introduction to functional analysis -...

Introduction to Functional Analysis

Yen Do

Fall 2015

Preface

This is the accompanying expository notes for an introductory course in Functional Analysisthat I was teaching at UVA. The goal of the course is to study the basic principles of linearanalysis, including the spectral theory of compact and self-adjoint operators. This is not amonograph or a treatise and of course no originality is claimed. The prerequisiteis some basic knowledge about real analysis and topology. Some preliminary understandingof functional analysis is beneficial but not required.

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Contents

I Linear Spaces 9

1 Basic facts 111.1 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.1.1 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.2 Linear spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.2.1 The Hahn-Banach extension theorem . . . . . . . . . . . . . . . . . . 151.2.2 The HB theorem with symmetry constraints . . . . . . . . . . . . . . 17

1.3 Topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.3.1 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.3.2 Continuous functions on LCH spaces . . . . . . . . . . . . . . . . . . 191.3.3 Proof of Urysohn’s lemma . . . . . . . . . . . . . . . . . . . . . . . . 201.3.4 Proof of Tietze’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 211.3.5 Partition of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2 Overview of normed linear spaces 232.1 Linear functionals and dual spaces . . . . . . . . . . . . . . . . . . . . . . . . 25

2.1.1 Linear functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.1.2 Dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.1.3 Reflexivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.2 (Non)compactness of the unit ball . . . . . . . . . . . . . . . . . . . . . . . . 28

3 Geometry and topology on NLS 313.1 Topologies on normed linear spaces . . . . . . . . . . . . . . . . . . . . . . . 31

3.1.1 The Banach–Alaoglu theorem . . . . . . . . . . . . . . . . . . . . . . 313.1.2 The sequential Banach-Alaoglu theorem . . . . . . . . . . . . . . . . 323.1.3 Weak compactness of the closed unit ball in reflexive spaces . . . . . 33

3.2 Uniform convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.3 Isometries between normed linear spaces . . . . . . . . . . . . . . . . . . . . 35

4 Basis on Hilbert spaces and Banach spaces 394.1 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2 Orthogonal basis for a Hilbert space . . . . . . . . . . . . . . . . . . . . . . . 404.3 Schauder basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

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6 CONTENTS

5 Bounded maps on Banach spaces 45

5.1 The Baire category theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5.2 The principle of uniform boundedness . . . . . . . . . . . . . . . . . . . . . . 45

5.3 The open mapping theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5.4 The closed graph theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5.5 Uniform boundedness of canonical Schauder projections . . . . . . . . . . . . 47

5.6 Lp spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.6.1 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.6.2 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.6.3 Basic facts about bounded operators on Lp . . . . . . . . . . . . . . . 50

6 Bounded continuous functions and dual spaces 53

6.1 Riesz representation theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6.1.1 The dual space of L∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6.1.2 Measures on a locally compact Hausdorff spaces . . . . . . . . . . . . 54

6.1.3 The dual space of Co(X) and Cc(X) . . . . . . . . . . . . . . . . . . 54

6.2 The Stone-Weierstrass approximation theorem . . . . . . . . . . . . . . . . . 56

7 Locally convex spaces 59

7.1 Two equivalent definitions of LCS . . . . . . . . . . . . . . . . . . . . . . . . 59

7.1.1 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

7.1.2 Linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

7.2 The Hyperplane separation theorem . . . . . . . . . . . . . . . . . . . . . . . 61

7.3 The Krein-Milman theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

7.4 Inductive limit and weak solutions . . . . . . . . . . . . . . . . . . . . . . . . 63

II Spectral analysis for linear operators 65

8 Elementary spectral theory 67

8.1 Spectrum and resolvent set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

8.2 Functional calculus and spectral mapping . . . . . . . . . . . . . . . . . . . . 68

8.3 Examples of operators and their spectra . . . . . . . . . . . . . . . . . . . . 69

8.4 Adjoint operators and spectrum . . . . . . . . . . . . . . . . . . . . . . . . . 70

9 Compact operators 73

9.1 Compact operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

9.2 Compactness of integral operators . . . . . . . . . . . . . . . . . . . . . . . . 75

9.3 Spectral properties of compact operators . . . . . . . . . . . . . . . . . . . . 76

9.3.1 Riesz’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

9.3.2 Spectral properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

CONTENTS 7

10 Bounded self-adjoint operators 7910.1 Diagonalization form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7910.2 Projection-valued measure and spectral projection . . . . . . . . . . . . . . . 8110.3 Spectral representation and decomposition . . . . . . . . . . . . . . . . . . . 82

11 Spectral theory for unbounded self-adjoint operator 85

12 Fredholm determinant 8912.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8912.2 Fredholm’s approach for integral operators . . . . . . . . . . . . . . . . . . . 90

12.2.1 Convergence and Continuity of the determinant . . . . . . . . . . . . 9112.3 Fredholm determinant for operators on Hilbert spaces . . . . . . . . . . . . . 95

8 CONTENTS

Part I

Linear Spaces

9

Chapter 1

Preliminaries

Generally speaking, a space is a set with some structures, which could be- algebraic (add, multiply)- geometric (distance, angle, convexity)- topological (open sets, continuity)Our goal is to study spaces of functions and their structures using analytic tools. One

could think of this study as one place where“analysis” meets “linear algebra” and “geome-try/topology”.

Our focus will be on linear spaces with some notion of geometry/topology.Some times further understanding of these structures could be obtained via looking at

functions on these spaces, which could takes value in R, C, or even in other spaces withsimilar structure. We refer to these functions are operators/maps/functionals depending onthe nature of the target spaces.

In this chapter we will recall several basic facts about metric spaces, linear spaces, andtopological spaces.

1.1 Metric spaces

. Here we have an additional geometric structure (distance) on the given set M , namely dmeasures the distance between two elements of M , such that

• (positive) d(x, y) ≥ 0 equality iff x = y, and

• (symmetric) d(x, y) = d(y, x), and

• (triangle inequality) d(x, y) + d(y, z) ≥ d(x, z).

A sequence (xn) converges to x in (M,d) if limn→∞ d(xn, x) = 0 A sequence (xn) is calleda Cauchy sequence in (M,d) if limmin(m,n)→∞ d(xn, xm) = 0.

It is clear that “convergence” ⇒ “Cauchy”.If the reverse holds then the metric space is called complete.

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12 CHAPTER 1. BASIC FACTS

Definition 1. A metric space (M,d) is complete if any Cauchy sequence is convergent.(Note that the sequence and the limit are required to be in M).

Examples: M = continuous functions on [0, 1], let

d1(f, g) = supx∈[0,1] |f(x)− g(x)|d2(f, g) =

∫ 1

0|f(x)− g(x)|dx

Then (M,d1) is complete but (M,d2) is not complete.

Question: if (M,d) is imcomplete can we add more stuffs to make it complete?

Definition 2. M1 ⊂ M2 is dense in M2 wrt metric d if every m ∈ M2 could be ap-proximated by a sequence in M1. In other words for some sequence (an) in M1 we havelimn→∞ d(an,m) = 0.

Theorem 1. If (M,d) is a metric space then there exists a complete metric space (M, d)

and a map h : M → M such that

d(m1,m2) = d(h(m1), h(m2))

and h(M) is dense in M .

(such a h is called an isometry between M and h(M).)

Ideas of the proof: One starts by considering the set of all Cauchy sequences in M .

Inside the new space M these sequences should converge to a limit, although two differentsequences may have the same limit. The idea is to identify these sequences together usingsome equivalence relation and build M from there.

More specifcally, for two Cauchy sequences x = (xn) and y = (yn) in M let d0(x, y) =limn→∞ d(xn, yn). It is clear that such limit exists since d(xn, yn) is a Cauchy sequence ofreal numbers and it is not hard to verify that this is indeed a metric.

If d0(x, y) = 0 we say that x and y are equivalent, and we then let M be the set of all

equivalence classes, and d be the induced metric on M .

Turns out (M, d) is complete (this requires some work using the Cantor diagonal trick),

and we could embed M inside M using the following isometry: for x ∈M ,

h(x) = [(x, x, . . . )]

the equivalence class of (x, x, x . . . ) in M . It is not hard to see that if the given metric has some reasonable symmetries then they

are preserved in the completion space. For instance, this applies if the space is linear andthe metric is dilation invariant or translation invariant (for instance if the metric is inducedfrom a norm). Thus we could easily see that the completion theorem also holds for normedlinear spaces.

1.1. METRIC SPACES 13

1.1.1 Separability

We recall that a topological space is separable if there exists a countable dense set. We saythat this set separates the space; for instance the rationals separate the real numbers.

A metric space is said to be separable if the topology induced from the metric is separable.Note that there is also the concept of second countable, which says that one could find a

countable collection open sets from which one could get all open sets in the topology simplyby taking union.

This would imply separability (just take one point from each such open set), however formetric/metrizable spaces these two are equivalent.

Separability is important in constructive proofs, for instance we could typically avoidZorn’s lemma/the axiom of choice if separability is available.

Basic properties:1. All compact metric spaces are separable.2. If the given topological space is an union of a countable number of separable subspaces,

then it is also separable.Combining Properties 1 and 2, it is clear that an Euclidean space RI (consisting of

functions from I to R) is separable if any only the dimension |I| is countable.3. A metric space is not separable if there is an uncountable collection of functions such

that the distance between any two is at least 1. (Indeed, if one could find a countable denseset then at least an element of this dense set has to be close to two functions in the givencollection, which by the triangle inequality would imply that the distance between the twohas to be small, contradictory to the given hypothesis).

Using Property 3, we could show that L∞[0, 1] (or L∞(R) etc.) is not separable. To seethis, one could simply take the collection C of characteristic functions of intervals [a, b] ⊂ [0, 1]where a, b are irrationals – there are uncountably many such functions and any two elementsof C are of L∞ distance at least 1, contradiction.

Using a similar line of reasoning, one could see that the space of all signed Borel measureson [0, 1] (with norm being the total mass) is not separable. Here simply use the (uncountable)collection of delta measures at points of [0, 1].

4. We could also define separability for measure spaces by defining a metric on theunderlying σ-algebra. Namely, given a σ-algebra A on X and a corresponding measure µ wecould define a metric on A by

p(A,B) = µ(A∆B)

the measure of the symmetric difference

A∆B = (A ∪B) \ (A ∩B)

and we say that (X,A, µ) is separable if the metric space (A, p) is separable.

Lemma 1. If a σ-algebra is generated from a countable collection of sets, then it is separable.

Proof. Let A be a countable generating set, let A′ consist of all finite intersections of elementsof A, and A′′ consist of all finite unions of element of A′, clearly A′′ is still countable and it


is an algebra of set (i.e. the union and intersection of any two elements are still inside A′′).Clearly A′′ also generates the given σ-algebra A, thus to show separability it suffices to showthat given any E ∈ A and any ε > 0 one could fine F ∈ A′′ such that µ(E∆F ) < ε. This isdone by looking at the set of all such E and show that it is actually a σ algebra containingA′′, thus by minimality it has to contain A and the desired claim follows.

Exercise: Rn with the usual Lebesgue measure is separable.

Theorem 2. Let (X,A, µ) be a separable measure space. Then for any 1 ≤ p <∞ the spaceLp(X,A, µ) is separable.

Proof: use simple functions. More details later.

1.2 Linear spaces

Here the additional structure is algebraic, namely we allow for addition of two elements anda somewhat limited notion of multiplication. Let F be a field (for us this is R or C), weallow for multiplication of an element of the space with an element of F, the space is thensaid to be linear over F.

Examples: 1. finite dimensional vector spaces (linear algebra).

2. Let p ∈ (0,∞). Then Lp(R) = all (complex valued Borel) measurable functions on Rsuch that

(

∫R|f(x)|pdx)1/p <∞

Similarly L∞(R) is all measurable f such that for some M > 0 finite the set x : |f(x)| > Mhas measure zero. Note that f could be unbounded.

Lp spaces are strong type spaces. We could also define weak Lp(R), denoted by Lp,∞(R)to be all measurable f such that

supλ>0

λ||f | > λ|1/p <∞

For p =∞, weak L∞ and L∞ are the same.

One could further generalize this to Lp on an arbitrary measure space. For instance wecould replace the Lebesgue measure with another Borel measure on R.

3. `p(N) = all complex sequences (a1, a2, . . . ) such that

(∑n≥1

|an|p)1/p <∞

Now `∞ = all bounded sequences.

Exercise: prove that these spaces are actually linear spaces, namely they are closedunder addition and scalar multiplication.

1.2. LINEAR SPACES 15

1.2.1 The Hahn-Banach extension theorem

Given a linear space X and p : X → R, we say p is homogeneously convex if

• (positive homogeneity) p(ax) = ap(x) for a > 0 and x ∈ X, and

• (subadditivity) p(x1 + x2) ≤ p(x1) + p(x2) for x1, x2 ∈ X.

Intuitively, the graph of p over X looks like a convex cone with top at the origin.Note that this is equivalent to

p(ax1 + bx2) ≤ ap(x1) + bp(x2)

for all a, b ≥ 0 with a+ b > 0 and x1, x2 ∈ X.The following is known as the HB extension theorem for linear spaces over R (there is a

version for C discussed later). Note that it does not requires geometric structures on X, allwe need is linearity.

Theorem 3. Let X be a linear space over R and let p : X → R be a homogeneously convexfunction. Let Y ⊂ X linear subspace and ` : Y → R linear such that

`(y) ≤ p(y) for all y ∈ Y .

Then there exists L : X → R linear that agrees with ` on Y such that

L(x) ≤ p(x) for all x ∈ X.

One could think of this theorem as an example of a local to global principle for linearmaps: as long as the local constraint is reasonable we could preserve it globally. Here weneed homogeneity and convexity.

Proof of the HB theorem in this generality uses Zorn’s lemma which is equivalent to theaxiom of choice. (for more concrete X such as Lp, `p, one could avoid the axiom of choice.)

A relation R on a set X is a subset of X×X (could be empty!). We say xRy if (x, y) ∈ R.R is an equivalence relation if three conditions holds:

• (reflexive) xRx for all x ∈ X,

• (symmetric) if xRy then yRx, and

• (transitive) if xRy and yRz then xRz.

We say R is a partial ordering if reflexive, transitive, and anti-symmetric (namely if xRyand yRx then x = y).

We say R is a linear ordering if x, y ∈ X with x 6= y then either xRy or yRx. In thiscase we sometimes say that X is a linearly ordered chain. Note that the size a chain couldbe uncountable.


We say α ∈ X is an upper bound for Y ⊂ X if yRα for all y ∈ Y .

Example: the inclusion order in P(R).

We say that an element m ∈ X is maximal if for every α ∈ X if mRα then α = m.

Lemma 2 (Zorn’s lemma). X 6= ∅, R is a partial ordering on X such that any linearlyordered chain Y inside X has an upper bound in X. Then for each such Y there is amaximal m ∈ X such that m is an upper bound for Y .

Note that it is very important that we could find an upper bound for a chain. Withoutthat we have a counter example: X = Z+ is the set of all positive integers, and the chain1 ≤ 2 ≤ . . . does not have any (maxinal) upper bound.

Technically speaking, this assumption is used in the proof of the Lemma (via say, trans-finite induction plus the axiom of choice); details could be found in most introductory text-books in mathematical logic.

Proof of the HB theorem. We divide the proof into two steps. In step 1, we show thatif Y is a proper subspace of X then we can increase the dimension of Y while preseving thegiven hypothesis. In step 2, by induction on the dimension of Y (iterating step 1) and usingZorn’s lemma we obtain the whole space X.

Step 1: Assume existence of z ∈ X \Y . Let Y = span(Y, z) = ay+bz, a, b ∈ R, y ∈ Y .Want to define `0 such that `0(y) = `(y) and

`0(ay + bz) = a`0(y) + b`0(z)

so only need `0(z). We also want

`0(ay + bz) ≤ p(ay + bz)

thusb`0(z) ≤ p(ay + bz)− a`(y)

all a, b ∈ R and y ∈ Y . Just need a = 1.By considering b > 0 and b < 0 we end up needing

supc>0,y∈Y

−1

c[p(y − cz)− `(y)]

≤ infc>0,y∈Y

1

c[p(y + cz)− `(y)]

After algebraic manipulations this follows from the given convexity assumption on p,more specifically the following estimate, c1, c2 > 0,

p(c1

c1 + c2

x+c2

c1 + c2

y) ≤ c1

c1 + c2

p(x) +c2

c1 + c2

p(y)

1.2. LINEAR SPACES 17

Step 2: Here we iterate step 1 and use Zorn’s lemma. If X is finite dimensional then itis clear that the process will stop after finitely many iterations. If X is infinitely dimensional(the dimension of X could even be uncountable), we need to be able to go beyond anylinearly ordered chain of these extension in order to apply Zorn’s lemma.

Formally, consider the set of all possible extensions of ` from Y to bigger subspaces ofX while remain dominated by p. Define a partial order where (Yα, `α) ≤ (Yβ, `β) if Yα ⊂ Yβand `α agrees with `β on Yα. Now any linearly ordered chain (Yα, `α)α∈I (here I could beuncountable) has an upper bound (Y ′, `′) defined by

Y ′ =⋃α∈I

Yα

and `′ is the same as `α inside Yα. Now apply Zorn’s lemma.

Corollary 1 (complex HB). X is a complex vector space, p : X → R+ such that

p(ax+ by) ≤ |a|p(x) + |b|p(y)

all |a| + |b| > 0 complex numbers. Assume that ` : Y → C linear functional such that|`(y)| ≤ p(y) all y ∈ Y . Then can extend ` linearly to X so that it is still dominated by p(y).

Note: the extension given in the HB theorems is not necessarily unique!

Proof. Idea of proof: `1(y) = Re(`(y)) and `2(y) = Im(`(y)) are real functionals boundedby p, and `2(y) = −`1(iy). So

`(y) = `1(y)− i`1(iy)

By real HB we could extend `1 to X, thus we extend ` to. But still need to show thatthe extended ` satisfies

|`(x)| ≤ p(x)

For any x, let α be such that |α| = 1 and `(x) = α|`(x)|, then the desired estimatefollows:

|`(x)| = `(α−1x) = `1(α−1x) ≤ p(α−1x) = p(x)

We emphasize again that technically speaking the HB extension theorem applies to any

linear spaces; although heuristically speaking the existence of p implies some implicit ge-omeric structures.

1.2.2 The HB theorem with symmetry constraints

One possible way to generalize the HB theorem is to introduce additional symmetries. Morespecifically, if the local space Y and the linear map ` (on Y ) and the convex function p(defined globally on X) are invariant under some “compatible” set of linear transformations,we would also like to impose this on our extension `.


More specifically, let A be a set of linear maps from X to itself that commute, thus ifT1, T2 ∈ A then T1T2x = T2T1x for all x ∈ X. Assume that p is invariant under elementsof A, thus p(x) = p(Tx) if T ∈ A and x ∈ X. Assume that (on Y ) ` is invariant under A.Then

Theorem 4. ` could be extended to all of X so that it remains invariant under A and stillcontrolled by p.

Idea of proof: We could add more to A all products of its elements, thus we may assumethat A contains 1 and is closed under composition (a semigroup). Let B contain all finiteconvex combination of elements of A. By definition, (inside Y ) the given ` is controlled byp0(x) = infT∈B p(Tx), which is a homogeneous convex function on X. Thus we could extend` to all of X while remains dominated by p0 (thus is dominated by p). We need to show that` is invariant under A. Let T ∈ A, it can be shown that

p0(x− Tx) ≤ 0

for every x ∈ X. Indeed, by definition p0(x) ≤ p0((1 + · · · + T n−1)x/n) for any n ≥ 1,therefore

p0(x− Tx) ≤ 1

np(x− T nx)→ 0 as n→∞.

Thus `(x − Tx) ≤ p0(x − Tx) and thus `(x) ≤ `(Tx) for all x. Since `(−x) = −`(x), itfollows that `(x) = `(Tx).

An application: The most typical applications of the HB extension theorem are hyper-plane separation theorems which require some local convexity of the underlying space. Wewill revisit these applications later, here we discuss a concrete example.

1.3 Topological spaces

We recall the notion of locally compact Hausdorff spaces (LCH) and discuss related results.

1.3.1 Compactness

X is compact if for any open covering there is a finite subcover. A space is locally compactif every point has a compact neighborhood.

Properties: 1. (finite intersection property) For any family of closed set with nonemptyintersection we could find a finite subfamily with nonempty intersection. In fact this isequivalent to compactness.

2. Tychonoff’s theorem: If (Xα)α∈A is a family of compact spaces then the productX =

∏α∈AXα with the product topology is compact.

(The product topology is the minimal topology such that all coordinate projections πα :X → Xα are continuous, equivalently speaking it is generated by π−1

α (open sets). )

1.3. TOPOLOGICAL SPACES 19

3. If there is some geometry (i.e. the topology is metrizable) then compactness is equiv-alent to sequential compactness, which states that for any sequence there is a subsequencethat converges.

4. Net convergence: Without geometry, one needs to use more than sequences. Anet (xα)α∈I is a collection indexed by some directed set I, i.e. a set with some partialordering < so that any two elements has at least one common upper bound. We say this netconverges to x if given any neighborhood of x there is β ∈ I such that xα ∈ P if β < α.

(be careful, this limit is not necessary unique in general).

Bolzano–Weierstrass’s theorem: X is compact iff every net has a convergent subnet.

1.3.2 Continuous functions on LCH spaces

Recall that a function f : X → R is continuous if for every open A ⊂ R the set f−1(A)is open in X. Note that a continuous function maps a compact subset of X to a compactsubset of R, thus continuous functions are always bounded on compact sets.

A space is Hausdorff if one could separate two points using two disjoint open sets. Thereare variants (both weaker/stronger) of this notion, but we won’t discuss them here. Themost important property of a Haudorff space is that limit is unique (if exists). Also a closedsubset of a compact subset is compact.

Basic questions:

1. Are there (a lot of) nonconstant continuous functions on a topological space? (We areinterested in bump functions, since this is related to the second question below.)

2. Can we extend continuously a given local continuous function (i.e. supported on acompact subset) to all of the space?

3. Can we construct partition of unity on such a space?

We will show that if the space is LCH then the answer is yes for all questions. (Onecould do better than LCH but we will not discuss that here.) These confirmative answerslead to the study of C(X) the space of continuous functions on X.

Urysohn’s lemma: Let X be an LCH space. Let K be compact and U be open in X,such that K ⊂ U . Then there exists a continuous function f : X → [0, 1] such that

• f = 1 on K and

• f = 0 outside a compact subset of U .

Tietze’s extension theorem: Let X be LCH and K ⊂ X compact subset. Then anyfunction f ∈ C(K) could be continuously extended to all of X. Furthermore the extendedfunction vanishes outside a compact set.

Partition of unity: this ia collection of nonnegative functions whose sum is 1 everywhereon the space, but locally at each point only finitely many of them are nonzero. One certainlycould only do this partition for a compact subset of the space.


1.3.3 Proof of Urysohn’s lemma

Proof consists of two steps.Step 1: First we consider a simpler setting when the space is actually compact Haus-

dorff. The idea is that a compact Hausdorff space has a nicer structure, namely one couldseparate two disjoint closed sets using two open sets. (also known as the “normal” property).Urysohn’s lemma works for normal spaces, here the assumptions would simply be K is closedinside an open set U .

To see why compact Hausdorff implies normal, take two closed disjoint sets C1, C2, theyare then compact. Given x ∈ C1 and y ∈ C2 we could use open sets Wx,y and Vx,y to separatethem, now (Vx,y) covers C2 so using compactness we could get a finite subcover Vx,yk , thuswe could separate x from C2 using two open sets ∩Wx,yk and ∪Vx,yk . Repeat this argumentto separate C1 from C2 using open sets.

Thus now X is a normal space, K closed, U open. We construct a large family of opensets (Ur)r that interpolates K and U . This family is indexed by dyadic rational numbersr = m

2kthat are in (0, 1), such that K ⊂ Ur ⊂ U and

Ur ⊂ Us

if r < s. One then definesg(x) = infr : x ∈ Ur

for all x ∈ X. Clearly g(x) = 0 if x ∈ K and g(x) = 1 if x ∈ U c, thus we could definef(x) = 1 − g(x) and f = 1 on K and vanish outside U . If we want f to vanish outsidea compact subset K ′ of U , we could apply this for K and U1/2, and notice that U1/2 is acompact subset of U .

Such f is continuous: it suffices to show

f−1((α,∞)) open

f−1((−∞, α)) openfor all α ∈ R. Wlog

assume 0 ≤ α ≤ 1. Thenf(x) < α⇔ x ∈ Ur for some r < α, thus

f−1(−∞, α) =⋃r<α

Ur is open.

f(x) > α ⇔ x 6∈ Ur for some r < α. Using the inclusion assumption, this is equivalentto existence of r > α such that x 6∈ U r, so

f−1(α,∞) =⋃r>α

Urc

is open.

Thus the remaining step is to construct the family. Here we use normality: to get W1

open and W2 open that surrounds K and U c (disjoint closed sets). Thus K ⊂ W1 ⊂ W c2 ⊂ U ,

and we define U1/2 = W1, now the closure of U1/2 is inside W c2 so inside U . Thus

K ⊂ U1/2 ⊂ U1/2 ⊂ U

1.3. TOPOLOGICAL SPACES 21

We repeat this with the new pairs (K,U1/2) and U1/2, U) and so on, get the family.

Step 2: Reduction to compact setting. The idea is to show that for some V ⊂ X openit holds that V is compact and K ⊂ V ⊂ V ⊂ U . Thus if the Lemma holds for compactHausdorff, we simply restrict to the subspace V and then extend the local function on V tothe whole of X by letting it be 0 outside V .

The existence of such a V can be done as follows: first we show that given each x ∈ Kwe could get a compact neigborhood Nx that remains inside U . Then the family of interiorof these Nx forms an open cover of K, thus using a finite subcovering we easily get a openset containing K such that its closure is inside U .

Now the existence of such a neighborhood follows from LCH property: the idea is forHausdorff space we could actually separate point and a compact set.

Now given each x we could get a neighborhood of x, called Mx, that is compact but notnecessarily inside U , then the set P = Mx ∩ U is compact and is a neighborhood of x, it isalmost contained inside U . One uses Hausdorffness to separate x further from the compactboundary of this set.

1.3.4 Proof of Tietze’s theorem

Essentially, Urysohn’s lemma implies Tietze’s theorem, at least for compact Hausdorff spaces(or more generally normal space). We’ll focus on this, the extension to the locally compactcase is similar to the last proof.

The idea is to approximate f by a sequence of continuous functions that have globalextensions, and this sequence converges uniformly to a continuous function on X.

To start, wlog assume 0 ≤ f ≤ 1 everywhere on K. Then by Urysohn’s lemma there is abump function 0 ≤ h1 ≤ 1 such that h1 = 1 on the closed set f−1([2

3, 1]) and h1 = 0 outside

the (bigger) open set f−1(13,∞). Clearly

0 ≤ f − 1

3h1 ≤

2

3on K.

Applying this argument to f1 = 32(f − h1) which takes values in [0, 1] on K, we get a

bump function h2 continuous on X such that locally on K it holds that 32(f− 1

3h1)− 1

3h2 ≤ 2

3,

or equivalently

0 ≤ f − 1

3h1 −

2

9h2 ≤

4

9.

Repeating this argument, we get continuous functions (globally on X) h1, h2, . . . suchthat hj takes value in [0, 1] and

0 ≤ f − 1

3h1 −

2

9h2 · · · −

2n−1

3nhn ≤

2n

3n

It is clear that the sequence 13h1,

13h1 + 2

9h2,

13h1 + 2

9h2 + · · ·+ 2n−1

3nhn converges uniformly to

a continuous function on the compact space X.


1.3.5 Partition of unity

Given K compact subset and a finite open cover (Uj)1≤j≤n can we find a partition of unityconsisting of compactly supported bump functions such that for each j at least one suchfunction is supported in Uj? Yes if LCH.

The idea is to find a compact subset Vj (could be empty) from each Uj such that theycover K. This follows from the fact that each point in K has a compact neighborhood insidesome Uj, thus by compactness one could refine this and get a finite covering consisting ofsets of this type, and let Vj be the union of those set that are strictly inside Uj. (Vj coversK since any set in the covering has to be a subset of some Vj.)

Then use Urysohn’s lemma to construct gj bump functions that equal 1 in Vj and vanishoutside Uj. (If Vj = ∅ simply take gj ≡ 0.)

Clearly g :=∑

j gj ≥ 1 on K and continuous, but this will be zero outside a compactset, so we can’t simply divide everything by g to get the partition of unity. The idea is touse Urysohn’s again ti get a bump function f that equals 1 in K and vanish outside g > 0(which is an open set). Now we could add to g the function 1 − f , which does not changeanything inside K but will be 1 as soon as g = 0. We get the partition of unity of Kconsisting of

gjg+1−f .

Chapter 2

Overview of normed linear spaces

Starting from this chapter, we begin examining linear spaces with at least one extra structure(topology or geometry). We assume linearity; this is a natural feature of functional spaces.

In this chapter, we start with spaces whose geometric and linear structures are compatible.More precisely, we assume that the metric is translation invariant d(x, y) = d(x + z, y + z)and homogeneous d(λx, λy) = |λ|d(x, y). We then define ‖x‖ = d(x, 0) the norm of x, andit is clear that ‖.‖ satisfies the following three properties (below x, y ∈ X and λ ∈ C):

• ‖x‖ ≥ 0 and equality holds iff x = 0.

• ‖x+ y‖ ≤ ‖x‖+ ‖y‖.

• ‖λx‖ = |λ|‖x‖.

A linear space equiped with such a norm is called a normed linear space (NLS).

Conversely, if such a norm exists we could always define a metric d(x, y) = ‖x− y‖ thatis translation invariant and homogeneous.

If the norm topology is separable then we say the NLS is separable.

A Banach space is a complete normed linear space, i.e. the induced metric space iscomplete. A Hilbert space has additional structures, where we could talk about “angle”through the scalar product. We’ll discuss these spaces further in separate chapters.

We recall that an incomplete metric space could be completed, and this applies to normedlinear spaces too: the metric remains invariant under dilation and translation in the com-pleted space thus it remains a norm.

Some examples:

1. Let (X,µ) be a measure space. Let 1 ≤ p ≤ ∞. Then Lp(X,µ) are normed linearspaces, with

‖f‖p = (

∫X

|f(x)|pdµ)1/p

23

24 CHAPTER 2. OVERVIEW OF NORMED LINEAR SPACES

In fact they are complete, this is a theorem of Riesz-Fisher. To see this, suffices to show thatif fn has

∑n ‖fn‖p <∞ then

∑n fn converges. 1

Now by the triangle inequality∫

(∑

n |fn|)pdµ <∞, thus h(x) =∑

n |fn(x)| is finite for µ-almost every x. Thus

∑n f(x) converges to some g(x) measurable, clearly ‖g‖p ≤ ‖h‖p <∞

so g ∈ Lp.Finally, |(

∑n fn) − g|p converges pointwise a.e. to 0 and dominated by (h(x) + |g(x)|)p

which are integrable, thus by dominated convergence∫|(∑

n fn)−g|pdµ→ 0, thus∑fn → g

in Lp.On the other hand, for p < ∞ weak Lp(X,µ) is not a normed linear space, since the

triangle inequality fails

‖f‖p,∞ = supλ>0

λµ(x ∈ X : |f | > λ)1/p

However, we still say “the weak Lp norm” in practice.Here is a counter example to the triangle inequality: use f(x) = x10≤x≤1 and g(x) =

(1− x)10≤x≤1.2. If X is compact Hausdorff then C(X) is a complete normed linear space if we use the

sup norm‖f‖ = sup

x∈X|f(x)|

If X is not compact then the sup is not necessarily finite and so this is not even a norm.For locally compact (Hausdorff) spaces we could instead look at Cc(X) consisting of

compactly supported continuous functions, and this space is a normed linear space. However,Cc(X) is not complete in the noncompact case, infact its completion is C0(X) the space ofcontinuous functions on X that vanish at ∞. (These are functions f ∈ C(X) such that forevery M > 0 the set |f | ≥M is compact.)

To see this, we first show that C0(X) is complete under the uniform norm. One way toshow this is to use one-point compactification. Alternatively, consider a Cauchy sequence(fn) in C0(X), then for each x ∈ X fn(x) is a Cauchy sequence of complex numbers, thusit converges pointwise to some f(x). Furthermore on any compact subset of X the uniformconvergence fn(x)→ f(x) implies continuity of f . Since X is locally compact and continuityis a local property, it follows that f ∈ C(X), then using fn ∈ C0 it is not hard to see thatf ∈ C0. (One way to check this is to use the net convergence characterization of continuity).

Now, given a function f ∈ C0(X) we could approximate it by a convergent sequence ofcompactly supported continuous functions on X using Urysohn’s lemma: let Kn = |f | ≥1/n which is compact and sits inside Un = |f | < 1/(n−1) an open set. Then by Urysohn’slemma we may find φn a bump function that vanishes outside Un but equals 1 on Kn, it isclear that φnf → f in the uniform metric.

3. L2-Sobolev spaces on R. If k ≥ 0 integer we could define

Hk := f ∈ L2 : f, . . . , f (k) ∈ L21For normed linear spaces, “ completeness” is equivalent to “every absolutely summable sequence is

summable”.

2.1. LINEAR FUNCTIONALS AND DUAL SPACES 25

be the space of L2 functions whose first k derivatives exists almost everywhere and are L2

integrable. Hk is complete and we may alternatively define Hk as the completion of thespace of locally compacted C∞ functions on R under the norm

‖f‖Hk = ‖f‖2 + · · ·+ ‖f (k)‖2

Using the Fourier transform, we could generalize this to allow for k fractional and evennegative, and we could also use Lp instead of L2. We’ll revisit this in the future if timepermits.

4. If Y is a closed subspace of X then the quotient space X/Y (the space of equivalentclasses where x1 ∼ x2 if x1 − x2 ∈ Y ) is a normed linear space with norm

‖[x]‖ = infy∈Y‖x− y‖

An equivalent definition is ‖[x]‖ = infz∈[x] ‖z‖. Note that closedness of Y is essential here toensure that ‖[x]‖ = 0 iff x ∈ Y .

5. If Y is a subspace of X then the closure Y of Y (with respect to the norm topology)is another subspace of X. If this closure is the same as X then we say that Y is dense in X.Note that this closure is not necessarily complete and Y is not the same as the completionof Y under the norm. For instance we could take X = Y incomplete NLS, then the closureof Y under the norm topology of X is the same as X, still incomplete.

2.1 Linear functionals and dual spaces

There are two basic notions of dual spaces:

• algebraic dual, which consists of all linear maps (aka linear functionals) ` : X → R (orC); and

• continuous dual, which consists of all continuous linear functionals.

We will be only interested in continous dual spaces, which will be implicitly understoodwhenever we refer to dual spaces.

Note that we could define continuous dual spaces even if X is only a topological linearspace, without any norm.

2.1.1 Linear functionals

For a normed linear space, we could talk about boundness of a linear functional. We saythat a linear functional ` is bounded if there exists C > 0 such that for every x ∈ X it holdsthat

|`(x)| ≤ C‖x‖Equivalently, this means supx∈X:‖x‖≤1 |`(x)| < ∞ (equivalently supx∈X:‖x‖=1 |`(x)| < ∞)which sometimes is used as the definition. Note that this fact also holds for linear transfor-mations from X to a Banach spaces.


Theorem 5. On normed linear spaces, a linear functional is continuous iff it is bounded.

Proof. In one direction, clearly being bounded implies being continuous. For the otherdirection, we’d like to show that the image of the closed unit ball under ` is a bounded setif ` is linear continuous. Note that ` maps compact sets to compact sets, but unfortunatelyas we’ll see the closed unit ball is not compact if X is infinite dimensional. So one has toexploit linearity of `. Assume towards a contradiction that xn is a sequence of unit vectors.t. |`(xn)| > n. Then xn/n converges to 0 in the norm, but |`(xn/n)| > 1 which violatescontinuity of `.

One of the most frequently/implicitly used theorems for bounded linear functionals onnormed linear space is the socalled B.L.T. (bounded linear transformation) theorem. Thetheorem applies even for bounded linear maps from X to another Banach space.

Theorem 6. Let D be a dense subspace of the normed linear space X. Let ` : D → Y whereY is a Banach space, and for some C > 0 it holds for all x ∈ D that

‖`(x)‖Y ≤ C‖x‖X

Then ` has an unique extension to a bounded linear map from X to Y and satisfies the aboveestimate for all x ∈ X.

Proof. It is not hard to see that if such ` exists it has to be unique. To define `, fix x ∈ X.Since D is dense in X there exists a sequence (xn) in D that converges to x. We then define`x = lim `xn. Note that this limit exists because `xn is a Cauchy sequence in Y which is acomplete space. One could easily show that the value of `x does not depend on the choiceof the sequence xn. Linearity and boundedness could be easily checked.

While working with singular integral operators on Lp spaces we typically invoke theabove theorem implicitly: these operators are explicitly defined only for a nicer dense subsetof functions (say sufficiently smooth and with sufficient decay), and so the theorem saysthat as long as we could bound the operators on these dense subspace we could extend theoperator to all of the corresponding Lp and get the same bound.

Example: the Hilbert transform Hf(x) = p.v.∫R

1yf(x − y)dy is defined for smooth

compactly supported functions on R which is dense in Lp. It turns out that ‖Hf‖p ≤ C‖f‖p,1 < p <∞, thus H extends to a bounded maps on Lp.

2.1.2 Dual spaces

For a bounded linear functional we could define ‖`‖∗ = supx∈X:‖x‖=1 |`(x)| and this defines anorm on the dual space of X. By definition |`(x)| ≤ ‖x‖‖`‖∗.

Theorem 7. The dual space of a normed linear space is a Banach space.

Proof. This is actually a special case of a more general fact: the space B(X, Y ) of boundedoperators from normed linear space X to a Banach space Y

‖Tx‖Y ≤ ‖T‖‖x‖X

2.1. LINEAR FUNCTIONALS AND DUAL SPACES 27

is in turn another Banach space. (For us Y = R (or C) with the distance norm). It is nothard to see that this is a normed linear space, the main thing is to show completeness. Givenany Cauchy sequence (Tn) in B(X, Y ) it is not hard to see that supn ‖Tn‖ < ∞. Now forany x ∈ X the sequence Tnx is a Cauchy sequence in Y therefore it converges (thanks tocompletness of Y ) in Y , and we let T∞x to be this limit. It is clear that T∞ is linear andbounded ‖T∞‖ ≤ supn ‖Tn‖ <∞. It remains to show that limn→∞ ‖Tn − T‖ = 0.

Examples:1. The dual space of a Hilbert space is itself.2. If 1 < p < ∞ then the dual space of Lp(X,µ) is Lq(X,µ) where q is the conjugate

exponent 1/p+ 1/q = 1. In particular,

‖f‖Lp(X,µ) = supg:‖g‖q=1

|∫fgdµ|

If µ is a σ-finite measure(namely one could break the space into countably many subsetswhere µ is finite) then the dual space of L1(X,µ) is L∞(X,µ). This may not be true withoutthe σ-finite assumption. ON the other hand, the dual space of L∞(X,µ) generally speakingnot L1(X,µ) (but there are examples when they are, say when X is a finite set with thecounting measure).

3. If X is a locally compact Hausdorff space, then the dual of C0(X) is the space ofregular Borel measures with finite total mass. This result is one of Riesz’s representationtheorems and we will prove them later in the course.

2.1.3 Reflexivity

If X is a normed linear space we let X∗ denote its dual and X∗∗ denote the dual of X∗.

Definition 3 (Reflexive spaces). We say that X is reflexive if X = X∗∗ (up to isomor-phism).

In particular, a reflexive space has to be a Banach space to begin with, but certainly notall Banach spaces are reflexive.

The interest in reflexive spaces is natural, since we always can isometricaly embed X intoX∗∗. To see this, fix x ∈ X. Let ‖.‖∗ and ‖.‖∗∗ be the norms on X∗ and X∗∗ respectively.We could map x ∈ X to the following linear map x on X∗:

x(`) := `(x) , ` ∈ X∗ .

Note that x is bounded on X∗, since by definition |x(`)| ≤ ‖x‖‖`‖∗. Thus ‖x‖∗∗ ≤ ‖x‖.On the other hand, using the Hahn Banach theorem it follows that we could find ` ∈

X∗ such that `(x) = ‖x‖ and |`(y)| ≤ ‖y‖ for all y ∈ X. It follows that ‖`‖∗ ≤ 1,therefore |x(`)| = ‖x‖ ≥ ‖x‖‖`‖∗, and consequently ‖x‖∗∗ ≥ ‖x‖. Thus ‖x‖ = ‖x‖∗∗ and theembedding x 7→ x is an isometric embedding of X into X∗∗.

As we’ll see, being reflexive helps in many situation; we’ll explore that gradually.


Examples:1. Hilbert spaces are reflexive.2. Lp are reflexive if 1 < p < ∞. As discussed above, generally speaking both L1 and

L∞ are not reflexive. For examples they are not reflexive when the underlying space is Rn

with Lebesgue measure or Z with counting measure.3. A closed linear subspace of a reflexive space is also reflexive.4. C[−1, 1] is not reflexive. Note that this space is separable but its dual is not (since

the dual of this space contains in particular all Borel measures). As we’ll see later, for areflexive space its dual space has to be separable too. One could prove this directly w/ousing separability.

2.2 (Non)compactness of the unit ball

One of the key themes in analysis is existence of limit of a sequence (in some functionalspaces) that are bounded uniformly in norm. By rescaling if necessary we may assume thebound is 1.

Sometimes, it suffices to get a convergent subsequence. In other words we are interestedin the sequential compactness of the closed unit ball.

Since NLS is in particular a metric space, sequential compactness = compactness. So thequestion is about compactness of the closed unitball in the norm topology.

For Euclidean spaces it is clear that for finite dimensional setting the closed unit ball iscompact. For infinite dimensional setting, the answer is no, for instance consider the sequencean = (0, . . . , 0, 1, 0, . . . ) (the nth coordinate is 1). If there is a limit x = (x1, x2, . . . ) for somesubsequence ank

, i.e.‖ank

− x‖ → 0

as k → ∞. Then clearly ‖x‖ = 1. On the other hand, given any j by taking k large it isclear that 0 ≤ |xj| ≤ ‖ank

− x‖ → 0, thus x1 = x2 = · · · = 0, which is a contradition.This suggests that the answer should be similar for a general normed linear space: positive

for spaces with finite dimensions and negative in the infinite dimensional setting (whetherthe space is complete or not).

In the finite dimension case, this follows from the fact that in finite dimensional linearspaces all norms are equivalent. In other words, if ‖.‖1 and ‖.‖2 are two norms then thereexists C1, C2 > 0 such that 1

C1‖x‖1 ≤ ‖x‖2 ≤ C2‖x‖1 for all x ∈ X. To see this, for simplicity

consider linear spaces over R, and assume that the given space is spanned by x1, . . . , xm,it suffices to show the conclusion for ‖.‖2 being the Euclidean norm ‖a1x1 + . . . amxm‖2 =√a2

1 + · · ·+ a2m the Euclidean norm. Then by the triangle inequality

‖a1x1 + . . . amxm‖1 ≤ (|a1|+ · · ·+ |am|) maxj‖xj‖1 ≤

√m√a2

1 + . . . a2m max

j‖xj‖1

so we could take C1 =√mmaxj ‖xj‖1. To show existence of C2, note that the identity map

` : (X, ‖.‖2) → (X, ‖.‖1) (i.e. `x = x) is continuous since it is Lipschitz. The unit ball

2.2. (NON)COMPACTNESS OF THE UNIT BALL 29

in (X, ‖.‖2) is compact, thus its image under this continous function is also compact, thusbounded in ‖.‖1, therefore for some C2 > 0 we have

sup ‖x‖1 ≤ C2

the sup is over x with ‖x‖2 = 1, which is equivalent to the desired estimate.In the infinite dimensional case, this is a theorem of Riesz.

Theorem 8 (Riesz). If X is an infinite dimensional NLS then the closed unit ball is notcompact with respect to the norm topology.

Proof. Our plan is to find a sequence of elements of unit vectors in X, x1, x2, . . . , such thatthe distance ‖xi − xj‖ between any two elments of the sequence is uniformly larger than0, for instance ‖xi − xj‖ ≥ 1

10for all i 6= j. If that could be constructed it is clear that

no subsequence of (xn) is Cauchy, thus no subsequence of this sequence is convergent andtherefore the closed unit ball is not (sequentially) compact.

To construct this sequence it suffices to show that if Y is a closed proper subspace of Xthen one could find x ∈ X \ Y with ‖x‖ = 1 such that dist(x, Y ) ≥ 1

10. Once this is proved

we could start with any point x1 of unit length and let Y be spaned by x1 (which is closed)and then select x2 of unit length as above (in particular ‖x2− x1‖ ≥ 1

10), then reset Y to be

spaned by x1, x2 (a closed subspace because it has finite dimension) and select x3, etc.Thus we only to show existence of x ∈ X of unit norm satisfying dist(x, Y ) ≥ 1

10whenever

Y is a closed proper subspace of X. Let z ∈ X \ Y and let

d := dist(z, Y ) > 0 .

(if d = 0 then the closedness of Y would imply that z ∈ Y , contradiction.) Now for somey ∈ Y we have ‖z−y‖ < 10d. Let x = z−y, it follows that dist(x, Y ) = dist(z, Y ) = d, thus

dist(x, Y ) > ‖x‖/10

now we just rescale x so that it has length 1. Because of this result, there is a natural question of determining a topology on X so

that the closed unit ball is compact with respect to this topology. One wants a weakertopology (with fewer open sets, too many open sets is probably the reason why the ball isnot compact).

There is a natural notion of weak-topology that is weaker than the norm topology. Herewe wants just enough open sets so that all bounded linear functionals are continuous.

Theorem 9. The closed unit ball is compact with respect to the weak-topology if and only ifX is reflexive.

Note that compactness of the closed unit ball is compact in this weak topology does notimply local compactness of X.

We’ll prove this result later when discussing topologies on Banach spaces; as we’ll seethis theorem is a consequence of the Banach Alaoglu theorem.


Exercises:1. Prove that for a normed linear space, “completeness” of the norm is equivalent to

“every absolutely summable sequence is summable”.2. Prove that for every x ∈ X a normed linear space it holds that ‖x‖ = sup`∈X∗: ‖`‖=1 |`(x)|.3. Prove that if K is compact Hausdorff then C(K) is complete with respect to the

sup norm. Use this to complete the proof that C0(X) is complete if X is locally compactHausdorff without appealing to the one-point compactification trick.

4. Let S be a subset of X a normed linear space, and let Y be the linear span of S(consisting of all finite linear combination of elements of S). Let L ⊂ X∗ to be the set of all` ∈ X∗ that vanishes on S. Prove that z ∈ Y the closure of Y if any only if `(z) = 0 forevery ` ∈ L. (Hint: one direction should be easier, for the other direction you should usethe BLT theorem somewhere.)

5. Prove that all closed linear subspaces of a reflexive (Banach) space are reflexive. [Hint.Use problem 4 at some point.]

Chapter 3

Basic geometrical and topologicalproperties of normed linear spacesand their duals

3.1 Topologies on normed linear spaces

Recall that the norm topology is induced from the norm and the weak topology is theminimal topology that makes all bounded linear functionals continuous.

If X = Y ∗ is the dual of another normed linear space Y (thus in particular X is a Banachspace) we could introduce another topology, called the weak* topology: this is the minimaltopology such that all bounded linear functionals y, y ∈ Y , are continuous. We could usethe finite intersections of the following sets as a neighborhod base at 0 in this topology:

x ∈ X : |y(x)| < ε

Note that the maps y 7→ y isometrically embedded Y inside X∗, the weak* topology issmaller than or equal to the weak topology.

Certainly if X is reflexive then the weak* and weak topologies are the same.

3.1.1 The Banach–Alaoglu theorem

Theorem 10 (Banach–Alaoglu, 1940). Let X be the dual of another normed linear space.Then the closed unit ball B of X is compact in the weak* topology.

Proof. Let X = Y ∗. For simplicity assume that the spaces are over R. For each y ∈ Y letIy = [−‖y‖Y , ‖y‖Y ] ⊂ R and consider

I =∏y∈Y

Iy

31

32 CHAPTER 3. GEOMETRY AND TOPOLOGY ON NLS

consisting of tuples indexed by Y . By Tychonoff’s theorem, I is compact in the producttopology and it is also Hausdorff (tensor product of Hausdorff spaces with the producttopology is Hausdorff). Now, for each elements x in the unit ball B of X consider the map

x 7→ Tx := (x(y))y∈Y ∈ I

(the fact that Tx ∈ I follows because for x ∈ B we have |x(y)| ≤ ‖x‖X‖y‖Y = ‖y‖Y ). Notethat elements of TB are special because the coordinates of Tx are actually related linearly.

Now, T is clearly linear and one to one and it embeds B into I, in fact it is not hardto see that if we use the weak* topology on B and the inherited product topology on TBthen T is indeed a topological isomorphism between B and TB. We now show that TB is aclosed subset of I, which together with compactness of I will then implies that TB (henceB) is compact.

Let z = (zy)y∈Y ∈ TB where closure taken inside I under the product topology. Thenwe want to show that z is “linear and bounded by 1”, namely

zαx+βy = αzx + βzy

Once that’s done it follows that the maps y 7→ zy defines a linear map on Y with norm atmost 1 and so z ∈ TB as desired.

To verify the above property, simply observe that there is a net in TB that converge to z(unique because of Hausdorffness), and linearity relationship between coordinates is actuallypreserved in the limit, so this property survives and z has it.

Examples:1. If X is the space of regular Borel measures on R with the norm = total mass, then its

is the dual space of C0(X) which is a separable space. Therefore the closed unit ball in X issequentially compact. In particular, given any sequence (µn) of probability measures on Rthere is a subsequence (µnk

) that converges vaguely to some Borel measures µ, i.e. for everycontinuous function f that vanishes at ∞ it holds that

limn→∞

∫f(x)dµn(x) =

∫f(x)dµ(x)

Note that µ is not necessarily a probability measures, in particular it is possible that µ = 0,which happens when the masses in µnk

escape to ∞ (for instance we could take µn to havethe density function 1[n,n+1]). To avoid this there is a notion of tightness, which basicallyassume that given any ε > 0 we could find a compact Kε such that µn(R \Kε) < ε for everyn. With this assumption we could show that µ is a probability measure, and basically wehave just shown the Helly selection theorem for probability measures on R.

3.1.2 The sequential Banach-Alaoglu theorem

Recall that compactness means any net has a convergent subnet. In practice we are interestedin the sequential variant of this theorem, which says that

3.1. TOPOLOGIES ON NORMED LINEAR SPACES 33

Theorem 11. If X = Y ∗ where Y is separable normed linear space, then the closed unit ballB in X is sequentially compact in the weak* topology. In other words given any sequence(xn) in B there exists a subsequence xnk

and x ∈ B such that

limk→∞

xnk(y) = x(y)

for every y ∈ Y .

Namely, since Y is separable the weak* topology is metrizable (i.e. it arises from somemetric on X), thus compactness and sequentially compactness are the same on this topology.

We could also prove this directly: let (yk) be a dense subset of Y , then from the sequencexn we could select a subsequence xnk

such that y1(xnk) converges. Keep doing this and

use a diagonal argument we get a sequence xnksuch that for any fixed j it holds that

limk→∞ yj(xnk) exists. Using the fact that (yj) is dense in Y it follows that for every y ∈ Y

the limit limk→∞ y(xnk) exists, let this limit be z(y), it is clear that z is a linear functional

on Y and also bounded with ‖z‖ ≤ 1. This implies xnkconverges weakly to z and element

of B.

3.1.3 Weak compactness of the closed unit ball in reflexive spaces

As a consequence of the Banach Alaoglu theorem, for any reflexive space the closed unit ballis compact in the weak topology. The reverse direction is true (Kakutani’s theorem) and ispart of the homework.

If the given space is furthermore separable then its dual is also separable, so (from thesequential BA theorem) the closed unit ball is sequentially compact in this weak topology.

It is surprising that this sequential compactness property still holds even if we don’tassume separability on the given reflexive space!

Theorem 12. Let X be a Banach space. If X is reflexive then the closed unit ball in X issequentially compact in the weak topology.

Notes: The reverse direction also holds, this is a result of Eberlein–Smulian: if everybounded sequence has a convergent subsequence then X is reflexive. In fact, ES showedthat a subset of a reflexive Banach space is weakly compact if any only if it is sequentiallycompact, which implies the above statement.

Weak convergence of sequences: In a normed linear space, we say that a sequenceconverges weakly if it converges in the weak topology. More precisely, (xn) converges weaklyto x if for every ` ∈ X∗ it holds that limn→∞ `(xn) = `(x). This is in contrast with strongconvergence of the sequence xn, which means ‖xn − x‖ → 0. Clearly strong convergenceimplies weak convergence, but the opposite direction is not true in general.

It follows from the above theorem that in a reflexive Banach space (for instance Lp with1 < p <∞) any norm bounded sequence has a subsequence that converges weakly.

Proof. We’ll show that reflexivity implies sequentially compactness of the closed unit ball.Let (xn) be the given sequence and let Y be the closure of the linear space spanned by (xn).


Then Y is a closed linear subspace of X so is also reflexive, on the other hand Y is separable,thus one could see that the closed unit ball inside Y is sequentially compact with respect tothe weak topology on Y . Thus there is a subsequence xnk

and x∞ ∈ Y such that for everybounded lienar functionals ` on Y we have

`(xnk)→ `(x)

Since every bounded linear functionals on X is also a bounded linear functional on Y , thisimplies xnk

converges weakly to x.

3.2 Uniform convexity

Uniform convexity is a geometric notion introduced by Clarkson (1936).

Definition 4 (Uniformly convex). We say that a normed linear space X is uniformly convexif for each 0 < ε ≤ 2 there exists δ = δ(ε) > 0 such that the following holds: If ‖x‖ = ‖y‖ = 1and ‖x− y‖ ≥ ε then

‖x+ y

2‖ ≤ 1− δ

Note that this is a property of the norm: there may be equivalent norms on the samespace that is not uniformly convex.

Uniform convexity does not imply completeness. For instance we could take the rational

numbers with the usual distance. Then |x+y2|2 + |x−y

2|2 = |x|2+|y|2

2which implies uniform

convexity.

Theorem 13 (Milman–Pettis). If X is uniformly convex Banach then it is reflexive.

Note that there are reflexive Banach spaces where it is not even possible to replace thegiven norm with an equivalent norm that is uniformly convex, this is due to M. Day (BAMS,1941). The idea is to use a vector valued Banach space with norm ‖.‖ = (

∑n ‖xj‖p)1/p where

xn belongs to a reflexive Banach space Bn (which makes it reflexive); now for each j onecould still replace the norm on Bj by an equivalent norm that is uniformly convex, but asn → ∞ these replacement couldn’t be done uniformly because the underlying equivalenceconstants blow up if the Bj are carefully chosen.

Properties:

1. Hilbert spaces are uniformly convex, since in Hilbert spaces we have the paralellogramlaw ‖x− y‖2 + ‖x+ y‖2 = 2‖x‖2 + 2‖y‖2.

2. C[−1, 1], L1(R), L∞(R) are not uniformly convex because they are not reflexive.

3. Clarkson(1936): if 1 < p <∞ then Lp and `p spaces are uniformly convex.

For p ≥ 2 this holds essentially some form of parallelogram law holds.

(‖x+ y‖p + ‖x− y‖p)1/p ≤ 21/p(‖x‖p′ + ‖y‖p′)1/p′

3.3. ISOMETRIES BETWEEN NORMED LINEAR SPACES 35

where 1/p + 1/p′ = 1. For p ∈ (1, 2) we have a variant (‖x + y‖p′ + ‖x − y‖p′)1/p′ ≤21/p′(‖x‖p + ‖y‖p)1/p. 1

4. (Radon–Riesz, aka property (H)) If xn converges weakly to x in a uniformly convexBanach space X, i.e. `(xn)→ `(x) for all ` ∈ X∗, and ‖xn‖ → ‖x‖, then ‖xn − x‖ → 0. 2

5. A space is called uniformly smooth if its dual is uniformly convex.6. A normed linear space is uniformly convex if the following holds for every bounded

sequences (xn) and (yn): if

limn→∞

‖xn‖2 + ‖yn‖2

2− ‖xn + yn

2‖2 = 0

then limn→∞ ‖xn − yn‖ = 0. Below are two notions weaker than uniformly convex:Local uniform convexity: A norm is locally uniformly convex if the following holds

for every x ∈ X and (xn) in X: if

limn→∞

‖x‖2 + ‖xn‖2

2− ‖x+ xn

2‖2 = 0

then limn→∞ ‖x− xn‖ = 0. It is clear that this is a consequence of uniform convexity.Strict convexity: A norm is strictly convex if the following holds for every x, y: if

‖x‖2+‖y‖22

= ‖x+y2‖2 then x = y. It is clear that this is a consequence of local uniform

convexity.It can be shown that if X is a separable normed linear space over R then there exists an

equivalent locally uniformly convex norm.

3.3 Isometries between normed linear spaces

We say that a map T : X → Y between normed linear spaces X and Y is an isometry if

‖Tx1 − Tx2‖Y = ‖x1 − x2‖X

for every x1, x2 ∈ X. Note that we do not assume that T is affine (i.e. T (αx + βy) =αTx+ βTy, which is the same as linear if we impose T0 = 0).

The Banach–Mazur distance: The distance is useful to compare two norms on Rn -it is known that all norms are equivalent so the point is to be more quantiative.

1Alternatively, use Hanner’s inequality which says that if ‖.‖ is the Lp or `p norm then

‖f + g‖p + ‖f − g‖p ≥ (‖f‖+ ‖g‖)p + |‖f‖ − ‖g‖|p (3.1)

if p ∈ [1, 2] and the reverse holds if p ∈ [2,∞).2To see this, it suffices to show the theorem for sequences with ‖xn‖ = 1. Hahn Banach gives us ` ∈ X∗

such that `(x) = 1 and ‖`‖ ≤ 1. Then by weak convergence of xn to x we obtain 2 ≥ ‖xn + xm‖ ≥`(xn + xm) → 2`(x) = 2 as min(m,n) → ∞. This easily implies ‖xm + xn‖ → 2, from there and uniformconvexity it follows that (xn) is a Cauchy sequence and converges to some y, clearly `(y) = lim `(xn) = `(x)for every ` ∈ X∗, thus y = x.


Analytic: Let X and Y be normed linear spaces. The multiplicative BZ distance is definedto be

d(X, Y ) = inf‖T‖‖T−1‖

the infimum is over all isomorphisms T : X → Y . Sometimes people also use log d(X, Y )which satisfies the usual (additive) triangle inequality.

Geometric: given two bounded convex sets with nonempty interiors (aka convex bodies)K and L in Rn that are symmetric (i.e. K = −K and L = −L) the Banach–Mazur distancebetween them is

d(K,L) = infC > 0 : L ⊂ TK ⊂ CL , T linear on Rn

For Rn these two notions are equivalent. Basically given any norm on R the unit ballis a symmetric convex body and conversely given any symmetric convex body K we couldconstruct a norm such that its unit ball is K, namely ‖x‖ = inft > 0 : x ∈ tK.

Theorem 14 (F. John). Let `n2 denote Rn with the Euclidean norm. If X is a n-dimensionalNLS over R then

d(X, `n2 ) ≤√n

(This is actually sharp; equality holds for instance if X = `n∞, for the geometric case takethe unit ball of this space.)

It follows from the theorem and the (multiplicative) triangle inequality that given anytwo n-dimensional normed linear spaces over R it holds that

d(X, Y ) ≤ n

This result is also sharp upto a constant (Gluskin, 1981, a randomized construction).Existence of nonlinear isometry: A natural question is whether non-affine isometry

exists (or equivalently nonlinear if we assume the normalization T0 = 0)?This is not true for complex NLS, even complex Banach spaces, see for instance the

conjugation mapping C to itself z 7→ z. We would need surjectivity, since otherwise the mapT : R→ R2 mapping x to (x, sin(x)) is not linear but is an isometry if we equip R2 with theL∞ norm ‖(x, y)‖ = max(|x|, |y|) (and distance norm in R).

It turns out that if the underlying normed linear spaces are over R then the answer is no.

Theorem 15 (Mazur-Ulam, 1932). If X and Y are NLS over R and T : X → Y surjectiveand isometric with T0 = 0 then T is linear.

If T0 = 0 is not given the conclusion could be equivalently changed to “T is affine”,namely T (αx+ (1− α)y) = αT (x) + (1− α)T (y).

Main ideas of proof: Isometries are continuous so it suffices to show T (kx + my) =kT (x) +mT (y) for all rational m, k, in fact suffices to show for k = m = 1/2. We constructa nested sequence of subsets A1 ⊃ A2 . . . each of them is symmetric around x+y

2such that

∩An = x+y2. This construction will be invariant under surjective isometries, in particular

3.3. ISOMETRIES BETWEEN NORMED LINEAR SPACES 37

to get Tx+Ty2

we could use the nested sequence TA1 ⊃ TA2 . . . , which are symmetric aroundTx+Ty

2. When that’s done we could take intersections to get Tx+Ty

2= T (x+y

2) as desired.

Now, let A1 contains all elements z of X such that ‖x − z‖ = ‖y − z‖ = ‖x−y‖2

. Toconstruct An+1 from An, simply let An+1 contains all z ∈ X such that for all w ∈ An it holdsthat ‖w − z‖ ≤ 1

2diam(An), one could check that An+1 is still symmetric around x+y

2and

diam(An+1) ≤ diam(An)/2 ≤ · · · ≤ diam(A1)2n

→ 0 as n → ∞. Note that we need TX = Ysince we want TAn+1 contains all z ∈ Y = TX such that “so and so” holds.

Exercises:1. For any 1 < p < ∞ and any measure space (X,µ) prove that Lp(X,µ) is uniformly

convex (use the hints from the lecture notes).2. Show that the Banach Mazur distance satisfies d(`n∞, `

n2 ) ≥

√n. [Hint: use the geomet-

ric formulation with corresponding convex bodies, and invoke a generalized parallelogramlaw.] (Note that together with F. John’s theorem this implies d(`n∞, `

n2 ) =

√n.) Then via

Holder’s inequality and the analytic formulation prove that d(`np , `nq ) = n|

1p− 1

q| if 1 ≤ p, q ≤ 2

or 2 ≤ p, q ≤ ∞.3. Prove F. John’s theorem for n = 2. [Hint. Use the geometric formulation, you may

want to use the fact that the Banach-Mazur distance is invariant under isomorphism of theplane.]

4. Let X be a normed linear space. Let B = x ∈ X : ‖x‖ ≤ 1 and B∗∗ = z ∈ X∗∗ :‖z‖∗∗ ≤ 1. We know that there is a canonical map x 7→ x that embeds B isometricallylinearly into B∗∗. Prove that the image of B under this map is dense in B∗∗ in the weak*topology of X∗∗ (i.e. the minimal topology on X∗∗ that makes all bounded linear functionalsconstructed from elements of X∗ continuous on X∗∗). Use this fact to show that if B isweakly compact then B = B∗∗ and therefore X is reflexive.

5. Find two examples demonstrating that ”compactness” and ”sequentially compactness”do not imply each other.

6. Let K be a convex subset of X a Banach space. Prove that if K is closed in the normtopology then it is closed in the weak topology.

Chapter 4

Basis on Hilbert spaces and Banachspaces

A Hilbert space is a complete normed linear space where the norm arises from an innerproduct ‖u‖ =

√〈u, u〉, and the inner product 〈u, v〉 is a bilinear form that satisfies the

following properties (we state the properties for Hilbert spaces over C):(i) (u, v) ≥ 0, equality iff u = 0.(ii) 〈u, v〉 = 〈v, u〉.(iii) 〈u, v〉 is linear in u (hence conjugate linear in v).Examples: H = L2(X, dµ) with 〈f, g〉 =

∫Xf(x)g(x)dµ; by Holder this is finite

|∫X

f(x)g(x)dµ| ≤ (

∫X

|f |2dµ)1/2(

∫X

|g|2dµ)1/2

4.1 Basic properties

Parallelogram law: ‖x+ y‖2 + ‖x− y‖2 = 2‖x‖2 + 2‖y‖2.Schwarz’s inequality:

| 〈u, v〉 | ≤ ‖u‖‖v‖

Proof. ‖x+ y‖ ≤ ‖x‖+ ‖y‖, therefore

‖x‖2 + ‖y‖2 + 〈x, y〉+ 〈y, x〉 = 〈x+ y, x+ y〉

≤ (‖x‖+ ‖y‖)2 = ‖x‖2 + ‖y‖2 + 2‖x‖‖y‖Therefore Re 〈x, y〉 ≤ ‖x‖‖y‖ thus get desired estimate if 〈x, y〉 ∈ R. In general multiply xwith α unimodular such that | 〈x, y〉 | = 〈αx, y〉 ≤ ‖αx‖‖y‖ = ‖x‖‖y‖.

Orthogonality: We say u ⊥ v if 〈u, v〉 = 0. A set E ⊂ H is orthonormal if ‖u‖ = 1 forall u ∈ E and 〈u, v〉 = 0 if u, v ∈ E with u 6= v.

Bessel’s inequality: If x1, . . . , xN orthonormal then∑N

n=1 | 〈x, xn〉 |2 ≤ ‖x‖2. Asequence (xn) is a Bessel sequence if

∑n | 〈x, xn〉 |2 <∞ for all x ∈ H.

39

40 CHAPTER 4. BASIS ON HILBERT SPACES AND BANACH SPACES

Theorem 16 (Riesz). For any ` ∈ H∗ there is h` ∈ H such that `(x) = 〈x, h`〉 for all x ∈ H,and the map ` 7→ h` is a surjective isometry (conjugate linear). Consequently H∗ ∼ H.

Proof. Clearly ker(`) is a closed linear subspace of H, let Y = y ∈ H : y ⊥ ker(`). Thenany x ∈ H has a unique decomposition into y + z with y ∈ Y and z ∈ ker(`). Sincd ` isnonzero on Y so there is y0 ∈ Y such that `(y0) = 1. We now decompose each x ∈ H usingx = `(x)y0 + (x − y0`(x)), now the first element is clearly in Y and the second is in ker(`)since `(y0) = 1. It follows that H = span(y0, ker(`)) and we could let h` = y0

‖y0‖2 , then

`(x) = `(〈x, h`〉 y0) = 〈x, h`〉

it is not hard to check isometry/(conjugate)linearity of the map ` 7→ h`. A generalization of Riesz’s theorem is Lax-Milgram’s theorem:

Theorem 17 (Lax-Milgram). Let B(x, y) : H2 → C be a bounded bilinear map (conjugatelinear in y), |B(x, y)| ≤ C‖x‖‖y‖ such that |B(x, x)| ≥ c‖x‖2. Then every ` ∈ H∗ has theform `(x) = B(x, y`) and the map `→ y` is bounded invertible (conjugate) linear.

Proof. Since B(x, y) is bounded linear in x for each y, we could find hy such that B(x, y) =〈x, T (y)〉 with ‖y‖ . ‖T (y)‖ . ‖y‖, furthermore y 7→ T (y) is linear. It follows that one couldinvert this map, now by Riesz representation theorem `(x) = 〈x, h`〉 = 〈x, T (T−1(h`))〉 =B(x, T−1h`), so we could let y` = T−1(h`).

4.2 Orthogonal basis for a Hilbert space

We say that a set E ⊂ H is an orthogonal basis if iE is an orthogonal set and whenverv ⊥ span(E) we must have v = 0. If furthermore every elements of E has unit norm thenwe say that E is an orthonormal basis.

Theorem 18. A Hilbert space is separable iff it has a countable orthonormal basis.

Proof. If there is a countable orthonormal basis (vn) then we simply use finite linear combina-tion of vn with (complex) rational coefficients and get a dense countable subset. Converselyif H has a countabler dense subset D we could apply Gram Schmidt to get a countableorthogonormal set E. Let v ⊥ span(E), we will show v = 0. If not, assume ‖v‖ = 1, thenfor all x ∈ E, exploiting orthogonality of v and x, we have

‖v − x‖2 = ‖v‖2 + ‖x‖2 ≥ 1

so span(E) is not dense in H, contradiction.Notes:1. If H is separable then we can extend any given basis to the full basis using Gram-

Schmidt and avoid axiom of choice for Hahn-Banach.2. If H is not separable and E is an orthogonal basis of H then for each x ∈ H the

set of elements v ∈ E such that x has nonzero projection onto that element 〈x, v〉 v 6= 0 is

4.3. SCHAUDER BASIS 41

countable. Using this fact and Zorn’s lemma we could show that every Hilbert space has anorthonormal basis.

3. If H is separable then we could define a natural isometry i : H → `2(Z).

4.3 Schauder basis

Schauder basis: A sequence E = (xn) is a (Schauder) basis for a normed linear space Xif for every x ∈ H there is a unique scalar sequence αn such that limn→∞ ‖x−

∑nk=0 αkxk‖ = 0.

Equivalently, (xn) is a Schauder basis if two conditions hold: the closure of the linearspan of xn is the whole space and

∑n=1 anxn = 0 iff an = 0.

Note that this is different from an algebraic basis (aka Hamel basis) which consists oflinearly independent vectors and any x ∈ X could be written as a finite linear combination ofthese elements. Note that this could be defined on any linear spaces, topology (convergence)is not needed, as we dont’ permit infinite sums. Also the order of the elements in a Hamelbasis is not important.

A Schauder basis is said to be an unconditional basis if it remains a basis even afterreordering. There are two other equivalent characterizations

Theorem 19. Let (xn) be a (Schauder) basis in a Banach space (over R or C). The followingare equivalent:

(i) (xn) is an unconditional basis.(ii) For any scalar sequence (αn), if

∑αnxn converges then it converges unconditionally

(i.e. it remains convergent even if we change the summation order).(iii) There exists C > 0 finite such that

‖n∑k=1

εkαkxk‖ ≤ C‖n∑k=1

αkxk‖

uniformly over n and all sequences εn with |εn| ≤ 1 and all scalar coefficients α1, . . . , αn.

Property:1. For NLS, existence of Schauder basis means the space is separable.2. Every orthogonal basis in a separable Hilbert space is an unconditional Schauder basis.Examples:1. Fourier basis: L2[0, 1] has (e2πinx/

√2π)n∈Z as an unconditional Schauder basis. This

remains a basis for Lp[0, 1] 1 < p <∞ but not unconditional for p 6= 2.2. Haar basis: 1[0,1] and hI(x) = 1√

|I|(1Il − 1Ir) indexed by dyadic subintervals I ⊂ [0, 1]

is a basis for L2 and all Lp[0, 1] with 1 ≤ p < ∞, note that p could be 1 here. This is anunconditional basis if 1 < p <∞.

3. Orthogonal polynomials: If w(x), x ∈ R has sufficiently fast decay (say subexpo-nential - Bernstein’s theorem) then the orthogonal polynomials with respect to dµ = wdx is


a basis for L2(R, µ). Here pn(x) = an,nxn + · · ·+ a0,n where an,n > 0 and∫

pn(x)pm(x)dµ = δmn

which is 1 if n = m and zero otherwise.

Proof of Theorem 19. It is clear that (i) and (ii) are the same. So we only show equivalenceto (iii).

We first observe that the following are equivalent for any given sequence (xn) in a Banachspace X:

(a) the series∑

n xn converges unconditionally;(b) for any ε > 0 there exists n = n(ε) such that ‖

∑k∈M xk‖ < ε for any finite subset

M ⊂ (n,∞) of the integers.Note that the direction (b) → (a) is a consequence of the completeness of the space.

(Basically (b) implies that for any permutation of N the sequence∑n

k=1 xσ(k) would be aCauchy sequence).

For the other direction (a) → (b), assume towards a contradiction that there is an ε > 0such that no such n could be found, thus we could find a sequence of sets M1, M2, ... suchthat maxMj < minMj+1 and ‖

∑k∈Mj

xk‖ > ε. It is not hard to build a permutation σ of Nsuch that each Mj appears as a consecutive block of σ(1), σ(2), . . . , thus

∑nk=1 xσ(k) is not a

Cauchy sequence in X so won’t be convergent, contradiction.Using this observation we claim that if a given series

∑n xn is unconditionally convergent

then∑

n xσ(n) is the same as∑

n xn for all permutation of N. Using this fact it is not hardto see that (i) and (ii) are equivalent. To see this claim, just note that for any σ there issome M such that σ(M + 1), · · · > n(ε), therefore ‖

∑k>M xσ(k)‖ ≤ ε and ‖

∑k>n(ε) xk‖ ≤ ε,

and by canceling out the common terms it follows that

‖∑k≤n(ε)

xk −∑k≤M

xσ(k)‖ = ‖∑

k≤M :σ(k)>n(ε)

xσ(k)‖ < ε

thus ‖∑

k xσ(k) −∑

k xk‖ < 3ε for all ε > 0, implying the desired claim.Now we show that (ii) and (iii) are equivalent. In fact for simplicity we will only show

the equivalence to the version with ε = ±1.It is clear that (iii) implies (ii): suppose that

∑k αkxk converges and (iii) holds true.

Then for any ε > 0 there exists n = n(ε) such that ‖∑

k>n αkxk‖ < ε/(2C). Given any finiteA ⊂ (n,∞) by choosing the sign sequence that equals 1 on A, equals 0 on 1, 2, . . . , n andequals ±1 constantly on N\A and using (iii) we obtain ‖

∑k∈A αkxk‖ ≤ 2C

∑k>n αkxk‖ < ε.

Thus by the observation above the series∑

k αkxk converges unconditionally.Now, to show that (ii) implies the sign sequence version of (iii), we will use the following

lemma

Lemma 3. Let Pn be the projection Pnx =∑

k≤n αkxk if x =∑

k αkxk is the expansion intothe Schauder basis (xn) for x. Then supn ‖Pn‖ <∞.

4.3. SCHAUDER BASIS 43

We first prove the above implication using the lemma. The proof of the lemma will bediscussed in the next class, after we have introduced several basic tools in Banach spaces,such as the principle of uniform boundedness and the open mapping theorem. Now, it sufficesto show that there exists N > 0 and a scalar CN > 0 such that for every sequence of scalarαj and signs εj and n ≥ N it holds that ‖

∑nj=N εjαjxj‖ < CN‖

∑nj=1 αjxj‖. Indeed, the

desired claim would follows from the fact that given any N and any sign sequence ε1, . . . itholds uniformly over n ≥ N that

‖N∑j=1

εjαjxj‖ .N ‖n∑j=1

αjxj‖

To see this, simply use Lemma 3 to obtain ‖αjxj‖ ≤ ‖(Pj−1−Pj)(∑n

k=1 αkxk)‖ ≤ C‖∑n

k=1 αkck‖and then use the triangle inequality.

Now assume towards a contradiction that for every N > 0 one can not find such CN .We then construct a sequence n1 < n2 < . . . and Aj ⊂ [nj, nj+1 − 1] and β1, β2, . . . re-cursively as follows, with the following property: for any j it holds that ‖

∑k∈Aj

βkak‖ ≥j2‖∑

nj≤k≤nj+1−1 βkak‖.Suppose that we have chosen n1 < · · · < nj and A1, . . . , Aj−1 and β1, . . . , βnj−1 satis-

fying the above requirements. Then given C > 0 by the assumption there exists nj+1 andβ1, . . . , βnj+1−1 and B ⊂ [nj, nj+1) such that

‖∑k∈B

βkxk −∑

k∈[nj ,nj+1)−B

βkxk‖ ≥ C‖∑

1≤k≤nj+1−1

βkxk‖

On the other hand, by the triangle inequality

‖∑

nj≤k≤nj+1−1

βkxk‖ ≤ ‖∑

1≤k≤nj+1−1

βkxk‖+ ‖Pnj−1(∑

1≤k≤nj+1−1

βkxk)‖

.nj‖

∑1≤k≤nj+1−1

βkxk‖

therefore by choosing C large enough we could ensure that

‖∑k∈B

βkxk −∑

k∈[nj ,nj+1)−B

βkxk‖ ≥ 2j2‖∑

nj≤k≤nj+1−1

βkxk‖

from here it is clear that either Aj = B or Aj = [nj, nj+1) − B will work, with αk = βk fork ∈ [nj, nj+1). This completes the selection of Aj and nj and αj.

Now by rescaling αk’s we may assume that ‖∑

k∈Ajαkxk‖ = 1 while ‖

∑nj+1−1k=nj

αkxk‖ ≤1/j2. Now, given any m1 < m2 we let ns < · · · < nt be the elements of (nj) inside [m1,m2),then using the uniform boundedness of Pn we obtain

‖m2∑

k=m1

αkxk‖ .t∑

j=s−1

‖∑

nj≤k<nj+1

αkxk‖ .1

s


and clearly if m1 → ∞ then so is s, thus∑

k αkxk converges. On the other hand it doesnot converge unconditionally since it violates the second property of the initial observation:given any n we could select j such that nj > n, and so Aj ⊂ (n,∞) and ‖

∑k∈Aj

αkxk‖ = 1.

Exercise:1. Prove that (e2πinx)n∈Z is an unconditional Schauder basis for L2[0, 1].2. Prove that the Haar functions are orthogonal in L2[0, 1] and form a Schauder basis

for L1[0, 1] but not unconditional in L1.3. (Radon–Nikodym) Let µ and ν two nonnegative σ-finite measures on the same σ

algebra (X,A) such that the following holds: for every measurable A if µ(A) = 0 thenν(A) = 0. Show that there exists g nonnegative such that for every measurable E it holdsthat ν(E) =

∫Egdµ.

Chapter 5

Basic results about Banach Spaces

Here we will prove several basic results about bounded linear maps between Banach spaces:the principle of uniform boundedness, the open mapping theorem, and the closed graphtheorem, and applications. We then focus on Lp spaces.

5.1 The Baire category theorem

We first recall the Baire category theorem for metric spaces, which will be used in the proofsof these theorems.

Below we say that a set A is nowhere dense if its closured A has empty interior.

Theorem 20 (Baire). A complete metric space can not be written as a countable union ofnowhere dense set.

Proof: If X =⋃An we can construct a Cauchy sequence xn that does not have a limit

as follows: Since A1 is nowhere dense there is an open ball of small radius B1 = B(x1, r1)such that B1 ⊂ X \A1. Then again find an open ball B2 = B(x2, r2) such that B2 ⊂ B1 \A2.Repeat this, make sure rk ≤ rk−1/2, we get a Cauchy sequence which can not converge toanything in any Ak, so no limit inside X.

5.2 The principle of uniform boundedness

Let X be a Banach space and Tα, α ∈ A is a family of bounded linear functionals from X tosome normed linear space Y .

Theorem 21 (P.U.B, aka Banach-Steinhaus). If supα∈A ‖Tαx‖ < ∞ for every x ∈ X thensupα∈A ‖Tα‖ <∞.

Remark: the opposite direction is clearly true.Example: If gn ∈ L∞(R) such that supn |

∫gn(x)f(x)dx| <∞ for every f ∈ L1(R) then

supn ‖gn‖∞ <∞.

45

46 CHAPTER 5. BOUNDED MAPS ON BANACH SPACES

Proof. We first show that for a linear operator T : X → Y (only need normed linear spacesX, Y ) T is bounded if any only if T−1(‖y‖Y ≤ 1) has nonempty interior in X.

Certainly if T is bounded then the set T−1(‖y‖Y ≤ 1) contains ‖x‖X ≤ 1/‖T‖ sononempty interior.

Conversely, if T−1(unit ball in Y ) contains an interior point x0, then for some ε > 0 itholds for every ‖x‖X ≤ ε that ‖T (x0 + x)‖ ≤ 1. It follows from the triangle inequality that

‖Tx‖ ≤ 1 + ‖Tx0‖ for such x, thus ‖T‖ ≤ 1+‖Tx0‖ε

.

Therefore in order to show P.U.B it suffices to show that the set

B1 =⋂α∈A

T−1α (‖y‖ ≤ 1)

has an interior point. In fact it suffices to show that for some n > 0 the (closed) set Bn

(replace the radius by n) has nonempty interior. But⋃Bn = X because supα ‖Tαx‖ < ∞.

Thus the desired claim follows from the Baire category theorem.

5.3 The open mapping theorem

Theorem 22 (Open mapping, aka Banach Schauder). Let T : X → Y linear map betweenBanach spaces. If T is bounded and surjective then T maps open sets to open sets.

Remark: as a corollary, if T : X → Y bounded bijective linear operators between Banachspaces then T−1 exists and is bounded linear.

Proof. Let Br = ‖x‖ < r ⊂ X.

By translation invariant if suffices to show that for every r > 0 the set T (Br) contains aneighborhood of 0. By dilation, if this holds for one r > 0 then it holds for all r > 0.

Now this in turns follows if we could show that T (Br) contains an interior point: indeedsuppose that for some ε > 0 and y0 we have |y−y0‖ < ε ⊂ T (Br), then by linearity T (B2r)contains T (Br)− T (Br) which in turn contains |y‖ < ε/2 a neighborhood of 0.

To show that T (Br) has nonempty interior, we first note that T (Br) has nonemptyinterior. This follows from surjectivity of T and the Baire category theorem

⋃n T (‖x‖ <

n) = Y . Therefore it suffices to show that T (Br) contains T (Br/10).

To see this, we note that T (Br) contains a neighborhood of 0 via the set differenceargument (as above). Assume that ‖y‖ ≤ r/C ⊂ T (Br) for some C > 0; by dilationinvariant this C could be chosen uniformly over r. Let y0 ∈ T (Br/10) then there is a pointx1 ∈ Br/10 such that ‖y − Tx1‖ < r

20C, thus y − Tx1 ∈ TBr/20, so again there is x2 ∈ Br/20

such that ‖y−Tx1−Tx2‖ < r40C

, thus by repeating this argument we obtain x1, x2, . . . suchthat ‖xn‖ ≤ r

2n5and T (x1 + · · · + xn) → y in Y as n → ∞. Since X is complete it follows

that x1 + · · · + xn → x ∈ X as n → ∞, thus using continuity of T if follows that y = Tx,and clearly ‖x‖ ≤

∑‖xk‖ < r, thus y ∈ TBr as desired.

5.4. THE CLOSED GRAPH THEOREM 47

5.4 The closed graph theorem

Graph of a function f : X → Y is defined to be Γ(f) = (x, f(x)), x ∈ X which is a subsetof X×Y equiped with the product topology. If X and Y are normed spaces then the producttopology on X × Y is the topology of the norm ‖(x, y)‖ = ‖x‖+ ‖y‖.

Theorem 23 (Close graph). Let T : X → Y be linear where X and Y are Banach spaces.Then T is bounded iff Γ(T ) is closed.

Remark: Since (X, Y ) is a Banach space also, Γ(T ) is closed iff whenever (xn, T (xn))converges to (x, y) one must have y = Tx. Comparing with the usual continuity of T(which is equivalent to boundedness of T ), which states that whenever xn → x we must haveTxn → Tx, we see that the extra thing we could assume is the convergence of Txn.

Proof. Clearly if T is bounded then T is continuous and the graph is closed. Assume nowΓ(T ) is closed, then Γ(T ) is a Banach subspace of X×Y . Let π1(x, y) := x and π2(x, y) := ybe the coordinate projections. By the open mapping theorem, the bounded bijective mapπ1(x, y) : Γ(T )→ X is boundedly invertible, therefore T = π2 π−1

1 is bounded.

Corollary 2 (Hellinger-Toeplitz). If A is linear operator on H a Hilbert space and 〈x,Ay〉 =〈Ax, y〉 then A is bounded.

Proof. Clearly Γ(A) is closed: if (xn, Axn) converges to (x, y) then for every z ∈ H we have〈z, y〉 = lim 〈z, Axn〉 = lim 〈Az, xn〉 = 〈Az, x〉 = 〈z, Ax〉, thus y = Ax.

5.5 An application: uniform boundedness of partial

sum projections for Schauder basis

Here we prove Lemma 3 from the last Chapter.Recall that Pn is the projection Pnx =

∑k≤n αkxk if x =

∑k αkxk is the expansion into

the Schauder basis (xn) for x ∈ X a Banach space.We need to show that supn ‖Pn‖ <∞.By the principle of uniform boundedness it suffices to show that the linear operator Pn

are bounded (the pointwise uniform boundedness of Pn follows from the fact that at eachx ∈ we have ‖Pnx‖ → ‖x‖ and a convergent sequence is always bounded). Now to showthat Pn are bounded, we consider the following norm on X

‖x‖1 = supn‖Pnx‖

it is not hard to see that this is a norm and ‖x‖ ≤ ‖x‖1. We will show that Pn is boundedin (X, ‖.‖1) and the norm ‖.‖1 is actually equivalent to ‖.‖, these two facts will take care ofthe lemma. Now the boundedness of Pn with respect to the new norm is clear

‖Pnx‖1 = supm‖PmPnx‖ = sup

m‖Pmin(m,n)x‖ ≤ ‖x‖1


Now we show that (X, ‖.‖1) is complete. Once we did that, the identity map from(X, ‖.‖1) → (X, ‖.‖) is a bijective bounded linear map thus by the open mapping theoremit is boundedly invertible, thus the two norms are equivalent and completes the proof of thelemma. Now, let X ′ be the completion of X under ‖.‖1, then it is not hard to see that xn isstill a Schauder basis for (X ′, ‖.‖1) (see the homework). So if a ∈ X ′ we have a =

∑j αjxj

convergence in ‖.‖1, but this series also converges in ‖.‖ too because ‖∑

m<j≤n αjxj‖ ≤‖∑

m≤n αjxj‖1 → 0 and X is complete. So, let b ∈ X be its limit under ‖.‖. Now, it is clearthat b = a because

‖b−∑j≤n

αjxj‖1 = supm‖Pm(b−

∑j≤n

αjxj)‖ = supm≥n‖(Pm − Pn)b‖ → 0

if n→∞.

5.6 Lp spaces

In this section we conduct a case study of Lp spaces, which are the most fundamental typeof Banach spaces.

5.6.1 Separability

Recall that if (X,A, µ) is separable then Lp(X,A, µ) is separable for 1 ≤ p <∞; this is partof the homework.

5.6.2 Duality

We’ll use Hilbert space techniques to show that

Theorem 24. For 1 < p <∞ the dual of Lp(X,µ) is Lq(X,µ) where 1/p+ 1/q = 1.If furthermore the measure µ is σ-finite then the dual of L1 is L∞.

Proof. We first consider the simpler case when µ is σ-finite, in this case we will show theresult for all 1 ≤ p <∞.

If µ is not necessary σ-finite we will need p > 1. Let ` ∈ (Lp)∗. Suppose that E ⊂ Xmeasurable such that µ|E is σ-finite. Clearly ` is also a bounded linear functional on Lp(E, µ)whose norm does not exceed ‖`‖. Thus, by the σ-finite case, for every f ∈ Lp(E) there isgE ∈ Lq(E, µ) such that

`(f) =

∫E

fgEdµ

‖gE‖q ≤ ‖`‖

Now if E ⊂ E ′ both measurable and σ-finite wrt µ then it is clear that gE = gE′ on E.

5.6. LP SPACES 49

Now, using the fact that a countable union of σ-finite sets is σ-finite one could find Fσ-finite wrt µ such that ‖gF‖q = supE σ−finite ‖gE‖q. We will show that

`(f) =

∫fgFdµ

for every f ∈ Lp(X). Observe that g|A\F = 0 for every A σ-finite, because otherwise∫|gA\F |qdµ > 0 and therefore ‖gA∪F‖q > ‖g‖F contradiction (note that here is where we

need p > 1 so that q <∞). In particular let A = |f | > 0 which is σ-finite wrt µ and so

`(f) =

∫fgAdµ =

∫fgFdµ

We now focus on the σ-finite case. For simplicity we will assume µ(X) < ∞, otherwisewe could decompose X into countably many subsets where µ is finte and apply this argumentto each subspace and add things up: because of continuity of `, for each f ∈ Lp(X) we have

`(f) = limn→∞

∑k≤n

∫Xk

fgkdµ

and use monotone convergence plus the fact that ‖∑

k≤n gk‖q = ‖∑

k≤n ‖gk‖qq)1/q ≤ ‖`‖ <∞to interchange the sum and define g =

∑k gk. (Note that the gk have disjoint support).

Now we assume µ(X) <∞. In this case it can be shown that any ` could be written asa linear combination of finitely many positive linear functionals, i.e. `(f) ≥ 0 if f ≥ 0.

For each E ⊂ X measurable construct a measure

ν(E) = `(1E)

it is clear that ν µ, therefore by the Radon Nikodym theorem there exists g ∈ L1, g ≥ 0,such that

`(1E) = ν(E) =

∫E

gdµ

for any measurable E ⊂ X. Now given any f ∈ Lp we will show that `(f) =∫fgdµ. We

may assume wlog that f ≥ 0, now by the monotone convergence theorem it suffices to showthis for f being simple functions, for which the claim holds. It remains to show that g ∈ Lq(note that since we assum µ(X) <∞ this is better than g ∈ L1 since L1 contains Lq for allq ≥ 1). Simply let f = |g|q−11|g|≤K , then∫

|g|q1|g|≤Kdµ = `(f) ≤ ‖`‖‖f‖p = ‖`(∫|g|q1|g|≤Kdµ)1/p

thus

(

∫|g|q1|g|≤Kdµ)1/p ≤ ‖`‖

therefore by monotone convergence we obtain ‖g‖q ≤ ‖`‖.


5.6.3 Basic facts about bounded operators on Lp

The most typical type of operator on Lp spaces are integral operators. Let (X,µ) and (Y, µ)be two measure spaces. Let

Tf(y) =

∫Y

K(x, y)f(x)dµ(x)

mapping from measureable functions on (X,µ) to measurable functions on (Y, ν). At thebeginning the operator T may not be defined for all f , as the integral may not be convergent.The adjoint of T is

T ∗g(x) =

∫X

K(x, y)g(y)dσ(y)

Think of this as a generalization of the matrix in finite dimensional setting. (integrationis like adding over indices and K gives the matrix entries.) (This turns out to be the caseif say X = 1, . . . , n and Y = 1, . . . ,m with the counting measures.) The function K iscalled the kernel (assumed measurable etc.).

Boundedness of integral operators

Theorem 25. For 1 ≤ p ≤ ∞ it holds that

‖T‖Lp→Lp ≤ [supx

∫X

|K(x, y)|dσ(y)]1/p[supy

∫X

|K(x, y)|dµ(x)]1/p′

In the special case p = 2 one also has

‖T‖L2→L2 . (

∫X×X

|K(x, y)|2dµ(x)dσ(y))1/2

Remarks: The dual estimates hold for T∗, namely we also have the same L2 → L2

esimates and

‖T‖Lp′→Lp′ ≤ [supx

∫X

|K(x, y)|dσ(y)]1/p[supy

∫X

|K(x, y)|dµ(x)]1/p′

In fact we will see that ‖T‖p→p = ‖T ∗‖p′→p′ .Operators for which (

∫X×X |K(x, y)|2dµ(x)dσ(y))1/2 < ∞ are called Hilbert-Schmid op-

erators and the right hand side is called the Hilbert Schmidt norm. In the finite dimensionalsetting, for instance if X = Y = 1, . . . , n with counting measures then this norm is thesame as (

∑1≤i,j≤n |K(i, j)|2)1/2, or equivalently the `2 norm of the sequence of eigenvalues

(if diagonalizable).For 1 < p <∞ to use the first estimate we need both factor to be finite. But if p = 1 or

p =∞ we only need one of them to be finite. More precisely

‖T‖L1→L1 ≤ supx

[

∫X

|K(x, y)|dσ(y)]

5.6. LP SPACES 51

‖T‖L∞→L∞ ≤ supy

∫X

|K(x, y)|dµ(x)

These turn out to be the easiest cases. For the L1 case this follows from

‖Tf‖L1(Y,ν) ≤∫Y

∫X

|K(x, y)||f(x)|dµ(x)dσ(y)

=

∫X

|f(x)|[∫Y

|K(x, y)|dσ(y)] dµ(x)

≤ ‖f‖1 supx

[

∫X

|K(x, y)|dσ(y)]

Similarly for the L∞ case we have, for each y ∈ Y

|Tf(y)| ≤∫X

|K(x, y)||f(x)|dµ(x)

≤ ‖f‖∞ supy

∫X

|K(x, y)|dµ(x)

To get the case 1 < p <∞ of the first estimate, we will need Riesz-Thorin interpolationtheorem (aka convexity theorem).

The Riesz-Thorin interpolation theorem

Theorem 26 (Riesz-Thorin). Let 1 ≤ p0lep1 ≤ ∞ and 1 ≤ q0 ≤ q1 ≤ ∞. Let T be a linearmap from (X,µ) to (Y, ν) such that

M0 := ‖T‖Lp0 (X,µ)→Lq0 (Y,σ) <∞

M1 := ‖T‖Lp1 (X,µ)→Lq1 (Y,σ) <∞Assume 1 < p, q <∞ be such that for some 0 < α < 1

(1

p,1

q) = (1− α)(

1

p0

,1

q0

) + α(1

p1

,1

q1

)

Then‖T‖Lp(X,µ)→Lq(Y,σ) ≤M1−α

0 Mα1

Proof. We will use the Hadamard three lines theorem: let h be analytic bounded in the stripRe(z) ∈ [0, 1]. For each α ∈ [0, 1] let H(α) = supRe(z)=α |h(z)|. Then

H(α) ≤ H(0)1−αH(1)α

The proof of this is an application of the maximum principle: if H(0) = H(1) = 1 thenthe claims follows by the maximum principle, in the general case modify h by a suitableexponential factor to reduce to this setting.


We now set up the bounded analytic function h(z). For Re(z) ∈ [0, 1] let pz and qz bedefined by

(1

pz,

1

qz) = (1− z)(

1

p0

,1

q0

) + z(1

p1

,1

q1

)

Then pz and qz are analytic functions. For any f ∈ Lp and g ∈ Lq′ (1/q′+1/q = 1) we define

h(z) =

∫Y

T (fz)gzdσ

where fz(x) = |f(x)|ppz−1f(x) and gz(x) = |g(x)|

qqz−1g(x). It is not hard to see that h(z)

is bounded analytic for Re(z) ∈ [0, 1] and h(α) =∫Tfgdσ, therefore using the duality

characterization for Lp we obtain

M = supf∈Lp,g∈Lq′

|H(α)|

Note that everything depends on f, g where f ∈ Lp and g ∈ Lq′ . We normalize ‖f‖p =‖g‖q′ = 1, then it is not hard to see that

|H(0)| ≤M0 , |H(1)| ≤M1

therefore the desired claims follows from Hadarmard three lines lemma. Exercises:1. Let X be a Banach space. Prove that separability of X∗ implies separability of X.2. Prove that the measure space (X,A, µ) is separable if any only if L2(X,A, µ) is

separable. Is this true if we replace 2 by any fixed p ∈ (1,∞)? Use this to construct a finitemeasure space (X,µ) (µ(X) <∞) such that L2(X,µ) is nonseparable.

3. Prove that if X is a compact metric space then C(X) with the sup norm is separable.4. Let X be a normed linear space and assume that (xn) is a Schauder basis, furthermore

the canonical projection Pn into the first n vectors is bounded for each n ≥ 1. Show that (xn)remains a Schauder basis for the completion of X. [Hint: show that Pn extends boundedlyto the completed space and use Pn to find the linear expansion/proving required propertiesfor Schauder basis.]

5. Let ` be a bounded linear functional on Lp(X,µ) where µ(X) < ∞. Show directly(without using the duality theorem) that ` could be written as a linear combination of finitelymany positive linear functionals on Lp(X,µ). Here 1 ≤ p <∞.

6. Show that every NLS is isomorphic to a subspace of the space of bounded continuousfunction on some complete metric space Y . 1

1if the given space is separable then one could take Y to be [0, 1] - this is the Banach-Mazur theorem.

Chapter 6

Bounded and continuous functions ona locally compact Hausdorff space anddual spaces

Recall that the dual space of a normed linear space is a Banach space, and the dual spaceof Lp is Lq where 1/p + 1/q = 1 if 1 < p < ∞. If the underlying measure space is σ-finitethen the dual of L1 is L∞. What about the dual of L∞? Or its subspace Co(X) and Cc(X)where say X is locally compact Hausdorff. In this chapter we discuss several representationtheorems related to these themes.

6.1 Riesz representation theorems

6.1.1 The dual space of L∞

The dual space of L∞(X,Σ, µ) consists of bounded (i.e. the total measure of X is finite) andfinitely additive signed/complex measures on Σ that is absolutely continuous with respectto µ.

Clearly if σ is a such a measure we may define a linear functional `σ(f) :=∫fdσ, it is

not hard to see that this is well defined as a bounded linear functional on L∞. (The factthat σ µ ensures that if f and f ′ equals almost everywhere with respect to µ, i.e. theyrepresent the same L∞ function, then the two integrals agree.) Note that it is possible todefine integration with respect to a finitely additive measure: start with positive functionand define the integral as the supremum over integration of simple functions dominated bythe given positive function. Then define integral of signed functions and complex valuedfunctions etc.

Conversely, given a bounded linear functional `, after several reductions we may assumethat ` is nonnegative, namely if f ≥ 0 then `(f) ≥ 0. Then we may define σ(E) = `(1E), itis not hard to see that σ is a finitely additive bouned measure, and if µ(E) = 0 then 1E ≡ 0in L∞(µ) thus σ(E) = `(1E) = `(0) = 0. Boundedness of the measures follows from choosing

53

54 CHAPTER 6. BOUNDED CONTINUOUS FUNCTIONS AND DUAL SPACES

the right input.

6.1.2 Measures on a locally compact Hausdorff spaces

There are several related and sometimes equivalent notions of measures on a given locallycompact Hausdorff space. Here we list and compare some of them.

Regular Borel measures and Radon measures

Recall that a Borel set is a set in the σ-algebra generated by the open sets.A regular Borel measure is a Borel measure (i.e. all Borel sets on X are measurable)

such that for any Borel set E the following holds.

µ(E) = inf µ(U) : U open, E ⊂ U

µ(E) = sup µ(K) : K compact, K ⊂ E

A Radon measure is a locally finite regular Borel measure, i.e. in addition to beingregular Borel it also assignes a finite value to all compact subsets of X.

Note that some textbooks define Radon measure as locally finite inner regular Borelmeasures, thus requiring only that measures of Borel sets could be approximated from insideusing measures of compact subsets.

Baire sets and Baire measures

A Baire set is an element of the sigma algebra generated by all compact subsets of X thatare at the same time countable intersections of open sets.

A Baire measure is a measure on the Baire sigma algebra.One could check that a Baire measure is also regular with respect to the Baire sigma

algebra.The Baire sigma algebra is the smallest sigma algebra such that all continuous functions

are measurable.It can be shown that any Baire measure extends uniquely to a Radon measure and vice

versa.

6.1.3 The dual space of Co(X) and Cc(X)

Let X be LCH. Since the closure of Cc(X) under the uniform norm is Co(X), it suffices toconsider the dual of Co(X). Note that if X is not compact then C(X) is not a normed linearspace with the sup norm, and we have to look at its dual as a topological dual. More on thisin later chapters.

We say that a signed measure is a signed Radon measure if it could be written as thedifference of two Radon measures. We say that a complex valued measure is a complexvalued Radon measure if its real and imaginary parts are signed Radon measures.

6.1. RIESZ REPRESENTATION THEOREMS 55

Theorem 27 (Riesz). Let X be a LCH. Then the dual of Co(X) is M(X) the space ofcomplex valued Radon measures (locally finite regular Borel measures) on X. Namely thefollowing map is an isometric isomorphism between this dual and M(X)

µ 7→ `µ(f) =

∫fdµ

and ‖`µ‖ = ‖µ‖ the total variation of µ defined by sup∑m

j=1 |µ(Aj)| supremum over allcollection A1, . . . , Am of disjoint subsets of X.

Step 1: We first show that if ` is a bounded positive linear functionals on Co,R(X), thespace of real valued continuous functions on X that vanish at ∞, then there is a Radonmeasure µ such that for any f ∈ Co,R(X) it holds that

`(f) =

∫fdµ

To see this, we would like to define µ by µ(E) = `(1E), however 1E is not continuous thereforeone can not apply ` to this function. To get around this we may define for each open setU ⊂ X

Defintion: µ(U) is defined to be the supremum over `(f) where f ∈ Co(R) and thesupport of f is inside U and 0 ≤ f ≤ 1 pointwise.

(Note that we may not be able to do this if U wasn’t open since such a continuousfunction may not exists; for open U the existence of f is due to Urysohn’s lemma).

Now we obtain a premeasure on open sets which is a ring of sets, thus by Caratheodorywe may exend µ to the sigma algebra generated by these sets, which are exactly the Borelsets.

We want to show that `(f) =∫fdµ for every f ∈ Co(X). Without loss of generality

assume 0 ≤ f ≤ 1. Since f vanishes at infinity the sets Ak = f ≥ k/n are compact, andwe may find continuous functions fk such that

1Ak≤ nfk ≤ 1Ak−1

and f =∑n

k=1 fk. We will use the following lemma

Lemma 4. If E ⊂ F are compact subsets and f is continuous on X such that 1E ≤ f ≤ 1Fthen µ(E) ≤ `(f) ≤ µ(F ).

Using the lemma and positivity of ` it follows that

1

n

n∑k=1

µ(Ak) ≤ `(f) ≤ 1

n

n∑k=1

µ(Ak−1)

consequently

|`(f)−∫fdµ| ≤ 1

nµ(A0) ≤ 1

nµ(supp(f)) = O(

1

n)


by sending n→∞ we obtain the desired claim.We now show the lemma.For the second estimate, recall that by definition µ is outer regular, thus µ(F ) =

infµ(U) : F ⊂ U , U open. Now for every U open containing F we have 0 ≤ f ≤ 1Uthus by definition of µ(U) it follows that `(f) ≤ µ(U). Consequently µ(F ) ≥ `(f).

For the first estimate, it suffices to show that for every ε > 0 we have µ(E) ≤ (1 + ε)`(f).Let Uε = f > 1

1+ε which is an open set that contains E. Thus

µ(E) ≤ µ(Uε) = sup`(g) : 0 ≤ g ≤ 1 , supp(g) ⊂ Uε

≤ sup`(g) : 0 ≤ g ≤ (1 + ε)f

≤ (1 + ε)`(f)

We now show regularity properties for µ. (It is clear that µ is finite on compact set usingUrysohn’s lemma.) Outer regularity of µ follows from construction using premeasure, andit remains to show inner regularity. Let E be Borel, it suffices to show that

µ(E) ≤ supµ(K) : K ⊂ E compact

now using outer regularity it suffices to do this for E open. Let U be an open set, we have

µ(E) = sup`(f) : 0 ≤ f continuous ≤ 1, sup(f) ⊂ U

≤ supµ((supp(f)) . . .

≤ supµ(K) : K ⊂ E compact

Step 2: We now reduce the desired result to the positive setting. This is done via writingbounded linear functionals on Co(X) as a linear combination of positive linear functionals. Tosee this note that without loss of generality it suffices to consider bounded linear functionalon C0,R(X) real valued members of C0(X). Let `+ be defined by

`+(f) = sup`(g) : 0 ≤ g ≤ f, g ∈ C0,R(X)

and let `− = `+−`, it is not hard to check that both `+ and `− are positive linear functionalson C0,R(X).

6.2 The Stone-Weierstrass approximation theorem

Let X be compact Hausdorff. Let P be a subspace of CR(X) such that P is also an algebrainside CR(X), i.e. p1p2 ∈ P if both p1, p2 ∈ P .

Theorem 28. Assume that P separates point, i.e. if x1 6= x2 are elements of X thenthere exists one element p ∈ P such that p(x1) 6= p(x2). Then one of the following holds:P = CR(X) or there is some x0 ∈ X such that P = f ∈ CR(X) : f(x0) = 0.

6.2. THE STONE-WEIERSTRASS APPROXIMATION THEOREM 57

Note that there are also versions for complex valued functions, in which case the sameconclusion holds if we assume further that P is closed under conjugation p ∈ P then p ∈ P .

There are also versions for C0(X) and C0,R(X) where X is locally compact Hausdorff.The original Weiertrass approximation theorem is for X being compact intervals, which

implies that polynomials are dense inside continuous functions. Applications in probablity(moment problems etc.)

Proof: Since P is also an algebra that separate points, without loss of generality we mayassume that P is closed. We divide the proof into two steps

Step 1: We will show that P is a lattice, namely if f, g ∈ P then so are max(f, g) andmin(f, g).

Step 2: Using the lattice property of P , we will show that if f ∈ CR(X) be such that forevery x 6= y in X there exists h ∈ P such that h(x) = f(x) and h(y) = f(y), then f ∈ P .

Combining these two steps, we prove the theorem as follows. Given any x 6= y considerthe algebra (h(x), h(y)), h ∈ P . Clearly this is a subalgebra subspace of R2 and it can’t be(0, 0) since P separate points. Thus it could be either 0× R or R× 0 or R2. Now in thelast case by step 2 any f ∈ CR(X) must be in P as desired. In the first case for instance,h(x) = 0 for all h ∈ P and so we must have h(y), h ∈ P = R for any other y. So by Step2 again it follows that any function f such that f(x) = 0 must be in P as desired.

Proof of step 1: since max(f, g) and min(f, g) could be written as linear combination of|f + g| and |f − g| and f and g, therefore it suffices to show that if f ∈ P then so is |f |. Theidea is to approximate |f | with a polynomial of f uniformly. Indeed, clearly f is boundedtherefore without loss of generality assume |f | ≤ 1, it then suffices to show that |x| could beapproximated uniformly by polynomials on [−1, 1]. To see this use the Taylor expansion of√

1− t = 1− 12t+ . . . which converges uniformly on 0 ≤ t ≤ 1, then write

|x| =√x2 =

√1− (1− x2) = 1− 1

2(1− x2) + . . .

Proof of step 2: Let ε > 0. It suffices to show that there exists g ∈ P such that|g(x) − f(x)| ≤ ε uniformly over x ∈ X. Assume that for every x ∈ X there exists gx ∈ Psuch that |gx(x)− f(x)| < ε and gx(y) ≤ f(y) + ε for every y ∈ X. Then the desired g couldbe constructed as follows. By continuity for each x ∈ X there exists Ux ⊂ X neighborhoodof x such that supUx

|gx − f | < ε. By compactness of X one could refine the collectionUx, x ∈ X to a finite subcollection Ux1 , . . . , Uxm , and we simply let

g = max(gx1 , . . . , gxm)

which is in P by the lattice property and clearly supX |g − f | ≤ ε.Now to show the existence of such a gx we fix x and notice that, using the given hypothesis,

for each y ∈ Y we may find hy ∈ P and two open neighborhoods of x and y respectively,denoted respectively by Uy and Vy (note that x is fixed so we ignore the dependence on x)such that

supUy

|hy − f | < ε , supVy

|hy − f | < ε


By compactness of X again we may refine the covering Vy, y ∈ Y to a finite subcoveringVy1 , . . . , Vyn , and we simply let gx = min(hy1 , . . . , hyn) which has the desired property withUx := Uy1 ∩ · · · ∩ Uyn .

Chapter 7

Locally convex spaces, the hyperplaneseparation theorem, and theKrein-Milman theorem

Recall that C(X) is not a normed linear space when X is not compact. On the other handwe could use semi norms on C(X): given any compact K ⊂ X let ‖.‖K be the sup nom onK. This family of seminorms determine C(X): together they provides a lot of properties sothat various theorems (for NLS) remains valid here.

7.1 Two equivalent definitions of LCS

Let X be a linear space over R (results over C are similar). We consider topologies on Xsuch that addition and scalar multiplication are continuous operations (which are assumedin the definition below), in which case we say that X is a topological linear space.

Now there are two equivalent definitions of local convexity for a topological linear space.In one definition, we use seminorms. A seminorm ρ on X is essentially a norm except for thenondegrenerate condition, namely it satisfies the triangle inequality ρ(x+ y) ≤ ρ(x) + ρ(y),nonnegativity ρ(x) ≥ 0, homogeneity ρ(λx) = |λ|ρ(x), however it is possible that ρ(x) = 0when x 6= 0. A typical example is ρ(x) = |`(x)| where ` is a linear functional on X. Now, atopological linear space X is said to be locally convex if there exists a family of seminormsρα, α ∈ I, such that the topology on X is the minimal topology such that (addition andmultiplication are continuous and) ρα are continuous.

Another definition uses convex sets. A subset A of X is called balanced if λx ∈ Awhenever x ∈ A and |λ| ≤ 1. We also say that A is absorbent if for every x ∈ X there issome t > 0 such that tA contains x. Then a topological linear space X is said to be locallyconvex if there exists a neighborhood base at 0 consisting of only convex balanced absorbent(open) sets.

To see the equivalence of two definitions, we start with the seminorm definition. Then a

59

60 CHAPTER 7. LOCALLY CONVEX SPACES

neighborhood base at 0 could be taken to consists of all open sets of the form

U = x ∈ X : ρα1(x) < ε, . . . , ραm(x) < ε

where ε > 0 and α1, . . . , αm are elements of I. (m ≥ 1 is arbitrary). It is clear that theseopen sets are convex and balanced and absorbent.

Conversely given a neighborhood base consisting of convex balanced absorbent sets wecould use the Minkowski gauge functional to construct the seminorm. Namely if A is convexbalance absorbent we let

ρA(x) = inft > 0 : x ∈ tA

(It is not hard to check that ρA is seminorm.)

7.1.1 Basic properties

A family of seminorm is said to be separated if whenever ρα(x) = 0 for every α ∈ I it mustfollow that x = 0. This is actually equivalent to the Hausdorffness of the topology.

If (xi)i∈D is a net in X then (xi)→ x if any only if ρα(xi− x)→ 0 (as a net in R) for allα ∈ A.

If there is a lot of seminorms one expects that the topology is very rich; on the otherhand if there are fewer seminorms one expects a more well-structured topology. In particularif there are only countably many seminorms involved then the topology is equivalent to thetopology of some pseudo-metric, i.e. a distance notion that resembles the metric notionexcept for the fact that two distinct points could have distance 0. To see this, enumeratethe seminorms ρ1, . . . , and let

d(x, y) =∞∑n=1

2−nρn(x− y)

1 + ρn(x− y)

If one assume that the (countable) family is separated then the above pseudo metric isactually a metric. If this metric is furthermore complete then the given space is called aFrechet space.

Examples: Recall that C(X) is a locally convex space with seminorms given by ρK(f) =supx∈K |f(x)| where K ⊂ X is compact. Other examples are

(i) the space of Schwartz functions on Rn: These are functions that are C∞ and theirderivatives decay faster than any polynomial. One could use ρα,β(f) = |x|α|Dβf(x)| whereα and β are nonnegative integer multi-indices.

(ii) if `α, αA is a family of linear functionals on X then the minimal topology such that`α are continuous and linear operations (addition/scalar multiplication) are continuous iscalled the weak topology induced by this set of linear functionals. Examples of these (fornormed linear spaces) are the weak topology and the weak* topology. It can be shown thatif ` is a linear functional that is continuous with respect to this topology it must be a finitelinear combination of the given linear functionals.

7.2. THE HYPERPLANE SEPARATION THEOREM 61

(iii) Given any open set Ω ⊂ Rn the space D(Ω) consisting of compactly supportedinfinitely smooth functions on U is also a locally convex space and actually complete (recallthat the version without infinite smoothness, i.e. Cc(U), is not complete).

7.1.2 Linear maps

Let X and Y be locally convex spaces with seminorms ρα and ρ′β where α ∈ I and β ∈ Jindex sets. Then it can be shown that a linear map ` : X → Y is continuous if and only ifgiven any β ∈ J there exists α1, . . . , αm ∈ I and C > 0 such that the following holds for allx ∈ X:

ρ′β(`(x)) ≤M(ρα1(x) + · · ·+ ραm(x))

(Note that this generalizes the usual equivalence of continuity and boundedness of linearmaps between normed linear spaces.)

7.2 The Hyperplane separation theorem

Let K be a convex subset of a topological linear space X over R and let y ∈ X \K. Assumethat K contains at least one interior point.

Theorem 29. There exists a continuous linear functional ` on X such that supx∈K `(x) ≤`(y). If furthermore K is closed then this inequality could be taken strict.

Proof. By translation invariant we may assume that 0 is an interior point of K. Considerthe Minkowski gauge functional

ρK(x) = inft > 0 : x ∈ tK

since 0 is an interior point of K it is clear that ρK(x) <∞ for any x ∈ X. Furthermore ρKis convex and positive homogeneous, and since y 6∈ K we have ρK(y) ≥ 1 ≥ ρK(x) for everyx ∈ K. Let ` be defined on the one dimensional subspace spanned by y using `(y) = 1. Then|`(z)| ≤ ρK(z) inside this subspace, so by Hahn Banach we may extend ` to all of X suchthat |`(x)| ≤ ρK(x) for all x. In particular `(x) is continuous since ρK is continuous at 0.

Now, if K is closed then one could see that ρK(y) > 1 ≥ supx∈K ρK(x) and we couldtherefore obtain a strict inequality.

For a locally convex Hausdorff space over R, it follows from the above theorem thatfor any two distinct points x 6= y there is a continuous linear functional ` on X such that`(x) 6= `(y). To see this, by Hausdorffness and local convexity we could find a convex open setA such that x ∈ A while y ∈ X \A. Since A is closed convex we could apply the hyperplaneseparation theorem and get a continuous linear function ` such that supz∈A `(z) < `(y). Inparticular `(x) < `(y).


7.3 The Krein-Milman theorem

An extreme point of a convex set K is a point x such that if x is a convex combination ofx1, x2 ∈ K then x1 = x2 = x.

Theorem 30 (Krein-Milman). Let X be locally convex Hausdorff and K is a nonemptycompact subset of X. Then

(i) K has at least one extreme point.(ii) K is the closure of the convex hull of its extreme points.

Remark: If X = Rn for some finite n then we don’t need to take the closure in (ii), infact Caratheodory showed that one could get any point of X from a convex combination ofat most n+ 1 extreme points. For infinite dimensional space the closure is essential.

Proof:We first generalize the notion of extreme points to extreme subsets of K. A subset A of

K is said to be extreme if it is nonempty convex and furthermore if any x ∈ A is a convexcombination of two points x1 and x2 in K then we must have x1, x2 ∈ A. It is clear thatif a family of extreme subsets of K has nonempty intersection then this intersection is alsoextreme. In particular K is an extreme subset of itself.

Now, consider the collection A of all closed extreme subsets of K, which is nonemptysince it contains K, and we may order this collection partially using set inclusion, namleyA ≤ B if A ⊃ B.

(i) We first show that there exists a maximal element in A, which we will show to be apoint later. In order to show existence of the maximal element we plan to use Zorn’s lemma,and what is needed here is the fact that any chain (i.e. a totally ordered subcollection ofA) has an upper bound. The idea is to take the intersection of the elements in this chain,and what needs to be shown is the fact that this intersection would not be empty. Assumetowards a contradiction that this intersection is empty, it follows from compactness of Kthat there exists a finite subcollection that has empty intersection, but this is a contradictionbecause the intersection of a finite chain is simply the smallest element, which is nonempty.

Now, let A be a close extreme subset of K and assume towards a contradiction that itis not a point. Say x 6= y are two elements of A, then there is a continuous linear functionalon X that separates x and y. We’ll show that the subset B of A where ` achieves maximumis an extreme subset of K, which would violate maximality of A. (Clearly B 6= A). SinceA is extreme in K it suffices to show that B is extreme in A. Now extremality of B in Afollows easily from linearity of `.

(ii) We now show that the closure of the convex hull E of extremal points of K is K, i.e.K = E. Suppose that z ∈ K \ E. Then by the hyperplane separation theorem there is acontinuous linear functional ` such that

supx∈E

`(x) < `(z)

Again the set of maximum of ` on K is a proper subset of K and also an extreme subset, and

7.4. INDUCTIVE LIMIT AND WEAK SOLUTIONS 63

this set is also closed and disjoint from E. So by repeating the above argument one couldfind one extremal point of K inside this set, which is therefore not inside E, a contradiction.

Examples: Let X be a compact Hausdorff space and consider CR(X) (we could do locallycompact Hausdorff too with CR,0(X)) and let A consists of all positive linear functional onC(X). Let A be the subset of A with `(1) = 1, it is not hard to see that A is convex and itsextreme pooints are the point evaluation linear functional exf = f(x).

7.4 Inductive limit and weak solutions

Let Ω be a domain inside Rn. As mentioned before the space D(Ω) consists of C∞ functionswhose supports are compact subsets of Ω. One way to construct a locally convex topologyon this space is to use inductive limit of topologies. Let X1, X2, . . . , Xn, . . . , be linear spacessuch that X1 ⊂ X2 ⊂ · · · ⊂ X =

⋃Xn. Each Xn has a locally convex topology that is

consistent with the topologies on other Xm in following sense: the topology of Xn is theinduced topology from Xn+1. If that is the case one could construct a limiting topology onX whcih should be thought of as limn→∞Xn.

In our context we let K1 ⊂ K2 . . . be compact subsets of Ω such that⋃Kj = Ω. Then

let Xn be the space of C∞ functions on Rn whose support are subsets of Kn. Note that Xn

is a complete metrizable locally convex space. We then define the topology on D(Ω) to bethe inductive limit of the topologies of Xn.

The dual space of D(Ω) is called the space of (tempered) distribution on Ω denoted byD′(Ω). This space contains D(Ω) as a dense subspace. We may define the action of a lineardifferential operator Dj (here j is a multiindex and the sum is over some finite collection)on any ` ∈ D′ by defining for each φ ∈ D(Ω)

Dj`(φ) = (−1)j`(Djφ)

A weak solution to a PDE is a distributional solution in the above sense. If this distributionarise from some sufficiently smooth functions then we say it is a classical solution.

Part II

Spectral analysis for linear operators

65

Chapter 8

Elementary spectral theory

Let B denote the space of all bounded linear operators on some given Banach space Xover C. The analysis here works for more general settings, say B could be a unital Banachsubalgebra of this space (i.e. an associative subalgebra that contains a unit and is a completesubspace with respect to the operator norm).

8.1 Spectrum and resolvent set

We say M ∈ B is invertible if its inverse exists and is in B. Observe that if M is invertiblethen so is all N with ‖N −M‖ < 1

‖M−1‖ . To see this by writing N = M + (N −M) =

M(I + M−1(N −M)) it suffices to consider the case M = I. In this case (i.e. M = I) usegeometric series to show that the inverse is exactly

I + (N − I) + (N − I)21 + . . .

Definition: The resolvent set of M in B consists of all complex number λ such thatλI −M is invertible in B. The resolvent set is denoted by ρ(M) and the spectrum of M isσ(M) := C \ ρ(M).

Properties:1. ρ(M) is open in C.(This is a consequence of the above observation.)2. The resolvent function (λI −M)−1 is an analytic function of λ on ρ(M).This is because the above observation also shows that the resolvent could be expanded as

a power series around each point λ in the resolvent set ρ(M)

(λ− h−M)−1 =∞∑j=0

(λ−M)n−1hn

which has positive radius of convergence |h| < ‖(λ−M)−1‖.3. ‖(λ−M)−2‖ ≥ 1/dist(λ, σ(M)).

67

68 CHAPTER 8. ELEMENTARY SPECTRAL THEORY

This is a corollary of the above analysis.4. (Gelfand) The spectrum σ(M) is a bounded nonempty closed subset of C. Further-

more, the spectral radius of M , defined by |σ(M)| = max |λ| over λ ∈ σ(M) satisfies

σ(M) = limk→∞‖Mk‖1/k

Closedness is clear, boundedness follows from the fact that if λ is large enough in modulusthen λ−1M has small norm and so (1− λ−1M) is invertible and so λ ∈ ρ(M). To show thatσ(M) is nonempty assume towards a contradiction that it is. Then consider the contourintegral along CR = |ξ| = R ∫

CR

(λ−M)−1dλ

clearly analyticity of the resolvent operator implies that this would be 0. On the other handwe know that (λ−M)−1 has the Laurent series expansion at ∞

(λ−M)−1 =∞∑n=0

Mnξ−(n+1)

therefore the contour integral will gives the residue term i.e. the term with n = 0 and weget I, contradiction.

To show the identity for the spectral radius, by elementary complex analysis it followsthat |σ(M)| ≥ lim sup ‖Mk‖1/k. So it suffices to show that |σ(M)| ≤ |Mk|1/k for all k ≥ 1.To see this, observe that

‖∑n≥0

Mnξ−(n+1)‖ ≤[ k−1∑j=0

‖M j‖|ξ|−(j+1)][∑

m≥0

(‖Mk‖|ξ|−k)m]

this follows from ‖Mmk+j‖ ≤ ‖M j‖‖Mk‖m. It follows that if |ξ| > ‖Mk‖ then the se-ries converges absolutely and therefore ξ ∈ ρ(M); this completes the proof that |σ(M)| =lim ‖Mk‖1/k.

8.2 Functional calculus and spectral mapping

We clearly could define a polynomial of a given element M ∈ B; in fact via power series thisworks for analytic functions provided that the spectral norm of M is smaller than the radiusof convergence at 0. In particular it works if the given funciton is analytic in some domainΩ that contains σ(M). We demonstrate that this later fact would be enough. Indeed, let Cbe a contour in the intersection of ρ(M) and the domain of analyticity of f that winds oncearound every point in σ(M) but zero time around Ωc. Then define

f(M) =

∫C

(ξ −M)−1f(ξ)dξ

8.3. EXAMPLES OF OPERATORS AND THEIR SPECTRA 69

note that by the Cauchy theorem this is independent of the choice of the contour. It turnsout that this is consistent with the usual polynomial case and furthermore

σ(f(M)) = f(σ(M))

(also known as the spectral mapping property). We also have the resolvent identity

(ξ1 − ξ2)[(ξ2 −M)−1 − (ξ1 −M)−1] = (ξ1 −M)−1(ξ2 −M)−1

which is useful to show that the functional calculus maps the algebra of analytic functionson open sets containing σ(M) into B is a homomorphism.

Also, if g is analytic on an open set containing f(σ(M)) and h = g f then h(M) =g(f(M)).

Spectral Projections: Assume the spectrum σ(M) could be decomposed into an unionof n disjoint closed components σ1 ∪ . . . σn.

Let Cj be a contour in ρ(M) that winds once around each point of σj but not othercomponents, and Pj =

∫Cj

(ξ −M)−1dξ.

Then Pj are disjoint projections P 2j = Pj and PjPk = 0 if j 6= k, and∑

j

Pj = I

and if σj 6= ∅ then Pj 6= 0.

8.3 Examples of operators and their spectra

1. Shifts

Let X = `2 consiting of x = (a0, a1, . . . ) such that∑|aj|2 <∞. The right and left shifts

R and L

Rx = (0, a0, a1, . . . , ) , Lx = (a1, a2, . . . )

Now LR = I but RL 6= I so these linear bounded maps are not invertible. The spectrum ofR and L consists of all points in the unit disk λ ∈ C : |λ| ≤ 1. Now R′ = L and L′ = Ras adjoint of each other.

The idea is the spectral radius of L is 1 so spectrum contained inside the disk.

2. The Fourier transform Ff(ξ) =∫f(x)e−2πixξdξ. Note that F 4 = I therefore the

spectum of F is a subset of x ∈ C : x4 = 1 which consists of ±1,±√−1. Let Hn denotes

the orthgonal polynomials wrt to e−x2dx∫

HmHne−x2dx =

√π2nn!δmn


then it could be shown that F [e−x2/2Hn(x)] = (−

√−1)ne−ξ

2/2Hn(ξ). This follows from

exp(−x2/2 + 2xt− t2) =∑n≥0

e−x2/2Hn(x)tn/n!

3. Volterra integral operator Let X = C([0, 1]) and let Tf(x) =∫ x

0K(x, t)f(t)dt,

also known as the Volterra integral operator, here K is continuous. The spectrumn of Tconsists of only one point λ = 0. This follows by computing T nf , for instance if K ≡ 1 thenT nf(x) = 1

(n−1)!

∫ x0

(x− t)n−1f(t)dt and show that ‖T nf‖ ≤ ‖f‖/n! in the sup norm, and usethe spectral radius identity.

4. Diagonal multiplication Let X = `p(Z+) and M acts diagonally (Mx)n = λnxn,then the spectrum of M is the closure of λn.

8.4 Adjoint operators and spectrum

For convenience of notation let 〈x, `〉 := `(x) for every ` ∈ X∗ and x ∈ X.Let T : X → X be a bounded operator on a Banach space X. Then its adjoint operator

T ∗ is defined to be the bounded operator from X∗ to X∗ such that for every x ∈ X and` ∈ X∗ it holds that

〈x, T ∗`〉 = 〈Tx, `〉

It is not hard to see that the adjoint operator exists (and is linear and bounded), and‖T ∗‖ ≤ ‖T‖.

Note that this definition differs the adjoint notion of an operator on a Hilbert space bya conjugate. Thus some texts use T ′ for the Banach space conjugate of T .

Theorem 31 (Phillips). If T : X → X bounded linear on a complex Banach space X thenσ(T ) = σ(T ∗) and for every λ ∈ ρ(T ) it holds that ([T − λ]−1)∗ = (T ∗ − λ)−1.

As a corollary, for Hilbert space we have σ(T ) = σ(T ∗).We will prove the first part, the second part follows as a by product.Step 1: ρ(T ) ⊂ ρ(T ∗). Take λ ∈ ρ(T ). It suffices to show the following lemma (then

apply it to T − λ)

Lemma 5. Let T : X → X be bounded linear and boundedly invertible. Then (T ∗)−1 existsand (T ∗)−1 = (T−1)∗.

Proof. : It suffices to show, using the open mapping theorem, that T ∗ : X∗ → X∗ boundedand injective and onto.

Injectivity: If T ∗` = 0 for some ` ∈ X∗ then we will show that ` = 0.

〈Tx, `〉 = 〈x, T ∗`〉 = 0

for every x, so using the fact that T is onto we obtain `(X) = 0 so ` = 0.

8.4. ADJOINT OPERATORS AND SPECTRUM 71

Onto: If ` ∈ X∗ then we want some `0 ∈ X∗ such that T ∗`0 = `. We equivalent transformthis equation into

〈x, T ∗`0〉 = 〈x, `〉 ∀ x ∈ X

or equivalently 〈Tx, `0〉 = 〈x, `〉 for all x ∈ X. Since T is invertible this is the same as〈y, `0〉 = 〈T−1y, `〉 for every y ∈ X, which is the same as `0 = (T−1)∗`.

Step 2: Here we show that ρ(T ∗) ⊂ ρ(T ). We will use the following Lemma

Lemma 6. Let T : X → X be bounded linear on a complex Banach space. Then(i) If range(T ∗) is dense in X∗ (in the weak* topology) then T is injective.(ii) If T ∗ is injective then range(T) is dense.

We first use the lemma to complete this step. If λ ∈ ρ(T ∗) we will show λ ∈ ρ(T ). Bythe lemma it follows that T − λ is injective and has dense range. It remains to show that(T − λ)−1 is bounded on its range (then we could invoke the BLT theorem). To see this forevery y ∈ range(T − λ), i.e. y = (T − λ)z for z ∈ X, and ` ∈ X∗ we have

〈y, `〉 = 〈z, (T − λ)∗`〉

and so〈z, `〉 =

⟨y, [T ∗ −−λ]−1`

⟩| 〈z, `〉 | ≤ ‖y‖‖`‖[T ∗ −−λ]−1‖

since this is true for all ` we obtain ‖z‖ ≤ ‖[T ∗ −−λ]−1‖‖y‖, as desired.It remains to show the lemma. For (i) note that if Tx = 0 then 〈Tx, `〉 = 0 for all ` ∈ X∗

and therefore 〈x, T ∗`〉 = 0, but the denseness of the range of T∗ implies that 〈x, `〉 = 0 forall ` ∈ X∗, and consequently x = 0. For (ii), assume that range(T ) is a strict subspace ofX, then we could find, using Hahn Banach, a nonzero bounded linear functional such that`(Tx) = 0 for all x ∈ X. Then 〈x, T ∗`〉 = 0 for all x therefore T ∗` = 0 but injectivity of T ∗

implies that ` = 0 contradition.

Chapter 9

Spectral theory for compact operatorson Banach spaces

Recall that a subset S of a metric space X is precompact if its closure is compact, orequivalently every sequence contains a Cauchy subsequence. Another characterization isthat S is totally bounded, namely for any ε > 0 one could cover S by finitely many ε-balls.

If X is a normed linear space we can add/multiply, and we have the following basicproperties:

(i) If S is precompact then so is αS for any α scalar.(ii) If S1 and S2 are precompact then S1 +S2 := s1 +s2 , s1 ∈ S1, s2 ∈ S2 is precompact.(iii) If S is precompact then so is the convex hull of S.(iv) Let T : X → Y where X and Y are say normed linear spaces. If S ⊂ X is precompact

then so is TS ⊂ Y .

9.1 Compact operators

Let T : X → Y a bounded linear map between Banach spaces X and Y . Let B1 be the unitball in X. We say that T is pre-compact if TB1 is a precompact subset of Y .

Properties:(i) A finite rank operator (namely the range of the operator is finite dimensional) is

compact.(ii) If T1 : X → Y and T1 : X → Y are compact operators then α1T1 + α2T2 is also

compact for any scalar α1, α2.(iii) If T : X → Y is compact and M : U → X and N : Y → V are bounded linear maps

between Banach spaces then NTM : U → V is compact.(iv) If T : X → Y is compact then it maps a weakly convergent sequence in X to a

strongly convergent sequence in Y . Such operator is called completely continuous, so wecould say that compactness implies complete continuity.1

1If T : X → X is completely continuous and X is reflexive then T is compact, this is left as an exercise.

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74 CHAPTER 9. COMPACT OPERATORS

To see property (iv), first note that Txn converges weakly to Tx. To see this, let ` ∈ X∗,then `(Txn) = (T ∗`)xn converges to T ∗`x which is the same as `(Tx). It then follows thatany strongly convergent subsequence of Txn has to converge to Tx. Now since xn convergesweakly to x in X it follows that xn is a family of bounded operator on X∗ that is uniformlybounded pointwise, consequently by the principle of uniform boundedness supn ‖xn‖ < ∞,thus (Txn) is precompact, so any subsequent contains a Cauchy subsequence; thus by aroutine argument it follows that Txn is convergent to Tx.

(v) If Tn is a sequence of compact operators from X to Y and Tn − T‖ → 0 as n → ∞for some T : X → Y bounded linear, then T is compact.

To see property (v), note that for any ε > 0 one could chooseN large such that ‖TN−T‖ <ε/2. Now, we could use finitely many ε/2 balls to cover TNB1. The ε balls with the samecenters will then cover TB1. (Recall that B1 is the unit ball in X.)

As a corollary, we have

Corollary 3. If T : X → Y is the limit of a sequence of finite rank operators in the normtopology then T is compact.

The converse direction of this corollary is not true in general (for Banach spaces X andY ), the first construction of counter examples is due to P. Enflo (’73). However, if Y is aHilbert space then the converse is true.

Theorem 32. Any compact operator T : X → Y where X is Banach and Y is Hilbert canbe approximated by a sequence of finite rank operators.

Proof. We sketch the main ideas of the proof. For every n the set TB1 is covered by⋃Mn

k=1 BY (yk,1n). We may assume that Mn is an increasing sequence. Let Pn be the projection

onto the span of y1, . . . , yMn , which is clearly a finite rank operator, thus PnT is also finiterank. It remains to show that ‖T − PnT‖ = O( 1

n). We observe that ‖Pny − yk‖ ≤ ‖y − yk‖

for every 1 ≤ k ≤Mn and every y ∈ Y , since projection are contractions. It follows that

‖PnTy − Ty‖ ≤ ‖PnTy − yk‖+ ‖Ty − yk‖ ≤ 2‖Ty − yk‖

and clearly given any y ∈ B1 there is a k such that ‖Ty−yk‖ ≤ 1/n, therefore ‖PnTy−Ty‖ ≤2n

for every y ∈ B1, thus ‖PnT − T‖ ≤ 2/n as desired.

Theorem 33 (Schauder). Let T : X → Y be a bounded linear operator between Banachspaces X and Y . Then T is compact if any only if T ∗ : Y ∗ → X∗ is compact.

Proof. It suffices to show the forward direction, namely if T is compact then T ∗ is alsocompact. For the other direction, apply the forward direction it follows that T ∗∗ is compactfrom Y ∗∗ to X∗∗, and by restricting T ∗∗ to X we obtain T therefore T is also compact.

Now, assume that T is compact. Given any sequence `n ∈ X∗ with ‖`n‖ ≤ 1 we willshow that (T ∗`n) has a Cauchy subsequence T ∗`nj

, in other words given any ε > 0 we have

sup‖x‖≤1

‖`nj(Tx)− `nk

(Tx)‖ ≤ ε

9.2. COMPACTNESS OF INTEGRAL OPERATORS 75

if j and k are large enough.

Let B be the unit ball in X and let K = TB which is a compact subset of X. Then theabove estimate is a consequence of supy∈K ‖`nj

y− `nky‖ ≤ ε. We may view `n as a sequence

of continuous funcitons on K, which are uniformly bounded pointwise and equicontinuouson K:

supn‖`ny‖ ≤ ‖y‖

supn‖`ny1 − `ny2‖ ≤ ‖y1 − y2‖

Thus by the Arzela Ascoli theorem the sequence `n has an uniformly convergent subse-quence, as desired.

9.2 Compactness of integral operators

We now discuss compactness of the integral operator T

Tf(y) =

∫U

K(x, y)f(x)dµ(x) , y ∈ V

where U and V are say compact metric spaces, viewing T as operator on different functionspaces.

Viewing T as a map from L2(U, dµ) to L2(V, dν) for some measure ν (note that bothspaces are separable Hilbert spaces), we know that one sufficient condition that guaranteeboundedness of T is

∫U×V |K|

2dµdν <∞. It turns out that this would also imply compact-ness of T . To see this, for each x we expand K(x, y) into the (countable) orthogonal basisof L2(Y, dν), which we may denote by φ1, φ2, . . .

K(x, y) =∞∑j=1

Kj(x)φj(y)

note that for almost every x ∈ X the function K(x.y) is L2(Y, dν) integrable in y and so∫ ∫|K(x, y)|2dµ(x)dν(y) =

∑j

∫X

|Kj(x)|2dµ(x)

thus we may approximate T with the finite rank operator

Tnf(y) =∑j≤n

∫Kj(x)f(x)dµ(x)uj(y)

so T is compact.


9.3 Spectral properties of compact operators

9.3.1 Riesz’s theorem

One of the main results about compact operators is the following fact: if T : X → X isa compact operator on a Banach space X and 1 − T is injective, then 1 − T is boundedlyinvertible. Note that it is possible for ‖T‖ to be large, so the basic theory about smallperturbation of 1 does not applies here. The key to showing this fact is the followingtheorem of Riesz.

Theorem 34 (Riesz). Let T : X → X be a compact operator on a Banach space X. Thenrange of 1−T is a closed subspace of X and furthermore dim(ker(1−T )) and codim(1−T )are finite and equal to each other.

Part of the proof of this theorem is the following lemma.

Lemma 7. Let T : X → X be a compact operator on a Banach space X. Then(i) ker(1− T ) is finite dimensional.(ii) There exists some k ≥ 1 such that ker((1−T )m) = ker((1−T )m+1) for every m ≥ k.(iii) range(1− T ) is a closed subspace of X.

Proof of Lemma 7. (i) If y ∈ ker(1− T ) then Ty = y. Assume towards a contradiction thatker(1− T ) is infinite dimensional. Then by another result of Riesz we could find an infinitesequence (yn) in this kernel such that ‖yn‖ = 1 and ‖yn − ym‖ ≥ 1/2 for all m 6= n. Thisimplies the set Tyn, n ≥ 1 is not precompact, contradiction.

(ii) Note that ker([1− T ]k) ⊂ ker([1− T ]k+1) for all k. Now, if ker([1− T ]k) = ker([1−T ]k+1) then ker([1− T ]m) = ker([1− T ]m+1) for all m ≥ k. Assume towards a contradictionthat we have the strict inclusion ker([1 − T ]k) ( ker([1 − T ]k+1) for all k ≥ 1. Sinceker([1−T ]k) are closed, by Riesz lemma we may find xn ∈ ker([1−T ]n) such that ‖xn‖ = 1and dist(xn, ker([1− T ]n−1)) ≥ 1/2. We will show that ‖Txn − Txm‖ ≥ 1/2 for all m 6= n,which will contradict compactness of T . Now, without loss of generality assume that m < n,then

Txn − Txm = xn − (1− T )xn − Txmand (1 − T )xn ∈ ker([1 − T ]n−1) and Txm ∈ ker([1 − T ]n−1) (since T commutes with(1− T )n−1). Therefore by choice of xn we obtain

‖Txn − Txm‖ ≥ dist(xn, ker([1− T ]n−1) ≥ 1/2

(iii) Assume that yk = (1 − T )xk is a sequence in range(1 − T ) that converges to somey ∈ Y , we will show that for some x ∈ X we have y = Tx. Certainly if xk has a convergentsubsequence we could simply take x to be the corresponding limit. Now, xk = yk +Txk so itsuffices to obtain convergence of some subsequence of Txk, and using compactness of T thiswould follow if xk is uniformly bounded. Unfortunately it is possible for xk to be unbounded,but we could correct this by modifying xk by an appropriate term in ker(1− T ) to make it

9.3. SPECTRAL PROPERTIES OF COMPACT OPERATORS 77

bounded. Note that this correction does not change yk and hence does not change the goalof this part. Let

dk = dist(xk, ker(1− T ))

By modifying xk by an amount inside ker(1 − T ) we may assume that dk ≤ ‖xk‖ ≤ 2dk.Thus it suffices to show that

supkdk <∞

Assume towards a contradiction that some subsequence of dk conveges to ∞. Without lossof generality we may assume lim dk =∞. We then have

ykdk

= (1− T )(xkdk

)

now yk/dk → 0 and xk/dk is uniformly bounded, so using compactness of T it follows thatsome subsequence of T (xk/dk) converges, which in turn implies that some subsequence ofxk/dk converges to some x ∈ X. We obtain 0 = (1 − T )x so x ∈ ker(1 − T ), on the otherhand it is clear that dist(xk/dk, ker(1− T )) ≥ 1, contradiction.

We now prove Riesz’s theorem using the lemma. It remains to show that ker(1 − T )has the same dimension as the codimension of 1− T . We first reduce the proof to the casewhen ker(1 − T ) is trivial. Let k be the index given by part (ii) of the lemma and letY = ker[(1− T )k+1] a closed subspace of X. Since (1− T )Y ⊂ ker[(1− T )k] = Y by choiceof k, it follows that TY ⊂ Y , thus T induces an operator TZ on Z := X/Y , which is aBanach space with the induced norm. It follows immediately that TZ is compact on Z, and1−TZ is also injective on Z. Now, the dimension of ker(1−T ) is the same as the dimensionof the kernel of 1− T viewing as an operator on Y , which by standard linear algebra is thesame as the codimension of the range of 1−T viewing as an operator on Y . Thus it remainsto show that the codimension of 1 − T on Z is 0. One could see that we have reduced theproof to the case of the compact operator TZ on the Banach space Z which is also injective.

Now, if (1 − TZ)Z is a strict subspace of Z it follows that (1 − TZ)nZ is a strictlydecreasing sequence of closed subspaces of Z, and again we may find zn ∈ (1− TZ)nZ suchthat ‖zn‖ = 1 and dist(zn, (1− TZ)n+1Z) ≥ 1/2. Then for n < m we have ‖TZzn− TZzm‖ =‖zn − (1− TZ)zn − TZzm‖ ≥ dist(zn, (1− TZ)n+1Z) ≥ 1/2, violating the compactness of TZ .It follows that the codimension of 1− TZ is 0 as desired.

9.3.2 Spectral properties

Theorem 35. Let T be a compact operator on a Banach space X. Then(i) its spectrum σ(T ) consists of at most countably many elements, all of them are eigen-

values with finite dimensional eiegenspace.(ii) 0 is the only possible accumulation point for the elements of σ(T ) if such an accu-

mulation point exists, and 0 will belong to σ(T ) if the dimension of X is infinite.

Note that if X is a (separable) Hilbert space then we could furthermore diagonalize Tusing the eigenvectors.


We now prove Theorem 35. First, given any λ ∈ σ(T ) such that λ 6= 0 we show that it isan eigenvalue with finite dimensional eigenspace. Clearly 1

λT is compact. Thus, if ker(1− 1

λT )

is trivial then by Riesz’s theorem it is also onto, therefore by the open mapping theorem1− 1

λT is boundedly invertible and therefore λ 6∈ σ(T ). Thus ker(1− 1

λT ) is nontrivial and

also finite dimensional by Riesz’ theorem, and so λ is an eigenvalue with finite dim eigenspace.

Now, we will show that 0 is the only possible accumulation point, which also implies thatσ(T ) is at most countable. It suffices to show that given any ε > 0 there is some C = C(T, ε)finite such that at most C(T, ε) elements of σ(T ) would be outside [−ε, ε]. Let λ1, . . . , λmbe a finite collection of distinct elements of σ(T ) with |λj| > ε, it suffices to show thatm < OT,ε(1).

The idea is to let Yn be the eigenspace associated with λn and observe that for any nit holds that Yn ∩ spanYm,m < n is trivial. Let Y<n = spanYm,m < n which is nowa strictly nested sequence. Then by Riesz’s lemma one could choose yn ∈ Yn such that‖yn‖ = 1 and dist(yn, Y<n) ≥ 1/2. We will show that

‖Tyn − Tym‖ ≥ ε/2

for any n 6= m. Indeed, without loss of generality assume n > m, then using the fact thatyn ∈ Yn and the fact that T leaves Ym invariant we have

‖Tyn − Tym‖ = ‖ − Tym‖ = ‖λnyn − Tym‖ ≥ |λn|dist(yn, Y<n) ≥ |λn|/2 ≥ ε/2

Consequently, m is bounded above by the maximum number of points in TB where B isthe unit ball inside X that are ε/2 apart. Since TB is precompact this is finite and dependsonly on T and ε (and certainly X), and independent of the sequence (λk).

Chapter 10

Spectral theorems for boundedself-adjoint operators on a Hilbertspace

Let H be a Hilbert space. For a bounded operator A : H → H its Hilbert space adjoint isan operator A∗ : H → H such that 〈Ax, y〉 = 〈x,A∗y〉 for all x, y ∈ H. We say that A isbounded self adjoint if A = A∗.

In this chapter we discussed several results about the spectrum of a bounded self adjointoperator on a Hilbert space. We emphasize that in this chapter A is bounded, there is alsoa notion of unbounded self adjoint operator which we will discuss in subsequent chapters.

10.1 Diagonalization form

The first result says that A could be diagonalized using some change of basis.

Theorem 36. Let A : H → H be a bounded self-adjoint operator on a Hilbert space H.Then there exists some L2(X,µ) and U : L2(X,µ) → H isometric isomorphism such thatfor some bounded M on (X,µ) and every f ∈ L2(X,µ) it holds that

(U−1AU)f(x) = M(x)f(x) , x ∈ X

Proof. We note that if H is the direct sum of subspaces H1 and H2 such that H1 and H2

are invariant under A then it suffices to prove the theorem for the restriction of A to eachsubspace. This applies to direct sums indexed by larger index sets.

Now it is not hard to see that H could be written as an orthogonal direct sum of subspacesof the form span(Anξ, n ≥ 0) where ξ ∈ H. (The proof uses Zorn’s lemma.) Note thatthese subspaces are invariant under A, therefore it suffices to show the theorem when H =span(Anξ, n ≥ 0) for some fixed ξ.

Now, recall from spectral caculus for bounded operators on a Banach space that if f isanalytic on a domain containing σ(A) then we could define f(A) and furthermore σ(f(A)) =

79

80 CHAPTER 10. BOUNDED SELF-ADJOINT OPERATORS

f(σ(A)). For polynomials we could do this directly using factorization into linear factors forpolynomials.

In the case of bounded self adjoint operator we will show below that σ(A) ⊂ R. (Notethat this fact also holds for unbounded case, but we will not discuss that in this section.)

Lemma 8. Let A be bounded self adjoint on a complex Hilbert space H. Then σ(A) ⊂ R.

To see this lemma, we will show that if λ ∈ C has nonzero imaginary part then λ ∈ ρ(A).To do this, we will show that

| 〈x, (λ− A)x〉 | ≥ c‖x‖2 > 0

for some c depending on λ. This would imply that λ − A is invertible using an applicationof the Lax-Milgram theorem: consider the bilinear form B(x, y) = 〈x, (λ− A)y〉, which isnondegenerate once we proved the above estimate, thus given any z we could find y suchthat 〈x, z〉 = B(x, y) for all x ∈ H, which implies that z = (λ−A)y, thus λ−A is bijectiveon H and so is boundedly invertible and so λ ∈ ρ(A) as desired.

To show the above estimate, simly write λ = a+ ib where a, b ∈ R and b 6= 0, then usingthe self-adjoint property of A it follows that 〈x, (a− A)x〉 is a real number, therefore

| 〈x, (a+ ib− A)x〉 | = | 〈x, (a− A)x〉+ ib‖x‖2| ≥ b‖x‖2

as desired. This completes the proof of the above lemma.We now discuss functional calculus for bounded self adjoint operators. Note that since

σ(A) is bounded and closed it will follow that σ(A) is a compact subset of R, and thus wecould define f(A) even if f is merely continuous (which would be weaker than the analyticassumption required by the complex method generally applied to all bounded operators).The idea is to use the Weierstrass theorem and define f(A) to be the limit in operator normof pn(A) where (pn) is a sequence of polynomials that approximates f . To see that this couldbe done, it suffices to show that if g is a polynomial then

‖g(A)‖ = supx∈σ(A)

|g(x)|

To see this last claim, we first show it for real polynomial. In fact we will consider g(x) = x.Then as we proved before

supx∈σ(A)

|g(x)| = limn‖An‖1/n ≤ ‖A‖

while ‖A‖ ≤ supx∈σ(A) |g(x)| using either the spectral theorem, or by elementary methods.1

The real polynomial case then follows from the spectral mapping theorem σ(g(A)) = g(σ(A))

1Here we could easily see that ‖Ax‖2 =⟨A2x, x

⟩≤ ‖A2‖‖x‖2 thus by repeating we obtain ‖A2n‖2−n ≤

‖A‖ as desired.

10.2. PROJECTION-VALUED MEASURE AND SPECTRAL PROJECTION 81

and the fact that g(A) which is self adjoint when g is real polynomial. To allow for complexpolynomial p, simply write

‖p(A)‖2 = ‖p(A)∗p(A)‖ = ‖(pp)(A)‖ = supλ∈σ(A)

(pp)(λ) = supλ∈σ(A)

|p(x)|2

We note that as a consequence of the definition we also have

Corollary 4. For all continuous g on σ(A) it holds that ‖g(A)‖ = supx∈σ(A) |g(x)|.

Now, we may define a linear functional on the space of polynomials on σ(A) as follows:given such a polynomial p, let L(p) = 〈p(A)ξ, ξ〉, it is clear that ‖L‖ ≤ supx∈σ(A) |p(x)|.Using Weierstrass’s theorem it follows that we could extend L to the space of continuousfunctions on σ(A). Using self-adjointness of A it is clear that L is positive: if f ≥ 0 thenwrite f = g2 and g = lim pn limit of polynomials then L(f) = L(g2) = lim 〈pn(A)2ξ, ξ〉 =lim ‖pn(A)ξ‖2 ≥ 0. Thus by the Riesz representation theorem we may write L(f) =

∫fdµ

where µ is some finite Borel measures on σ(A).We now construct the operator U , initially from the space of continuous functions on

σ(A) to H. If q is a polynomial then let Uq = q(A)ξ, clearly∫σ(A)

|q(x)|2dµ = L(qq) = 〈(qq)(A)ξ, ξ〉 = ‖q(A)ξ‖2 = ‖Uq‖2

thus the restriction of U to the polynomials is an isometry and the image of the polynomialsunder U is clearly dense inside H by the given assumption. Thus U extends to an isometricisomorphism from L2(σ(A), dµ) to H.

Finally we will show that U−1AUf(x) = xf(x) for all f ∈ L2, note that this equalityis understood in the almost everywhere since with respect to µ, furthermore m(x) := x isbounded in σ(A) thanks to compactness of σ(A). Thanks to Weierstrass’s theorem again itsuffices to show this equality for f being polynomials. In that case let g(x) = xf(x) also apolynomial, we then have

U−1AUf = U−1(Af(A)ξ) = U−1(g(A)ξ) = g(x) = xf(x)

10.2 Projection-valued measure and spectral projec-

tion

Recall that P : H → H is a projection is P 2 = P . We say that it is an orthogonal projectionif ker(P) and range(P) are orthogonal subspaces, which would be the case if and only ifP = P ∗ (the underlying assumption is that P is bounded). Recall a basic fact:

Theorem 37. Given any closed subspace K of H there is an orthogonal projection PK ontoK: range(P) is K and ker(P) is exactly K⊥.


In this section we take a closer look at the spectral representation of A.We say that a family PΩ indexed by Borel subsets Ω ⊂ R if a (compactly suppported)

projection-valued measure if(i) PΩ is an orthogonal projection on H(ii) P = 0 (and P[−M,M ] = I for some M sufficiently large).(iii) If Ω =

⋃n≥1 Ωn disjoint union then PΩξ =

∑PΩnξ convergence in the norm.

(iv) If Ω1 and Ω2 are Borel sets then PΩ1PΩ2 = PΩ1∩Ω2 .We note that property (iv) is a corollary of the first three properties.We first note that we could construct a measure out of (PΩ) when testing the projections

on some vector φ ∈ H. Namely let µφ1,φ2(Ω) = 〈PΩφ1, φ2〉 then this defines a compactlysupported complex Borel measure on R with finite total mass, in fact it is not hard to seethat

‖µφ1,φ2‖ ≤ ‖PR‖ = 1

therefore we could integrate any bounded Borel measurable function f and obtained abounded bilinear form

Tf (φ1, φ2) =

∫f(λ)dµφ1,φ2(λ)

Then by Riesz representation theorem we may write Tf (φ1, φ2) = 〈Bφ1, φ2〉 where B is abounded linear operator on H. If we approximate f using simple functions, it can be seenthat B is the limit in the weak operator topology of the corresponding linear combinationsof PΩ (note that this implies B is the limit in the strong operator topology too). We willlet∫f(λ)dPλ denote this operator B and we think of this as the integration of f over the

measure induced by the family (PΩ). In particular,∫λdPλ is a bounded self adjoint operator

on H.It turns out that the converse of this is also true:

Theorem 38. Given a bounded self adjoint operator A on a complex Hilbert space H letPΩ = 1Ω(A) then (PΩ) is a compactly supported projection valued measure, and

∫f(λ)dPλ

converges to f(A) (as defined by spectral caculus) in the strong operator norm (Tj → Tmeans ‖Tjx− Tx‖ → 0 for all x). Furthermore this is the unique projection valued operatorwith this property.

To check that∫f(λ)dPλ converges to f(A) in strong operator norm topology, it suffices

to show convergence in the weak operator norm, then it suffices to check that if f = 1Ω then⟨∫1ΩdPλφ1, φ2

⟩= 〈1Ω(A)φ1, φ2〉 for all φ1, φ2 ∈ H. By definition the left hand side the same

as∫

1ΩdPφ1,φ2 = 〈PΩφ1, φ2〉 which is the same as the right hand side.

10.3 Spectral representation and decomposition

Absolutely continuous, singular, and point spectralRecall that H has the spectral representation L2(X,µ) which comes from direct summing

the spectral representations L2(µj) of Kj. We may decompose µj into three parts (absolutely

10.3. SPECTRAL REPRESENTATION AND DECOMPOSITION 83

continuous, point, and singular parts) using the Radon Nikodym theorem. This leads todecomposition of Kj and also decomposition of H into the absolutely continous, singular,and point spectrum H(p), H(s), and H(c). Note that these are orthogonal subspaces and thecorresponding spectral measure of the cyclic subspace has the inherited properties. Say ifx ∈ H(c) then the specral measure of A on span(Anx) is absolutely continuous with respectto the Lebesgue measure.

Uniqueness of the spectrum Let A be bounded self adjoint on H and assume that Hbe separable. Recall from prior sections that there is a decomposition of H into orthogonaldirect sum of K1, K2, . . . where Kj are orthogonal closed subspaces of H and each of themis invariant under A, furthermore there is a linear isometry Uj mapping some L2(σ(A), µj)into Kj such that U−2

j AUj acts on this L2 space by multiplication, in fact U−1j f(A)Ujg(x) =

f(x)g(x) for all bounded measureable f and g ∈ L2(µj). One could certainly modify dµjmultiplicatively using a bounded positive function, this would affect the isometry part ofthe map U but the L2 space remains the same and the corresponding action of f(A) is stillpointwise multiplication.

Now if H has another decomposition into an orthgonal direction sum of closed subspacesL1, L2, . . . with spectral representations L2(Tj, νj) for Lj, then upto a set of (spectral) mea-sure 0 we have

⋃Sj =

⋃Tk, namely we will show that for any k the set difference Tk \

⋃Sj

has zero measure under νk. Note that if we assume furthermore that these measures areabsoluteluy continuous with respect to the Lebesgue measure then

⋃Sj =

⋃Tk up to set of

Lebesgue measure 0.To see this take f ∈ L2(Tk, νk), and let F be any bounded Borel measurable function.

Note that Sj and Tk are invariant under F (A), and f coressponds to some hf ∈ H. We nowdecompose hf into hj where hj ∈ Kj, and hj in turn corresponds to fj ∈ L2(Sj, µj). Then∫|F (x)f(x)|2dµk = ‖hf‖2 = ‖h1‖2+‖h2‖2+· · · =

∫|F (x)|2|f1|2dµ1+

∫|F (x)|2|f2(x)|2dµ2+. . .

therefore if Tk \⋃Sj has positive µk measure we may choose F to be the characteristic

function of this set and f ≡ 1, clearly the left hand side is now positive while the right handside is 0, contradiction.

Spectral multiplicityConsider the spectral representation of H, assumed separable, using the orthogonal direct

sum of K1, . . . with Kj = L2(Sj, µj where Sj is the support of µj.Given λ ∈ R we define its spectral multiplicity with respect to this representation as the

number of j such that λ ∈ Sj.Assume for simplicity that the spectrum of A is absutely continuous with respect to the

Lebesgue measure. Then it can be shown that the spectral multiplicity is independent ofthe spectral representation of H; this is a theorem of Hellinger.

Chapter 11

Spectral theory for unboundedself-adjoint operator

Let A be a linear operator on some dense subspace D of a Hilbert space H.Let A∗ be defined as follows: for every ω ∈ H, if there exists z ∈ H such that for every

x ∈ D it holds that 〈Ax, ω〉 = 〈x, z〉 then ω ∈ D∗ the domain of A∗ and A∗ω := z (it is clearthat such z is unique if exists).

We say that λ is in the resolvent set of A if A − z is bijective from D to H, andσ(A) = C \ ρ(A) as before.

Theorem 39. A be self-adjoint on H with dense domain D ⊂ H. Then σ(A) ⊂ R and thereis an orthogonal projection valued measure PΩ such that PΩ commutes with A on the domainD of A, which in turn consists of all vector φ such that∫

t2d 〈Ptφ, φ〉 <∞

and

Aφ =

∫tdPtφ

We first show that σ(A) ⊂ R. To do this we will show that for every z 6∈ R the rangeof A − z is closed and then we use the denseness of D to show that A − z is onto, as a byproduct of the proof we will also have A− z is injective, in fact (A− z)−1 is bounded fromH to D.

We now show the spectral theorem. To do that we first show that(i) R(z) = (A − z)−1 is a complex differentiable function of z ∈ C+ := Im(z) > 0 in

the strong topology, namely the topology on the space of bounded operators induced by theseminnorms Aφ, φ ∈ H. In other words, R(z)u is complex differentiable for every u ∈ H.

(ii) the adjoint of R(z) is R(z).To see (i), first note that R(z) is a bounded operator on H for those z, let u(z) =

(A− z)−1u for any u ∈ H. Then by simple algebra

u(z + ω)− u(z) = ωR(z)R(z + ω)u

85

86CHAPTER 11. SPECTRAL THEORY FORUNBOUNDED SELF-ADJOINT OPERATOR

so u(z) is complex differentiable for all u, thus R(z) is complex differentiable and thereforeis analytic in the strong operator topology.

To see (ii), note that 〈u,R(z)v〉 = 〈(A− z)R(z)u,R(z)v〉 = 〈R(z)u, (A− z)R(z)v〉 =〈R(z)u, v〉.

As a consequence, we know that Fu(z) := 〈R(z)u, u〉 is a complex valued analytic functionon the upper half plane and Fu(z) . 1

Im(z). Furthermore using the second property it follows

that F is rather symmetric accross the real line in the sense that F (z) = F (z).We now show that Fu(z) has nonnegative imaginary part for any u ∈ H and z in the

upper half plane and

limy→∞

yImFu(iy) = ‖u‖2

To see these points, let v = R(z)u, then Fu(z) = 〈v, (A− z)v〉 = 〈v, Av〉 − z 〈v, v〉, notethat self adjointness of A implies that 〈v, Av〉 ∈ R, therefore ImFu(z) = Im(z)‖v‖2 > 0 asdesired. Let z = iy where y ∈ R, then it folows that

yImFu(iy) = y2‖v‖2

on the other hand

‖u‖2 = 〈(A− iy)v, (A− iy)v〉 = ‖Av‖2 + y2‖v‖2

thus as y → ∞ we have y2‖v‖2 → ‖u‖2 as desired once we show that Av = AR(iy)uconverges to 0 as y → ∞ for any fixed u ∈ H. Note that this holds if u ∈ D: we couldcommute A and R(iy) and then ‖AR(iy)u‖ . ‖R(iy)‖‖Au‖ = O(1/y). If u 6∈ D we couldapproximate it using some thing in D, this could be done uniformly as long as we could showthat AR(iy) is uniformly bounded. Infact we will show that

‖AR(iy)‖ ≤ 2

To see this, simply write

‖AR(iy)u‖ = ‖u+ iyR(iy)u‖ ≤ ‖u‖+ y‖R(iy)‖‖u‖ ≤ 2‖u‖

Now one could think of Fu(z) as an analytic function that maps the upper half planeinto itself, thus using Poisson integral and the Riesz representation theorem we could showthat there is some nonnegative measure µu and some linear function l(z) of z such that

Fu(z) = l(z) +

∫dµu(t)

t− z

(this is basically Herglotz’s theorem) for all z in the upper half plane. Then combining withthe fact that yFu(iy) → ‖u‖2 it follows that the linear part is 0 and µu(R) = ‖u‖2 (this isbasically a result of Nevanlinna.)

87

We note that the values of Fu in the upper half plane determines its values in the lowerhalf plane via the relationship Fu(z) = Fu(z). Therefore the equation

Fu(z) =

∫dµu(t)

t− z

holds for all z ∈ C \ R. Then via algebras it follows that for all u, v ∈ H

〈R(z)u, v〉 =

∫dµu,v(t)

t− z

where µu,v is a signed measure, basically a linear combination of µu±v, µu±iv.

µu,v =1

4

(µu+v − µu−v + iµu+iv − iµu−iv

)It is not hard to check the following properties

(i) µu,u = µu is a nonnegative measure with total mass ‖u‖2;(ii) µu,v is linear in u and conjugate linear in v and µu,v = µv,u(iii) the total variation ‖µu,v‖ is controlled by ‖u‖‖v‖. (use Cauchy Schwartz)Then via repeated application of the Riesz representation theorem there exists a bounded

self adjoint operator PΩ for each Borel Ω ⊂ R such that µu,v(Ω) = 〈PΩu, v〉 for all u, v ∈ H.We will check that PΩ indeeds define a projection valued measure, and it will then followthat

〈R(z)u, v〉 =

∫1

t− zd 〈Ptu, v〉

Check:(i) Clearly P=0 since µu,v(∅) = 0 for all u, v, also PR = 1 since ‖u‖2 = µU(R) =∫〈Ptu, u〉 = 〈PRu, u〉 for all u ∈ H.

(ii) We will show that PΩ commutes with A. First we show that PΩ commutes with R(z).Note that R(z1) and R(z2) commutes with each other for all z1 and z2 non real. Using

the representation R(z)u, v >=∫

(t− z)−1d 〈Etu, v〉 we have∫(t− z1)−1d 〈R(z2)Ptu, v〉 =

∫(t− z1)−1d 〈Ptu,R(z2)v〉 =

= 〈R(z1)u,R(z2)v〉 = 〈R(z2)R(z1)u, v〉 = 〈R(z1)R(z2)u, v〉 = . . .

=

∫(t− z1)−1d 〈PtR(z2)u, v〉

Using the fact that the Cauchy transform determines uniquely the measure, it follows thatR(z2)PΩ = PΩR(z2) for any Borel Ω.

Let u ∈ H, then AR(z)u = u+ zR(z)u, thus for any Ω we have

PΩA(R(z)u) = PΩu+ zPΩ(R(z)u) = PΩu+ zR(z)PΩu = AR(z)PΩu = APΩR(z)u

88CHAPTER 11. SPECTRAL THEORY FORUNBOUNDED SELF-ADJOINT OPERATOR

since the range of R(z) is D it follows that PΩAu = APΩu for all u ∈ D.(iii) We will show that PΩ1PΩ2 = PΩ1∩Ω2 . This is a consequence of the resolvent identity

R(z)R(w) = (z − w)R(z)R(w). Indeed, this identity implies

〈R(z)R(w)u, v〉 =1

z − w

∫(

1

t− z− 1

t− w)d 〈Ptu, v〉 =

∫(

1

(t− z)(t− w))d 〈Ptu, v〉

On the other hand, as before

〈R(z)R(w)u, v〉 =

∫(t− z)−1d 〈R(w)Ptu, v〉

therefore for any Borel Ω1 ⊂ R and any w 6∈ R we have∫1

t− wd 〈1Ω1Ptu, v〉 = 〈R(w)PΩ1u, v〉 =

∫1

t− wd 〈PtPΩ1u, v〉

Using again the fact that the Cauchy transform determines the measure again, it followsthat given any Ω2 ⊂ R we have

〈PΩ2∩Ω1u, v〉 = 〈PΩ2PΩ1u, v〉as desired.(iv) Finally, assume φ ∈ D. Then φ = R(z)u for some u ∈ H, and z is arbitrary in the

upper half plane. We then have

〈PΩφ, φ〉 = 〈R(z)R(z)PΩu, u〉 =

∫Ω

1

t2 + |z|2d 〈Ptu, u〉

therefore d 〈Ptφ, φ〉 = 1t2+|z|2d 〈Ptu, u〉 as measures. In particular,∫

t2d 〈Ptφ, φ〉 .∫d 〈Ptu, u〉 <∞

Using orthgonality of PΩ on disjoint Ω’s one could construct∫tdPtu for all u ∈ D using

simple functions (the integral existrs in the strong operator topology) and we could showthat this is the same as Au. To see this, let v ∈ D and u = (A− z)v then similarly we couldshow for any φ ∈ H:

d 〈Ptv, φ〉 =1

t− zd 〈Ptu, φ〉

(t− z)d 〈Ptv, φ〉 = d 〈Pt(A− z)v, φ〉 = d 〈PtAv, φ〉 − zd 〈Ptv, φ〉td 〈Ptv, φ〉 = d 〈PtAv, φ〉

therefore∫td 〈Ptv, φ〉 = 〈Av, φ〉 for every φ, thus for any v ∈ D we have

Av =

∫tdPtv

Chapter 12

Fredholm determinant

12.1 Motivation

Fredholm determinant started in Fredholm’s investigation of the integral equation

(1 +K)u = f

where K is the integral operator Kf(y) =∫YK(x, y)f(y)dy maping functions on Y to

functions on X. (Here X, Y are compact metric spaces.) We know that K is compactfrom L2(Y ) to L2(X) if the kernel K is square integrable in X × Y . The setting originallyconsidered is for continuous functions, for which we have the following results:

(i) If K is a continuous function of x and y then K is compact from L1(Y ) to C(X).(ii) If K is a continuous function of x in the L1 norm wrt to y, i.e. ‖K(x, .)‖L1

yis

continuous wrt x, then K is compact from C(Y ) to C(X).Proof of these facts uses the Arzela Ascoli criteria: S haudorff compact, functions equicon-

tinuous and pointwise uniformly bounded, then the family is precompact in the sup normon S.

For simplicity, assume K : [0, 1]2 → C is continuous and f ∈ C[0, 1]. Note that this isnot a Hilbert space. By Riesz’s theorem the integral equation

u(x) +

∫ 1

0

K(x, y)u(y)dy = f(x)

is solvable on C[0, 1] (i.e. given f ∈ C[0, 1] one could find u ∈ C[0, 1]) iff the integral operatorK is injective on this space, or equivalently its range is the whole space. It can be shownthat a vector is in the range of K iff it is orthogonal to the null space of K∗ which has kernelK∗(x, y) = K(y, x).

Fredholm investigated the above equation by discretizing the equation and appeal tolinear algebra, and then taking a limit at the end. This give rises to a number called theFredholm determinant of 1 + K (we simply say the Fredholm determinant for K), whichdetermines whether the given integral equation is solvable or not. The determinant concept

89

90 CHAPTER 12. FREDHOLM DETERMINANT

has been extended to other settings, most commonly for K being a trace class operator onsome separable Hilbert space.

We first discuss Fredholm’s orginal approach for C[0, 1] (modulo some simplifications),and then we’ll discuss the Hilbert space theory later.

12.2 Fredholm’s approach for integral operators

We now investigate Fredholm’s approach towards solving the above equation. Fredholm’sidea is to use linear algebra, approximating the integral using discretized sums and takingan appropriate limit at the end.

ui + h∑

Kijuj = fi , i = 1, . . . , n ,

where h = 1/n, fi = f(ih) and Kij = K(ih, jh) and ui = u(ih). The determinant of thematrix acting on the vector (u1, . . . , un) is denoted by D(h)

D(h) = det(I + h(Kij))

which is clearly a polynomial in h of degree n

D(h) =n∑

m=0

cmhm

cm =1

m!(d

dh)mD(h)|h=0

We then use the product rule, which say that if C = det(C1, . . . , Cn) is the determinant ofa matrix with columns C1, . . . , Cn then by multilinearity d

dhC =

∑k det(C1, . . . ,

ddhCk, . . . , Cn).

In our case each column is linear in h, and Cj(0) = (0, . . . , 1, . . . , 0) the unit is in the jthposition. Therefore

D(h) = 1 + h∑j

Kjj +h2

2

∑i,j

det

(Kii Kij

Kji Kjj

)+ . . .

For convenience let K

(x1, . . . , xky1, . . . , yk

)denote the determinant of the matrix K(xi, yj), then

letting h = 1/n and send n→∞ we obtain

D =∞∑k=0

1

k!

∫. . .

∫K

(x1, . . . , xkx1, . . . , xk

)dx1 . . . dxk

This is called the Fredholm determinant of the integral operator K. Note that this is acomplex number.

12.2. FREDHOLM’S APPROACH FOR INTEGRAL OPERATORS 91

12.2.1 Convergence and Continuity of the determinant

We’ll show that the sum defining D converges. We’ll use Hadamard’s inequality: for columnvectors v1, . . . , vn ∈ Rn, it holds that

| det(v1|v2| . . . |vn)| ≤ nn/2n∏i=1

|vj|∞

This follows from the fact that the volume of the parallelopipde is at most the product ofthe side lengths.

Now, the given assumption on K implies that supx,y |K(x, y)| ≤ M for some finite M .So by Hadamard inequality we have

K

(x1, . . . , xky1, . . . , yk

)≤ (Mk1/2)k (12.1)

which implies the desired convergence.Let ‖‖ denote the sup norm on [0, 1]2 below, we obtain

Lemma 9. Let F and G be two functions on [0, 1]2. Then

| det(F (xi, xj))− det(G(xi, xj))| ≤ n1+n2 ‖F −G‖sup max(‖F‖, ‖G‖)n−1

here all matrices are n× n.

Proof. For convenience, let F be the n×n matrix with entries F (xi, xj) and G be the n×nmatrix with entries G(xi, xj). It is clear that

det(F )− det(G) = detM1 + detM2 + · · ·+ detMn

where Mk is the matrix whose first k − 1 rows are the same as G, and the kth row is thesame as F −G, and the last n− k rows are the same as F .

Apply Hadamard inequality we obtain the first desired estimate. We now discuss how to solve the integral equation using Fredholm’s determinant. The

idea is to solve it at the discrete level and then send h → 0 via n → ∞. Via this heuristicwe obtain

(I +K)(I + L) = (I + L)(I +K) = I

where

Lf =

∫L(x, y)f(y)dy

L(x, y) = −D−1∑k≥0

1

k!

∫. . .

∫K

(x, x1, . . . , xky, y1, . . . , yk

)dξ1 . . . dξk

We now ready to prove


Theorem 40. Let K acts on C[0, 1]. Let K be a continuous kernel.(i) If D = 0 then the operator I + K has a nontrivial null space and therefore is not

invertible.(ii) Conversely if D 6= 0 then the operator I +K is invertible, furthermore its inverse is

given by I + L where L is defined above.

Proof: We first note that if K1 and K2 are two integral operators with kernel K1(x, y)and K2(x, y) respectively then (1 + K1)(1 + K2) = 1 + K3 where K3 is another integraloperator whose kernel is given by

K3(x, y) = K1(x, y) +K2(x, y) +

∫K1(x, z)K2(z, y)dz

(i) Now, assume that D 6= 0. We need to show

K(x, y) + L(x, y) +

∫K(x, z)L(z, y)dz = 0

L(x, y) +K(x, y) +

∫L(x, z)K(z, y)dz = 0

We will show the first equality. For convenience of notation let

R(x, y) :=∑k≥0

1

k!

∫. . .

∫K

(x, x1, . . . , xky, y1, . . . , yk

)dξ1 . . . dξk

Then L = − 1DR.

Now, computing the determinant K

(x, x1, . . . , xky, y1, . . . , yk

)using its first row, we obtain

K

(x, x1, . . . , xky, y1, . . . , yk

)= K(x, y)K

(x1, x2, . . . , xky1, y2, . . . , yk

)+

+k∑j=1

(−1)kK(x, yj)

(x1, x2, . . . , xky, y1, . . . , (yj), . . . yk

)here (yj) means one omit yj.

Now let y1 = x1, ... yk = xk and integrate the above over x1, . . . , xk ∈ [0, 1]. Then theintegrals of the last k terms in the above expansion are actually the same, one could provethis by simple change of variable. It follows that∫

. . .

∫K

(x, x1, . . . , xky, x1, . . . , xk

)dx1 . . . dxk

= K(x, y)

∫. . .

∫K

(x1, x2, . . . , xkx1, x2, . . . , xk

)dx1 . . . dxk+

12.2. FREDHOLM’S APPROACH FOR INTEGRAL OPERATORS 93

+kK(x, x1)

(x1, x2, . . . , xky, x2, . . . , xk

)dx1 . . . dxk

Divinding by k! and summing over k ≥ 0, we obtain

R(x, y) = K(x, y)D −∫K(x, x1)R(x1, y)dx1

which implies the desired equality K(x, y) + L(x, y) +∫K(x, z)L(z, y)dz = 0.

For the second equality, one argues similarly, computing the determinant using the firstcolumn.

(ii) Since D = 0, by part (i) we have

R(x, y) +

∫K(x, z)R(z, y)dz = 0

for every x, y. If R 6≡ 0 then one could find one y such that g(.) = R(., y) is not the zerofunction (note that it is continuous), and it satisfies g + Kg = 0, therefore 1 + K is notinjective.

It is however possible that R ≡ 0. If so, consider the following functions

D(z) =∑k≥0

zk

k!

∫. . .

∫K

(x1, . . . , xkx1, . . . , xk

)dx1 . . . dxk

R(x, y, z) :=∑k≥0

zk+1

k!

∫. . .

∫K

(x, x1, . . . , xky, y1, . . . , yk

)dξ1 . . . dξk

One could think of D(z) and R(x, y, z) as the version of D and R with zK instead of K.It is clear that D and R are entire functions. Since D(1) = 0 the entire function D has azero of finite order n ≥ 1 at z = 1. By algebraic manipulation we have∫

R(x, x, z)dx = zD′(z)

therefore there is some (x, y) such that R(x, y, z) can not vanish at z = 1 with order morethan n− 1. Then for some 1 ≤ ` < n it holds that

R(x, y, z) = (z − 1)`g(x, y) +O((z − 1)`+1)

where g(x, y) 6≡ 0, and g is continuous in x, y (to see this note that g is the uniform limit ofa sequence of continuous functions on [0, 1]2). We recall that R(x, y, z) = zK(x, y)D(z) −z∫K(x, x1)R(x1, y, z)dx1 thus by dividing everything by (z − 1)` and then send z → 1 we

obtain

g(x, y) = −∫K(x, x1)g(x1, y)dx1

and since g 6≡ 0 continuous it follows that 1 +K is not injective.


Theorem 41. Assume that K is Holder c-continuous where c > 1/2. Then the nonzeroeigenvalues of K on C[0, 1] (only countablity many of them since K is compact) satisfies∑

j |λj| <∞ and for every z ∈ C we have

D(z) =∏j

(1 + zλj)

and we also have the trace formula∫K(x, x)dx =

∑j

λj

Proof. We recall Hadamard’s factorization theorem (or may be a consequence of this theo-rem): Let f be an entire function such that for some finite positive C1, C2 and ρ ∈ [0, 1) itholds for every complex number z that

|f(z)| ≤ C1 exp(C1|z|ρ)

Assume that f(0) 6= 0. Then f has at most a countable number of zeros, furthermore∑z: f(z)=0

1

|z|<∞

and for every complex number λ it holds that

f(λ) = f(0)∏

z: f(z)=0

(1− λ

z)

We plan to use this theorem to show the first part of the theorem. We note that thesecond part of the theorem, i.e. the trace formula would then follows from∫

K(x, x)dx = D′(z)|z=0

(which is part of the definiton of D) and the absolute convergence of the product∏

j(1+zλj)(viewed as an infinite power series for z).

First, we will show that for z 6= 0 it holds that: D(z) = 0 if and only if −1/z is aneigenvalue ofK. This is simply a consequence of our last theorem applied toD = det(1+zK).

Now, we will show that if K is Holder c-continuous then

|D(z)| . exp(O(|z|2

1+2c )

To see this, using Stirling’s formula it suffices to show that, for some C finite,

|D(z)| .∑n≥0

Cn|z|2n/(1+2c)

nn

12.3. FREDHOLM DETERMINANT FOR OPERATORS ON HILBERT SPACES 95

which is equivalent to showing that, for some C finite,

|D(z)| .∑k≥0

Ck|z|k

k( 12

+c)k

Using the definition of D it suffices to show that given any x1, . . . , xk, y1, . . . , yk it holds thatfor some C > 0 finite

|K(x1, x2, . . . , xky1, y2, . . . , yk

)| ≤ (Ck1/2)kk−c

Note that incomparision with (12.1) we gain a factor of k−c. To see this it suffices to showthat

|K(x1, x2, . . . , xky1, y2, . . . , yk

)| ≤ (Ck1/2)k

k−1∏j=1

|yj+1 − yj|c

(note that the constant C may be different in different display). Indeed, wlog we may assumey1 ≤ y2 · · · ≤ yk, then by the geometric arirthmetic mean inequality we have

∏k−1j=1 |yj+1 −

yj| ≤ ( 1k−1

)k−1, as desired.Now, note that if c1, . . . , ck are k column vectors (k × 1) then

| det(c1| . . . |ck)| = | det(c1, c2 − c1, . . . , ck − ck−1)|

therfeore using Hadamard’s inequality we obtain

|K(x1, x2, . . . , xky1, y2, . . . , yk

)| ≤

∏j

(k∑

n=1

|K(xn, yj+1)−K(xn, yj)|2)1/2

which implies the desired estimate using Holder continuity of K.

12.3 Fredholm determinant for operators on Hilbert

spaces

Let H be a separable Hilbert space over C. The Fredholm determinant det(1 + K) couldbe defined for trace class operators K on a Hilbert space H. Note that this is not the sameas the setting considered in the last section since C[0, 1] with the sup norm is not a Hilbertspace.

We first define the singular values of a compact operator T on H. Let T ∗ be its adjoint,clearly T ∗T is a nonnegative self adjoint operator on H, so one could define its square rootA =

√T ∗T using functional calculus. Note that T ∗T and A are both compact operators

with nonnegative eigenvalues.The singular values of T are defined to be the positive eigenvalues of A, counting with

multiplicity.


Definition: We say that T is a trace class operator if the sum of its singular values isfinite (counting with multiplicity). In that case the trace norm of T is defined to be thissum, denoted by ‖T‖tr.

Properties: T and T ∗ has the same trace norm, and if T is trace class then so is TBand BT where B is any bounded operator on H, and ‖.‖tr satisfies the triangle inequality.

‖T1 + T2‖tr ≤ ‖T1‖tr + ‖T2‖tr

This inequality is an immediate consequence of the following equivalent characterization ofthe trace norm:

Lemma 10. For any trace class operator T

‖T‖tr = supfn,en

∑n

| 〈Tfn, en〉 |

where the sup is taken over all (fn) and (en) orthonormal bases of H.

To show this characterization we will first derive the polar factorization of T : namelythere is a partial unitary operator U such that T = UA, here unitary of U simply meansthat U∗U when acting on the range of A is the identity operator. This should be thought ofas the operator analogue of the usual polar factorization of a complex number.

To define U , note that ‖Au‖ = ‖Tu‖ for all u ∈ H, therefore we may define an isometryU from range(A) to range(T ) by mapping Au to Tu.

One then extends U to H by letting U to be zero on the orthogonal complement ofrange(A). It follows immediately that U∗H ⊂ range(A): one simply notice that for every zin the othogonal complement of range(A) and z′ ∈ H it holds that 〈z, U∗z′〉 = 〈Uz, z′〉 = 0.

Now let z in the closure of the range of A. We want to show that (U∗U − 1)z = 0, whichis the desired local unitary property of U . Using the isometric property of U on range(A),for every z′ in the closure of the range of A we have

〈z′, z〉 = 〈Uz′, Uz〉 = 〈z′, U∗Uz〉

therefore (U∗U−1)z belongs to the orthogonal complement of range(A). But U∗U−1 leavesrange(A) invariant since the range of U∗ is inside range(A). This contradiction completesthe proof of the local unitary property of U .

We are now back to proving the above equivalent charactization of the trace class normof T . Let (Fn) be a complete set of normalized eigenvectors of A =

√T ∗T . (Note that since

A∗ = A we could find such a set.) Let Gn = UFn. (Note that ‖Fn‖ = ‖Gn‖ = 1. ) Then∑n

| 〈TFn, Gn〉 | =∑n

| 〈UAFn, UFn〉 =∑n

| 〈AFn, Fn〉 | = ‖T‖tr

therefore it remains to show that ∑n

| 〈Tfn, en〉 | ≤ ‖T‖tr


for any pair of orthogonal bases (fn) and (en). Let sn be the singular values of T , namelyAFn = snFn. Then expand fn into (Fn) we have

fn =∑k

〈fn, Fk〉Fk

and by an application of Fubini’s theorem∑n

| 〈Tfn, en〉 | ≤∑k

|sk|∑n

| 〈fn, Fk〉〈Gk, en〉 |

(it will be clear from the proof that the double sum is absolutely summable)

≤∑k

|sk|(∑n

| 〈fn, Fk〉 |2)1/2(∑n

| 〈Gk, en〉 |2)1/2

≤∑k

|sk|‖Fk‖‖Gk‖ =∑k

|sk| = ‖T‖tr

This completes the proof of the characterization. Trace: Given a trace class operator one may define a linear functional, namely the trace

of T

Trace(T ) =∑n

〈Tfn, fn〉

where (fn) is any orthonormal basis of H. This definition is independent of the choice of thebasis. Let (gn) be another orthonormal basis, then by expanding fn into this new basis wehave ∑

n

〈Tfn, fn〉 =∑n

∑k

〈fn, gk〉〈Tgk, fn〉

it is not hard to see that this double sum is abs convergence (using Cauchy Schwartz). Thenby Fubini ∑

n

〈Tfn, fn〉 =∑k

(∑n

〈fn, gk〉〈Tgk, fn〉)

using orthogonality of fn and normalization of fn we obtain

=∑k

(∑n

∑j

〈fn, gk〉〈fj, fn〉〈Tgk, fj〉)

=∑k

〈Tgk, gk〉

By definition it is clear that |Trace(T )| ≤ ‖T‖tr and the sum defining the trace isabsolutely convergent.

Now, Lidskii’s trace formula says that


Theorem 42 (Lidskii). If T is trace class on a separable Hilbert space H then

Trace(T ) =∑j

λj

where λj are the eigenvalues of T .

Note that T is compact so it has a countable set of nonzero eigenvalues. It is clear thatthe sum of the eigenvalues is bounded above by the trace norm.

Proof. The proof of this formula could be divided into two steps: first one consider the caseswhen T does not have any nonzero eigenvalues, and then in the second step we reduce thegeneral case to this setting.

Let’s assume for now that T does not have any nonzero eigenvalues, then we want toshow that Trace(T ) = 0. Let s1, s2, . . . be the singular values of T . Then we’ll show thatfor every λ > 0

eλ|Trace(T )| ≤ O(1 + |λ|)M)eλ∑

j>M sj (12.2)

(the implicit constant is independent of λ), from here by sending λ → ∞ we obtain|Trace(T )| ≤

∑j>M sj; and the desired estimate now follows from the fact that

∑j sj =

‖T‖tr < ∞. To show the above claim, we approximate T by finite rank operators, sayTn = PnTPn → T where Pn is the projection into the first n basis vectors of H, here one mayfix any orthonormal basis. Let Dn(λ) = det(1 + λTn), here the determinant is defined usinglinear algebra, in other words if Λn is the set of eigenvalues of Tn thenDn(λ) =

∏α∈Λn

(1+λα).We will show that uniformly on any compact subsets of the complex plane it holds that

eλTrace(T ) = limn→∞

Dn(λ) (12.3)

Note that by definition we have ‖Tn − T‖ → 0 (in operator norm) and Trace(Tn) →Trace(T ). Since the spectral radius of T is 0 it is clear that the spectral radius σ(Tn)of Tn converges to 0 too. In particular given any bounded set of the complex plane we maychoose n large such that this bounded set is contained inside the ball of radius 1/σ(Tn).Furthermore one could also show that sj(Tn) ≤ sj(T ) if the singular values of T and Tn areordered in decreasing order.

Now, it is not hard to see that

D′n(λ)

Dn(λ)=∑α∈Λn

α

1 + λα

= Trace(Tn) +O(∑k≥2

(|λ|σ(Tn))k−1‖Tn‖Tr)

therefore uniformly over λ in a bounded subset of C it holds that

limn→∞

D′n(λ)

Dn(λ)− Trace(T ) = 0


which implies the desired limiting equality (12.3).Now using (12.3), for any λ > 0 we have

eλ|Trace(T )| ≤ lim infn→∞

∏α∈Λn

(1 + λ|α|)

We will show that the last limit is bounded above by∏∞

j=1(1 + λsj(T )) where sj(T ) are thesingular values of T , which easily implies the desired estimate (12.2). To show this, usingsj(Tn) ≤ sj(T ) it suffices to show that∑

α∈Λn

log(1 + λ|α|) ≤n∑j=1

log(1 + λsj(Tn))

Thanks to convexity of log, one could show that this inequality is a consequence of the factthat ∏

α∈Λn

|α| ≤N∏j=1

sj(Tn)

which in turn could be easily verified (equality holds if Tn is nonsingular).Now to reduce the general case to the above setting, we consider the subspace K1 of H

spanned by the eigenfunctions and generalized eigenfunctions of T . Let K2 be the orthogonalcomplement of K1. Using linear algebra it is clear that the trace of the restriction of T toK1 is exactly the sum of the eigenvalues of T . More precisely if (gn) is a basis for K2 and(fm) is a basis for K1 then we may take the union of the two as a basis for H and

Trace(T ) =∑m

〈Tfm, fm〉+∑n

〈Tgn, gn〉

=∑

α∈Λ(T )

α +∑n

〈T ∗gn, gn〉

thus using the previous argument it suffices to show that T ∗ leaves K2 invariant and theonly eigenvalue of T ∗ on K2 is 0 (it is clear that T2 is also compact and trace class). Theseproperties could be easily checked.Determinant We now discuss det(1 + T ) for trace class operators T on a separable

Hilbert space H.Define an inner product on Hk by

〈(w1, . . . , wk), (v1, . . . , vk)〉 = det(〈wi, vj〉)i,j

Then T extends to a trace class operator Tk onHk, defined by Tk(w1, . . . , wk) = (Tw1, . . . , Twk),with ‖Tk‖tr ≤ ‖T‖ktr. Then define

det(1 + T ) =∑k≥0

Trace(Tk)


One could check that if T is finite rank then this determinant is the same as what wewould obtain from linear algebra, and det(1 + .) is locally Lipschitz wrt to the trace classnorm. Approximating T by a sequence of finite rank operators (for which det(1 + T ) couldbe defined using standard linear algebra), we could show that det(1 + T ) is the limit of thecorresponding determinants, and thus this limit is independent of the choice of the sequence.Now, one could use the polar decomposition T = UA and approximate A by projections intofinite dimensional subspaces of H spanned by eigenfunctions of A. By this approximationscheme it can be shown that

det(1 + T ) =∏

(1 + λj)

where λj are the eigenvalues of T and

det[(1 + T1)(1 + T2)] = det(1 + T1) det(1 + T2)

for any two trace class operators T1 and T2.Using exp(T ) = 1 + T + · · · + Tn

n1+ . . . we could also define det(expT ) too, and in fact

det(expT ) = eTrace(T ).As an example, if T is an integral operator on some L2(X) with kernel K(x, y) with mild

assumptions on K and X, then it could be shown that

Tr(Tk) =1

k!

∫. . .

∫K

(x1, x2, . . . xkx1, x2, . . . xk

)dx1 . . . dxk

and the definition of the determinant coincides with the previous section.

introduction to functional analysis -...

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