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LECTURE 1 Banach spaces

1.1. Introducton to Banach Spaces Definition 1.1. Let X be a K–. A functional p X 0, is called a , if (a) p λx λ p x , λ K, x X, ∶ → [ +∞) (b) p x y p x p y , x, y X. ( ) = S S ( ) ∀ ∈ ∈ Definition 1.2. Let p be a seminorm such that p x 0 x 0. Then, p is a ( + ) ≤ ( ) + ( ) ∀ ∈ (denoted by ). ( ) = ⇒ = Definition 1.3. A pair X, is called a normed linear space. Y ⋅ Y Lemma 1.4. Each normed space X, is a X, d with a metric given by ( Y ⋅ Y) d x, y x y . ( Y ⋅ Y) ( ) Definition 1.5. A x in a normed space X, is called a Cauchy se- ( ) = Y − Y n n N quence, if ∈ ε 0 N{ ε } N n, m N ε (xn Yx⋅ mY) ε. Definition 1.6. A sequence x converges to x (which is denoted by lim x x), ∀ > ∃ (n) n∈ N ∀ ≥ ( ) ⇒ Y − Y ≤ n n if ∈ →+∞ ε 0 { N}ε N n N ε xn x ε. = Definition 1.7. If every Cauchy sequence x converges in X, then X, is called ∀ > ∃ ( ) ∈ ∀ ≥n n (N ) ⇒ Y − Y < a complete space. { } ∈ ( Y ⋅ Y) Definition 1.8. A normed linear space X, which is complete is called a . ( 1 Y ⋅ Y) 2 LECTURE NOTES, FUNCTIONAL ANALYSIS

Lemma 1.9. Let X, be a Banach space and U be a closed linear subspace of X. Then, U, is a Banach space as well. ( Y ⋅ Y) 1.2. Examples( Y ⋅ Y) of Banach spaces Example 1. Let B T be a space consisting of all bounded maps x T K. For each x B T we ( ) x sup x t . ∶ Ð→ ∈ ( ) t T ∞ Then, B T , is a Banach space.Y Y To= prove∈ S ( the)S assertion we need to show that

(a) is a∞ norm, (b)( each( ) CauchyY ⋅ Y ) sequence converges to an element from B T . Y ⋅ Y∞ Concerning claim (a), let λ K and x B T . Then, ( ) (1.10) λx sup λ x t λ sup x t λ x . ∈ t T ∈ ( ) t T ∞ ∞ Let to T and x, y BY TY .= Then,∈ S ⋅ ( )S = S S ⋅ ∈ S ( )S = Y Y

x to y to x to y to sup x t sup y t x y . ∈ ∈ ( ) t T t T ∞ ∞ The right handS ( ) side+ ( of)S the≤ S inequality( )S + S ( above)S ≤ is∈ independentS ( )S + ∈ S on( )St.= Therefore,Y Y + Y Y taking supre- mum of both sides of the inequality over t T yields (1.11) x y x y . ∈ Finally, let x be such that x 0, which is equivalent to sup x t 0. This implies Y + Y∞ ≤ Y Y∞ + Y Y∞ t T that x t 0 for each t. Thus, x 0. Y Y∞ = ∈ S ( )S = Now,S we( )S shall= prove the statement= (b). Let xn be a Cauchy sequence. Then, for every ε 0 there exists N N ε such that for all n, m N it holds that xn xm ε. In particular, { } ∞ > = ( ) xn t xm t ε, ≥ t T. Y − Y <

Thus, for any t T the sequence xn t converges to some x t , due to the completeness S ( ) − n N( )S < ∀ ∈ of K (real and complex numbers are complete spaces). Define a candidate for a limit of the sequence x∈ , that is, x{ T ( )} ∈ as ( ) n n N K

∈ x t lim xn t . { } ∶ Ð→ n If follows from the statement above that( ) = there→+∞ exists( )No No ε, t such that (1.12) x t x t ε, n N . n =o ( ) Without loss of generality we can assume that N ε, t N ε for each t T . Then, for S ( ) − ( )S < o∀ ≥ n N it holds that ( ) ≥ ( ) ∈ x t x t x t x t x t x t ≥ n n No ε,t No ε,t xn xN ε,t( ) ε 2ε, ( ) S ( ) − ( )S ≤ T ( ) − o ( )T + T ( ) − ( )T ( ) ≤ Y − Y∞ + < LECTURE 1. BANACH SPACES 3 where we used the fact that x is a Cauchy sequence and (1.12). Moreover, for each n n N t T and N N ε we have ∈ x t { x} t x t x t x 2ε, ∈ = ( ) N N N which implies x x 2ε and so forth x B T . S N( )S ≤ S ( )S + S ( ) − ( )S ≤ Y Y∞ + Example 2. Let T be a metric space and C T be a space of bounded continuous func- Y Y∞ ≤ Y Y∞ + b ∈ ( ) tions on T . Then, Cb, is a Banach space. ( ) ∞ To prove the assertion,( Y it⋅ Y is) sufficient to show that Cb T is a closed subspace of B T (due to the Lemma 1.9), that is, to show that every sequence in Cb T which converges in B T converges to a point from C T . Let x be a( sequence) of bounded, continuous( ) b n n N functions convergent to x B T . We need to show that x is a continuous( ) . For ∈ any( )ε 0 there exists N N ε ( N), such{ that} xN x ε 3, since the sequence is convergent. Now, let to T∈. By( the) continuity of xN , there exists δ δ ε, to 0 such that ∞ > d t,= t ( )δ ∈ x t Y x −t Y <ε 3~. ∈ o N N o = ( ) > Therefore, for all t such that d t, t δ it holds that ( ) < o Ô⇒ S ( ) − ( )S < ~ x t x t x t x t x t x t x t x t o ( )

n n Co R f Cb R lim f t 0 t with a norm is a Banach( ) space.= › ∈ ( ) ∶ S S→+∞ S ( )S = Example 4. The following spaces Y ⋅ Y∞ co tn n tn , lim tn 0 , N K n ∈ c tn n tn , lim→+∞ tn exists = š{ } N ∶ ∈ K n = Ÿ ∈ with a norm are Banach= spaces.š{ } ∶ ∈ →+∞ Ÿ Remark 1. B is often denoted by l . Y ⋅ Y∞ N ∞ 1.3. lp spaces( ) Definition 1.13. Let 1 p . We define

p ≤ < +∞ p p p l x l xn and x xn . +∞ p ¿+∞ ∞ n 1 Án 1 ÀÁ = œ ∈ ∶ Q= S S < +∞¡ Y Y = Q= S S 1 1 Lemma 1.14. (H¨olderinequality for ). Let 1 p, q be such that p q 1. Then, for x lp and y lq it holds that ≤ ≤ +∞ + = ∈ ∈ 4 LECTURE NOTES, FUNCTIONAL ANALYSIS

(a) x y l1,

(b) x y 1 x p y q. ⋅ ∈ p LemmaY 1.15.⋅ Y ≤(MinkowskiY Y ⋅ Y Y inequality for sequences). Let 1 p and x, y l . Then x y x y . p p p ≤ ≤ +∞ ∈ p Example 5. Spaces l , p areY Banach+ Y ≤ spacesY Y + forY Y 1 p .

Space l , coincides‰ Y ⋅ withY Ž B N , , therefore≤ we≤ assume+∞ that p . Similarly as in the∞ Example 1, to prove the assertion we need to show that ( Y ⋅ Y∞) ( ( ) Y ⋅ Y∞) < +∞ (a) p is a norm, (b) each Cauchy sequence converges to an element from lp. Claim (a)Y ⋅ isY straightforward, when one uses the for the proof of the . For the proof of completness, consider a Cauchy sequence xn . n N n p n n n p Each element x l is a sequence given by x x1 , x2 ,... . Note that l l , what ∈ holds due to the following estimate {∞ } ∈ = ( ) ⊂

p p p p x p xk sup xk sup xk x , ¿+∞ k k Ák 1 ½ N N ÀÁ ∞ for x x , x ,... . Thus,Y Y = if weQ consider= S S ≥ the∈ sequenceS S = ∈xnS S =asY aY sequence of elements of 1 2 n N n l space, we conclude that there exists exactly one x l such that limn x x 0 ∈ (this∞ = follows( from) the completeness of l , ). In{ particular,∞ } →+∞ ∞ n ∞ ∈ Y − Y = (1.16) lim x xk, for∞ each k . n k ( Y ⋅ Y ) N p n We shall show that x x →is+∞ an element of l space and that x converges to x k k N = ∈ n N in lp. For any ε 0 there exists N N ε , such that for all n, m N it holds that = { } ∈ { } ∈ xn xm ε. > = ( ) p ≥ In particular, for every K N Y − Y < K ∈ p p xn xm xn xm ε. ¿ k k p Ák 1 ÀÁ Q= T − T m≤ Y − Y < Using (1.16) and passing to the limit with xk we obtain

K p p xn x ε. ¿ k k Ák 1 ÀÁ Since the estimate is valid for all K andQ= theT right− handT < side of the inequality is independent on K, it holds also that

p n p x xk ε. ¿+∞ k Ák 1 ÀÁ Q= T − T < LECTURE 1. BANACH SPACES 5

n p Therefore x x p ε, which proves that x is a limit of the sequence in l . Moreover, x x xN xN ε xN . Y − Y < p p p p 1.4. Minkowski functionalY Y ≤ Y − Y + Y Y ≤ + Y Y < +∞ Definition 1.17. Set A is called an absorbing set if for each x X there exists t K, such that t x A. ∈ ∈ Definition 1.18. Set A is called a if x A x A. ⋅ ∈ Definition 1.19. Let A be a convex, absorbing and balanced set. A functional µ X ∈ ⇒ − ∈ A 0, defined by x ∶ Ð→ (1.20) µ x inf t 0, A [ +∞) A t is called .( ) = › ∈ ( +∞) ∶ ∈ Lemma 1.21. Minkowski functional generates a seminorm on X. If additionally A is bounded in each direction, that is, for each x X a set A lin x is a bounded set, then it is a norm. ∈ ( ∩ { }) Proof. We shall concentrate on the essential part of the proof, that is, showing that µA fulfills the triangle inequality. Fix ε 0 and let t µA x ε, s µA y ε. Then, t 1x, s 1y A, what follows from the definition of the Minkowski functional. Set A is convex,− − therefore > = ( ) + = ( ) + t x s y x y ∈ A, t s t t s s t s which implies that + ⋅ + ⋅ = ∈ + x+ y + µ x y inf z 0, A t s µ x µ y 2ε. A z A A + Due to a freedom( + in) = the choice› ∈ ( of+∞ε the) ∶ assertion∈ is ≤ proved.+ = ( ) + ( ) +  1.4.1. Examples of normed spaces with a norm introduced by the Minkowski functional Let F F Ω be a space of real valued, Lebesgue measurable functions on Ω. Example 6. Orlicz spaces L Ω . = ( ) M Let M be a non-negative on 0, , such that M (0 ) 0 and lim M t . t [ +∞) ′ Define a set ( ) = →+∞ ( ) = +∞ (1.22) A f F M f x dx 1 . Ω LM Ω is the smallest= › ∈ linear∶ S space(S ( containing)S) ≤ A. It can be checked that Minkowski functional µA defines a norm on LM Ω . ( ) ( ) 6 LECTURE NOTES, FUNCTIONAL ANALYSIS

Example 7. Lebesgue spaces

Lp Ω f F f x pdx . Ω p The most important class of( Orlicz) = › spaces∈ arises∶ S S when( )S we< set+∞M x x , where 1 p . In this case we obtain Lebesgue spaces Lp Ω . Analogously as in the example above, ( ) = < < +∞ p A f F ( )f x dx 1 . Ω It turns out that Minkowski functional= › ∈ µ∶A Sis givenS ( )S by the≤ following formula

p p µA f f x dx. ½ Ω p Note that A is a convex, absorbing and( ) balanced= S set,( )S therefore µA is a seminorm on L Ω . p S Moreover, if µA f 0, then Ω f x dx 0, which implies f 0 a.e. Thus, µA defines a norm on Lp Ω . ( ) ( ) = ∫ S ( )S = = Example 8.( Generalized) Lebesgue spaces Lp Ω f F f x p x dx . Ω (⋅) ( ) p x The next important class of( Orlicz) = › spaces∈ ∶ S is createdS ( )S when< one+∞ sets M x x , where p x fulfills 1 p1 p x p2 for some p1, p2. In this case we obtain generalized( ) Lebesgue spaces Lp Ω . Similarly as before, ( ) = ( ) < ≤(⋅) ( ) ≤ < +∞ p x ( ) A f F f x dx 1 . Ω ( ) = › ∈ ∶ S ( )S ≤ 1.5. Lp spaces. S Definition 1.23. Let 1 p . The space Lp Ω consists of equivalence classes of Lebesgue measurable functions f Ω R such that ≤ < +∞ ( ) f x dx , ∶ →Ω where two measurable functions are equivalent if they are equal a.e. The Lp norm of S S ( )S < +∞ f Lp Ω is defined by

p ∈ ( ) Lp f x dx. ½ Ω For p the definition is slightly different. We say that a function f is essentially Y ⋅ Y = S ( )S bounded, if S = +∞ essup f inf sup f x . N N 0 Ω N

The space L Ω consists of equivalenceS S = ∶ S classesS= ( ƒ ) (twoS ( )S functions< +∞ are equivalent if they are equal a.e.) of∞ measurable, essentially bounded functions f Ω R with a norm ( ) ∞ essup f . L ∶ → Y ⋅ Y = S S LECTURE 1. BANACH SPACES 7

Remark 2. The reason to regard functions that are equal a.e. as equivalent is so that

f Lp 0 implies that f 0 and thus Lp is a norm. For example, the characteristic function of the rational numbers Q is equivalent to 0 in Lp R , for 1 p . Y Y = = Y ⋅ Y 1 1 Lemma 1.24. (H¨olderinequality for ). Let 1 (p ) and≤ ≤p+∞g 1. Let f Lp Ω , g Lq Ω . Then, f g L1 Ω and ≤ ≤ +∞ + = ∈ ( ) ∈ ( ) ⋅ ∈f g(L1) f Lp g Lq . p Theorem 1.25. Orlicz spaces LYM ⋅ΩY, Lebesgue≤ Y Y ⋅ spacesY Y L Ω and generalized Lebesgue spaces Lp Ω are Banach spaces. (⋅) ( ) ( ) We prove the( theorem) only for the Lebesgue spaces. In the proof we shall use the following lemma. Lemma 1.26. For any normed space X, the following conditions are equivalent (a) X, is a complete space. (b) If x is a sequence in X,( suchY ⋅ thatY) x , then there exists x X n n N n 1 n such( Y that⋅ Y) +∞ = { } ∈ N ∑ Y Y < +∞ ∈ lim xn x 0. N n 1 The condition (b) simply states that→+∞ anyY Q absolutely= − Y = convergent is convergent. Proof of Lemma 1.26. N (a) (b). The implication follows from the fact that SN n 1 xn is a Cauchy sequence. (b) (a). Let x be a Cauchy sequence. For each k there exists N , such that n n N N k ⇒ = = k ∑ ⇒ { } ∈ xm xn 2 , n, m Nk∈. − We choose a subsequence x such that nkY k N− Y < ∀ ≥

∈ k { } xnk+1 xnk 2 , k N − and denote y x , y x x for k 1. Therefore y . From 1 n1 k Ynk+1 −nk Y < ∀ ∈ i 1 i assumptions it follows that there exists y X, such that +∞ = = ( − ) > = Y Y < +∞ N ∑ ∈ lim yn y lim xnN+1 y 0. N n 1 N Therefore, the subsequence→+∞xY Q= −convergesY = →+∞ inY X. A− CauchyY = sequence, which has a nk k N convergent subsequence, converges as well, which ends the proof. { } ∈ Proof of Theorem 1.25 for the Lebesgue spaces, p . ◻

Checking that Lp is a norm, when one uses Minkowski inequality, is straightforward. Note that we have proved the assertion in an alternative way< +∞ for 1 p in the Example

7. For the proofY ⋅ ofY completeness we shall use claim (b) from Lemma 1.9. Let fn n N be < < +∞ ∈ { } 8 LECTURE NOTES, FUNCTIONAL ANALYSIS

p a sequence in L Ω such that M n 1 fn Lp . We need to construct a function p N +∞ f L Ω , such that limN fn f p 0. Defineg ˆn, gˆ Ω as following n 1 = L R ( ) n = ∑ Y Y < +∞ →+∞ = ∈ ( ) gˆn x Y fi x − Yand= g ˆ x fn x∶ . → ∑ +∞ i 1 n 1

From Minkowski inequality( we) = obtainQ= S ( )S ( ) = Q= S ( )S n n

gˆn p fi Lp fi p fn p M , L L +∞ L i 1 i 1 n 1

By construction,gY ˆn convergesY = Y Q= monotonicallyS SY ≤ Q= Y toY g ˆ≤.Q Therefore,= Y Y = from< the+∞ monotone conver- gence theorem and the inequality above it follows that

p p p p gˆ x dx lim gˆn x dx lim gˆn x dx M . Ω Ω n n Ω which implies thatg ˆ Lp Ω and in particularg ˆ is finite a.e. From the latter fact we S ( ( )) = S →+∞ ( ( )) = →+∞ S ( ( )) < conclude that ∈ ( ) f x fn x +∞ n 1 p is finite a.e. and f L Ω with f Lp( ) g∶ˆ= LQp=. Note( ) that p p p n ∞ p p 0 f x∈ ( fi) x Y Y ≤fiYxY fi x gˆ x M . +∞ i 1 i n 1 i n 1

Thus, by the≤ LebesgueW ( ) − Q= dominated( )W = W convergence=Q+ ( )W ≤ theoremŒ =Q+ S ( )S‘ ≤ ( ( )) < n p lim f x fi x dx 0, n Ω i 1 →+∞ which ends the proof due to the LemmaS W ( 1.26.) − Q= ( )W = ◻ LECTURE 2 Linear Operators

Definition 2.1. Let E, E and F, F be normed spaces. A space consisting of linear, bounded operators A E F is denoted as α E,F . A is understood here as the( operator,Y ⋅ Y ) which( mapsY ⋅ Y ) bounded sets onto bounded sets. ∶ → ( ) Remark 3. α E,F is a normed space with a norm given by

( ) A x F A α E,F sup . x E,x 0 x E ( ) Y ( )Y Y Y = ∈ ≠ Remark 4. For any A α E,F the following equalitiesY Y hold

(2.2) A∈ α (E,F ) sup A x F sup A x F . x x E 1 x x E 1 ( ) Y Y = ∶ Y Y ≤ Y ( )Y = ∶ Y Y = Y ( )Y Theorem 2.3. Let E, E and F, F be normed spaces and T E F be a linear operator. Then, the following conditions are equivalent: ( Y ⋅ Y ) ( Y ⋅ Y ) ∶ → (a) T is continuous, (b) T is continuous at 0,

(c) there exists M such that for each x E it holds that T x F M x E, (d) T is uniformly continuous. ∈ Y Y ≤ Y Y Proof of Theorem 2.3. c d . Condition c implies that T is Lipschitz continuous and thus, also uniformly continuous. (d) ⇒ (a) b . The( proof) is trivial. b c . We shall prove this implication by contradiction. Assume that T is continuous at( ) 0⇒ and( ) c⇒ does( ) not hold, which implies that there exists a sequence x such that n n N ( ) ⇒ ( ) ∈ ( ) T xn F n xn E. { } 9 Y Y > Y Y 10 LECTURE NOTES, FUNCTIONAL ANALYSIS

xn Define yn . Then, yn E 1 n and limn yn 0 E 0. Moreover, n xn E →+∞ = Y Y = ~ Y − Y = Y Y T xn F T yn F 1. n xn E Y Y Y Y = > Therefore, limn yn 0 E 0 and T yn TY 0Y F 1, which is a contradiction due to the fact that T is continuous. →+∞ Y − Y = Y − ( )Y > Lemma 2.4. In an infinite dimensional Banach space there exist unbounded operators,◻ which are defined everywhere.

Lemma 2.5. Let D E be a dense subset of a normed space E, E and F, F be a Banach space. Then, for each T α D,F there exists a unique Tˆ such that ⊂ ( Y ⋅ Y ) ( Y ⋅ Y ) ˆ ˆ ˆ T E F, T D∈ (T, and) T α E,F T α D,F .

( ) ( ) Proof of Lemma 2.5.∶ Ð→For eachS x= E there existsY Y a sequence= Y Y x D, such that n n N limn xn x E 0. Define an operator ∈ { } ∈ ⊂ →+∞ ˆ Y − Y = T x lim T xn. n ˆ = →+∞ The operator T is well defined, since T xn n N is a Cauchy sequence and F is a Banach ˆ space, thus the limit exists. Note that T is unique∈ (it does not depend on the choice of the { } sequence xn n N). Indeed, if there exist sequences xn n N and xn n N such that ′ ∈ ∈ ∈ { } lim xn lim xk{ x,} { } n k ′ →+∞ = →+∞ = then limn,k xn xk E 0 and from the boundedness of T , ′ →+∞ Y − limY = T xn T xk M lim xn xk 0. n,k F n,k E ′ ′ →+∞ Y − Y ≤ →+∞ Y − Y = Moreover, Tˆ is bounded since ˆ T x lim T xn lim M xn sup M xn C. F n F n E E n N It also holds thatY Y = →+∞ Y Y ≤ →+∞ Y Y ≤ ∈ Y Y <

Tˆ x Tˆ x T x ˆ F F F T α E,F sup sup sup T α D,F . x E, x 0 x E x D,x 0 x D x D,x 0 x D ( ) Y ( )Y Y ( )Y Y ( )Y ( ) Y Y = ∈ ≠ = ∈ ≠ = ∈ ≠ = Y Y Y Y Y Y Y Y ◻ LECTURE 2. LINEAR OPERATORS 11

Theorem 2.6. Let F, F be a Banach space. Then, α E,F , α E,F is a Banach space as well. ( Y ⋅ Y ) ( ( ) Y ⋅ Y ( ))

Proof of Theorem 2.6. We need to prove that a normed space α E,F , α E,F is complete. Let An n be a Cauchy sequence in α E,F , that is, N ( ( ) Y ⋅ Y ( )) ∈ { } ( ) Anx Amx F (2.7) ε 0 N N ε n, m N An Am α E,F sup ε, x E, x 0 x E ( ) Y − Y which implies∀ > that∃ A= x( ) is∀ a Cauchy≥ sequenceY − inY F for= each∈ fixed≠ x E as well.< Since n n N Y Y F is a Banach space, this sequence is convergent. Thus, an operator A E F given by { } ∈ ∈ Ax lim Anx n ∶ → is well defined. It is also linear and bounded.= →+∞ The latter claim holds due to the following Ax A x A x F lim n F sup n F . n x E x E n N x E Y Y Y Y Y Y = →+∞ ≤ ∈ < +∞ Now, we want to proveY thatY limn YAnY A α E,F Y Y0. Let x be such that x E 1.

Condition (2.7) implies then →+∞ Y − Y ( ) = Y Y = ε 0 N N ε n, m N Anx Amx F ε.

If we let m , then∀ > An∃x Ax= F( ) ε.∀This holds> forY each−x suchY that< x E 1, therefore

→ +∞ Y − Ysup< Anx Ax F ε, Y Y = x x E 1 Y − Y < which ends the proof due to the equality∶ Y Y = (2.2).

Lemma 2.8. Let X,Y,Z be normed spaces and S α X,Y , T α Y,Z . Then, ◻

T S α X,Z T α Y,Z ∈ (S α X,Y) . ∈ ( )

( ) ( ) ( ) Proof of Lemma 2.8. LetY x○ YX and ≤xY 0.Y Then,⋅ Y Y

T S x Z T Sx ∈Z T α Y,Z≠ Sx Y T α Y,Z S α X,Y x X .

Thus, Y( ○ )( )Y = Y ( )Y ≤ Y Y ( ) ⋅ Y Y ≤ Y Y ( ) ⋅ Y Y ( ) ⋅ Y Y

T S x Z T α Y,Z S α X,Y . x X Y( ○ )( )Y ( ) ( ) Taking supremum over x X ends the proof.≤ Y Y ⋅ Y Y Y Y ∈ ◻ 12 LECTURE NOTES, FUNCTIONAL ANALYSIS

Example 9. Examples of linear operators T X Y : (a) Identity : X Y , T Id. The norm T of T is equal to 1. ∶ α→X,Y (b) A between finite dimensional spaces : X m, Y n and A n m = = Y Y ( ) R R R R is a matrix A aij , 1 i n, 1 j m, = = ∈ × = { } ≤ ≤ T x≤ ≤A x. 1 (c) Differentiation : X C 0, 1 ,(Y) =C ⋅ 0, 1 , = ([ ])T f= ([f . ]) ′ (d) Let g Lq 0, 1 , µ and p, q be such that 1 1 1. X Lp 0, 1 , µ , Y ( ) = p q L1 0, 1 , µ , ∈ ([ ] ) 1 + = = ([ ] ) = Tg f fg dµ. ([ ] ) 0 (e) Fredholm operator : X Y C 0, 1 , ( ) = S 1 T f y = = k ([x, y f]) x dx, 0 where k C 0, 1 (0, 1( ))(. ) = S ( ) ( ) (f) An operator defined analogously as in the example (e), but with X Y L2 0, 1 , µ and k L∈2 ([0, 1 ] ×0[, 1 ,]) µ . = = ([ ] ) Remark 5. (to∈ example([ ] (e))× [ Note] that) continuity of T f follows directly from the uniform continuity of k. Indeed, from a definition of uniform continuity ( ) ε 0 δ 0 such that y y δ sup k x, y k x, y ε . ′ x 0,1 ′ Therefore,∀ it> also∃ holds> that ŒS − S < ⇒ ∈[ ] S ( ) − ( )S < ‘ 1 T f y T f y k x, y k x, y f x dx ε f . 0 ′ ′ ∞ Note that S( ( ))( ) − ( ( ))( )S ≤ S S ( ) − ( )S ⋅ S ( )S ≤ Y Y 1 T f sup k x, y f x dx f sup k , y f k . y 0,1 0 y 0,1 ∞ ∞ ∞ ∞ ∞ DefinitionY 2.9.Y =An∈[ operator] VS ( T ) α( X,Y) V ≤isY calledY ⋅ ∈[ an] Y ,(⋅ )Y ≤ Y Y if thereY Y exists an inverse operator T 1 α Y,X . Moreover, if ∈ ( ) − T T 1 1, ∈ ( ) α X,Y α X,Y − then T is called an andY XY ,( Y are) = Y saidY to( be) isometrically= isomorphic. Two isometrically isomorphic normed spaces share the same structure, so they are usually identified with each other. LECTURE 2. LINEAR OPERATORS 13

Theorem 2.10. Let X, X be a normed space and T α X,X α X . Then, the following implication holds ( Y ⋅ Y ) ∈ ( ) =∶ ( ) T n converges in α X Id T 1 exists and Id T 1 T n, +∞ +∞ n 0 − − n 0 n where T Q= T T . ( ) ⇒ ( − ) ( − ) = Q= n times = ○ ⋅ ⋅ ⋅ ○ m n Proof of Theorem´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ 2.10. Define Sm n 0 T . Then, Id T Sm Sm Id T and m m 1 = n n m 1 Id T Sm Id Sm T= ∑Sm T ( T − Id) T= .( − ) + n 0 n 1 + n n Since n 0 T converges( − ) in α= X○, we− have○ that= limQ= n − TQ= x 0= for− each x X. Therefore, passing+∞ to the limit in the equalities above yields = →+∞ ∑ ( ) = ∈ m 1 n Id x lim Id T x lim Id T Smx Id T lim Smx Id T T x. m m m +∞ + n 0 →+∞ →+∞ n →+∞ Similar( ) = argument( − proves) that= Id x ( − n )0 T =Id( T− x.) Therefore,= ( − )(Q= ) +∞ Id T 1x = T nx, x X. ( ) = (∑ +∞ )( − ) − n 0 ( − ) = Q= ∀ ∈

Remark 6. If X, X is a Banach space it is sufficient to assume that T α X 1. In◻ this case ( Y ⋅ Y) 1 Y Y ( ) < (2.11) Id T 1 . α X 1 T − α X Indeed, note that Y( − ) Y ( ) ≤ − Y Y ( ) T n T n , +∞ α X +∞ α X n 0 n 0 ( ) ( ) n since T α X 1. Since X isQ= aY BanachY space,≤ Q= Y thenY the< convergence+∞ of the series n 0 T follows from Lemma 1.26. The inequality (2.11) follows from a formula on a sum+∞ of a ( ) = geometricY Y series.< ∑ Exercise 1. Find a solution x C 0, 1 to the following equation 1 x ∈s ([ ])k s, t x t dt y s , 0 where k C 0, 1 0, 1 and y C 0, 1 are given. ( ) − S ( ) ( ) = ( ) ∈ ([ ] × [ ]) ∈ ([ ])

LECTURE 3 Dual Spaces

Definition 3.1. Let X, X be a normed space. Space α X, K consisting of linear, bounded functionals on X is called a to X. It is denoted by X or X . ( Y ⋅ Y ) ( ) ′ ∗ Remark 7. Dual space X is a Banach space even if X is not a Banach space.

′ 1 1 Theorem 3.2. Let 1 p, q be such that p q 1. Then, an operator

≤ < +∞q p + = T l l , T x y xnyn, +∞ ′ n 1 q p where x x1, x2,... l , y ∶ y1,→ y2(,...) l( , is)( an) isometric= Q= isomorphism.

Proof of= ( Theorem) 3.2.∈ Linearity= ( of )T∈is straightforward. T is bounded, which follows from H¨olderinequality

(3.3) T x y x q y p T x lp ′ x q.

To show that T is anS( isomorphism)( )S ≤ Y weY ⋅ Y needY to⇒ proveY Y that( ) ≤ itY isY injective and surjective. Indeed,

T x 0 xn T x en 0, n N, where e 0,..., 0, 1, 0,... , that is, the n-th coordinate is the only non-zero element of n = ⇒ = ( )( ) = ∀ ∈ en. Thus, ker T 0 , which proves that T is an injection. For the proof of surjectivity we have to= ( show that ) ( ) = { } y lp x lq such that T x y , i.e., T x y y y y lp. ′ ′ ′ ′ It is sufficient∀ ∈ to( ) prove∃ ∈ that the equality( holds) = for y e(n,)(n ) =N.( Indeed,) ∀ ∈ assume that T x en y en for each n N. Since T x and y are linear, then the equality holds for all y lin′ en n N . However, the operators are′ = also continuous∈ and therefore the equality( )( ) holds= ( for) ∈ ( ) ∈ { { } ∶ ∈ } p p lin en n N l SS⋅SS 15 { { } ∶ ∈ } = 16 LECTURE NOTES, FUNCTIONAL ANALYSIS

as well. Now, fix y lp and define s y e , x s . We prove that x lq. To n n n n N ′ this end, let tn n N′ be a sequence given by ′ ∈ ∈ ( ) s q= ( ) = { } ∈ ∈ n s for sn 0, { } tn n S S 0 for sn 0. ≠ Therefore, for each N it holds= thatœ N = N N N p p q 1 q ∈ tn sn sn n 1 n 1 ( − ) n 1

and Q= S S = Q= S S = Q= S S N N N N q sn tnsn tny en y tnen n 1 n 1 n 1 ′ ′ n 1 1 1 Q= S S = Q= = Q=N ( p ) = ŒQ= N‘ p p q y lp ′ tn y lp ′ sn . ′ n 1 ′ n 1 ( ) ( ) Therefore, ≤ Y Y ⋅ ŒQ= S S ‘ = Y Y ⋅ ŒQ= S S ‘ 1 N q q q sn y lp ′ x q y lp ′ x l , n 1 ′ ′ since the first inequality holds for all( )N . It is straightforward( ) now, that ŒQ= S S ‘ ≤ Y Y ⇒N Y Y ≤ Y Y ⇒ ∈ T x e s y e . ∈n n n ′ The fact that x y ′ together with (3.3) imply that T is an isometry. q lp ( )( ) = = ( ) ′ ( ) 1 1 Theorem 3.4.Y LetY ≤1Y Yp and q be such that p q 1 and Ω, Σ, µ be a measure◻ space, where µ is a σ-finite measure. Then, ≤ < +∞ + = ( ) T Lq Ω, µ Lp Ω, µ , T g f fg dµ ′ Ω is an isometric isomorphism. ∶ ( ) → ( ( )) ( )( ) = S Proof of Theorem 3.4. T is clearly a linear operator. It follows from H¨olderinequality that

T g f fg L1 f Lp g Lq T g Lp ′ g Lq . Thus, the norm of the operator T is at most 1. In particular, the function f defined as S ( )( )S ≤ Y Y ≤ Y Y ⋅ Y Y ⇒ Y Y( ) ≤ Y Y below q g¯ g p f , for 1 p , g g Lq S S (whereg ¯ is a complex conjugate= ofŒg) is such‘ that f< p < +∞1 and S S Y Y L T g f fg dµ Y Yg L=q , g¯ which implies that T g ′ g and thus T is an isometry. For p 1 we set f Lp ( L)(q ) = S = Y Y g and use similar arguments. Y Y( ) ≥ Y Y = = S S LECTURE 3. DUAL SPACES 17

Injectivity of T is straightforward. Indeed, assume that T g 0, which implies that

fg dµ 0, f Lp Ω, µ( .) = Ω q Thus, g 0 µ a.e. and g 0S in L Ω=, µ .∀ Now,∈ we( need) to show that the operator is surjective, that is, = − = ( ) y Lp g Lq such that T g y , i.e., T g f fg dµ y f , f Lp. Ω ′ ′ ′ ′ Assume∀ ∈ ( that) µ∃ Ω∈ and define( ) = ( )( ) = S = ( ) ∀ ∈ ν Σ , ν E y χ . ( ) < +∞ K E Since µ is finite, χ Lp Ω, µ and thus ν is well defined.′ It is clear that ν is additive and E ∶ → ( ) = ( ) in the case p it is also σ-additive (why?). Thus, ν is a signed (or complex) measure. From the construction∈ it( follows) that ν is absolutely continuous with respect to µ. Indeed, < +∞ p µ E 0 χE 0 µ a.e. χE L Ω, µ . Since χ 0 in Lp Ω, µ , it holds that E ( ) = ⇒ = − ⇒ ∈ ( ) ν E y χ y 0 0. = ( ) E By Radon-Nikodym theorem, there exists′ g L1 Ω′ , µ such that ( ) = ( ) = ( ) =

y χE ν E g dµ∈ ( χE)g dµ, E Σ. E Ω ′ p Characteristic functions( are) dense= ( ) in= SL , L=∞S (and in L ,∀ ∈Lp ), which implies that ∞ (3.5) y f ( fg Ydµ⋅ Y )f L Ω(, µ . Y ⋅ Y ) Ω ′ ∞ q Now, it remains to prove that(g) =LS Ω, µ . Cases∀ ∈ for q( ) and q are considered separately. Assume that q and define a µ- ∈ ( )q < +∞ = +∞ g x if g x 0, < +∞f x g x 0S ( )S if g x 0. ( ) ( ) ≠ q (p ) = œ Then, g x fg x f x . Define a µ-measurable( ) = set E x g x n . S ( )S = ( )( ) = S ( )S n Then, χ f L Ω, µ and thus we can plug it into (3.5). En = { ∶ S ( )S ≤ } ∞ ∈ χE(n f g )dµ y χEn f y Lp ′ χEn f Lp Ω ′ ′ 1 1 ( ) S ( ) = ( ) ≤ Y Y p ⋅ Yp Y q p y Lp ′ f dµ y Lp ′ g dµ , E E ′ n ′ n ( ) ( ) On the other hand = Y Y ‹S S S  = Y Y ‹S S S  q χEn f g dµ g dµ. Ω En S ( ) = S S S 18 LECTURE NOTES, FUNCTIONAL ANALYSIS

Combining the both expressions we obtain

1 q q g dµ y Lp ′ . E n ′ ( ) The inequality above holds also‹ whenS S theS integral ≤ Y isY taken over the whole Ω. Indeed, sup g q dµ lim g q dµ g q dµ, n n N En N En Ω due to the monotone convergence∈ S S S theorem.= → S Thus,S S = S S S

1 q q g dµ g Lq y Lp ′ . Ω ′ ( ) Now, consider the case q ‹S. DefineS S  = Y Y ≤ Y Y g x E x = +∞g x y ′ and f x χ x . L1 E g x ′ ( ) S ( )S It is clear that f L= ™Ω,∶ µ (and) > Yf Y µž E . If µ E( ) =0, then( ) L1 ( ) ∞ µ E∈ y( L1 )′ Yg Ydµ = ( fg) dµ (y )f> y L1 ′ f L1 E Ω ′ ′ ′ ( ) ( ) and µ E f L(1 ,)Y whichY is a< contradiction.S S S = S = ( ) ≤ Y Y ⋅ Y Y

Definition( ) < Y 3.6.Y Let Σ be a σ-algebra of subsets of Ω and K N, C . ν Σ K is called◻ a signed (or complex) finite measure, if (a) ν 0, = { } ∶ →

(b) ν i 1 Ai i 1 ν Ai , if Ai Aj for i j. (∅)+∞= +∞ = = Theorem(⋃ 3.7. Let) =Σ∑be a(σ-algebra) ∩ of subsets= ∅ of ≠Ω, µ be a finite measure and ν be a signed (or complex) measure on Σ. Moreover, let ν be absolutely continuous with respect to µ, that is, µ E 0 ν E 0. Then, there exists g L1 Ω, µ such that ( ) = ⇒ ( ) = ∈ ( ) ν E g dµ, E Σ. E ( ) = S ∀ ∈ 3.1. Extensions of functionals Definition 3.8. Let X be a vector space. Functional p X R is called sublinear, if (a) p λx λp x , λ 0, x X, (b) p x y p x p y , x, y X. ∶ → ( ) = ( ) ∀ ≥ ∈ Remark( 8.+Any) ≤ seminorm( ) + ( is) a sublinear∀ ∈ functional. LECTURE 3. DUAL SPACES 19

Theorem 3.9. (Hahn-Banach) Let X be a vector space and U be its subspace. If p X R is a sublinear functional, l U R is a linear functional such that l x p x , x U, ∶ → then there exists an extension L X , such that ∶ → R ( ) ≤ ( ) ∀ ∈ L l and L x p x , x X. U ∶ → In the proof we shall use theS following= lemma.( ) ≤ ( ) ∀ ∈ Lemma 3.10. (Zorn lemma) Let A, be a non-empty partially ordered set, such that every non-empty totally ordered subset has an upper bound in A. Then, the set A contains at least one maximal element. ( ≤) A totally ordered set is the set with partial order under which every pair of elements is comparable. Proof of Theorem 3.9. Assume that dim X U 1. With this assumption our task is to extend the functional l to X U Rxo, where xo X U. Note, that in this case any x X can be written as x u λxo, where u ( U~and) =λ R. Define = ⊕ ∈ ƒ (3.11) L x l u λr. ∈ = + r ∈ ∈ We claim that L is a linear extension of l, such that L x p x . If λ 0, we have r ( ) = ( ) + r that Lr x l u λr l u p u . Hence, we can assume that λ 0. Let λ 0. Then, inequality Lr x p x is equivalent to the following ( ) ≤ ( ) = ( ) = ( ) + = ( ) ≤ ( ) ≠ > λr p u λx l u ( ) ≤ ( ) o u u r p x l ≤ ( λ+ o ) − ( λ) r inf p v xo l v . ≤ v ‹U +  − ‹  Similarly, for λ 0 we obtain ≤ ∈ ( ( + ) − ( )) λr p u λx l u < o u u r p x l ≤ ( +λ )o − ( ) λ −r ≤ sup‹ l w− p−w‹ xo . w U− − Therefore, there exists r R such that≥ ∈Lr( (p,) if− ( − )) l w p w x p v x l v , ∈ o≤ o which is equivalent to ( ) − ( − ) ≤ ( + ) − ( ) l v l w p v xo p w xo , v, w U. But l v l w l v w p v w p v x p w x , since v w U and p is sublinear. ( ) + ( ) ≤ ( + )o+ ( − o) ∀ ∈ ( )+ ( ) = ( + ) ≤ ( + ) ≤ ( + )+ ( − ) + ∈ 20 LECTURE NOTES, FUNCTIONAL ANALYSIS

Now, consider the collection A V,L V X is a linear subspace,L V R is a linear extension of l such that L x p x , x V . = {( ) ∶ ⊂ ∶ → We define a partial order on A by setting ( ) ≤ ( ) ∀ ∈ }

V1,L1 V2,L2 V1 V2 and V2 Z1 V1. Now, we check that every chain (i.e., totally ordered set) ( ) ≤ ( ) ⇔ ⊂ S = Vi,Li i I ¯ ¯ ¯ in A has an upper bound. Let V i I Vi. Thus, V is a subset of X and V contains each ¯ ¯ {( ) ∶ ∈ } ¯ Vi. Let define L V R in the following way. If x V , then there exists i I such that ¯ ∈ x Vi and we set L x Vi x . The= ⋃ definition does not depend on the choice of i because ¯ ¯ ¯ of the chain definition.∶ → Moreover, it implies that L is∈ linear and Vi,Li V,∈L for every i ∈I. In other words( ) this= is( an) upper bound for the chain. Zorn’s lemma guarantees the existence of a maximal element V,L A. If V X, then it follows( from) ≤ ( the first) part of the∈ proof, that we can extend the functional and thus, it is a contradiction due to the fact that V,L is maximal. ( ) ∈ ≠

Theorem( ) 3.12. Let X be a C-vector space. Then, ◻ (a) If l X R is a R-linear functional, that is, l λ x λ x λ l x λ x , λ , λ , x X, ∶ → 1 1 2 2 1 1 2 2 1 2 R then, ( + ) = ( ) + ( ) ∀ ∈ ∈ h x l x il ix is a C-linear functional such that Re h l. ( ) = ( ) − ( ) (b) If h X C is a C-linear functional, then l Re h is a R-linear functional. = (c) If p is a seminorm and h X C is a C-linear functional, then for all x X the following∶ → equivalence holds = ∶ → ∈ h x p x Re h x p x . (d) If X, is a normed space and h X is a -linear continuous functional, X S ( )S ≤ ( ) ⇔ S (C)S ≤ C( ) then ( Y ⋅ Y ) h Re∶ h→ . α X,C α X,R Theorem 3.13. (Hahn-BanachY theoremY ( ) for= Y C) LetY ( X) be a C-vector space and U be its subspace. If p X R is a sublinear functional, l U C is a linear functional such that Re l x p x , x U, ∶ → then there exists an extension L X , such that ∶ → C ( ) ≤ ( ) ∀ ∈ L l and Re L x p x , x X. U ∶ → S = ( ) ≤ ( ) ∀ ∈ LECTURE 3. DUAL SPACES 21

Theorem 3.14. Let X, X be a normed space and U be its subspace. Then, for every continuous linear functional u U K there exists a continuous linear funtional x X K, such that ( Y ⋅ Y ′) ′ x ∶u →and x u . ∶ → U α X,K α U,K ′ ′ ′ ′ Proof of Theorem 3.14. S (only= for K YR)Y Define( ) =pY XY ( R)as following p x u x . = α U,K X∶ → ′ It is clear that p is a sublinear functional and(u )x u x , for all x U. Thus, ( ) = Y Y ⋅ Y Y α U,K X by Hahn-Banach theorem there exists x X R′ , such that′ ( ) ≤ Y Y ( ) ⋅ Y Y ∈ x u and′ x x p x , x X. U ∶ → Note that x x p x ′p x ,′ which implies′ that S = ( ) ≤ ( ) ∀ ∈ ′ x x u x , x X, (− ) ≤ (− ) = ( ) α U,K X ′ ′ which implies that x is continuous and x( ) u . For the proof of the inverse S ( )S ≤ Y Y α X,⋅ YK Y ∀α U,∈K inequality notice that′ ′ ′ Y Y ( ) ≤ Y Y ( ) u sup u u sup x u sup x x x , α U,K α X,K ′ u Bu ′ u Bu ′ x Bx ′ ′ ( ) ( ) where Y Y = ∈ S ( )S = ∈ S ( )S ≤ ∈ S ( )S = Y Y

Bu u U u X 1 and Bx x X x X 1 .

= { ∈ ∶ Y Y ≤ } = { ∈ ∶ Y Y ≤ }

Remark 9. Let X be a normed space. Then, for every xo X, xo 0 there exists x X◻ such that ′ ′ x 1 and x x x ∈ . ≠ ∈ α X,K o o X ′ ′ Indeed, let M Kxo. Define x M R by Y Y ( ) = ( ) = Y Y ′ x λx λ x , λ . = ∶ →o o X K One can easily check that ′ ( ) = Y Y ∀ ∈ x 1 and x x x . α M,R o o X The existence of the extension′ x X follows′ from the Hahn-Banach theorem. Y Y ( ) = K ( ) = Y Y ′ Remark 10. Let X be a normed∶ space.→ Then, for every x1, x2 X such that x1 x2 there exists x X such that x x1 x x2 . This is the immediate conequence of the previous remark.′ ′ ′ ′ ∈ ≠ ∈ ( ) ≠ ( ) Remark 11. For any normed space X, X it holds that

x X sup x x , x X. ( ′ Y ⋅ Y ) x Bx′ ′

Indeed, since x Bx′ x XY Y x= ∈1 Swe( have)S ∀ that∈ ′ ′ ′ ′ x x x x x , x B ′ . ∈ = { ∈ ∶ Y Y ≤ } X X x ′ ′ ′ S ( )S ≤ Y Y ⋅ Y Y ≤ Y Y ∀ ∈ 22 LECTURE NOTES, FUNCTIONAL ANALYSIS

On the other hand, it follows from the previous remark that there exists x such that x 1 (and thus x Bx′ ) and ′ ′ ′ x X x x x x sup x x . Y Y = ∈ ′ ′ ′ x Bx′ ′ Y Y = ( ) ≤ S ( )S ≤ ∈ S ( )S Lemma 3.15. Let X, X be a normed space and V X be its convex and open subset such that 0 V . Then, there exists x X such that Re x x 0, for each x V . ( Y ⋅ Y ) ′ ′ ⊂ ′ Proof of Lemma∉ 3.15. (only for K∈ R) Define sets A (B) and< A B as following∈ A B a b a A, b B . = + − Moreover, for x V define o ± = { ± ∶ ∈ ∈ } y x and U V x . ∈ o o o Set U is convex, open and y U, 0 U. For the set U define a Minkowski functional p . o = − = − { } U Note that pU is sublinear and pU yo 1. Now, consider a linear functional ∉ ∈ y lin y , y t y t p y . ( o ) ≥ R o U o It holds that y y p y ,′ for each y Y . Indeed,′ U ∶ { } → ( ⋅ ) = ⋅ ( ) ′ y t y 0 p t y , for t 0, ( ) ≤ ( ) o ∈ U o ′ y t yo pU t yo , for t 0. ( ⋅ ) ≤ ≤ ( ⋅ ) ≤ Let x be an extension of y such′ that x p . There exists ε 0 such that B 0, ε U ( ⋅ ) = ( U⋅ ) > such that′ ′ ′ 1 x x max x x , x x ≤ max p x , p x> x . ( ) ⊂ U U ε X ′ ′ ′ It is straightforwardS ( )S that= x{yo( ) pU(−xo )} ≤1 and{ for( each) x(− u)} ≤yo Y YV , where u U it holds that ′ x( x) =x (u ) x≥ yo pU u 1 =0(. − ) ∈ ∈ ′ ′ ′ ( ) = ( ) − ( ) ≤ ( ) − <

Theorem 3.16. (Hahn-Banach theorem) Let X be a normed space, V1,V2 X be convex◻ and V1 be open. If V1 V2 , then there exists x X such that ⊂ Re x v Re x v ′v V′ , v V . ∩ = ∅ 1 2 ∈1 1 2 2 ′ ′ Proof of Theorem 3.16. Let( define) < V ( V1) V∀2 ∈V1 ∈V2 , which is in fact given by V V x . = 1− = ( + (− )) x V2

Note that V is open and since V1 V2 = ∈ it( holds− { that}) 0 V . Therefore, by Lemma (3.15) there exists a functional x such that ′ ∩ = ∅ ∉ Re x v1 v2 0 for all v V and v V and thus, Re x ′ v Re x v . 1 1 2 2 ( 1 − ) < 2 ′ ′ ∈ ∈ ( ) < ( ) ◻ LECTURE 3. DUAL SPACES 23

Theorem 3.17. (Hahn-Banach theorem) Let X be a normed space, V X be a closed and and x V . Then, there exists a functional x X such that Re x x inf Re x v v V ′ , ⊂ ∉ ∈ that is, there exists ε 0 such that′ ′ ( ) < { ( ) ∶ ∈ } Re x x Re x x ε Re x v . > ′ ′ ′ Proof of Theorem 3.17. Since( ) V< is closed,( ) − there≤ exists( ) an open ball with radius r, such that B x, r V . From Theorem 3.16 if follows that ( ) ∩ ≤ ∅ Re x x u Re x v , u B 0, r , v V. Therefore, ′ ′ ( Re+ x)

LECTURE 4 Weak convergence and reflexivity

Let X, X be a normed space. X is defined as a space of linear functionals ′′ ( Y ⋅ Y ) i x X K ′ given by ( ) ∶ → (4.1) i x x x x . ′ ′ Remark 12. Operator i is a linear‰ isometry( )Ž( ) (in= general( ) not surjective). Indeed,

i x x x x x x X i x x X . ′ ′ ′ According to RemarkT‰ ( ) 9,Ž( for)T = everyS ( )Sx ≤ YX,Yx⋅ Y Y0, there⇒ existsY ( )Y a≤ functionalY Y x such that x x x X and x 1. Thus, ′ ′ ′ ∈ ≠ x x x i x x i x . ( ) = Y Y Y Y = X ′ ′ Definition 4.2. A BanachY spaceY =XS is( called)S = S ( reflexive)( )S ≤ ifY i( defined)Y as in (4.1) is surjective. We write then X X . ′′ Example 10. ≅ (a) Finite dimensional spaces are reflexive, (b) Banach spaces lp and Lp for 1 p are reflexive, (c) Banach spaces L1, L , C K , are not reflexive. ∞ < < +∞ ∞ Definition 4.3. Let xn n N be( a sequence( ) Y ⋅ Y in) a space X. We say that xn n N converges weakly to some x and denote it as xn x, if ∈ ∈ { } { } lim x xn x x in , x X . n ⇀ K ′ ′ ′ →+∞ Definition 4.4. Let yn n N be( a sequence) = ( ) in a space∀ Y∈ X . We say that yn n N ′ converges weakly to some x∈ and denote it as xn x, if ∈ ∗ { } ∗ ≅ { } lim yn x y x in , x X. n ⇀K →+∞ ( ) = ( 25) ∀ ∈ 26 LECTURE NOTES, FUNCTIONAL ANALYSIS

Theorem 4.5. (Sequential version of Banach-Alaoglu theorem) Let X, X be a sepa- rable normed space and Y X . If yn n N is a sequence such that supn N yn Y 1, then there exists a subsequence yn ′ k and an element y Y such that ( Y ⋅ Y ) k N ∈ ≅ { } ∈ Y Y ≤ ∈ y y. { } nk ∈ ∗ Proof of Theorem 4.5. If X is separable,⇀ then there exists a set L xj l N X which is linearly independent and linearly dense in X. Since yn x1 is uniformly bounded 1 ∈ with respect to n, then there exists a subsequence n such that y 1 x1 converges= { } to some⊂ k nk 1 2 element y Y . Similarly, y 1 x2 is bounded and thus, thereS ( exists)S a subsequence n nk k 2 ( ) such that yn2 x2 converges to some y Y . Iterating this procedure we obtain sequences ∈k S ( )S ynm ynm−1 yn1 yn. ( ) k ∈k k By a diagonal argument (setting yn ynk ) we chose a subsequence yn k such that k⊂ k ⊂ ⋅ ⋅ ⋅ ⊂ ⊂ k N l lim yn xl y , l . ∈ k = k N { } To complete the proof we need to show that y converges for all x X. To this end, fix →+∞ ( ) = nk ∀ ∈ ε 0. Since L is linearly dense, there exist N N ε and a sequence λli , xli i N, λli K, xl L, such that ∈ i ∈ > N ε = ( ) {( )} ∈ x λ x ε. ∈ ( ) li li X i 1 XX X X Define a linear functional y lin xlX x−l Q=L KXas< X X XN ε NXε ∶ {y ∶ λ∈ x} → λ yli . ( ) li li ( ) li i 1 i 1

Since the assumptions of Hahn-Banach( Q= theorem) = Q= are fulfilled, we can extend y on X pre- serving y Y 1. Note that for each ε 0 it holds that N ε N ε Y Y lim≤ y x y x 2ε >lim y λ x y λ x 2ε. nk nk ( ) li li ( ) li li k k i 1 i 1 R R →+∞ →+∞ R R S ( ) − ( )S ≤ + R ( Q= ) − ( Q= )R = R R Remark 13. If X is reflexive, then weak and weak convergence are equivalent. In such◻ a case, Theorem 4.5 can be formulated in terms of the∗ weak convergence as well. Remark 14. For a general predual space X there is a topological version of Theorem 4.5, which states that the unit ball in X is compact in the . ′ ∗ Definition 4.6. (a) A subset M of a metric space X is nowhere dense, if int M¯ ,

(b) A set M is of the 1-st category, if M n N Mn, where Mn are nowhere dense sets. ( ) = ∅ Theorem 4.7. (Baire category) A non-empty, complete∈ metric space is not of the 1-st = ⋃ category. LECTURE 4. WEAK CONVERGENCE AND REFLEXIVITY 27

Theorem 4.8. Let X be a Banach space and Y be a normed space. Let Ti α X,Y , where i I. If for each x X

sup Tix Y , ∈ ( ) ∈ ∈ i I then ∈ Y Y < +∞ sup Ti α X,Y . i I ( ) Proof of Theorem 4.8. Let n N∈ andY Y define < +∞ E x X sup T x n x X T x n . n ∈ i Y i i I i I

From the assumptions= › it∈ follows∶ ∈ thatY setsY ≤En are= ∈ closed{ ∈ and∶ YX Y ≤n N}En. Therefore, by Theorem 4.7 at least one of the sets En has a non-empty . Let no be such that ∈ N intEno . Then, for each xo intEno there exists ε 0 such= that⋃ B xo, ε Eno . Let x X be such that x X 1. Then, ∈ ≠ ∅ ∈ εx x B x , ε E >. ( ) ⊂ ∈ Y Y ≤ o o no If so, then + ∈ ( ) ⊂ Ti εx xo Y no, i I.

Therefore, for any x X such that x X 1 it holds that Y ( + )Y ≤ ∀ ∈ 1 1 1 T x T εx T εx x T x n T x . i Y ∈ε i Y Yε Y i ≤ o i o Y ε o i o Y Since the rightY handY = sideY ( of the)Y inequality= Y ( + above) − doesY not≤ depend( + Y on theY ) choice of x, it also holds that 1 T n T x . i α X,Y ε o i o Y Taking supremum over i I yields Y Y ( ) ≤ ( + Y Y ) 1 sup∈ Ti α X,Y no sup Tixo Y . i I ε i I ( ) ∈ Y Y ≤ ‹ + ∈ Y Y  < +∞ ◻

LECTURE 5 Open mapping and closed graph theorems

Definition 5.1. We say that T X Y is open, if for any U X a set T U Y is open. ∶ → ⊂ ( ) ⊂ Lemma 5.2. Let X, X , Y, Y be normed spaces and T X Y be a linear operator. Then, the following conditions are equivalent. (a) T is open, ( Y ⋅ Y ) ( Y ⋅ Y ) ∶ → (b) r 0 ε 0, such that BY 0, ε T BX 0, r , (c) ε 0, such that BY 0, ε T BX 0, 1 . ∀ > ∃ > ( ) ⊂ ( ( )) Proof of∃ Lemma> 5.2. i ( ii) ⊂and( ii ( iii)) are trivial. We shall prove ii i . Let U X be an open set and x U. It that ) ⇒ ) ) ⇔ ) ) ⇒ ) r 0, such that x B 0, r U ⊂ ∈ X Moreover, T x T U and due to the linearity of T ∃ > + ( ) ⊂ T x T B 0, r T x B 0, 1 T U . ∈ ( ) X X From the assumption we have that there exists ε 0 such that B 0, ε T B 0, r . + ( ( )) = ( + ( )) ⊂ ( ) Y X Therefore it holds that T x BY 0, ε T U and thus T U is open. > ( ) ⊂ ( ( )) Theorem 5.3. Let X, +X , (Y, ) ⊂Y be( Banach) spaces( and) T α X,Y be a surjective◻ operator. Then, T is open. ( Y ⋅ Y ) ( Y ⋅ Y ) ∈ ( ) To prove the theorem we need the following lemma.

Lemma 5.4. Let X, X , Y, Y be Banach spaces and T α X,Y be a surjective operator. Then, there exists ε 0 such that ( Y ⋅ Y ) ( Y ⋅ Y ) ∈ ( ) B 0, ε T B 0, 1 . > Y X

Proof of Lemma 5.4. Since T is( surjective) ⊂ ( it( holds)) that Y n N T BX 0, n . From linearity of T it follows that ∈ Y n T BX 0, 1 . = ⋃ ( ( )) n N

= ∈ ⋅ 29( ( )) 30 LECTURE NOTES, FUNCTIONAL ANALYSIS

Theorefore, int T BX 0, 1 due to Theorem 4.7 (Baire theorem). Thus, there exists yo Y and ε 0 such that Š ( ( )) ≠ ∅ BY yo, ε T BX 0, 1 . ∈ > Since T is surjective, there exists xo such that T xo yo and ( ) ⊂ ( ( )) B 0, ε B y , ε y T B 0, 1 T x T B 0, 1 T B 0, x Y Y o o X o= X X o X T B 0, 1 T B 0, x T B 0, 1 x . ( ) = ( X ) − ⊂ ( X ( o))X− ⊂ X( ( ))o−X ( ( Y Y )) ε Thus, form linearity of T it follows that BY 0, T BX 0, 1 . = ( ( )) + ( ( Y Y ))1 =xo (X ( + Y Y ))

( +Y Y ) Proof of Theorem 5.3. Due to Lemma 5.4Š there exists ⊂ εo( 0( such)) that ◻ B 0, ε T B 0, 1 . Y o X > Let y B 0, ε . Then, there exists ε 0 such that Y o ( ) ⊂ ( ( )) y ε ε . ∈ ( ) > Y o

Definey ¯ εo ε y. Then, y¯ Y εo andY y ¯Y

= Q= ∈ LECTURE 5. OPEN MAPPING AND CLOSED GRAPH THEOREMS 31 and by the continuity of T we obtainy ¯ T x¯. Denote x ε εo x¯. Then, y T x and

ε ε i ε i ε 1 x x¯ α xi α 1. X X +∞ = X +∞ = ( ~ ) = εo εo i 0 εo i 0 εo 1 α Y Y = Y Y ≤ Q Y Y < Q = ⋅ < = = − Remark 15. Let X,Y be Banach spaces and T α X,Y be bijective. Then, T 1 is◻ continuous. − ∈ ( ) Remark 16. If X, 1 and X, 2 are Banach spaces and norms 1, 2 are such that

( Y ⋅ Y ) ( xY ⋅1Y )M x 2, x X, Y ⋅ Y Y ⋅ Y then 1 and 2 are equivalent. Y Y ≤ Y Y ∀ ∈ Definition 5.5. Let X, Y be normed spaces, D be a subspace of X and T D Y be Y ⋅ Y Y ⋅ Y a linear functional. Then, T is closed if for every sequence xn n N D such that xn x and T xn n Y , T xn y, it holds that x D and T x y. ∶ → N ∈ { } ⊂ → (If T is continuous,∈ then it is also closed?) { } ⊂ → ∈ = Definition 5.6. For a linear functional T D Y we define its graph as following graph T x, T x x D X Y. ∶ → Lemma 5.7. Let X,Y,D,T, graph T be defined as above. Then, ( ) = {( ) ∶ ∈ } ⊂ × (a) graph T is a linear subspace of X Y , (b) T is closed if and only if graph( ) T is closed in X Y . ( ) × Lemma 5.8. Let X, Y be Banach spaces and D X be a linear subspace. If T D Y is ( ) ⊕ closed, then ⊂ ∶ → (a) D with a norm x x X T x Y is a Banach space, (b) T is continuous as a mapping from D, to Y . Y Y = Y Y + Y Y Proof of Lemma 5.8. a) Since T is closed,( weY ⋅ knowY) that for every sequences

xn n N D such that xn x X and ∈ { } ⊂ → ∈ T xn n N Y such that T xn y it holds that x D and T x y. xn∈ n N D is a Cauchy sequence with respect to the { } ⊂ → norm, since xn n is a Cauchy sequence in X and T xn n is a Cauchy sequence in Y . N ∈ N Moreover, its limit∈ in X is an= element{ } from⊂ the subspace D. Thus, Y ⋅ Y ∈ ∈ { } { } xn x xn x X T xn x Y 0, which implies that D, is complete. b) T is bounded, which holdsY − dueY = toY the− followingY + Y ( inequality− )Y → ( Y ⋅ Y) T x Y x X T x Y x . Thus, T is also continuous. Y Y ≤ Y Y + Y Y = Y Y ◻ 32 LECTURE NOTES, FUNCTIONAL ANALYSIS

Theorem 5.9. Let X, Y be Banach spaces and T X Y be a linear and closed. Then, T is continuous. ∶ → Proof of Theorem 5.9. T is continuous with respect to the norm defined in the previous lemma. Note that if we set D X in the previous lemma, then X is a Banach

space with norms X and . Moreover, x X x for all x Y X⋅ Y. Then, according to Remark 16 the norm X is equivalent to= . Therefore, T is continuous with respect to the norm X asY well.⋅ Y Y ⋅ Y Y Y ≤ Y Y ∈ Y ⋅ Y Y ⋅ Y Y ⋅ Y ◻ 5.1. Application of Banach -Steinhaus Theorem

Lemma 5.10. For a subset M of a normed space X, X , the following conditions are equivalent: (a) M is bounded, ( Y ⋅ Y ) (b) for every x X , x M K is bounded. ′ ′ ′ Proof of Lemma 5.10.∈ The( ) implication⊂ i ii is trivial. To show ii i consider i X X , ) ⇒ ) ) ⇒ ) such that i x is given by (4.1) for all x X. Note′′ that ∶ → sup x x sup i x x , x X. ( ) ∈ x M ′ x M ′ By Banach-Steinhaus theorem∈ S ( we)S have= ∈ S ( )( )S < +∞ ∀ ∈ sup x sup i x . x M x M

∈ Y Y = ∈ Y ( )Y < +∞ Remark 17. Any weakly convergent sequence is bounded. ◻

Lemma 5.11. Let X be a Banach space, Y be a normed space and Tn α X,Y for all n N. If for all x X there exists T x limn Tnx, then T α X,Y . ∈ ( ) →+∞ ∈ ∈ = ∈ ( ) LECTURE 6 Adjoint Operator

Definition 6.1. Let X, X , Y, Y be normed spaces and T α X,Y . Then, T Y X is defined by ′ ′ ′ ( Y ⋅ Y ) ( TYy⋅ Y )x y T x . ∈ ( ) ∶ → ′ ′ ′ One can check that T α Y ,X . ( )( ) = ( ) ′ ′ ′ Theorem 6.2. Let X,∈Y (be Banach) spaces and T α X,Y . Then, the following condi- tions are equivalent (a) Im T is closed, ∈ ( ) (b) Im T ker T , (c) Im(T) is closed,′ ⊥ (d) Im(T ′) = ker( T ) , where ( ′) ⊥ ( ) = ( ) U x X x x 0, x U ,V x X x x 0, x V . ⊥ ′ ′ ′ ′ ′ ⊥ Lemma 6.3.∶= {Let∈ X,Y∶ be( Banach) = ∀ spaces∈ } and T ∶= {α X,Y∈ ∶ have( ) a= closed∀ image.∈ } Then, there exists K 0 such that ∈ ( ) y Im Y x X s.t. T x y and x K y . ≥ X Y Proof of Lemma∀ 6.3.∈ (Define) ∃ a∈ canonical factorization= YTˆY(what≤ Y isY precisely a canonical factorization?) Tˆ X ker T Im T . Note that X ker T and Im T are Banach spaces and Tˆ is bijective. Then, according to ∶ ~ ( ) → ( ) Remark 15 there exists ~ ( ) ( )Tˆ 1 α Im T ,X ker T and K 0 such that Tˆ 1 K. − ∈ ( ( ) ~ ( )) − ≥ Y Y ≤ 33 ◻ 34 LECTURE NOTES, FUNCTIONAL ANALYSIS

Lemma 6.4. Let X, X , Y, Y be Banach spaces and T α X,Y . If there exists a constant C such that for all y Y ( Y ⋅ Y ) ( ′ Y ⋅ Y ′ ) ∈ ( ) C y T y ′ , ∈ α Y,K α X ,K ′ ′ ′ then T is surjective and open. Y Y ( ) ≤ Y Y ( )

Proof of Lemma 6.3. In order to prove that T is open we need to show that BY 0,C T BX 0, 1 (according to Lemma 5.2, claim (a)). In fact it is enough to prove ( ) ⊂ ( ( )) BY 0,C T BX 0, 1 D. Let y B 0,C and assume that y D (note that D is a convex set). According to o Y ( o) ⊂ ( ( )) =∶ Hahn-Banach theorem, there exists y Y , α R such that for each y D ∈ ( ) ∉ Re y y ′ α ′ Re y y y y . ∈ ∈ o o ∈ Note that 0 D, which implies′ Re y 0 0 ′ α. We′ can assume that α 0. Define ( ) ≤ < ( ) ≤ S ( )S y˜ y α. Then ′ ∈ ( ) = ≤ > ′ ′ Rey ˜ y Re y y α 1 Re y y α Rey ˜ y , y D. = ~ o o Thus, ′ ′ ′ ′ ( ) = ( ( )~ ) ≤ < ( ( )~ ) = ( ) ∀ ∈ y˜ y 1 y˜ yo , y D ′ ′ (note that y D λy D, λ such that λ 1). Moreover, for all x such that x X 1, T x y D and from the definitionS ( )S of≤T < S ( )S ∀ ∈ ∈ ⇒ ∈ ∀ S S > Y Y < T y˜ x′ y˜ T x 1, = ∈ which implies that T y˜ 1,′ but′ ′ α X,K S ( )S = S ( )S ≤ ′ ′ y˜( y ) y˜ y C y˜ Y Y o ≤ α Y,K o Y α Y,K ′ ′ ′ and 1 C y˜ , which leads to the contradiction.( ) ( ) α Y,K S ( )S ≤ Y Y ⋅ Y Y < Y Y ′ Lemma< 6.5.Y Y ( ) ◻ Im T ker T . ′ ⊥ Proof of Lemma 6.5. ( ) = ( ( )) ” ”. Let y Im T , which means that there exists x X such that y T x. If y ker T , then T y is a zero functional on X and thus ′ ′ ⊂ ∈ ( ) ∈ = ∈ ( ) ′ ′ 0 T′ y x y T x y y . ′ ′ ′ ′ Therefore, for each y Im T it= holds( )( that) =y y( 0,) = if y( )ker T . Since ker T y Y y ′ y 0, y ′ ker T ′ , ∈ ( ) ( ) = ∈ ( ) ′ ′ ′ ′ we conclude that Im T ker⊥T . ( ( )) = { ∈ ∶ ( ) = ∀ ∈ ( )} ′ ” ”. Let U Im T . Thus, U is a closed⊥ subspace of Y . We will show that ( ) ⊂ ( ( )) y U y ker T . ⊃ = ( ) ′ ⊥ ∉ ⇒ ∉ ( ( )) LECTURE 6. ADJOINT OPERATOR 35

By Hahn-Banach theorem there exists y Y such that ′ y U 0∈ and y y 0. ′ ′ In particular 0 y T x T y Sx =for all x X( ,) that≠ is, T y is a zero functional on X and thus y ker′ T . This′ proves′ that y ker T , since′ the′ opposire claim would imply that y ′y = 0.( ′ ) = ( )( ) ∈ ′ ⊥ ′ ∈ ( ) ∉ ( ( )) Proof of Theorem( ) = 6.2. ◻ b a is straightforward. b a follows from Lemma 6.5. Now,( ) ⇒ we( ) shall prove a d . Clearly Im T ker T (for this we do not need a ) (and) ⇐T y( )x y T x 0 for x ker T . Let x′ ker T ⊥and define a linear functional ′ ′ ′ ( ) ⇒ ( ) ( ′) ⊂ ( ( ))⊥ ( ) ( ) = ( ) = z Im ∈T (K), z y ∈x( x (, for)) y T x. ′ ′ ′ Note that z is continuous.∶ Indeed,( ) → using one( of) = the( previous) lemmas= and the open mapping theorem we have that y Im T x X s. t. y T x, and

z y x x x x x K y . ∀ ∈ ( ) ∃ α ∈X,K X = α X,K Y By Hahn-Banach theorem′ we′ can define′ y Y , which is′ an extension of z . Then, S ( )S = S ( )S ≤ Y Y ( ) ⋅ Y Y ≤ Y Y ( ) ⋅ Y Y ′ ′ ′ x x z T x y T∈ x T y x , x X, ′ ′ ′ ′ ′ that is, x T y . This finishes( ) = the( proof,) = ( since) = we( assumed)( ) ∀ that∈ x ker T . d c ′ is straightforward.′ ′ ′ ⊥ d a .= Let Z Im T Y and define S α X,Z by setting Sx∈ ( T x(. For)) y Y and x( )X⇒we( ) have ′ ( ) ⇒ ( ) = ( ) ⊂ ∈ ( ) = ∈ ∈ T y x y T x y Z Sx S y Z x , x X. ′ ′ ′ ′ ′ ′ Then, T y S y(Z . Therefore)( ) = ( Im) T= S (Im S) =.( To( proveS ))( the) opposite∀ ∈ inclusion assume that S z′ ′ Im ′S ′ for some z Z and′ consider′ any extension y of z (by Hahn-Banach theorem′ ′ such= extension( ′ S ) exists).′ Then,′ ( ) ⊂ ( ) ′ ′ ∈ ( ) ∈ S z T y Im T Im S . ′ ′ ′ ′ ′ ′ Then, by the assumption we have= Im S⇒ Im (S ).= Moreover,( ) Im S is dense in Z and thus S is injective (?). S is then a continuous′ bijection′ between Z and Im S . In fact it is an′ in isomorphism, in′ particular ( ) = ( ) (′ ) ′ ( ) C z S z , z Z . α Z,K α X,K ′ ′ ′ ′ ′ Then, by one of the LemmaY above,Y ( S) is≤ Y surjectiveY ( (and) ∀ open).∈ Thus, Im S Z Im T (?) and Im T Im T . ( ) = = ( ) ( ) = ( ) ◻ 36 LECTURE NOTES, FUNCTIONAL ANALYSIS

Lemma 6.6. Let P be a continuous projection on a normed space X (P is a projection if P 2 P ). Then, (a) P 0 or P 1, =(b) ker P and Im P are closed, (c) X = ker PY Y ≥Im P . ( ) ( ) Proof of Lemma= ( ) 7.11.⊕ ( ) a P P 2 P 2, which implies P 0 or P 1. b ker P P 1 0 is closed (since P is continuous), Id P is a continuous projection and( )Y ImY P= Y kerY−≤ YId Y P is closed. = Y Y ≥ (c) x( X) =it holds({ that}) x Id P x P x. − ( ) = ( − ) ( ) ∀ ∈ = ( − ) + ◻ LECTURE 7 Hilbert Spaces

Definition 7.1. Let X be a K- linear space. A mapping , X X K

is called a scalar product if < ⋅ ⋅ > × → (a) x1 x2, y x1, y x2, y , x1, x2, y X, (b) λx, y λ x, y , x, y X, λ K, (c) < x, y+ >y, = x < , x,> y+ ∀ ∈ (d) < x, x > =0, , ∀ ∈ ∈ (e) < x, x > = <0 >x∀ 0. ∈ < > ≥ ∀ ∈ Lemma< 7.2. >(Cauchy = ⇔ - Schwarz= inequality). Let X be a K-linear space with a scalar product. Then, x, y 2 x, x y, y , x, y X. The equality holds if and only if x λy for λ . S< >S ≤ < > ⋅ ∀ ∈ Lemma 7.3. Define a mapping = X X ∈ R by Y ⋅ Y ∶ x × →x, x . √ Then, is a norm. In particular, itY holdsY = that< > Y ⋅ Y x, y x y , x, y, X.

Definition 7.4. A normed spaceS< X,>S ≤ Y XY ⋅ YisY called∀ a∈ prehilbert space, if there exists a scalar product , such that ( Y ⋅ Y ) < ⋅ ⋅ > x X x, x , x X. √ Complete prehilbert space is calledY Y a= Hilbert< space.> ∀ ∈ Lemma 7.5. Let X, be a prehilbert space and U be its dense linear subspace. If for all u U it holds that x, u 0, then x 0. ( Y ⋅ Y) 37 ∈ < >= = 38 LECTURE NOTES, FUNCTIONAL ANALYSIS

Proof of Lemma 7.5. Define a set Y y X x, y 0 . Y is closed since the map y x, y is continuous. It also contains a dense subspace U. = { ∈ ∶ < >= } This implies that Y X. In particular, x Y and thus x 2 x, x 0. → < > If X, X is a normed space, we can introduce a scalar product by the norm X . More precisely, we set = ∈ Y Y = < > = ◻ ( Y ⋅ Y ) 1 2 2 Y ⋅ Y x, y x y x y , for 4 X X K R and 1 < > = ‰Y + Y − Y − Y Ž = x, y x y 2 x y 2 i x iy 2 i x iy 2 , for . 4 X X X X K C Lemma 7.6.< Scalar> = product‰Y + Y −, Y −X Y X+ Y K+isY a continuous− Y − Y mapping.Ž =

Proof of Lemma 7.6. Let2,∶ y1,× y2 X→. Then, x , y x , y x x , y x , y y 1 1 2 2 ∈ 1 2 1 2 1 2 x1 x2 y1 y1 y2 x2 . S< > − < >S = S< − > + < − >S ≤ Y − Y ⋅ Y Y + Y − Y ⋅ Y Y

Theorem 7.7. (Parallelogram equality) A normed space X, X is a prehilbert space◻ if and only if, for all x, y X the following inequality holds ( Y ⋅ Y ) x y 2 x y 2 2 x 2 2 y 2 . ∈ X X X X Proof of Theorem 7.7. Y(for+ KY +RY).− WeY introduce= Y Y + a scalarY Y product as following 1 2 2 x, y = x y x y . 4 X X 2 Clearly, it holds that x X < x, x> =. We‰Y need+ Y to− proveY − theY Ž properties of a scalar product form the Definition 7.1. Y Y = < > a Let x1, x2 X and define α x x y 2 2 x y 2 2 x 2 x x y 2 , ( ) ∈ 1 2 X 1 X 2 X 1 2 X β x x y 2 2 x y 2 2 x 2 x x y 2 . ∶= Y 1 + 2 + YX = Y 2 + YX + Y 1YX − Y −1 +2 Y X Therefore, ∶= Y + + Y = Y + Y + Y Y − Y − + + Y α β x x y 2 1 2 2 + 1 Y + +xY =y 2 x 2 x y 2 x 2 x x y 2 x x y 2 . 1 1 2 2 2 1 2 1 2 Similarly one= getsY + Y + Y Y + Y + Y + Y Y − ‰Y − + Y + Y − + + Y Ž 1 x x y 2 x y 2 x 2 x y 2 x 2 x x y 2 x x y 2 . 1 2 1 1 2 2 2 1 2 1 2 Y + − Y = Y − Y + Y Y + Y − Y + Y Y − ‰Y − − Y + Y − + − Y Ž LECTURE 7. HILBERT SPACES 39

Finally, 1 x x , y x x y 2 x x y 2 1 2 4 1 2 X 1 2 X 1 < + > = ‰Yx + y 2+ Y x − Yy 2+ x− Y yŽ2 x y 2 4 1 X 2 X 1 X 2 X x , y x , y . = ‰Y1 + Y +2Y + Y − Y − Y − Y − Y Ž Claim b follows from a for λ N. Form the construction of , we also have the = < > + < > m property b fulfilled for λ 0, λ 1 and thus for all λ Z. Let λ n Q. Then, ( ) ( ) ∈ x < ⋅ ⋅ > n λx, y n m , y m x, y nλ x, y . ( ) = = − n ∈ = ∈ By the continuiuty of< X> = K = < this result> = for< λ >R. Claims c , d and e are straightforward. Y ⋅ Y ∶ → ∈ Remark( ) 18.( )The following( ) claims hold: ◻ (a) A normed space is prehilbert if and only if all its 2-dimensional subspaces are prehilbert. (b) Any subspace of a prehilbert space is prehilbert. (c) of a prehilbert space is a . Example 11. Examples of Hilbert spaces. n n n ¯ (a) R and C with si, ti i 1 siti, 2 ¯ (b) l with si, ti i 1 siti, = 2 < +∞ > = ∑ (c) L Ω, µ with f, g = Ω f g¯ dµ. < > = ∑ Definition( 7.8.) Let X< be a> prehilbert = ⋅ space. We say that x and y are orthogonal and write x y, if x, y 0. Sets ∫A and B are orthogonal, if x, y 0 for all x A and y B. The set ⊥ < >= A y X x y, x A < > = ∈ is∈ called the ⊥ of A. = { ∈ ∶ ⊥ ∀ ∈ } Remark 19. For a Hilbert space X the following claims hold: (a) x y x 2 y 2 x y 2, (b) A is a closed subspace of X, ⊥ ⇒ Y Y + Y Y = Y + Y (c) A⊥ A , (d) A lin⊥ ⊥A . ⊂ ( ) ⊥ ⊥ Theorem 7.9.= ( Let{ })H, be a Hilbert space and K be its convex and closed subset. Let xo H. Then, there exists a unique x K such that ( Y ⋅ Y) x xo inf y xo . ∈ ∈ y K This statement still holds, if H isY a uniformly− Y = ∈ convexY − space.Y 40 LECTURE NOTES, FUNCTIONAL ANALYSIS

Proof of Theorem 7.9. It is trivial if xo K (we set x xo). Let xo K. Without loss of generality we can assume that xo 0. In this case we define d inf∈ y . = ∉ = y K

There exists yn n N K and limn y=n ∈ Yd.Y We will prove that yn n N is a Cauchy sequence. This is sufficient to show, since K, is a Banach space. By Parallelogram ∈ →+∞ ∈ equality we have{ } ⊂ Y Y = { } ( 1Y ⋅ Y) y y 2 y y 2 y 2 y 2 , m, n . n m n m 2 n m N 1 1 2 2 By convexity ofY(K +we have)~ Y that+ Y( 2 −yn )~ymY = K‰andY Y thus+ Y 2Y yŽn ∀ym ∈ d . Therefore,

2 1 2 2 d yn ym (2 + ) ∈yn ym ,Y (m,+ n )Y. ≥ 2 N 2 Since the right hand side+ Y( of the− inequality)~ Y ≤ converges‰Y Y + Y to dY Ž, we∀ have that∈ yn ym converges to zero as n, m tend to infinity. By the comleteness of K, there exists x K such that Y − Y x lim yn and x d. n ∈ Now, we shall show that the element= →+∞ is unique. LetY x,Y x=¯ K be such that x x¯ d. ∈ Then, by Parallelogram equality Y Y = Y Y = x x¯ 2 2 d2 x x¯ 2 2. Note that x x¯ 2 K, x x¯ 2 2 d2, which implies x x¯ 0 and thus x x¯. Y( − )~ Y = − Y( + )~ Y

Lemma 7.10.( + Let)~ K∈ beY( a closed+ )~ convexY ≥ subset of a HilbertY − spaceY = H, and= xo H◻. The, the following conditions are equivalent ( Y ⋅ Y) ∈ (a) xo x infy K xo y , (b) Re x x, y x 0, y K. o ∈ Y − Y = Y − Y Proof of Lemmaa − 7.10.− f ≤ ∀ ∈ b a x y 2 x x x y 2 ( ) ⇒ ( ) o o x x 2 2Re x x, x y x y 2 x x 2. Y − Y = Y( o − ) + ( − )Yo o a b . Let t 0, 1 , y K and define y 1 t x ty K. Then, = Y − Y + a t − − f + Y − Y ≥ Y − Y x x 2 x y 2 x x t x y , x x t x y ( ) ⇒ ( ) o∈ [ ] ∈ o t o = ( − ) + ∈o x x 2 2Re x x, t x y t2 x y 2. Y − Y ≤ Y o − Y = a − o + ( − ) − + ( − )f Finally, = Y − Y + a t − ( − )f + Y − Y Re x x, t y x x y 2, t 0, 1 . o 2 a − ( − )f ≤ Y − Y ∀ ∈ [ ] ◻ LECTURE 7. HILBERT SPACES 41

Theorem 7.11. Let U 0 be a closed subspace of a Hilbert space H, . Then, there exists a linear projection ≠ { } ( Y ⋅ Y) PU H U, s. t. Pu 1, ker PU U . ⊥ Moreover, Id PU is a projection∶ → such thatY Y = ( ) =

− Id PU H U , s. t. Id PU 1, if U H . ⊥ It also holds that H( −U )U∶ . → Y − Y = ( ≠ ) ⊥ Proof of Theorem= 7.11.⊕ Define

PU H U, PU xo x,

2 2 where xo H and x U is such that∶ x→o x inf( y )U=xo y . It is straightforward that PU is a projection. By the previous lemma ∈ ∈ ∈ Y − Y = Y − Y Re xo PU xo , y PU xo 0, y U.

Note that y PU xo U anda thus− ( ) − ( )f ≤ ∀ ∈

− ( ) ∈ Re xo PU xo , y 0, y U.

Using y and iy in the expressiona above− ( we) obtainf ≤ ∀ ∈

− xo PU xo , y 0, y D and xo PU xo xo x U . ⊥ Since U is a lineara − subspace,( ) f = it holds∀ ∈ that for all x1,− x2 (H, )λ=1, λ2 − K∈ ⊥ λ1x1 λ1PU x1 λ2x2 λ2PU ∈x2 U . ∈ ⊥ Therefore, λ1x1 λ2x2 ( z H− and P(U ))z +is( such− that z ( P))U z∈ U . Thus, ⊥ + PU= λ1∈x1 λ2x2 (P)U z λ1PU x1 − λ2(PU) ∈x2 , which proves that PU is( linear.+ From) = the( construction) = ( ) of+PU it( follows) that Im PU U and we have that ker PU U . Therefore, ⊥ ( ) = ( ) = PU xo 0 xo U . ⊥ We claim that Id PU is a projection( ) with= Im⇔ Id P∈U U and ker Id PU U. By Pythagoras theorem ⊥ − ( − ) = ( − ) = 2 2 2 2 xo PU xo xo PU xo PU xo Id PU xo and PU xo YU, Yxo= YPU (xo ) + U −. Finally,( )Y we= Y get(H )YU+ Y(U .− Moreover,)( )Y PU 1 and Id PU 1 (in fact equal to 1, if⊥ U 0 and U H). ⊥ ( ) ∈ − ( ) ∈ = ⊕ Y Y ≤ Y − Y ≤ ≠ { } ≠ ◻ 42 LECTURE NOTES, FUNCTIONAL ANALYSIS

Remark 20. Let H be a Hilbert space and U be its linear subspace. Then, U U . ⊥ ⊥ Indeed, for a closed set V it holds that Id P P ⊥ . Consider V U. It is clear that = ( V ) V U V and Id P ⊥ P ⊥⊥ . − = V V = ⊥ ⊥ Thus, P P ⊥⊥ and thus U U . V V = − = ⊥⊥ Theorem= 7.12. (Riesz theorem)= Let H be a Hilbert space. Then, mapping Φ H H ,H y , y ′ ¯ is bijective, isometric and conjugate∶ linear→ (i.e.,(Φ)λy= a⋅ fλΦ y ). It means that x H !y H, s. t. x x x, y , x H and x y . ( ) = ( ) ′ ′ ′ ′ Proof of Theorem∀ ∈ 7.12.∃ ∈ Clearly Φ( is) conjugate= a f ∀ linear.∈ It followsY Y = fromY Y the Cauchy- Schwarz inequality that Φ y y and for x y y we obtain Y ( )Y ≤ Y Y y, y = ~Y Y Φ y x y , y a f which implies that Φ is an isometric( )( and) = thus, it= isY injective.Y Let x H (without loss of generality we can assume that x 1) andY Y U ker x . By the′ previous′ theorem H U U , where dim U 1.′ Thus, there exists y ′ H such that∈ x y 1 and U lin y .⊥ For x u λy ⊥ U UY Y = = ( ) ′ ⊥ = ⊕ ( ) = ⊥ ∈ 2 ( ) = = ( ) = + x ∈x( ⊕λx y) λ and x, y λ y . Therefore, ′ ′ ( ) = ( ) = a f = Y Y Φ y y 2 x and Φ is surjective. Finally, y 1 and x 1. ′ ( ~Y Y ) = ′ RemarkY 21.Y = Each HilbertY Y = space is reflexive. ◻

7.1. Orthonormal basis Definition 7.13. Let H be a Hilbert space. A subset S H is an orthonormal system if and only if for all e, f S it holds that ⊂ e f 1 and e, f 0. ∈ A subset S H is called an orthonormal basis if Y Y = Y Y = a f = S T,T orthonormal system T S. ⊂ Example 12. Examples⊂ of orthonormal− systems: ⇒ = LECTURE 7. HILBERT SPACES 43

(a) H L2 0, 2π , R , 1 1 1 = S([ ]χ ) cos n , n sin n , n . 2π 0,2π π N π N [ ] (b) H L2 =0,›2π , C , ∪ œ√ ( ) ∈ ¡ ∪ œ√ ( ) ∈ ¡ 1 in S e , n Z . = ([ ] ) 2π

Theorem 7.14. (Gram - Schmidt theorem)= œ√ Let H∈ be¡ a Hilbert space and xn n N be a linearly independent set in H. Then, there exists an orthogonal system S such that linS lin xn n N . { ∶ ∈ } Example 13. Let H L2 1, 1 , and x t tn, for n . Then, = { ∶ ∈ } R n N 1 1 d n S n = P([−t ,] n ) , where( ) = P t ∈ t2 1 n. n N n n ⎧¾ 2 ⎫ 2 n! dt ⎪ ⎪ = ⎨ + ⋅ ( ) ∈ ⎬ ( ) = ‹  ( − ) Theorem 7.15.⎪ (Bessel inequality) Let⎪ H be a Hilbert space. If S en n N H is an orthonormal⎩ system and x H, then⎭ = { ∶ ∈ } ⊂ 2 2 x, en x . ∈ +∞ n 1 Ta fT ≤ Y Y Proof of Theorem 7.15. Let N Q=N. Define N x∈N x x, en en n 1

such that xN ek, for k 1,...,N. By= Pythagoras− Q= a theoremf N 2 N N 2 2 2 2 2 x ⊥ xN = x, en en xN x, en x, en . n 1 n 1 n 1

This estimateY Y holds= Y forY + allY QN= a N, thereforef Y = Y takingY + Q the= Ta limitfTN≥ Q= Ta endsfT the proof.

Lemma 7.16. Let H be a Hilbert∈ space, S H be an orthonormal→ +∞ system and x H◻. Then, the set Sx e S ⊂ x, r 0 ∈ is countable. = ™ ∈ ∶ a f ≠ ž Proof of Lemma 7.16. From Bessel inequality it holds that

Sx,n e S x, r 1 n is finite and thus S S is at most countable. x n N x,n = ™ ∈ ∶ a f ≥ ~ ž ∈ = ⋃ ◻

LECTURE 8

Definition 8.1. Let f L1 Rn . Define

1 ixξ n ∈ f (ξ ) f x e dx, ξ R . 2π n 2 n R − is called a Fourier transform.(F )( ) = ~ ( ) ∀ ∈ ( ) S Theorem 8.2. L1 n , C n , is a continuous linear mapping F R L1 o R such that F ∶ ( ( ) Y ⋅ Y ) → ( 2(π )n Y2.⋅ Y∞) n − ~ Here, Co R denotes a space of continuous functions vanishing at infinity. YFY ≤ ( ) Proof of( Theorem) 8.2. A linearity of is obvious. Note that e ixξ 1. Therefore, f 2π n 2 f 2π n 2. − FL1 S S = Now, we want to prove − ~ − ~ YF Y∞ ≤ ( ) Y Y ⇒ YFY ≤ ( ) k (a) limξk ξ f ξ f ξ 0, k (b) lim ξk f ξ 0. → S(F )( ) − (F )( )S = kS S→+∞ k a Let ξ k N beS( suchF )( that)S = limk ξ ξ. Then, for a fixed x we have k ∈ →lim+∞ e ixξ e ixξ 0. ( ) { } k = − − By Lebesgue dominated convergence→+∞ theoremU − it holdsU = that 1 k lim f ξk f ξ f x lim e ixξ e ixξ dx 0, k 2π n 2 n k R − − where we→ used+∞ T(F the)( latter) − theorem(F )( )T due≤ to the~ factS that( )S the⋅ →+∞ functionU under− U the= integral can ( ) S be estimated by 2 f L1 Rn . b Note that the smooth functions with compact support are dense in L1, that is, S S ∈ ( ) L1 n n R 1 n ( ) Cc R L R . ( ) ∞ 45 ( ) = ( ) 46 LECTURE NOTES, FUNCTIONAL ANALYSIS

n k Thus, it is enough to show that for all f Cc R it holds that lim ξk f ξ 0. ∞ k k k Let f be such a function and j be such that ξj max ξi i 1, .S . .S→ , n+∞, where ξj are k k k ∈ ( ) (F )( ) = coordinates of ξ . Then, ξj ξ n. We apply the formula for integration by parts and T T = {S S ∶ = } as a result obtain √ T T ≥ S S~ 1 ∂ 1 k f ξk f x e ixξ dx 2π n 2 n ∂x iξ R j j − − ~ T(F )( )T = W 1 ∂ ( ) n k W S f 0. ( )n 2 −k 2π ∂xj L1 √ξ →+∞ ≤ ~ ⋅ ] ] ⋅ Ð→ ( ) S S

Definition 8.3. We say that f Rn C is vanishing faster than polynomial, if ◻ α n lim x f x 0, α N , x∶ →

α S S→α+∞1 ( α)n = ∀ ∈ where α α1, . . . , αn and x x1 xn . A space

n n β n = (R f )C R = D ⋅f ⋅ ⋅ ⋅vanishes ⋅ faster that polynomial, β N ∞ is called SchwarzS( ) = space.™ ∈ ( ) ∶ ∀ ∈ ž Remark 22. Schwarz space consists of smooth functions, which vanish faster than poly- nomial together with all its derivatives. However, one has to note that

n n Cc R R . ∞ n Remark 23. If f R , then ( ) ⊊ S( ) α α n ∈ S( ) x f, D f R , n N. n Lemma 8.4. Let f R . Then, ∈ S( ) ∀ ∈ (a) f C n and Dα f i α xαf , R ∈ S( ) (b) Dαf∞ i α ξα f . S S F ∈ ( ) (F ) = (− ) F( ) S S Proof ofF( Lemma) = 8.4.⋅ F( ) a Formal calculations leads to ∂α 1 1 ∂α ( ) Dα f f x e ixξdx f x e ixξdx ∂ξα 2π n 2 n 2π n 2 n ∂ξα R − R − 1 ~ ~ (F ) = i ⋅α S f( x) xαe ixξd=x i αS xα(f) ⋅ξ . ( 2π) n 2 n ( ) S S R − S S = (− ) ~ ( ) = (− ) F( )( ) In the second equality we have( ) passedS with differentiation under the integral sign. Since xαf Rn L1 Rn this is justified by Lebesgue dominated convergence theorem.

∈ S( ) ⊂ ( ) LECTURE 8. FOURIER TRANSFORM 47

b If follows from the formula for integration by parts that 1 ( ) Dαf Dαf x e ixξdx 2π n 2 n R − F( ) = 1 α~ ( )(∂α) S f x e ixξdx 1 α i α ξα f . ( )n 2 α 2π S S Rn ∂x (− ) − S S S S Since xβf n boundary= terms~ vanished( ) ⋅ in the calculation= (− ) above.(− ) F( ) R ( ) S Lemma 8.5.∈ S(If f) Rn , then also f Rn . ◻

Proof of Lemma∈ 8.5.S( )In Lemma 8.4F we∈ S proved( ) that f C Rn . We need to show that f and all its derivatives vanish faster than polynomial, that∞ is, F ∈ ( ) ξαDβ f ξ 0 for ξ . F From Lemma 8.4 we know that (F )( ) → S S → +∞ ξαDβ f i β i α Dα xβf S S S S which implies that Dα xβf n and thus in L1 n . Fourier transform of a function (FR ) = (− ) (− ) F(R ( )) from L1 Rn is a continuous function vanishing at infinity, which ends the proof. ( ) ∈ S( ) ( ) Let us consider( ) a function ◻ n x2 2 γ R R, γ x e . It is a well known fact that − ~ ∶ 1→ ( ) = γ x dx 1. n 2 2π Rn ~ We also denote γa x γ ax , for a 0. ( ) = ( ) S Lemma 8.6. For(γ )and= (γa defined) as> above it holds that

2 1 ξ γ ξ e ξ 2 and γ ξ γ . a an a − ~ Proof of Lemma 8.6.(F )(The) = second statement(F is)( straightforward.) = (F ) ‹ To prove the first claim let n 1. Note that γ fulfills the following differential equation (8.7) y xy 0, y 0 1. = ′ By Lemma 8.4 + = ( ) = 1 0 γ xγ iξ γ γ γ ξ γ 0, i ′ ′ ′ which means that= F( γ +fulfills) = equation(F ) + (8.7)‹ (F with) the⇒ same( initialF ) + condition,(F ) = since − 1 2 F γ 0 e x 2dx 1. 2π 1 2 R − ~ (F )( ) = ~ = ( ) S 48 LECTURE NOTES, FUNCTIONAL ANALYSIS

Since the solution to (8.7) is unique we obtain γ γ. The case n 1 leads in fact to the previous case, since = F > 1 n x2 2 ix ξ n ξ2 2 γ ξ ... Π e k e k k dx ... dx Π e k . 2π n 2 k 1 1 n k 1 R R − ~ − − ~ (F )( ) = ~ = = = ( ) S S Lemma 8.8. For f Rn it holds that ◻ n ∈ S( ) f x f x , x R .

Proof of Lemma 8.8. We(FF proved)( that) = (−f ) ∀Rn∈ and thus f is well defined. Let f, g Rn . Since a function F ∈ S( ) F(F ) ixξ ∈ S( ) x, ξ f ξ g x e − is integrable, Fubini theorem yields( ) → ( ) ( ) (8.9) f x g x dx f x g x dx. Rn Rn ixξ n This implies that for g xS (eF )(o γ )ax( ,) where= Sξo (R )(andF )(a )0 are fixed, that −1 ixξo ixξ g ξ( ) = ( )e γ ax ∈e dx >γa ξ ξo . 2π n 2 n R − − ~ (F )( ) = S ( ) = (Fx ξo)( + ) We use Lemma 8.6 and the( following) change of variables u a , which leads to + 1 1 1 x ξo f x e ixξo γ ax dx f =x γ dx 2π n 2 2π n 2 an a R − R ~ 1 ~ + S (F )( ) ( ) = S f(au) ξ(Fγ )u ‹ du.  ( ) ( )n 2 o 2π R ~ = ( − ) (ixξ)o Now we want to pass to the limit with a 0.( Since) xS f x e γ ax is bounded 1 x ξo 1 − by f L and x f x an γ a is bounded by f , application of Lebesque convergence theorem ends the proof (the+ → limit of the left→ ( handF )( side) is equal( ) to f ξo ∞ andSF theS limit∈ on the→ right( hand) (F side) ‰ is equalŽ to f ξo ). Y Y (FF ) Theorem 8.10. The Fourier transform is a bijection(− on) Rn . The inverse operator ◻1 is defined by − S( ) F 1 1 ixξ n (8.11) f x f ξ e dξ, x R . 2π n 2 n − R (F n)( ) = ~ ( ) ∀ ∈ Moreover, for all f, g R it holds( ) thatS f, g f, g . ∈ S( ) L2 L2 aF F f = a f LECTURE 8. FOURIER TRANSFORM 49

Proof of Theorem 8.10. From the previous lemma we know that 4 Id. Thus, is bijective and 1 3 is bijective as well. It also holds − 1f x 2 f x f x , F = F F = F which proves that (8.11) holds.− From (8.9) we have that (F )( ) = (F (F ))( ) = (F )(− ) f ξ g ξ dξ f x g x dx. Rn Rn For transparency purposes let denote h g. Then, S (F )( )(F )( ) = S ( )(F(F ))( ) 1 1 h¯ x h ξ e ixξdξ h ξ eixξdξ 1h x g x n 2 = F n 2 2π Rn 2π Rn − − and thus(F )( ) = ~ S ( ) = ~ S ( ) = F ( ) = ( ) ( ) f, g f x( g)x dx f, g . L2 L2 Rn aF F f = ( ) ( ) = a f n S As a consequence, for all f R it holds that f L2 f L2 . The operator is thus n 2 n ◻ continuous as a mapping from R to L R , bijective and L2 -isometric. ∈ S( ) YF Y = Y Y F Remark 24. For f L2 n we define R S( ) ( ) Y ⋅ Y 1 ixξ gR ξ ∈ ( ) f x e dx and 2f ξ lim gR ξ , 2π n 2 B 0,R R − 2 n where a limit in( ) the= definition~ ( above) ( is) the limit in L (RF .)( ) = →+∞ ( ) ( ) S Theorem 8.12. (Hausdorff-Young inequality) Let 1 p 2 and 1 1 1. For f n ( ) p q R it holds that f Lq Rn and 1 ≤ ≤ + = ∈ S( ) f f . F ∈ ( ) Lq 2π n p n 2 Lp There exists an extension of YFtoY a continuous≤ ~ − operator~ ⋅ Y Y Lp Rn Lq Rn , which is defined by ( ) F f ξ lim gR ξ , F ∶ ( ) → ( ) R where a limit in the definition above is the limit in Lq n . (F )( ) = →+∞ ( )R Proof of Theorem 8.12. For p, q 1, and (p, q) 2, 2 wehave L1 L 2π n 2, ( ) = ( +∞) ( ) = ( ) L2 L∞2 1. − ~ YF ∶ → Y ≤ ( ) 2 For other case the proof follows from the Riesz interpolation theorem. One takes θ 2 p , 1 1 θ θ YF ∶ → Y ≤ so that q 2 . − = − = +∞ + ◻

LECTURE 9 Adjoint operators on Hilbert spaces

Definition 9.1. Let T be a linear bounded operator T α H,H . We say that T H H is an adjoint operator to T , if ∗ ∈ ( ) ∶ → T x, y x, T y , x, y H. ∗ Lemma 9.2. For each y, elementa Tf =yais uniquef and∀ T∈ y T x . Thus, T is well defined and its norm is bounded. ∗ ∗ ∗ Y Y ≤ Y Y ⋅ Y Y Proof of Lemma 9.2. Fix y H. Note that x T x, y is a linear bounded functional. From Riesz representation theorem ∈ → a f y H zy H s. t. T x, y x, zz , x H. We set T y z . Then, it also holds that z T x . y ∀ ∈ ∃ ∈ ay f = a f ∀ ∈ ∗ Lemma 9.3.= It holds that Y Y ≤ Y Y ⋅ Y Y ◻ (a) T S T S , (b) αT ∗αT¯ ∗, ∗ (c) (ST+ ∗ ) T= S∗ ,+ (d) T( )∗T=. ∗ ∗ ( ∗∗ ) = Definition 9.4.= Let T α H,H . Then, (a) T is normal if TT T T , (b) T is self adjoint∈ if (∗T T∗) , (c) T is unitary if T T = Id ∗ TT . ∗ = ∗ Theorem 9.5. If T α H,H= , then= T is normal if and only if, T x T x , x H. ∈ ( ) ∗ Proof of Theorem 9.5. ” ”Y It holdsY = Y thatY ∀ ∈ T x 2 T x, T x T T x, x TT x, x T x 2. ⇒ ∗ 51 ∗ ∗ Y Y = a f = a f = a f = Y Y 52 LECTURE NOTES, FUNCTIONAL ANALYSIS

” ” Before we prove this implication we shall show the following claim. Let T α H,H be such that ⇐ T x, x 0, x H. ∈ ( ) Then, T 0. Indeed, a f = ∀ ∈ T x y ,T x y 0 T x, y T y, x 0, = T x iy ,T x iy 0 T x, y T y, x 0, a ( + ) ( + )f = ⇒ a f + a f = and thus T y, x 0, for all x, y H, which implies T 0. Now we can proceed with the a ( + ) ( + )f = ⇒ −a f + a f = proof. We have that a f = ∈2 2 = 0 T x T x T T TT x, x . Thus, T T TT 0 due to the claim. ∗ ∗ ∗ = Y Y − Y Y = a( − ) f ∗ ∗ Definition− 9.6. For= an operator T we define its spectrum as ◻ σ T λ C T λId is not invertible . The condition which states that T λId is not invertible is equivalent to: ( ) = { ∈ ∶ ( − ) } (a) Im T λId H, (b) ker T λId 0 . ( − ) ( − ) ≠ Remark 25. If λ σ T , then T λId 1 is a linear bounded functional (due to open ( − ) ≠ { } mapping theorem). − ∉ ( ) ( − ) Lemma 9.7. If T H H is a bounded linear functional, then σ T B 0, T . ∶ → Proof of Lemma 9.7. Assume that( λ) ⊂ T( .Y Then,Y) T n Id TS Sλ> Y1 Y +∞ λ − n 0 is a convergent series and thus the( operator− ~ ) Id=TQ= λ‹ is invertible, which is a contradiction.

( − ~ ) ◻Definition 9.8. Operator T α H,H is a , if a closure of an image of the unit ball is compact, that is, ∈ ( ) T BX 0, 1 is compact. Lemma 9.9. If T is self adjoint, then σ T . ( ( )) R Proof of Lemma 9.10. We know that( T) ⊂T . Let λ α iβ σ T . Then, T λId x 2 T x αx iβx, T x αx ∗iβx = = − ∈ ( ) T x αx, T x αx T x αx, iβx iβx, T x αx iβx, iβx Y( − ) Y = a − + − + f T x αx 2 β2 x 2. = a − − f + a − f + a − f + a f = Y − Y + ⋅ Y Y LECTURE 9. ADJOINT OPERATORS ON HILBERT SPACES 53

Thus, T λId x 2 β2 x 2, which implies β 0.

TheoremY( − 9.10.) LetY ≥T α⋅ Y H,HY be a self adjoint= operator. Then, ◻ T sup T x, x . ∈ ( ) x x 1 Proof of Theorem 9.10. It is clearY Y = that∶Y Y= Ta fT sup T x, x sup T x x sup T x T . x x 1 x x 1 x x 1

∶Y Y≤ Ta fT ≤ ∶Y Y≤ Y Y ⋅ Y Y ≤ ∶Y Y≤ Y Y = Y Y To prove the opposite inequality define M supx x 1 T x, x . Then,

T x y , x y T x y , x y = 2 T x,∶Y yY≤ Ta2 T y,fT x a ( + ) + f − a ( − ) − f = 2aT x, yf + 2ax, T yf 2 T x, y 2 T y, x = 2aT x, yf + 2aT x, yf = 4Rea T x,f + y .a f In the calculation above some terms vanish due to the fact that T is self adjoint. By = a f + a f = a f parallelogram equality we have 4Re T x, y M x y 2 x y 2 2M x 2 y 2 Re T x, y M, x, y x , y 1 a f ≤ (Y + Y + Y − Y ) = (Y Y + Y Y ) Multiplication by λ such that λ 1 gives a f ≤ ∀ ∶ Y Y Y Y = T x, y M x, y x , y 1, S S = which proves that T M. Ta fT ≤ ∀ ∶ Y Y Y Y = Lemma 9.11. LetY TY = α H,H be a self adjoint and compact operator. Then, T or◻ T is an eigenvalue of T . ∈ ( ) Y Y Proof−Y Y of Lemma 9.11. There exists a sequence xn n N ( xn 1) such that T x , x T . ∈ n n { } Y Y ≤ This holds due to the definition of a norm of T . Since the T is compact one can chose a Ta fT → Y Y subsequence xnk which converges. For transparency purposes we omit the lower index k and denote this subsequence as xn n N . Denote

λ ∈ lim T xn, xn . { } n By compactness of T there exists y such= → that+∞ a f y lim T xn T n Therefore, = →+∞ ≤ Y Y 2 T xn λxn T xn λxn, T xn λxn T x 2 2λ T x , x λ2 x λ2 T x 2 2λ T x , x . Y − Y = a n − n− n f n n n n = Y Y − a f + Y Y ≤ + Y Y − a f 54 LECTURE NOTES, FUNCTIONAL ANALYSIS

2 2 A term T xn converges to λ and T xn, xn converges to λ. Thus, 2 Y Y aT xn λxf n 0

and y limn λxn. Since T is continuous and the sequence xn n N converges strongly, it holds that Y − Y → →+∞ ∈ = T y lim T λxn lim λT xn λy. { } n n Thus, λ σ T . Note that since the spectrum of T is contained in , the only possibility = →+∞ = →+∞ = R for a value of λ is T or T . ∈ ( ) Theorem 9.12. (SpectralY Y −Y theoremY for compact self adjoint operator) ◻ Let T α H,H be a self adjoint and compact operator. Then, there exists an orthonormal

system S ek k N and a sequence λi i N R converging to zero as i tends to infty, such that∈ ( ) ∈ = { ∶ ∈ } { } ⊂ (9.13) H ker T L2 lin S and = ( ) ⊕ ( ) T x λk x, ek ek, x H, +∞ i 1 where sup λk T . = Q= a f ∀ ∈ Proof ofS TheoremS = Y Y 9.12. We claim that dim ker λiId T . Indeed, if there were infinitely many v such that v ker λ Id T (which is equivalent to ( ( − )) < +∞ i T v λiv), then a dimension of the (scaled) unit ball would be infinite. However, in an infinite dimensional space the unit ball is not compact,∈ which( is− a) contradiction to the fact that=T is compact. In each space ker λiId T one can choose an orthonormal basis. Then, for k j we have T ek λkek and T ej λjej. Since ek ej, ker λ (Id T− )ker λ Id T . ≠ = k = j ⊥ Set λ1 to be equal to T or T . The corresponding space would be H1 ker λ1Id T . We make a decomposition ( − ) ⊥ ( − ) Y Y −Y Y H H1 L2 H1 = ( − ) ⊥ and restrict the operator to the subspace H1 . A norm of the restricted operator is not bigger that T . Iteration of this procedure= ⊕ leads⊥( to) the decomposition given by (9.13).

Now, the only fact which needs to be clarified( ) is that the sequence λi i N converges to 0. However,Y ifY λi i stabilized on some level, then it would be a contradiction with N ∈ compactness of T (argument similar to the one used for proving that{ a kernel} of λiId T ∈ is finite dimensional).{ } − ◻ LECTURE 10 Locally convex spaces

Definition 10.1. We define (a) Set A is called a disc set if λ λ 1 A A, (b) Set A is called an if it is convex and a holds. { ∶ S S ≤ } ⋅ ⊂ It can be shown that A is absolutely convex, if ( ) x1, x2 A, λ1 λ2 1 λ1x1 λ2x2 A.

Remark 26. In case∀K R∈a setS AS has+ S theS ≤ disc⇒ property,+ if it∈ is balanced. Thus, it is sufficient to check if 1 A A. = Remark 27. If a set−A⋅ is absolutely⊂ convex and absorbing, then pA x inf λ 0 x λA is a seminorm. ( ) = { > ∶ ∈ } Remark 28. If p is a seminorm, then A x X p x 1 is absolutely convex. = { ∈ ∶ ( ) ≤ } Definition 10.2. A topological linear space X, τ is a linear space X over a field K, which is endowed in a topology τ, such that vector addition and scalar multiplication are continuous (in the sense that a preimage of each( open) set is open). Definition 10.3. Let P be a family of on a linear space X and τ be generated by sets

UF,ε x p x ε, p F , where F is finite and F P . Then X, τ is called a locally convex space. Sets UF,ε form a basis of neighborhoods of 0. = { ∶ ( ) ≤ ∀ ∈ } ⊂ ( ) 55 56 LECTURE NOTES, FUNCTIONAL ANALYSIS

Theorem 10.4. A topological linear space X, τ is locally convex, if there exists a basis of neighborhoods of 0, which consists of absolutely convex and absorbing sets. ( ) Lemma 10.5. Let P be a family of seminorms which generates a locally convex topology τ on X. Then, the following conditions are equivalent (a) X, τ is a Hausdorff space, (b) x X, x 0 there exists p P such that p x 0, (c)( There) exists a family of open sets such that U U 0 . ∀ ∈ ≠ ∈ ( ) ≠ ∈U Proof of Lemma 10.5. Ua b Assume that ⋂X, τ is= a{ Hausdorff} space and let x 0. Then, there exist neighborhoods of zero U, V such that x U V . Without loss of generality we can assume( ) that⇒ ( ) ( ) ≠ ( + ) ∩ = ∅ V UF,ε u p u ε, p F . Since x V , there exists p such that p x ε p x 0. b c Let = = { ∶ ( ) ≤ ∀ ∈ } ∉ x UF,ε ( )x> p⇒x (ε,) ≠ p F . ( ) ⇒ ( ) F,ε F,ε Then, for all p P it holds that∈  p x = 0,{ which∶ ( implies) ≤ ∀c ∈. } c a Assume that x y. Then, there exists an open set U (which is an element of the basis of neighborhoods∈ of 0, see( ) Definition= 10.3) such( ) that x y U. Addition (and (subtraction)) ⇒ ( ) is a continuous≠ operation. Thus, the preimage of U is open and we can find two open sets W, V such that W V U, which implies x V −y ∉W .

Example 14. Below we give some− examples⊂ of families of( seminorms.+ ) ∩ ( + ) = ∅ ◻ n n (a) X Cb R , P pt, t R , where pt f f t (topology of pointwise conver- gence). (b) X =- normed( ) space,= { P X∈ , p}x′ x x (x ) (denoted= S ( )S as σ X,X ). (c) X - normed space, P X′, px x x′ x (denoted as σ X ,X′ ). = ( ) = S ( )S ( ) Lemma 10.6. Let X, τ be a locally′ convex′ space′ and P be a family′ of seminorms, which = ( ) = S ( )S ( ) generates a topology τ on X. Then, (a) For a seminorm( )q X 0, the following conditions are equivalent a.1 q is continuous, a.2 q is continuous∶ in→ 0[, +∞) (a.3) x q x 1 is an open neighborhood of 0. (b)( All p) P are continuous. (c)( The){ seminorm∶ ( )q< is} continuous, if and only if, there exists a finite family F P and a∈ constant M 0 such that q x M max p x x X. ⊂ > p F Proof of Lemma 10.6. Implications( ) ≤ a.∈1 ( )a.2∀ ∈ a.3 are straightforward. Assume that a.3 holds. Then, for x X and ε 0 we define U ε y q (y )1⇒ ( y) ⇒q (y )ε , ( ) ∈ > = ⋅ { ∶ ( ) < } = { ∶ ( ) < } LECTURE 10. LOCALLY CONVEX SPACES 57

which is an open set. Using the triangle inequality we obtain q x y q x q x y x q x q x q x y x , q x y q x q x y q x y q x y x q x y x , ( + ) − ( ) ≤ (( + ) − ) + ( ) − ( ) = (( + ) − ) which is equivalent( + ) − to( ) ≥ ( + ) − ( + ) − (( + ) − ) = − (( + ) − ) q x y q x q x y x . Therefore, S ( + ) − ( )S ≤ (( + ) − ) q x U a R a q x ε , which is in fact a definition of continuity. b According to the definition( + of )τ,⊂ sets{ ∈ x ∶ S p− x( )S <1 }are open and thus, all p are continuous (which follows from the claim a ). (c) According to the claim a , q is continuous,{ if∶ and( ) only< if,} there exists a finite set F P such that UF,ε x q x 1 holds, which( ) means that ( ) ( ) ⊂ q x 1 ε sup p x x X. ⊂ { ∶ ( ) ≤ } p F

( ) ≤ ( ~ ) ∈ ( ) ∀ ∈

Remark 29. Let P and Q be families of seminorms such that P Q q q is continuous◻. Then, topologies τP and τQ are equivalent. ⊂ ⊂ { ∶ } Definition 10.7. A space consisting of all linear and continuous functionals on a locally convex space X, τ is called a dual space X, τ (we shall also use a notation Xτ ). ′ ′ Theorem 10.8.( Let) X, τp and Y, τq be( locally) convex spaces and T X Y (be a) linear operator. Then, the following conditions are equivalent (a) T is continuous( (in) the sense( that) T 1 U is open for every open∶ set→ U), (b) T is continuous at 0, − (c) If q is a continuous seminorm, then q ( T)is a continuous seminorm, (d) For every q Q there exists a finite family F P and M 0 such that ○ q T x M max p x . ∈ p F ⊂ > The proof is similar to the proof of( the previous) ≤ ∈ theorem.( )

Remark 30. Let X, τp be a locally convex space and l X L be a linear operator. Then, l is continuous if and only if, there exist a finite subset p1, . . . , pn P and a constant M 0 such( that) ∶ → { } ∈ l x M max pi x x X. > i 1,...,n

= If τp1 τp2 , then we say that τ(p2 )is≤ stronger. In( ) particular,∀ ∈ it has more open sets, closed sets, continuous operators, but less convergent sequences (since it is harder to prove that n limn ⊂ p u u 0 for each seminorm p).

→+∞ ( − ) = 58 LECTURE NOTES, FUNCTIONAL ANALYSIS

Example 15. Let X Cb T and consider a topology τP generated by a family of semi- norms = ( ) pt x x t t T. On the other side consider a topology τ generated by a norm . One can easily show that ( ) = S ( )S ∀ ∈ ∞ ∞ pt x x pt P, Y ⋅ Y and thus the topology τ is stronger. ( ) ≤ Y Y∞ ∀ ∈ Remark 31. Consider a∞ norm topology on X and a weak topology σ X,X (or a weak topology σ X ,X ). We recall that σ X,X (respectively σ X ,X ) is generated′ by a∗ family of seminorms′ defined by linear functionals.′ One can show′ that( the unit) ball in the norm topology( is not) an open set in σ (X,X )(or σ X ,X ). ( ) Example 16. In the following example we′ shall show′ that if x is in a closure of a set A, ( ) ( ) it does not imply that there exists a sequence of elements converging to x. Let X, τ be 2 a locally convex space, where X l N and τ σ X,X σ X ,X . Let A e me 1 m n′ . ′ ( ) = (m ) n = ( ) = ( ) σ X,X′ Then, 0 A , but there is= no{ sequence+ ∶ xn≤n N < A} such that xn converges to 0 in σ X,X . ( ) ∈ Proof: Let′ ∈ U be any neighborhood of zero. If{ we} prove⊂ that U A , then 0 is in the closure( of) A indeed. Without loss of generality we can assume that 2 ∩ 2≠ ∅ U x l N x, yi ε i 1, . . . , r and yi l N . Let m be such that = { ∈ ( ) ∶ Ta fT ≤ } = ∈ ( ) yi m ε 2, for i 1, . . . , r and choose n m such that S ( )S < ~ = m y n ε 2 for i 1, . . . , r. > i Then, ⋅ S ( )S ≤ ~ = em men, yi yi m m yi n ε and therefore e me U. m n Ta + fT ≤ S ( )S + ⋅ S ( )S ≤ On the contrary, let emk mkenk k N be a sequence in A. If the sequence converged to zero in σ X,X ,+ it would∈ imply mk, nk . The norm of each element em mken ∈ k k should be bounded′ (uniformly{ + with} respect to k), since each weakly convergent sequence is bounded.( But,) → +∞ + 2 em men, em men 1 m , which is unbounded. a + + f = + Theorem 10.9. (Hahn-Banach theorem) Let X be a locally convex space and U be its subspace. Assume that l U . Then, there exists an extension L X . ′ ′ ∈ ∈