123 CONTENTS
Last Change: Tue 5 Feb 11:22:41 COT 2013 DRAFT 2 Index 5 21 37 55 79 105 116 119 ...... 26 ...... 120 122 ...... 32 ...... 71 ...... 87 ...... 65 79 ...... 18 ...... 121 ...... 38 1 ...... 50 69 ...... 45 ...... 47 ...... 99
...... 30 ...... 21 ...... 61 102
...... 52 ...... 89 ...... 9 37 ...... 5 ...... 55 58 80 DRAFT 1.1 Metric1.2 spaces Normed1.3 spaces and H¨older Minkowski inequality 2.1 Bounded2.2 linear maps The2.3 dual space and Examples the2.4 of Hahn-Banach dual theorem spaces The Banach space adjoint and the bidual 3.1 Baire’s3.2 theorem Uniform3.3 boundedness principle The3.4 open mapping theorem The3.5 closed graph theorem Projections3.6 in Banach spaces Weak convergence 4.1 Hilbert4.2 spaces Orthogonality 4.3 Orthonormal4.4 systems Linear4.5 operators in Hilbert Projections spaces4.6 in Hilbert spaces The adjoint of an unbounded operator 5.1 The5.2 spectrum of a The linear5.3 resolvent operator The5.4 spectrum of the Compact adjoint5.5 operators operator Hilbert-Schmidt5.6 operators Polar decomposition Problem Sheet 1. EspaciosProblem y m´etricos Sheet normados. 2. OperadoresProblem lineales. Sheet 3. Hahn Banach, Espacios duales. Contents 1 Banach spaces 2 Bounded maps; the dual space 3 Linear operators in Banach spaces 4 Hilbert spaces 5 Spectrum of linear operators A Exercises References Problem Sheets . The C CONTENTS or the complex field Last Change: Thu 7 Feb 17:21:44 COT 2013 R ). ∞
,
:= (0 + R usually denotes either the real field DRAFT K 4 Notation The letter positive real numbers are denoted by 3 y course lgebra and mbiguities. or changed notes, please opefully help istakes in the lombia. They e theory or an ey are rather a ach week there to prepare classes.
oo´,FebruaryBogot´a, 2013, M.W. Last Change: Tue 5 Feb 11:22:41 COT 2013 DRAFT to understand the material presented in the lecture. These lecture notes are work in progress. They may be abandoned advanced course in analysis is ofAn advantage important but part not necessary. ofis any a mathematics problem lecture sheet with are exercises exercises. (stolen For from e various books) which h radically at any moment. VeryIf likely you they contain find a mistakes lotlet of or me mistakes know. have and a suggestions how to improve the lecture collection of excerpts of the books of the list of references I used very basic notions of topology. Having attended a lecture in measur CONTENTS The notes areon written functional and analysis modifiedare at while by the no I Universidad means de am a los teaching new Andes, presentation an Co Bogot´a, Many of introductor thanks basic to functional the analysis, studentsfirst th from draft my of 2009 the class notes. who foundPrerequisites a for lot of the m course are a solid knowledge in analysis, linear a : . ) f is X y N ( M f N with → . The 0 such y if there → 6 = is called is closed M M x ) of : M is bounded ∈ x if and only N ε > c ( denotes the : M y of f N f U ∈ f if and only if ) is open in . M N n x, y if and only if ompleted in the ) U c M ( in n M 1 1.1. Metric spaces and x . Then − < ε. ∈ x f implies M ) ) = < ε. x m , where of y ⊆ ) ) for all M ( , x x M to , x induces a topology on N ϕ 6 = n neighbourhood U n x, y there exists an x ( d = x ( x ( d M U d Last Change: Thu 7 Feb 17:21:44 COT 2013 d N homeomorphism has the Hausdorff property, that ∈ ⇒ if if and only if ⇒ d Cauchy sequence = p = )) = converges such that M y is called a N ( is a . N in c M M ≥ U ) be topological spaces. N ) be metric spaces. A map ,f M ≥ ) ∈ Y N ⊆ ⊆ x is called a n n ⊆ → ( ) O x f n N continuous m,n Y U ( U dense x Y, ∈ N, d : M n N ∈ : ) → N d : n x ϕ 6 = 0, and that every isometry is continuous. ∈ x N X . ) : is complete. ∈ ) and ( ) and ( N is bounded. is called
. ) be a metric space. Then there exists a complete metric f X is called ∃ there exist neighbourhoods M } x, y N Y Y ) M . A subset ( O be a complete metric space and N | ∃ d M ) 0 M M → ∈
X, M, d such that M, d 0 ⊆ ∈ n ∈ N, d ( X x ( : ε> if and only if )) = M, d : x y U ( y N A sequence ( ∀ open in n Every convergent sequence is a Cauchy sequence. A metric space in which every Cauchy sequence is convergent, is Let ( Let ( ( ε> DRAFT are called isometric if there exists a bijective isometry f ; it is unique up to isometry. Let x 6 = . In particular, the open balls are open and closed balls are closed ∀ • { U ,f is open if and only if for every are contiunous. Let M N ) U and an isometry x . Let 1 x . is unique. A sequence ( . ) = 0, that is, ( ) of − M ⊆ ∅ b d M f N x f and isometry , x ) ( ) be a metric space. Recall that the metric ⊆ p = n d complete metric space of ( c x M, M y ε ( ( if and only if U for every and Every Cauchy sequence is bounded. Recall that a sequence ( if the set U d M, d B • ∩ M (i) A function x (ii) An bijective function →∞ lim It is easy to see that the topology generated by Recall that a set U because is, for every space that Theorem 1.7. subsets of exists an open set Note that an isometry is necessarily injective since 6 Let ( a set completion spaces closure Definition 1.2. n Definition 1.3. called a Definition 1.4. The limit called an if The following lemma is often useful. Lemma 1.5. following sense. Definition 1.6. Not every metric space is complete, but every metric space can be c in Remarks. 23 Jan 2012 5 ) is and y r, r, discrete M, d = r. x implies and radius d and radius x together with a map is called the x and radius ) = 0 for . d M. M x ∞ ∈ x, y ( < d } Last Change: Thu 7 Feb 17:21:44 COT 2013 by N M, ∈ R ∈ R is a metric space. . Usually the metric space ( → , x,y,z | x, y ) ). → y X ): − M x, z x, y ( ( × x ) is a metric space. | d d × x, y , x,y ) is a non-empty set ( 0 X d M : =: closed ball with centre =: open ball with centre =: sphere with centre { |≤ )= ≥ X, d if : ) )=2 } } } ) M, d d d r r ( x, y
y,z ( ≤ < r y, x ( x, y d ). , ( ( := sup d ) ) )= y d triangle inequality d : − N . Then ( bounded z,y =
) x, y x, y x, y ( y M )+ ( ( ( d x d d d ∈ 6 = : : : x, y . diam x, y ( . Then x with the ), )+ ( DRAFT d metric space | d M M M M M R ⇐⇒ . is called A y, x x, z ∈ ∈ ∈ ≤ x,y,z ∈ ( ( • X ) y y y d d M x { { { be a set and define on ≤ ⊆ x, x )= 1 for )=0 )= ) ( X d ) := ) := ) := N x x x 0 and x, y x, y x, y x, y ( ( ( ( ( ( ( r r r d Let metric d d d S B K r> • (i) (ii) (iii) Let Examples. Chapter 1. Banach spaces Chapter 1 Banach spaces 1.1 Metric spaces We repeat the definition of a metric space. Definition 1.1. Note that the triangle inequality together with the symmetry of A subset denoted simply by such that for all The last inequality is called It is easy to check that since 0 = ) . 2 ) is b d c , d M . N ]) c M, ]. For . )) a,b m → ∞ a,b ([ z ( C → ∞ . Then ) be complete , ϕ is a complete such that M N m } , n x ⊆ M (ˆ 0 1.1. Metric spaces b , d . Then the map N, d } ⊆ N ] by → N N , m,n ) ,...,n )+ ∈ 0 N a,b ) m m n [ x ) ˆ )= x ) is complete, ( ( ,z → → = 1 n ∈ , 1 n z ψ n M is a complete metric space. n z j x 1 ( c m , d ( M be a Cauchy sequence in . x : 2 : ψ (ˆ d ) (the set of all Lebesgue inte- | x. ]) = ( T x Last Change: Thu 7 Feb 17:21:44 COT 2013 | N b if it contains a countable dense N | d n d ) )) = j z ∈ | y )+ x ∈ )= y x a,b ) n a,b ( →∞ ( ( ) m →∞ x − g )+ ([ 1 M ( ˆ n ϕ n − m x n ( n ( g L C x , − j ˆ x ϕ x + lim n | ) x because , − : x ) x 1 n ) := lim {| ) , n (ˆ ) is + ( ]). ( separable n T n b ,T 2 x d 1 M z f n ) ( x . < ( , d ··· {| f a,b + ) ϕ N | < M ] because ( ]) ([ ( b z 1 b + n ) d ,z ∈ →∞ a C ψ ) n 2 n Z | a,b n n ( lim < ≤ ) = max 1 ,z z ([ → T y ( is unique (up to isometry). Let ( n ), )) ) C be a separable metric space and b ϕ x an isometry such that ) := ) := max x, y − = ( ( m M c ( M ) b b z d d M ( 1
d ( ( N x there exists a sequence ϕ f,g f,g ) is a dense subset of the complete metric space ( | T x ϕ ( ( ) is complete. Let (ˆ : , ϕ ∈ b 2 1 d M, d → p )). c M )) + ) M ( d d ( converges to [ n n
n T with ϕ z x z M c are metrics on N M, a,b ( ( n )= : ∈ 2 Let ϕ C n ]) let d ( , ϕ ψ ) A metric space is called b d n DRAFT x, y ]) be the set of all continuous functions on the interval [ n with • is a Cauchy sequence in ( x a,b x (ˆ d n z and b ([ d a,b )= x 1 C ([ is an isometry. m d ≤ The completion of ( ) is dense in C ∈ ) ϕ with ,z →∞ M n n ( ,z n z n ( ϕ metric space. C f,g Then Let is not complete. = lim d x (ˆ x • • b d metric space and can be extended to a surjective isometry Finally, we have to show that and that Since The sequence 8 Next we show that ( Examples. Proposition 1.9. The sequence (ˆ separable. We have shown that subset. for Remark. grable functions on ( Definition 1.8. 25 Jan 2012 ) 7 i ]). on y [ (1.1) | , ∼ ] ) | z ) m ([ n b ˜ d such that y ,y , . ]. is indeed a operties ( m n N y } x ])+ x ( , the limit in ∼ z (˜ . ∈ [ d M R d , → ∞ ] ] = [ − N − x x . ) [ ) ) ([ → ∞ n n b n d ,y ,y ,y is an isometry and that ]). n is a consequence of the n m )= x x x x [ . ϕ n ( (˜ ( ] ∼ , ⇐⇒ < ε. , m,n d ] , n → ∞ d d N | 0 | ,y y 0 ∈ ε 2 n n ([ + + z ) b → Last Change: Thu 7 Feb 17:21:44 COT 2013 , n d →∞ ( → | | y ≤ x 0 ) ) n d ) ) ) n n ∼ n m n )= )+ ˜ → y ,y ,y x , n ,y n , x ) ]) = lim n m ) = [( n n n ] it follows that y x N ,z , x y y x [ x (˜ y . Then ( ( ( x n ( , is a Cauchy sequence in [ n ,y d ] ( d d d x y n . It is easy to check that M x d N ( ( ∈ 0. Then there exists an . We define the equivalence relation − x d d ∈ − ([ ( ⇐⇒ M ∈ N ) b n d )+ )+ ) we define d M ) ∈ n n n →∞ m N n n ˜ n ε > x →∞ →∞ ) ∈ C ,y lim x c x ,y M ) , ]. Then n n n , x n c n n M, ϕ n y y N n : ( x [ , x x := ≤ ( ∈ x ( ,y (
and ]) = lim R ( n d ∈ → ) ) = lim z ⇐⇒ d d n x ) M d n n [ x N ]. On n , c → ( M ≤ y ∈ M x ⊆ ) ,y ,y ≤ d ] and (˜ | ≤ | n z : n | ≤ | n
∈ y ) . Then ) N c ( x M ) x x . Let [ n n ∈ is well-defined. ] ϕ = ( ϕ ( ( ) is a complete metric space, that c ˜ m y ∼ n y M ( x × →∞ b d b d N d ∈ ) d [ , b d n x ,y is a Cauchy sequence in the complete space n n ∈ N ≥ c DRAFT ∈ M , y x ] x m ∈ c M, . ( N (˜ : →∞ →∞ N z x n N [ { ∈ d n n ) ( ∈ the set of all equivalence classes. The equivalence class containing b ]) = lim , d c ∈ n M ] and ( ] n d n y n − ) [ m,n x y is a metric space. ∼ )) x := ) [ [ , n ) − , ] n n , ) / ]) = lim ]) = lim is denoted by [ ] x n ε 2 )= ) b ∈ x x d M y y in several steps. x ,y n [ [ M C ([ N ,y N n M , , is an isometry. b < C d ∈ ] ] n ∈ ( = ( ,y c x c M, M is well-defined. ) n x x n x ( n ϕ ( ϕ ) ) ( Let ( Let b := x d This follows immediately from the definition. Let [ m ([ ([ d x ) of the definition of a metric and transitivity of n n d b b ( d 0= d | ii x x )= d , x c | by M ) exists. • • • n M x = ( ( M ( 1.1 We will show that ( Let Hence equivalence relation (reflexivity and symmetry follow directly from pr C Proof. Step 3: for all Step 2: Let ( Chapter 1. Banach spaces Proof. Moreover, for (˜ ϕ Step 1: Proof. x We have to show that triangle inequality). Let ( and ( be the set of all Cauchy sequences in Proof. d Since ( . ) 1 ε − M M iii n + } , for = 0. ∈ x ∼ >N [ X/M ∼ } k y M 2 ] ∼ . ffices to . j k x M − sequence : Choose N ∈ ] [ ε − is Banach k·k . Since k n x x ∈ 2 k≤k [ . because for + x m k 1 y < ≥ m : +1 ∼ ]. Then − = X n k . m k k X n x j m ] ) e with x x : X/M K 2 → ∞ x k [ m k − . k ∈ X } y by [ < ) + 1.2. Normed spaces n } ∈ x m { k x < M x for all , n k n M k M . Next choose − 0 ∈ − x }| n x, M X, α ∈ 1 ∈ converges to y m x, y − . y z − m N ) ( y ∈ 2 N . Continuing this process, + {k : 2 k → ] such that ∈ ≥ − ≤ +1 x k n ,m x x N n [ ) k y x y,M Last Change: Thu 7 Feb 17:21:44 COT 2013 ∼ m m x n − := dist( containing . k . ∈ + inf k m {k ≥ ] } m n − : ∼ 1 } e x m m x k y − m k ] M x = 1.13 , x,y X [ M n y j ] x = 0. Now assume that for all + [ is Cauchy sequence in = inf = dist( ∈ X/M ≤k − ∈ m k 1 ≤ ] x } αx } N ∼ such that x − n ∼ m − ∈
k {k 2 k ] x : M , M n ] n . [ a closed subspace. Then m for all y is normed space. It remains to prove com- ) x k M x x k [ : R ∼ ≤ ∈ ∈ n 2 ] := [ 0, that is, ( is well-defined. For k k + k x ] m ∈ M x − − − ∼ . [ x x → = inf m y m . By definition of dist there exists a sequence ∼ 2 k [ n − ] : : m k m m +1 M x X/M ∼ k → [ M m x ≤ k k n k 0 there exists an n k ] k·k + − − x ∈ , α x [ m ] ∈ y m {k ∼ X/M m x x be a Cauchy sequence. 1 ∼ x y = m k − exists and : ] k x − − + − ] . Then ( = ] ε>
N − {k {k X 1 ∼ n + j n n m y x ∈ x x ∼ k [ [ N x x ] − x n
x [ inf k k k 1 x {k {k x ]) [ [ 2 ] = 0 implies ) := inf = k . Then n k·k − ≤ = = = inf
x →∞ x − k ] < n ] := [ 2 ∼ ] m y = inf k n n N x, M ] x (i) In the proof above we used that, by definition of ) is a normed space with and every ) = inf x := lim x − X/M y ) in the definition of a norm is easily checked. For property ( [ [ ∼ DRAFT − k ii k x ∈ ] + [ X such that n +2 ] x [ ] + [ such that ∈ x x, M y N be a Banach space and We already saw that k·k First we show that [ ∼ x k ∈ [ , x N k let dist( ] we denote the equivalence class of n ] k X ) x ∈ n>m dist( n X +1 X 1 n X/M, m ∈ ∈ we obtain a subsequenceconverges with if the desired and property.prove only Since convergence if of a the Cauchy itBy subsequence contains definition constructed a of above. the convergent quotient subsequence, norm it we can su assume that N Proof. First we show that we can assume such that pleteness. Let ([ Therefore Property ( x all ( Let Proof. let [ It is clear that [ we find We have to show( that space with the norm defined in Example is closed, it follows that every Equivalently, there exists an x x • • is a vector space if we set For 10 For Remark 1.14. 9 = N N k 1.10 ∈ ⊇ 0 there . Let k y 1. choose B a closed A M − closed. J = X = . The last ∈ k α ) ⊆ . A x is a countable k estriction of the 2 ) because is a map M . It follows that ii B such that n,m J ) of Definition X ) with ∈ N iii k ≤ ii . By assumption on ) . on x ⊆ k M X 2 such that − B n, x Banach space ) and ( . By definition of M k ) and ( X. ii norm 1 k iii ∈ M. . For every ( Last Change: Thu 7 Feb 17:21:44 COT 2013 . Obviously, < } } ⊆ } ) follows from ( . A ∈ i 1 ) m J N by setting K y ∈ ∈ in ( < b,y R converges in − X ( ) ) n d x it suffices to show that for every be a Banach space and ⇐ → X , x,y : ,y k because 0 = n X N n y X via the metric and we have the concept of n,m x . In particular, ( . x : ( k 1 X − k { 1 d : ( 2 X Let x ⇐⇒ ∈ < k := < k·k which satisfies only ( ) such that x n,m ) with A y R ,y y
{ ,y B K ) := n N n is dense in x ∼ → ∈ x ∈ ( := ( be a normed space. Then the following is equivalent: ∈ x d x, y B α b X d ( .
be a vector space over y induces a metric on , d k B X ∃ 0 for all ]: y )+ = 0, X · X : k X n x ∈ Let . As seen in the remark above for norms, a seminorm is non- , + N , x DRAFT . k k ≥ . k A complete normed space is called a Let k k × x 2 x Note that the implication x 0 we have the equivalence relation x, y k such that n, 1.2 k |k N • ⇐⇒ y α X ( | A ∈ and define d =2 k≤k there exists a ) . To show that ∈ = A norm on A function [ y seminorm k ≤ N =0 0 k n N N ) + is complete. · x k ∈ n,m We have to show that there exists a countable set Exercise ,y 2 x x αx ∈ ( k k Note that k k k inequality follows from the triangle inequality ( Every absolutely convergent series in X { 2 k • there exists a countable set n,m n, := (i) (i) y y (ii) (ii) ( (iii) where the closure is taken with respect to the metric on M J a such that for all exists a Remarks. d convergence etc. on a normed space. Definition 1.11. subset of subspace. On Chapter 1. Banach spaces Proof. 1.2 Normed spaces Definition 1.10. is called a Example 1.13 (Quotient space). Hence a norm induces a topology on negative and satisfies [0] = 0. Remark. Proposition 1.12. Obviously, every subspace of anorm. normed space A is subspace a of normed a space Banach by space r is a Banach space if and only if it is Remark. and Proof. . p − ee ℓ y . In ) let . We R ∞ N + ( ℓ ∈ n m such that x ) , } m N = y n ) are Banach y ∈ ∞ follows from the := ( ) is not a Banach N y y . ∞ 1.2. Normed spaces . k·k T , N . ) , p } k·k T ∈ , ( 0 and } d, <ε (e. g. an interval in , p p ) in the definition of a norm BC -th components is a Cauchy . be a Cauchy sequence in T k ε > ∞ ii } . k exists. Let m N p N Last Change: Thu 7 Feb 17:21:44 COT 2013 T , x < , m ∈ converges to } ∈ p n p ∈ , − ) k <ε ε N n,m } t is a non-closed subspace of k n . Let n ) and ( ∈ p x M ) and ( is continuous is bounded : | x x y i x ≤ n d | j ∞ ) exists =0 f f ) . p y − hold. Moreover, it can be shown that n t | p ) : : →∞ ( j n n x ≤k n − n T y f x x ∞ k·k k·k x K K ( p −−−→ is a vector space, we obtain ℓ | , k {| 6 = 0 for at most finitely many − ) B n,j := lim p n → → k,j ( n x ℓ T < →∞ →∞ x ∩ | For metric space ( m x and that x n n | n,j c T T ) y ) are Banach spaces, ( . For every x B : : : : lim : lim | =1 − T M . That ( ∞ j ∞ X := sup ( ( N k,m ε ℓ ) for a compact metric space f f x K K K =1 { { 0 C ∞ n,j T on the left hand side finally gives ∞ ≥ j X c ( ≤ . Since on the left hand side yields x k k·k − can be chosen arbitrarily. ⊆ ⊆ ⊆ | p f p
ℓ ) is complete, let ( ( and that c, ε N N N k k ) := ) := p ) := k,n BC =1 ∈ ∈ ∈ M ∈ d y p n,m j X T T T . Then the sequence of the n n n → ∞ ℓ ( ( ( x ) ) ) y → ∞ N | − k·k n n n )= C
∈ ). B ∈ k M Let , − n x x x m T p ) and ( BC x y . -vector space. Properties ( ) ( n ℓ for all because 1.5 k ∞ x = ( = ( = ( K C ∞ ℓ K n,m DRAFT x x x ). <ε x { { { k·k : See measure theory. and is a , k is complete, the limit 1.3 0 k p p ) let := := := = ( c ℓ ℓ x K T c 0 d n ( c ∈ − are closed subspaces of x spaces B ) n c p n x ∈ To show that ( Hence are easily verified.Section The triangle inequality is the Minkowski inequality (s Set inequality above since x Taking the limit sequence in Since k Taking the limit in particular, will show that L Spaces of continuous functions. Subspaces of f and 0 12 (iv) (v) (vi) space (see Exercise Obviously, the inclusions For c particular, ( spaces. Note that In Analysis 1 it was shown that ( 22 Jan 2010 30 Jan 2012 11 we p ℓ p with a ∈ |} n y | . x, y are complete , | ∞ n n . x o R and < } {| ) ∞ p p K k Minkowski inequality and < , y ∈ max . . p k n . | ,...,n 1 p ∞ p α } ℓ n ℓ C ∞ + =1 ∞ x p X | n ∈ ∈ p p | =1 < j k p j x =1 p x ∞ | | : X n k Last Change: Thu 7 Feb 17:21:44 COT 2013 n | ( =2 j : p n x =1 | j x X p is called the , x , x {| } K 1 p p =1 =2 ∞ . |} k X n T ) is a Banach space. = ) is a Banach space. p n ⊆ bounded map y p X | p p y | p k ∈ | | ∞ n N k n ∈ , K y t α ∈ | | x x | + = max : | k·k n x n k ) p , | k·k → + x = ) p k n , ∞ =1 t ∞ p ℓ ) {| x p x k T | X ( , n | ( x : T n x is a vector space. For n R ( n k x are complete normed spaces with {| x | ≤ k p ∞ { ℓ αx := → n ℓ | p
2max , =1 for every ∞ p R k n ) := X n R k y =1 ∞ C N ∼ x p =1 X ) := n ∞ ( : k := sup k X n + 2 → ] p T and
p ℓ ( x x n [ ∞ ≤ ≤ n k k ∞ . Then ( K p ℓ := C x | Finite dimensional normed spaces. k·k : k n p let ℓ DRAFT y k≥k . ∞ (i) x + ∞ ) is a Banach space. k ) is a vector space. Let 1.4 n ). N k·k T is a vector space and ( . Then ( x ( | p< p 1.3 ∞ ∞ First we show that ℓ ∞ ℓ ℓ =1 ≤ be a set and define ∞ X n := T p< Exercise supremum norm ∞ ≤ With the usual component-by-component addition and multiplication scalar, have Proof. and and For 1 ℓ Sequence spaces. • • (ii) Obviously, The triangle inequality normed spaces with Let 1 Examples 1.15. Obviously, (see Section Chapter 1. Banach spaces (iii) be the (ii) Let Proof. . . - ) . ∈ 1 2 1 ℓ X ℓ 1 1.2 such such (1.2) − = k·k m U , X 2 . 0, so by U 1 e ∈ U ⊆ − 1 1 + m m N N + 2 ∈ k·k V ∈ −−→ − ) is uniformly u n n V 2 such that ( ) . ) ) n if and only if it 1 n + ed by equivalent n be norms on y N if and only if it 6 = R u 1 − x 2 k·k 2 ∈ x 1 . The the following m ∈ 1 1.2. Normed spaces . . m th unit vector in 2 -limit and the n = ( } e 4 X X, n ]) k·k ] 1 ( k·k M u 1 k·k m , = . . Then , a,b a,b → n [ + 1 k·k X. ([ V ) x − and 0 such that 1 m ∈ 1 1 even odd ∈ be the 2 ∈ m 1 t C − , it follows that 1 for all 2 1 : n n N u Last Change: Thu 7 Feb 17:21:44 COT 2013 . Therefore k·k ∈ ℓ ∈ > 1 |} k·k n ℓ , n = (e + 2 := ) ) 2 X, , n t k 1 n n m ( 6 = n 0 m , x m,M > n ′ v 2 2 y 1 with respect to y . Let e u x ( k | 1 k 0 , x ℓ x mv , = ( = | k ) (1) v t n = and e . Let k ( , M ). It is not hard to see that 1 x n x m 7 U k − , hence ≤ ( 2 N {| 2 = and in this case the v m N 2 . + 2 ∈ 1 be norms on a vector space 2 k 2 . Let ≤ , x u = ∈ k if there exist 1 x V 2 converges with respect to N max ℓ 1.15 m n + { ∈ m is a dense subset of (2) x m converges to k·k k·k 2 ⊆ n 6 = 1 k k k·k ≤k , X ) } u x − 2 ]) define the norm n 1 X U
U N exists. Then there exists a sequence ( 1 m k + x m ⊆ >n 2 4 ⊆ x and be a normed space and ∈ ∈ + ≤k a,b := sup v ) is clear. m 0. This contradicts 2 k N M 1 − 2 N = ( 1 ([ n k ∈ V iii v X = 1 ∈ m
: 2 − n ( (1) (2) n x = n are equivalent. ) 1 k x k·k k C m ) n = are closed subspaces of 1 n k·k k 2 x − f 2 e n −−→ ⇒ x k − Let k { x m m V ( n ( Let 2 2 On m )= DRAFT y 2 equivalent norms x x k·k ii u be as in Example . It follows for all ( . Suppose that there exist = and u 1 ℓ 0= 2 (1) ⇒ and = 1 and )”: Obviously it suffices to show the existence of U . Then e + ) i ∈ ( 1 1 N v 1 m )= k i x k·k n ( ∈ = ⇒ (2 . Since span x limit are equal. A sequence converges with respect to converges with respect to A sequence k·k x k U m )= + (i) iii (ii) (iii) “( Clearly that Let 14 Obvioulsy, are equivalent: The theorem above implies in particular, that the topologies generat is true. Assume no such norms coincide. Moreover, the identity map id : ( V Proof. Theorem 1.18. impliying that continuous for equivalent norms. Example 1.19. assumption also Now we will show that Definition 1.17. and let They are called that . n is N ]). 13 f ∈ k Y with ) a,b k . N = ([ ]), so it a finite n 1 ∈ a k (1) = 1. Let Y C ) a,b N k . + ([ ∈ , see as the n N } k·k 15]. a C g / ). Then there N § . Then the in , 1 2 (1) ) x 2 a Cauchy sequence . − . N k·k → } n , ∈ Hal98 N n n ]) + ) x → ∞ ∈ z 2 , therefore t n -norm. But a,b in the supremum norm. A . } a Y ([ ∞ y , n N 1 ∞ + ∈ n k Last Change: Thu 7 Feb 17:21:44 COT 2013 n ′ C 2 ∈ → z , hence y ) := ( k·k f , n t y ] a real interval. We can define ′ n k 0 nx ( a closed subspace and converges because it is the sum of x , n n = ( = + = f a,b → N ′ Y N 1 ∈ x ∞ =0
, ∈ k in the . To proof that it is closed, we proceed and − k k n ) K n n | ) f n z is a closed subspace. In particular, every X t x 2 2 is continuously differentiable Let [ n | k k x x x x n k f → : : N → a n 1 : := 1 1 is continuous. a )= n 1] + ℓ ℓ − t ′ K ,
x ( (1) and ( k 1 f Y ∈ ∈ g k n = } → − N N f is bounded, it follows that x , ]
k ∈ ∈ K :[ K N k n n ]) will always be considered to be equipped with the ) ) n ∈ ∈ is closed, this implies = ( ) is a Banach space. Note that the right hand side is n a,b k n n y → shows that every finite-dimensional subspace is closed. f ) λ N . Then x x a,b k :[ Y (1) k
( ( : ∈ is a subspace of n 1 } be a Banach space, be a Cauchy sequence in ( ([ 1] n k { { X f a 0 1 , let ) x { ]) such that N { N 1 λz k C X k·k ∈ ) is not a Banach space. + := := { n N , − ]) let n + = y a,b DRAFT z ∞ ) ]) ]) is not closed as a subspace of the Banach space U = V :[ ∈
n ([ is unbounded. Then there exists a subsequence ( Y ]) := unless stated otherwise. Let Y a,b consider the subspaces x g is bounded. Then it contains a convergent subsequence ( C n N a,b N ([ a,b k·k 1 N . Since ( 1 , a,b (1) ∈ ([ ℓ ∈ ([ ∈ 1 n 1 ]) ([ C ) = 0. Since ) ∞ n 1 For Let ( C ) In C n . ∈ k·k n C x, y a = a,b a X . f | z, Y ([ ( k 1 N Obviously, n d such that C a Spaces of differentiable functions. | Hence Proof. converge to is not a Banach space. ( norm In the following, For Proof. Then ( finite because by assumption well-known theorem in analysis implies exist + X Y • • ase 2. ( ase 1. ( ∈ →∞ lim by induction. Therefore we can assume without restriction that dim k finite-dimensional subspace is closed. Proof. several norms on the vector space dimensional subspace of Then the sequence ( Example. z two convergent sequences. C Theorem 1.16. C in Chapter 1. Banach spaces (vii) Note that thefollowing sum example shows. of Another two example closed can subspaces be found is in not [ necessarily closed Hence Finally, choosing closed in . , X ε. := X for = 1 with 1 2 − (1.4) X = αB k is said 1 (or an 2 > } 6 linearly X x αB 2 . > k X k ∈ 1 X m , x ⊆ n − = 1 and set x 1 ). Therefore k x k Y − y { Schauder basis 1 with 0 also n . The existence 1.4 x + λ x is called k k x X ≥ ξ Y ,...,x is a k 1.2. Normed spaces Hamel basis 1 X α ∈ ∈ N x = is dense in ∈ z 2 := span with ⊆ with n x ∼ ) 2 Y . A subset Λ k n X X ] U ∈ x K ξ λ [ ∈ ⊆ ) k . λ 1 N 1 does not contain a convergent K Last Change: Thu 7 Feb 17:21:44 COT 2013 x x ∈ − . Set x n k ∈ ) . 1 2 y ) n n j . x α > + x that is linearly independent and such X, ( . , in contradiction to ( ξ k 2 1 ε 2 1 = ( 2 1.20 X B x = 1 and for every − . Choose x )= with ≥k > the following are equivalent: ⊆ k 1 , then, by Riesz’s lemma, there exists an − n Y =1 ) ∞
x [ j Λ 1 } X X z k > ∈ = x k ⊆ λ } contains a convergent subsequence. k n y x, U ) 6 = x span( Y X λ X is compact. Then there are + n X x Y U B ∈ {z ξ is compact. α ∈ X ) be a normed space over z } B : =1 ∞ 1
−k X n . If k y | k·k } is a (finite!) linear combination of the z if be a normed space. A family ( is not compact (Recall that a compact metric space is n k≤ . Therefore, the sequence + is compact, then obviously for every ). Obviously, − X, X x y
X X ξ , such that m X k x = 1 and dist( . By Riesz’s lemma there exists an X
∈ can be written uniquely as is a family ( , in particular : + B 1 6 = k B X {k . 2 1 x ξ Let Let ( − x X n ( X X ,...,x n )” follows from Theorem For a normed space k k , 1 = DRAFT 1 > ii y ∈ ≤ ∈ − x ( ,...,n is compact. Since every bounded sequence is a subset of some ) ∞ } 6 k { x ) of 1 + x 1 U y } if every finite subset is linearly independent. A ) = inf { with ⇒ x )”: If ξ =1 X )”: Assume that dim ,U { k + i )”: Assume that iii 2 N i j ( X < B := )= ( ξ ( i x x, Y = k ∈ ( ∈ 1, X be a vector space and Λ a set. A family ( such that = span “( k ⇒ = dim d ⇒ ⇒ if every = z x Every bounded sequence in B : X X U X x ) = − := span k≤ )= ) = n,m X j (i) dim ∈ ii ii iii x 1 (ii) αx x (iii) k Let of x Hence it must contain a convergent“( subsequence. Continuing this way, we obtain a sequence { to be a total subset of independent that every element Definition 1.24. “( that is, if the set of all linear combinations of elements of “( 16 Let Theorem 1.22. dim Let Proof. k of a Hamel basis can be shown withDefinition Zorn’s 1.23. lemma. U and dist( elgebraic basis all sequentially compact). subsequence, hence ). is is 15 S = 1 } X (1.3) ional / ∈ k x = 1 1. Note a closed k 2 = 1. By k is a norm, 2 x ∼ X k m of Heine- is bounded, k k x of we use the ] he unit ball is k at in a certain ξ ⊆ , the quotient S C [ k·k <ε< mplete because = 0 but 0 : k . Using triangle X -vector space are = Y n al complement in x e K 1 2 such with X n define S, 2 α X ∈ k n . Since ∈ e j ∈ =1 x e n j M n { such that X. x k α . P = 0, thus ∈ ε n . := =1 X . In addition, =1 k = j X x n j , x 1 Last Change: Thu 7 Feb 17:21:44 COT 2013 − x R . 2 ∈ S x k 1 1 2 k 2 1 P ξ in x k < = 2 2 } | | , x and j j k 2 1 x with M be a normed space, y { k α α | | X x S = + is uniformly continuous, so its restriction k X n n ∈ =1 =1 ξ and a maximum j j X X T . For M M x n m Let ), is closed and different from ≤k independent of K := are equivalent. k≤ k≤ ∼ 1 2 1.3 Y k≤ x x 2 and it suffices to show that every norm on k n ] j k . Then there exists an 2 e ξ such that K x 0
[ k X ≤k ≤k k k |k j Y be a norm on j is continuous, hence 2 . By( e ). m k k α ∈ k ε > | x 1, the assertion is clear. Now assume 0 x R 1=
?? = y k k 1. Since k·k n =1 =1 2 ≥ n j be a basis of j m X = → k and > ε } has a minimum x P ε. ≤ n k DRAFT ε X X e S or 1
− and inequality, H¨older’s we obtain − . Let m j 1 , T x : All norms on 2 is a norm on } e 1 6 = := R 0 j 2 2 { > ,..., C Y α k·k with the Euclidean metric). In particular, the unit ball in a finite k·k there exists 1 → ) is compact by the theorem of Heine-Borel. Now consider the map e = is not trivial. Hence there exists an ) n n =1 k·k { k·k 2 j X K S Y
x, Y 1.14 k·k Let If = X/Y k 0 (otherwise there would exist an X, dist( x with the Euclidean norm is complete and that a subset of a finite dimens k : ( n inequality for with constant equivalent to therefore T Obviously, Chapter 1. Banach spaces Theorem 1.20. The theorem above implies that all norms a a finite-dimensional Note that that in this case Proof. Theorem 1.21 (Riesz’s lemma). Borel for normed space is compact if and only if it is bounded and closed (Theore subspace with Proof. compact if and only iffollowing the theorem dimensions which of is the alsosense space of quotient independent is finite. interest, spaces as For caninner it the shows product pro work th spaces as (see a substitute for the orthogon dimensional space is compact. This is no longer true in infinite dimensional normed spaces. In fact, t and equivalent. Moreover, it followsK that every finite normed space is co space closed being the preimage of the closed set to the compact set and by the homogeneity of the norms Therefore m> Remark ∈ N =1 ∈ . (1.5) (1.6) n p ) ). n For the y = 1 1.5 . q 1 = ( ) , i. e., 0. Since the 1 ab + p − , y p 1 p yields ( N ∈ ) with = R n and ab > = ) q 1 ) = ln( n 1.6 ℓ b → q } } x | q q j ∈ R k y and y | . = ( N {z . k 2 ∈ x = k q q q n =1 ) + ln( ∞ | {z y k =1 j ) X | j a k y n ≤ ∞ y q q Last Change: Thu 7 Feb 17:21:44 COT 2013 | k 2 y For k q | p k 1 n k j y q . 1 . x x y k y q | q ≤ | k k ) clearly holds. Also the cases b + ) = ln( q 1 1 . y q = ( q ≤k 1 1.3. and H¨older Minkowski inequality p p p k = b | 1.6 +
k z p j j =1 + x =1 Let k } } p x y ln( | | k p x j q p p 1 j q 1 q a 1 k , b x 1 p x x | ) is clear. Now assume p | + 1 p {z + k ≤k j k , then =1 ∞ . The Young inequality ( = 1 p ≤ q )+ x p 1 x j X =1 1 ∞ 1.5 ≤ | {z =1
ℓ j p X | k k such that ∞ q a z p p | ) ∈ k ab k j = k := = 1 is follows that y 1 y y ln( x ∞ k a 1 q i| , k | | | p 1 p j
k b | ≤ ∈ p, q = j q 1 : = 0 then the inequality ( x y 0 p j + y Let x 6 =0 and 1 DRAFT | ≥ p . If a =1 ∞ 1 p j X x, y a,b = 0, then inequality ( are clear. = 0) =0or q ln ab k x 1 ∞ ∞ y k If If 1 = p k p x the inequality H¨older implies k 2 setting Then for all Taking the sum over gives ( and Now assume 18 1.3 and H¨older MinkowskiIn inequality this sectionproof we we prove use inequality H¨older’s Young’s and inequality. Minkowski’s inequality. Theorem 1.27. Theorem 1.28 inequality). (H¨older’s In the special case Proof. Corollary 1.29 (Cauchy-Schwarz inequality). yields Application of the monotonically increasing function exp : Proof. logarithm is concave and ℓ } 17 e Q ials ∈ , that X 0. Since , λ j such that N = a | ε> ∈ K j A ut not every λ n ∈ there exists an , it follows that : − n and -vector space. In A n j . span A C µ ∈ . We will show that X | λa ∈ p y { ℓ ∈ n x =1 is a → ∞ j X 6 = ,...,λ x := 1 X . Then x ε. λ , . . . , n. p B if + ℓ , n i s not countable. Obviously, = } ε 2 0 . Let Q ⊆ =1 N ε 2 and Last Change: Thu 7 Feb 17:21:44 COT 2013 }} . X N < 1 . ∈ → such that + i . Let ∈ + A , = 1 for ε 2 ∞ p n = n 0 Q X e j ε 2 ) Q ∈ : . Then obviously ∞ a j n < B k j n , j ∈ e n x < ∈{ := X p< k y 1 j λ th unit vector in x is separable if and only if it contains a n j k n x − n e Q a − ≤ ) n . Then for every x = ( n =1 j 7→ a j x X : k +1 λ ∞ x k j k n ℓ ∞ N ,...,a a − X k·k = , x 1 n ∈ =1 k n j =1 y j X ,...,µ n a j R X ) 1 X, | n , that is, there exists no countable dense subset ) be the =1 ( n . Let − j µ j X → p + = A x λ x ℓ . Since ( ] ... and p k { 1 2 − j ε 2 0 a j A j a,b , j < [ e := µ < 1 j | λ { . | x , ∞ A j ]. is separable for 1 -vector space and 0 X k n =1 span n λ =1 , there exist x j X n R =1 p j b max j X ℓ ∈ X a,b − , there exist − [ j is countable total subset of DRAFT − j + − K a ,..., C x is a µ x (i) j ε 2 | x A A normed space be a dense subset of k µ X ≤ B if := (0 =1 k≤ be a dense countable subset of is countable. We will show that n j is a total subset of y n Q A e } B P ] is separable since by the theorem of Weierstraß the set of polynom − N := is dense in x . Let is not separable. := such that k a,b ∈ Recall that the set Let Let e [ Q e ∞ Q y is a total subset of n ℓ ∞ B C . ℓ : p A ∈ has at least the cardinality of ⊆ ℓ is a total subset of n e x A countable total subset. Proof. Since Then b Now assume that Proof. is, Chapter 1. Banach spaces Theorem 1.25. Note that every normed space with a Schauder basis is separable, b (ii) (iii) separable normed space has a Schauder basis. Examples 1.26. where Proof. B A of both cases is a total subset of { DRAFT q 1 q { 1) p }| − p z ( 19 | are j (1.7) y ∞ + j = = 1. The ). : x Minkowski’s | p 1 q N p ℓ 1.7 =1 M ∈ + j X ∈ p 1 M 1 p x, y p = 1 and | j p y and | 1 p =1 M j X − k finally proves ( ) such that p x ≤ ∞ | . k j p ∞ p Last Change: Thu 7 Feb 17:21:44 COT 2013 + ) yield for all y , k + 1 q ≤ y + (1 p . 1.6 → ∞ k 1 j k 1 q q { ∈ x x + 1) M q p p | | p }| | − For j j p k y z ( ≤ k y | x | large enough. Hence the above inequality j 1 1 p + y =1 − M . Let j X j p M ≤ k p | + | x j ∞ | j + j p y y k x 1 | =1 y M + − + j X p j | j =1 + M j x j X 6 = 0 for x y | k p x j | and inequality H¨older’s ( = 1. Taking the limit + j y | | + x j j K | p + x x q 1 k | DRAFT | j 6 =0 and 1 x =1 x M − | j k X =1 =1 y M M j j 1 X X = 0 then ( =1 + M j p ≤ = ≤ ≤ y x p = P | + j q p x y − If + p j x | =1 M j X using Note that yields Chapter 1. Banach spaces Theorem 1.30 (Minkowski inequality). triangle inequality in easy to check. Now assume Proof. inequality holds: . ) T n . x k 1 , and − T X, Y ( k n L + := ∈ k n S = 0 and that y . k S,T T T . ). Then k≤k X. k Sx. . Let T k} ∈ + x N + Y,Z + k k ( ∈ S S . L . This is can be made linear. The following is M ) k n 2.1. Bounded linear maps Y ∈ λT x } } Y , x o T x 1 S k k≤ ( k≤k := x S → =1 we define ), Last Change: Fri 15 Feb 10:55:26 COT 2013 ) is a vector space. From defi- and x 6 =0 ) are obvious. T x T x ) k≤ k for all kk is indeed a norm. Note the the ii k K k := X x x ( S X = 0 if and only if T k k : X ∈ X, Y + X, Y = 1 yields + k ( ( ⇒ X, x > n kk k λ T ∈ k T k·k L X, X, L S k ∈ x k in contradiction to the continuity of x )= λT 2.3 , or similar notation. Sx n k ∈ i ∈ ∈ ( ) is a Banach space. x ( ∀ ) is a vector space. N : x x : T ) T x that : : Z, STx ) and ∈ k≤k ⇒ k X,Y k R ( k k k Y, with k≤k n L x → X, Y ∈ )= T x 2.6 is not bounded. Then there exists a sequence T x k X, Y ( k T x X → T x ( k X, Y T x iv L T X ( ( T kk L M n k ∈ : {k {k + L X { S x ) as composition of continuous linear functions and ⇒ ∈ = 1 and ): . To prove the triangle inequality let Sx ST 1 for all k k S K )= n = sup = sup = sup = inf k≤k be normed spaces, we have seen that . be normed spaces. X,Z iii > + x = ∈ S, T ( k 0 ) is an algebra. k k k L T because by Remark λ n 2.4 x X k λT ST x ) ( k ( ∈ k X, Y X, Y T L T Ty k be normed spaces and + (i) For ST DRAFT kk Let Let for all S S is a normed space. obviously is linear, ( k such that k )”: Assume that ) T S 0 but iii X is Banach space, then . ( |k k≤k + X,Y,Z λ → ⊆ | X X, Y is continuous. is bounded. is continuous in is uniformly continuous. Y ⇒ ( n (i) In Remark The implications ( ST N λT ∈ y = k explicit using the notation Obviously, operator norm depends on the norms on L If In particular, T T T It will be shown in Theorem Since the sum( and composition of continuous functions is continuous T ∈ x n ) = k ) ii n (i) (i) (ii) Let (ii) The following is easy to check: (ii) (ii) λT (iv) x (iii) in 0. Then ( and 22 Theorem 2.6. “ ( Taking the supremum over all k nition of the operator norm it is clear that Theorem 2.5. Proof. equivalent: Remark 2.4. Proof. 21 let } 0 . Note k \{ T X X ∈ kk x N ∈ . For a linear operator norm ∈ . The set of all x K n . ) K } n X k≤k <ε x e x is the k 6 = 0. 0 T k x T Y . kk T x k } x = 1. For → k open in − X. k =1 } ∈ x = and X ), i. e., k T x k U Last Change: Fri 15 Feb 10:55:26 COT 2013 : k x ∈ k T T x X, Y ) ⇒ k k ( x X, ( = , x L k f = 0 or ∈ linear and continuous : x x x <δ is unbounded and T : kk X : we find k k T T 0 ∈ k Y x T x T x k − { → {k k≤k x k X is denoted by for every convergent sequence ( T x )= : k n bounded linear operator Y U x 0 : T ( be normed spaces over the same field := sup operator norm 1 { be normed spaces over the same field → − k →∞ δ > T T X )= n k ∃ X, Y X, Y ). 0 ⇒ lim DRAFT is called a X, Y T ( Let Let is continuous . By definition of (i) For a continuous linear map T ε> define the L x X,X = ∀ 1 ( Y Y n − L open = The inequality is obvious for k X then → → x T x Y ∈ k X ∞ ) := 0 X ⊆ : = x : X →∞ < . ( e x ∀ Proof. that the inequality is also true if T n U k T k L T T k (i) k (ii) lim (iv) (iii) Definition 2.2. If linear continuous maps of Remark 2.3. map and Recall that the following is equivalent: Chapter 2. Bounded maps; the dual space Chapter 2 Bounded maps; the dual space 2.1 Bounded linearDefinition maps 2.1. N ∈ k≤ n be a ) x. ) ). It n n ξ nt X . because . To see − T x − k e or. X . The se- T ⊆ k n x T , ( M N x x is complete. of ∞ T D . ( exp( k ∈ k T Y 1 . By what was = n d equal to the and a sequence k≤ M T k n , x k≥k , it is continuous. k : T } X e k = T k N k | = T →∞ ∈ ∞ ) := j ∈ =1 ∞ k n t k k b α ( x T x. k b n | k T n n because n = lim x k : = which converges to x k n k =1 T x k Y , x 2.1. Bounded linear maps j X 1 1 n n n k D . For R X, b and that M T ξ T T x ⊆ . ∈ → Last Change: Fri 15 Feb 10:55:26 COT 2013 . which converges to T − N ≤ | T sup x all norms are equivalent, we can j ∈ n 1] →∞ : Y n D α , n = sup ≥ | ) a Banach space. Let k X k is bounded with = 0, the following operator is well which converges to n T x j ⊆ = lim k e n x =1 : [0 k } Y finite-dimensional. Then every linear e T Tx j X ) N D 0 n T n n ∈ {k ξ = X x n →∞ |k \{ T x ) n j = − n . Then there exists exactly one continuous α j x | ) 1]) } n e is linear and bounded by . Then , →∞ . Since on for any ( x j } n n ( e =1 T we find α X j X k =1 ([0 = lim 1 D, Y n , hence, by the uniform continuity of x . Then lim k ( n ≤ =1 n C x j X D Y L →∞ T x k lim Y ,...,n ∈ n Sx ∈ is a linear extension of k x j D, e →∞ : , and therefore it converges in be normed spaces, . The extension is bounded with =1 e T T T ∈ n →∞ k T Y T j n ∞ j be normed spaces, x ∞ : k := lim k = : α be a basis of of and X = lim k x X, Y Tx k k k n j k n ∈ =1 ) Y e e e choose a sequence ( T x X j X n Sx X, Y j T x ξ T DRAFT ) converges in is bounded. e Let → X n j is an arbitrary continuous extension of {k − {k let = Y sup α which converges to X ∈ is the unique continuous extension of ,..., S T ξ As in the example above let n be another Cauchy sequence in : 1 Y Y, x x b → =1 k D T ( N b n j T ∈ → X ⊆ n T x kk P : := max ) Let e For k N T X n follows that Proof. ∈ = k T ξ : n M x ) e T n →∞ x lim defined: It is not hard to see that n that indeed equality holds, we only need to observe that by definition 24 Lemma 2.8. Theorem 2.9. Finally we give a criterion for the invertibility of a bounded linear operat map Proof. is a Cauchy sequence in Proof. Let ( quence is a Cauchy sequence in extension said before, ( dense subspace of assume that Assume that Therefore, hence the supremasupremum of of both the closed sets). sets Since without the closure are equal (an ( Let for 6 Feb 2012 , . p ≤ < k T 23 ∞ ) k . It ′ ∞ (2.1) → k x =0 ... ′ k n . x < ε 0 p T k , N + k N ε 2 0 0 , implying , ∈ ∞ = k ∈ p 1 k ℓ follows as in n , , ). Obviously, x = ∞ ) : is an isometry k k ≤ ∈ k , the sequence T (0 . p nt is bounded and x ) k k 1 ∞ R X x , m n ) − := C k T x ,... → ε T 2 k , X n = k 2 1 p n ∈ n ,... C x k p − + Tx 3 ,... , x T ) shows that are defined by k exp( k → k k x x 1 0 k ) x ) is a vector space, also 1 p , , x n x k ℓ 2.1 T x 2 0 , x X k , m x ... ≤ k . ( (0 T k 0 sup (1 p X, Y x. is not bounded. N. , x ) := ( k L with n 1 because − 0 t 7→ 7→ ′ ≥ k ). For ( Last Change: Fri 15 Feb 10:55:26 COT 2013 , L ′ T ≥ = 1. It follows that x kk = 1. N N n x Lx (1 ∈ ∈ T x m k } k ∞ = k≤ k n n 0 = T , x that k L →∞ ) ) = 1. X, Y T k n n R ( = k − k \{ N such that L is bounded and ...... k≤k n p R T x , T x := lim → 3 3 , T x k x T ≥ ) k . k = 1. N 1]) T ) ) n , . , n,m . , x , x ∞ k n 1] X T ∈ ∞ p 2 2 , ε 2 ℓ T ... ([0 ∈ − k 1 N , x , x 0 =1 k≤k because < k·k k·k x , 1 1 : [0 x x , C , 0 ] x x Y, Tx k ] 6 = id m m Y , ( n ( + 1 and N ∈ m T 1 T x → , 1 T k n a,b a,b x ∈ RL − ([ : ([ (0 for all − + n X = , , is bounded with x C C L : : 1 1 ; in particular p k p n ∞ n 1 1 let it follows for k ℓ ℓ p C k but x T C T on the right hand side yields is linear. That T are well-defined and linear. Moreover, 1 C n 0 exists an k T k → → 1, so that altogether k k k k 1 m p x . It is easy to see that X x Tx ℓ ) x → → ) T k L n p T 1+ k k 1 k ≥ k x p p ∈ ∞ ε> is bounded and ) C ℓ ℓ k≥ k − T = ≤k x . The left shift and the right shift on : : k → ∞ is a bounded linear map. Since DRAFT = id L T and p k k·k k·k k ∞ L n 1. Since T x R ,... be a Cauchy sequence in ) is a bounded linear map. In addition, ( , 1]), , m k In the following examples, the linearity of the operator under con- R T 0 , ′ n LR = sup = sup , 1] 1] ≤ ∞ ): For N Rx x T , , 0 − 2 ∈ k k k p ([0 , ( k ≤ n be a normed space. Then the identity id : The operator − 1 k≤k T T ) ([0 ([0 L + (1 ≤ x k = 1. 1 C 1 n k T k n is a Cauchy sequence in X T k 1.13 ∞ C C T ∈ is complete, we can define k : N : + ( id . n x ∈ − n Y Obviously, T because The operator Note that Proof. The left shift is not an isometry because, e. g., 1 = To proof that we also have that k x k T n ) T x T x = n (i) Let k → ∞ (ii) Let 1 (iv) (iii) T In particular, for all Since Taking the limit It is easy to check that ( T Example n sideration is easy to check. Chapter 2. Bounded maps; the dual space (ii) Let ( Examples 2.7. follows that ∈ = of K x − 1] ∞ t , k d y [0 1 k t ∈ ∞ , d if , ! k n N k ∞ t , s dual space k 0 k ∈ Z x 1] , k s =0 , 0 ∞ n ′ , n ∞ Z ([0 k X P C ∞ k 1]) seminorm ) is the k k , ∈ ≤ x K ′ ∈ s k converges so that id with the norm ([0 such that x ≤ ∞ X, n C 2 ∞ ( k t is a a X R k y K d ∈ L k ∈ R | , x Last Change: Fri 15 Feb 10:55:26 COT 2013 . ) has exactly one solution ) . kk X. =0 } t := , x n ∞ n ( 1 ≤k M → ∈ ′ 2.4 x 1] K dual space defined above. t P , | | X X k d ) y. k≤ : [0 n 1 x =0 p t ∈ k s,t ∞ n K ( , x k ) d functionals P k | X, =0 1 . ∞ x s t X n ∈ ( 0 k X ≤ Z x topological = x , s X, dual space, i. e., the space of all linear maps ) M ∈ : 1 ∞ x X ≤ ∞ ∈ | ≤ k ) if there exists an sublinear functional t,t ∈ t y . In particular, x ) x ( 2.2. The dual space and the Hahn-Banach theorem n k ( x , x k ′ . ∞ . Hence equation ( ! ( ) d t k x 2 K , x n t K n ∞ p 2 , x,y k 0 s ( 0 k ) k {| Z x ∈ algebraic be a normed space. k y =0 be normed space. 2.10 ) ∞ ) ≥ ( k ∞ n k p bounded ! X x n X s,t s,t k≤ P n s k ( ( n = sup , λ The dual space of a normed space k k )+ ) , λ 2 ∞ k K s s x Let ) x Let ′ k k ( 0 0 ( |≤ = x x k ) DRAFT p p Z Z ( k | k s ( λ is called ∞ ≤ | λp is a well-defined linear operator and for all x = = = k implies immediately: ) p n | | . x ), then it is called a y ) ) i K k K s s ∞ )= )= | 2.6 + ( ( k x in general is larger than the y λx λx x 2 k ( ( ( Kx p p p K | 1]), given by K ∞ | , satisfies k → k p (i) ; elements in the dual space are called ([0 (i’) (ii) k instead of ( C e 26 Obviously, 2.2 The dualDefinition space 2.12. and the Hahn-Banach theorem Definition 2.14. Moreover, A seminorm If is a Banach space. Proposition 2.13. Repeating this process, it follows that which shows that Note that in general the X is invertible by Theorem Theorem X 8 Feb 2012 ) ∈ 25 SR y 2.2 such (2.3) (2.2) (2.4) ) 7→ ometric X ( . the series converges S L ) and 1] n 2 , ∈ T 2.6 1] [0 , and T ∈ =0 . ([0 ∞ n 0 C RS is invertible and N P . t, s is a Cauchy sequence 1] ∈ ∈ T 7→ , N and ) d . Then − k t ∈ [0 0 ) ( m id N x ∈ X ) , S ) = id and Theorem ( j ∈ . Let 1. Then ) T 1 L T , m s,t X X − < Last Change: Fri 15 Feb 10:55:26 COT 2013 ( m − ( is invertible and formula ( ) k . , +1 , then k =0 k L n m j s n 1 , s m (id T T T 0 ) T T k T s Z P − < → ) holds. ( n converges. y − =0 k y −k be a normed space and =0 ∞ T ) X n m n m 2.2 = T = ) := (1 X k X T =0 = ∞ s ( P t X n 1 L )( Kx ) = id =0 − Let := k≤ ) gives is invertible in ) d ) T ∞ m t 1 − ( = Kx T m − T − ( x 2.3 x P ) ? n S by assumption. ) ) can be written as an equation in the Banach − − y T T (id , in ( id − s,t 2.4 (id m ( =0 . In particular, ∞ S 1]) k X n n → ∞ , (id s ) k ) the maps k = 0 because T T m → ∞ ([0 Z X m − ( C − S m 1]). If a solution exists, is it unique? Can the norm of the L ) for ) , (id → s is invertible and that ( T → ∞ k≤k n is a Banach space and that ) is complete by assumption on DRAFT ( n is a Banach space and ([0 − ∈ T x T X m 1]) T X ( C , − k X R =0 (id L ∈ ∞ n converges. Then ([0 x n 0 for P C 1]): : T , Note that equation ( → → K =0 ([0 m ). Since The proof is analogous to the proof for the convergence of the ge m ∞ n C respectively are continuous. S T 1]). We ask if the equation X P , ( L (i) ([0 (ii) (iii) For fixed Proof. space Hence taking the limit Note that: where series. We define the partial sums In particular, if Chapter 2. Bounded maps; the dual space Theorem 2.10 (Neumann series). Application 2.11 (Volterra integral equation). in that converges. By the first part of the proof, id Solution. Now assume that holds. in norm because implying that id solution be estimated in terms of has solution C as )) ) of p y ) for )) ( R y x 0 ( ( ϕ 0 because p ) X, . ( y ϕ . ) X ) (i L x R | ≤ ( V ) ∈ bx p i ∈ 2 x i Re(i ( V − (i = )) + Re( ϕ − ) | . a,b x V as a vector space ) i ) ( x Y. )) 0 x ψ (i − y ( X ϕ α ∈ ( V i ) i 0 with . such that . 1 (e is defined on ϕ ) bx b ). In summary, we have Re( ∈ D − . ϕ R x z (i which is bounded by p ϕ ) α ( x )) (i + i y ∈ such that V 0 p y p ( ≤ V , y ( a ϕ 0 i α ) V ϕ )] ϕ Last Change: Fri 15 Feb 10:55:26 COT 2013 y ) )) = )+ = − ) for )= ( x x )) = Re( y . This -linear extension ( z p x . ) ζ . ( y ax )+ ( α and ) ϕ 0 x R i V − Y x (i ( (i x ψ ( ϕ ( 0 ( (e |≤ R X p ∈ V V ) and ϕ i V ) = Re( y y ζϕ ∈ ∈ ) + i y V ≤ )= )+ ( ( − x X . 0 x x x ) 0 α )= ) ) ( = ( (i z ϕ i Re( x y ∈ )) = 0 ( is an extension of ( )] = V y − ax ) [ 0 − vector space. Consider ( x ( 2 b and x c αϕ ϕ + (i | ≤ | ϕ V x, y )) ψ α i , V x y Y V )) , f i ( y ) R (e )] + 0 ∈ , ϕ (i( ( has the upper bound )= )) = x ψ − ϕ ) = 0 ϕ 2.2. The dual space and the Hahn-Banach theorem ( C y ⇐⇒ V ). By Step 1, Φ is not empty and partially ordered 0 z ) → (i , i . ϕ ( bx ( ) . Now define ϕ ) complex D x 0 )) = Re( c V To see this, let x, y 2 (i ( y → y Y i 0 − y ϕ ( ψ X ( ϕ . : Φ Re( ) V 0 0 ϕ − | − ∈ + X y [ 0 ∈ )[ V is a < ϕ ϕ ) To prove this, let ψ : ) = Re( b V ) =Re )+ x = x 1 + y -linear continuous extension of . x ( )+ αx ϕ | X ( ϕ p (i ( x )+ + i ) C ), x ax V ( 0 )= 0 ϕ . ( ( x y [ x a yields ( V ( f To show this, let ϕ ( a V ϕ ϕ α )) + i Im( i Z ( 0 i 0 p y z DRAFT is a V ( D = = = ( ∈ − | is bounded by the sublinear functional 0 − αV ) )= ϕ )= =e z |≤ 0 ϕ y y | ) = ( V ) ζx ), ) (the domain of x 0 ) = Re( ( + x z ( -linear. ϕ V ( y ( ϕ ( x C V p ϕ ( | Let Φ be the set of all proper extensions of | = Re( + and define the functional is bounded by ϕ is -linear because for all )= is an extension of ∈ D |≤ y ϕ R R αx ϕ ) ( ϕ ( x with z ϕ 0 has the following properties: ( 0 V ψ Step 2. all | by 28 Application to Every totally ordered subset Φ By Zorn’s lemma, Φ contains a maximal element In conclusion, otherwise, by Step 1, it wouldNow not we be assume maximal. that (iii) In addition, desired. over By what we have already shown, there exists an It is ϕ (i) V (ii) . p → 27 b. (2.6) (2.5) ). X Y, z . : , := can be ( ) that is, ) ∈ p p ( 0 0 } 0 ϕ Y as desired. λz λz Y )= 0 ∈ + vector space. + 0 6 = 0: ϕ y x y λ ( λz ( Y, p . We distinguish p dmits non-trivial ): real , that is, , y,x + ∈ ) 0 X Z . 0 y z )+ ( z )+ R Y 6 = ∈ p y is a y + ( ( + . ∈ ∈ 0 0 x ≤ Y is linear, then ,z X y is continuous in 0, hence ( Z ϕ ) ϕ ( p c z p − K λc − , y ∈ ( ψ , y,x p Y, λ R ) z = = )+ be normed space and 0 → )+ | ∈ Last Change: Fri 15 Feb 10:55:26 COT 2013 )+ x z ∈ ) 0 X. Y. ( y is an extension of ) |≤ which satisfies ) z X X x 0 ( and every sublinear functional ) . We will show that + 0 0 ( ∈ ∈ : c 0 a linear functional on ϕ z } z z + x ϕ X ( ψ ) clearly holds. For ϕ X ( ϕ c 0 Let + x + p {− , Y ( ψ ϕ ∈ , y | 2.6 y on 0 y 7→ | p ) )= z 1 λ 1 x λ inf 0 . Obviously 0 ϕ { )+ ( λc, λ ( ≤ , y , x R p x ) ) p λz ) λz ( ) holds for all y x y 0 ∈ }≤ ( − must be of the form ( )+ + + p ϕ p ) y )+ Y λ − , x 2.5 y ( ψ y y y ( 0 ( ≤ R x 1 ∈ λ such that 1 λ p c ). ( := span ϕ |≤ ) ( ( ≤− } p ψ c ) x 0 and y 0 → 0 0 for every ) is sublinear. If ( x Z ϕ ϕ 0 { |≤ ( ≤ 0 = 0 equation ( ): Y ) )= z X ≥ a subspace and ϕ ) ϕ 0 )= X 0 0 − | − = λ ) z y ∈ z + ( x and λz λz X c y ( such that ( y + X − λ λ ( p ψ ′ + + x x Y p ⊆ ≤ ( ≤ ( Z y y \ 0 p ( ( ). For − Y with c c ∈ ϕ DRAFT 0 ) λa λb with − there is nothing to show. Now assume ψ ψ X ϕ y 2.6 | Z ψ ) ) ( ≤ ≤ 0 x ∈ )= X ∈ ] arbitrary. We show that then ( ϕ y 0 . We have to find λc λc 0 0 ( = − ϕ z 0 R a,b linear [ λz ϕ Y Observe that ∈ {− ∈ + 0 : 0 : K − has an extension to a functional c Let c y ) . 0 For ′ → x ϕ ( = λ< λ> Z 0 Y z := sup ϕ : ∈ a seminorm. Let a c 0 We have to show ( ψ is sublinear. Remark. Now let Moreover, note that every seminorm is a sublinearThe functional. next fundamental theorem shows that every normed space a functionals (except when Let Theorem 2.15 (Hahn-Banach theorem). In both cases we obtain Chapter 2. Bounded maps; the dual space For example, every norm on implying so that By assumption on R for some Obviously, every linear extension of Proof. ϕ Then between the real and the complex case. First assume that extended to some We divide the proof inStep two steps. 1. . . = 0. with (2.7) 0 . We p c ℓ p . } U ℓ | 1 4 ′ ∈ ∈ x ) ) =1 < n n | k y y . is separable. ) x p n } . To this end, k x q = ( = ( : ( of ℓ ′ . Assume this is =1 6 = 0 and x X | ∈ X , y , y ′ k 1 q ′ x ∈ x + ℓ ℓ x = , x k = 0 this is clear. Now k ′ ∈ ∈ k { : ′ U th unit vector in x x ) ) ′ y . n n n p . := − X x x k ′ n y 2.3. Examples of dual spaces X ∈ x k k≥k such that S = ( = ( Last Change: Fri 15 Feb 10:55:26 COT 2013 . ′ a subspace. Then the following q , . . For is the q ′ ′ x k x x ∈ ℓ y { T x x X X n ¨le conjugate H¨older k n 6 =0 =0 |≤k ∈ S ) separable = x ⊆ for for := n n n ≤k ∈ =1 ′ such that x ′ X Y ( q 1 X T x n q ′ x , x . We will show n n S y , x x q } | + y y q n . and ⇒ k | , where e ′ n n n N x = n x 1 p + and and that x , x N x x X . x | ′ | ∈ 0 S ) =0 ∞ ∈ T is called the X =0 =0 n ∞ X n ( ∞ ∞ n . This leads to the contradiction X X n n of x n q : and that k≤k 4 1 ∈ ( ′ } := . n = = = q ), x p< ϕ ℓ define x N | T x < n n is well-defined by inequality H¨older’s and y y is clear. The inequality above gives { t k y ) ) − separable k ≤ (e ∈ =0 ) ∈ N ′ ′ T there exists an ) ′ T , 1 x y N n 1 n T x T x ∈ . the unit sphere ∞ T x ( ( ∈ X : x span ( − be a normed space, n n be a normed space. ( | ∞ := ) ′ n Let 2.19 ′ n ′ n =0 1.9 = n , , x n X x x ′ ′ X { x x ϕ ) ) k U (i) 0 p c ℓ Let ( ( )=0 ( ) Rud91 ∈ | π ϕ ϕ if and only if T x , X ( kk x x ( ′ ∈ X ( ( M x . To show that in fact we have equality, we recall that . In particular for all . ϕ T ϕ , p ) for all with be a normed spaces. } k ∞ X ϕ x ϕ ∈ { y X R x ′ and be the canonical projection. Then such that ( with kk be a normed space, ′ ∈ ′ , see [ be a normed space, = ϕ k be a normed space, by a sublinear functional such that p ′ ): ′ X ϕ } =1 holds because X → ϕ X X x p X X X k = = 1 and M X X, Y k X ∈ linear { T x = sup )= ∈ ∈ ∈ 0 ϕ ( ): X k k X/Y ∈ DRAFT is a normed space by Example x ∈ k 6 = 0 and ϕ k ϕ x y x ϕ ϕ ( Y ϕ : ϕ k≤k Let ( Let there exists a ϕ Let Let , { T k ϕ ϕ → ′ X )) p can be extented to a functional ϕ k kk → 0 T x { ) X 0 x X y = 1. By Corollary k≥k := span ( ⇒ ∃ ( ϕ X 2.17 : kk ∈ ϕ 0 ) is proved. Note the the supremum is in fact a maximum. π k = ⇐⇒ ∀ : exists i ϕ ( k 0 := sup Y π ϕ ϕ If in the Hahn-Banach theorem we consider only real normed space ) for all ϕ y ϕ T = sup x . Obviously, Y k ( ) For all k ): ≤k k M i such that such that \ =0 6 = q ( Let Let The map x x ) T , then ′ ′ 6 = 0. Since |≤k ( For k x x k . ) ≤ X Y ) ϕ and 0. Then there exists an X X y y T x ) 0 { ( = ∈ ( k x x ∈ ) Let (i) (i) ∈ 0 ∈ is arbitrary, it follows that − (ii) (ii) ( ( ϕ 0 x ii ϕ the desired properties. Statement ( ϕ replace the seminorm y desired properties. ε sup ϕ such that by Corollary k Proof. ( M Chapter 2. Bounded maps; the dual space Remark. The next corollary shows that Corollary 2.18. Corollary 2.19. π formula in ( Proof. ε > The Hahn-Banach theorem has some important corollaries. Corollary 2.16. x x there exists an extension Corollary 2.17. Proof. that it separates points in ϕ Proof. | other hand be extended to a q . ce o , hat is and , ) we have (Ω) ∞ p ′ K ( T L C ∈ Banach space p < ∈ ≤ X. , g , g 1 ) ∈ be a compact metric (Ω) ). The K , q ′ ( , x K ′ L Y M Y X, Y ∈ ∈ ∈ ( , Let ′ ∈ L ′ ∞ . ∈ Last Change: Fri 15 Feb 10:55:26 COT 2013 ) » T µ, f Y p< , y µ, µ » d , y ) d , is linear and isometric, that is, ′ k ′ ≤ y X, Y g g ( T T T x f 1 ( L Ω ′ in pairwise disjoint measurable sets Ω kk Z . y ′ Z 7→ ∈ ) y T . Then K K K T := , )= ( q )= g x =1 ,T g -finite measure space. Let ) L ) M 2.4. The Banach space adjoint and the bidual ′ )( k≤k ′ σ q 1 T and )( y ◦ ′ T ∼ = ∼ = ′ ) + y Tµ ′ ′ ,X T ◦ Tf ′ ( X ) ( = ′ ( 1 p ′ )) p be a Y y is clear. Immediately by the definition of x Y,Z ( k be normed spaces. K ) partition of L ′ ( ( ( L be normed spaces and T L = ,µ , C Z , ′ , ′ ( ′ : k ∈ Σ → ′ 7→ )) , | X y , Theorem 6.19]. ) S ) ′ X, Y T K (Ω such that X,Y,Z ( V (Ω)) T → ( p k ′ q . Then C for µ ( Let | X, Y L ′ Y Let Let the set of all regular Borel measures with finite variation, t ( ( . In general, it is not surjective. Rud87 DRAFT : S =1 k ′ → ) L ∈Z ′ and X → T ) 1 q T V is linear and continuous as composition of continuous functions, hen T K ) and the following diagram commutes k ′ ( is ′ K ∞ n = + ( T = ′ T M (Ω) p 1 ,X ) with q ′ k M ) Linearity of ′ p< i of : Y L ( T ∞ ( ST : The map k ≤ T := sup L < 1 T k ∈ k µ (i) ′ (ii) ( µ k that is an isometric isomorphism. For a proof, see [ such that T Theorem 2.26. 32 Theorem 2.23. The theorems above show that 2.4 The BanachDefinition 2.25. space adjoint and the bidual adjoint Proof. Obviously, Let space and k is an isometric isomorphism. Theorem 2.24 (Riesz’s representation theorem). , , } a is is 0 q 1 th 31 c N } 1 n T ). It ℓ = N ∈ n 1 p e p n ∈ ( ′ : for every − y n n is the n ) 1 e n : e k n orems. n . n { . Since x 1 α n t ) e 1 , that is, ℓ e ℓ k n { =1 x ∈ N α X ∼ = n k k≤k ′ ) = , x k ′ = exp(i 0 ) where e exp(i T x ∞ k , and y c | is a total subset of ℓ n k ) N . } =0 ∈ n (e ∞ T x X n N e ≤k n N k ∈ ( y ′ ′ ∈ ∈ n y y | n , . , := e n 1 p Last Change: Fri 15 Feb 10:55:26 COT 2013 n : ∞ n , y n is separable, in contradiction to e t . )= 1 ′ k x n q k ′ n | k e ′ y ∞ y n =1 x N { y p< ℓ (e , N X n k x ′ k = | y 1) ≤ ′ ∞ ≤k ) such that n = − k y =1 1 N q e n 1 q y ( and let X n R ∞ α p n k ′ . In particular, with the inequality above, | ) it follows that q are clear. Moreover ′ x ) | )= k n ∈ 0 ′ y = n n n x T c y 2.7 . , | n | = x ( 1 q = | (e exp(i n α ) e ′ ℓ ℓ n x ∈ =1 N n y | ≤k =1 X n =0 ′ N ∞ n ∼ = ∼ = α )e T x X n X n t y 1 p . To see this, assume that ( ′ ′ =0 = ∞ 1 ) ) X n = ℓ T x 0 p p =1 . As before, since p N | c ℓ | | choose X n k ∞ ( ( exp(i n ) ′ n ≇ t k t n y | N | = ′ y =0 ) ∞ would imply that also (e X n n ′ ∈ =1 N =1 t N ∞ ≤k y and the proof is complete. (Note however, that X n X n ℓ | n ≤k n ′ k 1 . Since ( x ′ , y k . Together with ( DRAFT k 2.21 q y =0 n ′ ∞ ℓ k x N X n y y =1 N .) = k we find n X n k , it follows that ∈ ∈ . For = ≤k ∞ is surjective, let T x x p = 1 is similar. q 0 ℓ k | ℓ = = x T x c p N T n . and =0 q = ∞ | = x X n 1 | p n Note that ( ℓ k ′ x 1.26 | − =0 y ∞ ∈ X n because x =1 large enough, the last factor in the line above is not zero, so, using 1 N pq X n ≤k 1 ℓ N q k ) Well-definedness and injectivity of ∈ x ii is not dense in it follows that Hence follows that Example unit vector in k x Other important examples are given without proof in the following the total subset of To show that ( Chapter 2. Bounded maps; the dual space Using The theorem above shows that Remark. an isometry. The proof for implying that we obtain For separable, Theorem Hence, for all 15 Feb . . . , ′ ∈ is ′′ ′′ 0 y T x )] ◦ U ) = S X ′ x Y J x ∈ J ( )( . )= ′′ ′′ 0 = 0 and 0 T Y because = u ′ x , ◦ U ( ) (but 0 | ⊆ X X is reflexive, ′′ x Y X ϕ J ′′ ) x K, x J J ∈ ◦ ( X ′ X X ′′ ). By definition ( ′′′ 0 , x ′ and let ′ → T x = [( x. ∼ = is surjective. Let , S ′ . ′ , X ′ ′ X x 2.16 X k y . ′′′ X X ′ X ∈ ) )) x ∈ = 0 )) = : ′ such that X , that is . Since ′ x x ′ such that x ′′ via the canonical maps ′′ kk ( . Then ( ′′ 0 x T x ′ = id ) ] u ′′ → ( X x X ′′ u X ′ X x u ′ Y J ∈ Y . X ( X J ∈ ′ ∈ is reflexive. X , x )= J J . For ′′′ 0 X, Y ϕ ) )= ′ 0 ′ 0 U (Corollary : ) = 0 by choice of such that ′′ Last Change: Fri 15 Feb 10:55:26 COT 2013 ( ◦ x 0 )) is linear and bounded, hence ¯ Y U x ′ x 0 k≤k x | ′ X X L ( . Let x J x ∈ and ′ ′′ ′ x X ¯ ) = ( ( ) if and only if ( U = ′′ X X » X ′ ( U x | Y » J ∈ Y ′′ ′ » ϕ u )) = [ ( J X J → ϕ » x ∈ x ′′ x T x J ′ 0 T X ( ( ( u ′′′ ′ x ϕ X, Y kk . Let )= X . Therefore ′′′ 0 y ( ′′ X ) = = J ′′ ′′′ 0 ϕ x L , u . Note that with this identification u : u ( ′ = x x ′′ ) ]( )= ( T ′ ′′ 0 T x 0 ∈ T ′ 0 ′ X ) x x X x x ′ x )= ′ 0 J ( J T y )= ′ x |≤k x ′ )( X 2.4. The Banach space adjoint and the bidual ) = ( ) ( x u ′ 0 x J ′ 0 T )= ( ( U ¯ X x | and U ( X ′ ′′ J | X ′ ¯ , x J x )= X J ϕ ′ )= ′′′ ( X X ) = [ ′ ( X K u ′′ J is closed, there exists a x J ′′ y )= X ( ′ ′ u ′ ]( u | ′′ )= → ′ U x such that T Y u ′ 0 ( → x = ). On the other hand x ) ′′ 0 is reflexive if and only if ′ is reflexive there exists an ∈ X ′ ( be normed spaces and ))( U | is linear and bounded because )= x ′ ′ : ) ′′ Every closed subspace of a reflexive normed space is reflex- x X X ′ be a normed space. Then y ( ′′ 0 u X X such that x ′ 0 ∈ J : is an extension of x ( 2.19 x ( x X ′ ′′ ◦ )= . Since (i) ′′ X = ′′ 0 u J X X, Y 0 u X x U ′ and T ′′ X x | J . It remains to be shown that ∈ DRAFT ( J x be a closed subspace of a reflexive normed space / ′ ∈ Let ) Let U X x [( x . Since ′ 0 ) = ( 0 U ′ ′′ x ∈ ∈ x are identified with subspaces of y ( ′ 0 X and choose an arbitrary extension , then x be reflexive. We have to show that ′ x X Y Y . The map ))]( ∈ J ) Let ) is adjoint operator of some U J . We will show that x X i ′ ′′′ ′ ′′ 0 ). Obviously, ( ( For Note that ∈ x U X ive. A Banach space it follows that X X and | ′ ,X ) = 1 (Corollary ′ J ′ 0 and u 0 ∈ ( x ∈ x X ( Y x ′′ ) Let ( (ii) ( ′′ ′ 0 ′′′ 0 X ii T Assume that that is, the following diagram commutes: hence ϕ We have to find a there exists an u Proof. 34 Note that there are non-reflexive Banach spaces Lemma 2.33. If Theorem 2.35. of not surjective). ( x x [ Proof. J Therefore L Let Lemma 2.34. Proof. , : ). k b- 33 p x l )= )) = ∈ 2.17 n . T x can be . Since x ( N T x ′ ′′ X n ( ∈ ) X k≤k n S X ) ) of ( ′ x n Ry ⊆ z ( ( x ε X, ∞ ′′ k y )( x X ′ ): ′ X Lx ∞ Ry ϕ X ∈ ), the closure T , hence S ( X | ,L ′ = ( ): k ). k n T x x p ( x T x y ∈ → = = ( ℓ 2.6 y ′ ( be a normed space. = p< k 1.7 y − − n chosen at the beginning of the proof. y N < = N )) n n ) ′ y ∈ ) | ∈ x x X ψ n is reflexive, there exists a ,j < converges. Now the “diagonal sequence” ε x 2 ) ∈ n ′ ( ( n 1 y : X is weakly convergent. Let k y 1 2 is separable. By theorem ( k n Y m j N ′ ( )) | + X k ′ ϕ ϕ y ψ x ′ y ∈ x n M ( ( ( ( j x ϕ x X ε 2 y 2 2 k ( is not separable. Let )) ′ ϕ ϕ − < ≤ ,j x →∞ converges weakly, define the map ′ lim ). Hence ( | ≤ | n X − x m n ) N DRAFT n k ) y ∈ m . Since ( x n . Then ′ n ) and reflexive (Theorem = y ( ′′ ′ y ) ( x be a dense subset of ( | is bounded, so it contains a convergent subsequence ′ m n ) k X X ′ x } y ϕ N ϕ 1.25 x | ∈ N ∈ ∈ ( →∞ − ′ n n ) ψ ∈ x | ψ has the desired property. )) n such that n y n ). Therefore we also have such that for every ( : ) = lim x ′ N 0 ,m ′ ( y x x n x m 1 | ( ∈ . Let ( ′ n ϕ ϕ Y Now assume that show that ( By what is already shown, x there exists a subsequence ( 36 First we assume that such that the bounded sequence in Hence ψ This implies that ( (Theorem { of ( Now the sequence ( quence Continuing like this, we obtain a sequence of subsequences x such that ( Now we will show that k ). . is is 35 ′ ′ 2.17 2.21 X X m. . Then 1 ) and the ℓ 2.35 ). is reflexive. ) belongs to ∼ = t 2.24 ′′ ( be a bounded onsider for ex- limit. ′ 0 y 3.12 X c 1] the sequence converges weakly . Sometimes the , X is separable. 7→ to the norm in the er than weak con- ∈ is reflexive by part N [0 ′ ion ∈ ⊆ 1]). Then the follow- n ϕ X X ∈ , t ) , points (Corollary N , 1] the point evaluation n K t , ∈ X x ([0 n . [0 J ) ′ → C n ∈ was shown in Theorem X . Let x 1] 0 0 , ∈ c t X Last Change: Fri 15 Feb 10:55:26 COT 2013 ′ = ( in norm convergent is separable if and only if x N 1)]. to obtain that ∈ , is separable. By Theorem X n , x 1)]. ) ′′ ) , [(0 n 0 2.21 X C x [(0 ( ′ ∈ ). C x y ∈ ?? . y via the canonical map )= x n ′′ ). By assumption, [0 x = t ( X ′ ( implies separability of be a bounded sequence in n x y ′ x N . X ∈ is used. Note, however, that in spaces of linear operators n ′′′ 0 be a normed space. A sequence ( →∞ → ∞ lim ) x n - lim n n X x w is reflexive. By what was is already proved, ′ )= ′ 0 DRAFT or Let A reflexive normed space X x Every bounded sequence in a reflexive normed space contains a )” It is easy to see that for every (i) If the weak limit of a sequence exists, then it is unique, be- Let ( ( x ii ′ ( X be a reflexive normed space and converges pointwise to converges weakly to w −→ J ⇒ N N n X ∈ ∈ if and only if )” follows from Riesz’s representation theorem (Theorem converges to some x n n i ) is a bounded linear functional. Hence for all )= i ) ) ( 0 is a closed subspace of ) = 0 but the sequence of the unit vectors does not converge in nor N t X n n n “( ∈ That separability of Let ( strongly convergent x x ⇒ n X x ∈ 1]). (e is separable and reflexive, then also ) , t 0 ϕ ( )= x 7→ N X n (i) ( ii ([0 (ii) ( ) of the theorem. ∈ x i lim Proof. Example 2.39. ing is equivalent: (ii) Every convergent sequence is weakly convergent(iii) with A the weakly same ample convergent sequence the is sequence not of necessarily convergent. the C unit vectors (e to n ( x sequence. Notation: separable. Proof. “( C Since given Banach space, then the sequence is called If it should be emphasised that a sequence converges with respect notion the term “strong convergence” has another meaning (see Defint Chapter 2. Bounded maps; the dual space hence indeed Corollary 2.36. Definition 2.37. Theorem 2.40. weakly convergent subsequence. Proof. Now assume that The next remark showsvergence. that strong convergence is indeed strong Remarks 2.38. cause, by the Hahn-Banach theorem, the dual space separates If Lebesgue convergence theorem (see ( reflexive, so we can again apply Theorem , a x n k·k Y A (3.2) ∈ and of maps x f . Λ 0 such that R ar bounded ∈ rtain ball. us functions λ ) λ r> f , . X such that = ( . . be a complete metric X x x ∈ such that N 0 F X R x ∈ be a Banach space, < C < C , . x ∈ . k k } N is closed because n Let Λ X ) ) < M. n } x x M ∈ k ∈ ( ( n ) . Last Change: Tue 26 Feb 11:55:21 COT 2013 f f . Let x is dense in N k≤ ( N ) f A k≤ such that does not contain an open set. x = ∈F ) A complete metric space is of second 3.2. Uniform boundedness principle ( X, λ is of second category in \ 0 x f A ( ∈ k R M X a family of continuous functions which ∈F k ∈F k f : . k ) , if r> f f : R ∈F k X ∀ ∀ X f X ∈ and X, Y in ∀ 0 0 x ( ∈ if there exists an { M, x , n ) be a metric space. A family λ X, Y n ) be a metric space. ) is satisfied with R R ( A the set ∈ f | A X X L ∈ ∈ x N 3.2 X, d ∈ ∈ X, d ∀ nowhere dense ∈ M M x x is of first category in n . By Baire’s theorem exists an ∀ ∀ F ⊆ n first category second category let Let ( Let ( Q is pointwise bounded, there exists an uniformly bounded DRAFT A N N F ∈ , that is, ( is of is of is called ∈ n is nowhere dense if and only if N n A X X X A a normed space and ⊆ ∪ ⊆ ⊆ ⊆ is called ⊆ For Y . Since X ) A A A R 0 X x • • • ( → ∈ r B Note that are continuous. Since all x hence X Note that for every Theorem 3.6 (Uniform boundedness principle). The next theorem showson that a a complete family metric space of is pointwise necessarily bounded uniformly continuo continuous on a ce 3.2 Uniform boundedness principle An equivalent formulation of Theorem 3.3 (Baire’s category theorem). category in itself. Examples 3.4. 38 Definition 3.2. Definition 3.5. The Banach-Steinhaus theoremfunctions. is obtained in the special caseTheorem of 3.7 line (Banach-Steinhaus theorem). Proof. space, Then there exists an Then there exists an normed space and pointwise bounded, i. e., is pointwise bounded, i. e., 22 Feb 2012 20 Feb 2012 . . n N } n , 37 . ∈ x ) 2 A n . 1 ) (3.1) N < r n ≥ X ,ε ∈ →∞ x k 0 n n and ), hence ξ x 1 T ( ε 2 we obtain ε − . n := lim B , ) − 1 x 1 y 2 ∩ k x , N ( ,ε 1 ) : , we have that ), hence < 0 1 B A x N ,ε is dense in n X ( ,ε 0 ∩ ∩ 0 n x be a family of closed B ∈ 2 2 x ( n,m > N <ε A ( N A ∩ ξ B . ∈ { B 1 ) =1 n ⊆ ∩ ) ...A A ∞ n ∩ ,ε ) is complete, 1 n 0 2 1 ∩ T ∩ A ) := B x with 0 . This implies that ( and all A 1 2 X ,ε ( ( 2 − ⊆ N A N ∩ B N Last Change: Tue 26 Feb 11:55:21 COT 2013 n x x, r ∈ ) 2 ( ( n be a complete metric space and 1 ∩ ⊆ A ) is open and not empty. Hence ≥ ∈ ) B B 1 ) 1 2 ) ,ε n . Then ε ⊆ n because for fixed 1 1 2 x A N , ε ) ...A X . Since ) is open and not empty. Hence there x 1 let , 1 ( N ⊆ x 1 X, d ∩ ,ε − +1 ( ( B and ) x 0 X n 1 ∈ ( N 1 B x and ( − has non-empty intersection with ( − N ,ε ∈ B ,ε N N ∩ 1 2 N Let such that B 1 X x ∈ ∩ ∈ − ) implies 2 A x n 2 ∩ n < 2 ( ) A 1 x A ⊆ N ) n 3.1 A B ( A ε ) contains an open subset. Then at least one of the N ∈ such that B 1 ⊆ ⊆ 0 and 2 − 1 n , x ) ) ∩ x 2 N ) for n 1 A B 2 n ε x because, for fixed N , hence 2 ,ε ,ε r > ( , ∈ A ε ) for every 2 =1 1 , hence 1 d X , 1 ). . Hence ( ∞ n x N X − x ⊆ ( x X ( ) and N S ,ε N N ) For ,ε ε be a complete metric space and x B 0 )+ B x n 2 . ( N . ( ) x ≥ N − DRAFT ( ⊆ x B ,ε X B 2 ( n ) n 3.1 B , , x ) and ∈ ∈ 2 B x ε 2 X, d ⊆ ε m ( ∩ (0 ( 1 0 ) n , x ) if ∈ x − B n ( 2 x ∈ N 2 such that 2 N 2 d x A ε 0 ε 2 , Let ⊆ ( , ε x , N ) X ≤ (0 B ∈ N N be a family of open dense subsets of n 2 n ) x contains a non-empty open subset. ε 0 and ∈ x ( n ( N , T 1 n ∈ n B ε B , x n ∈ B x ε> ) is open and dense in ( ∈ is open and dense in m ∈ y n x 2 1 B y n ( A Taking complements, it is easily seen that theTheorem. theorem above implies Proof of Theorem sets A Observe that subsets of ( A Chapter 3. Linear operators in Banach spaces Chapter 3 Linear operators in Banach spaces 3.1 Baire’s theorem Theorem 3.1 (Baire-Hausdorff). Let exist so We have to show that any open ball in d there exist x In this way we obtain sequences ( exists and is a Cauchy sequence in . , . ) ⊆ N X N for ∈ N ∈ ∈ ) m ∈ t ) equal. n x X, Y n ) , if and , if and ( m n x T L T , so that positivity T ( N ∈ s w −→ = ( −→ ∈ T ) = sin( n leading zeros) n operators on a x al. t T . ( ) a sequence of n T ′ n 2 . Now let = 1 for all x Y or converges strongly N or ) ( which implies that k , ) for ∈ N X, Y T ) ∈ x T ( t ∈ C n n n = L T ,... = ) 2 k ,... X. . n n ] ∈ n 0 X, ϕ T T k ≤ exists. Then π ∈ T . , , x ( N x 2 = 1 for all , if and only if 1 ∈ n T x ∈ , → ∞ for all T . ) = cos( n n → ∞ k kk t [0 , x - lim ) n T Last Change: Tue 26 Feb 11:55:21 COT 2013 ( - lim n n s 0 = n id N 1 ∈ T w =0 T the space of the continuous func- be a normed space and ∈ x n k , x < C − n ] . Then k T N ,...,x , 3.2. Uniform boundedness principle , x 2 Y π k 1 ∈ n T , on a function space is called 2 n sup n ,..., =0 x := lim T 1 , T − k =0 →∞ , k T k [0 n | n ). )=1 = ( T x ) , t,s t T C = (0 k ≤ T x ( = 0 k s x x 0 T x x = j n − n , denoted by 2 x ( − n X, Y . T , denoted by x ( T ϕ t →∞ X k lim 0 in the domain of n be a sequence of positivity preserving operators T L X n 2 − T and to ) X,T ≥ with such that k ∈ ∈ ) the limit X,T to converges weakly to 0 but →∞ Let x X f n x T R ( n → X X T , denoted by lim → be normed spaces, ( →∞ T L lim = lim ∈ ( ) = sin T → ∞ n ∈ be a Banach space, ∈ X s X ϕ k ⊆ is well-defined and linear. By the uniform boundedness ( | C x n j : t : to X x y for all N X, Y n T T x n ∈ k T . Then for does not converge strongly to 0. →∞ T x lim n Let N n ) j N ), ∈ 0 for every ), Let n → x ∈ N m T N n ( DRAFT ≥ ) ( x ( ) and let 2 → 2 n m n ℓ ℓ ] converges weakly converges converges strongly T converges strongly to id but j does not converge to id in norm. x T π Tf x N N N = 2 = N ). n T ∈ ∈ ∈ if , (i) The limits are unique if they exist. ∈ = ( . Let n n n T = 1. Then ] n X [0 ) ) X ) such that for all x ) π n n n k n ) We define the auxiliary functions It is clear that 2 X, Y T T T x , ( T only if only if k so that ( for ( Then Let Let , that is, L [0 id • • X, Y ∈ (i) ( ∈ ( (ii) ( (iii) ( Proof. to such that Proposition 3.13. t L tions on T Proof. principle, there exists an preserving Remark. The reverse implications are not true: (ii) Convergence inStrong convergence norm implies implies weak convergence strong and the convergence limits and are equ the limits are with Theorem 3.14 (Korovkin). bounded linear operators and space of continuous functions. An operator 40 Definition 3.12. We finish this section with a result on strong convergence of positive . X X 39 ∞ d. ℓ ∈ ⊆ , but ′ ) . x d 0 } ⊆ x 3.9 ( n define the r . Then the B . For N X ∈ ∈ ∈ F such that 0 m x ′ M, f X =: k is bounded. Therefore, is the Kronecker delta. . and ⊆ ′ ) ′ and Corollary ′ } . Then the following is N ) and r ′ M rx . Then the following are M such that N 0 ∈ )). For ) is not bounded in M 2 m,n n 5 x X A. δ X − m ∈ ) ∈ ( ( R 2.40 r n ϕ 0 ≤ ∈ n ⊆ ∈ x x ⊆ is a Banach space is necessary. ) B Last Change: Tue 26 Feb 11:55:21 COT 2013 ( ( k ′ 1.15 ): f ) ∈ A M X A ) } n where 0 − x is bounded. rx x is. , f be a weakly convergent sequence in x ( ) ( ) ′ is pointwise bounded by assumption. } } 0 r − 0 N {z is bounded. x m,n . x B x ∈ A ′′ 0 { M, a ( ( M } n ∈ 6 = 0 for at most finitely many x | ′ ) f M X ∈ f ( mδ ∈ for all n k A n f a x ′ ⊆ k≤ k x r 1 is bounded. ∈ m ) )= : )= ): A ′ a } + n the set n ( = a ∈ a ): x N )” is clear. The other direction follows directly N ′ k ( a ( X ′ < M , hence the family ( (e k ) x ∈ ii ) ( 0 J )) f x ( ): X ∈ n m k m ) k { x a x be a normed space, ) m ϕ rx ( ( n ( ∈ n x ′ ( ϕ ⇒ = ) : f = ′ x ( X →∞ a { f lim k x k f J n by { k n k ( ) = k a x m be Banach space and i = ( k·k r r 1 1 k K { we find )” the assumption that ϕ be a normed space and the set i k x ( X ′ = ≤ X, { → ( X the set k X . ⇒ d DRAFT the set ∈F ) = d and n : x Let ∈ f ( x ′ )” is clear. Let d )= X ′ Let ∈ Every weakly convergent sequence in a normed space is bounde m d f ii such that is uniformly bounded by , the set ii x ( k ϕ ∈ x ∈ is bounded and there exists a total subset → ∞ be a normed space and ( R F 3.8 x - lim ⇒ n N m ∈ X ∈ w )” The family ( ϕ ′ i n )= = 1 and ( is bounded. i ) is bounded. = is bounded. n M k ′ “( The implication “( Let By the uniform boundedness principle there exists an open ball 0 A x x ⇒ A For all x A For every k )= (i) (i) (i) is a non-complete normed space (see Example . By hypothesis, for every ii (ii) (ii) ( (ii) for every fixed equivalent: Obviously X Corollary 3.11. Theorem 3.10. Proof. Proof. by Corollary linear function following is equivalent: Proof. d equivalent: For example, let “( showing that Chapter 3. Linear operators in Banach spaces Proof. Corollary 3.8. Corollary 3.9. The following theorem follows directly from Theorem Note that for “( from the Banach-Steinhaus theorem. By the Banach-Steinhaus theorem there exists a Hence with and an . ) 0 s ( n (3.5) (3.3) (3.4) ) and ε> =1a D b x = , . − s a d ) s 6 =0 =0 ) ) let l sums of the 2 1 s s 2 )) ks ı 3.5 + ction. where. Finally, e kt , s ) n ) = cos( n s =0 b 2 sin 2 ) ) X k 2 1 s 2 . ) sin( + sin(( )) 1 2 ns n i . . ks 2 th partial sum ) sin( − kt x 2 sin( = n a + e sin(( s n ) = sin( )=1 1 2 ( s t k ( d + 2 s. b ) + sin( ks Last Change: Tue 26 Feb 11:55:21 COT 2013 ı n x )+sin( d . s/ e ı( ı )= b kt s n )) − − )+ )) ( t ) e e − )= =1 3.2. Uniform boundedness principle n n kt = kt X − s − − ks k D ) cos( )cos( x, t s 2 s we define the a ) ( 1 2 ( ) d 2 1 ks cos( sin( s/ cos(0). For the proof of ( n k ı s cos( N s -periodic continuously differentiable func- k k e + ( b a n π n =1 ∈ )= n =1 ( = with 2 ı( n k ), cos( X k (cos( D e n ks s i n =1 )+ π P 2 1 + π X k n n 3.3 =1 =1 − s converges uniformly to − X k X k kt ) d is continuous and + 1 2 be a Z + s = x ) d 1 2 ( +e + + n 0 1 π ) s 1 R n 2 s t ( a cos( ı 2 2 1 1 D 1 = − n D k . (e − + → 1 2 ) ) a π D π s ns -periodic continuously differentiable function. Let ( s s s ı ) n )= − =1 ( ( ( t R ı2 π e + X k Z n x x x e =1 : n X k 6 =0 2 1 + . Then, by ( x, t 1 π π π π π π π a 2 ( x ns s s R − − − i + n ( 2 + Z Z Z R − s )= x 0 1 e 2 2 s 1 1 1 a π π π Let π ( π DRAFT → − n = = = = Z R -periodic function and )= D )= : π 1 0 π Dirichlet kernel ks -periodic we obtain x → x, t π s ( )= n cos( s Let Using the trigonometric identity cos( is 2 x, t n =1 x ( X k is called n s + n 2 1 that Proof. Proof. tion. Then the Fourier series of constant function on Note that lim For a given 2 the Korovkin theorem impliesFourier that series the of arithmetic a means periodic of function the converges partia uniformly to the fun 42 functions whose Fourier series does not converge pointwise every Theorem 3.17. Lemma 3.16. D Now we calculate for ; δ 41 , in } . 2 0 such . , hence ∞ N. ] , x k 1 π N n ] ] ≥ 2 δ > T , π π , x ≥ e will prove sly differen- 0 2 2 t continuous [0 − , we can find , , x n , n ] { 0 function. Next N. ∈ [0 [0 π x 2 ∈ ∈ n ≥ by definition of , → ∞ span [0 xT for all , , n < ε. k ] , n ∈ ∈ ] <δ ε , , t | π )) t t . 0 ) π | + 2 y y ) s + 2 , N N kt ( s n , t ∞ , s,t ( ) for [0 x , t ∈ ∈ ) k [0 t 2 x 0 s αT αy sin( ). Hence we have that ∈ x − ( x ∈ ) s t − k αy ε, t ) + + t n ( b t ) Last Change: Tue 26 Feb 11:55:21 COT 2013 t ( T 0 t 0 because by the inequality above + αy ( x x )), hence 0 = 2 αy sin( x − t n + )+ | X s, k s, k 0 ε ε εx ≤ − < εx kt εT x ε, t , s,t ) d 1 t ) d ) ≤ ≤ y ) sin( x αδ ⇒ | kk )+ ≤ s + ) s ks n t ( ks cos( ) x = t t , then ( t x s = x k t ( -periodic integrable function. The Fourier n αT a αy αy sin( − x ( π ∞ ≤k T αy <δ cos( + )cos( ) sin( k 0 + − + − ) =1 s s ∞ − x x 0 + ∞ − ) s 0 a 2 ( ( X k ) ) ε k <δ k t x ( is uniformly continuous there exists a 0 t t ε x x t 2 n ( ( n x + εx (1 R s y x π π . T x x π π n 1 2 is a positive function |≤ 0 − − 2 εT T 2 − ) a |≤ t − | ≤ Z Z ) → s )cos( ≤ ≤ t ) ) t y | ≤ ( s → 0 t s ( 1 1 π π 2 t ) x x R ( → ∞ x x t )( s n x )= n : y ( x n − αy T t := := cos( n ≤ 0. Since n − ) x xT T ] t x, t k k − T − − ) t αy k ( ( b converges to 0 in norm in ( for π a t 0 y 0 ) = sin x DRAFT ( 2 S t | 0 (1 − ε> n s + x we obtain that , − Let x ( y are such that εx 1 2 x | and n ε t ) [0 n ∞ t T αT y ∞ 0 − ) k T → k follows now from and ∈ t )( δ x s,t 0 x − t )= ( 0 x k − is positive and s x y x 0 x 2 X n | ( 0 n → , then x s,t n t n ⇒ − is x δ T n y ∈ → T T 0 = xT = n x x )( x x ≥ εT α t n − large enough such that n xT ( ) − T T s x x N | ( k t y ∈ because either particular Now fix or Since Chapter 3. Linear operators in Banach spaces Note that Fourier Series Definition 3.15. and since that for all N where series of That Hence Setting we will use the uniform boundedness principle to show that there exis Note that the Fouriertheorems series on is convergence a of the formalFirst Fourier series we series. only. will use Intiable methods the periodic from following function Analysis w its 1 Fourier to series show converges that uniformly for to a the continuou )) s s d ( | n ) s s. ] D ( . n . 1 ) d 1 2) s D π, π − | ) . 2) − s ) − π 2 1 d s [ π i ns urier series + 1) s/ ns/ i − 1 | X σ e ( e ∈ ) + Z s k d ) = sign( 2 s ( ∈ t d 2 s s | k ) 1 π sin( i | ( π 2 1 σ x ) sin e ) y 1 s 2 , = σ ( − + =0 . and = s sin sin(( s n X k n N | n ns/ 1 2 i π D X 2) ) d − | =0 e = 1 such that π ∈ s/ → ∞ X k s i n = Im ∈ π k ( π ) − − n sin(( − n +1) = 4 e m ns/ π x 2 )) | k Z ( s 2 y t ( D − σ x k π kπ ks for i ns/ ∞ )) 0 d + i 2 Z e k s , n | Z 1 x s ( 1 s/ (e 0 x Last Change: Tue 26 Feb 11:55:21 COT 2013 σ i . ( 2 sin )= 2 − =0 2 n − k =0 ) e 2)) sin( x X k n with π X k n · → 1 π D ≥ 2 sin π − 2) s 2 ns/ π i x m | i 3.2. Uniform boundedness principle ( x, − ns/ e s ≤ y e n s < C ( x s/ π ≥ Z d d T n sign( | s | s σ ) ) 1 +1) π 1 nπ π s d s ]) : k ) d − ( ( ) − =0 = Im | s 2i sin( kπ i. e. X k Z n 1 2 n = ( σ s 2 = Im 2 Z n π, π D 1 1 π n s σ s | 2) + i sin( because + s/ ) i id ( D − sin 1 π 2 1 n ) π ) d = | − ([ → s − s 2 sin e − + + 1) ns/ ) ( 1 ( . On the other hand, the function x Z k s C 1 2 s k x k n − i( ns → ∞ ( . 1 i π d + sin(( ∈ e T π 2 ) d D π e | | π n 1 k 1 n s − ( ) ) x s/ ( = 1 t i − s =0 π π Z n e 2i(cos( ( X k 0 0 n − n =0 =1 k As before let + for n X k Z Z k n X D 1 π are well-defined and that for all 0) := ) s 2 D , ( s | n ( · = x X π T = Im = Im = 2 = 2 = 4 ≥ x ( π k π − π n )=Im π defined by s R π − s s 0) − k → ∞ d ) Z k ) | DRAFT Z 2 1 1 x, 0) ) X ( − , =0 k≤ s 1 converges strongly to π ( + ( n X · k n L s 0) n ( N k = 1 , n k n ∈ ∈ D · s n k | ( ) k n π n 0) π n sin(( T )= s , − Note that the T t 1 k ( Z ( m − =0 x y X k n ( n n T s However and let can be approximated by continuous functions k Hence the theorem is proved. hence 44 It is easy to see that Finally we show thatof the a arithmetic continuous mean function of converge. the partial sumsTheorem of 3.19 the (Fej´er). Fo We simplify the sum in the integrand: Then Proof. so that finally we obtain . , t 1 43 we M ε 2 . =: . Then < . x s such that 2 1 d <ε 2) t } + M ∞ | s | C s n h/ k . Integrating s ∞ 2 s ) x t d | whose Fourier k t uniformly in k ) d ( ) k | 2 2] we obtain s ′ · n x 2 sin( ( ... {z x ′ C t . ] x, k 2 sin π f ( =: ,π/ h ≤ → ∞ ) n h , π h Z [0 s s . − | ) n is bounded (because it [0 such that ∞ Z ∈ 1 2 − ) s k + 2 1 ) it follows that N ∈ t π σ ≤ x } s ∈ + N f ( k n d + ) d s k x 3.5 , t . ) n | . s ] d t )) n ( ( N 2 1 | n n ) for } Last Change: Tue 26 Feb 11:55:21 COT 2013 ... ) )= {z with ∈ B x, t . σ | + h, π D s M ( cos(( [ π ) x h =: n R n )) h 1 2 ) and ( K − π t ∈ − s ( sin( h ( . , that is, Z | ′ 1 x + Z x ) tend to 0 for 3.4 → R ≤ {z ≤ t + s X ) =: n ( + , n − 2 ∈ } σ X 0 n ) π h ]) : s 2 t π t ) d C C M , s d s ) ) ): ) N sin(( ) ) t s + t h ( | | ( 1 2 ,t π, π ( ∈ t s n ... · . When we choose | k≤ ( . Using ( x f − {z − -periodic continuous function t ) A ) ( n + h x ) t π s for all 2 ([ ) and 0) n ∞ π π ( ( − s − =: t 2 , n k s ) π − ( C ′ x ) π · | ε . its Fourier series converges pointwise to 1 2 t x ] the sequence ( n Z − s 2 ( + ( ε 1 2 2 k ∈ | − <ε | Z n B π ] and 1 + + 2 sin( X | ) s x . In particular, we have 1 + ) π t < π, π s k ) sin(( ), n M 1 π N ( t s ∈ n -periodic functions on − 2 + ( π, π π ( x 2 sin . Note that the bounds do not depend on [ (2 x, t h< t ∈ π n 2 = ≤ := s x ( − f ∞ 1 2 [ ( ∈ | n A n cos(( k M ) π x s ′ t X | )= 1 ∈ h + x DRAFT s − Z t h − ( k h =: x, t n t are continuously differentiable and − There exists a ) ( h ) is a Banach space. f and s Z 2 n t 2) ≤ = ( )). By the uniform boundedness principle there exists for all ∞ s f t does not depend on ∞ x )= t ( ≤ h/ | X k t − ′ ′ ) such that x ( )= C x t ) k·k ∈ n k ( s M , π ( C 2 sin( n x X, x (0 k≤ B | ) We identify the 2 ≤ ∈ ,t · h ∞ ( k ′ t n s f for every k converges to Assume that for every The functions Proof. is linear and bounded, hence an element in Define the auxiliary function Note that by parts, we find Note that for fixed Clearly ( Chapter 3. Linear operators in Banach spaces and Theorem 3.18. series does not converge everywhere pointwise to Using the mean value theorem and that k We have to show that obtain finally ) ). < < ,ε k k and and and ) k y,δ (0 0 z x N 0 such Y y − X ). Let k ( − ), so by ). Since B ∈ 0. Then = y δ Y =1 y,δ 2 δ > 0 − B n and ∞ k , 1)) is dense ( − , ,ε/n < ε, ( ∈ (0 P δ > T x − X k (0 Y (0 2 2 Y injective. Then Y X B 2) ). B z/ is continuous. B B T x ) U and z/ ( 1 k ( − T − T − Y y . Hence be Banach spaces and T y be Banach spaces and X, Y − ) because ⊆ ∈ ( 1 − 2 and L ( )) − y ) . → ∞ − ε X, Y ,ε ∈ k X, Y ) and 2 n − (0 < ε/ is closed. T T x Last Change: Tue 26 Feb 11:55:21 COT 2013 . . X k (1 Let 2)) is dense in T x 3.3. The open mapping theorem 1)) is dense in ) y,δ for , , Let 1)) is dense in )) k ( B , 2) =1 ∞ T , 0 X k (0 Y (0 (0 , k y + + z/ (0 B B X rg( = k B ) for some u (0 − X ( ( B ∈ δ < ε/n )), in other words, 2) X ( y B X k T ( 2 ( y,δ B T can be chosen arbitrarily small. x T . Then there exists an open ball ,n z/ ( ( − T z/ ε U Y is open, so its inverse 1 (0 T − =1 ∞ )) = X k B − ∈ X T y =1 T x exists and is continuous. ( ∞ y ,ε [ B u k is continuous, we know that k ( T 1 converges to − (0 1)) for all T − = T k 1 , X T ). )= ⊆ Y B (0 0 T x exists and lies in T x ( ) ,δ X be Banach spaces and x k T 2)) because 2) and ( be a linear map between normed spaces . By what was shown above, there exists an (0 B =1 , , x ( T y,ε n k U + Y . Since ( Y (0 (0 T k≤k 1 =1 B Y P ⊆ z X X X, Y 1) such that is open if and only if it is surjective. − ∞ k 0. Note that 1)) is dense in Tu ⊆ ) → B ) , B B , ε − ) is surjective. We use the notation of the preceding lemma. By ( P T (0 ⊆ ,ε ) (0 ∈ X T ε> Let − 2 ,δ ) T X is continuous if and only if (0 : 2 := B x DRAFT ∈ ( (0 B x ,δ be an open set and (1 X 0 ). By the remark above T X 0 such r + Y z X x B ∈ (0 + a bijection. Then 1 X T B . Then ) and = Y 2 1 → x + is open, then it is obviously surjective. ) ) δ ( Let x ⊆ ε> 1 B ) 2 , x u T T − , 1 T 1)) is dense in is complete, by Baire’s category theorem there must exist an + U k k 3.21 , x ) it follows that (0 If By the open mapping theorem Obviously it suffices to show that y/n, ε/n X, Y X, Y and rε ( Y ( ( (0 Y Tu : rg( Y L L 3.6 Y B B =1 1 ( B is linear, it follows immediately that ∞ k ∈ ∈ 2. Since ∈ ∈ − X Proof. Now assume that T assumption Proof. that P T in such that Lemma By ( 46 It follows that In the proof of the open mapping theoremRemark. we use the following fact. Theorem 3.22 (Open mapping theorem). The open mapping theorem has the following importantCorollary corollaries. 3.23 (Inverse mapping theorem). Corollary 3.24. Now let T y Choose ε/ it follows that T assume that the statement is proved. Since Proof. z T 27 Feb 2012 , . } ), r . 45 t := 3 R with < r (3.6) A if the for all <ε k X 0 0 k y <ρ T x 0 x − s , Y − is closed in j Y ) (0 x ) := ∈ x R ε B ( (0 x − between metric spaces } s )= B X 1 1) such that ( 3 t k , ). We have to show that there exists an 2 : Y T 0 )= , )= ( respectively. − n 1 is an open set in − n B x : f · y · ( , r B x n − ≥ be Banach spaces and F (0 π k = , , n Y (1 n T X X 2 1 (0 X 0 T = x x Y y ⊆ ,st ∈ B ( ( j <ε 0 k 0 B ) x DRAFT and x X, Y s s ∈ k n A map { , r ≥ ∈ as are positive functions, hence the n and T 1 X s x (0 0 , e´rkernel Fej´er 1 x x ) is contained in Let n k : y 0 )) is dense in Y . Then for every n − F ) := x ,ρ ) 0 T B ) such that R ,ε ε ) (Korovkin theorem). Using ( , r (0 t = 0 ,ε (0 × − Y 0 x ), there exists an r> X (0 ( 0 and R x r B (1 and that B 2 n X X Note that the assertion is equivalent to ( ∈ T 0 B B ,ε < ) T ε > N ) = sin( k (0 t ∈ 0 ( s,t ∈ Y 2 2 x ( k Since { k show the theorem, it suffices to show that Fix Here we can write Note that an open map does not necessarily map closed sets to close example, the projection Lemma 3.21. Note that all for some are open balls in Proof. The lemma says that if Chapter 3. Linear operators in Banach spaces If we define the 3.3 The openDefinition 3.20. mapping theorem image of an open set in the ball x there exists an Continuing in this way, we obtain a sequence ( that B x , . ): is = ∅ , ) = } y,y = Y ∈ D (3.8) (3.9) T 1 ( 0 − . Then x D 0 \D 6 T is linear. for some X, Tx ( x ) (choose converges such that D → his, fix an { X y : T iv . N x ( ∈ T = converges n x. = =0 is differentiable ) n with N 2.7 n = ⊇ D → n n n ∈ ˜ converge. From a x ˜ x X x T x n n ) X N T x x − n : ∈ is clear. By definition, n →∞ →∞ n →∞ is well-defined because T ) n n is closable and n x T x →∞ →∞ lim lim n T ( n n T 0 = 0 because T = lim := lim is continuous because it is . T T x the following is true: n 0 n ). which converges to and D ⇒ x implies closeness of = x | . with 3.4. The closed graph theorem 1]) and = x T x Last Change: Tue 26 Feb 11:55:21 COT 2013 a subspace. Then S y 3.9 , T Y = 0 and lim and ( 1 ⊆ D = × →∞ ⊆ D X − := N n N n →∞ n a dense subset of ⊆ D N X ) is the graph of a linear function. ∈ x ∈ ([ n x ∈ n n T N T C D n ) ) as in ( ( = lim ∈ ) D⊆ is closed. n n n }⊆ →∞ G n with lim ⊆ ′ n x x →∞ ) 1 ). Hence T lim n x ( n x n X were closed, this would imply that − T converges D x ) = 0. Linearity of ( ) converges, it follows by assumption that 1]) ) and T ( ∈ T , n n N G 1 ˜ X →∞ ˜ in x ∈ x x .) Then with lim ( n with n 0 . − − ) N L c is the closure of 0 − for ): ([ )= ). If n = lim ∈ n x 1 0 n ∈ n converge, it follows that ( ′ T T x ) ) ( x ( such that ( ⊆ D C ⊆ D T x ( S n n Sx ( T G D N x N x ) x, T x T = ( ∈ ∈ n be normed spaces. ∈ ), so { n n in . D is linear and continuous, then ⊆ D + 1))). ) ) ) normed space and →∞ T →∞ t T x and n Y ( n n ) is closed. N N is closable. Then ) = ( ( n ,y and (˜ Y x x is dense in ∈ ∈ G ) holds and define n n 1]), × ( = lim X, Y n n ˜ x T , N d ) (i) A continuous operator that is not closed. ) − = ( lim → is given by 1 n ∃ →∞ X, Y ∈ 3.8 X n n X, Y =0 n ˜ − n : x ′ 0 ( ˜ − x x ) is closed and not continuous. Let T and choose a sequence ( n x X T DRAFT L n ([ n T = lim exp( converges (to x X Let x T }⊆ x of = C ∈ is not continuous was already shown in Example ( ) 1 n N ∈ \ D 0 Closedness of Let ( →∞ T ∈ T x T lim be normed space, T = T D ⊇ →∞ x T n n lim X ) n and ( { : is closed and injective, then T x ). )= rg( X − X n t ∈ T N ( T n T . Then ∈ ( Assume that ∈ ′ n 0 )= T x it follows that (0 n That and Proof. y Let Proof. x contradicting the choice of Let x D well-known theorem in Analysis 1 it follows that ( (For example, the restriction of ax continuous function, but is not closed. To see t T x ) T ) is the closure of ( Y n T D (i) Every ∈ ( (ii) If (ii) A closed operator that is not continuous. →∞ T x (iii) If lim The closure n closable if and only if for every sequence Now assume that ( y G Hence for a sequence ( 48 Lemma 3.28. Remarks 3.29. Proof. for sequences ( to 0. Since Examples 3.30. ( 29 Feb 2012 , 0 N ∈ is 2 = Y 47 := ∈ k ) n n x 0 X such ) 0 (3.7) x n α > x kk inuous, ⊇ D → X → D → 1 , T x : T x ) 0 − →∞ X T n x ) is closed, T T : T sts an . ( T ) is complete, 0 and ( . The operator ≤ k : rg( G T and ) = ( T N 1 1 n T x . norms on ∈ k − X 2 n x = 2 T ) k 1 , T x n y ) be a sequence that n − , k n x . Since T converge, hence T k k·k ), so rg( x ( T x k ( y + the following is true: Y T ) this implies lim Y N k G 2 ∈ T = × k × ( and . Sometimes the notations n + →∞ ⊆ →∞ x ) lim 1 n T is surjective and bounded by n G k 1 X ⊆ D k k n X N x x Last Change: Tue 26 Feb 11:55:21 COT 2013 N a subspace. Then ∈ p k T . k and rg( a subspace of ∈ T x n k·k such that ( n Y . and X = = )) ) D x X N n n k k × ∈ ) ) x = n ( ∈D}⊆ and ( X ) ∈ D D⊆ , T x n x x, y x, y N n n x ( ( is a normed space with either of the norms ∈ x x k k ): n , T x ) converges in ) Y ) ) is the graph of an operator . Then the two norms are equivalent. n 2 . T →∞ × T converge ( x n ( X which shows that lim n , , x, T x G N X G 0 ( ∈ ). By definition of k·k T x R R ∈ { := lim ⊆ n if -vector space and T x ) ( 0 T x X, → → N , also denoted by dom K n then it is a Banach space. So by the previous lemma, ) are used. x ( →∞ G ∈ be normed spaces, T = n ) := Y Y Y n Y . T x ∈ → T n ) is a subspace of ( )) = lim T × × if its graph normed space and ( ⇒ be a → n ) for all ) T 0 = 0 ( G closable 1 is closed and let ( T x X, Y X X y 1 ) holds and let (( X X G : : k and ,y is ( T . Then both ( , T x DRAFT 0 = x is an isomorphism between X, Y 3.7 T N n k·k →∞ k x T Let Y 0 closed Let ∈ x n . α . ) is closed. T n k·k k·k or domain of × X, ) Let Y T T x Y ≤ n ( Y ) is closed in : ( ) has a continuous inverse. On the other hand, if × x X 2 ) = ( and lim G ( T k n T T X x k is called ) and be normed spaces. Then rg( ∈ D , T x T is complete with respect to both norms. Assume that there exi T Assume that If rg( Let n ( n → ⊇ D → . x x ( X ), hence X X, Y X X ∈ D is called the T is closed if and only if for every sequence is called the closure of : : ∈ ( , so it is continuous. By the open mapping theorem, its inverse is cont 0 →∞ →∞ lim lim G linear. Y n Proof. x continuous, then Note that the two norms defined above areDefinition equivalent. 3.26. n x is closed in hence closed in converge. Then (( converges in T Lemma 3.27. Chapter 3. Linear operators in Banach spaces Proof. Corollary 3.25. 3.4 The closedLet graph theorem T D Now assume that ( Obviously, the graph hence bounded. The statement follows now from T that α Proof. such that : n = = D Y Y := λ T (on 2 k n ) 7→ m x P x D → − must be − , n : ⊇ D → ) is closed T Λ n X, T . and and its limit T X x ( n ∈ : k projection → y G X n T N λ contradicting the x →∞ n n N λ is continuous. Then an algebraic basis of c is closed. because ∈ := lim Λ Λ X ∈ ∈ T 0 n is called a X λ D λ y ) be Banach spaces and → T x x, λ = converges in X Λ. Choose ) a subspace and x x N T ∈ = 1, )= )= → ∈ Last Change: Tue 26 Feb 11:55:21 COT 2013 a subspace and X X, Y n λ k ) with ) 3.5. Projections in Banach spaces X n X n : rg( T ) because for λ is a projection because (id : x x, T x x, T x Let 1 T x ( ( D ⊆ k = 1, − P P 2 1 n is bounded. rg( D⊆ T λ k − is open by the open mapping theorem x λ T ∈ n x 1 λ 0 k π while . Hence ( is continuous. nc n T x k Y, π X, π 1 N m − = 1 ∈ = y → → X n π n ) ) k is a Cauchy sequence in is closed. By the closed graph theorem − y ◦ n T T n 2 λ N ( ( T )= y π is bijective. By assumption the graph ∈ x is continuous. is closed. G G →∞ n ( nx ) are clear. 1 kk : : ) i n = ) be Banach spaces, be a vector space. k ( 1 T π n be Banach spaces, 2 1 T is continuous. − T = lim x ( π π = X ⇒ T 0 T D . y k X, Y be a Cauchy sequence in rg( n P X, Y )= Let λ X,T − N iii . An everywhere defined linear operator that is not closed. k≤k DRAFT . and ) follows from the closed graph theorem because by assumption ∈ Let T Let T x n ii m is a projection, then also id P → ). Hence k . Then ( ) ( y D n = id 1 N = P X y is closed and − ⇒ 2 2 : ) and ( 3.22 ∈ ) T , hence a Banach space, so P P iii T )= T be a closed linear operator. Then Y is closed and is closed and n i ( − ( + ( Let ( Note that the projections be an infinite dimensional Banach space and ( , n is closed. × D T T Y y n P ⇒ )) if ) 1 X y 2 1 belongs to X T P − → − (i) )= . Without restriction we can assume − (ii) 0 T (iii) ii is Banach space. ( x k Proof. linear. Then the following are equivalent: T 50 Lemma 3.34. Theorem 3.35 (Closed graph theorem). Lemma 3.36. Example 3.37. Let in bounded, but Proof. boundedness of rg( 3.5 ProjectionsDefinition in 3.38. Banach spaces X a closed injective linear operator such that Note that if is well-defined. Assume that rg( Proof. (Theorem be an injection. Then the operator id are continuous and that X 5 Mar 2012 22 Feb 2010 , ) N = ⊇ = Y Y 49 T ∈ 1]), U ( 1 n and , ) X e 1 − T of n ) : − 1 x ([ . Since − . T ⊆ D i )= 1 X, Tx ⊇ D → ⊇ D → Y ◦ U C N ( X X ∈ e ∈ D T ∈ ert spaces are T n : : ) the operator ) is continuous ) x g T n T T , . , , be a Cauchy se- = ( is injective, then x ⊇ D → 0 0 T 3.22 1 k k·k T − =0 , x X k T : ⊆ D D ( = N k T , t< ∈ 1 → a subspace and open with respect to the 1]. The sequence of the Y , t , t> . Then ( n , T x k ) 0 − 1 2 1 1 k Y n ) X : . Most of the time, the graph 2 − x T x [ + 1 − Last Change: Tue 26 Feb 11:55:21 COT 2013 k ⊆ D D a subspace and a subspace and k 1]) and n − ∈ , )= x + e t T U D ⊆ X X k on + ( t X ([0 ) let ( h 2 2 2 = k , is a Cauchy sequence in T t k | x t L T | . Moreover, the norm defined above is D⊆ D⊆ N → k T x ∈ is continuous, also D ⊆ k·k x k ≤k n i , k ) = ( ) 2 1 . t + 0 ) n Y D ) = ( 1]) is open. If, in addition, 1 t k x 2 n , , ( − k were closed, it would follow that T x g T R , is continuous. e n x t T Tx ([0 , x k T k 1 , is open. Let + R k·k T x is easier to use in calculations. However, the norm −−−→ C p 2 , is bounded, hence continuous. Hence is injective. Then t . Then, by definition of the graph norm, ( and ( T x D → n x = ( . → = = be Banach spaces, T where T : x 1). If is a norm on T = = 3.31 , ′ T X 1] T D e T x 1 k T , )= be Banach spaces, be Banach spaces, of x k·k t 1 − and e ( k ( X, Y T x and the open mapping theorem (Theorem 1), , ix Y, − k·k 2 ′ n ) , k·k → ∞ x 1 is also open with respect to the graph norm because obviously :[ L DRAFT → Y, X, Y X, Y ∈ D n k·k − Let 3.32 n ) ( U ∈ , 0 is not closed. 2 x T is a Banach space. → x D h Let Let ( for T L ) ) T =: T g k·k Let graph norm = → . , n ) are complete, the sequences converge. Hence, by the closeness . Then Y x D X → T k·k k·k ) To show completeness of ( X , i : ( a linear operator. Then Y , is continuous. . Then ( By Lemma n ′ D k·k →∞ e 1 D x derivatives converges Obviously a contradiction. Proof. x T , Y n is continuous. − - lim D : ( e and 1 T ) The statement follows from (i) ( − (ii) : ( ◦ e (iii) Let ii T It is easy to see that is called the D → equivalent to the norm i norm defined in Definition norm in Chapter 3. Linear operators in Banach spaces Definition 3.31. Lemma 3.33. quence with respect to i with the square root is sometimes more useful when operators in Hilb considered. Lemma 3.32. is open in Now assume in addition that a closed surjective operator. Then a closed linear operator. Then is a Cauchy sequence in by the inverse mapping theorem. Since ( Proof. X T Proof. i k·k ′ ) N ), U λ ∈ X U n , τ ( λ τ )) (3.10) on Y n . , hence and all λ . x ) for all ′ U ( 0 U and let N ϕ x X . . Then the ( n U ϕ ≥ Λ let ( ⊆ x X ) is called the is weakly con- ′ initial topology such that N ∈ n topology . ∈ → ∞ n X λ X ∗ n ) = - lim n ) J n ⊆ ) are open in n X,X w ) of functions. The x ( λ ϕ ( N σ ( . Y for all weak ∈ 3.6. Weak convergence ϕ := 3.10 n ) the system of sets de- N ,...,ϕ j ) 0 1 n the family → if and only if it converges U U n x and ≥ ( →∞ ϕ x ) n σ ( ∈ X n a family of subsets of sets in X : ) Last Change: Tue 26 Feb 11:55:21 COT 2013 2.37 Λ n are open is called the topology ⊆ λ U and ∈ x . f ( λ λ N } ) j ⊆ ∈ U ) is a topology containing for all U in the weak topology. Since by defi- λ λ , a family of subsets of n ϕ ) τ = ( U ) 0 U j ). Since lim ) is called the ( U n Λ 0 x γ,k V ∈ σ F x ∈ ( x U ( λ λ 1 ( = ( ,X by the canonical map ′ ) j are continuous, it follows that ( }⊆ − j λ n ) ′′ =1 are continuous, is called the ϕ ′ U . \ ϕ X , U j k U ( λ X X V Γ Λ . Then there exist k·k σ f ( n =1 such that ∈ 0 \ . 1 σ k [ ∈ ′ ∈ γ = ( x − j such that all N ∈ ϕ ⊆ X λ ϕ U ). 0 ) { converges to ∈ ∈ X ′′ x U . is weakly convergent with ): ( =1 ϕ ) X λ containing ′ n k τ N N ,X U ′ ⊆ ∈ T be a set and ( be a normed space. The topology R ). be a set, Λ be an index set and for every in the sense of Definition n 1 X ( N consists of all sets of the form ) ∈ ( such that all F − λ ∈ X be a set, X,X X n be a normed space. A sequence be a normed space, X σ n ( f n U x X ) X { x σ X, . The topology X ⊆ n X X ( ∈ Let Let Let x ) σ τ X ) for every 0 DRAFT ) all functionals 0 x ′ Let Let Let ). It is not hard to see that ,X x ) be the topology generated by ). On the other hand, all sets of the form ( , denoted by ′ on ) exists and is the intersection of topologies containing all ( )= ). U U U ϕ )-open set containing U U X ( F . Hence ( ′ ( ( ( ( 3.10 τ X,X τ τ τ τ σ ( σ is identified with a subset of open subsets in ⊆ , we can choose an X,X Assume that ( Let ′ j ) ( X V ,...,n , denoted by U σ X ( X σ ∈ . The smallest topology on =1 containing scribed in ( be a topological space. Consider a family 52 3.6 Weak convergence Definition 3.44. Definition 3.46. Lemma 3.49. X in the weak topology Obviously Proof. generated by so Lemma 3.45. Proof. smallest topology on that is, of arbitrary unions of finite intersections of sets i be a Definition 3.47. with topology generated by ϕ j on Note that nition of Now assume that ( when weak topology converges to Note that vergent to some Lemma 3.48. 7 Mar 2012 . is = V X 51 on ). . k ) is k and and U U k k P P u ψ U X ,...,n is called kk dim ) implies −k x P = 1 = 0, hence ) of X (1 k − n kk k≤ y k of P P ws that j, k ϕ k − k k , y Ui = . ker(id jk is a Banach space. ,...,u =1 2 δ 1 U = n k k with ∈ u V y P P x . U is not complemented as a projection. Then the ⊕ ) = . . − U 0 k ) k k U to c u u k≤ v k−k ( X ) k on Py ) is well defined, linear, bi- ( j P x X closed subspaces such that ( P x ϕ L P a finite dimensional subspace. + = X k k Last Change: Tue 26 Feb 11:55:21 COT 2013 of k X ∈ ≤k ϕ rg( y X u ). Indeed, P k ⊆ ) is closed. To see that rg( n P P =1 and ⊕ ⊆ id) } X k = ) − are continuous. By the inverse map- 0 U U − { k = P ( = 1 and ) P U, V 1 P k − , hence 0 k ker( 2 u, v P . ϕ ( k and can be extended to linear functionals is linear, continuous and bijective. Hence by ) k k ∈ P k P v P ) = ker(id ) = ϕ ] is linear, bijective and continuous. Since ) = − + P rg( k v X, Px with P [ u k k≤k ) implies ( ⊕ . u 2 . U → P x, P x 1 k 7→ 7→ P ) P ⊕ ) k X P rg( is the desired projection. are Banach spaces, their sum : V ker k≥ − be a Banach space. A closed subspace X/U = be Banach space, be a normed space, u ∈ u, v are closed. V P P k ( be a normed space and k X y ) X X P 7→ ((id . We define X/V, v k P X DRAFT v k X, or such that and let k 7→ . Let Let ′ rg( + 7→ ψ → k Note that not every closed subspace of a Banach space is comple- U Let k ) and U x x =0 . if there exists a continuous linear projection on V V k P ) are isomorphic. ). = is continuous, ker( and U,u ∞ is a Banach space. By the inverse mapping theorem it follows that n is a linear bounded projection on P P ℓ ⊕ P k are algebraically isomorphic. Then the following holds: ) rg( ) of k P P = → − are isomorphic. n U P rg( ϕ V ∈ k ) Note that ) Since is isomorphic to is isomorphic to is isomorphic to k i i X X/U x ( From linear algebra we know that there exist bases ( ( ⊕ ⊕ k : There exists a continuous linear projection of V X X Either ) P x X/U U ker(id P P =1 ,...,ϕ with ) Obviously ) The map n k 1 = (i) (i) ) Since ) ∈ (ii) (ii) ker( (iii) ϕ (iii) ii iii ii iii P Obviously X and subspace of Proof. closed, Remark 3.43. mented in the sense of the theorem above. For example, ( and the inverse mapping theorem, also( the inverse is continuous. By the Hahn-Banach theorem the following holds: ( ( Chapter 3. Linear operators in Banach spaces Lemma 3.39. ker( Theorem 3.40. Theorem 3.41. Definition 3.42. Proof. The map complemented ping theorem then also the inverse operator is continuous which sho Proof. closed, it suffices to show that rg( Then there exists a linear continuous projection jective and continuous because id x ( y , is A } g | ε. ∈ ) 3 y <ε ϕ ( < | g ) | ) y ( y . Since )+ ( ′ 1 ϕ g x ( K − g − ) ∩ ) y y ( U )+ ( a | y ϕ ∈ | ′ 1 is linear. Since + 3.6. Weak convergence + x K | ϕ ( ) < ε, g ∈ x | ( ) ). Similarly it can be shown g − g y x ( ( ) − Last Change: Tue 26 Feb 11:55:21 COT 2013 ϕ y ϕ ( ) − ϕ x )+ ( ) x − ϕ x ( | ( ) ϕ a x | + ( | ϕ ) . It follows that )= y y − X < ε, + ) . + 0. Then | ∈ y ′ 1 x ) x ( x ( y K + g ϕ ε> + x ∈ − ( x and . Hence there exists an ) ϕ ϕ ( | y ϕ and ϕ K = + − X ∈ | x ) ) ( λ y ∈ y ( ϕ ϕ 1, hence + ≤ | x, y x − ) for ( ) x a k≤ | DRAFT ( x ϕ ( : k ϕ λϕ A and let − ) ′ ∈ 1 )= y a K was arbitrary, this implies { λx + ( ∈ ε x ϕ := ( ϕ ϕ | U that 54 Let Since it follows that is an open neighbourhood of linear, it follows that j x if A λ 53 A ) is ε < X ⊆ x X is the ( Λ the ′ be the 1 ∈ a ⊆ . x A ) ∈ k− k K is called j 0 n -topology Q 7→ U X λ ϕ ∗ R ϕ ( , k k P := , a Λ . U. → . K A o ∈ k ⊆ Λ n →∞ X . Hence j is compact. Note ) → } n O ⊆ x O : λ j ∈ X k A A U f ( = lim inf ∈ : {z 1 j , λ k and let x x − λ λ lower semicontinuous →∞ x π π | , open in n X k} k kk = 1 such that , x x n =1 ∈ ) n \ = lim inf j k j ) ϕ there exist is exactly the weak ( k x with the product topology. λ k = Last Change: Tue 26 Feb 11:55:21 COT 2013 n f k . ( A |≤k = 1. Hence U |≤k x } n A f the map ) z | k . The product topology on ϕ x ∈ . : ′ 1 )= kk : →∞ ( ϕ X n be a normed space. Then the closed f n with a ϕ k u K x | lim inf ( λ k K ). It is called ∈ -compact. j . →∞ x ∈ X X ∗ X ,...,n x n ∈ ( lim inf Λ and ϕ f k≤ z →∞ ∈ ∈ ) with →∞ ) be topological spaces. Define [ 3.48 , π k { n λ =1 Let induced by x n λ lim inf ≤ 0 λ is lower semicontinuous in the weak topology. x is closed in lim inf ( k≤ ′ K 1 x . ) n , τ ′ 0 X 1 → := k n , j is weak λ K ϕ ϕ X j k·k K x k≤ x → k k≤ } X → ( 0 U ) :Λ A 1 f x for every )= n X 0 ∈ f X k x →∞ x x : ( ) n x n ( j k≤ is the weakest topology such that for every ϕ ⇒ k → λ ϕ λ Λ let ( ϕ n = = lim π ( := be a topological space. A function such that for every x k s X ⇒ k ∈ k λ : →∞ ) k≤k = lim : as above with the product topology. Let such that for every X ′ X λ n if lim sup X x x on n ( are maps k X Λ A 0 X ⊆ DRAFT X ϕ kk ∈ ϕ ). Let Y ∈ = For A n λ , such that define the set ϕ ∈ = 0 the assertion is clear. By the Hahn-Banach theorem there k U x x such that k ( ϕ s → ∞ ∈ Let 0 0 ′ { X := f - lim n x x a → ∞ ε< X k ∗ |≤k ∈ ∈{ ≥ 0. Then there exists an X - lim n := ) ∈ ) u ,...,n x w x ′ w 1 ( n k− ϕ K x ϕ 0 = ) For = is the product topology on | ε > ( i . The statement follows as above: =1 ϕ 0 ( For This is a special case of Lemma 0 k f k . So it suffices to show that j O ) ϕ x ′ 1 product topology x x ( K → 0 (i) ) Let n (ii) ϕ ii x continuous. Hence the topology on on weakest topology on because that elements together with the product topology. By Tychonoff’s theorem Proof. exists an ( Proof. Chapter 3. Linear operators in Banach spaces Definition 3.50. Theorem 3.53 (Banach-Alaoglu). k unit ball family of all sets Proof. upper semicontinuous Then Hence the lemma above states that is continuous. Lemma 3.52. lim inf The projection open, Definition 3.51. . K X X the k =0 y , . i k X R C λy + ∈ = = . + X k d with the 2 K K x k is a continu- lements of x, y k y . if if k λy , x X r product space = k + 2 + k y x ) i| 4.1. Hilbert spaces h i| 2 , = 0, then obviously kk ,y 1 . Hilbert space x, y 2 2 ( x 1 x,y . βy k ormula. ) k 2 x y |h k i k − + y y − 1 k are linearly dependent. − k y +2 1 αx be normed space. Then the y x y + 2 k k Last Change: Thu 4 Apr 11:51:53 COT 2013 i +2 k x X x 2 k−k be an inner product space over − k and k i − h 2 becomes a normed space by setting 2 2 x y ≤k x k y ) X k Let = ( y 2 − ·i − k 2 , 2 such that y k =2 2 Let + i x h· k y 2 with a semidefinite hermitian sesquilinear x K k , x k kk + k X, 1 y 1 ∈ + ( i X x , x k + i − |h are linearly dependent. y 2 2 x x,y k k ≤k = α, β y k h is equipped with the norm kk y y i| x + 2 X − − k 2 and k x x × ,y y x 1 +2 +2Re X y 2 2 −k −k + ) does not imply that k k 2 2 x x x k k k . k 4.1 i − h y y when ). On the other hand, if equality holds, then 2 X ≤k = + + K , x ∈ 2 ). If there4.1 exist A complete inner product space is called a is the triangle inequality. To prove the triangle inequality, choose x x 1 DRAFT k k k x → y x 4.1 An inner product space , Note that the scalar product on a inner product space ) is true also in a space |h 1 4 4 1 2 1 + is generated by an inner product if and only if for all X . Then k·k i x 4.1 × X = = k X i i X ∈ . Using the Cauchy-Schwarz inequality, we find x , x chosen as above, so h Straightforward calculation. The only property of a norm that does not follow immediately from the The statement follows from X x,y x,y λ h x, y h := ∈ k x Proof. and Proof. definition of x, y with ous map form but equality in ( In this case, the inner product is given by the polarisation f in terms of their norms. Theorem 4.7 (Polarisation formula). 56 which proves ( Note that ( In the following, we will always consider inner product spaces endowe The polarisation formula allows to express the inner product of two e A necessary and sufficient criterion for a normed space to be an inne topology induced by the norm. Definition 4.5. is the following. Theorem 4.8 (Parallelogram identity). norm on parallelogram identity is satisfied: Lemma 4.6. Proof. k Lemma 4.4. equality holds in ( 14 Mar 2012 = X 55 (or i (4.1) x , x h 6 = 0. For -vector y 2 . K i i i| y,y y,y x,y h h because 2 |h | -vector space with X λ K | + 2 + i inner product space i| i . be a } 0 y,y , x,y x,y h h X , |h K X \{ λ i| Last Change: Thu 4 Apr 11:51:53 COT 2013 , ∈ − ∈ X + Let X 2 2 λ x K i y,y ∈ i i , i| i| ∈ x → X ) is called an y , x i| |h y,y y,y h ·i y , x x,y ∈ h h X x, z , λ we obtain |h |h 0, h· × x , x i i + X 0, i X i− i− X, y ,y i≥ x,y ∈ h h : x,y,z ≤ |h , > i are linearly dependent. 2 . , i − x , x ·i i i x , x x , x h , x , x i| y h h x, y and ( = y , x h· = h -vector space. A map x , x = = λ x,z y,z i x,y X h h K i and = |h , for a hermitian sesquilinear form if for all λy ⇐⇒ h i + + x λy on X i i X + ⇐⇒ h + be a ∈ x,y = 0 there is nothing to show. Now assume that on x x,y x,z X h h y , λ DRAFT λ λy , x R λy , x . Then for all A positive definite hermitian sesquilinear form on a Let + = = ·i ⇐⇒ h ). + , i i x i ∈ x z h· = 0 or + ≤ h inner product y,z ≤ h x x , x h 0 + 0 K For λx x,λy . hermitian positive (definite) h positive semidefinite h ∈ i sesquilinear form • • • λ (i) (ii) x , x all with equality if and only if inner product Proof. The inner product is called is a Chapter 4. Hilbert spaces Chapter 4 Hilbert spaces 4.1 HilbertDefinition spaces 4.1. h Lemma 4.3 (Cauchy-Schwarz inequality). In particular, when we choose Definition 4.2. is called an Note that pre-Hilbert space . a y ple- ⊥ H pace. x ⊆ 1]) is an if , 1])). 2 , k M ([0 y ([0 k C 2 , , , if and only if , if and only if + L N y 2 B ∈ =1 , k k n k 4.2. Orthogonality x ) ) ⊥ ⊥ 1]) k k k , y y x . A is a closed subspace of = } ([0 2 = ( = ( ⊥ k M y M ∈ R ∈ , y , y + N x =1 Last Change: Thu 4 Apr 11:51:53 COT 2013 ∈ be a Hilbert space, k n k k is ) ) . Then there exists exactly one k k m, m H X H x x , denoted by , denoted by . ⊥ ⊆ = 0. ∈ t, f,g = ( = ( Let X x i is an inner product space. 0 } : 1]) which is not positive definite, since, M . 0 x ⊆ ) d , . ) { X t X ( A X ([0 g ,χ ∈ ) ,M } t R , x , x orthogonal ⊆ 0 0 orthogonal ( x . So with the Euclidean inner product k k . x { { f y y A of a set n X χ B k k 1 h 0 C x x := ∈ Z b ⊥ n =1 =1 ∞ to the space of the continuous functions = X k , X k = dist( its orthogonal complement and M i A ·i k = = are called n be an inner product space. , 0 are called X ∈ i i y R h· 6 = 0, but f,g X X for every subset a ⊆ X h } − 0 x,y (i) x,y 0 ⊥ ∈ { h (i) Pythagoras’ theorem holds: ⊆ h ) M x Let χ k A for every subset 1]. Then DRAFT The completion of an inner product space is an inner product s x, y , ⊥ 1]) be the vector space of the Riemann integrable functions on the ) , A, B ⊥ =0 ([0 = 0 for all A i ( orthogonal complement i ) with R = (span N such that ⊆ . ( of an inner product space ⊥ By continuity of the norm, the parallelogram identity holds on the com 2 x,y a,b h inner product which is not complete (its closure is the space The restriction of A X is an inner product space. for example, h interval [0 A defines a sesquilinear form on are inner product spaces. ℓ M X ∈ (i) Elements (ii) Subsets (ii) For every set (ii) 0 (iv) (iii) (iii) Let (iii) The 4.2 Orthogonality Definition 4.11. tion Proof. 58 Theorem 4.9. Examples 4.10. Theorem 4.13 (Projection theorem). Remarks 4.12. nonempty closed and convex subset and y 19 Mar 2012 21 Mar 2012 26 Mar 2012 . . i = 57 Q Q 2 k . ∈ i x,y x is an i + i h k z λ 2 λ ·i Q 2 + , x,y y , x z h y h 7→ h· ∈ 2 i 2 . − k 0 λ for all y and let − 2 Re =4 x , λ i i ≥ 2 x 2 ·i C − k − , 2 2i 2 y . k 2 z x 2 2 k h· i 2i x,y → k x − k y h + z 2 k y i − 2 k z z i − λ C x y + i 2 − 2 2 + − − − k = z =4 2 2 i 2 2 2 y y x i z k 2 2 k + k k i can be proved analogously. x,y x k − x z − 2 2 i z h y + 2 i x k 2 − + i 2 k i R 2 y − k − + i z = y − + λx,y z z + Last Change: Thu 4 Apr 11:51:53 COT 2013 i y 2 y − h = z x 2 i x x − k 2 2 k i + + + i − x + z k 2 y + K + y k , hence i y y x x x i i 2 + i y z k i i x,y + i 2 y 2i + i − − h i y − 2 2 + 2 − − − x z k − 2 x , k − k− k y x x k x x h x , 2 z 2 + 2i 2 + x . k 2 − − 2 i i y 2 + i + i z y + i +2Re =2 k x x + i 2 2 − − − z . The case =2 2 y 2 − x i − k k k + i x k 2 2 2 k i C y x 2 x y + k y −k − x k is such that the parallelogram identity holds and y z z z = − k − x,y 2 2 , hence homogeneity is proved for +2 + i h i x,z k − + i + y x 2 2 2 must be equal because they are equal on the dense 2 2 h − − − K − the two continuous functions x X z parallelogram identity holds: 2 2 2 k k + i i y y y + 2i x k k + z x 2 + 2 x,y C i −k −k − x 2 x k z i X h k 2 x k 2 2 − by the polarisation formula. We prove that k y y ∈ + i − 2 k k k ∈ − z x i +2 − . λx,y y x = =2 z z = i − − ) y . x,y = =4 2 2 2 i + C i h 2 x + 2 2 z + i + + i x, y −k k z x, y y + x,y 7→ h z z y x of = 0 we find y h DRAFT y y 2 2 in the case + 2 z x,z k k + k −k z 2 2 h + + x,y y − Q x , x z + i i y 2 2 X + , λ h = = h + i + i y y x k + x 4 + + C y i k y + i x x i x , + + x h x define + + + Q 4 k → x x x,y · 2 x x x,y h X + i + i + + + 2i C k h k 4 y ∈ 4( = 2 = = 2 = 2 = . Then for all Assume that the norm is generated by the inner product + 2 1 and subset Homogeneity. From the additivity we obtain If we choose Hence for fixed Note that Additivity. Positivity. Hermiticity. x i k x, y • • • • x , x h Chapter 4. Hilbert spaces Proof. Now assume that the norm on for inner product on . 6 = = U or H . ⊥ is a and and k U ∈ = U . By ) . U . V 0 U 1 and 2 U ⊥ y H U k ∈ P . 0 u, v U . = 1 ) 2 y if ( 6 = 0 k k ∈ ) = k ≤ k . ) x − ⊥⊥ ( 0 U U antilinear U U 0 )) X x U U is a subspace, =0 P P 2 x − . k P k )( , hence ∈ x k ⊥ = ( U U U U − U U ≥k P P U 4.2. Orthogonality P ∈ 0 P x, y 2 − x − = 1 if y is called a projection = k , U. is called − y 2 id k U ∈ (id 2 X ⊥⊥ Y and k U . − k P x V ) and ker(id . Since ( such that P 0 the unique element . Then 2 P K + → y K for all − x 0 ⊥ − k 2 H ( U a closed subspace with ∈ y and . orthogonal projection on y ∈ k X U id U )) ) such that + )= , y ⊥ λ Last Change: Thu 4 Apr 11:51:53 COT 2013 1 λ 0 k P U because ⊥ 0 H 2 on x U x U k ⊥⊥ ( 0 ) := ( ), for every closed subspace − ) = ⊥ ) 0 y 0 ⊆ ∈ V U U ) y U )= i≤ and U x H P 1 P 0 P ) ( − P ( U = 1, U x 0 x − k 4.16 − is equipped with the norm U 0 ( H L − − P x ) for all 0 k x )= P ( U = 1 y x ⊥ ∈ U (id ∈ ( U U x ). k 2 on a Banach space 2 ( U shows the statement. P f P P λP k − ⇐⇒ λ k ) U = , x then 0 X ⊕ − U P 3.38 1 2 x )+ )= k x and let U U = x = id ) 2 → )( . , y . ker( ( )) = y x U ) )=0 2 ⊥ H ⊥ V U ). Then rg( 0 X x x P λf V − − U ( ( x implies ∈ : . If ( P − 0 1 ,U U with norm as in the statement, and U 0 ⊕ is continuous projection with y )= 0 U P − H P be vector spaces. A map P x )= x λx U P ⊥ U U y ( , 3.39 − P 6 = U be a Hilbert space, P U − U ) = ) + ( + Fix = id . P − ) is the unique element in 1 0 0 P ) + (id ⊕ U X, Y . By assumption as in the theorem is called the be a subspace of a Hilbert space . 0 is linear. Let 0 y H x x (see Definition H H V ( h λx id x x U = dist( U if ( U ( U ( P − U U P = f P Let U P k ∈ U Re 0 is a projection with rg(id 4.16 Let = 0 P x P = λP DRAFT if y U y ( ( . Application to U k Let , . 2 = 1 , then k = 0. By linearity of H P . Also − U 1 2 1. Lemma − P i = k ⊥⊥ ) − = ⊥ 0 2 2 2 4.15 U ,y x V 2 x U k k = 0 if P k k 0 0 ) Let v i = y y − k k ≤ − x k ( k By the projection theorem (Theorem 1 U U )= − − V id + P P U ⇒ 0 0 k 2 λx P − − x x . Then there exists a projection k h k } )= u id id 0 ii k if and only if { conjugate linear We will show that ( ( Therefore id Theorem 4.16. 60 so Recall that a linear operator Lemma 4.18. Definition 4.19. Pythagoras’ theorem we obtain Definition 4.17. projection on and Hence, by definition of By Lemma Proof of Theorem such that we obtain k In particular, ker( We already know that rg( k Proof. hence is = 59 M is a d = 1, ∈ M . For . N k = t 2 . and fix ∈ y H y k 2 k n k 0 then the d 0 2 ) ∈ y ˜ H y . k n 2 0 k y y . − M k x k )= 0 0 2 = 2 | y x ∈ k λ k =0. If 0 ). Then = 0. | 0 by the convexity − → ∞ ˜ 0 0 y i , and since y 0 y k x + y k x X ,y k i≥k and fix i M − + 2 0 0 y t ∈ y y 2 2 ∈ y . Then there exists a H ( . If k } − i| + t 0 − 0 − } m 0 y 2 2 i y 0 y 2 M ,y 0 + 0 k k 2 k x M d + ,y ) 0 y 0 ( , n,m h 0 n 0 ∈ y 0 y 0 2 1 y y ) yields such that − y ∈ y − y , x y − 2 h − Last Change: Thu 4 Apr 11:51:53 COT 2013 = − y − := 4.8 λ : 0 M y , y −→ 2 y t : 0 x k . We will show that ( m k 0 ( t ∈ y 2 x 2 d |h y t y y h 0 k i− 0 d ) 0 i| ˜ y 2 y 2 y y + − − = {k because let ,y − − i≤ ,y , ˜ 2 0 n 0 − − 0 0 k ) 2 k 0 y 0 y x y 2 y y 0 d n y 0 M y h . k y k y +2Re − x . − − 2 k ≥ ∈ m 2 − 2 0 0 M Re + 0 y k converges to some {k k y 2 ) = inf 0 x y t − ) 2 x + k 0. , y y k ∈ 2 x k 2 h y N →∞ 2 = ˜ 0 k = 0, then obviously m λ + ∈ n − ,M y − = y m 0 − n y 2 0 i≤ 2 0 ). 0 λy y y + (1 − − |h + 2 ) 2 0 k 0 − x 0 y y n k n k y 2 2 2 2 y n k x y 1] and − − 0 t 0 y k k k ) := inf 2 y . If , +˜ , y y , y k x , 0 0 0 0 ) n − k + = h ) 0 0 U y y y y [0 ( y − − y 2 0 ) + ( ,M 2 1 k . Without restriction we may assume ∈ 0 ∈ , y − − − − 0 0 y 0 . By assumption ,U ,M =0, so 0 y i 0 0 x x such that lim 0 i≤ x t 2 Re := dist( 0 0 0 0 y ≤ = y be a closed and convex subset of a Hilbert space . By assumption M k and by assumption on 0 k y ≤ x x x x x x ,y − 0 be a closed subspace of a Hilbert space 2 y d k k k the following are equivalent: 2 / − ˜ ∈ y M M 0 − = ≤k 0 M DRAFT M 0 − y U 0 x 0 ≤k = = = 2 0 − m Assume that there are ∈ x ( ⊆ M x x . k Let y y 2 − k k 0 h ,y . 0 y ) For ) Let 1] k N y U Let ∈ = dist( = dist( 0 2 0 Let y 2 0 , ii ii . +˜ 0 − k ∈ = = . y ( ( x y 0 0 y k k 0 n ⊥ h n (0 − 2 y y 0 0 ) y 1 y M − k y 0 y − 0 ⇒ n ⇒ ∈ − y y 0 x y 0 ). The parallelogram identity yields ∈ k t − − ) Let k x is a Banach space, ( x i y ≤ )= )= − − 0 h 0 0 i i ( k k the following are equivalent: y . For ( ( Recall that dist( 2 0 0 ,M x x X d 0 x k x k = H ⇒ . Hence the parallelogram identity (Theorem U k x λ ∈ ∈ M (i) )= (i) (ii) (ii) Re 0 0 ii Now assume that Cauchy sequence. Note that y of assertion is clear (choose It follows that which implies Re Proof. sequence ( So for all Proof. by the convexity of Existence of ( Chapter 4. Hilbert spaces Proof. Lemma 4.14. Lemma 4.15. dist( x let Since Uniqueness of closed, , . ) a N t ∈ ( H } n and and n 6 = 0 . ) N P 2 2 } n 2 s 1 2 s n i| y ∈ . ,...,N n ⊥ o + n = ( a family of 1 N n : =1 s x,s N S ∈ . q n |h n ∈ . Application ,...,x s n n 1 n +1 { ) t =1 x N n n for X n { ): y ) = x · 1 t ( n . Then ( ≥ − 2 s n ) = . Note that n k y 2 t 1 s span 1 ( ⊥ s i| n +1 − i n ∈ cos( mal system n 1 ) are a orthonormal sys- x k N 4.3. Orthonormal systems y 2 is n π x N ,s o y k 1 s ( H. x,s a orthonormal system in k √ 1 2 1 n , } H ℓ |h ∈ x n := Λ N Last Change: Thu 4 Apr 11:51:53 COT 2013 := ∈ ∪ in =1 N is well-defined. By construction, =0 λ ∈ 2 − h +1 X n ) i| ≥ the set i 6 s o n . Since N λ 2 λ n be a Hilbert space, λ n ∈ + +1 s N x N defined by s n n 2 , x i ) s ∈ H 2 k ∈ x,s n n = ( x,s k H := h n N |h n x be a Hilbert space and . , s S x 2 ∈ } k Let k x,s th Legendre polynomial. ): y s Λ: h Λ: H N · n i n ≤k = k ∈ n x ∈ =1 ∈ 2 . 2 N n } ,s Let i| λ λ n { n k N n : n sin( P x s is the is at most countable. h ∈ i n π := − 1 let 1 n := n x x,s n √ n =1 x { x x,n |h 1)) and 1) : X k ≥ S n . Next set , S x,n = 1 x,s n k − − := =1 . Note that for every h ∪ ∞ S . Then x x X n 2 ((0 S =1 1 { ). An orthonormal system in t o 2 N =1 H ∞ n N +1 ( − the set be a Hilbert space, X n x L π n k n S n 2 t 1 x (i) The unit vectors (e 1 π, π d d H H ,...n span span x = are linearly independent, = √ are linearly independent. Let let − ! + k := ∈ ( n x 1 n = DRAFT 2 1 2 N H n = 1 Let +1 x S x k 2 L S := = n +1 j ∈ N n 1 x S = = 1. Now for y Let , hence s k and N )= for k S t span H 2 2 ( = j s x ,...,x n s k 2 1 By the Bessel inequality, for every Let For P k x tem. ⊥ span = x k k ∈ 1 +1 (ii) Let n s n of the Gram-Schmidt orthogonalisation yields polynomials Then for every is at most countable. Proof. Lemma 4.25 (Gram-Schmidt). s Since where 62 Examples 4.24. Example. ‘ Lemma 4.27. such that because linearly independent vectors. Then there exists a orthonor is finite. Hence Proof. Theorem 4.26 (Bessel inequality). x k Pythagoras’ theorem yields orthonormal system in Proof. 9 Abr 2012 . ∈ is = 61 0 S 2.40 2 k k λy 0 0 y y k + k = 1 and u k be a Hilbert x of vectors in = = k = 1. Choose k H Λ x ϕ ∈ ⊥ k λ ) ) Let ϕ = Φ(0). Otherwise λ x , H, ϕ ) . For x if and only if ∈ } ( = ( 0 ) if and only if for every ϕ . Then it is easy to check y H S H. { ) follows with Theorem . Note that H is reflexive. ∈ ii i y ∈ . A orthonormal system )= 4.20 1 ′ i 0 H u Last Change: Thu 4 Apr 11:51:53 COT 2013 − x ( λλ ,y = 0, then k y , x δ ϕ h y , x,y T. = span ϕ k k = , hence dim(ker = y , y = contains a weakly convergent subse- . So we have shown that Φ is well- ⊥ )+ 7→ h· i ′ K 0 = ) ′ k y is closed, there exists a decomposition S kk . If H λ ( y ϕ ′ i H x . x k } ) , x H λϕ y ϕ i → , y ⇒ ) = ′ ⊆ λ is an inner product space by = { = ∈ ϕ x 4.20 = ,y Φ( H as in Theorem h N k , 0 i|≤k H ) ϕ λ ∈ T ) ′′ x if y n → x ) = ⊆ H − x,y n Φ( i Φ( H . The Cauchy-Schwarz inequality yields x k n 0 S |h ′ h converges weakly to ( → y x h ′ 2 H = Φ: H complete orthonormal system − H k ∈ 6 = 0, then set k ) be a Hilbert space. A family ⊆ 0 x = 1. Since ker be a Hilbert space. y ) = 1. Then (ker y is surjective, implying that N Every Hilbert space is reflexive. 0 k )( . Since Φ is an isometry, it follows that 1 = as in Theorem k H ∈ y i . Note that rg( y If (or a H H ′ ( n ϕ 0 x , ) ⊥ J T k y (i) ϕ DRAFT k h ) H Φ( n 1 Let of a Hilbert space y k ϕ x Let − ( ′ → k , so , 0 , implying that H orthonormal system with H y ⊥ k H (ker J k ) = 1. y , k≤k ⊥ ϕ k ⊕ k ) ) 0 Φ: h· y ϕ y ϕ ) is clear. Let Ψ : = Φ= ) follows from theorem, the and Riesz-Fr´echet ( k = ii i Φ( (ker ◦ Obviously Φ(0) = 0 ( ( k k ϕ x The dual quence. A sequence with Every bounded sequence (ker ⊕ ) , so y ϕ orthonormal basis = ker ∈ k is called an (i) 0 1 (ii) (ii) y Φ( 0 k is an isometric antilinear bijection. Proof. space. Then the map an H hence Chapter 4. Hilbert spaces Theorem 4.20 representation (Fr´echet-Riesz theorem). Corollary 4.21. Corollary 4.22. 4.3 Orthonormal systems Definition 4.23. Proof. k that Ψ Proof. orthonormal system defined and an isometry.To In show particular, that Φ Φ is is injective. surjective, fix an ker hence y we can assume that H := ′ S ) are a (Γ) is a . with re- 2 Γ and N -dense in o ℓ ). ( 2 } N 2 ∈ ) N ℓ ( π , then γ ∈ 2 ( } k·k ℓ in f 0 n is = } . Then for every ): \{ } π N S · ) = 2 . ⊥ N n π C are complete. ∈ ) S H ∈ − ′ ( n ∈ ∈ n S cos( f : : 4.24 4.3. Orthonormal systems x π n 1 n . e √ ]) : e a orthonormal system. Then . Since { { 2 n Last Change: Thu 4 Apr 11:51:53 COT 2013 , contradicting the maximality it can be shown that =0 (Kronecker delta) is a complete is the orthogonal projection on ′ H , x ∪ π, π 2 2 S . ]). N s ⊆ k·k − i i| o i| λγ o δ ( ([ N S ,s such that ∞ π, π C ′ S x,s x,s ∈ s − h )= < H ) because |h ∈ . |h n S ([ γ 2 S N 2 ∈ ( | S ∈ f H ( s ∈ ) λ ): ∈ L 2 { X s . 6 = 0 and the sum is absolutely convergent X s γ f · ℓ x ∈ P ( . By the theorem of the Fej´er, trigonometric = n 6 = 0 for at most countably many f = ) is a well-defined inner product (note that := R | 2 ) γ 4.18 7→ 2 Γ ( k π γ k = sin( 2 ∈ ( ′ g x x . where X ) γ s . f k C , π K (Γ). k γ 1 H Γ ( 2 : H √ ∈ ℓ , x,y f ∈ λ i n ∈ 4.30 . ) Γ K 1= x y λ ∈ ∪ holds: f γ be a Hilbert space and s,y → = o , is the set of all trigonometric polynomials. Without restric- ih P π x H S 2 :Γ (i) The set of the unit vectors is a total subset of 1) and 1 , hence also with respect to =0 , = s, x √ f x,s ∞ S i x h Let (0 . n ) If there exists an n DRAFT i S 2 ii S ]), ∈ ( L = is a orthonormal system with k·k X s ⇒ x,s h f,g = } S = S h = ⇒ . ) By theorem π, π x we get the contradiction ∈ span ) follows from Lemma ) straightforward. ) Choose i ) Assume there exists an orthonormal system (Γ) := 1 X s S − H i iv H v 2 − iii ( ( S vi ) = ( = ℓ ([ is a complete orthonormal system. i ( ( k ⊥ = 2 \ ( = x,y x ′ ⇒ ⇒ x S H Hilbert space and ( h x only countably manyby terms inequality). H¨older’s As are in the case Γ = tion we can assume that Note that span polynomials are dense in spect to L complete orthonormal system in orthonormal system in Then ⇒ ⇒ ⇒ S S . ) = )= ) = S ∈ (i) )= ∪ {k )= (v) (ii) (ii) Let Γ be a set and define ′ (iv) (vi) Parseval’s equality ii iii iv v vi (iii) (iii) Let of ( S the following is equivalent. Proof. ( ( Examples 4.32. 64 Theorem 4.31. span Now we show that the orthonormal systems in Example ( ( s 11 Abr 2012 : . } a 63 H N π . con- H ∈ ∈ λ ∞ n x ⊆ x , the set : < Λ S n is at most H 2 ∈ s k λ . } { ∈ z . Let } S. k P . x N = 2 i exists. Let rgence is equiv- 6 = 0 ∈ k verge absolutely x ∈ 4.29 h space, however, ,z k y 0 λ is at most count- S → ∞ s exists. We have to n π x i y ) : } k h ≤k k ( n . Then π Λ: λ = s 2 x,s = 0 ). Let , s X { i i h ∈ is unconditionally conver- i| ) s . For fixed i 6 ). This is clear because i = k ,z ⊆ λ =1 ,z P ( =0 H 4.27 ∞ k 0 a orthonormal system. Then { π H. π Λ is a Cauchy sequence because n s E x,s ∈ ,M,K s ∈ x,s 4.16 Last Change: Thu 4 Apr 11:51:53 COT 2013 ∈ P 0 h N H h 0 ih λ := |h x ) ∈ S π : x,s ,s 0 n λ h ⊆ be a Hilbert space and ∈ := i =1 X s ∞ x S X n −→ k S =1 y,s y s H = ∞ k h . Therefore the series in the definition 2 i x,s ∈ , x π ). Fix k i| h 2 P =1 y 2 ∞ x k s is well-defined. k X P n S Let (Theorem i| { x = 1 follows from Corollary ∈ := n x,s X = P s is absolute convergent and can therefore be ⊥ h x,s y = π and the series is unconditionally convergent. i i S |h − y =1 y,s k≤ x ,z S n k ,z x N i|≤k |h S P n H, Px D n M = if and only if Λ X P k is complete, s k s span =1 ∞ = ih x,s → ih X n span H = ∈ H n H for every enumeration Λ n |h E 2 0 H ∈ S ∈ x : ∈ ≤ P x ,s y,s k be a normed space, ( y,s z x X s h is at most countable (Lemma = s h i 2 P , the set i − be a Hilbert space and . Then } n k =1 X to x , it follows that ∞ =1 x H λ i| X n S ∞ n H x,s x H h =0 DRAFT ,z ∈ = x,s P S Let ∈ i 6 h n ), so the claim follows from the Bessel inequality for countable =1 i ∈ is a linear and Let x s . For all ∞ n X s z N π , M = ih P X x,s − y P z k y,z n 4.27 h h x : = ⊥ D y,s y S π y |h ∈ be a permutation. Then also − For fixed First we proof that the series in the definition of s =1 ∞ y { X n N is unconditionally convergent and = P → x It is clear that We have used that Since show that Proof. rearranged, because, by inequality H¨older’s and Bessel’s inequality of N able (Lemma verges unconditionally is an orthogonal projection on by Bessel’s inequality. Since orthonormal system. Then Proof. S countable and Chapter 4. Hilbert spaces Definition 4.28. orthonormal systems. Theorem 4.30. alent to absolute convergence. Inthere every exists infinite a dimensional Banac unconditionally convergent series that does not con Recall that in finite dimensional Banach spaces unconditional conve (Dvoretzky-Rogers theorem). Corollary 4.29 (Bessel inequality). gent (this proves then well-definedness of be an enumeration of We have to show that . , so 2 such 2 k 1 H T . =0 ces also H k 1 i H y )= ∈ = ∗ . 2 k ∗ . H i T = id x, y , x,T TT 1 T h kk ξ , η ∗ H ∗ = h . T =1 T ) ). 2 ∗ ∈ 2 2 = k T H is surjective. Now we ) = rg(id i x ∗ k ∗∗ and ,H ,H ∗ ∈ . 1 1 k≤k T 1 T 2 y 1 , TT T H H H i H ( ( ∗ = H L L T rg( 2 ∈ Tx,Ty , x,y . ∈ h Last Change: Thu 4 Apr 11:51:53 COT 2013 H . x,y ∈ ∈ = id T ⊇ h ∗ ∗ = ∗ =0 ) T T T T . Then there exist i = x, y k≤k ⇐⇒ ∀ i T 2 , x,y = 1 shows the desired equalities. i = TT i = . H k T x 1 ∗ x,y T 4.4. Linear operators in Hilbert spaces ∗ x ) note that for h − . ∈ k ” follows from the polarisation formula v x,y T ) . . In particular ⊥ h are selfadjoint. ∗ Tx,y TT = ⊥ =0 1 Tx,Ty ⇒ ∗ )) and kk T i H normal. i H = )) T ξ, η . T x T 2 ∗ i with h k T ∈ T Ty and T = ( H i − h T ∗ x = 2 H = id ∗ k ) = i Tx,y i≤k . ⇒ H 1 h ∈ ) = T we conclude that id x,y 1 = (rg( and − ∗ ∗ h = (rg( x = be Hilbert spaces, k Tx,Ty T x ∗ H ∗ . T Tx,T T T is invertible and T ∗ 2 ∗ 2 ⊥ 1 = ( 2 T ∗ ” is clear; “ . if k T H be Hilbert spaces and is a surjective isometry. i ∗ )) T H T TT rg( k Tx,Ty T ⇐ h ,H 2 T h ∈ ker ∗∗ , so k if x,T if T 1 Ty . It follows that ⊥ h y ⇐⇒ h , = y T ∗ + because for η ⇒ H ,H = as a surjective isometry. Since x and i ⊥ ∗ T 1 = = ⊥ η k = ) . )) T i selfadjoint = ) ∗ H 1 − ) = ∗ ∗ ⇐⇒ T is unitary, if follows that rg( ∗ λS Let is an isometry. Let 2 ∗ is an isometry because for all T Ty R T T = (rg( Ty selfadjoint DRAFT normal unitary ∗ ∗ T ∗ ,H T = Let T ξ,T ∗ 2 ,H k T x,y T T ∗ ∗ ⇐⇒ ⇐⇒ ∀ 1 h (i) . T ) H ). ) are clear. For the proof of ( T = Tx,Tx and ( H = T T h h ( = (rg iv = (rg( L k ξ ∗ is invertible, then 4.7 ∗ L ) = + = T T ∈ )–( ) The direction “ = is unitary is an isometry is called is called is called T i i ∈ 2 =0 ∗ ∗∗ ” Since ( ( TT k RT λS T T T T T T T If T k ⇒ T x T x T x ” Assume that ) ker k is surjective. (i) (i) (i) ( )“ (v) (ii) (ii) (ii) (ii) ( ⇐ (iv) (vi) ker (iii) vi ii (iii) (vii) Remarks. Proof. ( Next we showpreserves that angles. a length preserving linearLemma map 4.41. between Hilbert spa Taking the supremum over all 66 Definition 4.40. (Theorem ( it follows that Proof. By the same argument as for T will show that “ that Then also ker ) . . ) | if ∈ S 3 s = 1 65 by ) S ′ | | s s T ,H ( ,H S a H . For 2 | y 2 = . Then H 6 = S H = H | ( ( ∈ H s s N S L L a ). Let || because by P ∈ follows from S ∈ ∅ the canonical is well-defined R 4.20 span 2 ′ . j := H , . ) Φ ) T H ′ x | ≤ | N 2 H ) = ( . ′ T rem T 2 s | 1 → . S ). ℓ ,H ( is at most countable ) − 1 1 ε . Now assume that . j . Then ∈ N } ∼ = 2 B H ( H ∞ ( H s 2.25 2 H : ℓ H ∩ L , := Φ = 0 < ∈ j ). In particular, ) ∼ = H ∈ , hence s | s . i 6 ∗ be a dense subset of ( ( x S ,y ε ε T ) and define ∈ | 1] H T Last Change: Thu 4 Apr 11:51:53 COT 2013 1 , be complete orthonormal system . A B B S x,y S x ∈ ( S,T H is h [0 an orthonormal basis of ∈ 2 T ∈ , 2 x is countable. : ℓ i ∈ 1.25 S s L S x, y a ∈ T H and , ) is linear and isometric by Parseval’s x,s i 2. Let y ⊆ h (see Definition ∈ S S . Then is not separable. ( √ y . 2 S T T , x ℓ x,y { y i H = h ) = ) y ∈ s ii = 2 ∗ ( → ( = := ′ k such that is an uncountable complete orthonormal system ′ s i s x T x H . T x k A T H x,T . By the theorem Schr¨oder-Bernstein then be Hilbert spaces, 6 = adjoint operator : 4.31 | )) h 3 be Hilbert spaces and Φ + ∈ ii ⊆ ) T s ( T | 2 = s 2 ) by k S ,H a i Tx,Ty ) and S s 2 = h is surjective, let ( ,H k | 2 the set 4.32 1 be a Hilbert space and ℓ ,H T p N is a separable Hilbert space, then ) and 4.30 H 1 be an infinite dimensional Hilbert space. Then the following S || 1 2 be Hilbert space and Tx,y is separable by Theorem H . h → H = | T − H ∈ DRAFT H be a countable orthonormal system in 2 H k T Let H , ′ | H If x Let Let S s Hilbert space : ) Assume (0 = . By Theorem Let | ≤ | Let ii − | ( and T S ∈ | s S (Theorem | cannot be countable, thus there exists an k see Example 4.27 ε ⇒ ) The existence of a complete orthonormal system in . is characterised by ) Let S H A is the Banach space adjoint of 4.31 ) ( i ). Its ( iii )= ( is separable. K ∗ ( 2 ′ i S ∈ ∈ Define The statement is proved in linear algebra if ( ( T T ∈ 2 s x , so Every complete orthonormal system in H ⇒ There exists an countable complete orthonormal system in ,H ⇒ ′ ℓ . Let . Then λ 1 s . H H )= ∼ = H | (i) ) = 6 = ( (ii) ii iii (iii) T isomorphism in the representation theorem Fr´echet-Riesz (Theo 4.4 Linear operatorsDefinition 4.38. in Hilbert spaces Corollary 4.37 (Fischer-Riesz theorem). Proof. by Bessel’s inequality. Then L ( where Corollary 4.36. s Pythagoras Proof. H in equality. To show that every in by Lemma is equivalent. Zorn’s lemma. By assumption, it( must be complete. Chapter 4. Hilbert spaces Lemma 4.33. Lemma 4.34. Theorem 4.35. Note that by construction Theorem is not finite. For Then Hence | and Analogously, Theorem 4.39. Proof. ∈ = i =0 . It T . i i . if and H 1 . M ∈ 1 Tx,x h k≤ 1, M. x Tx,y Tx,x . We write xample, let x h h k≤ } k k≤ ≤ if and only if 0 T y ) k≤ k k is selfadjoint in 2 the assumption y \{ i k k increasing , y T 1 = 4Re H H k 1, 0 for all i i x,y 6 = 0 but ) is ) i| ∈ + y k≤ m 2 T H x k≤ x k ( T decreasing i ≥ selfadjoint. If − because for k x , so x Ty,x y,Tx L ) k h h k − i| k ( , x ∈ . H ) is n T ) ( k 2 1. This shows T +2 +2 Tx,x y N M H a normal operator. Then ( H, L Last Change: Thu 4 Apr 11:51:53 COT 2013 h . Then T i i ( ∈ h ◦ ), for ≤k ) 0 for all Tx,y − n ∈ ∈ L = M, ) . |h k≤ x H 2 ≤k n > 4.8 ( ∈ T ( M x )= 2 k 0, if T 2 = k i L T k Tx,y Tx,y x N k i| ≤ h h i ∗ y |h ∈ ≥ x 4.4. Linear operators in Hilbert spaces ∈ x, y n T ,y ( . Observe that − ) , x + kk ) T T =2 =2 n k 1 for x H Tx,x i| T nm i x T −k h k Tx,y λx y ∗ y h ∈ ( 2 λ T M k k≤ + T − k h . Obviously + 0, if i|≤k x, y T x T x , x i| = , x converges strongly. , s 2 Re k = 0 automatically implies that k ) ) k | k K y i y y 0. A sequence ( H = T > = the rotation about 90 + i T x − Tx,x → + , so ≥ . Tx,x i , denoted by k x 2 |h ∗ x x |h ( ( M H be a Hilbert space. A bounded selfadjoint operator ,y T k R Tx,x T . A sequence ( 1 be a Hilbert space and x , x be a Hilbert space. Every monotonic bounded sequence of T h ) T ∗ × N − |h M H = 1 such that k≤ → be a Hilbert space, H λx H x | S ∈ . H ( ik ≤ 2 be a bounded monotonic sequence of selfadjoint operators. 1 k 4 4 1 TT λ i − h : H = ker | T R n Let y h N − Let , ≤ : Let , =0 T ∈ ) is increasing. , T x i≤ DRAFT C + nm , denoted by = n Let non-negative T +1 s ) H T x ∈ := sup n ( kh· i| ker n ∗ , x . T T L λ 2 ) T h M y Tx,y R ≤ ∈ h 1 0 positive + , then 0 1 ∈ n N Tx,y − Re if and only if ∈ T Let ( 0= Let is selfadjoint is necessary for the statement in the corollary. For e x |h H x ( n ) S T ∈ ) is called T n h T x H ≤ = ( , − Without restriction we assume that it is increasing. Let Proof. ( Theorem 4.49. selfadjoint linear operators on T only if is called L the case of a complex Hilbert space. In a real Hilbert spaces Proof. 68 Proof. Corollary 4.46. Definition 4.48. To show the reverse inequality fix in particular, In particular, Now choose that 0 Note that the condition T Hence, by the parallelogram identity (Theorem Lemma 4.47. for all 16 Abr 2012 a 67 n< H . , i i = , → 2 . H H =1 =0 Ty,x x,Ty s, s. : n ij . h h i ) λ λ H T ji ) d ) d is closed. Let the following is a s s + + ,Ty = dim ( ( ) i i 0 )= f 1 f H h ) H ) T ( H has a representation ( = L , s,t 2 s,t D i ( i ( y,Tx Tx,y = i we obtain , H. ∈ ) = H h h k is then ( k i λ λ λ 1 T 1 ∈ ∗ T ( 0 0 ,Ty → T + + Z Z H. D n x,Ty 1 selfadjoint. Then i i h x ∈ x,Ty ) H Last Change: Thu 4 Apr 11:51:53 COT 2013 h . + : )= )= H t t i| i ( + , x →∞ = 1 and T lim be a Hilbert space, i )( )( Ty,y Ty,y i L n h h f h f λ 1]) define k k ∈ H , + + = T T Tx,x y,Tx i i ( ( [0 T i |h , x,y Tx,x y,Tx i h Let 1 −h h × . Observe that = k≤ ,Ty y = = 1] x , Tx,x Tx,x , , i i i n k h h x,Ty 1] → x 1] 2 2 . is closed because be Hilbert spaces with dim | | h h ([0 , , = sup C n λ λ 2 | | [0 [0 ∞ k = T 2 2 ∈ x,Tx Ty,x Ty,x L →∞ i T x T h h = = ,H L L n λ k is closable, hence closed since 1 ∈ i i = + y y = lim H → → k T i − h i i . i i + + Tx,y and 1] 1] . 0 and h , be a Hilbert space, be a complex Hilbert space. For , ,y H H ). The matrix corresponding to n , so in the special cases [0 [0 1] and → Tx,y Tx,y 2 Tx,x 2 H H C , , λx , λx ∈ . h h B x,Ty h ∈ (i) Let ( n DRAFT 1]. For L L h k ) ) T x [0 , n h x : : y y T = 2 Let Let = , x [0 ∗ k k M L x, y ) follows from 2 + + = A i i R T T →∞ ( L ∈ ∈ ∗ k n with λx λx k T = ( ( i ∈ ⇒ = lim =1 T H T T Tx,y n ij 2 h h H h ) Let ) )= k ⊆ is bounded, hence selfadjoint. ii is selfadjoint. ii . After choice of bases, a linear operator y ij ( ( It suffices to show that := := k N Tx,x a T that is h T ∞ ( Then ∈ = 0. This implies that A ⇒ B n ) y n (i) )= (ii) (ii) Let i x so finally Chapter 4. Hilbert spaces Examples 4.42. Theorem 4.44. Theorem 4.45. so linear operator such that equivalent. Theorem 4.43 (Hellinger-Toeplitz). ( Proof. By assumption, ( Proof. Then , ) 2 is 2 + ii P P the 2 y and , 1 ) P ), so } is an 1 2 1 P 1 1 0 P P P , hence P P P , so also 2 = . . , − \{ ⊥ 1 2 2 2 P ∗ U U H P U ) 2 4.50 ∩ ∈ = ⊕ = P =0 = (id 1 = 0. Since ( 1 . In this case, 1 y 1 2 ⊥ 1 = 0. Therefore U 2 y and by Proposi- P U ) P ) P P 1 P . 2 ) = rg( ) 2 y ∈ 2 1 2 P 2 1 P P P P is selfadjoint, hence, P P is a projection. By x . P )= P . = 1 2 2 = 2 and ( + = − is a projection because P P P P 2 P 2 2 1 2 2 P orthonormal projection P = (rg + P + = ing is equivalent: P )=0 orthogonal projections on 1 P P 1 1 ) orthonormal projection 2 2 1 1 1 2 P 2 2 P P = , hence P P + closed subspaces and P P P P is selfadjoint, hence 1 P P )=(1 2 1 1 for every 2 = 0. , then . Note that for + 1 = = 0, hence P Last Change: Thu 4 Apr 11:51:53 COT 2013 P P . On the other hand, if H rg( − P x P 1 ker P )+ 1 x 1 2 H y 2 1 1 ) P P ) 4.5. Projections in Hilbert spaces = and 2 + P ⊆ P − 2 ( 2 P P ⊆ = 2 ∈ P 1 1 ( 1 1 2 P P P 2 = ker x − P − P (1 P 1 P 2 ) 4.43 − 2 2 2 P 2 P ,U 2 ⊥ = = 0, so 1 P ) 1 (1 P P P 1 x, y P 1 2 U 1 ∗ 1 U . P P + P P ⊆ = rg 2 2 ∗ 2 + . P 2 P = 0 and 2 P 2 4.50 1 1 U U ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ for all U = P y P = (id + = ) = 2 i 2 ∩ ∗ + x P 2 = ( ) P 1 2 2 1 1 2 P 1 P U − P P P ) P 1 1 y,Px 2 . = h P is an orthogonal projection, it follows that P P } 2 are linearly independent, hence (id + , it follows also that 2 0 2 2 ) P = + { y we obtain P = ( 2 U 2 2 + is an orthogonal projection. Then i be a Hilbert space, be a Hilbert space and P . 2 + P 2 P P )= 1 1 P + P . 1 holds. an orthogonal projection. H H 2 4.52 . On the other hand, 1 is an orthonormal projection, then, by Proposition P 2 and =0 P 1 P 2 2 P P 2 U and 1 1 P x,Py U ) By assumption, DRAFT U ( h P P U y Let Let ii 4.50 1 2 ) ∩ ∩ ( is an orthogonal projection if and only if is an orthogonal projection. 2 P . ) Observe that P and 1 1 2 is an orthogonal projection. In this case, rg( P 2 2 is selfadjoint. U and iii U ⇒ 1 is an orthogonal projection if and only if is an projection on = U ( P P − 2 ker(id + ) Since U orthonormal projection 2 2 2 P ) By assumption, rg i ⊆ ⊇ P P ⊥ i − + ) = ) If ( . P P P 2 ⇒ 1 ⊆ i i ( ) ) 1 2 1 1 1 1 1 Assume that ( ( 2 2 P P 2 = is zero if and only if (id P P P ⇒ P P U U P P )= ⇒ P 4.50 1 1 − 2 1 ii y commute, then it is easy to verify that ( ⊥ P P 1 )= 2 P P (i) (i) )= ) Using Lemma 1 (ii) ),( (ii) 2 P P ii i iii ii (iii) rg 2 vectors (id tion In particular 0 = ( the corresponding orthogonal projections. Then the follow subspaces Since the sum of two selfadjoint operators is selfadjoint, is symmetric in 70 Proof. Lemma 4.52. Lemma 4.53. ( the theorem of Hellinger-Toeplitz (Theorem ( rg( Proof. If one of the equivalent conditions above hold, then If ( rg( orthogonal projection by Proposition selfadjoint, that is by Proposition U ( hence P Proof. , R 69 for 6 = 0 i . P . ) H x,y i ) h converges x ∈ 0. ) H. T 2 1 Im , x,y is an orthog- m i i h λ ∈ T be a bound of x x, y λ 2 1 i≥ 2 H ) ) − y , x x M k + 2i h m ) n → λ 2 T m T m k ( . T , H T + for all i − x i + 2 Re( : because Px,Px x, 2 − i a projection. If n h − 2 ( , x,y . Let k i T n k K P y ≤k ) n H ( m = y T k nm T is selfadjoint follows from ∈ k k H i 2 ∈ ( s 2 ( x , 1 2 ≥ . λ | y,Px λ Px,Py ) T Px,Py L h k h = R h ⊥ λ x,Ty x, P x n Last Change: Thu 4 Apr 11:51:53 COT 2013 | m ) h = ) i m x, y ∈ = T ∈ = = . i + , and 0 P x if m T + = ) i 2 1 i i T λ R − 2 P i i − λy m k H y n P x − ∈ T n x Py ). That n + T . k T n for all ( = (rg λ − x , x − + h k x,Py T ) is a Cauchy sequence, hence i ∗ h 2 2 = 1 1 n . Then, using Cauchy-Schwarz inequality, Px,Py (( x,T 3.13 m i i N 4.16 P T 2 h Py T ∈ ( λy , x =1 k M + h n ). Then for all nm . A linear operator and 2 = 0. Px,x ) k − − for all s x,Py x , x x , x x , h λy P →∞ x x i = h H ) ) ) i n n P x 2 1 P k n = ker i + = T m m m ) . Since for all T rg( k≤ = ( − = lim T T T = i x h P P x,y i λy m x,y ∈ i 3 h 2 Px,y x 2 h T − − − x, x h h ) with rg is a monotonically increasing bounded sequence in + y ( P ≤k n n n − Re be a Hilbert space, = = N M ∈ x,y T T , ker T 2 n H Px,x , x nm ( ( ∈ ( λ i i n Px,y ) (Proposition = Im k h T y h s (2 h ) n h ) and . T H ∈ i k ) : , H h i 4.47 = ≤ h ≤ ≤ = ( +2 λy H x λy P P Hilbert space H L 2 2 = 0. Let ∈ x,y + Px,y x,Py k →∞ k + DRAFT ∈ h h h x , x H ∈ i n ker x ker x x x x and n ) x ( ( , T T ∈ ∈ = lim ) follows from Theorem m P h 0 P Let m ≤k x,y i ii T x x k h ( . ≥ ≤ h − = i≥ ⇒ n ) is clear. 0 n 2 =1 ) Observe that ) By Lemma Tx,y T k , hence Re ) Let iv ) For all i ) Let h ( i i )= k ( ( is an orthogonal projection. is selfadjoint. is normal. iii v i ( . Note that then ( k R ( ( i λy ( P N Px,x k P k P P h ⇒ ⇒ ∈ ∈ ⇒ ⇒ ⇒ ⇒ n ) λ )= )= n (i) )= )= (v) (ii) )= )= (iv) (iii) ii i iii iv i v T then the following is equivalent. Proof. By assumption ( ( strongly to some ( hence convergent. It follows that ( onal projection if and only if Chapter 4. Hilbert spaces is a positive semidefinite sesquilinear form on 4.5 ProjectionsProposition in 4.50. Hilbert spaces Lemma 4.51. ( ( ( we find for ( it follows that In particular, 0 ( all . ) S ′ is to S ) . ′ . If + ) = X ) and Y TS TS T ) ′ T , so it , hence ) S (( ( ). Note − X S ), which X → . ). For all D ′ x | from T ′ 0 ( + ). Now let ( . For every T ∈ D T X 2.26 ϕ T ϕ , x ′ = id = id D T ′ ( ′ 0 z ) (( y ) ) S,T 1 ′ X = T x − ′ ( ) = ) ) and ∈ D T ′ D Y , S ∈ D T ) = id ) = ( x T x ′ ( ′ 6 = ′ + ′ n ) y , hence = ( 1 ) a dense subspace and ) T x , 1 ′ T − = id ( , x η X − ′ 0 ⊆ D ′ n T ) (Theorem T X (( ′ X ′ y ( y T Sx ) ) ( ( y iii is not necessarily true. For − ϕ ′ S T be the graph of TT ) ′ ′ ⊆ ( ( 1 z ) for all ′ ) →∞ ) and T − D →∞ ′ ) Sx S )= n ( Last Change: Thu 4 Apr 11:51:53 COT 2013 ⊆ X . n T x T T T ′ ) = ′ is closed. )( ( )= ( D + ( ( ) ′ ′ × Sx . ′ T z ) and ( ′ y ( ′ ′ T ) = lim D ′ T ′ T x is invertible and ) a densely defined linear operator in Sx ) ⊆ T ii 1 y Y 1 T ) is dense in ′ ). )( + ∈ D = )( ′ ( ′ − T x ′ T − ) is dense in ) T = ( ϕ ( is closed. z S ′ 1 ϕ T x T S T ′ ′ ′ TS n ′ ( ) )+ ( ′ . Then 7→ are densely defined such that also are densely defined such that also − }⊆ ( ′ y 1 , that is, if x S T . ) + T D ′ ′ ( ) ) T ⊆ ). Then the map − D = ( S Y ) and the map ) ′ T x 4.6. The adjoint of an unbounded operator ( + ′ ′ T ) ( Z T T Y 1 ′ ′ ′ S ( ( →∞ S 7→ 6 = T ( , x y − )= ( ). Let . Then TS n T T ′ ′ ) ) D ) ( → → ′ ′ ) K be densely defined linear operators T 1 + = lim 7→ ∈ D T T X ( ∩ D ⊆ Y ′ X − ( ∈ D ( ( )) → , x ( x ′ ( ) ) = (( ϕ D S ′ ′ L T D ′ n ) T ( T K T z T ( ( ′ ′ x T ′ be Banach spaces. S ): ∈ T x , x ( ′ ( S be Banach spaces, be Banach spaces. For linear operators ( T D T )= → 1 y ( ′ K D η ) we find ′ D ) it follows that ( ) ) and − and →∞ ii T, ) if and only if D ii . 1 be a Banach space, ′ ( ′ ( n ( T ) ) → . ,T − ′ ′ T S )= TS X, Y ) ) is a densely defined unbounded linear operator such that Y ′ is an extension of Y X, Y ( y X = lim y T ( X,Y,Z ) ( ( ) a convergent sequence with lim . S S ′ ⊆ Y ′ ′ G ′ 4.59 D x T { be a linear operator. Then and ) which implies that 4.59 ). Then → ′ ′ → z S ′ 0 ′ T T + Let ′ . + Let if ∈ ( Let ( x ) T T Let → ′ Y T )) ′ T DRAFT + ( ), hence X ′ X T ) ′ G D )= ′ T T T ( ′ ( S ′ ( ′ ( S G ( ( S X → T S ( S T D T ⊆ D ( ( ⊆ , x D ) ( ∈ ′ then = are bounded, then “=” holds in ( T N y ) S T G ′ ∈ D = ( S, ∈ ( ′ 0 ∈ D T z ( ′ T ∈ D ′ n = id ) it follows that ′ ′ ) is clear from the definition of the adjoint operator. y ) ) it follows that y ′ , x i z ⊆ T η ′ )) is densely defined with By Theorem Let ( ′ 0 ) ( T T ′ n ′ ⊇ D y ( ′ S densely defined. Then Assume Let Assume is densely defined. Then S TS and ) T , x 1 X + ′ with bounded inverse n ) Let − S we write ∈ D : (i) ) Let ∈ D y (ii) T ii iii (iii) T Corollary 4.60. thus ( also Again by Theorem ( example, if x X suffices to show Proof. x Theorem 4.59. Note that ( (( ( T Y Proof. implies is continuous. Since by assumption that for unbounded linear operators 72 Theorem 4.57. Definition 4.58. If is continuous. Then also its restriction Proof. ( is continuous. Note that by assumption and ( ( . , N 0 = = 71 P ∈ i n H 0 , ) = 1, = x. P } n 2 ) n 2 k . P ) n P T ( ′ P 0, hence ker ( Px,x P T h + k ( D 0 ⊆ x P ′ . ) i ≥ ⊥ 1 k n T ∈ D x f an unbounded P ker H 1 . Then x , x P − a dense subspace. 0 ⊆ , ϕ because ) P ) P ⊥ 0 1 X exists and is a selfad- ( T x ( P ) n H k≤k P n m , it follows that ( X T ⊆ T ∈ P ) P . . P we have defined the adjoint of bounded linear opera- P n P ( H ( ( (id ϕ 2 be Banach spaces and k 0 T 0 T H k D P x − ( = ker ∈ , 0, : ′ (rg H P 4.4 7→ ∈ x ∈ D P ⊥ , x H → P x X 1 i ≤ h we already know that , the map x i → k )( be a Hilbert space and ⊇ D ϕ X, Y x H ∈ be a Hilbert space and n X : → n = ⇐⇒ ⇐⇒ ⇐⇒ P x X x , x , x ∈ ) ′ P H x , x 4.49 DRAFT P x for i ′ H k : N 1 0 ) m Let + ′ Y ) Let x x T = n P P P ( ∈ 1 T n h X iii P ∈ Let . x P Let P ( n = 2 0 n and section , ∈ ϕ − 1 P 1 − P { ⊇ D i ≤ h Px,Px ϕ P ′ ′ P H h = 0 2.4 P and ⇐⇒ ) Let ) For all k≤k ) i ) is dense in Y T P 1 ii ( x ) ⊆ = = 0. It follows that x , x : iv ) := ( T 0 P H ′ 0 . Then the following is equivalent. ii = ( ( i ( ′ 0 i 0 ( By Theorem P P T ⇒ ∈ T x H D . Since H h k P ⇒ ( 2 N x D x , x . ⊆ x , x P ⇐⇒ ) = )= 2 0 (i) ⊥ ∈ 0 1 ) (ii) (iv) P i iv iii P (iii) h ( Since projections with Chapter 4. Hilbert spaces Lemma 4.54. Lemma 4.55. 4.6 The adjointIn sections of an unbounded operator H Proof. ( converges strongly to an orthogonal projection. Proof. extension For a linear map ators defined on a subspace ( H h tors between Banach or Hilbert spaces. Now we define the adjoint o linear operator. Recall that Note that ( Definition 4.56. is well-defined. For joint operator. It remains to be shown that n 1 is is 1, − x T T < (4.2) k 7→ . ( . Then ′ 0 . Then ′ ′ ′ x is closed. x a densely x k X . , and since T ) 0 such that = 1). Assume } . Since is a Banach ◦ = ′ , 0 ′ ) X, x Y { ′ y Y T (0 ′ T y r> ′ → Y ) with = ∈ T → ′ rg T ′ ′ ) B T y (ker T X T ( . ⊇ rg( ⊆ T T contains the open unit .) )) is open as composition ) k·k ∈ 1 ′ is bounded on , r ) with , r − T ′ ′ 0 ) ker T ) is proved analogously. T x ) 1 be reflexive Banach spaces x ′ (0 T T . . Define a linear functional iv ∈ − rg ( ( ◦ ) such that ( = T ) X ′ ) ). ′ ( x ′ = 0 for all D . Note that B T T → Last Change: Thu 4 Apr 11:51:53 COT 2013 T < r ) is a Banach space and ( T . This is the case if and only if ( e x Y ( T x : ( ∈ D X, Y ) ) T k T Y ′ ⇐⇒ ′ i ′ 0 ′ T y (rg ) → ) ( y ′ y . Then ∈ D ◦ 1 ii k Let k·k ) T ′ T, Y )= − , ′ ∈ y is open by the open mapping theorem T ) ) = (ker T ⊇ D ( rg T x ) T ′′ and T = 0, hence )= ( ). ii ) = ( T X ( → , similar for = ( ii ′ 0 T D 4.6. The adjoint of an unbounded operator ) be a Banach spaces, rg : ( ⊇ D ) x (ker 1 T x T x T ◦ 1 0 such that T ( − is closed and . Choose ), hence ( rg( , that is, if and only if ker ′ ′ (ker → = ) − 4.64 X ′ ◦ i y Y . Choose ) T rg( 4.64 ′ T 0 : T is open, then it is surjective, hence rg ◦ X, Y k·k y T r > that T ( ∈ 4.57 ′ with centre 0 and radius 1 , rg e → is open. T ) T )= y T − ′ ) i ) T ∈ ker X Let k·k ′ is open. is closed, ( rg T (0) = 0. It follows that rg( is open if and only if there exists an , = ( T ◦ ′ , 4.63 ′ (0) = 0, hence ( T ) ∈ y ′ D ). Observe that also x ). Then T and T T e D T → rg( y T D x ′ T a closed densely defined linear operator. The following is : ( ) rg( : as above. : ( ( ) with if and only if T = ′ e T rg( 3.32 ) = r T . . e T Y ( )= T T → ) ◦ Y ′ T ( ) ∩ D ) and D ) ker → T x ′ T → ) and T x : ) ( T T = ′ ) ′ ′ ( ∈ T ) ) Since ′ y ( T T ∈ D DRAFT T T is open, we have to show that for every y T T ( ϕ iii ′ ′ ( 0 (ker ( ker( = ), then also ◦ rg( y = rg T (rg is closed (Theorem ⊇ D is closed. x ◦ ∈ x ′ ∈ is closed. ⊇ D ) Recall that ′ ), ) ) ) follows from theorem ′ ⊇ D ) = (ker ) ′ ′ 3.22 T ii ∈ ′ )= ′ ) ⇐⇒ x iv ) follows from theorem Y y y is bounded if and only if x vi ′ X ( . That is, there exists a ′ v ) T T is closed, the statement is proved. T T : ( x X i : ( 1 T ′ is open and let is closed, then T ( Y ◦ : − ) T T T T T T ⇐⇒ T = ( = ( ⇐⇒ ) = 0 for all ⇐⇒ ) By( ) ) Let y ) (i) rg( ) Let x x (v) rg( ) (ii) rg( ( ′ ′ (vi) rg( (iv) iii iv v i ii iii (iii) ( there exists a ( defined injective linear operator with and (Theorem Proof. space. If rg of open maps. If the image of the open ball in denotes the inverse of Proof. To show that ( ( 74 ( Theorem 4.65 (Phillips). Theorem 4.66 (Closed range theorem). that x ball in ( ϕ Note that x is continuous (Lemma (ker Now let continuous and that closed, it follows by Lemma equivalent: and . ′ is ) 73 A )). we Y ) ′ such X ,y ∈ T × ( ) X ′ x ((0 because X T ⊆ . If ( ψ ( a densely . ∈ D ). Assume A B B ∈ ◦ 0)) ) ◦ X x ). Hence, by J A ψ ∈ D Y x, ( T ∈ ) = ) = ( ◦ 6 = 0. Therefore and ϕ y (( ◦ be a convergent n ( G ) → ψ A ⊆ x a X ◦ ( = 0 for all ( 6 = ◦ − A , ϕ A X ϕ ⊆ ) ( x →∞ 0 )) K , ′ n ′ T n ⊆ A )=0 x ,y X . N )=0 X. → := lim =0 ∈ y T x ), also ) there exists x, T x ( ( n and →∞ ◦ 0 ϕ ) Y ϕ n 4.62 }⊆ ϕ (( ) because for all ′ }⊆ x = 0 and A ′ ′ n } : ( ψ A 0 x Last Change: Thu 4 Apr 11:51:53 COT 2013 2.19 . B T ◦ = lim A ) { ) ) ( | ,T ′ ∈ ϕ ) ∈ T x T ϕ ⊆ ′ T ( ′ 0 6 = T x ∈ D A 0)) = , x ( rg( . Then there exists a , ϕ Y } using the canonical map ϕ ker( ∈ D ∈ x, is closed and (0 0 ( ′′ ∈ D ∈ y x ) = 0. . T )=0 a \{ } − X )=0 ϕ ϕ ( x 0 ) ) and Remark x ( 6 = 0. Let such that Y ϕ { i ( because are closed subspaces and ϕ ) of ′ ϕ ∈ : 6 = ◦ )) T : with X . By a corollary to the Hahn-Banach theorem ′ ( B 0 0 ) A x, T x ′ ◦ y . A G ∈ X ) by ( ,y X X ⇐⇒ ∀ ⇐⇒ ∀ ⇐⇒ ⇐⇒ (( T ′ ∈ B ), also \ ( ϕ ψ ∈ are subspaces. Let ( ◦ be Banach spaces. For a densely defined closed linear ∈ T . due to Phillips. ) ′ . ′ n ((0 ∈ D and . Since obviously ◦ A 6 = 0. Moreover, ϕ x x ψ T ( Y B B . ◦ ϕ { ◦ { T. A A ) ◦ ) ◦ ( (ker be Banach space, 0 is injective. )) = A = 4.65 ◦ be a Banach space. For subspaces T ◦ y 0)) ′ →∞ := := ( ◦ and )= n ) ∈ T ◦ x, X ϕ ◦ = ker rg( and , T x B . . )= B X A ◦ (( a ◦ A the following holds: ) X, Y ◦ ◦ DRAFT ◦ ∈ ◦ ) = ker := lim ′ ) ( be a convergent sequence. Then ) ) ψ ((0 ) ◦ Let A T = 0 and and ′ 0 T T ϕ T T ψ ( − Let Y x ′ B ⇐⇒ = 0 for all The sets ) Let ◦ rg( = ) there exists a T Y n ( , in particular, → (ker ∩ D )= ((rg (ker ⊆ G = ∈ ◦ Y ◦ ) | ϕx ′ X follows if we identify ◦ ⊆ 2.19 N ϕ T x ψ 6 =0 ( = = ) T = ′ ( ∈ ) T B ) The first equality is clear. The second equality follows from T n ϕ T T T 0 i →∞ T ) Obviously, ( By assumption, the graph , so by definition of y n (rg n = ( ◦ ◦ rg rg rg rg x ϕ ◦ = lim A ) 0 ) (i) rg( ∈ B (v) (ii) (iv) ii (iii) ◦ Now we show that Let ( ϕx sequence. Then reflexive, then also ( Proof. Obviously (Corollary that there exists an Proof. Chapter 4. Hilbert spaces More general is Theorem Lemma 4.63. Theorem 4.64. operator ( such that ϕ defined closed linear operator and ( Definition 4.61. that define the annihilators Remark 4.62. Proof. a corollary to the Hahn-Banach theorem (Corollary , . . 1 1 H , so H H ) i . T ) . For = , so it = ( ∗ y = id ∗ H G S x,y T ) ∗ h ( 1 TS + = H G − = = id T T i ∈ ⊥⊥ ) ) ) = id . T yields ( ) = ( T ∗ ( ∗ . ∗ D y,z ) T Tx,y ) z ( 1 ( T G 1 y − . G − − = to T = id T a densely defined closed h TT ⊥ x, ) , ∗ ( )= = ) and 2 )] ) 2 = ). This implies ∗ T 4.69 . ⊥ H T H ( ⊆ ∗∗ i T ) ) T 1 × ( ) = ( y ( G . T 1 ∗ T − iii ∗ → ( Last Change: Thu 4 Apr 11:51:53 COT 2013 ( ) H T T = −∗ i 1 1 ∗ ( G y, x ) ∈ D ) ( − ( ∈ D T H 1 y is invertible and ( T − ⊆ and ⊥⊥ are densely defined with a densely defined linear operator in − ). y ( y,z ) =: ∗ T ∗ ) are densely defined with ( ∗ ) T T ) = [ T ii T ∗ 3 T ) ( be densely defined linear operators. If 1 ( ) ( 2 1 . Let H ⊥ T x , − G , U , V ) D h − H )] 2 2 ) Y T ). 4.6. The adjoint of an unbounded operator = 0 ( T ). Let → ⊥ = H H ∗ ) ∗ → )= ,y i ⊥ 2 4.3 = 0 for all T T . Then )= T × ∗ 1 = ( ( ) ( ) ∗ H → (0 i ))] 1 1 ( h H 1 G ∗ D T H ( 1 . − be Hilbert spaces. ,y ( H ( − T ( T ∗ = ) T . Tx,z 0 T H ( ) 3 L D ∗ h ∗ ( is densely defined and y ( 2 → h T G ∗ VU T T ) = ∈ S H ∗ ( be Hilbert spaces, 1 ( and application of Lemma ,H and ∗ T 1 × and 2 1 + ) it follows that ( ) T ( 2 1 2 ) we find H ) = [ − 1 . . ii − . ) 2 S ii ) D be a Hilbert space, H H ( ∗ ∗ 2 − ,H ( ( U and ∗ × i T ⊥ ,H ) [ × . H S 1 ) = ker T 1 T H 1 2 ) ( H ( , . Then ⊆ ⊥ . ))] ∈ 0=0, so H 2 H ∗ ⊥ ) H ⊆ H D TS ∗ 4.71 ⊥ ) → ) T ∗ ) → 4.71 z,y T ( ) : H T id ∗ ∗ 0 T T ( ∗ 1 T 1 . − Let ( 4.69 S T Let = ⊆ ) Let ( − V ,y T DRAFT D H = id H G → ( 0 ∗ ( T rg( + ( , ( (0 = ∗ ( y . ) it follows that = ker ) S 1 T ∗ S V 0 = T ∗ G T T then T H ⊥ ∗ ∈ D ( S ( are bounded, then “=” holds in ( ) ) ,y ⊥ = VU = is closable, then 0 ) ) = (ker ∗ 1 T T y ) which implies ) = [ (0 T − ) follows immediately from ( T T T ∗ h T ∈ D ∗∗ i ⊆ ∗∗ T ( By Theorem As is in the Banach space case. T T x ( Then Assume Then Let If S Assume rg( T and ( 0= G with bounded inverse S (i) rg( (i) ) Let ∈ D (ii) (ii) (ii) (iii) ii (iii) rg( z If every hence linear operator. Then the following holds. suffices to show ( 76 Theorem 4.71. Corollary 4.72. Theorem 4.73. Proof. Again by Theorem hence ( Proof. Proof. H It follows that Obviously Hence by Lemma X 75 sub- (4.3) 2 ) ) H T T ( ( are called as desired. ⊆ 1 ) such that . Note that , 1 ) H } ∈ D T ∈ D S T H ) ( ( is defined by ) T ) . By the Riesz → ∈ D ( ′ ∗ 1 T . The first equal- tor is not unique; T a dense subspace. . ) ∗ , ( ( D ) , x . T , x ⊥ ∈ D 1 H y S 1 ) S ( x H ( ))] x H ∈ x ∈ D =0 =0 ∈ D − , by the reflexivity of ′ 0 T ⊆ ∗ ( ⊆ y 1 1 x ). ⊇ D . y, y ) ∈ D T ϕ H ) H G ( T . Note that 2 ⊥ ( T is formally adjoint to every × × ( T , x r = . ( linearly to rg 2 2 = , x ( H )= U , y } ))] 7→ ⊥ ) y H H : D X 0 ′′ D ϕ i i ) ∗ T i { =0 ∈ D Last Change: Thu 4 Apr 11:51:53 COT 2013 T ) ) 1 choose T x ( T k≤ S 0 ))] ( ( 0 0 i ′ 0 x ) = [ G = T 0 ϕ y x, y ( < is continuous and densely defined ( , x ⊥ , x ( k ) 0 0 x , x U D ∈ D i G and is a bounded on k h y y ( T x , x )= ( ( y ,T 2 ( i 1 k U = , , [ for all H G with Hilbert space adjoint ) = [ , T x ) is its maximal formally adjoint operator. H ) ( Y i ( ′ ∈ 1 Tx,y x ⊥ i − h i ∗ ∗ ′ 0 U and extend → ) 0 0 ) Y , x H its → y − T 0 ) with Tx,y 1 T x ) there exists exactly one ) 2 ( ∈ 7→ h × ′ 0 T x, T x H ∗ x,y , x 7→ ( ( i is bounded, so by the theorem of Hahn-Banach T ′ x h 0 G H 2 0 T x, x T ( 7→ h be Hilbert spaces. For a densely defined linear y y Tx,y Tx,y U ( ( 4.20 ϕ H rg ( = U x → 2 , ⊇ D i be Hilbert spaces and define k ) = ) , : → ∈ H be Hilbert spaces and x,Sy be Hilbert spaces and 1 y y ⊇ D 2 h T ( 2 )= K k 2 ( 2 y 2 H 2 ∗ H r ϕ = H : H H ). T Tx,y → and ,H ,H 2 ⇐⇒ ⇐⇒ h ⇐⇒ h ⇐⇒ h ⇐⇒ h ( h : × 1 T 1 T ) ⊇ D ∈ H ) the map 1 ( k≤ G ∗ and 1 is unitary, hence i ∗ H T H 1 y x H ( T 1 H . Set T { ) H U the following holds: ( y : DRAFT D ∗ kk ∈ D H : ′ 0 ) Let Let T ′ 0 Let U x Tx,y ( = T Y y if h ) := ∈ D is a densely defined linear operator, then G Let ∗ such that ) y → T T x ∈ 1 2 ( |≤k ) ) Follows analogously if we note that x H X D H 0 ′ 0 ( ) as follows: for iii ) follows from ∈ x ( is closed. and T T , x | → 0 ∗ ∗ Observe that 4.3 1 y y = . ( T | H is densely defined, then its adjoint < r Y ⇐⇒ ) ( y ) on rg( k T T (i) ( x iv ϕ representation theorem (Theorem and can therefore be extended uniquely to an element is continuous, so ( Note that for Proof. If | ity in ( Lemma 4.69. For a linear map where k and operator it can be extended to a functional linear operator. If Chapter 4. Hilbert spaces ϕ Definition 4.67. Theorem 4.70. formally adjoint spaces. The linear maps Definition 4.68. Note that the formal adjoint of a non-densely defined linear opera in particular, the operator trivial operator with is symmetric. 3 T a dense subspace and H ⊆ ) Last Change: Thu 4 Apr 11:51:53 COT 2013 T ( . D ∗ T = T 4.6. The adjoint of an unbounded operator if . . ∗ ∗ T T = ⊆ T T if if be a Hilbert spaces, H a linear map. Let H in the example above is selfadjoint, the operator DRAFT selfadjoint essentially selfadjoint symmetric 2 → T ) T ( is called is called is called ⊇ D T T T H : (i) (ii) (iii) The operator 78 Definition 4.75. T 77 )= t , } 1] , ). This is [0 ). We will 2 T ) ( ?? L 1 t . Application ) T ∗ ∈ 3 T t ( . Define Φ( ′ ) d T ∈ D t g ( = ( ∗ 1 ) d ′ x t , x ∈ D T y , ( ∗∗ , ) ) x ) ⊆ D t ϕ 3 T t = ( ) 1 ) t. T (Theorem t x . T ( 1 ( i is absolutely continuous ϕ ′ ( ) d T x for all 1 x t H ) d i ( g 0 t . For . i 1 . ∈ D Z D are closed. Φ( 0 ϕ ∈ D i ⊥ = i ) . Φ( ) Z . t y k − ) ) . ) and x ( Last Change: Thu 4 Apr 11:51:53 COT 2013 ). 1 = } t 2 T T ′ 2 , y ∗ 1 0 k 1 3 ( = ′ ) T ′ x T Tx,y T T i ) ( 1 i h x ( ( t i T 1 x,T ( } = ( 0 h 1 , x y ) 0 Z ) and ) ∈ D 3 x , ϕ t (1) (1) = 0 ∈ D ∈ D ∗ Z 2 ) is dense in T h ( − x x ( ) = (ker ∈ D =0 k T g g x − ∗ H,T absolutely continuous ( 1 0 D . = T t | (0) + ( T ∗ ) ⊥ i y x 2 = i t is chosen to be a constant function. Hence ) → . It follows that D T ∗ t } ⊆ D )) d ) let ) rg( (0) = (0) = x Φ(1) (0) Φ( 0 3 k T x 1] : ) x,g x x , x , x,y ) x { can be shown. T , 2 1 T i t ) d i i . if and only if ( t ( ( (1) and T 1 T = 0. , in particular all [0 ( T ( = x h x − 2 T ⊥ 2 )iΦ( 1] : 1] : 1 y 1]. Let y t ) (1) = 0, so , , T ) D , T shows L ∗ ( ⊥ = = = t ⊆ D = ⊇ D ϕ T = (ker ) ( [0 [0 g x,Ty x,Ty (1) [0 ∗ T t ′ = 1 ( h h ) 2 2 = ∗ 2 y 3 ∈ , ) x ∗ T is closed and densely defined and = ker T H ∗ 1 ) i ) L L rg( 2 t T L = = 3 ⊥⊥ ) d : x ( ∗ T 1 ∗ 1 T ′ t ) (1) ⊥ T ( { ∈ ∈ i i ∈ 0 ( ( rg( k T ) = , x x T . T i (1) = i ( Z ∗ g D 1 ) and ) to T D x x ). Then, using integration by parts, y ) and | g ⊥ ∗ 1 ∈ T ii 1 T t H ) ∗ 0 3 = i = rg( ( 2 T DRAFT T ′ 1) = Z ∩{ ∩{ T T ( = i ( T ⊆ D 4.70 Tx,y Tx,y , x ) ) are well-defined and h h , i ∗ ) Let 3 1 1 3 )= (0 = 1 3 + iΦ k T T T T 1 0 ∈ D x,y ∈ D 2 T ∗ T ( ( T 2 g , 1 Z ( (0) = 0, y 3 = T D W D T 3 let D T ϕ h rg( shows rg( ) , ) = (ker x, y 3 2 ) ∗ ∗ T = i . Then Φ is absolutely continuous and Φ , ( ) := ) Note that ) := ) := s T T ∗ i 1 D 2 1 3 | ( Let T T T T ) d ∗ =1 1 rg( (0) = i ( ( ( s g T ( ) to k D D D i ) By Theorem ) Application of ( ϕ ) By( t 0 (iv) ii iii iv implying that ( ( ( equivalent to This shows that Proof. and Analogously, Chapter 4. Hilbert spaces Example 4.74. and R In particular we obtain of ( To prove the inclusion For Note that Φ(1) = 0 as can be seen if Proof. Obviously, the show: = ) = ) = µ be a λ ( − ) with X ) if and P λ, T ) X ( ). Hence, T ( X − ( R λ ( L ρ ) Let . P − } X ∈ ∈ 7→ ) ( T λ X P S )( ) is the set of all ( 5.2. The resolvent , λ T T L ) ( ( . p ∈ X Q σ ( +1 1 1 such that ) is not bijective which n L − − ) λ )= )) C k ( a closed linear operator. λ and ) . T such that → ) P ( ∈ ,T ∅ − T ) ,T 0 Q X Q − n 0 ) λ T . T ( ) must belong to the spectrum of λ ( ) Last Change: Wed 17 Apr 11:33:04 COT 2013 . λ . ) the resolvent has a power series ( → ) = ρ and T R j n T )) T ( ( − ( R ). T λ S 0 ( ( T X n k ρ P T r ρ ( λ ( ) T ). With this definition it follows that ( ,...,λ σ =0 ∈ σ λ is closed if it exists. Therefore, by the < ∞ T ρ 1 ∈ ( X n X | 0 1 ( − 0 0 ∈ P λ − = ) = ( λ L 0 λ ); ) = ) is not invertible, at least one of the terms λ / ) because one of the following cases holds: a, λ a polynomial. Then is defined everywhere and not closed, so it λ 1 λ λ T T ( ∈ ) and it is given explicitly by the Neumann ( − ] µ − ( ( 1 X ) )) = c 1 P ρ − λ X − X =0 | ∞ − σ [ ( ) S − T X n → T is injective and ( for all ) ∈ − ( ) L C T − ) T P T λ / λ X ∈ ∈ ( ( − T ( )= )) with − ( (id σ − 1 P λ T T P ⇒ λ , then P − ( C T ) λ, T ∞ : ∈ S ( , σ )). 1 )). 0 and ) is not bijective, so − R λ T )). There exist λ C λ T ( ) ( T σ ∈ ). Since ( X < X − ( Recall that for a bounded linear operator ) is a densely defined linear operator. Then P be a Banach space and ( n ) and ( λ P P dist( . is closed, then ( T X λ L ( σ { ) X σ . Then there exists a polynomial ∈ T ∈ T − DRAFT 5.7 ⊆ 2.10 → ( ) C . ∈ k≥ If dim j )= Often the resolvent set of a linear operator is defined slightly differ- If ρ T X Let ) T )) λ ∈ ( T X µ is bijective and ( ( ∈ ). In particular, T ( ( ) shows that locally around a is bijective, then ( is not bijective = ,T ( ρ λ 0 P ii T T 0 ), then ( X ··· σ . We will show that its domain is open and that it is analytic. λ T T ( for every non-closed ( λ ) T 1 = − ( 1 ( Q P ∅ − − − cannot be invertible, that is at least one Let ) λ λ 1 the operator (id σ R µ ) λ cannot be bounded, which implies λ λ For k T T − ∈ < )= − − − k and λ (i) (i) X T (ii) (ii) j ( ( S X λ Theorem 5.6 (Spectral mapping theorem for polynomials). eigenvalues of Remark 5.5. λ Banach space, T Proof. a ρ implies Note that ( Now assume ( if Remark 5.4. ently: Let only if closed graph theorem, expansion with coefficients depending only on Proof of Lemma ( Lemma 5.7. 80 Remark 5.3. 5.2 The resolvent In this section we will study the resolvent map k series (Theorem 79 , } X ), and the T . id) = ( . T, λ T λ } his chapter are ). The map N of N of − T. , ∈ λ T of λ, T n continuous spectrum ( in rg( . R ), } T T ) ) a densely defined linear ( X X, := . p of . 1 X σ 6 = 6 = T ) − eigenvalues λ, T spectrum resolvent set ) T − ( → ( T id) id) r R Last Change: Wed 17 Apr 11:33:04 COT 2013 id σ X − λ λ ( λ = 0 for some ˙ 7→ ∪ } id T − − x ) λ , n T T T ) } ( λ c , λ is ( rg( rg( ), by ) are called σ ) − ˙ λ T , , T ∪ X point spectrum ( ( ( T ) , instead of λ p ) in L T is bijective σ A λ : ( ( T T , p ∈ T geometric eigenspace → − λ σ − ): X ) λ − of T T λ T in ∈ ( id ( r )= is not injective is injective is injective ρ x σ λ T T { T T T ( : ) of σ − − − . T C ( be a Banach space and ) := ker( ) := id id id ρ resolvent ∈ T T λ λ λ \ X ( ( λ : DRAFT : : λ λ (i) Elements { C is further divided in ) we define the A N ) the C C C T Let T ( T p ∈ ∈ ∈ ( ) := ) := σ ρ λ λ λ T T ( ( { { { resolvent map ∈ ∈ ρ σ residual spectrum λ λ ) := ) := ) := T T T algebraic eigenspace is the ( ( ( ) and r c p T σ σ σ ( (ii) For c (iii) For Definition 5.2. operator. In the following, we often write assumed to be complex vector spaces. 5.1 The spectrumDefinition 5.1. of a linear operator Chapter 5. Spectrum of linear operators Chapter 5 Spectrum of linear operators If not stated explicitely otherwise, all Hilbert and Banach spaces in t The spectrum of σ It follows immediately from the definition that ′ . ′′ C Ω X . X <ε <ε } with → (5.2) | | 1 ∈ X ) ) ). By ′ → }⊆ ϕ m m r → :Ω X the map x x X k≤ 0. Then ( ( λ, T ′ f : ϕ ( ∈ ϕ ϕ |≤ :Ω k is a Cauchy X 0 R X open. A map − − f ϕ , z J ′ ) ε > ) ∈ X C n n if and only if it − 7→ . X ) x x ϕ . It follows that z ∈ ( 0 ⊆ ( 5.2. The resolvent | ∈ z ϕ : N ϕ ( | Ω , λ N | ) (5.1) ϕ r ) ∈ 0 C and all : ≥ n B . More generally, for z open and let X ) ) ( if and only if it is holo- ∈ | ( 0 r < ε. n ∈ N C ϕ L z z x B ) } ( { ( ∈ r 1 ≥ m,n m ∈ → x K Ω such that such that ) ) ) X is uniformly Cauchy for ) := for k≤ Last Change: Wed 17 Apr 11:33:04 COT 2013 0 0 N T N J 0 z z ( ϕ m,n Let z, a z X ( ( 0 0 a Banach space and ∈ ρ k ( ∈ − d f z z f r , . ⊆ weakly holomorphic n < ε ′ holomorphic N X N z, a K − − − − x 1. Recall that the map N +1 1) Ω if and only if the limit d X Ω. Hence, for every k ) ) z z n ∈ X ) z z ) for all n ∈ m a J ) z ∈ ( ( ∈ a ( ( x z 0 )) k≤ f f k≤ ( − ϕ f z 0 {| n − ϕ ϕ 0 0 f − z z : z z x k k exists a is called z < ε Ω if and only if the limit is called ( ) be a Banach space and let is a Cauchy sequence and let ( n N | → → lim lim 0 0 such that ϕ ) z z f x k ∈ ) f z ( ( 0 k ≥ X m 0 = sup 0 m X r z with with ( x z Γ x k r ε> ( ′ ′ Z Γ ⊆ ϕ m Let be an open set, − i r> Z X X m,n x π Ω. 1 i N − n ! C ∈ 2 X ∈ ∈ π x ) ∈ n J n 2 and ∈ n ϕ )) is holomorphic in the usual sense. ϕ ) 0 kk x z − n such that z be a Banach space. A sequence ( )= ( Ω ϕ x n a ϕ f )= ( N ( ∈ x X a {| f ( ϕ 0 Let Ω X ) ∈ z J and all n and all weakly holomorphic in DRAFT holomorphic in ( k≤k k 7→ Let ) f N that is, for every N N is the positively oriented boundary of = = sup m , z x 0 and assume that there exists an ) k ≥ we already now that its domain is open and that it is analytic, that is, ≥ 1 ( ( is holomorphic if and only if it is weakly holomorphic. 0 C m ϕ z ( x X ε> r 5.7 → is called is called − k≤ Assume that ( m,n , 1. Γ m,n − ) ϕ 0 morphic in every f exists in the norm topology. exists in the weak topology. Ω is weakly holomorphic in every f → n k n N x x k≤ ( k (i) :Ω ∈ (ii) ϕ ϕ for all Now let is an isometry. It follows for k k for all with Proof. there exists a Lemma 5.11. sequence if and only if the sequence f Theorem 5.13 (Dunford). Lemma 82 Next we study properties of the resolvent map Recall the following fundamental theorem of complex analysis. Theorem 5.12 (Cauchy’s integral formula). Then locally it has a power series representation. Definition 5.10. where holomorphic. Let n 81 . ) a linear ), hence . S ) ( T . ) ( ) ρ T ρ X ( T 1 ∩ 1 ρ ∈ ) − − − ) → ) ∈ T µ 0 1 T ( S ) It follows that λ − ρ X ( + n T − ( − ( 1 ) is invertible since 0 − ∈ 1 S ρ ) µ λ − λ T − ( ( ) , 1 ) T ) ∈ T ) ) − a closed linear operator. T − λ T X T , λ,µ ) − λ ) 0 − − , 0 . − − λ X , λ , then ) 1 0 1 ) ( → ) λ 1 λ λ − λ +1) − n (id ( ) − ( ) µ,T )( n → ) )( ( ( λ k Last Change: Wed 17 Apr 11:33:04 COT 2013 X λ λ, S S T λ λ µ,T λ, S ( 1 − − ( − R . X ( ( ) ) − − ( T − − = − R S − T ) R R T ) ), hence 0 0 T 0 ) 5.7 ) λ µ 0 S − ( λ − ( T λ λ λ S − ( λ, T ( ( λ ( µ 0 − − − 2.10 − − − ) is proved. − λ, T − =0 i λ R ∞ 1 1 ( 1 1 T ( X ) n λ T id T − − − − id ( R n λ ) ) )( ) ) ) ) k )( 1 1 T T λ λ T T − is closed and bounded, it is compact. − − < ) ) µ − − − − − λ, T − | σ λ, T )= ( T T ( 0 λ λ λ µ λ µ λ | R λ , so also ( . Then R − − ( 1 − k ) = ( 0 0 = = ( = ( = ( is compact. − , then the term in brackets is invertible, hence so is 0 T =0 λ λ ∞ k ) = ( ) = ( 1 ) . X λ k n )= 1 with ) ( − T µ,T − ). Since k we find ( = ( = = ( ( ) is open by Lemma T ) > C 1 − λ, S ( µ,T T σ 0 λ, S 1 T ( R be a Banach space and ( − ( C | ( be a Banach space and ( λ D ∈ T ) − ρ λ R R ρ ) − | R ∈ T µ − X − ) X DRAFT T − − λ − 0 )= − T ) ) )= ) , then − ( λ S k} ⊇ 0 ) ( Let T Let λ, T with λ λ ). If T ( ( = ( X ( D λ, T ii k λ, T ≥k σ ( λ, T k ( R C ( T ( ). For L \ > R 1 (Neumann series, Theorem )) R R is closed. T < ∈ − | ( ∈ C T | ) λ ( ρ λ < | ) ) follows from a straightforward calculation: λ λ T i i T : ( ( ( ∈ , σ k and we obtain − 0 If σ 1st resolvent identity: In particular, the resolvents commute. 2nd resolvent identity: 0 C T λ 0 T λ 1 ∈ λ − | ) Let ) is shown similarly: (i) (i) − (ii) (ii) λ λ ii ii dist( Proof. Chapter 5. Spectrum of linear operators Let As a corollary we obtain the following theorem. Theorem 5.8. Next we prove the so-called resolvent identities. Theorem 5.9. k If λ ( operators with ( which proves ( Proof. { and 0 such X . ∈ is to the hy in the ∅ x ) 6 = C > x ) ) ), there exists a T ( ( T σ 3.8 , 5.2. The resolvent ), it is continuous, x − 0 z x 1 ( 5.11 ) is analytic locally at ) x r . For fixed +1 − x h ) k n ) a densely defined closed ) +1 . . Then T )) + ) ) n ) ), there exists 0 , ,T a − )) = 1. Analogously as in the ) is continuous. Since the last z X . 0 λ, T ,T X ( ( ( 0 ) ) ( 0 λ λ k 3.7 r . z ϕ T 0 ,T L x ( λ ( x R ( z Γ ) 0 ( → )) ( r Last Change: Wed 17 Apr 11:33:04 COT 2013 ( k R 1 a h ∈ λ a r Γ R ϕ k ∈ ( ( ( 0 ( ) = ( Γ X T T R T n ∈ ( < → → ( ) ∈ lim h T | with 0 − − λ 0 λ, T λ ϕ ) x ( λ n ) ϕ X h , z ) − R h − 0 ∈ = + λ , z λ ( + x,ϕ λ 7→ x a | x ( a ) − =0 ( ∞ 1. Therefore, by Lemma x T X n λ ) < C ( T < C T lim L h , 5.7 x ′ k T )= T λ ) ( k x 5.14 → X be a Banach space, h ) ϕ be a Banach space and | ) ∈ + T X X it follows the existence of ( a λ, T ϕ ) and such that ( ( ρ ) is holomorphic in a neighbourhood of Γ T T R Let ( Let x ( ( ) DRAFT 5.11 x,ϕ ρ ϕ ϕ z C ( ∈ 1 h T 0 ( 0 λ ϕ → lim h 7→ we have by Lemma Let ′ z 0 such that X > , hence holomorphic. Since weak holomorphy is equivalent to holomorp ∈ x 0 exists, uniformly for so there exists λ and by the theorem of Banach-Steinhaus (Theorem operator norm (Theorem By a corollary to the theorem of Banach-Steinhaus (Corollary sum is absolutely convergent, it follows that where we used that the operator series converges and C resolvent map. Theorem 5.16. 84 Since Theorem 5.15. The preceding theorem allows us to apply theorems of complex analys linear operator. Then the resolvent map exists and convergence is uniform for This implies that is holomorphic. Proof. that proof of Lemma ϕ : ). is z. ) 83 → 0 we d a C z z f , ( ( ) ) z f r ∈ :Ω a h ( )) d K ′ − 5.13 z T ) z − )) d ) { ( f ) z h a f ( x ◦ ( ) f + − ), hence with ϕ ( z ϕ . 0 ) = a i ( ), there exists z ′ ϕ ( z 0 ( 2 i T ( z 2 C f ) ( = ( r . ( ) 2 3.8 k a ) ) r ϕ open and a 1 h 0 ϕ K h a ), it is in particular − z K k 0 )) 0 − 1 C ( ∈ ) follows form Theo- h − z z a r → z ( ( ( h ( z ⊆ h iii r B ≤ f ( ( , lim ) ( − z. + 0 Ω ∈ . ϕ z d ) − a a ( ) a r . 0 ) ) 5.11 − 1 ) Γ − z ⇐⇒ )( h h 0 ( Z f z )) z r ) Last Change: Wed 17 Apr 11:33:04 COT 2013 i − ( − h ◦ Γ ii 0 such that r π h − a a 2 Γ z, a ) yields ϕ + ∈ )) ( h d − − z a ∈ z 1 ( ( 5.1 d z z r > − d f )) )= ( f a ( )( ( 1 2 a z x ). As in the proof of Theorem ) ( a ) i ϕ ϕ − ) − it follows that ( a z − f , z a ( ( | − z ( ϕ z 1 h − ′ a ϕ z )) ⇒ ) h be a Banach space, ( 0 z ) ϕT a − f 1 h ( 0 h ( ( → z < C such that 0 a ◦ ) ) ) = lim ( | z C r r r − | Z f z z ( Let z Γ Γ Γ ∈ ( ( − i d − r d 1. Therefore, by Lemma Z Z Z f f π ′ 1 0 ) ( i i i )) k 2 < r z a − ϕ X h π π π 1 1 h | )) ( | 2 2 2 k≤ a h f ∈ | + ( ) ϕ x f = = = − a k )) = ϕ < ) ( ( a , ) DRAFT a ′ ( ϕ f h ( ( f X Cauchy’s integral formula ( ( − T ) follows from the definition. ( ϕ and + )) is holomorphic in a neighbourhood of Γ ′ Ω. and let Γ ii ϕ ∈ − z a ( )) 1 X h ( ( X h x ϕ ) and 0 f f ) 0 ∈ ( ∈ ⇒ } ∈ + h z r ϕ ( 1 x h ϕ a r + ( . It remains to prove ( ) = is holomorphic in the operator norm. is strongly holomorphic. is weakly holomorphic. a B i f 7→ ϕ ( Clearly, holomorphy of ( ( | ≤ 0 z ∈ 0 T T T T ϕ . Then the following is equivalent: 0 such that 5.13 z ( ) → a lim h ϕ 1 h X − (i) ( (ii) (iii) z 1 h C > rem Hence we obtain weakly holomorphic. Let This implies that exists. By a corollary to the theorem of Banach-Steinhaus (Corollary Chapter 5. Spectrum of linear operators Proof. Theorem 5.14 (Dunford). For every continuous. Hence there exists For Cauchy’s integral formula we obtain L | obtain for Proof. Since uniformly for k ), is is . . n T N A T ( /m with (5.3) ∈ 1 k =0 ρ n ) is k 0 ) T ). Now +1 n m N X n ( T such that such that }⊆ T T z ( ) . ∈ L k r ) ). Hence R T 1 m ), the series =0 +1) ∈ n ( ). It follows T in the largest k ∞ n , in particular n ( ≤ ∈ , then ular, for every T ( T →∞ , q ( T σ , m λ ( N 4.39 ). Observe that 0 − ). In particular, n P 4 > r m . r T M p . Then µ k T . | ∈ of T 5.2. The resolvent n its spectral radius. ∈ k ( ) ( p λ T )=0 > r > r | k} ) n k , T k ) = lim 1 = λ ) T : ( . | | T m ( > r − T k ( (Neumann series), T λ λ = r ( C | | m T r ( | m ). Let T ) that r q nm 2 λ > r > r k µ . k , ii ∈ T | k | − } (Theorem ) for all k 3.9 C > r ) M λ 2.10 1 λ ∗ m | and | { 1 T T n k )= there exist ) λ ( ∈ ) holds for all | . m T TT σ k≤ 1+ , hence it is weakly convergent, X ( Last Change: Wed 17 Apr 11:33:04 COT 2013 N ( 1 n ′ λ T n ,..., ( r µ ) k spectral radius 5.3 ∈ q k , k k 1 L n ∈ ) n 1 2 T X λ T = n , ( T ∈ : k n n k T k L M kk | , ( 2 is normal with T T , in particular 1 λ ) 1 r n k )) T ( 1 > r λ p T . Then − T ( n 2 − . Let |≤ | ) − defines an analytic function for k be a Banach space The : r T N n ) . T λ T µ λ be a Banach space, |≤ arbitrary. For every T | k q | ( | 1 µ k ( = T ≤k ∈ λ : converges in norm. By Theorem X →∞ λ ϕ = − | ϕ − = + N k n X ) ≤ = − : n {| C ∗ A ) and by the formula of Hadamard λ λ 1 k /m 2 , 1 ) converges for every +1) λ ( n m ∈ 7→ 1 n C Let − with ∈ T k ( . Because n n n k ) k − Let ( =: 2 ( TT T DRAFT λ ϕ p T n ∈ λ ) = lim sup 1 T n k C m m − k R , in particular T n ( { T T k ) T − ∈ . and 1 = s x C X ′ n x ) . x } TT C (0) = 0. Hence ′ n 0 k ) d ∈ y ) and x ∩ +1) { s = x )= n ( T x s. ( T = y ( X , so constant by the Liouville 6 = ) can be seen immediately: ( is closed. d − ). s. λ p C D . k ∈ λs T (0) + X T σ ∅ ( Last Change: Wed 17 Apr 11:33:04 COT 2013 T − λs λ − x ) d p e = 0 has the solution { s − |k σ t T ( e λx n X, Tx )= )= 0 y 1 ∈ λ λx Z T , it follows that T | X, Tx 0 )= ( ( → λs λ Z λt − σ T σ − =0 } e → ∞ = 0 for all ( ′ e X (0) = n λ 1]) . Hence for every x ) t D ′ λ e , x n 0 T , ≤ x ) | → ∞ , ( Z 1]) and } (0) = 0, so in λ ([0 1 , | n 1 = 0, so { ) and using the Neumann series λt x λ, T C T ([0 λ | → ∞ the initial value problem ⊇ D ( +1) ( = sup x λ . By a corollary to the Hahn-Banach theorem n C | ρ ′ R ) ( ⊇ max which implies that X X →∞ 0 for )=e ∞ λ (0) = 0 − X = ∈ t n , then, by a theorem of Analysis 1, ). : y k ∞ x X ( ∈ T − λ k T λ → : λ X n T ( X , then, by a theorem of Analysis 1, ∈ = y x y 0 for x (0) = 0 and ) λ T ρ k ′ ∈ T λ X x 1]) : ϕ x ) ∈ x y (0) = lim , . Observe that this implies =0 ≤k = ∞ given by is unbounded and densely defined. If ( ∈ X x ∅ n = λ k → λ ([0 ) → ∞ 1], y T . Then λ, T ) is holomorphic and bounded in and x k , (i) Let k DRAFT ) and ( C ( . This contradicts the fact that 1 = n x T x y )= = [0 ) T T → 1 C R ∈ λ, T 1 T is unbounded and densely defined. If ( ( k X ( k and every ( T x − ( the differential equation ) n ) C x ∈ ϕ σ ) it follows that R T is closed. > { λ ∈ are possible. λ, T C C k is unbounded and closed and is unbounded and closed and ∈ ( λ | ∈ D T x ) and and T λ, T − = ∈ λ C x ∈ R λ λ | ( T T T 2.16 ( . Moreover, x ( x x , T λ λ R X ϕ y ( D k and C ) Obviously, )= k i → = Assume ( → T ) = 0, x with n ∈ ( ′ Then Then λ x x σ C λ → λ, T ) Obviously, ∈ ( (ii) Let n ii ( Note that λ bijective, in particular exactly one solution For every hence in that Obviously x Chapter 5. Spectrum of linear operators Proof. The following example shows that for unbounded linear operators th Note that in the last example the continuity of ( and For every Examples 5.17. implies that and (Corollary R all theorem. Since Proof. In particular, 0. map . . . . ). ). ⊆ H H H , it T T )= N ∗ ( 5.22 ( 5.22 ∈ T is an ρ T n )= )= )= − ) ) . H. i ⊆ ∈ D n T T T T T ± x ( y = . T − − − σ ) ⊥ = ± a symmetric ± ± T } z ( ∈ z z x . ) 0 σ { erators. λ belongs to H H rg( rg( . Since | are bijective. For = = y ± → ) ⊥ z T ⊥ = 0. The set of all ) T ) densely defined and } , so it suffices to show H |ℑ T − 0 ∗ − n ( X { x λ λ T − < T ) is closed by Theorem such that | such that ∗ → = λ is closed by Theorem ⊆ λ ± 0 such that rg( = ( which implies ), in particular ker( 0 0 H ⊥ z − and X T ) T T } x ( < < < ( ∗ ) 6 = 0 − Last Change: Wed 17 Apr 11:33:04 COT 2013 T ρ ∗ T T − { T ) ( − − − λ i T z | z z = ker( T ⊆ − − ± if there exists a sequence ( − i )= →∞ ⊥ λ − R Im Im ∗ λ ) n ). ∓ that \ ∗ λ T with T C T ( is injective and its inverse is bounded by − and C and is an approximate eigenvalue of 5.3. The spectrum of the adjoint operator ap and lim is selfadjoint, then every − 4.73 λ . 0 and Im T σ 0 0 ∈ ∗ C = ker( N . . Then H T λ > − λ > > ⊆ ker( R R ∈ + ⊥⊥ ± + ) + \ \ . ) ∈ z z )= . Then rg( z z ) such that ( T ∗ n T T C y ( is symmetric, hence R C T = ker( T σ ( Im Im \ ∈ − − T ∈ , every i − x z C i ⊥⊥ ± λ ∈ D ) Every approximate eigenvalue belongs to , it is closed and 5.7 be a Banach space and ∈ ± with Im ∗ with with be a complex Hilbert space and T y λ is closed, the range of T . C X C (i) C approximate eigenvalue − H ker( . . H = T for all ∈ , hence ∈ ∈ λ ) = rg( , repeating the argument above finitely many times shows that = 1 for all ∗ ). Let H H ) is obvious. T ± Let ± ± R T H T Let T z k z )= z iv , it follows that ( DRAFT \ ) Let ( ∗ n − = )= )= ii . . i x C T ). It follows by Theorem ( )= is called k T T ) = rg( ⇒ R R 5.22 ± T ∈ T − ⊆ D Ty T ( ) Since C ⇒ − − is a Hilbert space and if i ⊆ ⊆ p λ )= ) Let ) − i i ) Since ) is obvious. iii − ) By assumption, ) ) ± σ ∈ ) there exists a ∗ v ( i iii ii ± )= λ ± ( ∗ X ( vi is essentially selfadjoint. is closed and i λ ( T T ( ). T / λ ∈ ( ( ( ( ( T T ⇒ ( ∗ approximate eigenvalue of If Every boundary point of T There exist rg( rg( σ T There exist σ ⇒ . Hence, by Lemma ( ⇒ D ⇒ rg( ⇒ | ⇒ λ ρ . ± such that )= ) = } z (i) ) = )= )= ∈ D ∈ (v) (v) (ii) (ii) )= 0 (iv) (vi) (iv) (vi) ker( (iii) (iii) rg( ii ii iv v vi i (iii) ( follows that x ( λ Given any By Theorem ( operator. Then the following is equivalent. Therefore rg( 88 and Analogously, we find a characterisation of essentially selfadjointTheorem op 5.24. that Proposition 5.26. ( ( { Proof. Definition 5.25. approximate eigenvalues is denoted by closed. ( |ℑ X ∈ = 87 for λ . It ∗ n ) 2 X k for T 6 = λ, T i ) k ′ ( ◦ if and only ) R = a symmetric T X . λ, T k ) λx , x ( k n 2 and H R T a densely defined ker( )= Im T k a symmetric oper- h } k ) = T ) → ⊆ ) = ) is closed. ′ X T + i ) − ). ) k H T ( ′ H i T λ ( σ T T ( → − k → T p λ, T densely defined and closed. λ − rg( ( ∈ σ x , x X ) ) H ( R λ ( ∈ →∞ λ T T X ). n T ′ λ / : rg( 4.64 a densely defined closed linear ) = 0 for every nilpotent linear Last Change: Wed 17 Apr 11:33:04 COT 2013 − T = lim and → , ( T ) λ C } ( . p ) n . 1 ) 0 r 2 X H σ is invertible with continuous inverse T { ( ∈ k T ( (Re ( ∈ ) n T h σ 4.65 2 ) → λ λ λ { T ) . = − ∈ D k ) T H ) = ′ ( ′ x T i T − T . T ( ) = →∞ ( . By Theorem r ). rg( R is closed. λ n ′ T σ . σ x , x ( is real. X \ ) k ) T = lim , for example σ → ∪ ( x T T . C for all k r 6 = T ) 1 ) ) n |k ′ σ ′ ) T − k − ∈ k T λ k T T x ( T n be a Banach space and ∈ . λ ( z ( λ ) = ) ). ) p ( p T D ∗ < |k Im λ − h T σ T k iv ) σ : X rg( T ( be a complex Hilbert space and ( ) λ be a complex Hilbert space, λ ( ), then ker( ρ ∈ λ ∈ T ≥ | σ ( T ) or be a Banach space and H →∞ λ rg( ∈ H λ Let ′ ( k≥ r − for all ∈ D Im( n is injective, which implies that is not injective, hence p DRAFT x T λ . By induction, it can be shown that hence X T ′ x σ ( . 2 H ⇒ T ⇒ (i) Let p Let T kk k R ) = lim ∈ σ x for − k ≥ | Let T \ ) − T ) = ) = k )= ) ∈ x λ ), then ( λ T ) ∗ C λ be a Hilbert space and r T T T T λ ( ) is continuous and closed, to its domain rg( ( = T ( − ∈ p r = r ,T is closed, then − T . k H σ σ ∗ ′ σ − λ ) λ 2 , implying that ) λ ) If ) For all ( λ − is selfadjoint. T i i T λ ∈ ∈ T ( k ∈ N ( The assertions follow from Theorem ( T ( ( k λ operator. Then ρ Let closed linear operator. Then and the point spectrum of In particular T If R k λ λ λ ∈ − n λ (i) (i) (i) ) follows directly from ( (ii) (ii) rg( ) If ) If ( (ii) (ii) v i i operator. 5.3 The spectrumLemma 5.20. of the adjoint operator Proof. if ( follows that ( operator. Then the following is equivalent. Chapter 5. Spectrum of linear operators hence Note that in general Lemma 5.21. Theorem 5.22. Theorem 5.23. ( ator and Proof. In particular, ( Proof. all . N ⊆ X T ∈ . n and A ⊆ ) → , also n N ε M , . Then T x . Then n ∈ 3 ) ) T n N k ) ) is called < n m ) a Cauchy ≥ Y,Z x X, Y ( k such that ( Ty k , the sequence L converges. Let X, Y L T ( X, Y N − N m,n M ∈ ( L ∈ ∈ − is bounded, hence m ∈ ). n T K ∈ y ) is a closed subspace ) such that T T N ) k k is finite-dimensional. k ∈ k ⊆ T is compact. ) T ( n k k , hence convergent. converges. X, Y and ) X x N 5.4. Compact operators T ( n k X, Y ∈ ) Y completely continuous N is compact, there exists a + ( n T K X, Y ∈ ) 1 k L ( Sx k n ) T is compact, then there exists ) is relatively compact. The is compact. m ∈ T 1)) is relatively compact. k K X, Y y A ) implies that the linear com- , n ( S k ( T ). The . Then, for all ii T Last Change: Wed 17 Apr 11:33:04 COT 2013 such that L T ( (0 T T N or X ∈ N . If − T Sx be a compact metric space and ≥ . Since B converges. By continuity of n S S X ( 5.30 ∈ ) y N N ) such that ( T k k ε. ∈ ∈ T 1) is denoted by the set k is bounded, ( m,n converges. Continuing like this, for every k ) − = be Banach space and n M, d k k ( Y , N S ( n + X n ∈ for x M k n M to ⊆ such that ( Mε 3 n ε 3 ) Sx Let X, Y y N k A X + (1) n ∈ )of( ) is compact if and only if k≤ T x ) k ε 0. Choose 3 k≤ n 1 ) k Let X ) and Remark is compact. ( n − x k T m ( ). Take an arbitrary bounded sequence ( contains a convergent subsequence. + x ′ k n x n be normed spaces. An operator L x k be Banach spaces. Then be Banach spaces, T N ε > T ∈ ∈ M Ty X, Y n Mε 3 X, Y ( such that ( ) ( − ) is complete, there exists a n X, Y K N ≤ n K X, Y the diagonal sequence. Then, for every ) with finite dimensional rg( ∈ x is compact. Since k≤k ∈ k T x k ∈ be a bounded sequence in (i) Every compact linear operator is bounded. X,Y,Z ) N such that m Let X, Y T k T ∈ ) is compact if and only if for every bounded sequence ( Let N ( T ) such that ( ) is compact if and only if k n n DRAFT is Cauchy sequence in the Banach space R X, Y ∈ Sometimes compact operators are called L Ty ) x ( Let n ) (1) n N ∈ ) L n − x ∈ ( n n X, Y converges. X, Y n ( n x ∈ x ( ) M converges. Let . N L n L ) T ∈ Ty if for every bounded set N = ( k ∈ is bounded, k ∈ ∈ Ty ) such that Obviously, 0 Let ( a family of real or complex valued continuous functions on n k N the sequence ( T ) T A n ∈ X, Y ) N n 2 we find a subsequence ( ( n is compact if at least one of the operators y ) ∈ L M is compact if and only if k (i) n ≥ ( (v) The identity map id (ii) (iv) Let (iii) y T T Sx of Proof. Theorem 5.31. Hence ( ( Proof. ( k compact subsequence ( bination of compact operators is compact. Now let ( 90 Definition 5.28. Lemma 5.32. Theorem 5.33 (Schauder). a subsequence ( Remarks 5.30. TS set of all compact operators from ( Remark 5.29. and choose Now assume that T there exists a subsequence ( For the proof we use the Ascoli-Arzel´atheorem. Theorem 5.34 (Arzel´a-Ascoli). sequence. Since We have to show N C 30 Abr 2012 . . . ∈ ). is ⊆ ⊆ ⊆ } H 89 M H T ) = ) N k ≥ = 1 N = 1 ( m ) (5.4) M ∈ → ∈ = ρ T T k k = 1 → ∞ n n ( ( ) . ) n ) n = 1 for ⊥ ,T ∈ σ σ ε n k , y ) T n losure is n x n k k x λ x k λ T λ n − 2 1 k ( . . x ) =0 λ − : 2 1 R k n X i λ n k i x n x ) ∈ ) and ) : rg( x → ∞ if and only if its ) m x T T defines a positive 1 ( . m − i 2 − − Tx,x such that ) k − ) = ker( λ T T x {h ) is shown. ∈ D ( tends to 0 for such that X k T x,y (( and the second term is T − T , n ) ) ( s k n ( N 0 λ , ∈ − 1 2 y 0. Assume that n m ρ ∈ →∞ is complete. In particular, m selfadjoint. Then ) n x n n λ n ) n ∈ ) − x ) we know that ) lim − → := sup x k it suffices to show that 2 −→ M n if and only if every for every 1 ) → ∞ m λ n ) T rg( ) is analogous. , x H n x T ( 1 . Choose a sequence ( − ( y n Last Change: Wed 17 Apr 11:33:04 COT 2013 ( x n ∞ T ) k m − 5.7 M − is compact. It can be shown that − h T ) L ( x n k ( . Then T 5.26 )= σ λ T T ) − x i≥ s n ( ( ∈ M relatively compact ) M, T − T choose x ∈ k − T and R ) T ( ) := − n T λ − h |≤ N if and only if every open cover of x , x } λ λ = ) M − 1 2 ) ,T ( n n i . ∈ n k n x, y n T λ x n ( ( λ x λ N ) ). Then there exists a sequence ( = 1 s ( n ) ( + − and ( is injective because for all R , x ) is closed because ( T m ∈ T R k k R k λ ( 1 totally bounded n N 1 T n k ( x n σ − − x − with finitely many open balls of radius − y k ) compact ) ∈ ) − . From Lemma k + +2 λ T : ) n k T | | ) if and only if m ( n λ ) is M )( λ n n , i ). Hence rg( k ≥ h . Since − ). Analogously ( ,T − λ λ n T , it follows that x λ − ,T ). By Proposition m n T x is compact if and only if M 5.4 n ( T ( − − λ is called − ∞ (( kk T there exists a sequence ( λ ). The proof of is bounded, therefore ( s ( ρ λ . For every ( ( | h x λ λ ( λ D Tx,x T | 1 λ ) M σ < R ( ( m = 1 for all . Then ∈ the spectrum of a selfadjoint operator is real, so M i ց R − k = = . T {h 2 3 be Hilbert space and R ) ) = ≤ | k k n ). ) ∈ ap 2 ) )= 1 2 n precompact T σ = 1 for all − k≤k T T k := , x x ( ( T DRAFT H m n m n , m k λ 5.23 ∈ k →∞ − n y σ σ ( n x n λ q ker( T := { x = rg + U T T ∞ min )= q p min by step 1. Therefore ∞ ∈ be a vector space and : be a Banach space, q +1 rg( ( . . T ( +1 = } 2 n )= be a vector space and T q p T k . Then 0 . Then and ker and rg X x n ∈ T X = . { y and the descent T = 0. +1 : − ) X T n X x ) := p ) and n x ) ) := ), hence ker( )= )= y λ + rg . ) + ker( ∈ Let n p p T T X T T is finite dimensional for every q T Let T )= T T n k . We have to show ∩ ( ( ( ( T ker( DRAFT T ( ( p x T T n > k n > k ) and n ∈ Let ) is a closed subspace with ). α δ α T α rg δ α p ) T +1 ≥ p q ∩ ker( = q x = T T ) T ≤ T n X { ≤ := := := T p x rg( x ⊕ ) ) − . q p T rg( ) = T T ⊆ = rg( rg( ∩ T } ) = rg( T p λ T ( ker 0 ( ) ∈ ,so0= of U of ⊆ T X α δ rg( +1 +1 y Let We prove the lemma by induction. The case when 1 p ∈ ) n +1 \{ p ) + ker( is closed and Assume that If integer q q x T T n T x ( C T T T = rg( R (i) ker( = ∈ (ii) ) Assume that ) Assume that (ii) ker ascent i i rg For the proof fix ( ker( To see this, choose x such that Step 2. follows that 92 Definition 5.37. Theorem 5.39. Step 3. rg( Let Step 4. By step 1 and step 2, we have that define both the ascent Step 1. X Proof. Proof. descent exist ker( tion. ( λ Lemma 5.38. ′′ ∈ 91 for Y <δ | ≤ x 5.32 ) | k { 1)) is 2 2 → , T t y . ( k (0 Mε. X − f : m 1 X < ε. t ≤ n 0 such that − | 1) := ) K T f = ker , ( y 1]) a bounded ) } ∞ ◦ ( s. , 1 − n k T (0 f y Y 1) n δ > k ( T , ) d X J ([0 x n − f s := | k f ( K C (0 ) 1] with k ker x , X x ) ⊆ K ( ), so [0 = K f N be a bounded sequence s<ε s,t } ∈ ∈ ′ ∈ ( d n converges because 2.33 2 k K ) | Y x ..., ⇒ | . ) 1 is relatively compact, hence n N ,t : ) . Then is compact, so there exists a ∈ = 0 s 1 ⊆ ∈ x a linear operator. 0 y ( ) is compact. By Lemma ⊆ N k Z t k ( y N ′′ ) n N ∈ ) N 3 .... n : ∈ X x n k Last Change: Wed 17 Apr 11:33:04 COT 2013 ∈ n T ) ϕ n ∈ || k n <δ , Y ) T x )= ⊇ ) n ) ) is compact. ( t ′′ ϕ n ) → (Lemma 2 k ′ y x 3 n f m ( ker )( k X is compact. Let Y ϕ T T n X ) := ( X x T m s,t x, y J y ⊆ ϕ k : ( n L ( . By assumption T is bounded. To show that it is equicon- ( rg → . Now for ◦ T k 2 ϕ n ( − T ∈ N X T T ⊇ ) X − ∈ − . We define the functions ′′ <δ a linear operator. Then obviously 2 : n ) ) = N ) T for some 1 , f k T T x ker y n T ) ( ( X , T ′ A d ∈ k K x k k , rg . Now let ( ⊆ s,t ◦ ]. and equicontinuous because k n n ′′ T ,t ( n ∈ k ) and ′ 1]) → T Y → k . Then also ( ϕ ϕ T ⊇ 2 , , s Y | T f J ( N k M 1] X 1 {k {k T ∈ ∀ K M ([0 , 0 − : k ker : Yos95 Z ) ) = ker C rg k T ([0 n is uniformly continuous, there exists a 0 n f k≤ ⊇ s,t C → }⊆ +1 f k |≤ n = sup = sup ( k First assume that ] or [ 0 ) . Hence ( k ∈ ϕ 2 k X { δ > a vector space and T . By the ( Ascoli-Arzel´a, . t . k 1]) if M k m ( ∃ , k is closed in is well-defined and bounded. Let ( K is compact. Then n DRAFT n X ′ ker k ϕ x ≥ ([0 Y 5.33 ∈ ′ k Rud91 T 0 T <ε Let 2 C n T T | 0. Since Let ) : is bounded by be the closed unit ball in ′ ,y and bounded by − − k ε> 1 N ,t } ) k y T ′ such that ∀ Y 1 1 ∈ ε > n s we obtain , t n ( ϕ ( k R ) ′ is compact. < k 2 n N is compact. n is closed, is equicontinuous, that is, is compact. Recall that T y ∈ x k f See, e. g., [ Obviously k be vector space and − k ∈ k T A X Assume that all integer A A ) x − T k J M n | X 1 ◦ : y s,t (i) ( k ′′ (ii) (iii) k Lemma 5.36. is compact. Since Then Proof. X and Then Chapter 5. Spectrum of linear operators Proof of Theorem T Example 5.35. Let Proof. Then ( compact in tinuous fix sequence with bound C | convergent subsequence ( Now assume that and By the Ascoli-Arzel´atheorem it follows that ( it contains a convergent subsequence. . } is ( n 0 . λ ) 1 . ). If 1 such \{ (5.5) T U , that ) − T . ( . Let +1 N ( n ) we can . Then T T n λ ∗ ρ ∈ ( m ) is finite. is closed U C n A . σ T ) T ∈ n } ( ∈ is finite. If ... ∈ T ) . n 6 = σ λ j . − y j 2 1 T T ) 1 − x n α > ε λ = ) − 2 − λ − | n ) , where the bar n } λ } . Since λ λ ) . rg( λ k≥ T U | for is the Riesz index 0 ∗ 2 1 } N be a Banach space. m ∈ − ∈ − of T p x ( m ∈ j ): with eigenvalues \{ z z λ ) p X λ σ λ T ) n is a Hilbert space then k≥ ( for some T ( ∗ T 5.4. Compact operators j , ∈ } rg( j σ m 6 = T − α Let H x λ ( where of ) j ∈ p λ : >ε (step 2), hence n for all Ty n for all =1 ) } α σ | n λ j X λ -invariant. N 0 C 1 2 T n x − 2 1 +1 + ( { . If T n =1 λ = ) = 1 ∈ n ∈ Last Change: Wed 17 Apr 11:33:04 COT 2013 | ) or it is an eigenvalue of n j ′ − n is the only possible accumulation 5.38 \{ − y -invariant, closed and that j T x } n ) n T ) λ ) λ 0 m either belongs to x ( T \{ T k λ ) C ⊆ m p Ty ∈ ∈ ) λ 5.39 ′ . By Theorem T σ − rg( ) by the proof of Lemma N − = ∈ n } . Note that all is at most countable and ε > Ty ∈ T T T λ , then the dimension of the algebraic eigenspace ∪ λ 0 − } k ⊕ ( n } λ ) T ) n p − p ) = ker( − n m 0 { ). n ) σ rg( does not contain a convergent subsequence in contradiction to \{ T λ , y n T λ ( , then x ker( T ( are linearly independent because T ( N ∈ T x C ( \{ ρ = } = n λ such that ∈ p Ty 0 ) DRAFT − n n =1 α ∈ } n − } k σ A N N x ) T is not finite. Since in this case rg( 0 . Using the Riesz Lemma, we can choose a sequence ( 0 and ,...,x k λ ∈ . Note that n = rg( \{ ( λ }⊆ = n 1 ∈ n ∈ n δ by Theorem 0 σ ) x y T n C \{ , x \{ ... λ X n T x { k n y T x N ) ) ∈ k ) \{ − λ ∈ ( T T T ) Let ∈ =1 = ker( λ λ m , T 0 . Obviously 0 λ ( n X + ( n U n ( ∈ { { δ x | − = ) λ k δ ) ) − ⊆ λx n n is closed. Moreover, ) x λ T = T k = rg( λ = ( N T )= such that j ) = ∈ − U be a Banach space, )= T x n − − for all = U = ( ϕ X T n ) T N λ it suffices to show that λ T ) λ k a basis of ∈ = y ∩ T : rg( − n X be a Banach space and kj ) − n =1 m − ) δ k x ker( n j ) λ n must be closed. does not contain a convergent subsequence in contradiction to 1 − DRAFT ) | T n ( λ ker( n T λ X j − 5.38 T N ( T x λ P α U x Let x − | ∈ ∈ is closed for every λx − ( α because ) U n − λ k = ker( | is not finite. Since in this case ker( Let n ) = rg(( ) n λ T ). Hence ( ) this is case if and only if ker( ) = ϕ n ) U n v T ∩ ,...,x α T X ( p − T 1 is closed for all there exists a closed subspace ∈ ( − T x U ker( x ) we already know that dim ker( T x , x λ . Then − i n − y P k } λ ) X,Px ∈ λ λ 0 5.30 is compact, there exists a convergent subsequence ( T = ( =1 5.40 Note that ( Let By Lemma y → \{ T k − = 1 and M n C λ k X can be extended to functionals x ) By( k being compact. : ) Since k ∈ ) Observe that k ϕ (iii) rg( i ii iii ( Remark ( domain rg( pact. Hence it suffices to show the assertions for Assume that Hence Hence rg( Chapter 5. Spectrum of linear operators Since Lemma 5.40. Theorem 5.41. lemma Hence Note that exists a sequence ( ker( show that its inverse is bounded. Assume ( The last condition can be satisfied by the Riesz lemma (Theorem can find a sequence ( Proof. Therefore ( T Then there exists a closed subspace Proof. λ k P ϕ The number that Proof. ) ) n ⊥ ). = = 2 1 2 } . X Let T T ) ⊥ ( ( X . In (5.6) (5.7) ) T p p ( M T T σ σ L − ). Note → ∞ 1 2 ∈ − ∈ ∈ T λ . Define and 1 2 2 with eigen- 1 λ λ ). Note that − . X X λ } rators). 2 | 0 T 1 ker( 1 T ∈ , is compact and λ T H. , n ) − x 0 = 1 = 0. Recall that 2 T T is an orthonormal ∈ := λ := ker( k 6 2 k − . The only possible 2 ). 1 x T a selfadjoint compact 1 B 0 . T k k T λ X ) ( ∪ k −→ ). Let H. 5.4. Compact operators +1 p H. > 1 n x H X, . Since n ∈ σ , x ( B |k X ∈ ker( T n is an eigenvalue of X ( L { ∈ e +1 L k ). Let . ∈ ). i ∈ n λ k} ∩ x 0 n T ∈ n λ is selfadjoint and consequently e T Last Change: Wed 17 Apr 11:33:04 COT 2013 T of eigenvectors of e − : | ≥ · · · n is , x 5.42 4.45 i 2 −k n , x T n i T x , n λ N , then ). Hence =1 := span n λ e h e 6 = 0, then there exists a k ≤ | {±k ∈ e N n i T 3 or ) n 2 i , and n ) =1 6 = 0, then there exists a x , +1 n N n } X e T -invariant, hence k h 5.42 X n n n e )= 0 Tx,x 1 | ≥ | ker (e n such that λ T x T { 1 T x , T ( + k {h h λ ( x , | +1 h ...B 6 = x is σ (Theorem . Let n 0 ∪ n =1 N } 1 2 T X . If } n λ P X ) a compact selfadjoint operator. B k 1 be an orthonormal basis of ker( X T = inf T ∪ {∞} + ∈ ) − k} ∩ =1 , N be an orthonormal basis of ker( 1 n = =1 X n λ x e − N = T H 2 B 0 k k m ( 2 = = 1 B − ∈ x P n L λ belong to the spectrum of 5.42 k T e x x {±k has at least one eigenvalue distinct from } = ∈ i N 1 is compact (Theorem T x n X, − = ker( x = T T e ) T , ). Here we used that and =1 ∅ 6 ...... ) ∈ T T +1 x , k be Hilbert space, T = 0. Let . x n h is compact (Theorem = 0. Let X x − where : n x − k 6 − 6 =0 k k 6 H T 1 λ , then 1 1 2 = i| ( λ 5.44 T =1 ...B the numbers λ λ T X, k N n DRAFT , 1 ∪ . Continuing like this we obtain a sequence of Banach spaces k ) ) Let 1 6 =0 3 ∈ X X . By Theorem n = B = 0, the assertion is clear. Now assume that T can be chosen such that is selfadjoint and compact. If X = ker( λ ( T | x R | ∈ Tx,x { | ( 5.27 2 k 2 ρ 1 n : n 2 is the orthogonal projection on e T T λ ⊆ T λ i {|h λ | ∈ ) = rg( | 0 k ) T ) Let is finite because T λ P i := T If ( − 1 ( 3 − values accumulation point of the sequence There exists an orthonormal system The If If p B Tx,x = sup = 0, it follows that T 1 σ is finite because T x λ be a Hilbert space and k k 6 {h k (i) ∈ 1 (ii) T T (iii) and basis of span Obviously, such that that k λ B and a sequence of compact selfadjoint operators It follows that Proof. operator. Then at least one the values such that particular, if 96 Proof of Theorem Now we return to the spectrum of compactLemma operators. 5.46. By Lemma Theorem 5.47 (SpectralH theorem for compact selfadjoint ope rg( Proof. inf k . . 95 be a ) := X 6 = 0, )) = . T T λ ∈ X x − it follows (rg 5.42 λ Y Let . C X } ∈ 0 ). Since y). Fredholm oper- λ { Fredholm index X T )= have exactly one ′ ( . p , . η, T dim(ker( σ )= )= ′ =0 = − (C) ) := codim . Then exactly one of (B) ∈ T (D) T } λ ϕ ϕ T λ 0 ) ) ( − − a compact operator and of ′ of ′ d ) is called and λ λ T T ) x \{ ker( ϕ ) Hence for X ◦ X − − ( C ( ) is called the and (A) L λ λ L . ∈ T 5.33 Last Change: Wed 17 Apr 11:33:04 COT 2013 ( ∈ have solutions if and only if ∞ λ has exactly one solution X d and ∈ and rg( T y ∈ } < − and ker( . . have only the trivial solutions). T } 0 ) (C) ) ′ x (C) ( (D) ( = 0 { T T { X (D) ( x (B) ) (D) n ) is equivalent to consider the equations and )= of T ′ of ii )= ′ )= for all solutions for all solutions ′ . T the equations ϕ − T x ) X and ) := T ′ (A) λ − T − T ( ∈ ( , − ( 5.42 X λ ρ y, λ ). The assertion follows from Theorem ) and ( χ (B) λ and ∈ T ∈ T =0 = )=0 rg( )=0 ( ϕ, η ( ) := dim(ker . . y 0 η x ker( is compact (theorem rg( λ ◦ σ x ρ x ( x ϕ ( T ) ) ′ ( be Banach spaces. ϕ η > ∈ ∈ T T T n and λ and λ )) a compact operator and be a Banach space, − ′ − X have non-trivial solutions. In this case ) T λ λ X, Y X the equation ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ X ∈ X − ( ∈ does not contain a convergent which contradicts the assumption DRAFT for every solution for every solution X has a non-trivial solution . In this case, (D) λ L Let y N ) Let ∈ the following is equivalent: x, y, ∈ ∞ ∈ the following is equivalent: (B) ( (A) ( T n ′ y ( =0 ) has a solution and has a solution T < ρ )=0 )=0 n X X x y x )) ) ∈ ( ( ) is closed and ∈ ∈ ) follow from Theorem . For T x T T η dim(ker( For all ϕ solution (in particular λ T } η y iv 0 ) is equivalent to − i rg( is compact. ( (a) (C) (a) (a) (A) λ (b) (b) (b) (B) \{ For every For Fredholm alternative: Exactly one of the following holds: For if rg( T Y/ C ) and ( (i) (i) ) By Schauder’s theorem ∈ (ii) ( (ii) iii v (iii) Banach space, that Chapter 5. Spectrum of linear operators Therefore ( Theorem 5.43 (Fredholm alternative; Riesz-Schauder theor A more precise formulation of the Fredholm alternative is the following Theorem 5.44. ( that ator the latter is equivalent to dim( ( Proof. the following is true: λ Definition 5.45. Then . . a n ∗ n s are Q and and n TT n 1 µ 1 are the n H H | T ϕ . and n P ∈ T 1 ⊆ s | nm T x δ = . ∗ of = 2 =1 k T . = S N n n n ) a compact oper- n s ψ . n . ϕ T x ) , 2 nm i ϕ R 2 k δ ( n H and for 2 n = , = . ,H s 5.47 1 ∈ 2 k 1 ) a positive selfadjoint i S 5.4. Compact operators 2 n y x , ϕ 1 H y and H H s h ∗ H ). ( ( 2 n ∈ ∈ T T L s = L , so k 5.8 Tx,x . i T ∈ ∗ ∈ =1 N = m X n T because T Last Change: Wed 17 Apr 11:33:04 COT 2013 of . and for h =0 T k . Moreover, the 2 n , ϕ = , x , y n coincide and are equal to the k ∗ y n is an eigenvalue of = | n n Q n H s λ y | ∗ i 2 k n ∗ ψ ϕ ∗ T x s i i T TT µ T T ϕ ∗ T n n | ∗ k →∞ lim k | x , x n n T 2 X = h | and . By assumption x , ϕ y , ψ 2 n k T . The non-zero eigenvalues 1 h h and n = and s T 1 2 n n y h| | ∗ ,T k Q | ) s s k ∗ n = . T , then T S = T | ϕ n T i ∗ =1 =1 is an ONS in N N i i T x s = X X n n m x n T k k | N | if and only if 5.61 ) exists for arbitrary compact operators. is compact, it has a representation T ∈ = = T be the singular values of = n | be Hilbert spaces and T a ONS such that, see Theorem S ) y := ( 5.6 , T ϕ x , ϕ 0 ∗ k n 2 1 h | T x x , n | n be a Hilbert space and T x ψ T H | s > 1 | T ,H . T T ϕ 1 are positive selfadjoint compact operators. H H ⊆ h| th roots the eigenvalues ∗ h =1 N | k H k | X n 2 n N T ∗ ∈ 1 = . Since such that s ∈ . T 2 x T Let = n | ≥ · · · 2 T ) (i) k Let . Then ( = DRAFT x H n 2 = | n x , see Definition i | of s ϕ T and k ⊆ m can be defined more generally for positive selfadjoint operators on are the | T H are the eigenvalues of T ϕ | ≥ S | . n k | n , ψ 2 2 n 1 T =1 T 1 x µ | | s N n n s H ) ψ ) For all ) Let ( n i i h is a singular value of ∈ := ( ( ψ y s The eigenvectors of If there are infinitely many Let The non-zero eigenvalues of ( n ψ such that (i) ) follows from the uniqueness of the representation ( (ii) (ii) ii Let Obviously singular values Hence the Lemma 5.51. Hilbert space Proof. An analogous calculation shows Theorem 5.52. Proof. ator. compact operator. Then A representation similar to ( 98 S Definition 5.50. Note that ( with pairwise orthogonal projections he 97 . 0 (5.8) ≥ 2 is normal the claim i| . To show n n oduli of the are pairwise e T P n n x , = λ λ → ∞ mpact operators |h is an infinite se- . k k | ) follows from n k i = ive linear operator R λ n ∞ n ) +1 k n a compact selfadjoint n . P λ λ X | ) 0 for n ( a compact operator. e is positive if and only if = = H i ) ( n E } ). Then ( T L | → e H n ( i are strictly positive. e +1 L ∈ 5.7 +1 x , is unique if the , k h Last Change: Wed 17 Apr 11:33:04 COT 2013 T . Obviously k ) ∈ λ n n T n | we will show that every bounded e λ e ≥ the operator such that K 5.8 i n x, ( n n . =1 as in ( N are positive. P h N e X n : 5.60 n n T , of pairwise orthogonal projections with n ... | T λ λ ∈ = n h· = k λ =1 n ). Since n =1 N k |≥ /k e N n X X n {| 1 n 2 ) and e R λ λ T n + = i 5.19 n n P then there exists exactly one positive compact n x λ ( T e 0 P , | ≥ | = sup in the operator norm. If all eigenvalues of N 0 1 x , P , P = λ h T | n ∈ . In Corollary n P such that =1 e R N T k X ⇐⇒ n n i λ be a Hilbert space and R n . Let all eigenvalues of . The representation = = e be a Hilbert space and Recall that a linear operator k =1 H 2 ) allows us to define the root of a positive compact selfadjoint H . X n x , H =0 A T x h ∈ 5.8 DRAFT n − n ⇐⇒ ) define x Let λ 5.49 λ ii Let T n X and a sequence →∞ lim n D is positive and = 0 for all ) such that i T is strictly positive is positive H If the series is a finite sum, the assertion is clear. Now assume that t ∪ {∞} i≥ ( selfadjoint operator If T T N L Tx,x ) ∈ h ∈ ) Note that . . . (i) (ii) ii iii Tx,x and that the normelements of of a its normal spectrum operator is (Theorem equal to maximum of the m positive selfadjoint operator has a unique positive root. Proof of Theorem series is an infinite. Note that for every follows from Note that the theorem does not imply that there cannot be non-co A distinct. Proof. uniqueness, assume that there exists a compact selfadjoint posit quence, then ( ( Chapter 5. Spectrum of linear operators This implies that Corollary 5.48. The representation ( For the proof of ( where the series converges to N operator. There exists a sequence h operator. Theorem 5.49. 2 0 = 6 = of an H λ N , S Λ e ∈ . For 2 H of ∈ λ n k be the λ ) K ψ ) B n is also a M i perators. n λ 2 ∈ ≥···≥ ϕ S n ∗ β P containing i| 2 is compact −→ ∞ ) λ s − K with 1 , ϕ β e 2 . S , let ψ λ S H h· ≥ k k β n , 1 n N . s K ψ s e ≤k k ∗ ∞ . For the proof of ∞ ∈ 2 =1 1 K K = < N n k k < H |h n } 2 n M 2 P . s B +1 where S k n ∈ λ = ∞ X ∞ n =1 β = 1 can be extended to a are compact because they =1 = X N − ψ X n k Λ S m k < i n S y ∈ Kψ x n ). For k P = X k λ S k = k 5.5. Hilbert-Schmidt operators we can choose ONSs ( 2 H S , Last Change: Wed 17 Apr 11:33:04 COT 2013 ). In particular, Λ S = k 1 does not depend on the chosen 2 H ∈ x , ϕ 4.27 2 H . Then the following is equivalent: ) First we show that H X 2 λ k h ) H K ( k ∗ ) with n 5.52 S K λ = H ∈ s n , it follows that the Hilbert-Schmidt K . Then e ( 1 such that 4.31 a subset containing all e ∗ H 2 ≤k P N y n i L H ∈ i| . For an arbritray ONB ( N 2 ≤k : N λ K − ψ ∈ n λ e i k ∈ ∈ ∈ 2 k ∈ e , ∞ n n n n k ) is a Hilbert-Schmidt operator, we find that ) S P (id T β n x n n < , ϕ Ky ψ . Note that all K ,K = Kϕ ψ K ∗ 2 } be a Hilbert-Schmidt operator and (e h· β k {k k Kϕ n and n K ψ and choose an arbitrary ONB of k e ≤k s λ h K Kx =1 N |h e Λ and (e 1) 2 sup X n and ( =1 N , ∈ ∞ k =1 B 1 X n K X ) λ M n ∈ = N . (0 k ,..., n X H β < ∈ 2 . 1 k≥ = 2 n P P Λ n B Λ e L s ) 2 2 n ∈ ) Let K ∈ { ∈ − i k s λ n = X X β λ Kx ( = → ∞ λ =1 N ϕ such that X n P . (id M =1 = = is compact. By Theorem H 2 N n M ≥k S 2 K Kϕ does not depend on the chosen ONB of H k k K k P 5.55 β an ONS of Let Λ HS is a Hilbert-Schmidt operator. of DRAFT ∗∗ = ψ ∈ k is compact because it is the norm limit of compact operators. we observe that every The finite-rank operators are dense in the Hilbert-Schmidt o Λ ∗ 0 for X λ 2 N be a Hilbert space and K K ∈ K k ∈ K λ K k n → λ ) such that k HS = ) . Applying the same proof to H . Hence λ k 2 n 2 1 B n 1 s ∈ KP ψ K K H H H X . It follows that β let us define Let N +1 − ∈ N is a Hilbert-Schmidt operator, then M n K and ( ) = k≤k ∈ k ) Assume that N n n K ) Let(e 1 K ii iii ϕ we find, using Parseval’s equality (Theorem ONB of and there exist ONBs ( Proof. Hilbert-Schmidt operator. H In particular, the Hilbert-Schmidt norm of 100 Proof of Theorem Lemma 5.56. An important class of examples is given inTheorem the 5.57. following theorem. Now assume that M P ONB of ( If are the singular values of orthogonal projection on have finite-dimensional range. Since implying that 0 (this family is at most countable by Lemma norm of in particular ( k ( ONB of ) 2 ,H ). is ≥ 99 1 2 ∗ and 2 H such x . ( ,H K k 1 1 o L K ≥ is called H H k ∈ ( 1 =1 S of K x = is a Hilbert- K k Λ 1 x ∗ ). ∈ s k 2 λ . K ∞ ) , A } λ ,H ∈ < n 1 . α } 2 ) H . 0 k α ( . Then . Similarly the reverse n λ L } e ψ \{ 0 ,...,x 2 n ... ) ∈ 1 s K is denoted by H ∗ x k \{ ≥ . 2 { K = be Hilbert spaces, 2 ) Λ 3 TT H ∗ ∈ 2 n ( s H X λ Last Change: Wed 17 Apr 11:33:04 COT 2013 p . to . span σ ≥ TT of T ϕ , H = ) the Hilbert-Schmidt norm ( 1 1 ∞ 2 n 2 n ∞ 2 p ⊥ 2 B s s s H k σ for an ONB (e H ∈ }⊆ < < ,H β β 0 ≥ , x = 2 ) 1 e 2 n 1 2 1 k β Let 1 s n H } ⊆ \{ s H λ K ( (e ) e k 2 n N S ∈ T ϕ T k X B ∗ K H ∗ ≤ x α ∈ k and X T e β ∈ ( : n Λ TT 1 p is a Hilbert-Schmidt operator if and only if ∈ K = k K X n σ λ H ≤ k 1 ) 2 s be Hilbert spaces and 2 k A be Hilbert spaces. of 2 Kx ∈ : 1 α = X k α if and only if there exists an ONB (e ,H 2 e A be a compact operator with singular values 2 n n 1 n ,H ∈ s ψ 1 K ) H ,H α { is a Hilbert-Schmidt operator if and only if ∗ k 2 ( ) 1 H := sup α L A = H K 1 ,H ∈ TT S (e ∈ DRAFT } X 1 H α H Let 0 ∈ k H Let n K ( K \{ ) shows that for L i k ) ,...x ( . Then 1 ∈ T x ∗ ... T K = inf 5.55 ( ≥ p 2 ... σ 3 +1 a Hilbert-Schmidt operator. In this case: for all ONBes Schmidt operator if and only if Every Hilbert-Schmidt operator is compact. Let A operator s n ≥ s n )... (i) Hilbert-Schmidt operator (ii) (iii) ii ( a compact operator with singular values Theorem for inclusion can be shown, so that Hence is well-defined. Proof. Chapter 5. Spectrum of linear operators Moreover Theorem 5.53 (Min-Max-Principle). 5.5 Hilbert-Schmidt operators Definition 5.54. Theorem 5.55. a that The set of all Hilbert-Schmidt operators from ) ≤ = 2 b . m S + ∞ H (5.11) X . Note 2 k n a ∗ < R is also a X T 2( 2 − . ∗ ). To show ∗ k ) 2 T ≤ are positive. A λ ∗ k e . . 2 n = T b S ,H := id R ∗ T id. Now assume 1 2 H X = k + A , k H R X ≤ 2 S ( T + by induction. S . a 0 . 2 H S 2 T H 2 N k = ( N + k kk k H ≥ . ∈ and X ∗ T ≤ 2 λ A n ) ∈ ) T k ∈ e ) R ∗ 5.6. Polar decomposition + T b 2 is a non-negative solution n T A a selfadjoint operator with √ k S T − ≤k − k X ∗ ) X λS A R 2 a . Using the above remark is S k Λ , n A H ( 1 ) ( ). ∈ H − ( 1 commutes with 2 k X λ = k Last Change: Wed 17 Apr 11:33:04 COT 2013 H − L := id − 2 2 2 2 S 1, hence 0 n . ,H b ∈ = k A such that ) of T A 1 = id ≤ X 2 λ k ) + T 2 Λ S e 2 H X k≤ ) + 2 ( ) ∈ id. Hence H H T a k λ ( T k S X + A ) , then k ). k ≤ λ L ( . H λ = e Λ T , ) − A 2 1 T A ∈ are Hilbert-Schmidt operators and with positive coefficients and that ∈ S ( X ∈ T X k H λ ab 5.11 H 2 1 and is denoted by k ( A R 2 k . Then obviously T := ≤ exists. Let L k T = (1 + T A a bounded linear operator between appropriate = T C n +2 T A is positive, this implies that all 2 k ). 2 for ∈ of X kk b A ii λ R = S A e λ and A ≤k + 2 H = S i 2 2 . R root k and k a , X S T ≤k ( is a non-negative solution of AT λ H ) Λ S = ) and ( e ) fix an ONS (e commutes with ) and 2 = k i ∈ T,T X ) of 2 H 2 be a Hilbert space and 2 X λ h ) . Since T k ) A := id is well-defined. The properties of an inner product are clear. H ,H b N AT H ,H ≤ H = ,H 1 0 − kk ( k S 1 1 L( + 2 AT is a polynomial in ∈ S H id if and only if 0 X H L A Λ id k k H ( a H H k ∈ n ( ∈ ·i Let ( λ ∈ DRAFT S ≤ k X , λ S S X is called the = R H T S )e h· R H n,m H k ≤ is a two-sided ideal in R T ∈ ≤ ∈ ∈ ) S . + if and only if H T H T R k ( S T ∗ ( is a Hilbert-Schmidt operator. It follows that for all S ∈ S,T + Note that ( Without restriction we can assume k T . Construction of a solution of ( = . Then there exists exactly one Let Hilbert spaces. Then n Λ S 0 2 kk ∈ X a,b AT X ∗ λ ) is a consequence of ( R ≥ ) Note that m (ii) ) Let A i ii iii (iii) H follows that Step 1 Proof. that Note that every T that for In particular, It follows that ( Note that 0 102 5.6 Polar decomposition Theorem 5.59. We define that a solution so ( In addition, if Proof. of We will show the following properties of the sequence ( Hilbert-Schmidt operator withk norm ( X the operator s. ). ) d s 101 ( (5.9) . f 4.31 t (5.10) in the 2 d N 1) ) , 2 t ∈ ( i| (0 n is an ONB n 2 ) n he sum and it suffices to y L e n n . ) , k ) S , s ) k n ) ( e k 2 n ,t 5.56 · x ( . = rg(Ψ) is Cauchy sequence ,H 2 k 0 1 . =1 S ∈ n 1) N |h X , k n H H ∈ 1 T ( k (0 n 0 be of finite rank. Then 2 n ) S H T Z 1 L = S n k T 0 H s. k 1 ∈ S k Z =1 k ∞ − 1 . ∈ X n f k Ψ 2 ) d − Ψ = s 1) = , ( = 1). Then also ( ). By Lemma s Last Change: Wed 17 Apr 11:33:04 COT 2013 , t x (0 2 H ) 2 →∞ d lim ( (0 ) d , n L t t 2 S 2 ( k s,t d , S,T L ( i n k 1)) s y , k k α s. ) e 1 ) and choose a sequence ( d . t s (0 ) d =Ψ = 0 ( 1 2 d s 2 H Z t n | ( n ,T ( L H complex conjugated function) and we find 2 ) d x n ( S α S ) ) we used Parseval’s equality (Theorem i| 1 n e s S s of so that for every )e )= n H s,t − ( d t e H ( S n f A h , k 2 ∈ t such that )( 5.10 s,t y | | ∈ 1 ) d ( i ΨΨ ) → α 0 2 1 α S k n ,t X ) 0 2 T x Z be an ONB of 2 · 1) ( 1 . k α Z ( s,t be Hilbert spaces. , x , 0 ) 0 1) ( n = is a normed spaces. The norm is induced by the inner →∞ k 1 T =1 (e ) , n Z k n 2 0 h· X ,t n (0 |h | S n · Z 2 (0 ( 1 1 H S = lim 0 =1 2 ) and in ( for i 0 ,H 0 L k =1 ∞ H n n 1 Z X k Z n n L )= k ∈ t ?? = 1 1 1 S H rg Ψ. Fix denotes the to e ( T is an isometry, it follows that (Ψ P 0 0 0 =1 k ∞ k DRAFT n S,T k·k X Z Z n Z Ψ: n n ∈ h y , = T e i ) →∞ Let ) Let(e k is a Hilbert-Schmidt operator and that = = = = 2 n T i n T ( 2 T k = lim ,H 1 f , x n ⇒ S h e H 0 ( ) By the proof we have an isometry =1 T n )= ii S X n k 1) (where is a Hilbert-Schmidt operator. ( ii , ( H ) we have used the monotone convergence theorem to exchange t =1 for an arbitrary ONB T There exists a product ∞ (0 X n )= , hence its limit exists. Using the continuity of Ψ we find ⇒ 2 t 5.9 ( H L is of the form (i) (i) )= (ii) i In ( the integral (Theorem We will show that the range of Ψ is dense in H Chapter 5. Spectrum of linear operators Theorem 5.58. of It follows that show that rg(Ψ) contains the finite-rank operators. Let This shows that Tf In this case we write If one of the equivalent conditions holds, then ( T range of Ψ. Since Ψ Proof. in . . . | ∈ 2 = S for k T k | T k U . T x x Ty | k = − T . = 0 is called a T = 0 for all k | i T x U k = i≥ Ux k ). 2 T x initial space T x Tx,Tx √ ,H h k such that 1 ) = H 2 ( T x , i L 5.6. Polar decomposition . √ T x. ,H by setting 1 2 be Hilbert spaces and 1 S is unique. ∈ it follows that ) h 1 ) H 2 Tx,x y T ( )= U U H U | is called its = ∗ ∗ x L | ,H T i T T Last Change: Wed 17 Apr 11:33:04 COT 2013 | 1 because ⊥ to rg( h T ∈ | 1 H = 1 and has a unique continuous ( = U = U U → := ( H , then k x i | | Tx,x ⊥ ⊥ x U ∈ Let ) ) T T 1 2 √ | k | x T ) U , U T TS ) ∗ exists, is selfadjoiunt and commutes with √ is and isometry because T with (ker T h ( : (ker 1 T for all is U rg( = H ⊥ x , 2 . ) i 1 2 U k ∈ U → ) ) we define . Ty ) T be Hilbert spaces and | T x H T ∗ (ker is an isometry. ker k ( 2 T = x, y | ). Now we extend | Tx,x T L ( ⊥ T U = h √ ) ,H ∈ the root of T x 2 1 U T : rg( rg( k = H T √ ) = ker i x | | U (ker S x → | | 5.59 h T T | ) Let U For T | | = k | H if DRAFT T i | ∈ x , | = 0 hence x rg( = ker( is an partial isometry if and only if T k h| y as shown above. In particular, ⊥ | U . Then there exists a partial isometry ) | ) = STx,x T 2 h H T 2 Note that By Theorem | k − | ∈ ,H x 1 rg( x | | x H is well-defined because for T ∈ T ( k | Proof. Hence for all Definition 5.62. If in addition the initial space of extension to x partial isometry Note that 104 Proof. Definition 5.61. all is unitary. Theorem 5.63 (Polar decomposition). L k| We define U is ≤ we 103 and , k ≤ . H 1 (see ST +1 . n . AS ∈ mials in =0 ) , so that → ∞ with 0 commute X 2 k≤ k x N k = ′ n x T X . ) R )=0 ∈ n X ′ 1 + k = , n n SA − R 0 ws that both of 2 n 2 SX A . ( − and )=0 4 R X ′ →∞ , then also − 2 1 k n ) R R R ′ − x )( n for all k −→ TS 2 R n − 2 )= . For every ′ x ) because 2 X 2 n − R A X R = ( ) SX ( . n lim inf ′ X ) follows from k ′ 2 1 R , then also X − ii ( 5.11 X R A 2 ) k . := 0). Now assume that the ST → ∞ ′ − AR ) 0. Then k≤ →∞ = R )= - lim 1 n 1 n R = n 1 s Last Change: Wed 17 Apr 11:33:04 COT 2013 = X − − 4 ≥ X − X − ′ n k = lim ( k 2 + n +1. ( X n ′ ) solves ( k R ) = ( X ′ k R ′ X 2 2 n R X A ( ) R ( R + ′ X and + Sx 1 2 − n − R n n − ) k≤ A ′ X ( X x X with R R ) R ( = ( 1 2 )= ( )( ′ k − 2 1 n are polynomials in T 2 X − → ∞ = 0 (with ) − R with positive coefficients, so that in particular − x X ′ ) - lim n n n ) are positive and + n ′ R = n s . By definition of R A 2 ( n + X X ( n R . Note that ) 2 ′ . Therefore R = X SX N A X TS ) − 2 H − R n R ′ ( k ( ) is proved for + )( ∈ X n i , it follows that 2 1 X R R = L = X T A X n ( ( − →∞ A → ∞ ∈ → ∞ ′ n − 1 1 2 2 ST - lim R n - lim ) + ( n since the R n ′ s s = lim = = R n X k = ( = ( = n S,T 1. X k − DRAFT 4 0. we obtain a bounded selfadjoint solution of 2 n X ) ′ ) such that If ≤ ) with R = X X ≥ − ( X XSx is a polynomial in R ) H k H 1 ( k R is uniformely bounded monotonically increasing sequence in, there 1 − ( x ) be solution of − − − → ∞ ) +1 L 1. ) − ′ for all L is normal, it follows that n N n +1 commute with - lim 2 ). n n H R ′ R n ∈ s ∈ ( X ( X S ∈ n X X n X = id R L X ) n k≤ − = SXx − X X k − 4.25 n S n − ∈ − X R ∈ R n n 2 n X X X + ( x . Uniqueness of the solution. ′ R ≤k X X k = k X id. R ( 0 A S k with positive coefficients, ( ≤ k (i) n (ii) ( 1 2 Since both operators onthem the are left 0 hand and side therefore are non-negative, it follo Since Setting It follows that because which shows that positive. R X Step 2 Let assertions are true for some All assertions are clear in the case Chapter 5. Spectrum of linear operators Corollary 5.60. Since ( exists an for all therefore obtain Exercise Now let Since all A Since by induction hypothesis both terms in the second line are polyno = n ) . . } λ T x V. n = ∈ ϕ . V k complejos. T U, v X. k ∈ muestre que: , ∈ ) n u X, x x X t definido por ( ∈ (i p ϕ , entonces con i ℓ ) d es un subespacio tal que →∞ lim ϕ t n v ( , x . definido como en el punto − ) → x U = ) y calcule ϕ ) + x p b ϕ 7→ p ( x λ ℓ , x a u ℓ V ( = p Last Change: Tue 5 Feb 10:58:45 COT 2013 ( N Z : /p ϕ , ∈ L ϕ = 1 n ϕ = V , i. e., T ) |≤ ∈ x t ) n y X d . Muestre que x ) := T x ( x sea p ∞ X ϕ ( | en de finita. dimensi´on Muestre que toda ) para V ϕ acotado? Si es as´ı,calcule su norma. = ( t ∞ con Re ( ℓ U k·k X T x , T x | -lineal, entonces X -lineal con Re ∈ C b , V , x R a C N ⇐⇒ | C n 2 Z K ∈ → u ) n ) → x → X =0 ( n ∞ : X z . Muestre que n p := c X . Muestre que p : : p ℓ es acotada. } T k con la norma 0 0 = ( = 0 excepto para un finito n´umero de ´ındices ϕ |≤ )= ∈ x { ϕ Y ) V -lineal sobre z x n k X x N ( x . ¿Sigue siendo = ]) con la norma ( C un funcional un funcional ∈ → : ϕ n | ∞ V ) el conjunto de las sucesiones convergentes. Muestre que el R C N a,b X n . , ϕ ∈ ∩ : ([ x k n → ∞ → . Para K ) p< ϕ C ℓ U T n espacios normados con DRAFT V ) considere el subespacio ∞ X = ( X x y ≤ → = k ⊆ un funcional sublineal sobre : N ( : x ( Y { c = 2 p X X X ϕ λ ℓ p < y : k = = el complemento algebraico de ϕ = ϕ ≤ para X U para 1 es un operador lineal y acotado. es ¿Cu´al su norma? k funcional anterior, entonces es continuo y calcule su norma. es un funcional V V n X z + n (c) Sea (a) Sea (a) Sea (a) Sea (b) Ahora considere (d) (b) Sea x U Sea uc´nlineal funci´on Sea 1 es un funcional lineal bien definido y no acotado. En Sean Demuestre el teorema de Hahn-Banach paraSugerencia: espacios vectoriales Para un espacio vectorial sobre los complejos ...... 9 8 2 3 7 1 Exercises for Chapter 106 M 105 tales → . 0 X x M : f )= with 0 x ( K has exactly one f . → . M T } X 1.15 : M. is a closed subspace of W . x ∈ ∈ ∞ W such that < . n operadores lineales en M } Last Change: Tue 5 Feb 10:58:45 COT 2013 and V, w T T ∈ ∈ X converges in 0 is closed. ∈ S,T , x,y v x t ) + 1) 1 such that : X : X be a metric space. A map n | ) are Banach spaces. 1y w ) x, y t ∞ γ < ( M ( ≥ + = ( x v S γ d X { {| k·k Let +1 ≤ c, 1 n := be defined as in Example on a complete normed space )) T . y W f := sup ( ,c − X 0 + ∞ , f ) and ( +1 k ) ) is a normed space, but that it is not complete. ) is a Banach space. ) be the space of all functions V n d, c ∞ x x ∞ T ( ∞ k if there exists a ( f ST ( ∞ k·k d ℓ k·k k·k , , 0 . Muestre que al menos uno de estos operadores no es acotado. ) d, c DRAFT id T ( = ∞ ℓ contraction is a finite-dimensional subspace of TS is complete. V − , then be a normed space. Then the following is equivalent: be a normed space. Show: un espacio normado con dim be a set and X is a closed subspace of X X ST X X T (i) (ii)Every absolutely convergent series in (a) Show that ( (a) Every finite-dimensional subspace of (b) Show that ( (b) If is called a Show that every contraction fixed point, that is, there exists exactly one Ayuda: Muestre que que Sea Let Let Show that ( Let the sequence spaces Let . . Banach’s fixed point theorem. . . . . 6 1 2 3 5 4 Chapter A. Exercises Appendix A Exercises Exercises for Chapter . ). ⊇ Y es no : punto X, Y ,...,n ( S . ) tal que L uniforme y ) , =1 ⊆ existe y es Y K producto de T x k N ( X, Y T x. , ∈ Y ( → → R n ] K + L ) ) existe. Entonces un operador lin- n T := ∈ → ⊆ ( T a,b x suma de operadores Sx ) Y x n ) N n ) :[ ( k T . Entonces existe un ∈ S := ′ p N α n ⊇ D ]. → ) ∈ Y x RT y n ) n ( ) X a,b T b ∈ T [ : rg( T : . ( ) := lim C ≤ ) + 1 ϕ T n Last Change: Tue 5 Feb 10:58:45 COT 2013 ) , − ∈ ( k S y n t } ), ( S T x ( n ( ) f ⊇ D f R X ( ) x ) ∞ t, n t, n ) := { → R T < ( ) ) d y tome ) d | t S t ) + ( ( ( ) := N n converge para todo f composici´on p ( k S ⊇ D b ( b ) tal que ). ∈ α o N espacios de Banach, a | ]) se define a RT ⊇ D D ∈ Z ( n espacios de Banach y Z Y espacios de Banach, n , DRAFT =1 D a,b : X )) n k es un operador lineal cerrado. X, Y X, Y → un espacio de Banach, se define como → R : ([ ( ( operadores lineales cerrados. Muestre que: es invertible cont´ınuamente (i. e., x X, Y ) ) P n T R L L T C X es cerrado. ⊆ S p f Z T S ( ( ] N ( ∈ ∈ + ∈ + n n ∈ ϕ X, Y, Z → X, Y, Z f sup S n Q Q T Suponga que para todo T operadores aislado es no enumerable. ( enumerable. SR R otıu) entonces cont´ınuo), a,b ) S ( (c) Si (a) Sea (a) (a) (a) Sea (a) Todo espacio completo m´etrico con infinitos elementos y ning´un (b) Sea (b) (b) Sean (b) Toda base algebraica de un espacio de Banach infinito dimensional (b) eal. Sean Muestre que los siguientes enunciados son equivalentes: Muestre que la de hip´otesis es completitud necesaria. en el principio de acotaci´on Para D Sea [ Sean . . . . . 3 4 5 2 1 108 Exercises for Chapter , . ′ = Y X X 107 y K con la X R . Muestre . Y | ∞ e Tietze: Sea f ℓ ∈ tal que )= N . ′ ∈ f es isom´etricamente } ( n )) ′ ) ρ R M n , M x ). ∈ un subconjunto cerrado. N ( X X. tal que x . Sea ( = ( ∞ ′ X R ∈ ℓ X es un homeomorfismo lineal. tiene una continua extensi´on una acotada sucesi´on en ( C X y que es cerrado. ⊆ Last Change: Tue 5 Feb 10:58:45 COT 2013 N ∈ ′ Y R ∈ f ∈ Y ′ 0 ϕ X y n , x ) x → ) definido por n , x ′ n un funcional lineal no nulo y ) X y x x Y Y x ( N ( K : . sea separable? R ′ 0 ) ∈ k f C x Y ) es un isomorfismo [isom´etrico]. Si →∞ ( X → ) el conjunto de funciones continuas real- k n )=0 para todo ′ C ′ n → lim sup x k X )= X x ( X ) ( f un subespacio de x ≤ f : R k ( X es isomorfo a un espacio de Banach reflexivo ) | k → ) el mapa inducido en el espacio cociente. Pruebe ( C x f M ′ ′ n = R ′ ( un subconjunto cerrado. Y Y x ) ) = 1. ( C X ϕ Y R X : X : ( ∈ ≤ C ) es completo. ′ ρ C →∞ lim ρ ⊂ f k k n X/K T e { → f x Y k es continuo si y solo si ker es un isomorfismo entre [isom´etrico] los espacios normados = ) es un subespacio cerrado de /I y la convergencia y d´ebil la convergencia en norma coinciden. f ρ ) ) el conjunto de todas las sucesiones acotadas en L Y es un espacio de Banach reflexivo. →∞ DRAFT 1 con . X R X l es un subespacio cerrado de n ( , lim inf → Y R R /L N L ′ ( C , entonces X es una isometr´ıa. := ker( → : X ∞ : Y ℓ e I ρ e ρ X un espacio compacto de Hausdorff y y T un espacio compacto, un espacio normado, un espacio normado separable y ( : un espacio normado y e Entonces cada continua funci´on X X f que que son espacios de Banach, el converso vale. tambi´en entonces norma del supremo. Muestre que existe X X X X isomorfismo entre espacios normados f (c) Demuestre que rg( (a) Si (a) Considere el mapa (a) Muestre que dim( (b) Sea (d) Use el teorma de Stone-Weierstraß para concluir el teorema d (b) Si un espacio normado (b) Muestre que (b) Sea Es cierto esto sin la de hip´otesis que Pruebe las siguientes afirmaciones. Muestre que isomorfo a Entonces existe una ( subsucesi´on ker Un evaluadas sobre Sea Muestre que en Sea Sea Sea ...... 5 7 9 8 4 6 Chapter A. Exercises . . j i } δ R . = no es ∈ j i tal que e λ X para todo : X sea un pro- λ 2 ∞ ∈ f w k es equivalente . Muestre que 0 { x ·i x . x , k w . ·i h· = 1 , ℓ ) = i h· y ,... = span 6 el vector usual e + s x, x es una seminorma en X. 3 h i . Then the following is X 1 x e 2 ( t. ) d M H } 0 para que s y p 0 + ( P . funciones casi peri´odicas ) d 2 g ∈ 1 N 5 w t l , ) Last Change: Tue 5 Feb 10:58:45 COT 2013 ℓ ( 2 y sea s 0 ,e ∈ ( = w L 4 x = 0. ) f n e t λs 1] tal que i H ⊕ 0 ( : 2 T , 1 T y 2 x 1 − ) [0 and t L +1 Z ( + C un subconjunto convexo, balanceado y se llaman n un subespacio denso y x 2 3 )=e 1]) se define T H → 1 e 1 s , 2 X en X { H ,e ( 0 2 2 λ ⊆ Z ·i L ([0 e ⊆ f . Muestre que la de completaci´on ⊆ , 2 1 convergente en N →∞ C 4 , . ⊕ := X span ∈ T h· U } . . ∗ M C 1 ∈ + n 0 w . ) w H H i L 0 1 { := := lim n . Muestre que e : x → 1 x i = 6 = no es convergente en d´ebilmente es = x, y U 0 L x,y 2 2 R 2 w N N P h −→ ∈ L L ∈ ∈ L : ? f,g =?) n n n u h 2 ) ) ⊕ ⊕ ∩ λ x span k L n n ′ f 1 1 1 e e λ ). Para L L L := f R and 2 , − 0 L 1] λ . x DRAFT defina , f 0 k x R ([0 k → C 1] ? es una no proyecci´on acotada. → un espacio vectorial y ∈ n , be Hilbert space, ( = 0 para todo un espacio pre-Hilbert, 0 n ∈ x i [0 λ x k P X H X w C ,u 0 ∈ x Sea (c) Muestre que (a) (a) Muestre que ( (a) Muestre que (b) (b) Muestre que ( (d) Defina el operador (b) Muestre que Defina equivalent: a la norma usual de Let h Ejemplo de una no proyecci´on acotada. Sea ducto interno. Bajo la qu´econdici´on norma inducida por separable. ( Los elementos en la de completaci´on absorbente. Muestre que el funcional de Minkowski y define un producto interior en Sea Sea Halle una necesaria condici´on y suficiente spobre Muestre que x ¿Existe producto alg´un interno Para ...... 3 4 2 1 6 5 Exercises for Chapter 110 13 . ′ } y y X X 0 n 109 si y x = 1 ∈ 0 k tal que w −→ x ϕ ′ k·k . k , n . N X : x } X ∈ biando “cer- ). n n X de ) ∈ 0 n ). es una sucesi´on ′ x . 0 . ∈ ( n x k ueio ´blde d´ebil sucesi´on x x U 0 Y . n x K ′ ) es cerrado en ϕ { ϕ un operador lineal tal que = ( U F , y k ′ ′ X, ( = x ( 1 Y ∈ )= X , X − L es una n } ϕ K ) ∈ →∞ x T ⊆ → ∈ ( n X N 0 es una de sucesi´on Cauchy lim inf N n ( ) T ϕ para 2 ∈ ϕ ⊆ N T ℓ n , ∈ ( ′ Last Change: Tue 5 Feb 10:58:45 COT 2013 ) N . Muestre que k≤ n ∈ n N ∈ 0 . Es decir: un funcional lineal X →∞ ′ ) es cerrado en ) es cerrado? )) ∈ n X ϕ N ϕ n ) 6 = 0 para solo finitos n n entonces ∈ )) A ⊆ k ⊇ D K ) ′ n , ( ( x n n ( n ( x ) es acotada. N Y x T X T n ϕ ∈ X nx un operador lineal densamente definido n X ⊆ ) : X,X , nx ): ( n T k N = ( N ϕ n ) la que sucesi´on tiene 1 en la posici´on T ∈ ( y x la esfera unitaria y n 2 . ¿Siempre son cerradas d´ebilmente (prueba ) : ( ) ℓ k N y ( X, σ } n X N ,... ∈ ( x ∈ 0 2 X . Muestre que lim , N ℓ = ( 0 = 1 n es cerrado si y solo si →∞ 1 la ( sucesi´on ∈ ∈ ϕ n ′ k ∈ n ′ , es cierto que lim inf ) T 0 ) x X, Tx N ∗ X x n X k ∈ w x ,..., , : −−→ n ∈ k≤ es un compacto en → k·k ) 0 n ueio acotadauna en sucesi´on X n ) n ϕ X = ( x F ϕ N x X, T k = (0 ⊆ ∈ x ( ∈ y n { n N compacto. Muestre que es continua con respecto a la topolog´ıainducida por es una de sucesi´on Cauchy para todo ) DRAFT x e 0 = ( ∈ n { . Muestre = x N n x K ⊇ D X 0 x ) espacios normados y ∈ ) y d { = ϕ n n sea . Muestre que ⊂ N X x → es cerrado en k·k ( Y = ( )) n −−→ S : ∗ N 2 es cerrado con: n )= )= K ℓ x w A n y K X −−→ x T si para todo ∈ T T un espacio normado. un espacio de Banach, ( T x ( : ( ( = un espacio normado. Una ( sucesi´on un espacio de Banach, n n ϕ en X la bola unitaria cerrada en o contraejemplo)? ϕ ´l sis´olo es continua con respecto a la topolog´ıad´ebil. ϕ ( D D ´blded´ebil Cauchy si y solo si existe un subconjunto denso X X . X X X X K Para Sea Sea (c) Sean (c) ¿Si (a) Muestre que ( (a) Sea (a) (a) Sea (b) Sean ( (b) Muestre que si (b) de Toda d´ebil sucesi´on Cauchy en (b) 0 en el resto. de tal que cerrado. Diga si Sea Sean Cauchy en rado” por “clausurable” Muestre que adem´as estas afirmaciones siguen siendo cam v´alidas Sea Sea ...... 7 8 9 6 12 11 10 Chapter A. Exercises ) k es H H y sus ,y 0 T × kk ⊆ x x W ( k H n P B ) , n M V x 7→ P y donde, k ≤ ) , entonces m ) . H fijo, V m x, y ( y 0 P ∈ x B W x k P , H. 2 sesquilineal. En ∈ k = tal que para todo conjunto x K ) es cont´ınua. . Las siquientes son equiva- k N W 0 X P m Last Change: Tue 5 Feb 10:58:45 COT 2013 tal que → ⊆ ) tal que V . , ∈ x, y < ε. ≥ ( k x P R A H H x . n ( ) es un operador lineal, las siquientes B 2 x , x,y x = ∈ y L subespacios cerrados y × k i H y − V 7→ ∈ x, x =1 m k X H M ( H b X k P tal que la ( sucesi´on x → x T : B + si alguna de las condiciones anteriores se ⊆ 1 k ⊆ m Tx,y B ) 2 h H N B ∈ k k X fijo, b : ∈ = W n x . n x 0 k i ) P )= m ⊕ ⇐⇒ y V, W n . y p 1 x V x, y − . 0 tal que ( := N x,Py m H W B h k ) = ⊆ ) ∈ ⊆ = W m > B k≤ i P x, y 1 V ( ⊆ − x, y + 0 existe un conjunto finito es una ortogonal. proyecci´on k T A V k . W = 0. Px,y P h ε> P converge incondicionalmente a W W con y DRAFT + n P es acotado, es decir, existe is parcialmente es cont´ınua, decir, para cada es cont´ınua. ⊥ x B P V V es entonces cont´ınuo, existe N P V P es para cont´ınua cada B para todo B B ∈ = un espacio de Hilbert. Muestre que para toda ( sucesi´on un espacio de Hilbert and un espacio de Hilbert, B un espacio normado, ( es una ortogonal. proyecci´on n un espacio de Hilbert. Si 2 H P P P finito invertible y H (i) H (i) X H (ii) (ii) (iii) (iii) Sea Sea Sea Sea Sea (c) Si existe adem´as (a) (a) (a) Muestre que (a) Muestre que las siguientes son equivalentes: (b) Muestre que las siguientes afirmaciones son equivalentes: (b) (b) Para todo (b) Si acotada, existe una ( subsucesi´on tiene. son equivalentes: lentes: Muestre que rg( correspondientes proyecciones ortogonales. converge. considere la norma . . . . . 16 15 13 14 112 12 y f n 111 . ? α m ∞ ) como + W tenga? = R ·i ( , p ··· p ´sm de -´esima h· L + convergen en α . es linealmente . (Ω) 1 s f . → } ) α T } D } α ) ( m . Show that there R = ′ funciones de prueba D ( . x, y . . | p Y { α k α |≤ L | . ϕ ∈ (Ω) := α 0. Los (Ω) 2 : k | i m , 0 0 . D s g = 2 ) → ϕ T H k ∈ α derivada d´ebil k ( s (Ω) y 0 2 k Last Change: Tue 5 Feb 10:58:45 COT 2013 ϕ Ω es compacto L k se define + f,D ⊆ ∈ 2 defina ) n ) k α f , ϕ N ( ) x ϕ i R id para with k α ϕ D ( ∈ 0 h ∈ son isometrias lineales. α s para todo escalar un subespacio. Muestre que = −→ D ) ϕ m s s k 2 n s T X (Ω) son espacios de Hilbert. |≤ x k of T X a subspace and (Ω) se llama la (Ω) and α y definimos el conjunto de f,D m | ′ 0 2 ⊆ ) y h m (Ω) : | R + L H H H 2 R muestre que el conjunto α U | := (Ω) : supp( ( W k≥k x L ) ∈ p ⊆ ∈ y ,...,α ⊆ k ∞ m , entonces ∩ 1 − L g ∈ αy ϕ y C α W ⊥ Y f i ∈ (Ω) y + ∈ ⊥ = ( { (Ω) x espacio de Sobolev m f x i = ( x ϕ m k Para Ω H f,g { h α C : . Muestre que si la derivada existe. g , ϕ con h (Ω) := ). Claramente los ∞ ϕ 6 =0y (Ω), s n 4.10 X m . Para DRAFT m X α y n (Ω) := − . ¿Se tiene alguna contenencia? ∈ (Ω) := W p < t W 6 = D ( m definimos el convergen en norma o convergen fuertemente en el caso ⊥⊥ ≤ ∞ f ⊥ . . . ∂ ≤ (Ω). Una funci´on H s , si y solo si x, y 6 = 0, 1 N U 2 U p y T α 1 L x ∈ ∂ 6 = ⊕ be a Hilbert space, ⊥ ≤ ) := un espacio pre-Hilbert y un espacio pre-Hilbert. Muestre los siguientes resultados t ∈ m = ¿El converso es cierto en general? ¿Hay caso alg´un para el que se independiente. ¿Como se puede generalizar estex resultado? norma? U U )( H (Ω) es un producto interior con X X f ϕ f m s α T Sea 1 Muestre que (c) (a) (a) Sean (a) Sea 1 (b) Si (b) Los (b) exists exactly one extension ( Para el problema Sea Let Sea donde la clausura es tomada con respecto a la norma inducida por Para un multi-´ındice si D Sea Note que la derivada es si d´ebil ´unica existe; se denota por Para W dma,definimosAdem´as, los espacios . . . . . 9 8 7 Chapter A. Exercises 10 11 , . - ). 0 ∗ = T 1 H ∗ − y, x ) S − ∗ = R . En este ∗ H ) ) = ( ) operadores 3 ∈ TS H x, y x . ( n converge en el T U → , 2 . = 1 ) H H ⊥ ( ∗∗ es densamente definido ) es inyectivo y ( T T × T ∗ 2 para todo ( un operador lineal acotado. TS S )y H G || 2 ( una base ortonormal de H, y, H Last Change: Tue 5 Feb 10:58:45 COT 2013 H U → T x N 1 || 2 → ∈ = → n < H = ) 1 ⊥ H || n × converge en el sentido fuerte a una : || H n x ))] ( 1 x ) entonces ) entonces y N las proyecciones ortogonales sobre T ∗ 1 T S 1 una acotada sucesi´on y mon´otonamente H ∈ ( = 0. − T 1 : n si alguna de las condiciones anteriores se N es densamente definido y ( G || ) ,H ,H P z n ∈ ( 2 2 n U n x , U W ) P 0 || H . H TS n ( ( P ⊕ T H . L L =1 ∞ 2 es creciente. es decreciente. X n ) = [ V ∈ . || ∈ ∈ es densamente definido y ∗ n n T 1 H 1 T ∗ P P || ( − , entonces − ) = T ∈ si si S G S N , x = y adem´as i W n n ∈ || ) entonces P . 2 P 3 ∗ n , x ) es densamente definido, + T x , x es una ortogonal. proyecci´on k rgP rgP espacios de Hilbert y T 2 1 || ,H ∗ x V N N . Muestre que ( 3 2 P H W 1 S P ∈ ∈ es normal si y solo si . H H P H P n n ( 0 una de mon´otona sucesi´on proyecciones ortogonales en un espa espacios de Hilbert y → , = y S T DRAFT T + P L 1 1 2 i ≤ h ∗ 2 N ) ∗ V H ∈ = = = ∈ H H ) es inyectivo y es clausurable, es inyectivo y H k≤k n ( P 1 , , 1 ) es cerrado. x ⊆ , para todo una tal sucesi´on que: x , x TS un espacio de Hilbert y 1 P . Entonces las siguientes afirmaciones son equivalenteas: T P S 1 S T T un espacio de Hilbert separable, ( − es unitario. un espacio de Hilbert y ( 0 un espacio de Hilbert complejo y P n 0 n ∗ 0 N 0 H y P R H P H P y ( ( U T ∈ H H h k H P H H n (iii) ⊥ ⊆ ) (i) n z 1 (ii) (iv) (iii) y Sea Sea Sea ( Sea Sea Sean Sean (c) (c) Si (a) rg (a) Si (a) (b) Si (d) Si (b) Si (b) rg caso, muestre que Muestre que cio de Hilbert creciente de operadoressentido autoadjuntos. fuerte a Muestre un que operador la autoadjunto. sucesi´o rycio ortogonal proyecci´on y H ( lineales densamente definidos. tiene. Entonces Muestre que rg( ...... 24 25 26 23 22 28 114 27 , . 0 } sus 113 H 0 N W . ∈ P f . En este , r , V H 1) P , ∈ is a orthonor- [0 Z x ) with ∈ . ∈ k define . t V − ) n,k P ) Z n W ∈ for − n,k P . t + 1)2 ) ψ k t r = para todo ( ( n, k un operador lineal acotado. (2 is a orthonormal projection , || W ψ k k,n , j una base ortonormal de H, y, H ) P ). 2 − Last Change: Tue 5 Feb 10:58:45 COT 2013 ψ h t T x 2 i N ( V 1 R k/ − || converges uniformely to 1] and ( r . For j → ( 1]. ∈ 1 P , 2 , n j < 1) k,n = ) → [0 , L h H = [0 t ψ || 2 subespacios cerrados y n 2 i 2 f,h : || / j )=2 n x h L let V x t L y [1 1 ( las proyecciones ortogonales sobre T ∗ H P } )= χ − = 0. t − T 1 1 k =0 f,h ( (1) = lim n,k ⊆ j 2 || − h P z n n n − x , + + P 2) =0 k . 0 || k k / ∞ j 2 2 2 1 P = H ⇐⇒ , . h h V, W =1 P ∞ 2 [0 X , ψ n ∈ . || χ R T H ,..., const. in intervals [ , Tf = || 2 W , entonces → ∈ f 1] ). , , x ψ N = , , i 1 ⊆ R R ( R , ∈ [0 || : 1]. 1] : 2 2 2 V 0 , , Let , x n x , x L L T → k [0 1 n,k [0 || 1], the series x ∈{ 2 P ψ 1] is an orthonormal basis of , 1 . → DRAFT = 0. ∈ L , n . es normal si y solo si P [0 Z 0 W is an orthonormal basis of 1] is a orthonormal system in , ∈ W ∈ , t C , P : [0 1 T i ≤ h P 0 0 f ⊥ and [0 N N ∈ k,n n H V { = 2 ) k≤k ∈ ∈ + 0 P V f j j 1 L x k ⊆ x , x = , para todo una tal sucesi´on que: )=1 ) ) un espacio de Hilbert, N un espacio de Hilbert y . Entonces las siguientes afirmaciones son equivalenteas: 2 un espacio de Hilbert separable, ( 0 P k,n t : 0 un espacio de Hilbert complejo y j j n 0 ( 0 h N P h ψ h P y U ∈ H 0 on the subspace T mal system in k h H P ∈ (i) H H H H (ii) h n k ⊥ ⊆ ) n z 1 y Sea Sea Sea Sea (c) For (e) ( (c) (a) ( (a) Muestre que (a) (d) ( (b) (b) Muestre que las siguientes afirmaciones son equivalentes: (b) (d) and caso, muestre que Muestre que correspondientes proyecciones ortogonales. For y H ( . . . Haar functions. . . 21 19 20 18 Chapter A. Exercises 17 DRAFT \ ). C ) 1 y A 115 ⊆ ( r TS < σ ( K || σ S )y , . = ) A } ( || || K , . c 0 ) ) T σ ,... || 3 \{ ), ) ,... ,... tro residual de los A , x 4 4 . ( 2 s. es invertible, encon- N p ) ST , x t ∈ ( σ , x ( 3 ) d , x n 3 σ s x 1 ) ), TS ( ) , x n t , x A 2 x − x ( , x ( 2 t . Suponga x a n x σ 1 0 λ , Z − ) k Last Change: Tue 5 Feb 10:58:45 COT 2013 ) cuyo espectro es )= ) = ( A t )=(0 1 ) = ( k )= ℓ ST ) = ( )( t ( N − L )( ,... ∈ ,... 3 Ax ,... ). n 3 ( ∈ ) 3 Sx S ). Muestre que y , x n ( ( T 2 r , x y (id x X , x } 2 ( σ 2 ). Muestre que, adem´as para todo 1 (( , x L 1 T − , x 1]). Muestre que ( , x ) x 1 ) y r ∈ X, , 1 ( x 5 σ S x H, ( ( ( TS (0) = 0 ([0 → c x − C → σ ,T S,T ) y ,T X 1 ) T ℓ ∈ es invertible si y solo si id ), : , R H ,L ( N ) ) c S a : 1]) : ( A ( → σ , N N p ( ST ∞ ( S 1 2 ℓ σ 2 ), ℓ ([0 − ℓ ℓ : T 1]) y ), C ( → , → S p → T ) ∈ una acotada, sucesi´on y, ( σ ) ) DRAFT ([0 N σ f N ( C N ), { C ( ( 2 ∞ 2 T ⊆ ℓ ℓ ℓ ( = = : : : N σ ∈ H X T R L n ) un espacio de Banach n Muestre que id Encuentre es un operador lineal acotado. Encuentre λ X . } 0 (a) Sea (b) Sea Encuentre { Hint. compacto no existe vac´ıo, un operador Sea ( trando una entre relaci´on (id Sea mire si la en relaci´on este caso es en v´alida general. Encuentre el espectro puntual,operadores: el espectro continuo y el espec . . . . 2 3 1 4 Chapter A. Exercises Exercises for Chapter DRAFT , , . eory, York, Grad- . Wiley . Theorie dex.html Applied Math- , volume 253 of , volume 68 of . Classics in Mathe- , volume 109 of . McGraw-Hill Book Co., New Linear operators. Part I . Classics in Mathematics. Springer- . International Series in Pure and Ap- . Mathematische [Mathemat- Leitf¨aden. . Springer-Verlag, Berlin, extended edi- Methods of modern mathematical physics. . Springer, New York, 2009. 117 . Springer-Verlag, New York, 1980. Translated Elementary functional analysis Linear operators in Hilbert spaces Introduction to Hilbert space and the theory of spectral Applied functional analysis . Springer-Verlag, New York, 1995. Main principles and Functional analysis Real and complex analysis Functional analysis Funktionalanalysis Funktionalanalysis Perturbation theory for linear operators . AMS Chelsea Publishing, Providence, RI, 1998. Reprint of DRAFT . Academic Press Inc. [Harcourt Brace Jovanovich Publishers], New from the German by Joseph Sz¨ucs. uate Texts in Mathematics tion, 2000. plied Mathematics. McGraw-Hill Inc., New York, second edition, 1991 I York, third edition, 1987. ical Textbooks]. B. G.und Anwendung. Teubner, [Theory Stuttgart, and fourth application]. edition, 2006 matics. Springer-Verlag, Berlin, 1995. Reprint of the 1980 edition. multiplicity the second (1957) edition. Verlag, Berlin, 1995. Reprint of the sixth (1980) edition. http://www.mathematik.uni-regensburg.de/broecker/in 2004. Classics Library. John Wiley &With Sons the Inc., New assistance York, of 1988.the William General 1958 G. th original, Bade A and Wiley-Interscience Robert Publication. G. Bartle, Reprint of second edition, 1980. Functional analysis. Graduate Texts in Mathematics ematical Sciences their applications. [Wer00] Dirk Werner. [Yos95] Yosida. K¯osaku [Wei80] Joachim Weidmann. [RS80] Michael Reed and Barry Simon. [Rud91] Walter Rudin. Bibliography [Den04] Robert Denk. Funktionalanalysis I. Universit¨at Konstanz [Kat95] Tosio Kato. [Les] Jaime[Mac09] Lesmes. Barbara funcional. An´alisis D. 2 edici´on. MacCluer. [Heu06] Harro Heuser. [Zei95] Eberhard Zeidler. [DS88] Nelson Dunford and Jacob T. Schwartz. [Hal98] Paul R. Halmos. [Rud87] Walter Rudin. . 1.15 con K → T : x } Problem Sheet 1 W . es un subespacio cerrado ∈ ∞ W < es un espacio de Banach si y y } Last Change: Tue 5 Feb 10:58:45 COT 2013 V, w X X T es cerrado. ∈ ∈ v X t definidos como en el Example : : | ) son espacios de Banach. w ) , c t ∞ 0 ( + x v . { {| d, c k·k X c, := W := sup ) y ( ) es un espacio normado, pero no es completo. ) es un espacio de Banach. + ) el conjunto de todas la funciones ∞ ∞ ∞ ∞ k T V ( x k k·k ∞ k·k k·k , ℓ , 0 ) c d, T ( ∞ ℓ ) un espacio normado. Muestre que DRAFT , entonces es un subespacio finito-dimensional de X k·k V un espacio normado. Muestre que: un conjunto y de es un subespacio cerrado de X, T X (a) Todo subespacio finito-dimensional de (a) Muestre que ( (b) Si (b) Muestre que ( Sea Muestre que ( Sea Considere el espacio de sucesiones Sea ( solo si toda serie absolutamente convergente es convergente. . . . . 2 1 3 4 120 Espacios y m´etricos normados. 119 DRAFT Problem Sheets = K . . con la } λ V. n R = ∈ . ϕ ∞ V ℓ complejos. ∈ U, v X. tal que N ∈ muestre que: , ∈ ′ ∈ ) n u n )) x x ) X R n (i Problem Sheet 3 , x ϕ , entonces con i N es un subespacio tal que →∞ lim ϕ ( n v , x . definido como en el punto − = ( ) ∞ U = ϕ ) + x ℓ ϕ 7→ ( ( x λ u es cerrado. V ( = p Last Change: Tue 5 Feb 10:58:45 COT 2013 N ∈ ϕ , ∈ f ϕ = n ϕ V ϕ , i. e., ) |≤ , x x ) n y n X un funcional lineal no nulo y x ) := x x ( x X ϕ ( K en para V ϕ = ( con Re U →∞ → n -lineal, entonces X -lineal con Re lim sup , V , x R C X ⇐⇒ | C n ≤ K : u ) ) → x x f → =0 ( ( ∞ ) = 1. X n p c X ϕ . Muestre que : : } ≤ 0 0 = 0 excepto para un finito n´umero de ´ındices ϕ |≤ )= { ϕ ) n X/K V -lineal sobre x n x x ( x = ( C un funcional un funcional es continuo si y solo si ker : ϕ | V el conjunto de las sucesiones convergentes. Muestre que el R f C N ) el conjunto de todas las sucesiones acotadas en . →∞ , ϕ ∈ ∩ R n k n → ∞ → K , lim inf ) ϕ ℓ U n DRAFT N V ) considere el subespacio X X ( x y → k ⊆ un funcional sublineal sobre : N ( : ∞ ( { ℓ c = 2 p X X ϕ λ ℓ : k = un espacio normado, = el complemento algebraico de ϕ = ϕ U k funcional anterior, entonces es continuo y calcule su norma. norma del supremo. Muestre que existe es un funcional V X V X f + (c) Sea (a) Sea (a) Muestre que dim( (a) Sea (d) (b) Muestre que (b) Sea (b) Sea U Sea es un funcional lineal bien definido y no acotado. En ker Demuestre el teorema de Hahn-Banach paraSugerencia: espacios vectoriales Para un espacio vectorial sobre los complejos Sea . . . . 2 3 1 4 122 Hahn Banach, Espacios duales. = 121 n ) T x ejercicio . Muestre Y | . f k T k )= f ( X, ρ t definido por ( ∈ Problem Sheet 2 p ). ℓ ) d t X ( ( → x ) y calcule R p b p C ℓ , x a ℓ Last Change: Tue 5 Feb 10:58:45 COT 2013 ( Z : /p L 1 = T ∈ t ) definido por d . Muestre que T Y sea p ∞ ( | de finita. dimensi´on Muestre que toda ) R acotado? Si es as´ı,calcule su norma. t ∞ ( C ℓ k·k X T x , T x ) el conjunto de funciones continuas real- | ∈ → C b X a ) N ( Z ∈ R → X n ) el mapa inducido en el espacio cociente. Pruebe ( C ) X R n un subconjunto cerrado. Y : z . Muestre que ( := C p R : p X ℓ es acotada. T C ) es completo. k con la norma = ( ρ ρ ∈ x ⊂ Y z k X → N . ¿Sigue siendo Y ]) con la norma ∈ → ) es un subespacio cerrado de n /I y ∞ ) ρ ) a,b X n = 1). DRAFT : ([ X X x . Para ( p p< C T R espacios normados con ∞ = ( C ≤ = es una isometr´ıa. := ker( : x Y e I ρ X e ρ p < y un espacio compacto, ≤ para X (Si no han visto claramente medida, teor´ıade y ind´ıquenlo solo hagan el para para 1 es un operador lineal y acotado. es ¿Cu´al su norma? que que X n z n (c) Demuestre que rg( (a) Sea (a) Considere el mapa (b) Ahora considere (b) Sea x Sea 1 uc´nlineal funci´on Sean evaluadas sobre Sea . . . . 4 3 2 1 Operadores lineales. 58 87 94 95 56 , 60 51 114 87 79 40 56 , 61 , , , 78 58 40 64 89 ∼ 61 6 ∼ 69 78 33 55 108 53 55 , , , , 104 53 16 98 86 ∼ 9 6 9 21 ∼ 15 79 60 23 108 , , , 84 , 79 68 26 89 ∼ 93 ∼ ∼ , 5 , 58 50 9 8 79 26 79 Last Change: Tue 5 Feb 10:58:45 COT 2013 33 positive, selfadjoint spectrum of a convergent metric, normed compact operator, Cauchy Banach complemented operator norm, operator product, operator sequences, operator sum, orthogonal, orthogonal complement, orthogonal projection, orthonormal basis, orthonormal system, Parallelogram identity, Parseval’s equality, partial isometry, Polarisation formula, positivity preserving, pre-Hilbert space, precompact, product topology, projection, projection theorem, reflexive, relatively compact, resolvent, resolvent map, resolvent set, Riesz index, Riesz’s lemma, Riesz-Schauder theory, Schauder basis, selfadjoint operator, semicontinuous, seminorm, separable, sequence sesquilinear form, shift operator, singular values, space spectral radius, spectrum, sublinear, subspace 88 , 78 , 46 99 ∼ 19 46 18 , 27 99 , 112 11 62 ∼ 55 19 99 40 14 , 11 18 30 63 95 95 6 , ∼ s, 25 , DRAFT 90 18 67 , , 52 18 47 113 ∼ 38 , ∼ , 15 , 104 55 , 9 , ∼ ∼ 56 62 47 , 5 93 , ∼ 82 ∼ ∼ 16 ∼ , , ∼ , ∼ ∼ ∼ ∼ 45 ∼ 6 Riesz Cauchy-Schwarz H¨older’s Minkowski Minkowski’s Young’s partial closed compact essentially selfadjoint Fredholm theorem Fredholm Bessel Riesz’s equivalent closable 124 conjugate, H¨older inequality, H¨older Haar functions, Hahn-Banach theorem, Hamel basis, Hellinger-Toeplitz Hilbert space, Hilbert-Schmidt norm, Hilbert-Schmidt operator, holomorphic, homeomorphism, index inequality initial topology, inner product, inner product space, Inverse mapping theorem, isometry, Korovkin theorem, Lax-Milgram theorem, Legendre polynomial, Lemma metric space, Min-Max-Principle, Minkowski inequality, Neumann series, norm normed space, nowhere dense, open map, Open mapping theorem, operator 61 88 61 38 18 , 50 , ∼ 75 6 65 47 74 83 95 63 , , 95 90 40 82 47 14 6 38 62 , 95 42 , 47 40 90 , second 62 40 35 44 , 94 ∼ 41 , ∼ , ∼ 6 49 105 ∼ 26 79 26 89 10 92 33 spectrum, weakly uniformly strongly Bessel inequality, bidual, bounded category, first Cauchy sequence, Cauchy-Schwarz inequality, closable operator, closed graph theorem, closed operator, closed range theorem, compact, compact operator, complete metric space, complete orthonormal system, contraction, convergence, convergent descent, Dirichlet kernel, distance, dual space, Dunford theorem, eigenspace, equicontinuous, equivalent norms, essentially selfadjoint operator, theorem, Fej´er Fischer-Riesz theorem, formal adjoint operator, Fourier series, representation theorem, Fr´echet-Riesz Fredholm alternative, Fredholm index, Fredholm operator, functional, Gram-Schmidt, graph norm, 123 38 38 37 75 , 71 65 , DRAFT , ∼ ∼ 16 16 , , 21 16 16 9 ∼ ∼ ), 99 , 73 47 ∼ X ), 30 60 ( 13 2 90 ), , 30 12 47 53 79 111 73 Y , ,L 99 ), 72 ]), 92 64 ,H 11 ) , ), 111 ), 47 , ]), 1 11 92 39 , 52 92 , 12 58 S → T B 10 , , Banach space Hilbert space algebraic Hamel Schauder T H , a,b ), 11 ◦ , ), ∞ p H ), (Ω), ), ( 11 0 a,b 58 X, Y , (Γ), ([ , T λ, T 12 X ⊥ X, Y T ⊆ U ( , (Ω), T ( 1 ∞ , ([ ( ◦ ( 0 ( ( ( , ( ∞ ∞ p Index ⊥ c, c d k·k k·k k·k A G HS K M R S T dom( C C L C D ℓ ℓ ℓ α δ τ dist, adjoint operator Alaoglu, algebraic basis, annihilator, antilinear, ascent, Baire’s category theorem, Baire-Hausdorff theorem, Banach space, Banach-Steinhaus theorem, basis 125 Last Change: Tue 5 Feb 10:58:45 COT 2013 61 , ∼ 38 112 67 38 , , 46 50 38 37 109 , , ∼ , ∼ , 46 27 65 ∼ 112 ∼ , ∼ 78 47 74 , 38 , DRAFT , 18 , ∼ 5 , 90 35 ∼ 58 ∼ ∼ 107 83 ∼ 40 ∼ 89 , 52 , 53 , 52 ∼ 90 52 111 , ∼ 52 82 , 74 52 44 ∼ , 107 , , ∼ ∼ ∼ ∗ ∼ 52 topology, ∗ Ascoli-Arzel´a, Baire’s category Baire-Hausdorff Banach-Steinhaus closed graph closed range Dunford, Fej´er Fischer-Riesz representation Fr´echet-Riesz Hahn-Banach Hellinger-Toeplitz inverse mapping Korovkin Lax-Milgram open mapping Phillips, projection Schauder, Tietze, initial product weak weak symmetric operator, test functions, theorem Tietze’s theorem, topology, totally bounded, triangle inequality, unconditionally convergent, Uniform boundedness principle, uniformly bounded, weak Cauchy sequence, weak convergence, weak topology, weak Young’s inequality,