|SP |i = h{|SP |i= 4. We have already seen, Ran(SP )  P and SP { = 0 i { = {  0 5 P B > i.e. Nul(SP ) = P B = Corollary 14.14. If P  K is a proper closed subspace of a Hilbert space K> then K = P  P B =

has a minimum at w = 0 and therefore 0 = j 0 (0) = 2Reh{  ||zi= Since z 5 P is arbitrary, this implies that ({  |) B P= Finally suppose | 5 P is any point such that ({  |) B P= Then for } 5 P> by Pythagorean’s theorem,

Proof. Given { 5 K> let | = SP { so that {  | 5 P B = Then { = | + ({  |) 5 P + P B = If { 5 P _ P B > then { B {, i.e. k{k2 = h{|{i = 0= So P _ P B = {0} =

k{  }k2 = k{  | + |  }k2 = k{  |k2 + k|  }k2  k{  |k2

Theorem 14.15 (Riesz Theorem). Let K  be the dual space of K (Notation 7.9). The map

which shows g({> P )2  k{  |k2 = That is to say | is the point in P closest to {= Definition 14.11. Suppose that D : K $ K is a bounded operator. The adjoint of D> denote D > is the unique operator D : K $ K such that hD{||i = h{|D |i= (The proof that D exists and is unique will be given in Proposition 14.16 below.) A bounded operator D : K $ K is self - adjoint or Hermitian if D = D = Definition 14.12. Let K be a Hilbert space and P  K be a closed subspace. The orthogonal projection of K onto P is the function SP : K $ K such that for { 5 K> SP ({) is the unique element in P such that ({  SP ({)) B P= Theorem 14.13 (Projection Theorem). Let K be a Hilbert space and P  K be a closed subspace. The orthogonal projection SP satisfies: 1. SP is linear and hence we will write SP { rather than SP ({)= 2 2. SP = SP (SP is a projection).  3. SP = SP > (SP is self-adjoint). 4. Ran(SP ) = P and Nul(SP ) = P B =

Exercise 14.1. Suppose P is a subset of K> then P BB = span(P )=

m

} 5 K $ h·|}i 5 K 

Proof. The map m is conjugate linear by the axioms of the inner products. Moreover, for {> } 5 K> |h{|}i|  k{k k}k for all { 5 K with equality when { = }= This implies that km}kK  = kh·|}ikK  = k}k = Therefore m is isometric and this implies m is injective. To finish the proof we must show that m is surjective. So let i 5 K  which we assume, with out loss of generality, is non-zero. Then P =Nul(i ) — a closed proper subspace of K= Since, by Corollary 14.14, K = P  P B > i : K@P  = P B $ F is a linear isomorphism. This shows that dim(P B ) = 1 and hence K = P  F{0 where {0 5 P B \ {0} =2 Choose } = {0 5 P B such that i ({0 ) = h{0 |}i> i.e.  = i¯({0 )@ k{0 k2 = Then for { = p + {0 with p 5 P and  5 F> i ({) = i ({0 ) = h{0 |}i = h{0 |}i = hp + {0 |}i = h{|}i which shows that i = m}= 1

Proof.

showing SP {1 + SP {2 = SP ({1 + {2 )> i.e. SP is linear.

Recall that m is conjugate linear if m (}1 + }2 ) = m}1 + m} ¯ 2

1. Let {1 > {2 5 K and  5 F> then SP {1 + SP {2 5 P and SP {1 + SP {2  ({1 + {2 ) = [SP {1  {1 + (SP {2  {2 )] 5 P B

(14.5)

is a conjugate linear1 isometric isomorphism.

2

for all }1 > }2 M K and  M C= Alternatively, choose {0 M P z \ {0} such that i({0 ) = 1= For { M P z we have i ({ 3 {0 ) = 0 provided that  := i({)= Therefore { 3 {0 M P K P z = {0} > i.e. { = {0 = This again shows that P z is spanned by {0 =

14 Hilbert Spaces Basics

203

Proposition 14.16 (Adjoints). Let K and N be Hilbert spaces and D : K $ N be a bounded operator. Then there exists a unique bounded operator D : N $ K such that hD{||iN = h{|D |iK for all { 5 K and | 5 N=

(14.6)

Moreover, for all D> E 5 O(K> N) and  5 C> ¯ > 1. (D + E) = D + E 2. D := (D ) = D> 3. kD k = kDk and 4. kD Dk = kDk2 =  5. If N = K> then (DE) = E  D = In particular D 5 ¡O (K) ¢ has a bounded  inverse i D has a bounded inverse and (D )1 = D1 =

Proof. For each | 5 N> the map { $ hD{||iN is in K  and therefore there exists, by Theorem 14.15, a unique vector } 5 K such that hD{||iN = h{|}iK for all { 5 K=

This shows there is a unique map D : N $ K such that hD{||iN = h{|D (|)iK for all { 5 K and | 5 N= To see D is linear, let |1 > |2 5 N and  5 C> then for any { 5 K> ¯ hD{||1 + |2 iN = hD{||1 iN + hD{|| 2 iN   ¯ = h{|D (|1 )iN + h{|D (|2 )iN   = h{|D (|1 ) + D (|2 )iN and by the uniqueness of D (|1 + |2 ) we find

This shows D is linear and so we will now write D | instead of D (|)= Since hD ||{iK = h{|D |iK = hD{||iN = h||D{iN 

¯  is Exercise it follows that D = D= The assertion that (D + E) = D + E 14.2. Making use of the Schwarz inequality (Theorem 14.2), we have

=

sup sup

kD nk sup

n5N:knk=1 k5K:kkk=1

=

sup

sup

k5K:kkk=1 n5N:knk=1

and kDk2 =

sup k5K:kkk=1

sup k5K:kkk=1

kDkk2 =

sup k5K:kkk=1

|hk|D Dki| 

|hDk|Dki|

sup k5K:kkk=1

kD Dkk = kD Dk =

Now suppose that N = K= Then hDEk|ni = hEk|D ni = hk|E  D ni which shows (DE) = E  D = If D1 exists then ¡ 1 ¢  ¡ 1 ¢ D D = DD = L  = L and ¡ ¢ ¡ ¢   D D1 = D1 D = L  = L= This shows that D is invertible and (D ) invertible then so is D = D =

1

¡ ¢ = D1 = Similarly if D is

Exercise 14.2. Let K> N> P be Hilbert spaces, D> E 5 O(K> N)> F 5 ¯  and (FD) = D F  5 O(N> P ) and  5 C= Show (D + E) = D + E O(P> K)= Exercise 14.3. Let K = Cq and N = Cp equipped with the usual inner products, i.e. h}|ziK = } · z ¯ for }> z 5 K= Let D be an p×q matrix thought of as a linear operator from K to N= Show the matrix associated to D : N $ K is the conjugate transpose of D=

1. Nul(D ) = Ran(D)B = 2. Ran(D) = Nul(D )B = 3. if N = K and Y  K is an D — invariant subspace (i.e. D(Y )  Y )> then Y B is D — invariant. Proof. An element | 5 N is in Nul(D ) i 0 = hD ||{i = h||D{i for all { 5 K which happens i | 5 Ran(D)B = Because, by Exercise 14.1, Ran(D) = Ran(D)BB > and so by the first item, Ran(D) = Nul(D )B = Now suppose D(Y )  Y and | 5 Y B > then hD ||{i = h||D{i = 0 for all { 5 Y

|hD n|ki| |hn|Dki| =

2

kD Dk  kD k kDk = kDk

Lemma 14.17. Suppose D : K $ N is a bounded operator, then:

D (|1 + |2 ) = D (|1 ) + D (|2 )=

n5N:knk=1

14 Hilbert Spaces Basics

=



kD k =

204

sup k5K:kkk=1

kDkk = kDk =

The last item is a consequence of the following two inequalities;

which shows D | 5 Y B =

14.1 Hilbert Space Basis

205

14.1 Hilbert Space Basis Proposition 14.18 (Bessel’s Inequality). Let W be an orthonormal set, then for any { 5 K> X |h{|yi|2  k{k2 for all { 5 K= (14.7) y5W

In particular the set W{ := {y 5 W : h{|yi 6= 0} is at most countable for all { 5 K= Proof. Let   W be any finite set. Then X X X h{|yiyk2 = k{k2  2Re h{|yi hy|{i + |h{|yi|2 0  k{  y5

= k{k2 

showing that

P

y5

X

y5

y5

y5

|h{|yi|2 2

206

14 Hilbert Spaces Basics

° ° ° X °2 X ° ° y° = kyk2 ? %2 = ° ° ° y5 y5 P Hence by Lemma 7.16, y5W y exists. P For item 1, let % be as above and set v% := y5% y= Then |kvk  kv% k|  kv  v% k ? %

and by Eq. (14.8), 0

y5W

|h{|yi|  k{k = Taking the supremum of this inequality over

Similarly if {yq }4 q=1 is an orthogonal set, then v =

q=1

kyq k2 ? 4= In particular if

rearrangements of

{yq }4 q=1 =

Proof. Suppose v =

P

y5W

X

y5

4 P

q=1

4 P

q=1

yq exists in K

yq exists, then it is independent of

X

y5 @ %

kyk2  %2 =

ular {vQ }4 Q=1 is Cauchy. So for Q A P= Q X

q=P+1

which shows that

° ° ° X °2 ° ° kyk = ° y°  1 ° ° 2

y5

P

kyq k2 = kvQ  vP k2 $ 0 as P> Q $ 4 kyq k2 is convergent, i.e.

4 P

q=1

kyq k2 ? 4=

Alternative proof of item 1. We could use the last result to prove P Item 1. Indeed, if y5W kyk2 ? 4> then W is countable and so we may writer W = {yq }4 q=1 = Then v = limQ$4 vQ with vQ as above. Since the norm, k·k > is continuous on K> °Q ° Q ° X °2 X ° ° yq ° = lim kyq k2 kvk2 = lim kvQ k2 = lim ° ° Q $4 Q $4 ° Q $4 q=1

=

4 X

q=1

2

kyq k =

X

y5W

q=1

kyk2 =

Corollary 14.20. Suppose K is a Hilbert space,   K is an orthonormal set and P = span = Then X SP { = h{|xix> (14.9) X

X

x5

2

for all {> | 5 K=

x5

2

(14.10)

h{|xihx||i = hSP {||i

(14.11)

x5

y5

Conversely, suppose that y5W kyk ? 4= Then for all % A 0 there exists %  W such that if   W \ % >

4 P

q=1

y exists. Then there exists   W such that

for all   W \ > wherein the first inequality we have used Pythagorean’s P theorem. Taking the supremum over such  shows that y5W \ kyk2  1 and therefore X X 2 2 kyk  1 + kyk ? 4= y5W

2

kyk2  kv% k =

q=1

2

P Proposition 14.19. Suppose W  KP is an orthogonal set. Then v = y5W y 2 exists in K (see Definition 7.15) i y5W Pkyk ? 4= (In particular W must be at most a countable set.) Moreover, if y5W kyk2 ? 4> then P 2 2 1. kvk = Py5W kyk and 2. hv|{i = y5W hy|{i for all { 5 K= i

X

Letting % & 0 we deduce from the previous two equations that kv% k $ kvk and P P kv% k2 $ y5W kyk2 as % & 0 and therefore kvk2 = y5W kyk2 = Item 2. is a special case of Lemma 7.16. For the final assertion, let Q P yq and suppose that limQ $4 vQ = v exists in K and in particvQ :=

  W then proves Eq. (14.7).

4 P

(14.8)

2

|h{|xi| = kSP {k and

14.1 Hilbert Space Basis

207

P 2 2 Proof. By Bessel’s inequality, Px5 |h{|xi|  k{k for all { 5 K and hence by Proposition 14.18, S { := x5 h{|xix exists in K and for all {> | 5 K> X X hS {||i = hh{|xix||i = h{|xihx||i= (14.12) x5

x5

Taking | 5  in Eq. (14.12) gives hS {||i = h{||i> i.e. that h{  S {||i = 0 for all | 5 = So ({  S {) B span  and by continuity we also have ({  S {) B P = span = Since S { is also in P> it follows from the definition of SP that S { = SP { proving Eq. (14.9). Equations (14.10) and (14.11) now follow 2 from (14.12), Proposition 14.19 and the fact that hSP {||i = hSP {||i = hSP {|SP |i for all {> | 5 K= Definition 14.21 (Basis). Let K be a Hilbert space. A basis  of K is a maximal orthonormal subset   K= Proposition 14.22. Every Hilbert space has an orthonormal basis. Proof. Let F be the collection of all orthonormal subsets of K ordered by inclusion. If   F is linearly ordered then ^ is an upper bound. By Zorn’s Lemma (see Theorem B.7) there exists a maximal element  5 F= An orthonormal set   K is said to be complete if  B = {0} = That is to say if h{|xi = 0 for all x 5  then { = 0= Lemma 14.23. Let  be an orthonormal subset of K then the following are equivalent: 1.  is a basis, 2.  is complete and 3. span  = K= B

Proof. If  is not complete, then there exists a unit vector { 5  \ {0} = The set  ^ {{} is an orthonormal set properly containing > so  is not maximal. Conversely, if  is not maximal, there exists an orthonormal set 1  K such that  & 1 = Then if { 5 1 \ > we have h{|xi = 0 for all x 5  showing  is not complete. This proves the equivalence of (1) and (2). If  is not complete and { 5  B \ {0} > then span   {B which is a proper B subspace of K= Conversely if span  is a proper subspace of K>  B = span  is a non-trivial subspace by Corollary 14.14 and  is not complete. This shows that (2) and (3) are equivalent. Theorem 14.24. Let   K be an orthonormal set. Then the following are equivalent: 1.  is complete, i.e.  is an orthonormal basis for K= P 2. { = h{|xix for all { 5 K. x5 P 3. h{||i = h{|xi hx||i for all {> | 5 K= x5

208

14 Hilbert Spaces Basics

4. k{k2 =

P

x5

|h{|xi|2 for all { 5 K=

= SP = Proof. Let P = span  and S P (1) , (2) By Corollary 14.20, h{|xix = SP {= Therefore x5

{

X

x5

h{|xix = {  SP { 5 P B =  B = {0} =

(2) , (3) is a consequence of Proposition 14.19. (3) , (4) is obvious, just take | = {= (4) , (1) If { 5  B > then by 4), k{k = 0> i.e. { = 0= This shows that  is complete. Suppose  := {xq }4 q=1 is a collection of vectors in an inner product space (K> h·|·i) = The standard Gram-Schmidt process produces from  an orthonormal subset,  = {yq }4 q=1 > such that every element xq 5  is a finite linear combination of elements from = Recall the procedure is to define yq inductively by setting y˜q+1 := yq+1 

q X hxq+1 |ym iym = yq+1  Sq yq+1 m=1

q

where Sq is orthogonal projection onto Pq := span({yn }n=1 )= If yq+1 := 0> let y˜q+1 = 0> otherwise set yq+1 := k˜ yq+1 k1 y˜q+1 = Finally re-index the resulting sequence so as to throw out those yq with yq = 0= The result is an orthonormal subset,   K> with the desired properties. Definition 14.25. As subset, > of a normed space [ is said to be total if span( ) is a dense in [= 4 Remark 14.26. Suppose that {xq }4 q=1 is a total subset of K= Let {yq }q=1 be 4 the vectors found by performing Gram-Schmidt on the set {xq }q=1 = Then  := {yq }4 q=1 is an orthonormal basis for K= Indeed, if k 5 K is orthogonal 4 to  then k is orthogonal to {xq }4 q=1 and hence also span {xq }q=1 = K= In particular k is orthogonal to itself and so k = 0=

Proposition 14.27. A Hilbert space K is separable i K has a countable orthonormal basis   K= Moreover, if K is separable, all orthonormal bases of K are countable. (See Proposition 4.14 in Conway’s, “A Course in Functional Analysis,” for a more general version of this proposition.) Proof. Let D  K be a countable dense set D = {xq }4 q=1 = By GramSchmidt process there exists  = {yq }4 q=1 an orthonormal set such that span{yq : q = 1> 2 = = = > Q }  span{xq : q = 1> 2 = = = > Q }= So if h{|yq i = 0 for all q then h{|xq i = 0 for all q= Since D  K is dense we may choose {zn }  D such that { = limn$4 zn and therefore h{|{i = limn$4 h{|zn i = 0= That is to

14.1 Hilbert Space Basis

209

say { = 0 and  is complete. Conversely if   K is a countable orthonormal basis, then the countable set ; < ?X @ D= dx x : dx 5 Q + lQ : #{x : dx 6= 0} ? 4 = > x5

is dense in K= Finally let  = {xq }4 q=1 be an orthonormal basis and 1  K be another orthonormal basis. Then the sets Eq = {y 5 1 : hy|xq i 6= 0} are countable for each q 5 N and hence E :=

4 S

q=1

Proposition 14.28. Suppose [ and \ are sets and  : [ $ (0> 4) and  : \ $ (0> 4) are give weight functions. For functions i : [ $ C and j : \ $ C let i j : [ × \ $ C be defined by i j ({> |) := i ({) j (|) = If   c2 () and   c2 () are orthonormal bases, then   := {i j : i 5  and j 5 } is an orthonormal basis for c2 ( ) = Proof. Let i> i 0 5 c2 () and j> j 0 5 c2 () > then by the Tonelli’s Theorem 4.22 for sums and Hölder’s inequality, X X X |i j · i 0 j 0 |   = |i i 0 |  · |jj 0 |  [

\

 ki kc2 () ki 0 kc2 () kjkc2 () kj 0 kc2 () = 1 ? 4=

So by Fubini’s Theorem 4.23 for sums, X X i i 0 · j¯ j0  hi j|i 0 j 0 ic2 ( ) = [

\

= hi |i 0 ic2 () hj|j 0 ic2 () = i>i 0 j>j0 = 2

Therefore,   is an orthonormal subset of c ( )= So it only remains to show   is complete. We will give two proofs of this fact. Let I 5 c2 ( )= In the first proof we will verify item 4. of Theorem 14.24 while in the second we will verify item 1 of Theorem 14.24. First Proof. By Tonelli’s Theorem, X X  ({)  (|) |I ({> |)|2 = kI k2c2 ( ) ? 4 {5[

|5\

14 Hilbert Spaces Basics

and since  A 0> it follows that X |I ({> |)|2  (|) ? 4 for all { 5 [> |5\

i.e. I ({> ·) 5 c2 () for all { 5 [= By the completeness of > X X¯ ¯ 2 ¯hI ({> ·) |jic2 () ¯2 |I ({> |)|  (|) = hI ({> ·) |I ({> ·)ic2 () = j5

\

and therefore,

2

kI kc2 ( ) =

Eq is a countable subset

of 1 = Suppose there exists y 5 1 \ E> then hy|xq i = 0 for all q and since  = {xq }4 q=1 is an orthonormal basis, this implies y = 0 which is impossible since kyk = 1= Therefore 1 \ E = > and hence 1 = E is countable.

[×\

210

=

X

 ({)

{5[

X

2

 (|) |I ({> |)|

|5\

X X¯ ¯ ¯hI ({> ·) |jic2 () ¯2  ({) =

(14.13)

{5[ j5

and in particular, { $ hI ({> ·) |jic2 () is in c2 () = So by the completeness of  and the Fubini and Tonelli theorems, we find ¯ ¯2 ¯ X ¯¯X X¯ ¯2 ¯ ¯hI ({> ·) |jic2 () ¯  ({) = ¯ hI ({> ·) |jic2 () i ({)  ({)¯ ¯ ¯ [ i 5 [ ¯ ¯2 ! à ¯ X ¯¯X X ¯ = I ({> |) j (|)  (|) i ({)  ({) ¯ ¯ ¯ ¯ \ i 5 [ ¯ ¯2 ¯ X ¯¯ X ¯ = I ({> |) i j ({> |)   ({> |)¯ ¯ ¯ ¯ i 5 [×\ X¯ ¯2 ¯hI |i jic2 ( ) ¯ = = i 5

Combining this result with Eq. (14.13) shows X ¯ ¯ ¯hI |i jic2 ( ) ¯2 kI k2c2 ( ) = i 5> j5

as desired. Second Proof. Suppose, for all i 5  and j 5  that hI |i ji = 0> i.e. X X  ({)  (|) I ({> |)i¯({)j(|) 0 = hI |i jic2 ( ) = =

X

{5[

Since

{5[

|5\

 ({) hI ({> ·)|jic2 () i¯({)=

(14.14)

14.2 Some Spectral Theory

X¯ X X ¯ ¯hI ({> ·)|jic2 () ¯2  ({)   ({) |I ({> |)|2  (|) ? 4>

{5[

{5[

211

(14.15)

|5\

it follows from Eq. (14.14) and the completeness of  that hI ({> ·)|jic2 () = 0 for all { 5 [= By the completeness of  we conclude that I ({> |) = 0 for all ({> |) 5 [ × \= Definition 14.29. A linear map X : K $ N is an isometry if kX {kN = k{kK for all { 5 K and X is unitary if X is also surjective. Exercise 14.4. Let X : K $ N be a linear map, show the following are equivalent: 1. X : K $ N is an isometry, 2. hX {|X {0 iN = h{|{0 iK for all {> {0 5 K> (see Eq. (14.31) below) 3. X  X = lgK = Exercise 14.5. Let X : K $ N be a linear map, show the following are equivalent: 1. X : K $ N is unitary 2. X  X = lgK and X X  = lgN = 3. X is invertible and X 1 = X  = Exercise 14.6. Let K be a Hilbert space. Use Theorem 14.24 to show there exists a set [ and a unitary map X : K $ c2 ([)= Moreover, if K is separable and dim(K) = 4> then [ can be taken to be N so that K is unitarily equivalent to c2 = c2 (N)=

212

14 Hilbert Spaces Basics

Definition 14.30. Suppose [ is a Banach space over F (F = R or C) and D 5 O ([) = We say  5 F is in the spectrum of D if D  L does not have a bounded3 inverse. The spectrum will be denoted by  (D)  F= The resolvent set for D is  (D) := F\ (D) = Remark 14.31. If  is an eigenvalue of D> then D  L is not injective and hence not invertible. Therefore any eigenvalue of D is in the spectrum of D= If K is a Hilbert space ant D 5 O (K) > it follows from item 5. of Proposition ¯ 5  (D ) > i.e. 14.16 that  5  (D) i  © ª ¯ :  5  (D) =  (D ) = 

Exercise 14.8. Suppose [ is a Banachnspace and D 5 O ([) = Use Corollary o 7.20 to show  (D) is a closed subset of  5 F : ||  kDk := kDkO([) =

Lemma 14.32. Suppose that D 5 O(K) is a normal operator, i.e. [D> D ] = 0= Then  5 (D) i inf k(D  1)#k = 0= (14.16) k#k=1

In other words,  5  (D) i there is an “approximate sequence of eigenvectors” for (D> ) > i.e. there exists #q 5 K such that k#q k = 1 and D#q  #q $ 0 as q $ 4= Proof. By replacing D by D  L we may assume that  = 0= If 0 5 @ (D)> then ° ° kD#k k#k inf kD#k = inf = inf = 1@ °D1 ° A 0= k#k kD1 #k k#k=1

Now suppose that inf k#k=1 kD#k = % A 0 or equivalently we have kD#k  % k#k

14.2 Some Spectral Theory

for all # 5 K= Because D is normal, For this section let K and N be two Hilbert space over C= Exercise 14.7. Suppose D : K $ K is a bounded self-adjoint operator. Show: 1. If  is an eigenvalue of D> i.e. D{ = { for some { 5 K \ {0} > then  5 R= 2. If  and  are two distinct eigenvalues of D with eigenvectors { and | respectively, then { B |= Unlike in finite dimensions, it is possible that an operator on a complex Hilbert space may have no eigenvalues, see Example 14.35 and Lemma 14.36 below for a couple of examples. For this reason it is useful to generalize the notion of an eigenvalue as follows.

kD#k2 = hD D#|#i = hDD #|#i = hD #|D #i = kD #k2 = Therefore we also have kD #k = kD#k  % k#k ; # 5 K=

(14.17)

This shows in particular that D and D are injective, Ran(D) is closed and hence by Lemma 14.17 Ran(D) = Ran(D) = Nul(D )B = {0}B = K= Therefore D is algebraically invertible and the inverse is bounded by Eq. (14.17). 3

It will follow by the open mapping Theorem 35.1 or the closed graph Theorem 35.3 that the word bounded may be omitted from this definition.

14.2 Some Spectral Theory

213

Lemma 14.33. Suppose that D 5 O(K) is self-adjoint (i.e. D = D ) then i h (D)   kDkrs > kDkrs  R= Proof. Writting  =  + l with >  5 R> then

k(D +  + l) #k2 = k(D + )#k2 + ||2 k#k2 + 2 Re((D + ) #> l#) = k(D + )#k2 + ||2 k#k2

(14.18)

214

14 Hilbert Spaces Basics

Hence if  5 G> we may let $1 = 1 above to find V  (1> > 2 > = = = ) = (1> > 2 > = = = ) where (1> > 2 > = = = ) 5 c2 = Thus we have shown  is an eigenvalue for V  for all  5 G and hence G  (V  )= Lemma 14.36. Let K = c2 (Z) and let D : K $ K be defined by Di (n) = l (i (n + 1)  i (n  1)) for all n 5 Z=

wherein we have used Re [l((D + ) #> #)] =  Im((D + ) #> #) = 0 since ((D + ) #> #) = (#> (D + ) #) = ((D + ) #> #)= Eq. (14.18) along with Lemma 14.32 shows that  5 @h (D) if  6= 0> i i.e. (D)  R= The fact that  (D) is now contained in  kDkrs > kDkrs is a consequence of Exercise 14.8.  Remark µ 14.34. ¶ It is not true that (D)  R implies D = D = For example let 01 D= on K = C2 > then (D) = {0} yet D 6= D = 00

Example 14.35. Let V 5 O(K) be a (not necessarily) normal operator. The proof of Lemma 14.32 gives  5 (V) if Eq. (14.16) holds. However the converse is not always valid unless V is normal. For example, let V : c2 $ c2 be the shift, V($1 > $2 > = = = ) = (0> $1 > $2 > = = = )= Then for any  5 G := {} 5 C : |}| ? 1} > k(V  ) #k = kV#  #k  |kV#k  || k#k| = (1  ||) k#k and so there does not exists an approximate sequence of eigenvectors for ¯ (V> ) = However, as we will now show,  (V) = G= To prove this it su!ces to show by Remark 14.31 and Exercise 14.8 that ¯   (V  )  G ¯ and hence  (V) = G ¯ G   (V  ) = For if this is the cae then G ¯ is invariant under complex conjugation. since G A simple computation shows, V  ($1 > $2 > = = = ) = ($2 > $3 > = = = ) and $ = ($1 > $2 > = = = ) is an eigenvector for V  with eigenvalue  5 C i 0 = (V   L) ($1 > $2 > = = = ) = ($2  $1 > $3  $2 > = = = )= Solving these equation shows $2 = $1 > $3 = $2 = 2 $1 > = = = > $q = q1 $1 =

Then: 1. D is a bounded self-adjoint operator. 2. D has no eigenvalues. 3.  (D) = [2> 2] = Proof. For another (simpler) proof of this lemma, see Exercise 23.8 below. 1. Since kDi k2  ki (· + 1)k2 + ki (·  1)k2 = 2 ki k2 >

kDkrs  2 ? 4= Moreover, for i> j 5 c2 (Z) > hDi |ji = =

X

X n

=

l (i (n + 1)  i (n  1)) j¯ (n)

n

X n

li (n) j¯ (n  1) 

X

li (n) j¯ (n + 1)

n

i (n) Dj (n) = hi |Dji>

which shows D = D = 2. From Lemma 14.33, we know that  (D)  [2> 2] = If  5 [2> 2] and i 5 K satisfies Di = i> then i (n + 1) = li (n) + i (n  1) for all n 5 Z=

(14.19)

This is a second order dierence equations which can be solved analogously to second order ordinary dierential equations. The idea is to start by looking for a solution of the form i (n) = n = Then Eq. (14.19) beocmes, n+1 = ln + n1 or equivalently that 2 + l  1 = 0= So we will have a solution if  5 {± } where s l ± 4  2 = ± = 2

14.2 Some Spectral Theory

215

For || 6= 2> there are two distinct roots and the general solution to Eq. (14.19) is of the form n i (n) = f+ n+ + f  (14.20)

216

14 Hilbert Spaces Basics

we see P  kW k= Conversely let i> j 5 K and compute hi + j|W (i + j)i  (i  j|W (i  j)i = hi |W ji + hj|W i i + hi |W ji + hj|W i i

for some constants f± 5 C and || = 2> the general solution has the form n i (n) = f+ + gnn+ (14.21) ¡ 2 ¢ 1 Since in all cases, |± | = 4  + 4  2 = 1> it follows that neither of these functions, i> will be in c2 (Z) unless they are identically zero. This shows that D has no eigenvalues. 3. The above argument suggest a method for constructing approximate eigenfucntions. Namely, let  5 [2> 2] and define iq (n) := 1|n|q n where  = + = Then a simple computation shows

lim

q$4

k(D  L) iq k2 =0 kiq k2

= 2[hi |W ji + hW j|i i] = 2[hi |W ji + hi |W ji] = 4Rehi |W ji=

Therefore, if ki k = kjk = 1> it follows that |Rehi |W ji| 

ª P© ª P© ki + jk2 + ki  jk2 = 2ki k2 + 2kjk2 = P= 4 4

By replacing i be hl i where  is chosen so that hl hi |W ji is real, we find |hi |W ji|  P for all ki k = kjk = 1=

(14.22) Hence

kW k =

and therefore  5  (D) = Exercise 14.9. Verify Eq. (14.22). Also show by explicit computations that lim

q$4

k(D  L) iq k2 6= 0 kiq k2

if  5 @ [2> 2] = The next couple of results will be needed for the next section. Theorem 14.37 (Rayleigh quotient). Suppose W 5 O(K) := O(K> K) is a bounded self-adjoint operator, then kW k = sup i 6=0

|hi |W i i| = ki k2

Moreover if there exists a non-zero element j 5 K such that |hW j|ji| = kW k> kjk2

|hi |W i i| = 2 i 6=0 ki k

P := sup We wish to show P = kW k= Since

|hi |W i i|  ki kkW i k  kW kki k2 >

|hi |W ji|  P=

If j 5 K \ {0} and kW k = |hW j|ji|@kjk2 then, using the Cauchy Schwarz inequality, kW jk |hW j|ji|  kW k =  kW k= (14.23) kjk2 kjk

This implies |hW j|ji| = kW jkkjk and forces equality in the Cauchy Schwarz inequality. So by Theorem 14.2, W j and j are linearly dependent, i.e. W j = j for some  5 C= Substituting this into (14.23) shows that || = kW k= Since W is self-adjoint, ¯ kjk2 = hj|ji = hW j|ji = hj|W ji = hj|ji = hj|ji> which implies that  5 R and therefore,  5 {±kW k}=

Lemma 14.38 (Invariant subspaces). Let W : K $ K be a self-adjoint operator and P be a W — invariant subspace of K> i.e. W (P )  P= Then P B is also a W — invariant subspace, i.e. W (P B )  P B = Proof. Let { 5 P and | 5 P B > then W { 5 P and hence

then j is an eigenvector of W with W j = j and  5 {±kW k}= Proof. Let

sup ki k=kjk=1

0 = hW {||i = h{|W |i for all { 5 P= Thus W | 5 P B =

14.3 Compact Operators on a Hilbert Space

217

14.3 Compact Operators on a Hilbert Space In this section let K and E be Hilbert spaces and X := {{ 5 K : k{k ? 1} be the unit ball in K= Recall from Definition 11.16 that a bounded operator, N : K $ E> is compact i N(X ) is compact in E= Equivalently, for all 4 bounded sequences {{q }4 q=1  K> the sequence {N{q }q=1 has a convergent subsequence in E= Because of Theorem 11.15, if dim(K) = 4 and N : K $ E is invertible, then N is not compact. Definition 14.39. N : K $ E is said to have finite rank if Ran(N)  E is finite dimensional. The following result is a simple consequence of Corollaries 11.13 and 11.14. Corollary 14.40. If N : K $ E is a finite rank operator, then N is compact. In particular if either dim(K) ? 4 or dim(E) ? 4 then any bounded operator N : K $ E is finite rank and hence compact. Lemma 14.41. Let K := K(K> E) denote the compact operators from K to E= Then K(K> E) is a norm closed subspace of O(K> E)= Proof. The fact that K is a vector subspace of O(K> E) will be left to the reader. To finish the proof, we must show that N 5 O(K> E) is compact if there exists Nq 5 K(K> E) such that limq$4 kNq  Nkrs = 0= First Proof. Given % A 0> choose Q = Q (%) such that kNQ  Nk ? %= Using the fact that NQ X is precompact, choose a finite subset   X such that min{5 k|  NQ {k ? % for all | 5 NQ (X ) = Then for } = N{0 5 N(X ) and { 5 > k}  N{k = k(N  NQ ){0 + NQ ({0  {) + (NQ  N){k  2% + kNQ {0  NQ {k= Therefore min{5 k}  NQ {k ? 3%> which shows N(X ) is 3% bounded for all % A 0> N(X ) is totally bounded and hence precompact. 4 Second Proof. Suppose {{q© }q=1ª is a bounded sequence in K= com© By1 ª 4 4 pactness, there is a subsequence {1q q=1 of {{q }4 such that N { 1 q q=1 q=1 is convergent in E= Working inductively, we may construct subsequences © ª4 © ª4 4 4 {{q }q=1  {1q q=1  {2q q=1 · · ·  {{p q }q=1  = = = 4

such that {Np {p q }q=1 is convergent in E for each p= By the usual Cantor’s 4 diagonalization procedure, let |q := {qq > then {|q }q=1 is a subsequence of 4 4 {{q }q=1 such that {Np |q }q=1 is convergent for all p= Since kN|q  N|o k  k(N  Np ) |q k + kNp (|q  |o )k + k(Np  N) |o )k  2 kN  Np k + kNp (|q  |o )k > lim sup kN|q  N|o k  2 kN  Np k $ 0 as p $ 4> q>o$4 4

which shows {N|q }q=1 is Cauchy and hence convergent.

218

14 Hilbert Spaces Basics

Proposition 14.42. A bounded operator N : K $ E is compact i there exists finite rank operators, Nq : K $ E> such that kN Nq k $ 0 as q $ 4= Proof. Since N(X ) is compact it contains a countable dense subset and from this it follows that N (K) is a separable subspace of E= Let {!q } be an orthonormal basis for N (K)  E and SQ | =

Q X

h||!q i!q

q=1

be the orthogonal projection of | onto span{!q }Q q=1 = Then limQ $4 kSQ |  |k = 0 for all | 5 N(K)= Define Nq := Sq N — a finite rank operator on K= For sake of contradiction suppose that lim sup kN  Nq k = % A 0> q$4

in which case there exists {qn 5 X such that k(N  Nqn ){qn k  % for all qn = Since N is compact, by passing to a subsequence if necessary, we may assume 4 {N{qn }qn =1 is convergent in E= Letting | := limn$4 N{qn > k(N  Nqn ){qn k = k(1  Sqn )N{qn k  k(1  Sqn )(N{qn  |)k + k(1  Sqn )|k  kN{qn  |k + k(1  Sqn )|k $ 0 as n $ 4= But this contradicts the assumption that % is positive and hence we must have limq$4 kN  Nq k = 0> i.e. N is an operator norm limit of finite rank operators. The converse direction follows from Corollary 14.40 and Lemma 14.41. Corollary 14.43. If N is compact then so is N  = Proof. First Proof. Let Nq = Sq N be as in the proof of Proposition 14.42, then Nq = N  Sq is still finite rank. Furthermore, using Proposition 14.16, kN   Nq k = kN  Nq k $ 0 as q $ 4 showing N  is a limit of finite rank operators and hence compact. 4 Second Proof. Let {{q }q=1 be a bounded sequence in E> then

kN  {q  N  {p k2 = ({q  {p > NN  ({q  {p ))  2F kNN  ({q  {p )k (14.24) 4 where F is a bound on the norms of the {q = Since {N  {q }q=1 is also a bounded sequence, by the compactness of N there is a subsequence {{0q } of the {{q } such that NN  {0q is convergent and hence by Eq. (14.24), so is the sequence {N  {0q } =

14.3 Compact Operators on a Hilbert Space

219

14.3.1 The Spectral Theorem for Self Adjoint Compact Operators For the rest of this section, N 5 K(K) := K(K> K) will be a self-adjoint compact operator or S.A.C.O. for short. Because of Proposition 14.42, we might expect compact operators to behave very much like finite dimensional matrices. This is typically the case as we will see below. Example 14.44 (Model S.A.C.O.). Let K = c2 and N be the diagonal matrix 4 3 1 0 0 · · · E 0 2 0 · · · F F E N = E 0 0 3 · · · F > D C .. .. . . . . . . . . where limq$4 |q | = 0 and q 5 R= Then N is a self-adjoint compact operator. This assertion was proved in Example 11.17 above.

220

14 Hilbert Spaces Basics

Theorem 14.46 (Compact Operator Spectral Theorem). Suppose that N : K $ K is a non-zero S.A.C.O., then 1. there exists at least one eigenvalue  5 {±kNk}= 2. There are at most countable many non-zero eigenvalues, {q }Q q=1 > where Q = 4 is allowed. (Unless N is finite rank (i.e. dim Ran (N) ? 4)> Q will be infinite.) 3. The q ’s (including multiplicities) may be arranged so that |q |  |q+1 | for all q= If Q = 4 then limq$4 |q | = 0= (In particular any eigenspace for N with non-zero eigenvalue is finite dimensional.) 4. The eigenvectors {!q }Q q=1 can be chosen to be an O.N. set such that K = span{!q }  Nul(N)= 5. Using the {!q }Q q=1 above, N# =

Q X

q=1

The main theorem (Theorem 14.46) of this subsection states that up to unitary equivalence, Example 14.44 is essentially the most general example of an S.A.C.O. Theorem 14.45. Let N be a S.A.C.O., then either  = kNk or  =  kNk is an eigenvalue of N= Proof. Without loss of generality we may assume that N is non-zero since otherwise the result is trivial. By Theorem 14.37, there exists iq 5 K such that kiq k = 1 and |hiq |Niq i| = |hiq |Niq i| $ kNk as q $ 4= kiq k2

(14.25)

By passing to a subsequence if necessary, we may assume that  := limq$4 hiq |Niq i exists and  5 {±kNk}= By passing to a further subsequence if necessary, we may assume, using the compactness of N> that Niq is convergent as well. We now compute: 2

2

2

0  kNiq  iq k = kNiq k  2hNiq |iq i +   2  2hNiq |iq i + 2 2

2

2

$   2 +  = 0 as q $ 4= Hence Niq  iq $ 0 as q $ 4 and therefore

(14.26)

1 lim Niq  q$4 exists. By the continuity of the inner product, ki k = 1 6= 0= By passing to the limit in Eq. (14.26) we find that Ni = i= i := lim iq = q$4

q h#|!q i!q for all # 5 K=

6. The spectrum of N is, (N) = {0} ^ {q : q ? Q + 1} = Proof. We will find q ’s and !q ’s recursively. Let 1 5 {±kNk} and !1 5 K such that N!1 = 1 !1 as in Theorem 14.45. Take P1 = span(!1 ) so N(P1 )  P1 = By Lemma 14.38, NP1B  P1B = Define N1 : P1B $ P1B via N1 = N|P1B = Then N1 is again a compact operator. If N1 = 0> we are done. If N1 6= 0> by Theorem 14.45 there exists 2 5 {±kNk1 } and !2 5 P1B such that k!2 k = 1 and N1 !2 = N!2 = 2 !2 = Let P2 := span(!1 > !2 )= Again N(P2 )  P2 and hence N2 := N|P2B : P2B $ P2B is compact. Again if N2 = 0 we are done. If N2 6= 0. Then by Theorem 14.45 there exists 3 5 {±kNk2 } and !3 5 P2B such that k!3 k = 1 and N2 !3 = N!3 = 3 !3 = Continuing this way indefinitely or until we reach a point where Nq = 0, we construct a sequence {q }Q q=1 of eigenvalues and orthonormal eigenvectors {!q }Q q=1 such that |l |  |l+1 | with the further property that |l | =

sup !B{!1 >!2 >===!l1 }

kN!k k!k

(14.27)

If Q = 4 then liml$4 |l | = 0 for if not there would exist % A 0 such that 1 |l |  % A 0 for all l= In this case {!l @l }4 = l=1 is sequence in K bounded by % By compactness of N> there exists a subsequence ln such that !ln = N!ln @ln is convergent. But this is impossible since {!ln } is an orthonormal set. Hence we must have that % = 0= Let P := span{!l }Q l=1 with Q = 4 possible. Then N(P )  P and hence N(P B )  P B = Using Eq. (14.27), kN|P B k  kN|PqB k = |q | $ 0 as q $ 4 showing N|P B  0. Define S0 to be orthogonal projection onto P B = Then for # 5 K>

14.3 Compact Operators on a Hilbert Space

# = S0 # + (1  S0 )# = S0 # +

221

Q X h#|!l i!l

222

14 Hilbert Spaces Basics

Let D be the positive square root of N  N defined by

l=1

and N# = NS0 # + N

D# :=

Q Q X X h#|!l i!l = l h#|!l i!l = l=1

q=1

l=1

Since {q }  (N) and (N) is closed, it follows that 0 5 (N) and hence {q }4 @ {q }4 q=1 ^ {0}  (N)= Suppose that } 5 q=1 ^ {0} and let g be the ^{0}= Notice that g A 0 because limq$4 q = distance between } and {q }4 q=1 0= A few simple computations show that: (N  }L)# =

Q X l=1

(N  })

1

Q p X q h#|!q i!q for all # 5 K=

h#|!l i(l  })!l  }S0 #>

A simple computation shows, D2 = N  N> and therefore,

® kD#k2 = hD#|D#i = #|D2 #

for all # 5 K= Hence we may define a unitary operator, x : Ran(D) $ Ran(N) by the formula xD# = N# for all # 5 K= We then have

exists,

N# = xD# =

(N  }L)1 # =

Q X h#|!l i(l  })1 !l  } 1 S0 #>

l=1

|h#|!l i|2

#q = 1@2 xD!q = q1@2 xD!q = 1@2 N!q q q

1 1 + kS0 #k2 |l  }|2 |}|2 !

µ ¶2 ÃX Q 1  |h#|!l i|2 + kS0 #k2 g l=1

and

1 = 2 k#k2 = g

We have thus shown that (N  }L)1 exists, k(N  }L)1 k  g1 ? 4 and hence } 5 @ (N)= Theorem 14.47 (Structure of Compact Operators). Let N : K $ E be a compact operator. Then there exists Q 5 N^ {4} > orthonormal subsets Q Q {!q }Q q=1  K and {#q }q=1  E and a sequences {q }q=1  R+ such that 1  2  = = = > limq$4 q = 0 if Q = 4> k#q k  1 for all q and Ni =

Q X

q=1

q hi |!q i#q for all i 5 K=

(14.28)

Proof. Since N  N is a selfadjoint compact operator, Theorem 14.46 implies there exists an orthonormal set {!q }Q q=1  K and positive numbers Q {q }q=1 such that N  N# =

Q X

q=1

q h#|!q i!q for all # 5 K=

(14.29)

s which proves the result with #q := x!q and q = q = It is instructive to find #q explicitly and to verify Eq. (14.29) by bruit 1@2 force. Since !q = q D!q >

and Q X

Q p X q h#|!q ix!q

q=1

l=1

k(N  }L)1 #k2 =

2

= h#|N  N#i = hN#|N#i = kN#k

hN!q |N!p i = h!q |N  N!p i = q pq = Q

This verifies that {#q }q=1 is an orthonormal set. Moreover, Q p Q p X X q h#|!q i#q = q h#|!q iq1@2 N!q

q=1

q=1

=N

Q X

q=1

h#|!q i!q = N#

P B since Q q=1 h#|!q i!q = S # where S is orthogonal projection onto Nul(N) = Second Proof. Let N = x |N| be the polar decomposition of N= Then |N| is self-adjoint and compact, by Corollary 37.12, and hence by Theorem Q 14.46 there exists an orthonormal basis {!q }q=1 for Nul(|N|)B = Nul(N)B such that |N| !q = q !q > 1  2  = = = and limq$4 q = 0 if Q = 4= For i 5 K> Ni = x |N|

Q X

hi |!q i!q =

q=1

Q X

hi |!q ix |N| !q =

q=1

which is Eq. (14.28) with #q := x!q =

Q X

q=1

q hi |!q ix!q

14.4 Weak Convergence

223

14.4 Weak Convergence Suppose K is an infinite dimensional Hilbert space and {{q }4 q=1 is an orthonormal subset of K= Then, by Eq. (14.1), k{q  {p k2 = 2 for all p 6= q and in 4 particular, {{q }q=1 has no convergent subsequences. From this we conclude that F := {{ 5 K : k{k  1} > the closed unit ball in K> is not compact. To overcome this problems it is sometimes useful to introduce a weaker topology on [ having the property that F is compact. Definition 14.48. Let ([> k·k) be a Banach space and [  be its continuous dual. The weak topology, z > on [ is the topology generated by [  = If z 4 {{q }q=1  [ is a sequence we will write {q $ { as q $ 4 to mean that {q $ { in the weak topology. 

Because z =  ([ )  k·k :=  ({k{  ·k : { 5 [} > it is harder for a function i : [ $ F to be continuous in the z — topology than in the norm topology, k·k = In particular if ! : [ $ F is a linear functional which is z — continuous, then ! is k·k — continuous and hence ! 5 [  = Exercise 14.10. Show the vector space operations of [ are continuous in the weak topology, i.e. show: 1. ({> |) 5 [ × [ $ { + | 5 [ is (z z > z ) — continuous and 2. (> {) 5 F × [ $ { 5 [ is (F z > z ) — continuous. z

4

Proposition 14.49. Let {{q }q=1  [ be a sequence, then {q $ { 5 [ as q $ 4 i !({) = limq$4 !({q ) for all ! 5 [  = z



Proof. By definition of z > we have {q $ { 5 [ i for all   [ and % A 0 there exists an Q 5 N such that |!({)  !({q )| ? % for all q  Q and ! 5 = This later condition is easily seen to be equivalent to !({) = limq$4 !({q ) for all ! 5 [  = The topological space ([> z ) is still Hausdor as follows from the Hahn Banach Theorem, see Theorem 7.26. For the moment we will concentrate on the special case where [ = K is a Hilbert space in which case K  = {!} := h·|}i : } 5 K} > see Theorem 14.15. If {> | 5 K and } := |  { 6= 0> then 2 0 ? % := k}k = !} (}) = !} (|)  !} ({)= Thus

224

14 Hilbert Spaces Basics

Remark 14.50. Suppose that K is an infinite dimensional Hilbert space 4 {{q }q=1 is an orthonormal subset of K= Then Bessel’s inequality (Propoz sition 14.18) implies {q $ 0 5 K as q $ 4= This points out the fact z that if {q $ { 5 K as q $ 4> it is no longer necessarily true that k{k = limq$4 k{q k = However we do always have k{k  lim inf q$4 k{q k because, k{k2 = lim h{q |{i  lim inf [k{q k k{k] = k{k lim inf k{q k = q$4

q$4

q$4

Proposition 14.51. Let K be a Hilbert space,   K be an orthonormal basis for K and {{q }4 q=1  K be a bounded sequence, then the following are equivalent: z

1. {q $ { 5 K as q $ 4= 2. h{||i = limq$4 h{q ||i for all | 5 K= 3. h{||i = limq$4 h{q ||i for all | 5 = Moreover, if f| := limq$4 h{q ||i exists for all | 5 > then P z and {q $ { := |5 f| | 5 K as q $ 4=

P

|5

2

|f| | ? 4

Proof. 1. =, 2. This is a consequence of Theorem 14.15 and Proposition 14.49. 2. =, 3. is trivial. 3. =, 1. Let P := supq k{q k and K0 denote the algebraic span of = Then for | 5 K and } 5 K0 > |h{  {q ||i|  |h{  {q |}i| + |h{  {q ||  }i|  |h{  {q |}i| + 2P k|  }k =

Passing to the limit in this equation implies lim supq$4 |h{  {q ||i|  2P k|  }k which shows lim supq$4 |h{  {q ||i| = 0 since K0 is dense in K= To prove the last assertion, let   = Then by Bessel’s inequality (Proposition 14.18), X X 2 2 2 |f| | = lim |h{q ||i|  lim inf k{q k  P 2 = |5

q$4

q$4

|5

P 2 Since    was arbitrary, P we conclude that |5 |f| |  P ? 4 and hence we may define { := |5 f| |= By construction we have h{||i = f| = lim h{q ||i for all | 5  q$4

z

and hence {q $ { 5 K as q $ 4 by what we have just proved. Y{ := {z 5 K : |!} ({)  !} (z)| ? %@2} and Y| := {z 5 K : |!} (|)  !} (z)| ? %@2}

are disjoint sets from z which contain { and | respectively. This shows that (K> z ) is a Hausdor space. In particular, this shows that weak limits are unique if they exist.

4

Theorem 14.52. Suppose {{q }q=1 is a bounded sequence in a Hilbert space, K= Then there exists a subsequence |n := {qn of {{q }4 q=1 and { 5 [ such z that |n $ { as n $ 4= Proof. This is a consequence of Proposition 14.51 and a Cantor’s diagonalization argument which is left to the reader, see Exercise 14.11.

14.5 Supplement 1: Converse of the Parallelogram Law

225

Theorem 14.53 (Alaoglu’s Theorem for Hilbert Spaces). Suppose that K is a separable Hilbert space, F := {{ 5 K : k{k  1} is the closed unit ball in K and {hq }4 q=1 is an orthonormal basis for K= Then 4 X 1 |h{  ||hq i| ({> |) := q 2 q=1

Proof. The routine check that  is a metric is left to the reader. Let  be the topology on F induced by = For any | 5 K and q 5 N> the map { 5 K $ h{  ||hq i = h{|hq i  h||hq i is z continuous and since the sum in Eq. (14.30) is uniformly convergent for {> | 5 F> it follows that { $ ({> |) is F — continuous. This implies the open balls relative to  are contained in F and therefore   F = For the converse inclusion, let } 5 K> { $ !} ({) = P h{|}i be an element of K  > and for Q 5 N let }Q := Q q=1 h}|hq ihq = Then PQ !}Q = q=1 h}|hq i!hq is  continuous, being a finite linear combination of the !hq which are easily seen to be  — continuous. Because }Q $ } as Q $ 4 it follows that sup |!} ({)  !}Q ({)| = k}  }Q k $ 0 as Q $ 4=

{5F

Therefore !} |F is  — continuous as well and hence F =  (!} |F : } 5 K)   = The last assertion follows directly from Theorem 14.52 and the fact that sequential compactness is equivalent to compactness for metric spaces.

14.5 Supplement 1: Converse of the Parallelogram Law Proposition 14.54 (Parallelogram Law Converse). If ([> k·k) is a normed space such that Eq. (14.2) holds for allp{> | 5 [> then there exists a unique inner product on h·|·i such that k{k := h{|{i for all { 5 [= In this case we say that k·k is a Hilbertian norm. Proof. If k·k is going to come from an inner product h·|·i> it follows from Eq. (14.1) that 2Reh{||i = k{ + |k2  k{k2  k|k2 and 2

2

2Reh{||i = k{  |k  k{k  k|k = Subtracting these two equations gives the “polarization identity,” 4Reh{||i = k{ + |k2  k{  |k2 =

14 Hilbert Spaces Basics

Replacing | by l| in this equation then implies that 4Imh{||i = k{ + l|k2  k{  l|k2 from which we find h{||i =

(14.30)

defines a metric on F which is compatible with the weak topology on F> F := (z )F = {Y _ F : Y 5 z } = Moreover (F> ) is a compact metric space. (This theorem will be extended to Banach spaces, see Theorems 35.14 and 35.15 below.)

2

226

1X %k{ + %|k2 4

(14.31)

%5J

where J = {±1> ±l} — a cyclic subgroup of V 1  C= Hence if h·|·i is going to exists we must define it by Eq. (14.31). Notice that 1X h{|{i = %k{ + %{k2 = k{k2 + lk{ + l{k2  lk{  l{k2 4 %5J ¯ ¯ ¯ ¯ 2 = k{k2 + l ¯1 + l|2 ¯ k{k2  l ¯1  l|2 ¯ k{k2 = k{k =

So to finish the proof of (4) we must show that h{||i in Eq. (14.31) is an inner product. Since X X %k| + %{k2 = %k% (| + %{) k2 4h||{i = %5J

=

X

%5J

2

2

%k%| + % {k

%5J

= k| + {k2 + k  | + {k2 + lkl|  {k2  lk  l|  {k2 = k{ + |k2 + k{  |k2 + lk{  l|k2  lk{ + l|k2 = 4h{||i it su!ces to show { $ h{||i is linear for all | 5 K= (The rest of this proof may safely be skipped by the reader.) For this we will need to derive an identity from Eq. (14.2). To do this we make use of Eq. (14.2) three times to find k{ + | + }k2 = k{ + |  }k2 + 2k{ + |k2 + 2k}k2

= k{  |  }k2  2k{  }k2  2k|k2 + 2k{ + |k2 + 2k}k2 = k| + }  {k2  2k{  }k2  2k|k2 + 2k{ + |k2 + 2k}k2 = k| + } + {k2 + 2k| + }k2 + 2k{k2

 2k{  }k2  2k|k2 + 2k{ + |k2 + 2k}k2 = Solving this equation for k{ + | + }k2 gives

k{ + | + }k2 = k| + }k2 + k{ + |k2  k{  }k2 + k{k2 + k}k2  k|k2 = (14.32)

Using Eq. (14.32), for {> |> } 5 K>

4 Reh{ + }||i = k{ + } + |k2  k{ + }  |k2

= k| + }k2 + k{ + |k2  k{  }k2 + k{k2 + k}k2  k|k2 ¡ ¢  k}  |k2 + k{  |k2  k{  }k2 + k{k2 + k}k2  k|k2 = k} + |k2  k}  |k2 + k{ + |k2  k{  |k2 = 4 Reh{||i + 4 Reh}||i=

(14.33)

14.6 Supplement 2. Non-complete inner product spaces

227

14 Hilbert Spaces Basics

P

2. k{k2 =

Now suppose that  5 J> then since || = 1>

1X 1X %k{ + %|k2 = %k{ +  1 %|k2 4 4 %5J %5J 1X 2 = %k{ + %|k = 4h{||i 4

228

4h{||i =

(14.34)

%5J

where in the third inequality, the substitution % $ % was made in the sum. So Eq. (14.34) says h±l{||i = ±lhl{||i and h{||i = h{||i= Therefore Imh{||i = Re (lh{||i) = Rehl{||i which combined with Eq. (14.33) shows

x5

|h{|xi|2 for all { 5 K.

Moreover, either of these two conditions implies that   K is a maximal orthonormal set. However   K being a maximal orthonormal set is not su!cient to conditions for 1) and 2) hold! Proof. As in the proof of Theorem 14.24, 1) implies 2). For 2) implies 1) let    and consider ° °2 ° ° X X X ° ° h{|xix° = k{k2  2 |h{|xi|2 + |h{|xi|2 °{  ° ° x5 x5 x5 X 2 2 = k{k  |h{|xi| = x5

Imh{ + }||i = Rehl{  l}||i = Rehl{||i + Rehl}||i = Imh{||i + Imh}||i and therefore (again in combination with Eq. (14.33)), h{ + }||i = h{||i + h}||i for all {> | 5 K= Because of this equation and Eq. (14.34) to finish the proof that { $ h{||i is linear, it su!ces to show h{||i = h{||i for all  A 0= Now if  = p 5 N> then hp{||i = h{ + (p  1){||i = h{||i + h(p  1){||i so that by induction hp{||i = ph{||i= Replacing { by {@p then shows that h{||i = php1 {||i so that hp1 {||i = p1 h{||i and so if p> q 5 N> we find h

q 1 q {||i = qh {||i = h{||i p p p

so that h{||i = h{||i for all  A 0 and  5 Q= By continuity, it now follows that h{||i = h{||i for all  A 0=

14.6 Supplement 2. Non-complete inner product spaces

2

Since k{k =

x5

x5

2

|h{|xi| > it follows that for every % A 0 there exists %  

such that for all    such that %  > ° °2 ° ° X X ° ° h{|xix° = k{k2  |h{|xi|2 ? % °{  ° ° x5

showing that { =

P

x5

x5

h{|xix= Suppose { = ({1 > {2 > = = = > {q > = = = ) 5  B = If 2)

is valid then k{k2 = 0> i.e. { = 0= So  is maximal. Let us now construct 2 a counter example to prove the last assertion. Take K = Span{hl }4 l=1  c and let x ˜q = h1  (q + 1)hq+1 for q = 1> 2 = = = = Applying Gramn-Schmidt to 4 {˜ xq }q=1 we construct an orthonormal set  = {xq }4 q=1  K= I now claim that   K is maximal. Indeed if { = ({1 > {2 > = = = > {q > = = = ) 5  B then { B xq for all q> i.e. 0 = h{|˜ xq i = {1  (q + 1){q+1 = 1

Therefore {q+1 = (q + 1) {1 for all q= Since { 5 Span{hl }4 l=1 > {Q = 0 for some Q su!ciently large and therefore {1 = 0 which in turn implies that {q = 0 for all q= So { = 0 and hence  is maximal in K= On the other hand,  is not maximal in c2 = In fact the above argument shows that  B in c2 is given by the span of y = (1> 12 > 13 > 14 > 15 > = = = )= Let S be the orthogonal projection of c2 onto the Span() = y B = Then

Part of Theorem 14.24 goes through when K is a not necessarily complete inner product space. We have the following proposition. Proposition 14.55. Let (K> h·|·i) be a not necessarily complete inner product space and   K be an orthonormal set. Then the following two conditions are equivalent: P 1. { = h{|xix for all { 5 K=

P

4 X h{|yi h{|xq ixq = S { = {  2 y> kyk l=1

so that

4 P

h{|xq ixq = { i { 5 Span() = y B  c2 = For example if { =

l=1

@ y B and hence (1> 0> 0> = = = ) 5 K (or more generally for { = hl for any l)> { 5 4 P h{|xq ixq 6= {= l=1

14.7 Exercises

229

14.7 Exercises Exercise 14.11. Prove Theorem 14.52. Hint: Let K0 := span {{q : q 5 N} — a separable Hilbert subspace of K= Let {p }4 p=1  K0 be an orthonormal basis and use Cantor’s diagonalization argument to find a subsequence |n := {qn such that fp := limn$4 h|n |p i exists for all p 5 N= Finish the proof by appealing to Proposition 14.51. z

4

Exercise 14.12. Suppose that {{q }q=1  K and {q $ { 5 K as q $ 4= Show {q $ { as q $ 4 (i.e. limq$4 k{  {q k = 0) i limq$4 k{q k = k{k = z

Exercise 14.13 (Banach-Saks). Suppose that {{q }4 q=1  K> {q $ { 5 K as q $ 4> and f := supq k{q k ? 4=4 Show there exists a subsequence, |n = {qn such that ° ° Q ° 1 X ° ° ° lim °{  |n ° = 0> ° Q $4 ° Q n=1

1 Q

PQ

i.e. n=1 |n $ { as Q $ 4= Hints: 1. show it su!ces to assume { = 0 1 and then choose {|n }4 (or even smaller if you like) n=1 so that |h|n ||o i|  o for all n  o= Exercise 14.14 (The Mean Ergodic Theorem). Let X : K $ K be a unitary operator on a Hilbert space K> P = Nul(X  L)> S = SP be orthogoPq1 nal projection onto P> and Vq = q1 n=0 X n = Show Vq $ SP strongly, i.e. limq$4 Vq { = SP { for all { 5 K= Hints: 1. verify the result for { 5 Nul(X  L) and { 5 Ran(X  L)> 2. show Nul(X   L) = Nul(X  L)> 3. finish the result with a limiting argument making use of items 1. and 2. and Lemma 14.17.

4

The assumption that f ? " is superfluous because of the “uniform boundedness principle,” see Theorem 35.8 below.