Functional Analysis I, Study notes of Functional Analysis

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Functional Analysis I

Term 1, 2009–

Vassili Gelfreich

• 1 Vector spaces
  • 1.1 Definition.
  • 1.2 Examples of vector spaces
  • 1.3 Hamel bases
• 2 Normed spaces
  • 2.1 Norms
  • 2.2 Four famous inequalities
  • 2.3 Examples of norms on a space of functions
  • 2.4 Equivalence of norms
  • 2.5 Linear Isometries
• 3 Convergence in a normed space
  • 3.1 Definition and examples
  • 3.2 Topology on a normed space
  • 3.3 Closed sets
  • 3.4 Compactness
• 4 Banach spaces
  • 4.1 Completeness: Definition and examples
  • 4.2 The completion of a normed space
  • 4.3 Examples
• 5 Lebesgue spaces
  • 5.1 Integrable functions
  • 5.2 Properties of Lebesgue integrals
  • 5.3 Lebesgue space L^1 (R)
  • 5.4 Lp spaces
• 6 Hilbert spaces
  • 6.1 Inner product spaces
  • 6.2 Natural norms
  • 6.3 Parallelogram law and polarisation identity
  • 6.4 Hilbert spaces: Definition and examples
• 7 Orthonormal bases in Hilbert spaces
  • 7.1 Orthonormal sets
  • 7.2 Gram-Schmidt orthonormalisation
  • 7.3 Bessel’s inequality
  • 7.4 Convergence
  • 7.5 Orthonormal basis in a Hilbert space
• 8 Closest points and approximations
  • 8.1 Closest points in convex subsets
  • 8.2 Orthogonal complements
  • 8.3 Best approximations
  • 8.4 Weierstrass Approximation Theorem
• 9 Separable Hilbert spaces
  • 9.1 Definition and examples
  • 9.2 Isometry to `
• 10 Linear maps between Banach spaces
  • 10.1 Continuous linear maps
  • 10.2 Examples
  • 10.3 Kernel and range
• 11 Linear functionals
  • 11.1 Definition and examples
  • 11.2 Riesz representation theorem
• 12 Linear operators on Hilbert spaces
  • 12.1 Complexification
  • 12.2 Adjoint operators
  • 12.3 Self-adjoint operators
• 13 Introduction to Spectral Theory
  • 13.1 Point spectrum
  • 13.2 Invertible operators
  • 13.3 Resolvent and spectrum
  • 13.4 Compact operators
  • 13.5 Spectral theory for compact self-adjoint operators
• 14 Sturm-Liouville problems

Preface

These notes follow the lectures on Functional Analysis given in the Autumn Term of

2009. If you find a mistake or misprint please inform the author by sending an e-mail to [email protected]. The author thanks James Robinson for his set of notes and selection of exercises which significantly facilitated the preparation of the lectures. The author also thanks all students who helped with proofreading the notes.

1 Vector spaces

1.1 Definition.

A vector space V over a field K is a set equipped with two binary operations called vector addition and multiplication by scalars. Elements of V are called vectors and elements of K are called scalars. The sum of two vectors x, y ∈ V is denoted x + y, the product of a scalar α ∈ K and vector x ∈ V is denoted αx. It is possible to consider vector spaces over an arbitrary field K, but we will con- sider the fields R and C only. So we will always assume that K denotes either R or C and refer to V as a real or complex vector space respectively. In a vector space, addition and multiplication have to satisfy the following set of axioms: Let x, y, z be arbitrary vectors in V , and α, β be arbitrary scalars in K, then

• Associativity of addition: x + (y + z) = (x + y) + z.
• Commutativity of addition: x + y = y + z.
• There exists an element 0 ∈ V , called the zero vector, such that x + 0 = x for all x ∈ V.
• For all x ∈ V , there exists an element y ∈ V , called the additive inverse of x, such that x + y = 0. The additive inverse is denoted −x.
• “Associativity” of multiplication:^1 α(β x) = (αβ )x.
• Distributivity:

α(x + y) = αx + αy and (α + β )x = αx + β x.

• There is an element 1 ∈ K such that 1x = x for all x ∈ V. This element is called the multiplicative identity in K. (^1) The purist would not use the word “associativity” for this property as it includes two different

operations: αβ is a product of two scalars and β x involves a vector and scalar.

7. The space C[ 0 , 1 ] of all real-valued continuous functions on the closed interval [ 0 , 1 ] is a vector space. The addition and multiplication by scalars are defined naturally: for f , g ∈ C[ 0 , 1 ] and α ∈ R we defined by f + g the function whose values are given by

( f + g)(t) = f (t) + g(t) , t ∈ [ 0 , 1 ] ,

and α f is the function whose values are

(α f )(t) = α f (t) , t ∈ [ 0 , 1 ] ,

We will always use these definitions on all spaces of functions to be considered later.

8. The set ˜L^1 ( 0 , 1 ) of all real-valued continuous functions f on the open interval ( 0 , 1 ) for which ∫ (^1)

0

| f (t)| dt < ∞

is a vector space. If f ∈ C[ 0 , 1 ] then f ∈ L˜^1 ( 0 , 1 ). Indeed, since [ 0 , 1 ] is compact f is bounded (and attains its lower and upper bounds). Then ∫ (^1)

0

| f (t)| dt ≤ max t∈[ 0 , 1 ]

| f (t)| < ∞ ,

i.e. f ∈ L˜^1 ( 0 , 1 ). We note that ˜L^1 ( 0 , 1 ) contains some functions which do not belong to C[ 0 , 1 ]. For example, f (t) = t−^1 /^2 is not continuous on [ 0 , 1 ] but it is continuous on ( 0 , 1 ) and (^) ∫ 1 0

| f (t)| dt =

∫ (^1)

0

|t|−^1 /^2 dx = 2 t^1 /^2

1 0

so f ∈ L˜^1 ( 0 , 1 ). We conclude that C[ 0 , 1 ] is a strict subset of f ∈ L˜^1 ( 0 , 1 ).

1.3 Hamel bases

Definition 1.1 The linear span of a subset E of a vector space V is the collection of all finite linear combinations of elements of E:

Span(E) =

x ∈ V : x =

n ∑ j= 1

α (^) je (^) j, n ∈ N, α (^) j ∈ K, e (^) j ∈ E

We say that E spans V if V = Span(E), i.e. every element of V can be written as a finite linear combination of elements of E.

Definition 1.2 A set E is linearly independent if any finite collection of elements of E is linearly independent:

n ∑ j= 1

α (^) je (^) j = 0 =⇒ α 1 = α 2 = · · · = αn = 0

for any choice of n ∈ N, e (^) j ∈ E and α (^) j ∈ K.

Definition 1.3 A Hamel basis E for V is a linearly independent subset of V which spans V.

Examples:

1. Any basis in Rn^ is a Hamel basis.
2. The set E =

1 , x, x^2 ,...

is a Hamel basis in the space of all polynomials.

Lemma 1.4 If E is a Hamel basis for a vector space V then any element x ∈ V can be uniquely written in the form

x =

n ∑ j= 1

α (^) je (^) j

where n ∈ N, α (^) j ∈ K and e (^) j ∈ E.

Exercise: Prove the lemma.

Definition 1.5 We say that a set is finite if it consists of a finite number of elements.

Theorem 1.6 If V has a finite Hamel basis then every Hamel basis for V has the same number of elements.

Definition 1.7 If V has a finite basis E then the dimension of V (denoted dimV ) is the number of elements in E. If V has no finite basis then we say that V is infinite- dimensional.

Example: In Rn^ any basis consists of n vectors. Therefore dim Rn^ = n.

Let V and W be two vector spaces over K.

Definition 1.8 A map L : V → W is called linear if for any x, y ∈ V and any α ∈ K

L(x + αy) = L(x) + αL(y).

Definition 1.9 If a linear map L : V → W is a bijection, then L is called a linear isomorphism. We say that V and W are linearly isomorphic if there is a bijective linear map L : V → W.

2 Normed spaces

2.1 Norms

Definition 2.1 A norm on a vector space V is a map ‖ · ‖ : V → R such that for any x, y ∈ V and any α ∈ K:

1. ‖x‖ ≥ 0 , and ‖x‖ = 0 ⇔ x = 0 (positive definiteness);
2. ‖αx‖ = |α| ‖x‖ (positive homogeneity);
3. ‖x + y‖ ≤ ‖x‖ + ‖y‖ (triangle inequality).

The pair (V, ‖ · ‖) is called a normed space.

In other words, a normed space is a vector space equipped with a norm.

Examples:

1. Rn^ with each of the following norms is a normed space:

(a) ‖x‖ =

n ∑ k= 1

|xk|^2

(b) ‖x‖p =

n ∑ k= 1

|xk|p

) 1 /p , 1 ≤ p < ∞

(c) ‖x‖∞ = max 1 ≤k≤n

|xk|.

2. `p(K) is a vector space with the following norm (1 ≤ p < ∞)

‖x‖`p =

∞ ∑ k= 1

|xk|p

) 1 /p .

3. `∞(K) is a vector space with the following norm

‖x‖`∞^ = sup k∈N

|xk|.

In order to prove the triangle inequality for the `p^ norm, we will state and prove several inequalities.

2.2 Four famous inequalities

Lemma 2.2 (Young’s inequality) If a, b > 0 , 1 < p, q < ∞, (^1) p + (^1) q = 1 , then

ab ≤

ap p

bq q

Proof: Consider the function f (t) = t^

p p −t^ +^

1 q defined for^ t^ ≥^ 0. Since^ f^

′(t) = t p− (^1) − 1

vanishes at t = 1 only, and f ′′(t) = (p− 1 )t p−^2 ≥ 0, the point t = 1 is a global minimum for f. Consequently, f (t) ≥ f ( 1 ) = 0 for all t ≥ 0. Now substitute t = ab−q/p:

f (ab−q/p) =

apb−q p

− ab−q/p^ +

q

Multiplying the inequality by bq^ yields Young’s inequality. 

Lemma 2.3 (H¨older’s inequality) If 1 ≤ p, q ≤ ∞, (^1) p + (^1) q = 1 , x ∈ p(K), y ∈q(K), then (^) ∞

∑ j= 1

|x (^) jy (^) j| ≤ ‖x‖p^ ‖y‖q^.

Proof. If 1 < p, q < ∞, we use Young’s inequality to get that for any n ∈ N

n ∑ j= 1

|x (^) j| ‖x‖`p

|y (^) j| ‖y‖`q

n ∑ j= 1

p

|x (^) j|p ‖x‖ `pp

q

|y (^) j|q ‖y‖q `q

p

q

Therefore for any n ∈ N n ∑ j= 1

|x (^) jy (^) j| ≤ ‖x‖p^ ‖y‖q^.

Since the partial sums are monotonically increasing and bounded above, the series converge and H¨older’s inequality follows by taking the limit as n → ∞. If p = 1 and q = ∞: n ∑ j= 1

|x (^) jy (^) j| ≤ max 1 ≤ j≤n

|y (^) j|

n ∑ j= 1

|x (^) j| ≤ ‖x‖1 ‖y‖∞^.

Therefore the series converges and H¨older’s inequality follows by taking the limit as n → ∞. 

Lemma 2.4 (Cauchy-Schwartz inequality) If x, y ∈ `^2 (K) then

∞ ∑ j= 1

|x (^) jy (^) j| ≤

∞ ∑ j= 1

|x (^) j|^2

∞ ∑ j= 1

|y (^) j|^2

1. the “sup(remum) norm” ‖ f ‖∞ = sup t∈[ 0 , 1 ]

| f (t)| ;

2. the “L^1 norm” ‖ f ‖L 1 =

∫ (^1)

0

| f (t)| dt ;

3. the “L^2 norm”

‖ f ‖L 2 =

0

| f (t)|^2 dt

Exercise: Check that each of these formulae defines a norm. For the case of the L^2 norm, you will need a Cauchy-Schwartz inequality for integrals.

Example: Let k ∈ N. The space Ck[ 0 , 1 ] consists of all continuous real-valued func- tions which have continuous derivatives up to order k. The norm on Ck[ 0 , 1 ] is defined by

‖ f ‖Ck =

k ∑ j= 0

sup t∈[ 0 , 1 ]

| f (^ j)(t)| ,

where f (^ j)^ denotes the derivative of order j.

2.4 Equivalence of norms

We have seen that various different norms can be introduced on a vector space. In order to compare different norms it is convenient to introduce the following equivalence relation.

Definition 2.6 Two norms ‖ · ‖ 1 and ‖ · ‖ 2 on a vector space V are equivalent if there are constants c 1 , c 2 > 0 such that

c 1 ‖x‖ 1 ≤ ‖x‖ 2 ≤ c 2 ‖x‖ 1 for all x ∈ V.

In this case we write ‖ · ‖ 1 ∼ ‖ · ‖ 2.

Theorem 2.7 Any two norms on Rn^ are equivalent.^3

Example: The norms ‖ · ‖L 1 and ‖ · ‖∞ on C[ 0 , 1 ] are not equivalent.

(^3) You already saw this statement in Analysis III and Differentiation in Year 2. The proof is based

on the observation that the unit sphere S ⊂ Rn^ is sequentially compact. Then we checked that f (x) = ‖x‖ 2 /‖x‖ 1 is continuous on S and consequently it is bounded and attains its lower and upper bounds on S. We set c 1 = minS f and c 2 = maxS f.

Proof: Consider the sequence of functions fn(t) = tn^ with n ∈ N. Obviously fn ∈ C[ 0 , 1 ]. We see that

‖ fn‖∞ = max t∈[ 0 , 1 ]

|t|n^ = 1 ,

‖ fn‖L 1 =

∫ (^1)

0

tn^ dt =

n + 1

Suppose the norms are equivalent. Then there is a constant c 2 > 0 such that for all fn:

‖ fn‖∞ ‖ fn‖L 1

= n + 1 ≤ c 2 ,

which is not possible for all n. This contradiction implies the norms are not equivalent. 

2.5 Linear Isometries

Suppose V and W are normed spaces.

Definition 2.8 If a linear map L : V → W preserves norms, i.e. ‖L(x)‖ = ‖x‖ for all x ∈ V , it is called a linear isometry.

This definition implies L is injective, i.e., L : V → L(V ) is bijective, but it does not im- ply L(V ) = W , i.e., L is not necessarily invertible. Note that sometimes the invertibility property is included into the definition of the isometry. Finally, in Metric Spaces the word “isometry” is used to denote distance-preserving transformations.

Definition 2.9 We say that two normed spaces are isometrically isomorphic (or simply isometric), if there is an invertible linear isometry between them.

A linear invertible map can be used to “pull back” a norm as follows.

Proposition 2.10 Let (V, ‖ · ‖V ) be a normed space, W a vector space, and L : W → V a linear isomorphism. Then ‖x‖W := ‖L(x)‖V

defines a norm on W.

Proof: For any x, y ∈ V and any α ∈ K we have:

‖x‖W = ‖L(x)‖V ≥ 0 , ‖αx‖W = ‖L(αx)‖V = |α| ‖L(x)‖V = |α| ‖x‖W.

If ‖x‖W = ‖L(x)‖V = 0, then L(x) = 0 due to non-degeneracy of the norm ‖ · ‖V. Since L is invertible, we get x = 0. Therefore ‖ · ‖W is non-degenerate.

3 Convergence in a normed space

3.1 Definition and examples

The norm on a vector space V can be used to measure distances between points x, y ∈ V. So we can define the limit of a sequence.

Definition 3.1 A sequence (xn)∞ n= 1 , xn ∈ V , n ∈ N, converges to a limit x ∈ V if for any ε > 0 there is N ∈ N such that

‖xn − x‖ < ε for all n > N.

Then we write xn → x.

We note that the sequence of vectors xn → x if and only if the sequence of non- negative real numbers ‖xn − x‖ → 0.

Proposition 3.2 The limit of a convergent sequence is unique.

Exercise: Prove it.

Proposition 3.3 Any convergent sequence is bounded.

Exercise: Prove it.

Proposition 3.4 If xn converges to x, then ‖xn‖ → ‖x‖.

Exercise: Prove it.

It is possible to check convergence of a sequence of real numbers without actually finding its limit: it is sufficient to check that it satisfies the following definition:

Definition 3.5 (Cauchy sequence) A sequence (xn)∞ n= 1 in a normed space V is Cauchy if for any ε > 0 there is an N such that

‖xn − xm‖ < ε for all m, n > N.

Theorem 3.6 A sequence of real numbers converges iff it is Cauchy.

Proposition 3.7 Any convergent sequence is Cauchy.

Exercise: Prove it.

Proposition 3.8 Any Cauchy sequence is bounded.

Exercise: Prove it.

Example: Consider the sequence fn ∈ C[ 0 , 1 ] defined by fn(t) = tn.

1. fn → 0 in the L^1 norm. Proof: We have already computed the norms:

‖ fn‖L 1 =

n + 1

Consequently, fn → 0. 

2. fn does not converge in the sup norm. Proof: If m > 2 n ≥ 1 then

fn( 2 −^1 /n) − fm( 2 −^1 /n) =

2 m/n^

Consequently ( fn) is not Cauchy in the sup norm and hence not convergent.

This example shows that the convergence in the L^1 norm does not imply the point- wise convergence and, as a results, does not imply the convergence in the sup norm (often called the uniform convergence). Note that in contrast to the uniform and L^1 convergences the notion of pointwise convergence is not based on a norm on the space of continuous function.

Exercise: The pointwise convergence does not imply the L^1 convergence. Hint: Construct fn with a very small support but make the maximum of fn very large to ensure that ‖ fn‖L 1 > n. Therefore fn is not bounded in the L 1 norm, hence not convergent. We can also make supp fn ∩ supp fm = 0 for all/ m, n such that n 6 = m. Then for any t there is at most one n such that fn(t) 6 = 0. The last property guarantees pointwise convergence: fn(t) → 0 for any t. 

Proposition 3.9 If fn ∈ C[ 0 , 1 ] for all n ∈ N and fn → f in the sup norm, then fn → f in the L^1 norm, i.e.,

‖ fn − f ‖∞ → 0 =⇒ ‖ fn − f ‖L 1 → 0.

Proof:

0 ≤ ‖ fn − f ‖L 1 =

∫ (^1)

0

| fn(t) − f (t)| dt ≤ sup 0 ≤t≤ 1

| fn(t) − f (t)| = ‖ fn − f ‖∞ → 0.

Therefore ‖ fn − f ‖L 1 → 0. 

We have seen that different norms may lead to different conclusions about conver- gence of a given sequence but sometime convergence in one norm implies convergence in another one. The following lemma shows that equivalent norms give rise to the same notion of convergence.

Definition 3.12 An open neighbourhood of x is an open set which contains x.

Lemma 3.13 A sequence xn → x if and only if for any open neighbourhood X of x there is N ∈ N such that xn ∈ X for all n > N.

Proof: ( =⇒ ). Let xn → x. Take any open X such that x ∈ X. Then there is ε > 0 such that B(x, ε) ⊂ X. Since the sequence converges there is N such that ‖xn − x‖ < ε for all n > N. Then xn ∈ B(x, ε) ⊂ X for the same values of n. (⇐=). Take any ε > 0. The ball B(x, ε) is open, therefore there is N such that xn ∈ B(x, ε) for all n > N. Hence ‖xn − x‖ < ε and xn → x. 

3.3 Closed sets

Definition 3.14 A set X ⊂ V is closed if its complement V \ X is open.

Example: In any normed space V :

1. The unit sphere S = { x : ‖x‖ = 1 } is closed
2. Any closed ball B(x, ε) = { y ∈ V : ‖y − x‖ ≤ ε } is closed.
3. V is closed.
4. The empty set is closed.

Lemma 3.15 A subset X ⊂ V is closed if and only if any convergent sequence in X has its limit in X.

Proof: The proof literally repeats the proof given in Differentiation. 

Definition 3.16 We say that a subset L ⊂ V is a linear subspace, if it is a vector space itself, i.e., if x 1 , x 2 ∈ L and λ ∈ K imply x 1 + λ x 2 ∈ L.

Proposition 3.17 Any finite dimensional linear subspace L of a normed space V is closed.

Proof: Since L is finite-dimensional, it has a finite Hamel basis

E = { e 1 , e 2 ,... , en } ⊂ L

such that L = Span(E). Suppose L is not closed, then by Lemma 3.15 there is a convergent sequence xk → x∗, xk ∈ L but x∗^ ∈ V \ L. Then x∗^ is linearly independent from E (otherwise it would belong to L). Consequently

E˜ = { e 1 , e 2 ,... , en, x∗^ }

is a Hamel basis in ˜L = Span( E˜). In this basis, the components of xk are given by (α 1 k ,... , αnk , 0 ) and x∗^ corresponds to the vector ( 0 ,... , 0 , 1 ). We get in the limit as k → ∞ (α 1 k ,... , αnk , 0 ) → ( 0 ,... , 0 , 1 ) ,

which is obviously impossible. Therefore L is closed.^4 

Example: The subspace of polynomial functions is linear but not closed in C[ 0 , 1 ] equipped with the sup norm.

3.4 Compactness

Definition 3.18 (sequential compactness) A subset K of a normed space (V, ‖ · ‖V ) is (sequentially) compact if any sequence (xn)∞ n= 1 with xn ∈ K has a convergent subse- quence xn (^) j → x∗^ with x∗^ ∈ K.

Proposition 3.19 A compact set is closed and bounded.

Theorem 3.20 A subset of Rn^ is compact iff it is closed and bounded.

Corollary 3.21 A subset of a finite-dimensional vector space is compact iff it is closed and bounded.

Example: The unit sphere in `p(K) is closed, bounded but not compact.

Proof: Take the sequence e (^) j such that

e (^) j = ( 0 ,... , 0 , (^) ︸︷︷︸ 1 jth^ place

We note that ‖e (^) j − ek‖`p = 21 /p^ for all j 6 = k. Consequently, (e (^) j)∞ j= 1 does not have any convergent subsequence, hence S is not compact. 

Lemma 3.22 (Riesz’ Lemma) Let X be a normed vector space and Y be a closed linear subspace of X such that Y 6 = X and α ∈ R, 0 < α < 1. Then there is xα ∈ X such that ‖xα ‖ = 1 and ‖xα − y‖ > α for all y ∈ Y.

Proof: Since Y ⊂ X and Y 6 = X there is x ∈ X \Y. Since Y is closed, X \Y is open and therefore d := inf{ ‖x − y‖ : y ∈ Y } > 0. (^4) This proof implicitly uses equivalence of norms on Rn+ (^1) to establish that convergence in the norm

obtained by restricting the original norm ‖ · ‖V onto ˜L implies convergence of the components of the vectors.