Lecture Notes in Functional Analysisaswiercz/AnalizaF/lecture.pdf · Lecture Notes in Functional...

Lecture Notes in Functional Analysis

Please report any mistakes, misprints or comments [email protected]

LECTURE 1Banach spaces

1.1. Introducton to Banach Spaces

Definition 1.1. Let X be a K–vector space. A functional p ∶ X → [0,+∞) is called aseminorm, if

(a) p(λx) = ∣λ∣p(x), ∀λ ∈ K, x ∈X,(b) p(x + y) ≤ p(x) + p(y), ∀x, y ∈X.

Definition 1.2. Let p be a seminorm such that p(x) = 0 ⇒ x = 0. Then, p is a norm(denoted by ∥ ⋅ ∥).Definition 1.3. A pair (X, ∥ ⋅ ∥) is called a normed linear space.

Lemma 1.4. Each normed space (X, ∥ ⋅ ∥) is a metric space (X,d) with a metric given byd(x, y) = ∥x − y∥.

Definition 1.5. A sequence xnn∈N in a normed space (X, ∥ ⋅ ∥) is called a Cauchy se-quence, if

∀ε > 0 ∃N(ε) ∈ N ∀n,m ≥ N(ε) ⇒ ∥xn − xm∥ ≤ ε.Definition 1.6. A sequence xnn∈N converges to x (which is denoted by limn→+∞ xn = x),if

∀ε > 0 ∃N(ε) ∈ N ∀n ≥ N(ε) ⇒ ∥xn − x∥ < ε.Definition 1.7. If every Cauchy sequence xnn∈N converges in X, then (X, ∥ ⋅ ∥) is calleda complete space.

Definition 1.8. A normed linear space (X, ∥ ⋅ ∥) which is complete is called a Banachspace.

1

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.

1.2. Examples of Banach spaces

Example 1. Let B(T ) be a space consisting of all bounded maps x ∶ T Ð→ K. For eachx ∈ B(T ) we set

∥x∥∞ = supt∈T

∣x(t)∣.

Then, (B(T ), ∥ ⋅ ∥∞) is a Banach space. To prove the assertion we need to show that

(a) ∥ ⋅ ∥∞ is a norm,(b) each Cauchy sequence converges to an element from B(T ).

Concerning claim (a), let λ ∈ K and x ∈ B(T ). Then,

∥λx∥∞ = supt∈T

∣λ ⋅ x(t)∣ = ∣λ∣ ⋅ supt∈T

∣x(t)∣ = λ∥x∥∞.(1.10)

Let to ∈ T and x, y ∈ B(T ). Then,

∣x(to) + y(to)∣ ≤ ∣x(to)∣ + ∣y(to)∣ ≤ supt∈T

∣x(t)∣ + supt∈T

∣y(t)∣ = ∥x∥∞ + ∥y∥∞.

The right hand side of the inequality above is independent on t. Therefore, 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 supt∈T ∣x(t)∣ = 0. This impliesthat ∣x(t)∣ = 0 for each t. Thus, x = 0.

Now, we shall prove the statement (b). Let xn be a Cauchy sequence. Then, for everyε > 0 there exists N = N(ε) such that for all n,m ≥ N it holds that ∥xn − xm∥∞ < ε. Inparticular,

∣xn(t) − xm(t)∣ < ε, ∀t ∈ T.Thus, for any t ∈ T the sequence xn(t)n∈N converges to some x(t), due to the completenessof K (real and complex numbers are complete spaces). Define a candidate for a limit ofthe sequence xnn∈N, that is, x ∶ T Ð→ K as

x(t) = limn→+∞

xn(t).

If follows from the statement above that there exists No = No(ε, t) such that

(1.12) ∣xn(t) − x(t)∣ < ε, ∀n ≥ No.

Without loss of generality we can assume that No(ε, t) ≥ N(ε) for each t ∈ T . Then, forn ≥ N it holds that

∣xn(t) − x(t)∣ ≤ ∣xn(t) − xNo(ε,t)(t)∣ + ∣xNo(ε,t)(t) − x(t)∣≤ ∥xn − xNo(ε,t)∥∞ + ε < 2ε,

LECTURE 1. BANACH SPACES 3

where we used the fact that xnn∈N is a Cauchy sequence and (1.12). Moreover, for eacht ∈ T and N = N(ε) we have

∣x(t)∣ ≤ ∣xN(t)∣ + ∣xN(t) − x(t)∣ ≤ ∥xN∥∞ + 2ε,

which implies ∥x∥∞ ≤ ∥xN∥∞ + 2ε and so forth x ∈ B(T ).

Example 2. Let T be a metric space and Cb(T ) be a space of bounded continuous func-tions on T . Then, (Cb, ∥ ⋅ ∥∞) is a Banach space.

To prove the assertion, it 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 inB(T ) converges to a point from Cb(T ). Let xnn∈N be a sequence of bounded, continuousfunctions convergent to x ∈ B(T ). We need to show that x is a continuous function. Forany ε > 0 there exists N = N(ε) ∈ N, such that ∥xN − x∥∞ < ε/3, since the sequence isconvergent. Now, let to ∈ T . By the continuity of xN , there exists δ = δ(ε, to) > 0 such that

d(t, to) < δ Ô⇒ ∣xN(t) − xN(to)∣ < ε/3.Therefore, for all t such that d(t, to) < δ it holds that

∣x(t) − x(to)∣ ≤ ∣x(t) − xN(t)∣ + ∣xN(t) − xN(to)∣ + ∣xN(to) − x(to)∣≤ 2∥x − xN∥∞ + ∣xN(t) − xN(to)∣ ≤ 2 ⋅ ε/3 + ε/3 = ε,

which ends the proof.

Example 3. A space of continuous functions vanishing at infinity

Co(Rn) = f ∈ Cb(Rn) ∶ lim∣t∣→+∞

∣f(t)∣ = 0

with a ∥ ⋅ ∥∞ norm is a Banach space.

Example 4. The following spaces

co = tnn∈N ∶ tn ∈ K, limn→+∞

tn = 0 ,

c = tnn∈N ∶ tn ∈ K, limn→+∞

tn exists

with a ∥ ⋅ ∥∞ norm are Banach spaces.

Remark 1. B(N) is often denoted by l∞.

1.3. lp spaces

Definition 1.13. Let 1 ≤ p < +∞. We define

lp = x ∈ l∞ ∶+∞∑n=1

∣xn∣p < +∞ and ∥x∥p =p

¿ÁÁÀ

+∞∑n=1

∣xn∣p.

Lemma 1.14. (Holder inequality for sequences). Let 1 ≤ p, q ≤ +∞ be such that 1p +

1q = 1.

Then, for x ∈ lp and y ∈ lq it holds that


(a) x ⋅ y ∈ l1,(b) ∥x ⋅ y∥1 ≤ ∥x∥p ⋅ ∥y∥q.

Lemma 1.15. (Minkowski inequality for sequences). Let 1 ≤ p ≤ +∞ and x, y ∈ lp. Then

∥x + y∥p ≤ ∥x∥p + ∥y∥p.

Example 5. Spaces (lp, ∥ ⋅ ∥p) are Banach spaces for 1 ≤ p ≤ +∞.

Space (l∞, ∥ ⋅ ∥∞) coincides with (B(N), ∥ ⋅ ∥∞), therefore we assume that p < +∞. Similarlyas in the Example 1, to prove the assertion we need to show that

(a) ∥ ⋅ ∥p is a norm,(b) each Cauchy sequence converges to an element from lp.

Claim (a) is straightforward, when one uses the Minkowski inequality for the proof of thetriangle inequality. For the proof of completness, consider a Cauchy sequence xnn∈N.Each element xn ∈ lp is a sequence given by xn = (xn1 , xn2 , . . . ). Note that lp ⊂ l∞, whatholds due to the following estimate

∥x∥p =p

¿ÁÁÀ

+∞∑k=1

∣xk∣p ≥ p

√supk∈N

∣xk∣p = supk∈N

∣xk∣ = ∥x∥∞,

for x = (x1, x2, . . . ). Thus, if we consider the sequence xnn∈N as a sequence of elements ofl∞ space, we conclude that there exists exactly one x ∈ l∞ such that limn→+∞ ∥xn − x∥∞ = 0(this follows from the completeness of (l∞, ∥ ⋅ ∥∞)). In particular,

(1.16) limn→+∞

xnk = xk, for each k ∈ N.

We shall show that x = xkk∈N is an element of lp space and that xnn∈N converges to xin lp. For any ε > 0 there exists N = N(ε), such that for all n,m ≥ N it holds that

∥xn − xm∥p < ε.

In particular, for every K ∈ N

p

¿ÁÁÀ K

∑k=1

∣xnk − xmk ∣p ≤ ∥xn − xm∥p < ε.

Using (1.16) and passing to the limit with xmk we obtain

p

¿ÁÁÀ K

∑k=1

∣xnk − xk∣p < ε.

Since the estimate is valid for all K and the right hand side of the inequality is independenton K, it holds also that

p

¿ÁÁÀ

+∞∑k=1

∣xnk − xk∣p < ε.


Therefore ∥xn − x∥p < ε, which proves that x is a limit of the sequence in lp. Moreover,

∥x∥p ≤ ∥x − xN∥p + ∥xN∥p ≤ ε + ∥xN∥p < +∞.

1.4. Minkowski functional

Definition 1.17. Set A is called an absorbing set if for each x ∈X there exists t ∈ K, suchthat t ⋅ x ∈ A.

Definition 1.18. Set A is called a balanced set if x ∈ A⇒ −x ∈ A.

Definition 1.19. Let A be a convex, absorbing and balanced set. A functional µA ∶X Ð→[0,+∞) defined by

(1.20) µA(x) = inf t ∈ (0,+∞) ∶ xt∈ A

is called Minkowski functional.

Lemma 1.21. Minkowski functional generates a seminorm on X. If additionally A isbounded in each direction, that is, for each x ∈X a set (A ∩ linx) is a bounded set, thenit is a norm.

Proof. We shall concentrate on the essential part of the proof, that is, showing that µAfulfills 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 isconvex, therefore

t

t + s⋅ xt+ s

t + s⋅ ys= x + yt + s

∈ A,which implies that

µA(x + y) = inf z ∈ (0,+∞) ∶ x + yz

∈ A ≤ t + s = µA(x) + µA(y) + 2ε.

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 Minkowskifunctional

Let F = F (Ω) be a space of real valued, Lebesgue measurable functions on Ω.

Example 6. Orlicz spaces LM(Ω).Let M be a non-negative convex function on [0,+∞), such that

M(0) = 0 and limt→+∞

M ′(t) = +∞.

Define a set

(1.22) A = f ∈ F ∶ ∫ΩM (∣f(x)∣)dx ≤ 1 .

Orlicz space LM(Ω) is the smallest linear space containing A. It can be checked thatMinkowski functional µA defines a norm on LM(Ω).


Example 7. Lebesgue spaces

Lp(Ω) = f ∈ F ∶ ∫Ω∣f(x)∣pdx < +∞ .

The most important class of Orlicz spaces arises when we set M(x) = xp, where 1 < p < +∞.In this case we obtain Lebesgue spaces Lp(Ω). Analogously as in the example above,

A = f ∈ F ∶ ∫Ω∣f(x)∣pdx ≤ 1 .

It turns out that Minkowski functional µA is given by the following formula

µA(f) = p

√∫

Ω∣f(x)∣pdx.

Note that A is a convex, absorbing and balanced set, therefore µA is a seminorm on Lp(Ω).Moreover, if µA(f) = 0, then ∫Ω ∣f(x)∣pdx = 0, which implies f = 0 a.e. Thus, µA defines anorm on Lp(Ω).

Example 8. Generalized Lebesgue spaces

Lp(⋅)(Ω) = f ∈ F ∶ ∫Ω∣f(x)∣p(x)dx < +∞ .

The next important class of Orlicz spaces is created when one sets M(x) = xp(x), wherep(x) fulfills 1 < p1 ≤ p(x) ≤ p2 < +∞ for some p1, p2. In this case we obtain generalizedLebesgue spaces Lp(⋅)(Ω). Similarly as before,

A = f ∈ F ∶ ∫Ω∣f(x)∣p(x)dx ≤ 1 .

1.5. Lp spaces.

Definition 1.23. Let 1 ≤ p < +∞. The space Lp(Ω) consists of equivalence classes ofLebesgue measurable functions f ∶ Ω→ R such that

∫Ω∣f(x)∣dx < +∞,

where two measurable functions are equivalent if they are equal a.e. The Lp norm off ∈ Lp(Ω) is defined by

∥ ⋅ ∥Lp = p

√∫

Ω∣f(x)∣dx.

For p = +∞ the definition is slightly different. We say that a function f is essentiallybounded, if

essup∣f ∣ = infN ∶ ∣N ∣=0

sup(Ω/N)

∣f(x)∣ < +∞.

The space L∞(Ω) consists of equivalence classes (two functions are equivalent if they areequal a.e.) of measurable, essentially bounded functions f ∶ Ω→ R with a norm

∥ ⋅ ∥L∞ = essup∣f ∣.


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 characteristicfunction of the rational numbers Q is equivalent to 0 in Lp(R), for 1 ≤ p ≤ +∞.

Lemma 1.24. (Holder inequality for integrals). Let 1 ≤ p ≤ +∞ and 1p +

1g = 1. Let

f ∈ Lp(Ω), g ∈ Lq(Ω). Then, f ⋅ g ∈ L1(Ω) and

∥f ⋅ g∥L1 ≤ ∥f∥Lp ⋅ ∥g∥Lq .

Theorem 1.25. Orlicz spaces LM(Ω), Lebesgue spaces Lp(Ω) and generalized Lebesguespaces Lp(⋅)(Ω) are Banach spaces.

We prove the theorem only for the Lebesgue spaces. In the proof we shall use the followinglemma.

Lemma 1.26. For any normed space (X, ∥ ⋅ ∥) the following conditions are equivalent

(a) (X, ∥ ⋅ ∥) is a complete space.(b) If xnn∈N is a sequence in X, such that ∑+∞

n=1 ∥xn∥ < +∞, then there exists x ∈ Xsuch that

limN→+∞

∥N

∑n=1

xn − x∥ = 0.

The condition (b) simply states that any absolutely convergent series is convergent.

Proof of Lemma 1.26.(a) ⇒ (b). The implication follows from the fact that SN = ∑N

n=1 xn is a Cauchy sequence.(b) ⇒ (a). Let xnn∈N be a Cauchy sequence. For each k ∈ N there exists Nk, such that

∥xm − xn∥ < 2−k, ∀n,m ≥ Nk.

We choose a subsequence xnkk∈N such that

∥xnk+1− xnk

∥ < 2−k, ∀k ∈ N

and denote y1 = xn1 , yk = (xnk+1− xnk

) for k > 1. Therefore ∑+∞i=1 ∥yi∥ < +∞. From

assumptions it follows that there exists y ∈X, such that

limN→+∞

∥N

∑n=1

yn − y∥ = limN→+∞

∥xnN+1− y∥ = 0.

Therefore, the subsequence xnkk∈N converges in X. A Cauchy sequence, which has a

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 Example7. For the proof of completeness we shall use claim (b) from Lemma 1.9. Let fnn∈N be


a sequence in Lp(Ω) such that M = ∑+∞n=1 ∥fn∥Lp < +∞. We need to construct a function

f ∈ Lp(Ω), such that limN→+∞ ∥∑Nn=1 fn − f∥Lp = 0. Define gn, g ∶ Ω→ R as following

gn(x) =n

∑i=1

∣fi(x)∣ and g(x) =+∞∑n=1

∣fn(x)∣.

From Minkowski inequality we obtain

∥gn∥Lp = ∥n

∑i=1

∣fi∣ ∥Lp ≤n

∑i=1

∥fi∥Lp ≤+∞∑n=1

∥fn∥Lp =M < +∞,

By construction, gn converges monotonically to g. Therefore, from the monotone conver-gence theorem and the inequality above it follows that

∫Ω(g(x))p dx = ∫

Ωlimn→+∞

(gn(x))p dx = limn→+∞∫Ω

(gn(x))p dx <Mp.

which implies that g ∈ Lp(Ω) and in particular g is finite a.e. From the latter fact weconclude that

f(x) ∶=+∞∑n=1

fn(x)

is finite a.e. and f ∈ Lp(Ω) with ∥f∥Lp ≤ ∥g∥Lp . Note that

0 ≤ ∣f(x) −n

∑i=1

fi(x)∣p

= ∣+∞∑i=n+1

fi(x)∣p

≤ (∞

∑i=n+1

∣fi(x)∣)p

≤ (g(x))p <Mp.

Thus, by the Lebesgue dominated convergence theorem

limn→+∞∫Ω

∣f(x) −n

∑i=1

fi(x)∣p

dx = 0,

which ends the proof due to the Lemma 1.26. ◻

LECTURE 2Linear Operators

Definition 2.1. Let (E, ∥ ⋅ ∥E) and (F, ∥ ⋅ ∥F ) be normed spaces. A space consisting oflinear, bounded operators A ∶ E → F is denoted as α(E,F ). A bounded operator isunderstood here as the operator, which maps bounded sets onto bounded sets.

Remark 3. α(E,F ) is a normed space with a norm given by

∥A∥α(E,F ) = supx∈E,x≠0

∥A(x)∥F∥x∥E

.

Remark 4. For any A ∈ α(E,F ) the following equalities hold

(2.2) ∥A∥α(E,F ) = supx ∶ ∥x∥E≤1

∥A(x)∥F = supx ∶ ∥x∥E=1

∥A(x)∥F .

Theorem 2.3. Let (E, ∥ ⋅ ∥E) and (F, ∥ ⋅ ∥F ) be normed spaces and T ∶ E → F be a linearoperator. Then, the following conditions are equivalent:

(a) T is continuous,(b) T is continuous at 0,(c) there exists M such that for each x ∈ E it holds that ∥Tx∥F ≤M∥x∥E,(d) T is uniformly continuous.

Proof of Theorem 2.3.(c)⇒ (d). Condition (c) implies that T is Lipschitz continuous and thus, also uniformlycontinuous.(d)⇒ (a)⇒ (b). The proof is trivial.(b)⇒ (c). We shall prove this implication by contradiction. Assume that T is continuousat 0 and (c) does not hold, which implies that there exists a sequence xnn∈N such that

∥Txn∥F > n∥xn∥E.9


Define yn =xn

n∥xn∥E. Then, ∥yn∥E = 1/n and limn→+∞ ∥yn − 0∥E = 0. Moreover,

∥Tyn∥F =∥Txn∥Fn∥xn∥E

> 1.

Therefore, limn→+∞ ∥yn − 0∥E = 0 and ∥Tyn − T (0)∥F > 1, which is a contradiction due tothe fact that T is continuous. ◻

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 T such that

T ∶ E Ð→ F, T ∣D = T, and ∥T ∥α(E,F ) = ∥T ∥α(D,F ).

Proof of Lemma 2.5. For each x ∈ E there exists a sequence xnn∈N ⊂ D, such thatlimn→+∞ ∥xn − x∥E = 0. Define an operator

T x = limn→+∞

Txn.

The operator T is well defined, since Txnn∈N is a Cauchy sequence and F is a Banach

space, thus the limit exists. Note that T is unique (it does not depend on the choice of thesequence xnn∈N). Indeed, if there exist sequences xnn∈N and x′nn∈N such that

limn→+∞

xn = limk→+∞

x′k = x,

then limn,k→+∞ ∥xn − x′k∥E = 0 and from the boundedness of T ,

limn,k→+∞

∥Txn − Tx′k∥F ≤M limn,k→+∞

∥xn − x′k∥E = 0.

Moreover, T is bounded since

∥T x∥F = limn→+∞

∥Txn∥F ≤ limn→+∞

M∥xn∥E ≤ supn∈N

M∥xn∥E < C.

It also holds that

∥T ∥α(E,F ) = supx∈E, x≠0

∥T (x)∥F∥x∥E

= supx∈D,x≠0

∥T (x)∥F∥x∥D

= supx∈D,x≠0

∥T (x)∥F∥x∥D

= ∥T ∥α(D,F ).

◻

LECTURE 2. LINEAR OPERATORS 11

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

Proof of Theorem 2.6. We need to prove that a normed space (α(E,F ), ∥ ⋅ ∥α(E,F )) is

complete. Let Ann∈N be a Cauchy sequence in α(E,F ), that is,

(2.7) ∀ε > 0 ∃N = N(ε) ∀n,m ≥ N ∥An −Am∥α(E,F ) = supx∈E, x≠0

∥Anx −Amx∥F∥x∥E

< ε,

which implies that Anxn∈N is a Cauchy sequence in F for each fixed x ∈ E as well. SinceF is a Banach space, this sequence is convergent. Thus, an operator A ∶ E → F given by

Ax = limn→+∞

Anx

is well defined. It is also linear and bounded. The latter claim holds due to the following

∥Ax∥F∥x∥E

= limn→+∞

∥Anx∥F∥x∥E

≤ supn∈N

∥Anx∥F∥x∥E

< +∞.

Now, we want to prove that limn→+∞ ∥An −A∥α(E,F ) = 0. Let x be such that ∥x∥E = 1.

Condition (2.7) implies then

∀ε > 0 ∃N = N(ε) ∀n,m > N ∥Anx −Amx∥F < ε.

If we let m→ +∞, then ∥Anx −Ax∥F < ε. This holds for each x such that ∥x∥E = 1, therefore

supx ∶ ∥x∥E=1

∥Anx −Ax∥F < ε,

which ends the proof due to the equality (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. Let x ∈X and x ≠ 0. Then,

∥(T S)(x)∥Z = ∥T (Sx)∥Z ≤ ∥T ∥α(Y,Z) ⋅ ∥Sx∥Y ≤ ∥T ∥α(Y,Z) ⋅ ∥S∥α(X,Y ) ⋅ ∥x∥X .

Thus,

∥(T S)(x)∥Z∥x∥X

≤ ∥T ∥α(Y,Z) ⋅ ∥S∥α(X,Y ).

Taking supremum over x ∈X ends the proof. ◻


Example 9. Examples of linear operators T ∶X → Y :

(a) Identity : X = Y , T = Id. The norm ∥T ∥α(X,Y ) of T is equal to 1.

(b) A linear map between finite dimensional spaces : X = Rm, Y = Rn and A ∈ Rn×Rm

is a matrix A = aij, 1 ≤ i ≤ n, 1 ≤ j ≤m,

T (x) = A ⋅ x.

(c) Differentiation : X = C1([0,1]), Y = C([0,1]),

T (f) = f ′.

(d) Let g ∈ Lq([0,1], µ) and p, q be such that 1p +

1q = 1. X = Lp([0,1], µ), Y =

L1([0,1], µ),

Tg(f) = ∫1

0fg dµ.

(e) Fredholm integral operator : X = Y = C([0,1]),

(T (f))(y) = ∫1

0k(x, y)f(x)dx,

where k ∈ C([0,1] × [0,1]).(f) An operator defined analogously as in the example (e), but withX = Y = L2([0,1], µ)

and k ∈ L2([0,1] × [0,1], µ).

Remark 5. (to example (e)) Note that continuity of T (f) follows directly from the uniformcontinuity of k. Indeed, from a definition of uniform continuity

∀ε > 0 ∃δ > 0 such that (∣y − y′∣ < δ ⇒ supx∈[0,1]

∣k(x, y) − k(x, y′)∣ < ε) .

Therefore, it also holds that

∣(T (f))(y) − (T (f))(y′)∣ ≤ ∫1

0∣k(x, y) − k(x, y′)∣ ⋅ ∣f(x)∣ dx ≤ ε ∥f∥∞.

Note that

∥Tf∥∞ = supy∈[0,1]

∣∫1

0k(x, y)f(x)dx∣ ≤ ∥f∥∞ ⋅ sup

y∈[0,1]∥k(⋅, y)∥∞ ≤ ∥f∥∞∥k∥∞.

Definition 2.9. An operator T ∈ α(X,Y ) is called an isomorphism, if there exists aninverse operator T −1 ∈ α(Y,X). Moreover, if

∥T ∥α(X,Y ) = ∥T −1∥α(X,Y ) = 1,

then T is called an isometry and X, Y are said to be isometrically isomorphic.

Two isometrically isomorphic normed spaces share the same structure, so they are usuallyidentified 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, thefollowing implication holds

+∞∑n=0

T n converges in α(X) ⇒ (Id − T )−1 exists and (Id − T )−1 =+∞∑n=0

T n,

where T n = T ⋅ ⋅ ⋅ T´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶n times

.

Proof of Theorem 2.10. Define Sm = ∑mn=0 T

n. Then, (Id − T )Sm = Sm(Id − T ) and

(Id − T )Sm = Id Sm − T Sm =m

∑n=0

T n −m+1

∑n=1

T n = Id − Tm+1.

Since ∑+∞n=0 T

n converges in α(X), we have that limn→+∞ T nx = 0 for each x ∈X. Therefore,passing to the limit in the equalities above yields

Id(x) = limm→+∞

(Id − Tm+1)x = limm→+∞

(Id − T )Smx = (Id − T ) limm→+∞

Smx = (Id − T )(+∞∑n=0

T n)x.

Similar argument proves that Id(x) = (∑+∞n=0 T

n)(Id − T )x. Therefore,

(Id − T )−1x =+∞∑n=0

T nx, ∀x ∈X.

◻

Remark 6. If (X, ∥ ⋅ ∥)X is a Banach space it is sufficient to assume that ∥T ∥α(X) < 1. Inthis case

(2.11) ∥(Id − T )−1∥α(X) ≤1

1 − ∥T ∥α(X).

Indeed, note that+∞∑n=0

∥T n∥α(X) ≤+∞∑n=0

∥T ∥nα(X) < +∞,

since ∥T ∥α(X) < 1. Since X is a Banach space, then the convergence of the series ∑+∞n=0 T

n

follows from Lemma 1.26. The inequality (2.11) follows from a formula on a sum of ageometric series.

Exercise 1. Find a solution x ∈ C([0,1]) to the following equation

x(s) − ∫1

0k(s, t)x(t) dt = y(s),

where k ∈ C([0,1] × [0,1]) and y ∈ C([0,1]) are given.

LECTURE 3Dual 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 dual space to X. It is denoted by X ′ or X∗.

Remark 7. Dual space X ′ is a Banach space even if X is not a Banach space.

Theorem 3.2. Let 1 ≤ p, q < +∞ be such that 1p +

1q = 1. Then, an operator

T ∶ lq → (lp)′, (Tx)(y) =+∞∑n=1

xnyn,

where x = (x1, x2, . . . ) ∈ lq, y = (y1, y2, . . . ) ∈ lp, is an isometric isomorphism.

Proof of Theorem 3.2. Linearity of T is straightforward. T is bounded, which followsfrom Holder inequality

(3.3) ∣(Tx)(y)∣ ≤ ∥x∥q ⋅ ∥y∥p ⇒ ∥Tx∥(lp)′ ≤ ∥x∥q.

To show that T is an isomorphism we need to prove that it is injective and surjective.Indeed,

Tx = 0 ⇒ xn = (Tx)(en) = 0, ∀n ∈ N,where en = (0, . . . ,0,1,0, . . . ), that is, the n-th coordinate is the only non-zero element ofen. Thus, ker(T ) = 0, which proves that T is an injection. For the proof of surjectivitywe 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 = en, n ∈ N. Indeed, assume thatT (x)(en) = y′(en) for each n ∈ N. Since T (x) and y′ are linear, then the equality holdsfor all y ∈ linen ∶ n ∈ N. However, the operators are also continuous and therefore theequality holds for

linen ∶ n ∈ N∣∣⋅∣∣p = lp

15


as well. Now, fix y′ ∈ (lp)′ and define sn = y′(en), x = snn∈N. We prove that x ∈ lq. Tothis end, let tnn∈N be a sequence given by

tn = ∣sn∣qsn

for sn ≠ 0,0 for sn = 0.

Therefore, for each N ∈ N it holds that

N

∑n=1

∣tn∣p =N

∑n=1

∣sn∣p(q−1) =N

∑n=1

∣sn∣q

andN

∑n=1

∣sn∣q =N

∑n=1

tnsn =N

∑n=1

tny′(en) = y′ (

N

∑n=1

tnen)

≤ ∥y′∥(lp)′ ⋅ (N

∑n=1

∣tn∣p)1p

= ∥y′∥(lp)′ ⋅ (N

∑n=1

∣sn∣q)1p

.

Therefore,

(N

∑n=1

∣sn∣q)1q

≤ ∥y′∥(lp)′ ⇒ ∥x∥q ≤ ∥y′∥(lp)′ ⇒ x ∈ lq,

since the first inequality holds for all N ∈ N. It is straightforward now, that

T (x)(en) = sn = y′(en).The fact that ∥x∥q ≤ ∥y′∥(lp)′ together with (3.3) imply that T is an isometry. ◻

Theorem 3.4. Let 1 ≤ p < +∞ and q be such that 1p +

1q = 1 and (Ω,Σ, µ) be a measure

space, where µ is a σ-finite measure. Then,

T ∶ Lq(Ω, µ)→ (Lp(Ω, µ))′ , (Tg)(f) = ∫Ωfg dµ

is an isometric isomorphism.

Proof of Theorem 3.4. T is clearly a linear operator. It follows from Holder inequalitythat

∣T (g)(f)∣ ≤ ∥fg∥L1 ≤ ∥f∥Lp ⋅ ∥g∥Lq ⇒ ∥Tg∥(Lp)′ ≤ ∥g∥Lq .

Thus, the norm of the operator T is at most 1. In particular, the function f defined asbelow

f = g

∣g∣( ∣g∣∥g∥Lq

)qp

, for 1 < p < +∞,

(where g is a complex conjugate of g) is such that ∥f∥Lp = 1 and

(Tg)(f) = ∫ fg dµ = ∥g∥Lq ,

which implies that ∥Tg∥(Lp)′ ≥ ∥g∥Lq and thus T is an isometry. For p = 1 we set f = g∣g∣

and use similar arguments.

LECTURE 3. DUAL SPACES 17

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

∫Ωfg dµ = 0, ∀f ∈ Lp(Ω, µ).

Thus, g = 0 µ − a.e. and g = 0 in Lq(Ω, µ). Now, we need to show that the operator issurjective, that is,

∀y′ ∈ (Lp)′ ∃g ∈ Lq such that T (g) = y′, i.e., (Tg)(f) = ∫Ωfg dµ = y′(f), ∀f ∈ Lp.

Assume that µ(Ω) < +∞ and define

ν ∶ Σ→ K, ν(E) = y′(χE).

Since µ is finite, χE ∈ Lp(Ω, µ) and thus ν is well defined. It is clear that ν is additive andin 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,

µ(E) = 0 ⇒ χE = 0 µ − a.e. ⇒ χE ∈ Lp(Ω, µ).

Since χE = 0 in Lp(Ω, µ), it holds that

ν(E) = y′(χE) = y′(0) = 0.

By Radon-Nikodym theorem, there exists g ∈ L1(Ω, µ) such that

y′(χE) = ν(E) = ∫Eg dµ = ∫

ΩχEg dµ, ∀E ∈ Σ.

Characteristic functions are dense in (L∞, ∥ ⋅ ∥L∞) (and in (Lp, ∥ ⋅ ∥Lp) ), which implies that

(3.5) y′(f) = ∫Ωfg dµ ∀f ∈ L∞(Ω, µ).

Now, it remains to prove that g ∈ Lq(Ω, µ). Cases for q < +∞ and q = +∞ are consideredseparately. Assume that q < +∞ and define a µ-measurable function

f(x) = ∣g(x)∣qg(x) if g(x) ≠ 0,

0 if g(x) = 0.

Then, ∣g(x)∣q = (fg)(x) = ∣f(x)∣p. Define a µ-measurable set

En = x ∶ ∣g(x)∣ ≤ n .

Then, χEnf ∈ L∞(Ω, µ) and thus we can plug it into (3.5).

∫Ω(χEnf)g dµ = y′(χEnf) ≤ ∥y′∥(Lp)′ ⋅ ∥χEnf∥Lp

= ∥y′∥(Lp)′ (∫En

∣f ∣p dµ)1p

= ∥y′∥(Lp)′ (∫En

∣g∣q dµ)1p

,

On the other hand

∫Ω(χEnf)g dµ = ∫

En

∣g∣q dµ.


Combining the both expressions we obtain

(∫En

∣g∣q dµ)1q

≤ ∥y′∥(Lp)′ .

The inequality above holds also when the integral is taken over the whole Ω. Indeed,

supn∈N

∫En

∣g∣q dµ = limn→N∫En

∣g∣q dµ = ∫Ω∣g∣q dµ,

due to the monotone convergence theorem. Thus,

(∫Ω∣g∣q dµ)

1q

= ∥g∥Lq ≤ ∥y′∥(Lp)′ .

Now, consider the case q = +∞. Define

E = x ∶ g(x) > ∥y′∥(L1)′ and f(x) = χE(x)∣g(x)∣g(x)

.

It is clear that f ∈ L∞(Ω, µ) and ∥f∥L1 = µ(E). If µ(E) > 0, then

µ(E)∥y′∥(L1)′ < ∫E∣g∣ dµ = ∫

Ωfg dµ = y′(f) ≤ ∥y′∥(L1)′ ⋅ ∥f∥L1

and µ(E) < ∥f∥L1 , which is a contradiction. ◻

Definition 3.6. Let Σ be a σ-algebra of subsets of Ω and K = N,C. ν ∶ Σ→ K is calleda 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 asigned (or complex) measure on Σ. Moreover, let ν be absolutely continuous with respectto µ, that is,

µ(E) = 0 ⇒ ν(E) = 0.

Then, there exists g ∈ L1(Ω, µ) such that

ν(E) = ∫Eg dµ, ∀E ∈ Σ.

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.


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 → R, such that

L∣U = l and L(x) ≤ p(x), ∀x ∈X.

In the proof we shall use the following lemma.

Lemma 3.10. (Zorn lemma) Let (A,≤) be a non-empty partially ordered set, such thatevery non-empty totally ordered subset has an upper bound in A. Then, the set A containsat least one maximal element.

A totally ordered set is the set with partial order under which every pair of elements iscomparable.

Proof of Theorem 3.9. Assume that dim(X/U) = 1. With this assumption our task isto extend the functional l to X = U ⊕ Rxo, where xo ∈ X/U . Note, that in this case anyx ∈X can be written as x = u + λxo, where u ∈ U and λ ∈ R. Define

(3.11) Lr(x) = l(u) + λr.

We claim that Lr is a linear extension of l, such that Lr(x) ≤ p(x). If λ = 0, we havethat 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 + λxo) − l(u)

r ≤ p(uλ+ xo) − l (

u

λ)

r ≤ infv∈U

(p(v + xo) − l(v)).

Similarly, for λ < 0 we obtain

λr ≤ p(u + λxo) − l(u)

−r ≤ p( u

−λ− xo) − l (

u

−λ)

r ≥ supw∈U

(l(w) − p(w − xo)).

Therefore, there exists r ∈ R such that Lr ≤ p, if

l(w) − p(w − xo) ≤ p(v + xo) − l(v),

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+xo)+p(w−xo), since v+w ∈ U and p is sublinear.


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)

(Vi, Li) ∶ i ∈ Iin A has an upper bound. Let V = ⋃i∈I Vi. Thus, V is a subset of X and V contains eachVi. Let define L ∶ V → R in the following way. If x ∈ V , then there exists i ∈ I such thatx ∈ Vi and we set L(x) = Vi(x). The definition does not depend on the choice of i becauseof the chain definition. Moreover, it implies that L is linear and (Vi, Li) ≤ (V , L) for everyi ∈ I. In other words this is an upper bound for the chain. Zorn’s lemma guarantees theexistence of a maximal element (V,L) ∈ A. If V ≠ X, then it follows from the first part ofthe proof, that we can extend the functional and thus, it is a contradiction due to the factthat (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(λ1x1 + λ2x2) = λ1l(x1) + λ2(x2), ∀λ1, λ2 ∈ R, x ∈X,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 thefollowing equivalence holds

∣h(x)∣ ≤ p(x) ⇔ ∣Re h(x)∣ ≤ p(x).(d) If (X, ∥ ⋅ ∥X) is a normed space and h ∶X → C is a C-linear continuous functional,

then

∥h∥α(X,C) = ∥Re h∥α(X,R).

Theorem 3.13. (Hahn-Banach theorem for C) Let X be a C-vector space and U be itssubspace. 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 → C, such that

L∣U = l and Re L(x) ≤ p(x), ∀x ∈X.


Theorem 3.14. Let (X, ∥ ⋅ ∥X) be a normed space and U be its subspace. Then, for everycontinuous linear functional u′ ∶ U → K there exists a continuous linear funtional x′ ∶ X →K, such that

x′∣U = u′ and ∥x′∥α(X,K) = ∥u′∥α(U,K).

Proof of Theorem 3.14. (only for K = R) Define p ∶X → R as following

p(x) = ∥u′∥α(U,K) ⋅ ∥x∥X .

It is clear that p is a sublinear functional and u′(x) ≤ ∥u′∥α(U,K) ⋅ ∥x∥X , for all x ∈ U . Thus,by Hahn-Banach theorem there exists x′ ∶X → R, such that

x′∣U = u′ and x′(x) ≤ p(x), ∀x ∈X.Note that x′(−x) ≤ p(−x) = p(x), which implies that

∣x′(x)∣ ≤ ∥u′∥α(U,K) ⋅ ∥x∥X , ∀x ∈X,

which implies that x′ is continuous and ∥x′∥α(X,K) ≤ ∥u′∥α(U,K). For the proof of the inverseinequality notice that

∥u′∥α(U,K) = supu ∈Bu

∣u′(u)∣ = supu ∈Bu

∣x′(u)∣ ≤ supx ∈Bx

∣x′(x)∣ = ∥x′∥α(X,K),

whereBu = u ∈ U ∶ ∥u∥X ≤ 1 and Bx = x ∈X ∶ ∥x∥X ≤ 1.

◻

Remark 9. Let X be a normed space. Then, for every xo ∈ X,xo ≠ 0 there exists x′ ∈ X ′

such that∥x′∥α(X,K) = 1 and x′(xo) = ∥xo∥X .

Indeed, let M = Kxo. Define x′ ∶M → R by

x′(λxo) = λ∥xo∥X , ∀λ ∈ K.One can easily check that

∥x′∥α(M,R) = 1 and x′(xo) = ∥xo∥X .The existence of the extension x′ ∶X → K follows from the Hahn-Banach theorem.

Remark 10. Let X be a normed space. Then, for every x1, x2 ∈X such that x1 ≠ x2 thereexists x′ ∈X ′ such that x′(x1) ≠ x′(x2). This is the immediate conequence of the previousremark.

Remark 11. For any normed space (X, ∥ ⋅ ∥X) it holds that

∥x∥X = supx′∈Bx′

∣x′(x)∣, ∀x ∈X.

Indeed, since x′ ∈ Bx′ = x′ ∈X ′ ∶ ∥x′∥ ≤ 1 we have that

∣x′(x)∣ ≤ ∥x′∥ ⋅ ∥x∥X ≤ ∥x∥X , ∀x′ ∈ Bx′ .


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)∣ ≤ supx′∈Bx′

∣x′(x)∣.

Lemma 3.15. Let (X, ∥ ⋅ ∥X) be a normed space and V ⊂X be its convex and open subsetsuch that 0 ∉ V . Then, there exists x′ ∈X ′ such that Re x′(x) < 0, for each x ∈ V .

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 xo ∈ V define

yo = −xo and U = V − xo.Set U is convex, open and yo ∉ U , 0 ∈ U . For the set U define a Minkowski functional pU .Note that pU is sublinear and pU(yo) ≥ 1. Now, consider a linear functional

y′ ∶ linyo→ R, y′(t ⋅ yo) = t ⋅ pU(yo).It holds that y′(y) ≤ pU(y), for each y ∈ Y . Indeed,

y′(t ⋅ yo) ≤ 0 ≤ pU(t ⋅ yo), for t ≤ 0,

y′(t ⋅ yo) = pU(t ⋅ yo), for t > 0.

Let x′ be an extension of y′ such that x′ ≤ pU . There exists ε > 0 such that B(0, ε) ⊂ Usuch that

∣x′(x)∣ = maxx′(x), x′(−x) ≤ maxpU(x), pU(−x) ≤1

ε∥x∥X .

It is straightforward that x′(yo) = pU(xo) ≥ 1 and for each x = (u − yo) ∈ V , where u ∈ U itholds 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 convexand V1 be open. If V1 ∩ V2 = ∅, then there exists x′ ∈X ′ such that

Re x′(v1) < Re x′(v2) ∀v1 ∈ V1, v2 ∈ V2.

Proof of Theorem 3.16. Let define V = V1 −V2 = (V1 + (−V2)), which is in fact given by

V = ⋃x∈V2

(V1 − x).

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 v1 ∈ V1 and v2 ∈ V2 and thus, Re x′(v1) < Re x′(v2). ◻


Theorem 3.17. (Hahn-Banach theorem) Let X be a normed space, V ⊂X be a closed andconvex set and x ∉ V . Then, there exists a functional x ∈X ′ such that

Re x′(x) < infRe 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′(x) +Re x′(u) < Re x′(v)and

Re x′(x) + ∥Re x′∥X′ ⋅ r ≤ Re x′(v)Re x′(x) + ∥Re x′∥X′ ⋅ r ≤ infRe x′(v) ∶ v ∈ V .

◻

LECTURE 4Weak convergence and reflexivity

Let (X, ∥ ⋅ ∥X) be a normed space. X ′′ is defined as a space of linear functionals

i(x) ∶X ′ → K

given by

(4.1) (i(x))(x′) = x′(x).

Remark 12. Operator i is a linear isometry (in general not surjective). Indeed,

∣(i(x))(x′)∣ = ∣x′(x)∣ ≤ ∥x′∥ ⋅ ∥x∥X ⇒ ∥i(x)∥ ≤ ∥x∥X .

According to Remark 9, for every x ∈ X, x ≠ 0, there exists a functional x′ such thatx′(x) = ∥x∥X and ∥x′∥ = 1. Thus,

∥x∥X = ∣x′(x)∣ = ∣i(x)(x′)∣ ≤ ∥i(x)∥.

Definition 4.2. A Banach space X is called reflexive if i defined 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 xnn∈N be a sequence in a space X. We say that xnn∈N convergesweakly to some x and denote it as xn x, if

limn→+∞

x′(xn) = x(x) in K, ∀x′ ∈X ′.

Definition 4.4. Let ynn∈N be a sequence in a space Y ≅ X ′. We say that ynn∈Nconverges weakly∗ to some x and denote it as xn

∗ x, if

limn→+∞

yn(x) = y(x) in K, ∀x ∈X.

25


Theorem 4.5. (Sequential version of Banach-Alaoglu theorem) Let (X, ∥ ⋅ ∥X) be a sepa-rable normed space and Y ≅ X ′. If ynn∈N is a sequence such that supn∈N ∥yn∥Y ≤ 1, thenthere exists a subsequence ynk

k∈N and an element y ∈ Y such that

ynk

∗ y.

Proof of Theorem 4.5. If X is separable, then there exists a set L = xjl∈N ⊂ Xwhich is linearly independent and linearly dense in X. Since ∣yn(x1)∣ is uniformly boundedwith respect to n, then there exists a subsequence n1

k such that yn1k(x1) converges to some

element y1 ∈ Y . Similarly, ∣yn1k(x2)∣ is bounded and thus, there exists a subsequence n2

k

such that yn2k(x2) converges to some y2 ∈ Y . Iterating this procedure we obtain sequences

ynmk⊂ ynm−1

k⊂ ⋅ ⋅ ⋅ ⊂ yn1

k⊂ yn.

By a diagonal argument (setting ynk= ynk

k) we chose a subsequence ynk

k∈N such that

limk→+∞

ynk(xl) = yl, ∀l ∈ N.

To complete the proof we need to show that ynkconverges for all x ∈ X. To this end, fix

ε > 0. Since L is linearly dense, there exist N = N(ε) and a sequence (λli , xli)i∈N, λli ∈ K,xli ∈ L, such that

XXXXXXXXXXXx −

N(ε)

∑i=1

λlixli

XXXXXXXXXXXX< ε.

Define a linear functional y ∶ linxl ∶ xl ∈ L→ K as

y(N(ε)

∑i=1

λlixli) =N(ε)

∑i=1

λliyli .

Since the assumptions of Hahn-Banach theorem are fulfilled, we can extend y on X pre-serving ∥y∥Y ≤ 1. Note that for each ε > 0 it holds that

limk→+∞

∣ynk(x) − y(x)∣ ≤ 2ε + lim

k→+∞

RRRRRRRRRRRynk

(N(ε)

∑i=1

λlixli) − y(N(ε)

∑i=1

λlixli)RRRRRRRRRRR= 2ε.

◻

Remark 13. If X is reflexive, then weak and weak∗ convergence are equivalent. In sucha 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 weak∗ topology.

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∈NMn, where Mn are nowhere dense sets.

Theorem 4.7. (Baire category) A non-empty, complete metric space is not of the 1-stcategory.

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

supi∈I

∥Tix∥Y < +∞,

thensupi∈I

∥Ti∥α(X,Y ) < +∞.

Proof of Theorem 4.8. Let n ∈ N and define

En = x ∈X ∶ supi∈I

∥Tix∥Y ≤ n =⋂i∈I

x ∈X ∶ ∥Tix∥ ≤ n .

From the assumptions it follows that sets En are closed and X = ⋃n∈NEn. Therefore, byTheorem 4.7 at least one of the sets En has a non-empty interior. Let no ∈ N be such thatintEno ≠ ∅. Then, for each xo ∈ intEno there exists ε > 0 such that B(xo, ε) ⊂ Eno . Letx ∈X be such that ∥x∥X ≤ 1. Then,

εx + xo ∈ B(xo, ε) ⊂ Eno .

If so, then∥Ti(εx + xo)∥Y ≤ no, ∀i ∈ I.

Therefore, for any x ∈X such that ∥x∥X ≤ 1 it holds that

∥Tix∥Y = 1

ε∥Ti(εx)∥Y = 1

ε∥Ti(εx + xo) − Tixo∥Y ≤ 1

ε(no + ∥Tixo∥Y ) .

Since the right hand side of the inequality above does not depend on the choice of x, italso holds that

∥Ti∥α(X,Y ) ≤1

ε(no + ∥Tixo∥Y ) .

Taking supremum over i ∈ I yields

supi∈I

∥Ti∥α(X,Y ) ≤1

ε(no + sup

i∈I∥Tixo∥Y ) < +∞.

◻

LECTURE 5Open mapping and closed graph theorems

Definition 5.1. We say that T ∶X → Y is open, if for any open set U ⊂X a set T (U) ⊂ Yis open.

Lemma 5.2. Let (X, ∥ ⋅ ∥X), (Y, ∥ ⋅ ∥Y ) be normed spaces and T ∶ X → Y be a linearoperator. Then, the following conditions are equivalent.

(a) T is open,(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). LetU ⊂X be an open set and x ∈ U . It means that

∃r > 0, such that x +BX(0, r) ⊂ UMoreover, Tx ∈ T (U) and due to the linearity of T

Tx + T (BX(0, r)) = T (x +BX(0,1)) ⊂ T (U).From the assumption we have that there exists ε > 0 such that BY (0, ε) ⊂ T (BX(0, r)).Therefore it holds that Tx +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 surjectiveoperator. Then, T is open.

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 surjectiveoperator. Then, there exists ε > 0 such that

BY (0, ε) ⊂ T (BX(0,1)).

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

Y = ⋃n∈N

n ⋅ T (BX(0,1)).

29


Theorefore, int (T (BX(0,1))) ≠ ∅ due to Theorem 4.7 (Baire theorem). Thus, there exists

yo ∈ Y and ε > 0 such that

BY (yo, ε) ⊂ T (BX(0,1)).Since T is surjective, there exists xo such that Txo = yo and

BY (0, ε) = BY (yo, ε) − yo ⊂ T (BX(0,1)) − Txo ⊂ T (BX(0,1)) − T (BX(0, ∥xo∥X))= T (BX(0,1)) + T (BX(0, ∥xo∥X)) = T (BX(0,1 + ∥xo∥X)).

Thus, form linearity of T it follows that BY (0, ε(1+∥xo∥X)) ⊂ T (BX(0,1)). ◻

Proof of Theorem 5.3. Due to Lemma 5.4 there exists εo > 0 such that

BY (0, εo) ⊂ T (BX(0,1)).

Let y ∈ BY (0, εo). Then, there exists ε > 0 such that

∥y∥Y < ε < εo.

Define y = (εo/ε)y. Then, ∥y∥Y < εo and y ∈ T (BX(0,1)). There also exists yo such that

yo = Txo ∈ T (BX(0,1)), ∥y − yo∥ < αεo, where 0 < α < 1

and α fulfillsε

εo⋅ 1

1 − α< 1.

Note thaty − yoα

∈ BY (0, εo).

By the same argument we can find y1 = Tx1 ∈ T (BX(0,1)) such that

∥ y − yoα

− y1∥Y< αεo.

Therefore,

∥y − (yo + αy1)∥Y < α2εo.

Iterating the procedure we obtain

∥y −n

∑i=0

αiyi∥Y ≤ αn+1εo


∥y − T (n

∑i=0

αii)∥Y < αn+1εo.

Note that xii∈N ⊂ BX(0,1) and 0 < α < 1, hence ∑ni=0α

ixi converges. Since X is a Banachspace we set

x =+∞∑i=0

αixi ∈X

LECTURE 5. OPEN MAPPING AND CLOSED GRAPH THEOREMS 31

and by the continuity of T we obtain y = T x. Denote x = (ε/εo)x. Then, y = Tx and

∥x∥X = ε

εo∥x∥X ≤ ε

εo

+∞∑i=0

αi∥xi∥X < ε

εo

+∞∑i=0

αi = ε

εo⋅ 1

1 − α< 1.

◻

Remark 15. Let X,Y be Banach spaces and T ∈ α(X,Y ) be bijective. Then, T −1 iscontinuous.

Remark 16. If (X, ∥ ⋅ ∥1) and (X, ∥ ⋅ ∥2) are Banach spaces and norms ∥ ⋅ ∥1, ∥ ⋅ ∥2 are suchthat

∥x∥1 ≤M∥x∥2, ∀x ∈X,then ∥ ⋅ ∥1 and ∥ ⋅ ∥2 are equivalent.

Definition 5.5. Let X, Y be normed spaces, D be a subspace of X and T ∶ D → Y bea linear functional. Then, T is closed if for every sequence xnn∈N ⊂ D such that xn → xand Txnn∈N ⊂ Y , Txn → y, it holds that x ∈D and Tx = y.

(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,Tx) ∶ 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 isclosed, then

(a) D with a norm ∥x∥ = ∥x∥X + ∥Tx∥Y is a Banach space,(b) T is continuous as a mapping from (D, ∥ ⋅ ∥) to Y .

Proof of Lemma 5.8. a) Since T is closed, we know that for every sequences

xnn∈N ⊂D such that xn → x ∈Xand

Txnn∈N ⊂ Y such that Txn → y

it holds that x ∈D and Tx = y. xnn∈N ⊂D is a Cauchy sequence with respect to the ∥ ⋅ ∥norm, since xnn∈N is a Cauchy sequence in X and Txnn∈N is a Cauchy sequence in Y .Moreover, its limit in X is an element from the subspace D. Thus,

∥xn − x∥ = ∥xn − x∥X + ∥T (xn − x)∥Y → 0,

which implies that (D, ∥ ⋅ ∥) is complete.b) T is bounded, which holds due to the following inequality

∥Tx∥Y ≤ ∥x∥X + ∥Tx∥Y = ∥x∥.Thus, T is also continuous. ◻


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 theprevious lemma. Note that if we set D = X in the previous lemma, then X is a Banachspace with norms ∥ ⋅ ∥X and ∥ ⋅ ∥. Moreover, ∥x∥X ≤ ∥x∥ for all x ∈ X. Then, according toRemark 16 the norm ∥ ⋅ ∥X is equivalent to ∥ ⋅ ∥. Therefore, T is continuous with respect tothe norm ∥ ⋅ ∥X as well. ◻

5.1. Application of Banach -Steinhaus Theorem

Lemma 5.10. For a subset M of a normed space (X, ∥ ⋅ ∥X), the following conditions areequivalent:

(a) M is bounded,(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

supx∈M

∣x′(x)∣ = supx∈M

∣i(x)(x′)∣ < +∞, ∀x ∈X.

By Banach-Steinhaus theorem we have

supx∈M

∥x∥ = supx∈M

∥i(x)∥ < +∞.

◻

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 alln ∈ N. If for all x ∈X there exists Tx = limn→+∞ Tnx, then T ∈ α(X,Y ).

LECTURE 6Adjoint Operator

Definition 6.1. Let (X, ∥ ⋅ ∥X), (Y, ∥ ⋅ ∥Y ) be normed spaces and T ∈ α(X,Y ). Then,T ′ ∶ Y ′ →X ′ is defined by

(T ′y′)(x) = y′(Tx).

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. Tx = y and ∥x∥X ≤K∥y∥Y .

Proof of Lemma 6.3. Define a canonical factorization T (what is precisely a canonicalfactorization?)

T ∶X/ker(T )→ Im(T ).

Note that X/ker(T ) and Im(T ) are Banach spaces and T is bijective. Then, according toRemark 15 there exists

T −1 ∈ α(Im(T ),X/ker(T ))

and K ≥ 0 such that ∥T −1∥ ≤K. ◻

33


Lemma 6.4. Let (X, ∥ ⋅ ∥X),(Y, ∥ ⋅ ∥Y ) be Banach spaces and T ∈ α(X,Y ). If there existsa constant C such that for all y′ ∈ Y ′

C∥y′∥α(Y,K) ≤ ∥T ′y′∥α(X′,K),

then T is surjective and open.

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 yo ∈ BY (0,C) and assume that yo ∉ D (note that D is a convex set). According toHahn-Banach theorem, there exists y′ ∈ Y ′, α ∈ R such that for each y ∈D

Re y′(y) ≤ α < Re y′(yo) ≤ ∣y′(yo)∣.Note that 0 ∈ D, which implies Re y′(0) = 0 ≤ α. We can assume that α > 0. Definey′ = y′/α. Then

Re y′(y) = Re (y′(y)/α) ≤ 1 < Re (y′(yo)/α) = Re y′(yo), ∀y ∈D.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,Tx = y ∈D and from the definition of T ′

∣T ′y′(x)∣ = ∣y′(Tx)∣ ≤ 1,

which implies that ∥T ′y′∥α(X,K) ≤ 1, but

∣y′(yo)∣ ≤ ∥y′∥α(Y,K) ⋅ ∥yo∥Y < C∥y′∥α(Y,K)

and 1 < C∥y′∥α(Y,K), which leads to the contradiction. ◻

Lemma 6.5.

Im(T ) = (ker(T ′))⊥.

Proof of Lemma 6.5.” ⊂ ”. Let y ∈ Im(T ), which means that there exists x ∈X such that y = Tx. If y′ ∈ ker(T ′),then T ′y′ is a zero functional on X ′ and thus

0 = (T ′y′)(x) = y′(Tx) = 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′(Tx) = (T ′y′)(x) for all x ∈ X, that is, T ′y′ is a zero functional onX and thus y′ ∈ ker(T ′). This proves that y ∉ (ker(T ′))⊥, since the opposire claim wouldimply 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′(Tx) = 0 for x ∈ ker(T ). Let x′ ∈ (ker(T ))⊥ and define a linear functional

z′ ∶ Im(T )→ K, z′(y) = x′(x), for y = Tx.

Note that z is continuous. Indeed, using one of the previous lemmas and the open mappingtheorem we have that

∀y ∈ Im(T ) ∃x ∈X s. t. y = Tx, and

∣z′(y)∣ = ∣x′(x)∣ ≤ ∥x′∥α(X,K) ⋅ ∥x∥X ≤ ∥x′∥α(X,K) ⋅K∥y∥Y .

By Hahn-Banach theorem we can define y′ ∈ Y ′, which is an extension of z′. Then,

x′(x) = z′(Tx) = y′(Tx) = (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 = Tx. For y′ ∈ Y andx ∈X we have

(T ′y′)(x) = y′(Tx) = y′∣Z(Sx) = (S′(y′∣Z))(x), ∀x ∈X.

Then, T ′y′ = S′(y′∣Z). Therefore Im(T ′) ⊂ Im(S′). To prove the opposite inclusion assumethat S′z′ ∈ Im(S′) for some z′ ∈ Z ′ and consider any extension y′ of z′ (by Hahn-Banachtheorem such extension 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 andthus S′ is injective (?). S′ is then a continuous bijection between Z ′ and Im(S′). In factit is an in isomorphism, in particular

C∥z′∥α(Z,K) ≤ ∥S′z′∥α(X,K), ∀z′ ∈ Z ′.

Then, by one of the Lemma above, S is surjective (and open). Thus, Im(S) = Z = Im(T )(?) and Im(T ) = Im(T ). ◻


Lemma 6.6. Let P be a continuous projection on a normed space X (P is a projection ifP 2 = P ). Then,

(a) P = 0 or ∥P ∥ ≥ 1,(b) ker(P ) and Im(P ) are closed,(c) X = ker(P )⊕ 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 projectionand Im(P ) = ker(Id − P ) is closed.(c) ∀x ∈X it holds that x = (Id − P )x + Px. ◻

LECTURE 7Hilbert 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 ∈X,(d) < x,x > ≥ 0, ∀x ∈X,(e) < x,x > = 0 ⇔ x = 0.

Lemma 7.2. (Cauchy - Schwarz inequality). Let X be a K-linear space with a scalarproduct. Then,

∣< x, y >∣2 ≤ < x,x > ⋅ < y, y >, ∀x, y ∈X.The equality holds if and only if x = λy for λ ∈ K.

Lemma 7.3. Define a mapping ∥ ⋅ ∥ ∶X ×X → R by

∥x∥ =√< x,x >.

Then, ∥ ⋅ ∥ is a norm. In particular, it holds that

∣< x, y >∣ ≤ ∥x∥ ⋅ ∥y∥, ∀x, y, ∈X.

Definition 7.4. A normed space (X, ∥ ⋅ ∥X) is called a prehilbert space, if there exists ascalar product < ⋅, ⋅ > such that

∥x∥X =√< x,x >, ∀x ∈X.

Complete prehilbert space is called a Hilbert space.

Lemma 7.5. Let (X, ∥ ⋅ ∥) be a prehilbert space and U be its dense linear subspace. If forall u ∈ U it holds that < x,u >= 0, then x = 0.

37


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 . Moreprecisely, we set

< x, y > = 1

4(∥x + y∥2

X − ∥x − y∥2X) , for K = R

and

< x, y > = 1

4(∥x + y∥2

X − ∥x − y∥2X + i∥x + iy∥2

X − i∥x − iy∥2X) , for K = C.

Lemma 7.6. Scalar product < ⋅, ⋅ > ∶X ×X → K is a continuous mapping.

Proof of Lemma 7.6. Let x1, x2, y1, y2 ∈X. Then,

∣< x1, y1 > − < x2, y2 >∣ = ∣< x1 − x2, y1 > + < x2, y1 − y2 >∣≤ ∥x1 − x2∥ ⋅ ∥y1∥ + ∥y1 − y2∥ ⋅ ∥x2∥.

◻

Theorem 7.7. (Parallelogram equality) A normed space (X, ∥ ⋅ ∥X) is a prehilbert space ifand only if, for all x, y ∈X the following inequality holds

∥x + y∥2X + ∥x − y∥2

X = 2∥x∥2X + 2∥y∥2

X .

Proof of Theorem 7.7. (for K = R). We introduce a scalar product as following

< x, y > = 1

4(∥x + y∥2

X − ∥x − y∥2X) .

Clearly, it holds that ∥x∥X = < x,x >2. We need to prove the properties of a scalar productform the Definition 7.1.

(a) Let x1, x2 ∈X and define

α ∶= ∥x1 + x2 + y∥2X = 2∥x1 + y∥2

X + 2∥x2∥2X − ∥x1 − x2 + y∥2

X ,

β ∶= ∥x1 + x2 + y∥2X = 2∥x2 + y∥2

X + 2∥x1∥2X − ∥ − x1 + x2 + y∥2

X .

Therefore,

∥x1 + x2 + y∥2 = α + β2

= ∥x1 + y∥2 + ∥x1∥2 + ∥x2 + y∥2 + ∥x2∥2 − 1

2(∥x1 − x2 + y∥2 + ∥ − x1 + x2 + y∥2) .

Similarly one gets

∥x1 + x2 − y∥2 = ∥x1 − y∥2 + ∥x1∥2 + ∥x2 − y∥2 + ∥x2∥2 − 1

2(∥x1 − x2 − y∥2 + ∥ − x1 + x2 − y∥2) .

LECTURE 7. HILBERT SPACES 39

Finally,

< x1 + x2, y > = 1

4(∥x1 + x2 + y∥2

X − ∥x1 + x2 − y∥2X)

= 1

4(∥x1 + y∥2

X + ∥x2 + y∥2X − ∥x1 − y∥2

X − ∥x2 − y∥2X)

= < x1, y > + < x2, y > .Claim (b) follows from (a) for λ ∈ N. Form the construction of < ⋅, ⋅ > we also have theproperty (b) fulfilled for λ = 0, λ = −1 and thus for all λ ∈ Z. Let λ = m

n ∈ Q. Then,

n < λx, y > = n <mx

n, y > =m < x, y > = nλ < x, y > .

By the continuiuty of ∥ ⋅ ∥ ∶X → K we can extend this result for λ ∈ R.

Claims (c), (d) and (e) are straightforward. ◻

Remark 18. The following claims hold:

(a) A normed space is prehilbert if and only if all its 2-dimensional subspaces areprehilbert.

(b) Any subspace of a prehilbert space is prehilbert.(c) Closure of a prehilbert space is a Hilbert space.

Example 11. Examples of Hilbert spaces.

(a) Rn and Cn with < si, ti > = ∑ni=1 siti,

(b) l2 with < si, ti > = ∑+∞i=1 siti,

(c) L2(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 andwrite x ⊥ y, if < x, y >= 0. Sets A and B are orthogonal, if < x, y > = 0 for all x ∈ A andy ∈ B. The set

A⊥ = y ∈X ∶ x ⊥ y, ∀x ∈ Ais called the orthogonal complement 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,

(c) A ⊂ (A⊥)⊥,(d) A⊥ = (linA)⊥.

Theorem 7.9. Let (H, ∥ ⋅ ∥) be a Hilbert space and K be its convex and closed subset. Letxo ∈H. Then, there exists a unique x ∈K such that

∥x − xo∥ = infy∈K

∥y − xo∥.

This statement still holds, if H is a uniformly convex space.


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

d = infy∈K

∥y∥.

There exists ynn∈N ⊂ K and limn→+∞ ∥yn∥ = d. We will prove that ynn∈N is a Cauchysequence. This is sufficient to show, since (K, ∥ ⋅ ∥) is a Banach space. By Parallelogramequality we have

∥(yn + ym)/2∥ + ∥(yn − ym)/2∥ = 1

2(∥yn∥2 + ∥ym∥2) , ∀m,n ∈ N.

By convexity of K we have that 12(yn + ym) ∈K and thus ∥1

2(yn + ym)∥2 ≥ d2. Therefore,

d2 + ∥(yn − ym)/2∥ ≤ 1

2(∥yn∥2 + ∥ym∥2) , ∀m,n ∈ N.

Since the right hand side of the inequality converges to d2, we have that ∥yn − ym∥ convergesto zero as n,m tend to infinity. By the comleteness of K, there exists x ∈K such that

x = limn→+∞

yn and ∥x∥ = d.

Now, we shall show that the element is unique. Let x, x ∈K be such that

∥x∥ = ∥x∥ = d.Then, by Parallelogram equality

∥(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. ◻

Lemma 7.10. Let K be a closed convex subset of a Hilbert space (H, ∥ ⋅ ∥) and xo ∈ H.The, the following conditions are equivalent

(a) ∥xo − x∥ = infy∈K ∥xo − y∥,

(b) Re ⟨xo − x, y − x⟩ ≤ 0, ∀y ∈K.

Proof of Lemma 7.10.(b) ⇒ (a)

∥xo − y∥2 = ∥(xo − x) + (x − y)∥2

= ∥xo − x∥2 + 2Re ⟨xo − x,x − y⟩ + ∥x − y∥2 ≥ ∥xo − x∥2.

(a) ⇒ (b). Let t ∈ [0,1], y ∈K and define yt = (1 − t)x + ty ∈K. Then,

∥xo − x∥2 ≤ ∥xo − yt∥2 = ⟨xo − x + t(x − y), xo − x + t(x − y)⟩= ∥xo − x∥2 + 2Re ⟨xo − x, t(x − y)⟩ + t2∥x − y∥2

.

Finally,

Re ⟨xo − x, t(y − x)⟩ ≤ t

2∥x − y∥2

, ∀t ∈ [0,1].◻


Theorem 7.11. Let U ≠ 0 be a closed subspace of a Hilbert space (H, ∥ ⋅ ∥). Then, thereexists a linear projection

PU ∶H → U, s. t. ∥Pu∥ = 1, ker(PU) = U⊥.

Moreover, Id − PU is a projection such that

(Id −PU) ∶H → U⊥, s. t. ∥Id −PU∥ = 1, (if U ≠H).

It also holds that H = U ⊕U⊥.

Proof of Theorem 7.11. Define

PU ∶H → U, PU(xo) = x,

where xo ∈ H and x ∈ U is such that ∥xo − x∥2 = infy∈U ∥xo − y∥2. It is straightforward that

PU is a projection. By the previous lemma

Re ⟨xo − PU(xo), y − PU(xo)⟩ ≤ 0, ∀y ∈ U.

Note that y − PU(xo) ∈ U and thus

Re ⟨xo − PU(xo), y⟩ ≤ 0, ∀y ∈ U.

Using −y and iy in the expression above we obtain

⟨xo − PU(xo), y⟩ = 0, ∀y ∈D and xo − PU(xo) = xo − x ∈ U⊥.

Since U⊥ is a linear subspace, it holds that for all x1, x2 ∈H, λ1, λ2 ∈ K

(λ1x1 − λ1PU(x1)) + (λ2x2 − λ2PU(x2)) ∈ U⊥.

Therefore, λ1x1 + λ2x2 = z ∈H and PU(z) is such that z − PU(z) ∈ U⊥. Thus,

PU(λ1x1 + λ2x2) = PU(z) = λ1PU(x1) + λ2PU(x2),

which proves that PU is linear. From the construction of PU it follows that Im(PU) = Uand we have that ker(PU) = U⊥. Therefore,

PU(xo) = 0 ⇔ xo ∈ U⊥.

We claim that Id − PU is a projection with Im(Id − PU) = U⊥ and ker(Id − PU) = U . ByPythagoras theorem

∥xo∥2 = ∥PU(xo) + xo − PU(xo)∥2 = ∥PU(xo)∥2 + ∥(Id − PU)(xo)∥2

and PU(xo) ∈ U , xo − PU(xo) ∈ U⊥. Finally, we get H = U ⊕ U⊥. Moreover, ∥PU∥ ≤ 1 and∥Id − PU∥ ≤ 1 (in fact equal to 1, if U ≠ 0 and U ≠H). ◻


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 − PV = PV ⊥ . Consider V = U . It is clear that

U⊥ = V ⊥ and Id − PV ⊥ = PV ⊥⊥ .Thus, PV = PV ⊥⊥ and thus U = U⊥⊥.

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) = λΦ(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 linear. It follows from the Cauchy-Schwarz inequality that

∥Φ(y)∥ ≤ ∥y∥and for x = y/∥y∥ we obtain

Φ(y)(x) =⟨y, y⟩∥y∥

= ∥y∥,

which implies that Φ is an isometric and thus, it is injective. Let x′ ∈ H ′ (without lossof generality we can assume that ∥x′∥ = 1) and U = ker(x′). By the previous theoremH = U ⊕ U⊥, where dim(U⊥) = 1. Thus, there exists y ∈ H such that x′(y) = 1 andU⊥ = lin(y). For x = u + λy ∈ (U ⊕U⊥)

x′(x) = λx′(y) = λ and ⟨x, y⟩ = λ∥y∥2.

Therefore,

Φ(y/∥y∥2) = x′ and Φ is surjective.

Finally, ∥y∥ = 1 and ∥x′∥ = 1. ◻

Remark 21. Each Hilbert space is reflexive.

7.1. Orthonormal basis

Definition 7.13. Let H be a Hilbert space. A subset S ⊂ H is an orthonormal system ifand 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

S ⊂ T, T − orthonormal system ⇒ T = S.

Example 12. Examples of orthonormal systems:


(a) H = L2([0,2π],R),

S = 1

2πχ[0,2π] ∪ 1√

πcos(n), n ∈ N ∪ 1√

πsin(n), n ∈ N .

(b) H = L2([0,2π],C),

S = 1√2π

ein, n ∈ Z .

Theorem 7.14. (Gram - Schmidt theorem) Let H be a Hilbert space and xn ∶ n ∈ Nbe a linearly independent set in H. Then, there exists an orthogonal system S such thatlinS = linxn ∶ n ∈ N.

Example 13. Let H = L2([−1,1],R) and xn(t) = tn, for n ∈ N. Then,

S =⎧⎪⎪⎨⎪⎪⎩

√n + 1

2⋅ Pn(t), n ∈ N

⎫⎪⎪⎬⎪⎪⎭, where Pn(t) =

1

2nn!( ddt

)n

(t2 − 1)n.

Theorem 7.15. (Bessel inequality) Let H be a Hilbert space. If S = en ∶ n ∈ N ⊂ H isan orthonormal system and x ∈H, then

+∞∑n=1

∣⟨x, en⟩∣2 ≤ ∥x∥2

.

Proof of Theorem 7.15. Let N ∈ N. Define

xN = x −N

∑n=1

⟨x, en⟩en

such that xN ⊥ ek, for k = 1, . . . ,N . By Pythagoras theorem

∥x∥2 = ∥xN∥2 + ∥N

∑n=1

⟨x, en⟩en∥2

= ∥xN∥2 +N

∑n=1

∣⟨x, en⟩∣2 ≥

N

∑n=1

∣⟨x, en⟩∣2.

This estimate holds for all N ∈ N, therefore taking the limit N → +∞ ends 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⟩ ≠ 0is countable.

Proof of Lemma 7.16. From Bessel inequality it holds that

Sx,n = e ∈ S ∶ ⟨x, r⟩ ≥ 1/nis finite and thus Sx = ⋃n∈N Sx,n is at most countable. ◻

LECTURE 8Fourier Transform

Definition 8.1. Let f ∈ L1(Rn). Define

(Ff)(ξ) = 1

(2π)n/2 ∫Rnf(x)e−ixξdx, ∀ξ ∈ Rn.

F is called a Fourier transform.

Theorem 8.2. F ∶ (L1(Rn), ∥ ⋅ ∥L1) → (Co(Rn), ∥ ⋅ ∥∞) is a continuous linear mappingsuch that

∥F∥ ≤ (2π)−n/2.Here, Co(Rn) denotes a space of continuous functions vanishing at infinity.

Proof of Theorem 8.2. A linearity of F is obvious. Note that ∣e−ixξ ∣ = 1. Therefore,

∥Ff∥∞ ≤ (2π)−n/2∥f∥L1 ⇒ ∥F∥ ≤ (2π)−n/2.Now, we want to prove

(a) limξk→ξ ∣(Ff)(ξk) − (Ff)(ξ)∣ = 0,(b) lim∣ξk ∣→+∞ ∣(Ff)(ξk)∣ = 0.

(a) Let ξkk∈N be such that limk→+∞ ξk = ξ. Then, for a fixed x we have

limk→+∞

∣e−ixξk − e−ixξ∣ = 0.

By Lebesgue dominated convergence theorem it holds that

limk→+∞

∣(Ff)(ξk) − (Ff)(ξ)∣ ≤ 1

(2π)n/2 ∫Rn∣f(x)∣ ⋅ lim

k→+∞∣e−ixξk − e−ixξ∣dx = 0,

where we used the latter theorem due to the fact that the function under the integral canbe estimated by 2 ∣f ∣ ∈ L1(Rn).(b) Note that the smooth functions with compact support are dense in L1, that is,

C∞c (Rn)

L1(Rn)= L1(Rn).

45


Thus, it is enough to show that for all f ∈ C∞c (Rn) it holds that lim∣ξk ∣→+∞(Ff)(ξk) = 0.

Let f be such a function and j be such that ∣ξkj ∣ = max∣ξki ∣ ∶ i = 1, . . . , n, where ξkj are

coordinates of ξk. Then, ∣ξkj ∣ ≥ ∣ξk∣/√n. We apply the formula for integration by parts and

as a result obtain

∣(Ff)(ξk)∣ = ∣ −1

(2π)n/2 ∫Rn

∂

∂xjf(x) 1

−iξje−ixξ

k

dx∣

≤ 1

(2π)n/2⋅ ∥ ∂

∂xjf∥

L1

⋅√n

∣ξk∣k→+∞Ð→ 0.

◻

Definition 8.3. We say that f ∶ Rn → C is vanishing faster than polynomial, if

lim∣x∣→+∞

xαf(x) = 0, ∀α ∈ Nn,

where α = (α1, . . . , αn) and xα = xα11 ⋅ ⋅ ⋅ ⋅ ⋅ xαn

n . A space

S(Rn) = f ∈ C∞(Rn) ∶ Dβf vanishes faster that polynomial, ∀β ∈ Nn

is called Schwarz space.

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

C∞c (Rn) ⊊ S(Rn).

Remark 23. If f ∈ S(Rn), then

xαf,Dαf ∈ S(Rn), ∀n ∈ N.

Lemma 8.4. Let f ∈ S(Rn). Then,

(a) Ff ∈ C∞(Rn) and Dα(Ff) = (−i)∣α∣F(xαf),

(b) F(Dαf) = i∣α∣ ⋅ ξαF(f).

Proof of Lemma 8.4.(a) Formal calculations leads to

Dα(Ff) = ∂α

∂ξα⋅ 1

(2π)n/2 ∫Rnf(x)e−ixξdx = 1

(2π)n/2 ∫Rnf(x) ⋅ ∂

α

∂ξαe−ixξdx

= (−i)∣α∣ 1

(2π)n/2 ∫Rnf(x)xαe−ixξdx = (−i)∣α∣F(xαf)(ξ).

In the second equality we have passed with differentiation under the integral sign. Sincexαf ∈ S(Rn) ⊂ L1(Rn) this is justified by Lebesgue dominated convergence theorem.

LECTURE 8. FOURIER TRANSFORM 47

(b) If follows from the formula for integration by parts that

F(Dαf) = 1

(2π)n/2 ∫Rn(Dαf)(x)e−ixξdx

= (−1)∣α∣(2π)n/2 ∫Rn

f(x) ⋅ ∂α

∂xαe−ixξdx = (−1)∣α∣(−i)∣α∣ξαF(f).

Since xβf ∈ S(Rn) boundary terms vanished in the calculation above. ◻

Lemma 8.5. If f ∈ S(Rn), then also Ff ∈ S(Rn).

Proof of Lemma 8.5. In Lemma 8.4 we proved that Ff ∈ C∞(Rn). We need to showthat Ff and all its derivatives vanish faster than polynomial, that is,

ξαDβ(Ff)(ξ)→ 0 for ∣ξ∣→ +∞.

From Lemma 8.4 we know that

ξαDβ(Ff) = (−i)∣β∣(−i)∣α∣F(Dα(xβf))

which implies that Dα(xβf) ∈ S(Rn) and thus in L1(Rn). Fourier transform of a functionfrom L1(Rn) is a continuous function vanishing at infinity, which ends the proof. ◻

Let us consider a function

γ ∶ Rn → R, γ(x) = e−x2/2.

It is a well known fact that1

(2π)n/2 ∫Rnγ(x)dx = 1.

We also denote γa(x) = γ(ax), for a > 0.

Lemma 8.6. For γ and γa defined as above it holds that

(Fγ)(ξ) = e−ξ2/2 and (Fγa)(ξ) =

1

an(Fγ) (ξ

a) .

Proof of Lemma 8.6. The second statement is straightforward. To prove the first claimlet n = 1. Note that γ fulfills the following differential equation

(8.7) y′ + xy = 0, y(0) = 1.

By Lemma 8.4

0 = F(γ′ + xγ) = iξ(Fγ) + ( 1

−i(Fγ))

′⇒ (Fγ)′ + ξ(Fγ) = 0,

which means that Fγ fulfills equation (8.7) with the same initial condition, since

(Fγ)(0) = 1

(2π)1/2 ∫Re−x

2/2dx = 1.


Since the solution to (8.7) is unique we obtain γ = Fγ. The case n > 1 leads in fact to theprevious case, since

(Fγ)(ξ) = 1

(2π)n/2 ∫R. . .∫

RΠnk=1e−x

2k/2e−ixkξkdx1 . . .dxn = Πn

k=1e−ξ2k/2.

◻

Lemma 8.8. For f ∈ S(Rn) it holds that

(FFf)(x) = f(−x), ∀x ∈ Rn.

Proof of Lemma 8.8. We proved that Ff ∈ S(Rn) and thus F(Ff) is well defined.Let f, g ∈ S(Rn). Since a function

(x, ξ)→ f(ξ)g(x)e−ixξ

is integrable, Fubini theorem yields

(8.9) ∫Rn

(Ff)(x)g(x)dx = ∫Rnf(x)(Fg)(x)dx.

This implies that for g(x) = e−ixξoγ(ax), where ξo ∈ Rn and a > 0 are fixed, that

(Fg)(ξ) = 1

(2π)n/2 ∫Rne−ixξoγ(ax)e−ixξdx = (Fγa)(ξ + ξo).

We use Lemma 8.6 and the following change of variables u = x+ξoa , which leads to

1

(2π)n/2 ∫R(Ff)(x)e−ixξoγ(ax)dx = 1

(2π)n/2 ∫Rf(x) 1

an(Fγ) (x + ξo

a)dx

= 1

(2π)n/2 ∫Rf(au − ξo)γ(u)du.

Now we want to pass to the limit with a → 0. Since x → (Ff)(x)e−ixξoγ(ax) is boundedby ∣Ff ∣ ∈ L1 and x → f(x) 1

an (Fγ) (x+ξoa

) is bounded by ∥f∥∞, application of Lebesqueconvergence theorem ends the proof (the limit of the left hand side is equal to (FFf)ξoand the limit on the right hand side is equal to f(−ξo)). ◻

Theorem 8.10. The Fourier transform is a bijection on S(Rn). The inverse operator F−1

is defined by

(8.11) (F−1f)(x) = 1

(2π)n/2 ∫Rnf(ξ)eixξdξ, ∀x ∈ Rn.

Moreover, for all f, g ∈ S(Rn) it holds that

⟨Ff,Fg⟩L2 = ⟨f, g⟩

L2 .

LECTURE 8. FOURIER TRANSFORM 49

Proof of Theorem 8.10. From the previous lemma we know that F4 = Id. Thus, F isbijective and F−1 = F3 is bijective as well. It also holds

(F−1f)(x) = (F2(Ff))(x) = (Ff)(−x),which proves that (8.11) holds. From (8.9) we have that

∫Rn

(Ff)(ξ)(Fg)(ξ)dξ = ∫Rnf(x)(F(Fg))(x)dx.

For transparency purposes let denote h = Fg. Then,

(F h)(x) = 1

(2π)n/2 ∫Rnh(ξ)e−ixξdξ = 1

(2π)n/2∫Rnh(ξ)eixξdξ = F−1h(x) = g(x)

and thus⟨Ff,Fg⟩

L2 = ∫Rnf(x)g(x)dx = ⟨f, g⟩

L2 .

◻As a consequence, for all f ∈ S(Rn) it holds that ∥Ff∥L2 = ∥f∥L2 . The operator F is thuscontinuous as a mapping from S(Rn) to L2(Rn), bijective and ∥ ⋅ ∥L2-isometric.

Remark 24. For f ∈ L2(Rn) we define

gR(ξ) =1

(2π)n/2 ∫B(0,R)f(x)e−ixξdx and (F2f)(ξ) = lim

R→+∞gR(ξ),

where a limit in the definition above is the limit in L2(Rn).

Theorem 8.12. (Hausdorff-Young inequality) Let 1 ≤ p ≤ 2 and 1p +

1q = 1. For f ∈ S(Rn)

it holds that Ff ∈ Lq(Rn) and

∥Ff∥Lq ≤1

(2π)n/p−n/2⋅ ∥f∥Lp .

There exists an extension of F to a continuous operator F ∶ Lp(Rn) → Lq(Rn), which isdefined by

(Ff)(ξ) = limR→+∞

gR(ξ),

where a limit in the definition above is the limit in Lq(Rn).

Proof of Theorem 8.12. For (p, q) = (1,+∞) and (p, q) = (2,2)wehave∥F ∶ L1 → L∞∥ ≤ (2π)−n/2,∥F ∶ L2 → L2∥ ≤ 1.

For other case the proof follows from the Riesz interpolation theorem. One takes θ = 2− 2p ,

so that 1q =

1−θ+∞ + θ

2 . ◻

LECTURE 9Adjoint operators on Hilbert spaces

Definition 9.1. Let T be a linear bounded operator T ∈ α(H,H). We say that T ∗ ∶H →His an adjoint operator to T , if

⟨Tx, y⟩ = ⟨x,T ∗y⟩, ∀x, y ∈H.

Lemma 9.2. For each y, element T ∗y is unique and ∥T ∗y∥ ≤ ∥T ∥ ⋅ ∥x∥. Thus, T ∗ is welldefined and its norm is bounded.

Proof of Lemma 9.2. Fix y ∈H. Note that x→ ⟨Tx, y⟩ is a linear bounded functional.From Riesz representation theorem

∀y ∈H ∃zy ∈H s. t. ⟨Tx, y⟩ = ⟨x, zz⟩, ∀x ∈H.

We set T ∗y = zy. Then, it also holds that ∥zy∥ ≤ ∥T ∥ ⋅ ∥x∥. ◻

Lemma 9.3. It holds that

(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,

∥Tx∥ = ∥T ∗x∥, ∀x ∈H.

Proof of Theorem 9.5. ”⇒ ” It holds that

∥Tx∥2 = ⟨Tx,Tx⟩ = ⟨T ∗Tx,x⟩ = ⟨TT ∗x,x⟩ = ∥T ∗x∥2.

51


”⇐ ” Before we prove this implication we shall show the following claim. Let T ∈ α(H,H)be such that

⟨Tx,x⟩ = 0, ∀x ∈H.Then, T = 0. Indeed,

⟨T (x + y), T (x + y)⟩ = 0 ⇒ ⟨Tx, y⟩ + ⟨Ty, x⟩ = 0,

⟨T (x + iy), T (x + iy)⟩ = 0 ⇒ −⟨Tx, y⟩ + ⟨Ty, x⟩ = 0,

and thus ⟨Ty, x⟩ = 0, for all x, y ∈ H, which implies T = 0. Now we can proceed with theproof. We have that

0 = ∥Tx∥2 − ∥T ∗x∥2 = ⟨(T ∗T − TT ∗)x,x⟩.Thus, T ∗T − TT ∗ = 0 due to the claim. ◻

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 openmapping 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 ∥. Then,

(Id − T /λ)−1 =+∞∑n=0

(Tλ)n

is a convergent series and thus the operator (Id−T /λ) is invertible, which is a contradiction.◻

Definition 9.8. Operator T ∈ α(H,H) is a compact operator, if a closure of an image ofthe 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 = ⟨Tx − αx + iβx, Tx − αx + iβx⟩= ⟨Tx − αx,Tx − αx⟩ + ⟨Tx − αx, iβx⟩ + ⟨iβx, Tx − αx⟩ + ⟨iβx, iβx⟩= ∥Tx − αx∥2 + β2 ⋅ ∥x∥2

.

LECTURE 9. ADJOINT OPERATORS ON HILBERT SPACES 53

Thus, ∥(T − λId)x∥2 ≥ β2 ⋅ ∥x∥2, which implies β = 0. ◻

Theorem 9.10. Let T ∈ α(H,H) be a self adjoint operator. Then,

∥T ∥ = supx∶∥x∥=1

∣⟨Tx,x⟩∣.

Proof of Theorem 9.10. It is clear that

supx∶∥x∥≤1

∣⟨Tx,x⟩∣ ≤ supx∶∥x∥≤1

∥Tx∥ ⋅ ∥x∥ ≤ supx∶∥x∥≤1

∥Tx∥ = ∥T ∥.

To prove the opposite inequality define M = supx∶∥x∥≤1 ∣⟨Tx,x⟩∣. Then,

⟨T (x + y), x + y⟩ − ⟨T (x − y), x − y⟩ = 2⟨Tx, y⟩ + 2⟨Ty, x⟩

= 2⟨Tx, y⟩ + 2⟨x,Ty⟩ = 2⟨Tx, y⟩ + 2⟨Ty, x⟩

= 2⟨Tx, y⟩ + 2⟨Tx, y⟩ = 4Re ⟨Tx, y⟩.In the calculation above some terms vanish due to the fact that T is self adjoint. Byparallelogram equality we have

4Re ⟨Tx, y⟩ ≤M(∥x + y∥2 + ∥x − y∥2) = 2M(∥x∥2 + ∥y∥2)Re ⟨Tx, y⟩ ≤ M, ∀x, y ∶ ∥x∥, ∥y∥ = 1

Multiplication by λ such that ∣λ∣ = 1 gives

∣⟨Tx, y⟩∣ ≤M ∀x, y ∶ ∥x∥, ∥y∥ = 1,

which proves that ∥T ∥ =M . ◻

Lemma 9.11. Let T ∈ α(H,H) be a self adjoint and compact operator. Then, ∥T ∥ or−∥T ∥ is an eigenvalue of T .

Proof of Lemma 9.11. There exists a sequence xnn∈N (∥xn∥ ≤ 1) such that

∣⟨Txn, xn⟩∣→ ∥T ∥.This holds due to the definition of a norm of T . Since the T is compact one can chose asubsequence xnk

which converges. For transparency purposes we omit the lower index kand denote this subsequence as xnn∈N . Denote

λ = limn→+∞

⟨Txn, xn⟩.

By compactness of T there exists y such that

y = limn→+∞

Txn ≤ ∥T ∥

Therefore,

∥Txn − λxn∥2 = ⟨Txn − λxn, Txn − λxn⟩= ∥Txn∥2 − 2λ⟨Txn, xn⟩ + λ2∥xn∥ ≤ λ2 + ∥Txn∥2 − 2λ⟨Txn, xn⟩.


A term ∥Txn∥2converges to λ2 and ⟨Txn, xn⟩ converges to λ. Thus,

∥Txn − λxn∥2 → 0

and y = limn→+∞ λxn. Since T is continuous and the sequence xnn∈N converges strongly,it holds that

Ty = limn→+∞

Tλxn = limn→+∞

λTxn = λy.Thus, λ ∈ σ(T ). Note that since the spectrum of T is contained in R, the only possibilityfor a value of λ is ∥T ∥ or −∥T ∥. ◻

Theorem 9.12. (Spectral theorem for compact self adjoint operator)Let T ∈ α(H,H) be a self adjoint and compact operator. Then, there exists an orthonormalsystem S = ek ∶ k ∈ N and a sequence λii∈N ⊂ R converging to zero as i tends to infty,such that

(9.13) H = ker(T )⊕L2 lin(S)and

Tx =+∞∑i=1

λk⟨x, ek⟩ek, ∀x ∈H,

where sup ∣λk∣ = ∥T ∥.

Proof of Theorem 9.12. We claim that

dim(ker(λiId − T )) < +∞.Indeed, if there were infinitely many v such that v ∈ ker(λiId − T ) (which is equivalent toTv = λiv), then a dimension of the (scaled) unit ball would be infinite. However, in aninfinite dimensional space the unit ball is not compact, which is a contradiction to the factthat T is compact. In each space ker(λiId−T ) one can choose an orthonormal basis. Then,for k ≠ j we have Tek = λkek and Tej = λjej. Since ek ⊥ ej,

ker(λkId − T ) ⊥ ker(λjId − T ).Set λ1 to be equal to ∥T ∥ or −∥T ∥. The corresponding space would be H1 = ker(λ1Id− T ).We make a decomposition

H =H1 ⊕L2 (H1)⊥

and restrict the operator to the subspace (H1)⊥. A norm of the restricted operator is notbigger 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 λii∈N converges to0. However, if λii∈N stabilized on some level, then it would be a contradiction withcompactness of T (argument similar to the one used for proving that a kernel of λiId − Tis finite dimensional). ◻

LECTURE 10Locally 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 absolutely convex set if it is convex and (a) holds.

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 set A has the disc property, if it is balanced. Thus, it issufficient to check if −1 ⋅A ⊂ A.

Remark 27. If a set A is absolutely convex and absorbing, then

pA(x) = infλ > 0 ∶ x ∈ λAis a seminorm.

Remark 28. If p is a seminorm, then

A = x ∈X ∶ p(x) ≤ 1is 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 arecontinuous (in the sense that a preimage of each open set is open).

Definition 10.3. Let P be a family of seminorms on a linear space X and τ be generatedby 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 forma basis of neighborhoods of 0.

55


Theorem 10.4. A topological linear space (X,τ) is locally convex, if there exists a basisof 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 topologyτ 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 U of open sets such that ⋂U∈U U = 0.

Proof of Lemma 10.5. (a) ⇒ (b) Assume that (X,τ) is a Hausdorff space and letx ≠ 0. Then, there exist neighborhoods of zero U,V such that (x + U) ∩ V = ∅. Withoutloss 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 ∈⋂F,ε

UF,ε =⋂F,ε

x ∶ p(x) ≤ ε, ∀p ∈ F.

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 ofthe basis of neighborhoods of 0, see Definition 10.3) such that x − y ∉ U . Addition (andsubtraction) is a continuous operation. Thus, the preimage of U is open and we can findtwo 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.

(a) X = Cb(Rn), P = pt, t ∈ Rn, where pt(f) = ∣f(t)∣ (topology of pointwise conver-gence).

(b) X - normed space, P =X ′, px′(x) = ∣x′(x)∣ (denoted as σ(X,X ′)).(c) X - normed space, P =X ′, px(x′) = ∣x′(x)∣ (denoted as σ(X ′,X)).

Lemma 10.6. Let (X,τ) be a locally convex space and P be a family of seminorms, whichgenerates 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 maxp∈F

p(x) ∀x ∈X.

Proof of Lemma 10.6. Implications (a.1)⇒ (a.2)⇒ (a.3) are straightforward. Assumethat (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),


∣q(x + y) − q(x)∣ ≤ q((x + y) − x).Therefore,

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 ∶ p(x) < 1 are open and thus, all p arecontinuous (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 ⊂ Psuch that UF,ε ⊂ x ∶ q(x) ≤ 1 holds, which means that

q(x) ≤ (1/ε) supp∈F

p(x) ∀x ∈X.

◻

Remark 29. Let P andQ 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 locallyconvex 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 linearoperator. 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(Tx) ≤M maxp∈F

p(x).

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 aconstant M > 0 such that

l(x) ≤M maxi=1,...,n

pi(x) ∀x ∈X.

If τp1 ⊂ τp2 , then we say that τp2 is stronger. In particular, it has more open sets, closedsets, continuous operators, but less convergent sequences (since it is harder to prove thatlimn→+∞ p(un − u) = 0 for each seminorm p).


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 showthat

pt(x) ≤ ∥x∥∞ ∀pt ∈ P,and thus the topology τ∞ is stronger.

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 afamily of seminorms defined by linear functionals. One can show that the unit ball in thenorm 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) bea locally convex space, where X = l2(N) and τ = σ(X,X ′) = σ(X ′,X). Let

A = em +men ∶ 1 ≤m < n.

Then, 0 ∈ A σ(X,X′), but there is no sequence xnn∈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 theclosure of A indeed. Without loss of generality we can assume that

U = x ∈ l2(N) ∶ ∣⟨x, yi⟩∣ ≤ ε i = 1, . . . , r and yi ∈ l2(N).Let m be such that

∣yi(m)∣ < ε/2, for i = 1, . . . , r

and choose n >m such that

m ⋅ ∣yi(n)∣ ≤ ε/2 for i = 1, . . . , r.

Then,

∣⟨em +men, yi⟩∣ ≤ ∣yi(m)∣ +m ⋅ ∣yi(n)∣ ≤ εand therefore em +men ∈ U .On the contrary, let emk

+mkenkk∈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 emk+mkenk

should be bounded (uniformly with respect to k), since each weakly convergent sequenceis bounded. But,

⟨em +men, em +men⟩ = 1 +m2,

which is unbounded.

Theorem 10.9. (Hahn-Banach theorem) Let X be a locally convex space and U be itssubspace. Assume that l ∈ U ′. Then, there exists an extension L ∈X ′.

Lecture Notes in Functional Analysisaswiercz/AnalizaF/lecture.pdf · Lecture Notes in Functional...

Documents

Transcript of Lecture Notes in Functional Analysisaswiercz/AnalizaF/lecture.pdf · Lecture Notes in Functional...