Math 206: Banach Algebra and Spectral Theory

Marc Rieffel, UC Berkeley Fall 2015Scribe: Christopher Eur (any error is due to the scribe)

Contents

1 Overview 31.1 Lecture 8/26/15 — Overview I: Banach algebras . . . . . . . . . . . . . . . . . . . . 31.2 Lecture 8/28/15 — Overview II: Banach ∗-algebras . . . . . . . . . . . . . . . . . . . 4

2 Basic Gelfand Theory 62.1 Lecture 8/31/15 — The spectrum of unital Banach algebras . . . . . . . . . . . . . . 62.2 Lecture 9/2/15 — The resolvent map and 6= ∅ spectrum in C . . . . . . . . . . . . . 72.3 Lecture 9/4/15 — Ideals and quotients in Banach algebras . . . . . . . . . . . . . . . 82.4 Lecture 9/9/15 — Maximal ideal spans and Gelfand transforms . . . . . . . . . . . . 92.5 Lecture 9/11/15 — Recovering the space from the algebra of functions . . . . . . . . 112.6 Lecture 9/14/15 — Gelfand’s spectral radius formula . . . . . . . . . . . . . . . . . . 122.7 Lecture 9/16/15 — Isometric embedding and C∗-algebras . . . . . . . . . . . . . . . 132.8 Lecture 9/18/15 — (Little) Gelfand-Naimark Theorem . . . . . . . . . . . . . . . . . 15

3 Basic Pontryagin duality 173.1 Lecture 9/21/15 — ((Cc(−)/`1(−), ∗), Fforgetful : Alg(Banach) → SemiGrp) . . . . . . . 17

3.2 Lecture 9/23/15 — The correspondence `1(S) ' HomSemiGrp(S,D) . . . . . . . . . . 18

3.3 Lecture 9/25/15 — The transform `1(G)→ G (f 7→ f) is faithful . . . . . . . . . . . 203.4 Lecture 9/28/15 — A first sight of Pontryagin duality . . . . . . . . . . . . . . . . . 213.5 Lecture 9/30/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.6 Lecture 10/2/15 — Haar measure and modular function . . . . . . . . . . . . . . . . 243.7 Lecture 10/5/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.8 Lecture 10/7/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.9 Lecture 10/9/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.10 Lecture 10/12/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.11 Lecture 10/14/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.12 Lecture 10/16/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.13 Lecture 10/19/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 Functional calculus 314.1 Lecture 10/21/15 — Holomorphic functional calculus . . . . . . . . . . . . . . . . . . 31

5 Operators on Hilbert spaces 335.1 Lecture 10/23/15 — Polarization identity and consequences . . . . . . . . . . . . . . 335.2 Lecture 10/26/15 — Normal and self-adjoint operators . . . . . . . . . . . . . . . . . 345.3 Lecture 10/28/15 — The GNS construction I . . . . . . . . . . . . . . . . . . . . . . 355.4 Lecture 10/30/15 — The GNS construction II . . . . . . . . . . . . . . . . . . . . . . 36

1

5.5 Lecture 11/2/05 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.6 Lecture 11/4/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.7 Lecture 11/6/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

6 Compact and Fredholm operators 406.1 Lecture 11/9/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.2 Lecture 11/13/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.3 Lecture 11/16/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.4 Lecture 11/18/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.5 Lecture 11/20/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.6 Lecture 11/23/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.7 Lecture 11/30/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466.8 Lecture 12/2/15 — . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2

Chapter 1

Overview

1.1 Lecture 8/26/15 — Overview I: Banach algebras

In this section we give few basic definitions and many examples without proof.

Definition 1.1.1. Let k be a field. A k-algebra A is a ring A with action of k on it (i.e.k ×A → A making A as a vector space over k satisfying a~x · b~y = (ab)~x · ~y). If A is unital, notethis action is given by a ring homomorphism k → A .

Remark 1.1.2. In other words, a k-algebra is a vector space V over k with a multiplication mapV × V ·→ V that respects the scalar multiplication.

Definition 1.1.3. A normed algebra is an algebra A over a normed field (e.g. R, C) that isequipped with a vector space norm ‖ ‖ satisfying ‖ab‖ ≤ ‖a‖‖b‖ ∀a, b ∈ A . A is a Banachalgebra if it is complete with respect to the metric given by the norm.

Remark 1.1.4. Let A is a normed algebra. Since the operations +, · as maps A × A → A areuniformly continuous, these operations extend to A , the completion of A . Thus, a completion Aof a normed algebra A is a Banach algebra.

Example 1.1.5. Here are some examples of Banach algebra that arise as a space of functions ona topological space. Note that all these examples are commutative Banach algebras:

(a) Let M be a compact topological space. Then C(M) (either C(M,R) or C(M,C)) is a Banachalgebra with norm ‖f‖∞ := sup |f |.

(b) Let M be a locally compact space. Then the space of functions vanishing at infinity,

C0(M) := f ∈ C(M) : ∀ε > 0, f−1((−∞, ε]) is compact

is a Banach algebra with the same norm ‖ ‖∞ (is in fact closed subalgebra of Cb(M), thespace of continuous bounded functions). Note that C0(M) is a non-unital ring.

(c) Let (M,d) be a (compact) metric space, and for f ∈ C(M) define the Lipschitz constantas

Ld(f) := sup

|f(x)− f(y)|

d(x, y): x 6= y ∈M

.

Then L (M,d) := f ∈ C(M) : Ld(f) < ∞ is a dense subalgebra of C(M) with the norm‖ ‖∞ (dense by Stone-Weierstrass). In fact, with norm ‖f‖L := ‖f‖∞+Ld(f), L (M,d) is infact complete, so it is a Banach algebra. (Note: Ld is a semi-norm on L (M,d), and in factwe can recover the metric d from Ld).

3

(d) Let M be a (compact) differentiable manifold (for example S1). We can consider a wholefamily of function spaces C(M) ⊃ C1(M) ⊃ · · · ⊃ C∞(M). If M has a Riemannian metric,then for C1(M) one can define ‖f‖(1) := ‖f‖∞ + ‖df‖ (where ‖df‖ is the norm of a linearmap df : TM → R). For example on S ' R/Z, we have ‖f‖(1) = ‖f‖∞ + ‖f ′‖∞. Further, forC2(S) we have ‖f‖(2) := ‖f‖∞ + ‖f ′‖∞ + 1

2‖f′′‖∞ (turns out the 1/2 factor is necessary). In

all these cases we have a Banach algebra.

(e) Let U ⊂ Cn be a open region, then H∞(U), the space of all bounded holomorphic functionson U is also a Banach algebra.

Example 1.1.6. A major example of a non-commutative Banach algebra is the following: Let Xbe a Banach space. Then B(X), the algebra of bounded operators on X, is a Banach algebra with

the operator norm (‖T‖ := sup‖T (x)‖‖x‖ : x ∈ X) as its norm.

Proof. The usual operator norm makes B(X) into a normed vector space, and if S, T ∈ B(X)

then ‖ST‖ = sup‖TS(x)‖‖x‖ = sup‖T (S(x))‖

‖S(x)‖‖S(x)‖‖x‖ : S(x) 6= 0 ≤ ‖T‖‖S‖. Thus B(X) is a normed

algebra. For completeness, if∑

n ‖Tn‖ converges then so does 1‖x‖∑

n ‖Tn(x)‖ for x 6= 0, and

so∑

n Tn(x) exists since X is a Banach space. Now, it is clear that T (x) :=∑

n Tn(x) definedpointwise is in fact bounded and linear.

1.2 Lecture 8/28/15 — Overview II: Banach ∗-algebras

When X is a Hilbert space in the above example, we observe an additional nice structure:

Example 1.2.1. Let H be a Hilbert space, and consider the Banach algebra B(H ). For anyT ∈ B(H ), there exists an adjoint operator T ∗ ∈ B(H ) such that 〈Tξ, η〉 = 〈ξ, T ∗η〉 ∀ξ, η ∈H .(Note that H ' H ∗ via the inner product H × H → C, so given η, the linear functionalξ 7→ 〈Tξ, η〉 is given by 〈−, T ∗η〉 for some T ∗η ∈ H . The linearity of the assignment η 7→ T ∗η iseasy to check).

Remark 1.2.2. Note that the map ∗ : T 7→ T ∗ is conjugate-linear, T ∗∗ = T , and (ST )∗ = T ∗S∗.This operation deserves a definition:

Definition 1.2.3. An algebra A over a normed field is a ∗-algebra if it has a map ∗ : A → Asatisfying: (i) conjugate-linearity, (ii) ∗∗ = Id, and (iii) (ST )∗ = T ∗S∗. A is a normed ∗-algebraif ‖a∗‖ = ‖a‖ ∀a ∈ A , and moreover Banach ∗-algebra if A is complete with respect to the norm.

Example 1.2.4.

(a) B(H ) with the adjoint ∗ is a Banach ∗-algebra. In fact, it satisfies the property ‖T ∗T‖ =‖T‖2, which is equivalent to ‖T‖ = ‖T ∗‖.

(b) For M locally compact, C∞(M,C) with f∗ := f is a Banach ∗-algebra. This space alsosatisfies ‖f∗f‖ = ‖f‖2.

Definition 1.2.5. A C∗-algebra is a Banach ∗-algebra over C that satisfies ‖a∗a‖ = ‖a‖2.

The next theorem classifies the kind of Banach ∗-algebras given in the above example:

Theorem 1.2.6 (Little Gelfand-Naimark Theorem). Let A be a commutative Banach ∗-algebrasatisfying ‖a∗a‖ = ‖a‖2. Then A ' C0(M) (as a Banach ∗-algebra) for some locally compact spaceM . In particular, any C∗-algebra is ' C0(M,C) for some M locally compact.

4

Example 1.2.7. Let T ∈ B(H ) with T ∗ = T for a H a Hilbert space. Let A be the closedsubalgebra of B(H ) generated by T and Id (i.e. take all polynomials in T ). Since T ∗ = T , Ais a Banach ∗-subalgebra of B(H ), and moreover ‖S∗S‖ = ‖S‖2. Applying Little G-N, we havethat A ' C(M) for M ⊂ R compact (since A is unital, ⊂ R needs work). This is the generalizeddefinition of spectrum of the operator T . (Recall that for H finite dimensional, M is the set ofeigenvalues of T . In this case, we have T 7→ (f : λ 7→ λ), Id 7→ (1 : λ 7→ 1))

Theorem 1.2.8 (Big Gelfand-Naimark Theorem). Any C∗-algebra is isomorphic to a closed ∗-subalgebra of B(H ) for some Hilbert space H .

5

Chapter 2

Basic Gelfand Theory

2.1 Lecture 8/31/15 — The spectrum of unital Banach algebras

Definition 2.1.1. Let k be a field, X a set. Denote by F (X, k) the k-algebra of functions f :X → k.

Remark 2.1.2. Note that for f ∈ F (X, k), λ ∈ f(X) if and only if f − λ1 is not invertible (as anelement in F (K, k)).

N.B. 2.1.3. We will assume that a normed algebra with unit satisfies ‖1‖ = 1 in this course unlessstated otherwise.

Definition 2.1.4. For a unital algebra A over a field k, the spectrum of a ∈ A is defined as:

σA (a) := λ ∈ k : a− λ1 is not invertible in A

Example 2.1.5. It is sometimes important to note in what algebra we are taking the spectrumof an element. For example, let A = C([0, 1]), and B = P (space of polynomials) (so B is adense subalgebra of A ), and let p ∈ B a polynomial of degree ≥ 1. Then, σA (p) = p([0, 1]), butσB(p) = R or C.

Proposition 2.1.6. Let A be a unital Banach algebra, and let a ∈ A . If ‖a‖ < 1, then 1 − a isinvertible, with ‖(1− a)−1‖ ≤ 1

1−‖a‖ .

Proof. Consider∑∞

n=0 an. The series absolutely converges, so the series converges (∵ A is com-

plete), and if Sn is the partial sum, then (1 − a)Sn = 1 − an which → 1 as n → ∞. Thus,1

1−a :=∑∞

n=0 an is the inverse of 1−a (since multiplication is uniformly continuous), and the claim

about its norm clearly follows.

Corollary 2.1.7. Let A be a unital Banach algebra. If a ∈ A and ‖1−a‖ < 1, then a is invertible.In other words, the open unit ball around 1 consists of invertible elements.

Definition 2.1.8. For a ∈ A , denote by La, Ra the left, right multiplication by a.

Remark 2.1.9. Note that a 7→ La is a algebra homomorphism A → L(A ), so LaLb = Lab (andlikewise RaRb = Rba). If A is normed, then we have algebra homomorphism A → B(A ) since‖La‖ = ‖a‖ (need ‖1‖ = 1 here). Thus, if a is invertible, we conclude that La is a homeomorphismA → A . In particular we arrive at:

6

Lemma 2.1.10. For A a unital Banach algebra, if a ∈ A is invertible, then Lab : ‖1 − b‖ < 1is an open neighborhood of a consisting of invertible elements.

Corollary 2.1.11. Let GL(A ) := invertible elements of A . Then for A unital Banach algebra,GL(A ) is an open subset of A ; moreover, for a ∈ A , σA (a) is a closed subset of A . Moreover, ifλ ∈ σA (a) then |λ| ≤ ‖a‖.

Proof. The first part is immediate from the lemma. For the second part, note that A \ σA (a) =λ ∈ k : a − λ1 ∈ GL(A ) = ϕ−1(GL(A )), where ϕ : k → A given by λ 7→ a − λ1 is continuous.If |λ| > ‖a‖, then λ− a = λ(1− a/λ), and ‖a/λ‖ < 1, and hence λ− a is invertible.

2.2 Lecture 9/2/15 — The resolvent map and 6= ∅ spectrum in C

We further note that GL(A ) is a topological group:

Proposition 2.2.1. The inverse map GL(A )−1

→ GL(A ) is continuous in the norm of A . Thus,GL(A ) is a topological group with the topology given by the norm on A .

Proof. Fix arbitrary a ∈ GL(A ). We’ll show the inverse map is continuous at a. First, note that ifb is close to a, then b−1 not much larger than a−1. More precisely, let O := c : ‖1− c‖ < 1/2, so cis invertible, and ‖c−1‖ ≤ 1

1−‖1−c‖ ≤ 2. Let b ∈ aO, so b = ac for c ∈ O, then ‖b−1‖ = ‖c−1a−1‖ ≤2‖a−1‖.

Now, for a, b ∈ GL(A ), writing b−1 − a−1 = b−1(a− b)a−1, we have

‖b−1 − a−1‖ ≤ ‖b−1‖‖a− b‖‖a−1‖

Thus, for b close enough to a, we have ‖b−1−a−1‖ ≤ 2‖a− b‖‖a−1‖, which we can make arbitrarilysmall.

Remark 2.2.2. These two tricks come handy at times: b−1−a−1 = b−1(a−b)a−1 and (equivalently)b−1 = (1 + b−1(a− b))a−1

Remark 2.2.3. Over R, we can have σA (a) = ∅. For example, if A is a rotation matrix on R2, thenit has no eigenvalues. We will see that the C case is much nicer.

Definition 2.2.4. The complement of the spectrum is called the resolvent set of a ∈ A for aunital Banach algebra A . I.e. the resolvent set of a ∈ A is:

ρA (a) := λ ∈ k : (a− λ) ∈ GL(A ).

Definition 2.2.5. On ρ(A ), define the resolvent of a as the function R(a, λ) = (λ− a)−1.

Proposition 2.2.6. R(a, λ) is an analytic function ρ(a)→ GL(A ).

Proof. We first check differentiability. Let f(z) = R(a, z), and consider:

f(z + h)− f(z)

h=

(z + h− a)−1 − (z − a)−1

h=

(z + h− a)−1((z − a)− (z + h− a))(z − a)−1

h

which approaches −(z−a)−2 as h→ 0. So f ′(z) = −(z−a)−2. Likewise, we have f ′′(z) = 2(z−a)−3,and so forth. To get a power series, let z0 ∈ ρ(a), and we compute that:

(z−a)−1 = (z−z0−(a−z0))−1 = (z0−a)−1(1−(z−z0)(a−z0)−1)−1 = (z0−a)−1∞∑n=0

(a−z0)−n(z−z0)n

7

Since z0, a are fixed, for a small open ball around z0, the above series absolutely converges andthe series is well-defined (cf. Proposition 2.1.6). Note that commutativity is not an issue sincez − z0 ∈ C.

Remark 2.2.7. We examine the behavior of R(a, z) at ∞. Note that R(a, z−1) = (z−1 − a)−1 =z(1 − az)−1 for all z so that ‖az‖ < 1. So, as z → 0, we see that R(a, z−1) → 0. DefiningR(a, 0−1) = 0, we have that R(a, z) is analytic at ∞.

Theorem 2.2.8. For a unital Banach algebra A over C, and for any a ∈ SA, σA (a) 6= ∅.

Proof. Suppose σ(a) = ∅. Then R(a, z) is defined on all C and 0 at ∞, and hence it is entire andbounded. By Liouville, R(a, z) is constant 0, i.e. (a − z)−1 = 0 ∀z ∈ C, contradiction. Liouvilleactually applies by the following observation:

Let A ∗ be the (bounded) dual Banach space to A , then for φ ∈ A ∗, the map z 7→ φ(R(a, z)) isa C-valued analytic function. So for any φ ∈ A ∗, we get φ(R(a, z)) = 0 ∀z, and hence R(a, z) = 0.(note we need Hahn-Banach theorem to make nontrivial A ∗.)

Theorem 2.2.9 (Gelfand-Mazur). Let A be a unital Banach algebra over C. If every nonzeroelement is invertible, then z 7→ z1A is an isomorphism from C onto A .

Proof. Given a ∈ A , let z ∈ σ(a) 6= ∅, so (z − a) is not invertible, so z − a = 0.

Remark 2.2.10. Note that this fails over R. Consider C or the quarternions over the reals.

2.3 Lecture 9/4/15 — Ideals and quotients in Banach algebras

Let A be an algebra. Recall what left/right/two-sided ideals of A are. In the case where I is aleft ideal, A /I is a left A -module. If two-sided, then again an A -algebra.

N.B. 2.3.1. Let A a normed algebra, and I an ideal in A , then the closure I is again an idealin A . To see this, note that if an → a ∈ I , then ban → ba ∈ I and so forth.

Proposition 2.3.2. Let A be a unital Banach algebra. If I is a proper ideal in A , then I isproper.

Proof. Use GL(A ) is open: since I proper, I ∩GL(A ) = ∅, and A \GL(A ) is closed.

Example 2.3.3. For a counter-example, consider Cc(R) ⊂ C0(R), where Cc(R) is compactlysupported functions. Cc(R) is an ideal but it is dense in C0(R) (note that C0(R) is non-unital). Foranother example, for A as polynomials in C([0, 1]), and consider I = p(x) : p(2) = 0 (here Ais not complete). Lastly, let A be C(R) (including unbounded ones), and let I = Cc(R) compactopen topology. (Here we have a Frechet algebra where GL(A ) is not open).

Corollary 2.3.4. Let A be a unital Banach algebra, then every maximal ideal is closed.

Aside: quotient norm

N.B. 2.3.5. Let’s recall few facts about quotients of normed spaces. If X is a normedvector space, and Y a subspace, we can define the quotient norm (really a seminorm) as

‖π(x)‖X/Y = inf‖x− y‖ : y ∈ Y = “d(x, Y )”

If Y is closed, then ‖ ‖X/Y is actually a norm.Moreover, if X is a Banach space, and if Y is a closed subspace, then X/Y with the

quotient norm ‖ ‖X/Y is complete (thus is a Banach space).

8

Proposition 2.3.6. Let A be a normed algebra and let I be a closed ideal in A . Then if I isa two-sided ideal, then A /I with the quotient norm is a normed algebra. If just left/right ideal,then A /I is a left/right A -module where ‖aπ(b)‖ ≤ ‖a‖‖π(b)‖.

Proof. In the case where I is two-sided, then ‖π(a)π(b)‖A /I ≤ ‖(a− c)(b−d)‖A since π(a)π(b) =π(ab) and (a− c)(b− d) = ab− (something ∈ I ), and ‖(a− c)(b− d)‖ ≤ ‖a− c‖‖b− d‖. Take infover c, d ∈ I , and we are done.

Proposition 2.3.7. If A is a Banach algebra and if A is a closed two-sided ideal, then A /I isa Banach algebra.

Definition 2.3.8. An algebra or a ring is simple if it contains no proper two-sided ideals except0.

Example 2.3.9. One can check that Mn(k) for any field k is simple. For proper left ideal, considermatrices with last column all zero.

Proposition 2.3.10. Let A be an algebra or ring, and let I be a maximal two-sided ideal in A .Then A /I is a simple algebra or ring.

Corollary 2.3.11. If A is a Banach algebra and I is a maximal closed two-sided ideal, thenA /I is a simple Banach algebra.

Let A be a unital commutative algebra or a ring, and if a ∈ A and a is not invertible, thenaA is a proper ideal.

Corollary 2.3.12. If A is a unital commutative algebra or a ring, and if A is simple, then A isa field.

Theorem 2.3.13. Let A be a unital commutative Banach algebra over C/R, and let I be amaximal in A . Then A /I ' C/R. (cf. Gelfand-Mazur).

How should the R case goes: check the complexification later

Remark 2.3.14. Let A be a unital commutative Banach algebra over C. From a maximal ideal,we get φ : A → C (unital ring homomorphism) such that I is the kernel of this homomorphism.Conversely, let A be a unital Banach algebra, and let ψ : A → C be a (unital) homomorphism,then ψ is continuous and the kernel is a maximal (surjective map) two-sided ideal in A . Forcontinuity of ψ, we need a quick but important lemma:

Lemma 2.3.15. For any given a ∈ A , for A a unital Banach algebra and ψ : A → C a (unital)homomorphism, we have ψ(a) ∈ σ(a), and furthermore ‖ψ‖ = 1.

Proof. Note that ψ(a − ψ(a)1) = 0, implying that a − ψ(a)1 is not invertible. Then |ψ(a)| ≤ ‖a‖by Corollary 2.1.11, implying ‖ψ‖ ≤ 1, but we do have ψ(1) = 1, so that ‖ψ‖ = 1.

2.4 Lecture 9/9/15 — Maximal ideal spans and Gelfand trans-forms

Notation. Today A will be a commutative unital Banach algebra over C.

Recall: For m ⊂ A maximal, we have a unital homomorphism A → A /m ' C, and if ϕ : A → Cunital homomorphism, the kernel is a maximal ideal. In other words, there is a natural bijectionbetween the maximal ideals of A and nonzero (unital) homomorphisms A → C.

9

Definition 2.4.1. Define A , the maximal ideal span of A , to be the set of maximal ideals (orequivalently the set of nonzero unital homomorphisms ϕ : A → C, called multiplicative linearfunctionals).

For ϕ ∈ A , recall that ϕ(a) ∈ σ(a), and ‖ϕ‖ = 1. In fact, A ⊂ unit ball of A ∗. On A ∗, havethe weak-∗ topology. Moreover, note that a unit ball is compact in the weak-∗ topology.

Aside: weak-∗ topology

Definition 2.4.2. Given a family of R/C-valued functions F on a set X, we define theF-weak topology on X be the weakest (coarsest) topology on X such that f ∈ F are allcontinuous. In other words, the basis is given by:

Bf1,...,fm,ε(x0) = x ∈ X : |fi(x)− fi(x0)| < ε ∀ = 1, . . . ,m

Definition 2.4.3. Given a normed space X, consider the family F of functions on X∗

given by “evaluation at x ∈ X” That is, evx(ϕ) = ϕ(x). The weak-∗ topology on X∗ isthe F-weak topology on X∗.

Remark 2.4.4. In this topology, ψα → ψ iff ψα(x) → ψ(x) ∀x ∈ X. In other words,convergence of a net in weak-∗ topology is the same as pointwise convergence.

Theorem 2.4.5 (Alaoglu’s Theorem). The unit ball in X∗ (with usual sup-norm asseminorm) is compact in the weak-∗ topology.

Proposition 2.4.6. A is closed in the weak-∗ topology, and so is compact.

Proof. Let ϕα be a net of elements of A that converges to ψ ∈ A ∗ (since the unit ball is compactψ exists). We now check that ψ ∈ A . Well, for a, b ∈ A , ψ(ab) = limϕα(ab) = limϕα(a)ϕα(b) =ψ(a)ψ(b) (since multiplication is continuous in C). Moreover, ψ(1A ) = limϕα(1A ) = 1.

Definition 2.4.7. For a ∈ A , define a ∈ C(A ) by a(ϕ) := ϕ(a). The mapping a 7→ a is called theGelfand transform for A .

Note that a is continuous (∈ C(A )) because if ϕα → ϕ in weak-∗ topology then a(ϕ) = ϕ(a) =limϕα(a) = lim a(ϕα).

Proposition 2.4.8. The Gelfand transform a 7→ a is in fact a unital homomorphism A → C(A )(of Banach algebras).

Proof. For a, b ∈ A we have (ab)(ϕ) = ϕ(ab) = ϕ(a)ϕ(b) = a(ϕ)b(ϕ), so (ab) = ab, and moreover,1A (ϕ) = ϕ(1A ) = 1. Lastly, note that ‖a‖∞ ≤ ‖a‖ because for any ϕ, a(ϕ) = ϕ(a) ∈ σ(a), so that|a(ϕ)| = |ϕ(a)| ≤ ‖a‖ (∵ Lemma 2.3.15).

Remark 2.4.9. Gelfand transforms certainly need not be surjective. Note, for A = C0(Z). Forn ∈ Z, define ϕn(f) = f(n). limn→∞ ϕn = 0 in weak-∗ topology. In fact, any X be any Banachspace, and define a product on X by setting all products = 0.

Proposition 2.4.10. The image of A under the Gelfand transform is a unital subalgebra thatseparates points of A (i.e. if a(ϕ) = a(ψ) for all a ∈ A then ϕ = ψ).

Proposition 2.4.11. Suppose that unital A is generated by one element, a0 (i.e. polynomials ofa0 is dense in A ). Then A ' σ(a0) is a homeomorphism.

10

Proof. Consider a0 : A → C. This map really maps into σ(a0). If ϕ,ψ ∈ A and ϕ(a0) = ψ(a0),then by density argument ϕ = ψ. So a0 is injective. Continuity should be easy (check). Forsurjectivity, consider z ∈ σ(a0). Then a0 − z is not invertible, and thus (a− z)A is a proper ideal.Let I be its closure. WE claim that I is maximal: consider A /I , C1A + I ⊃ poly(a0). Theimage of C1A + I in A /I , is C1A ⊂ A /I dense, so that A /I ' C. Let ψ.

finish proof as exercise

2.5 Lecture 9/11/15 — Recovering the space from the algebra offunctions

Question. Let X be a compact Hausdorff space, and let A = C(X), then what is A ?

For each x ∈ X, define ϕx : A → C by ϕx(f) = f(x). Since each ϕx is surjective, we have maximalideals mx := kerϕx. Thus, this gives us a map µ : X → A . We now show that this map is in facta homeomorphism:

Lemma 2.5.1. Let X be a compact Hausdorff space. Let I be a proper ideal of A = C(X).Then there is at least x ∈ X such that f(x) = 0 ∀f ∈ I .

Proof. If not, use compactness to construct a function f = |f1|2 + · · ·+ |fn|2 ∈ I such that f > 0on X. That is, 1/f ∈ C(X) and hence f is invertible. Thus, I = (1); contradiction.

This implies that the map X → A is surjective, and it is moreover injective by Urysohn’slemma.

and continuous (need check continuity). Thus µ is a homeomorphism.

Proposition 2.5.2. Let X be compact Hausdorff space. Then the weak-∗ topology on A is the

same as the topology on X via the bijection µ : X → C(X).

Let A be a commutative unital Banach algebra. Recall the Gelfand transform a 7→ a (A →C(A ). This really may not be a bijective in general: consider the nilpotent matrices.

Remark 2.5.3. Note that for A commutative unital Banach algebra, and for any ϕ ∈ A , if a isnilpotent then 0 = ϕ(an) = (ϕ(a))n so ϕ(a) = 0. Thus, a 7→ a = 0.

Proposition 2.5.4. Let a ∈ A for A commutative unital Banach algebra. Then, range(a) =σA (a).

Proof. Recall that for a ∈ A , the range of a ⊂ σ(a). For the converse, if λ ∈ σ(a), then λ − a isnot invertible, so (λ − a)A is a proper ideal, which is contained in a maximal ideal m. So, thereexists ϕ ∈ A such that kerϕ = m (∵ Theorem 2.3.13), so ϕ(λ− a) = 0, so that ϕ(a) = λ.

Example 2.5.5. Let A = Cb(Z+) be the algebra of all bounded C-valued sequences, with norm‖ ‖∞. And let B = C∞(Z+) be the ideal consisting of sequences that converge to 0. Then B iscontained in a maximal ideal of A , so there is a ϕ ∈ A with ϕ(B) = 0. Well, can you describe whatϕ is? No: this formulation is non-constructive. Note that A = βZ+ where βZ+ is the Stone-Cechcompactification of Z+.

Definition 2.5.6. The spectral radius r(a) of a ∈ A for A a Banach algebra is

r(a) := max|λ| : λ ∈ σ(a)

11

Remark 2.5.7. From the previous proposition, note that for a commutative unital Banach algebra,we have r(a) = ‖a‖∞.

Proposition 2.5.8. Let A commutative unital Banach algebra. For any a, b ∈ A , we have:

r(ab) ≤ r(a)r(b), r(a+ b) ≤ r(a) + r(b)

Proof. r(ab) = ‖ab‖∞ = ‖ab‖∞ ≤ ‖a‖∞‖b‖∞ = r(a)r(b). Similar for the sum case.

Remark 2.5.9. Above can fail if a, b does not commute.

We will work towards the first form of “holomorphic functions calculus” next time.

2.6 Lecture 9/14/15 — Gelfand’s spectral radius formula

Proposition 2.6.1. Let A be unital Banach algebra (over C/R), and let a ∈ A . Let f be analyticfunction with power series expansion

∑n αnz

n that converges (absolutely uniformly) on some opensubset of C/R containing z : |z| ≤ ‖a‖. Then f(a) =

∑n αna

n ∈ A and if λ ∈ σ(a) thenf(λ) ∈ σ(f(a)).

Proof. f(λ) − f(a) =∑

n αnλn −

∑n αna

n =∑

n=1 αn(λn − an) =∑

n=1 αn(λ − a)Pn(λ, a), andone checks that ‖Pn(λ, a)‖ ≤ n‖a‖n−1. So, we have (λ − a)

∑n=1 αnPn(λ, a). Now consider

f ′(z) =∑∞

n=1 αnnzn−1, which converges absolutely uniformly on |z| ≤ ‖a‖. And thus, calling

quantity∑

n=1 αnPn(λ, a) as b ∈ A , we have that f(λ) − f(a) = (λ − a)b. Thus, we have that iff(λ)− f(a) has an inverse, then so does (λ− a). Thus, we have f(λ) ∈ σ(f(a)), as desired.

Corollary 2.6.2. For A a unital Banach algebra (not necessarily commutative) we have that

r(a) ≤ infn‖an‖1/n

Proof. Consider the map f(z) = zn. Then if λ ∈ σ(a) then λn ∈ σ(an). Thus, |λn| ≤ ‖an‖,implying that |λ| ≤ ‖an‖1/n. Consequently, we have that |λ| ≤ infn‖an‖1/n.

For the rest of the class, we will give a proof of the Gelfand’s spectral radius formula:

Theorem 2.6.3 (Gelfand’s spectral radius formula). If A is a unital Banach algebra over C, thenwe have

r(a) = lim ‖an‖1/n

(in particular, the limit exists).

Proof. Consider the resolvent of a at ∞, which is R(a, z−1) = 1z−1−a = z(1 − az)−1 = z

∑n a

nzn

(which converges for ‖az‖ < 1, i.e. |z| < ‖a‖−1). However, since R(a, z) is analytic for |z| > r(a),if we look at R(a, z−1) is analytic for |z| < r(a)−1.

Thus, for any ϕ ∈ A ∗ (continuous linear functional) define fϕ(z) := ϕ(z(1 − az−1)). Thisis a ordinary holomorphic function (this can fail for real analytic) in z : |z| < r(a)−1. So thepower series expansion about 0 converges for |z| < r(a)−1. But then the power series for fϕ isz∑

n ϕ(an)zn, which converges for |z| < r(a)−1. So if one chooses r(a) < r, then fϕ will convergeabsolutely uniformly for |z| < r−1. So, we conclude that there is Mϕ such that |ϕ(an)||zn| ≤Mϕ∀n, |z| < r−1. So, |ϕ(an)r−n| ≤Mϕ for all n. For each n, define Fn ∈ A ∗∗ by Fn(ϕ) = ϕ(an)r−n.Thus, |Fn(ϕ)| ≤Mϕ ∀ϕ.

12

Aside: functional analysis theorems

Theorem 2.6.4 (Uniform Boundedness Principle). Let X be a Banach space, and Y anormed space. Let F ⊂ B(X,Y ) be a collection of continuous linear maps X → Y suchthat

∀x ∈ X, supT∈F‖T (x)‖Y <∞

thensupT∈F‖T‖B(X,Y ) <∞

Proof. Baire category theorem.

Theorem 2.6.5 (Hahn-Banach Theorem). If ϕ is a continuous linear functional on V0 thatis subspace of V satisfying ϕ(v) ≤ p(v) ∀v ∈ V0, then ϕ can be extended to a continuouslinear functional on V such that ϕ(v) ≤ p(v) ∀v ∈ V .

Proof. Can extend to dimension one more than V0 easily. Then Zorn’s lemma.

Here the theorem basically says that ∃M such that ‖Fn‖ ≤M ∀n. But is clear that ‖Fn‖ = ‖an‖r−n(this requires Hahn-Banach), and thus ‖an‖r−n ≤M . In other words, ‖an‖1/n ≤M1/nr → r.

Thus, we havelim sup ‖an‖1/n ≤ r(a) ≤ inf

n‖an‖1/n

from which we conclude r(a) = limn ‖an‖1/n.

Corollary 2.6.6. So, r(a) does not depend on the containing algebra.

Next time:

Corollary 2.6.7. Let A be a commutative unital Banach algebra over C, then a 7→ a from A toC(A ) is isometric exactly if ‖a2‖ = ‖a‖2 for all a ∈ A .

2.7 Lecture 9/16/15 — Isometric embedding and C∗-algebras

Hint for #2 on the homework: If Z is a closed subset of X, then consider X/ ∼ Z, with all thepoints of Z identified to one point, with quotient topology.

Proposition 2.7.1. Let A be a commutative unital Banach algebra (over C?). Then, the Gelfandtransform is isometric iff ‖a2‖ = ‖a‖2 for all a ∈ A .

Proof. If ‖a2‖ = ‖a‖2, then for any a we have ‖a4‖ = ‖a‖4, and so forth, to get ‖a2n‖ = ‖a‖2n .Thus, r(a) = lim ‖an‖1/n = lim ‖a2n‖1/2n = ‖a‖. Now, recall that ‖a‖∞ = r(a).

Conversely, if for some a we have ‖a2‖ = s2 < ‖a‖2, then we have ‖a2n‖ < s2n , so thatr(a) < ‖a‖, and hence not an isometry.

Remark 2.7.2. Now, let U be open bounded region in C, and A := continuous functions on Uholomorphic on U , and note that ‖f2‖∞ = ‖f‖2∞. Thus, the Gelfand transform isometricallyembeds into A ∗, but is not surjective.

How do we get surjectivity? This rises when we have convolution:

13

Definition 2.7.3. Let A an algebra over C/R. An involution to A is a mapping a 7→ a∗ suchthat:

(a+ b)∗ = a∗ + b∗, (αa)∗ = αa∗, (ab)∗ = b∗a∗, (a∗)∗ = a

An element in A is self-adjoint if a∗ = a. Moreover, A is a normed ∗-algebra if it is normedalgebra with ‖a∗‖ = ‖a‖.

Definition 2.7.4. A ∗-algebra is symmetric if for any a ∈ A we have a∗ = a.

Remark 2.7.5. There many self-adjoint elements. Given a ∈ A , consider a = a+a∗

2 + ia−a∗

2 , thenthe real and imaginary parts are self-adjoint. Note that if A is symmetric, then σ(a) ⊂ R.

Definition 2.7.6. A commutative algebra is semi-simple if the intersection of all its maximalideals is 0.

Proposition 2.7.7. Let A be a commutative unital Banach algebra, then the Gelfand transformfor A is injective exactly if A is semi-simple.

Proof. Exercise

Proposition 2.7.8. Let A be a commutative unital Banach algebra. If it is symmetric, then theimage of A under the Gelfand transform is a dense ∗-subalgebra of C(A ).

Proof. The image of A always separates the points of A . The image contains 1. By the symmetry,the image is closed under complex conjugation since if a = b+ic with b, c self-adjoint, then a = b+icand a∗ = b− ic = a. So, we can apply the Stone-Weierstrass theorem.

Remark 2.7.9. So, if in addition, ‖a2‖ = ‖a‖2, then the Gelfand transform is an isometric ∗-isomorphism A

∼→ C(A ).

Definition 2.7.10. By a C∗-algebra we mean a Banach ∗-algebra such that ‖a∗a‖ = ‖a‖2 for alla ∈ A .

Example 2.7.11. C(X) for X compact, C∞(X) for X locally compact.

Example 2.7.12 (Bounded operators on a Hilbert space). Let H be a Hilbert space, and considerT ∈ B(H ), then ‖T ∗T‖ = ‖T‖2 (check: ‖T ∗T‖ ≤ ‖T‖2 since ‖T‖ = ‖T ∗‖, the other direction byCauchy-Schwartz. Moreover, Any closed ∗-subalgebra of B(H ) is a C∗-algebra.

Lemma 2.7.13. Let A be a unital C∗-algebra, and let a ∈ A with a∗ = a. Then ‖a2‖ = ‖a‖2,and all powers of a are self-adjoint, so r(a) = ‖a‖.

From this lemma, we have that

Corollary 2.7.14. For a unital C∗-algebra, its norm is determined by its ∗-algebra structure.

Proof. Given a ∈ A , we have ‖a‖2 = ‖a∗a‖ = r(a∗a). Note that the spectral radius is entirely ofthe algebraic structure of A .

Example 2.7.15 (∗ is crucial). Let V be a finite dimensional vector space over C. Consider thealgebra A := L(V ). For each inner product on V , we get an involution ∗ on A , and a norm on Vand so a norm on A . This gives us many C∗-algebras on A .

14

2.8 Lecture 9/18/15 — (Little) Gelfand-Naimark Theorem

Proposition 2.8.1. Let A be a (unital) C∗-algebra. If A is symmetric then σA (a) ⊂ R. (In fact,we can change σA to σB where B is a smallest C∗-subalgebra generated by a and 1A ).

Proof. (Arens’ trick). Let a ∈ A such that a∗ = a, and let r + is ∈ σ(a). Need show s = 0 Givent ∈ R, let b = a+ it. Then note that b∗b = (a− it)(a+ it) = a2 + t2. Thus, ‖b∗b‖ ≤ ‖a2‖+ t2. Buton the other hand, ‖b∗b‖ = ‖b‖2. Since σ(a) + it = σ(a+ it), we have that r + i(s+ t) ∈ σ(b), andso |r + i(s+ t)|2 ≤ ‖b‖2. Now, putting the two inequalities together, we have

r2 + (s+ t)2 ≤ ‖b‖2 ≤ ‖a2‖+ t2

and thus,r2 + s2 + 2st ≤ ‖a‖2 for any t ∈ R

which implies that s = 0.

Theorem 2.8.2 (Little Gelfand-Naimark). Let A be a unital commutative C∗-algebra. Then theGelfand transform is an isometric ∗-isomorphism of A onto C(A ).

Proof. Recall that the image of the Gelfand transform is a (unital) ∗-subalgebra of C(A ) thatseparates points. Applying Stone-Weierstrass, we have that the image of Gelfand tranform isdense in C(A ). But then isometric implies that the image is closed, so the image is the whole ofC(A ).

So far, we’ve been doing what can be called the “holomorphic functional calculus.” Now, ontothe “continuous functional calculus”:

Definition 2.8.3. An element a in a ∗-algebra is normal if a∗ commutes with a.

Example 2.8.4. a is unitary, i.e. a∗a = 1 = aa∗.

Proposition 2.8.5. Let A be a unital C∗-algebra (e.g. C∗-subalgebra of some B(H )). Let a ∈ Abe normal. Let f ∈ C(σB(a)), and let B be C∗-subalgebra generated by a and 1A . Then B ' σ(a).

If ϕ is a multiplicative linear functional on A then ϕ(a) ∈ σ(a). So the map B → σ(a) byϕ 7→ ϕ(a).

huh? This needs work.In terms of the isomorphism B

ˆ→ C(σ(a)), there is a unique element b ∈ B such that b = f .Call this element f(a) := b. Then f 7→ f(a) is an isometric ∗-isomorphism of C(σB(a)) with B. Inother words, we just built an inverse of the Gelfand transform.

Theorem 2.8.6 (Spectral permanence). Let A be a unital C∗-algebra, and let B be any unitalC∗-subalgebra of A . Then for any b ∈ B, σB(b) = σA (b).

Proof. That is, we must show that if b is invertible in A then it is invertible in B. First, considerthe case b = b∗, and let C be the unital C∗-subalgebra generated by b and 1A . It suffices to showthat σC (b) = σA (b). Suffices to show that if b is not invertible in C then it is not invertible in A .So suppose d ∈ A is an inverse for b. Since b is not invertible in C ' C(σC (b)). Then b must takevalue 0 at some point of σC (b). Then there is g ∈ C(σC (b)) such that ‖g‖∞ = 1 but ‖gb‖∞ < 1

2‖d‖ .

Now, letting c = g(b), so c ∈ C , we have ‖c‖ = 1, ‖cb‖ ≤ 12‖d‖ . Then, with bd = 1 = db, we have

1 = ‖c‖ = ‖c(bd)‖ = ‖(cb)d‖ ≤ ‖cb‖‖d‖ ≤ 1

2

15

which is a contradiction.For the general case, if b is not invertible in B, then b∗b is not invertible in B, so that b∗b is

not invertible in A , and so b is not invertible in A .

16

Chapter 3

Basic Pontryagin duality

3.1 Lecture 9/21/15 — ((Cc(−)/`1(−), ∗), Fforgetful : Alg(Banach) → SemiGrp)

Notation. By S a semigroup we mean a set with binary operation and a unit. That is, by asemigroup we’ll really mean a monoid throughout this lecture and the following lectures to come.

Remark 3.1.1. We first note the forgetful functor U from unital algebras to semigroups. Given aunital algebra A , denote by G(A ) the semigroup where operation is multiplication (so just forgetthe additive structure of A ). For example, for a vector space V , we can consider End(V ) as asemigroup by composition.

The purpose of this lecture is to construct a left adjoint F to the functor U , given a semigrouphomomorphism π : S → U (A ), in whichever categories the forgetful functor U is acting on.That is, if U : Alg/Algnormed/AlgBanach → SemiGrp, then we have a functor F : SemiGrp →Alg/Algnormed/AlgBanach such that (F ,U ) is an adjoint pair.

N.B. 3.1.2. Suppose we have a semigroup homomorphism π : S → A (technically, π : S →U (A )). Consider the subalgebra of A generated by the image of S under π. That is, consider the

elements of A of the form∑finite

cxπ(x) for cx in the field and x ∈ S. Note that there is an obvious

surjection to this set from Cc(S) where S is given the discrete topology. With an appropriatemultiplication, Cc(S) is in fact an algebra of that surjects (homomorphically) to the subalgebra ofA that π(S) generates:

Lemma 3.1.3. Let π : S → A be a semigroup homomorphism, and let B ⊂ A be the subalgebraof A generated by π(S). With discrete topology on S, consider Cc(S), where multiplication isdefined by convolution:

(f ∗ g)(z) :=∑xy=z

f(x)g(y)

Then (Cc(S), ∗) is a unital algebra (commutative if S is), and the natural map (Cc(S), ∗) → Bdefined by f 7→ πf :=

∑x∈S f(x)π(x) is an algebra homomorphism.

Proof. That ∗ makes Cc(X) into a unital algebra is obvious, and the linearity of the map is obvious.To check multiplicative, note that

πfπg =∑x,y

f(x)g(y)π(xy) =∑z

(∑xy=z

f(x)g(y))πz

Indeed, the map is also surjective.

17

Proposition 3.1.4. We have an adjoint pair (Cc(−),U ):

HomSemiGrp(S,U (A )) ' Homalg((Cc(S), ∗),A )

Proof. Note that S → U (Cc(S)) where x 7→ δx (since δx ∗ δy = δxy). This indeed gives anisomorphism HomSemiGrp(S,U (A )) ' Homalg((Cc(S), ∗),A ), and that this isomorphism is naturalin both factors is a routine check.

Let’s try to enlarge this adjoint pair to normed algebras or Banach algebra.

Remark 3.1.5. If V is a normed space, then End(V ) is a semigroup with composition as product,and consider a map of semigroups S → End(V ). We want the πx’s to be the bounded operators,i.e. we want πx ∈ B(V ) ∀x ∈ S. More generally, if A is a unital normed algebra, consider S

π→ A ,so ‖πx‖ is defined. Can ask, is there an upper bound M such that ‖πx‖ ≤ M for all x ∈ S? In

general, the answer is no (e.g. embed (R,+) in GL2(R) via a 7→[1 a0 1

]).

Remark 3.1.6. For S = Z, can use Jordan canonical form (if over C). Now, consider Z2 → End(Cn);can you classify this? These questions are interesting problems. When we get to Z3 (just 3!), theproblem is already the edge of what is known (at least, according to Rieffel).

N.B. 3.1.7. Suppose for Sπ→ A , there exists M such that ‖πx‖ ≤M ∀x ∈ S. Then for f ∈ Cc(S)

we have ‖πf‖ = ‖∑

x f(x)πx‖ ≤∑

x |f(x)|M = ‖f‖1M . Hence, re-normalizing A , we may assumethat ‖πx‖ ≤ 1 ∀x, in which case, ‖πf‖ ≤ ‖f‖1. Since ‖f ∗ g‖1 ≤ ‖f‖1‖g‖1, we thus have a normedalgebra Cc(S) with ‖ ‖1. For completeness, allowing countable support with finite norm, we have`1(S) as a Banach algebra with convolution and norm ‖ ‖1. In other words,

Proposition 3.1.8. We have an adjoint pair (`1(−),U ):

Hom‖·‖≤1SemiGrp(S,U (A )) ' Hom

‖·‖≤1

AlgBanach(`1(S),A )

whereHom

‖·‖≤1SemiGrp(S,U (A )) = π : S → U (A ) | ‖π(s)‖ ≤ 1

Hom‖·‖≤1

AlgBanach(`1(S),A ) = π : `1(S)→ A | ‖π(f)‖ ≤ ‖f‖1 (i.e. ‖π‖ ≤ 1)

3.2 Lecture 9/23/15 — The correspondence `1(S) ' HomSemiGrp(S,D)

Recall we defined the Banach algebra `1(S) as the completion of (Cc(S), ∗) with convolution as theproduct.

Example 3.2.1. If V is a Banach space, and we have a semigroup map π : S → End(V ) with‖π(x)‖ ≤ 1 for all x ∈ S, then we have a map π : `1(S)→ B(V ) by πf =

∑f(x)π(x). This element

π is called the integrated form of π. In other words, π is a “representation” of S on V , and π isa representation of `1(S) on V .

Definition 3.2.2. For any unital algebra A , and any vector space V , a (unital) homomorphismπ : A → End(V ) is called a representation of A on V .

Remark 3.2.3 (Restatement of Proposition 3.1.8). Let V be a Banach space. Here is what Propo-sition 3.1.8 says in this new terminology: Any “representation” π of S on V satisfying ‖π(x)‖ ≤ 1gives a representation π of `1(S) on V , and conversely, any representation π of `1(S) on V with‖π(f)‖ ≤ ‖f‖1 gives a “representation” π of S on V by π(x) = π(δx).

18

Proposition 3.2.4. Suppose S is commutative (so `1(S) is commutative). Then

`1(S) ' homomorphisms ϕ : S → D s.t. ϕ(1S) = 1 = HomSemiGrp(S,D)

where D is the closed unit disk in C.

Proof. We can look at `1(S), the maximal ideal space, in terms of (unital) homomorphisms ϕ :`1(S) → C. Recall that ‖ϕ‖ = 1 (Lemma 2.3.15), and thus |ϕ(δx)| ≤ ‖δx‖1 = 1. Since ϕ ismultiplicative, ϕ(δx)ϕ(δy) = ϕ(δxy). Thus, we have a semigroup homomorphism S → `1(S) → Dfrom ϕ. C-linearity and continuity of ϕ (i.e. ϕ ∈ `∞(S)) implies that ϕ(f) =

∑x f(x)ϕ(δx). Hence,

if conversely we have a map ϕ : S → D ⊂ C, the usual adjunction ϕ : `1(S) → C gives the samemap back by ϕ(f) :=

∑x f(x)ϕ(x).

Example 3.2.5. Let S = N. And consider `1(N) ⊂ `1(Z) (a unital Banach subalgebra). Considerδ1 ∈ `1(Z). This has an inverse in `1(Z) (i.e. δ−1), but not in `1(N). This is an example of failureof spectral permanence.

Let’s go further with this example. Let A = `1(N) is generated by δ1. Then if ϕ ∈ A , then ϕ isdetermined by ϕ(1) = ϕ(δ1) ∈ D. Conversely, for any z ∈ D, we can define ϕz ∈ A by ϕz(n) = zn;in other words, for f ∈ `1(N), z ∈ D, we have ϕz(f) =

∑m f(m)zm = f(z), which is an absolutely

convergent power series (for |z| ≤ 1). Furthermore, one can show that ϕ 7→ ϕ(1), z 7→ ϕz is ahomeomorphism A ' D.

Here is a neat application of the previous example along with the Gelfand theory:

Theorem 3.2.6. Let f be a function on D with absolutely convergent power series. If f(z) 6= 0for all z ∈ D, then 1/f ∈ C(D) has an absolutely convergent power series.

Proof. Recall that for any commutative unital Banach algebra A , and any a ∈ A , if a never takesvalue 0 on A , then 0 /∈ σ(a), so a is invertible. Now, for f ∈ `1(N), if f(z) 6= 0 ∀z ∈ D, then

f−1 ∈ `1(N), so that 1/f(z) = f−1(z).

Now, let’s move on to a case where we have S = G groups, again with discrete topology. When wehave a inverses, the picture becomes quite nicer:

Proposition 3.2.7. LetG be a discrete group and V a Banach space. Then (unital) representationsπ of `1(G) on V with ‖π‖ ≤ 1 correspond to isometric representations π of G on V (i.e. we haveπ : G→ Isom(V )).

Proof. Given a representation π of `1(G) on V , we have ‖π(δx)‖ ≤ 1, but since ‖π(δx)π(δx−1)‖ = 1,we have ‖π(δx)‖ = 1. To see that πx := π(δx) is in fact an isometry of V , note that if ‖πxv‖ < ‖v‖for some v ∈ V then ‖πx−1(πxv)‖ = ‖v‖ > ‖πxv‖, contradicting ‖πx−1‖ = 1. Conversely, if wehave such π : G → Isom(V ), it obviously extends to π of `1(G) on V with ‖π‖ ≤ 1 as integratedform.

Scholium. For V normed space, a representation π : G→ Aut(V ) is isometric iff ‖π(x)‖ = 1 ∀x ∈ G.

The following is immediate from the previous proposition and Proposition 3.2.4

Corollary 3.2.8. Let G be a commutative discrete group, then we have 1(G) ' Hom(G,T ) whereT = z ∈ D : |z| = 1.

Notation. We denote by G := Hom(G,T ), called the characters of G. That is, by definition1(G) := G. For ϕ ∈ G, f ∈ `1(G), we have ϕ(f) =

∑f(x)ϕ(x).

19

3.3 Lecture 9/25/15 — The transform `1(G)→ G (f 7→ f) is faith-ful

Recall that for G discrete and commutative, we had G := Hom(G,T ) '1(G).

Example 3.3.1. If G = Z, then under T ' R/2πZ via eit ↔ t, we have that Z ' Hom(Z, T ) ' Tvia eit ↔ (m 7→ eimt). Thus, viewing eit ∈ Z, we have f(eit) =

∑m f(m)eimt. In other words, for

f ∈ `1(Z), we get an absolutely convergent Fourier series f (defining f(t) = f(eit) =∑

m f(m)eimt).

Theorem 3.3.2. If f ∈ `1(Z), and if f(t) 6= 0 for all t ∈ T , so that the function f is invertible inC(T ), then 1/f has an absolutely continuous Fourier series.

Proof. The proof is more or less identical to the proof of Theorem 3.2.6, which is the `1(N) case.

Remark 3.3.3. We can take this technique further. For example, if G = Zn, we look at G =Hom(Zn, T ) = Tn, and the same result holds.

Question. For general abelian G, if f ≡ 0, then is f ≡ 0? In other words, is Hom(G,T ) big enough?We tackle this question by looking at factoring the argument through a Hilbert space, whose

richer structure makes the question easier to answer.

Aside: Sequence spaces `p

Definition 3.3.4. For 1 ≤ p <∞, and a set X, define

`p(X) = (ax)x∈X | ax = 0 for all but countably many x ∈ X and∑x

|ax|p <∞

The space `p(X) is a Banach space with the norm ‖(ax)‖p := (∑

x |ax|p)1/p. `∞(X) isdefined as the bounded sequences.

Proposition 3.3.5. Dual thing

Definition 3.3.6. Let G be a discrete group and 1 ≤ p < ∞. Define `p(G) to be the completionof Cc(G) with the norm ‖f‖p := (

∑x |f(x)|p)1/p. Note that `p(G) is a Banach space.

N.B. 3.3.7. There is a natural representation λ of G on `p(G) given left translation; that is, wehave λ : G → End(`p(G)) given by λxξ(y) = ξ(x−1y) (for ξ ∈ `p(G)). Clearly, λx is an isometryfor all x ∈ G (and hence ‖λx‖ ≤ 1 ∀x ∈ G). Thus, by our usual adjunction we get a naturalrepresentation of `1(G) on `p(G) given by the integrated form: for f ∈ `1(G) and ξ ∈ `p(G), wehave (λfξ)(y) =

∑x f(x)(λxξ)y =

∑x f(x)ξ(x−1y).

Definition 3.3.8. Let G be a discrete group and 1 ≤ p < ∞. Then the natural left regularrepresentation λ of `1(G) on `p(G) (i.e. λ : `1(G)→ End(`p(G)) is given by convolution:

(λfξ)(y) =∑x∈G

f(x)ξ(x−1y)

Remark 3.3.9. Note that this representation is unital (obvious). Moreover, λ is faithful, since ifλf = 0, then λfδe = f ∗ δe = f = 0.

20

Now, consider λ of `1(G) on `2(G), which is a Hilbert space. The Banach algebra B(`2(G))has a natural ∗-structure by adjoints. Since 〈f, g〉 =

∑y f(y)g(y), we have (λx)∗ = λx−1 since∑

y f(x−1y)g(y) =∑

y f(x)g(xy). In other words, the representation λ of G on `2(G) is unitary.

This tells us how to lift λ to a ∗-representation of `1(G) on `2(G):

Proposition 3.3.10. For f ∈ `1(G), define f∗ ∈ `1(G) by f∗(x) = f(x−1). This gives a ∗-structure on `1(G), and this structure lifts the left regular representation λ of `1(G) on `2(G) to a∗-representation.

Proof. That ∗ as defined gives a ∗-structure on `1(G) is easy to check. To see that this lifts λ to a∗-representation, we compute that (λf )∗ =

∑x f(x)λx−1 =

∑x f(x−1)λx = λf∗ .

Remark 3.3.11. In general, given a unitary representation π of G on a Hilbert space H , we define∗ on `1(G) in the same way, and we can lift π to ∗-representation (i.e. (πf )∗ = πf∗).

We are now ready to answer the question (the answer is yes, G is big enough):

Theorem 3.3.12. Let G be a commutative discrete group. If f ≡ 0 for f ∈ `1(G), then f = 0.

Proof. Have this proof reviewed. Consider the natural ∗-representation λ of `1(G) on `2(G). DefineC∗(G) to be the norm closure of λf : f ∈ `1(G) in B(`2(G)) (Aside: we also denote this as C∗r (G)where “r” stands for “reduced” but since G is commutative it is amenable so C∗(G) = C∗r (G) here).This C∗(G) is a unital commutative C∗-subalgebra of B(`2(G)), and we have λ : `1(G) → C∗(G),and from faithfulness we know this is injective with a dense image.

By Little Gelfand-Naimark we have C∗(G) ' C(X) for some X compact Hausdorff; in fact,

recall that X = C∗(G). Any point of X gives a multiplicative linear functional on C∗(G), and so

on `1(G) by precomposing with λ. This map λ∗ : C∗(G) → G is injective since `1(G) is dense inC∗(G). Now, if f ∈ `1(G) and f ≡ 0, then f(λ∗ϕ) = 0 for all ϕ ∈ X, so that ϕ(λf ) = 0 ∀ϕ ∈ Xand hence λf = 0 (∵ for unital commutative C∗-algebra the Gelfand transform is faithful), andthus f = 0 (∵ λ is faithful).

(One can of course show that in fact (C∗(G)) = G)

Remark 3.3.13. Some bad news. Consider `1(G) for G abelian. What are its closed ideals? Isevery closed ideal in `1(G) exactly the intersection of the maximal ideals that contain it? (well,that is the case for C(X) when X compact). In other words, does spectral synthesis hold? Answeris unfortunately No. Not even for `1(Z).

Example: G = Rn for n ≥ 3. Laurant Schwartz. Then 1959 Mallioven.

Exercise 3.3.14. Check that for a discrete set X, `∞(X) with weak-∗ topology on `∞(X) is thetopology of uniform convergence on compact (finite) subsets.

3.4 Lecture 9/28/15 — A first sight of Pontryagin duality

Recall that for G discrete commutative group, we have G := Hom(G,T ). On G we have weak-∗topology for G ⊂ `∞(G) ' (`1(G))∗. Note that the weak-∗ topology here is in fact the topologyof uniform convergence on compact sets. Moreover, note that with pointwise operations, G =Hom(G,T ) is a Abelian group.

Proposition 3.4.1. For G discrete commutative, G = 1(G) with topology of uniform convergenceon compact sets, which in this case equal weak-∗ topology, is a compact topological group, calledthe dual group of G.

21

Proposition 3.4.2. Let G,H be discrete commutative groups, and let f : G → H be a homo-morphism. Then precomposition − f defines a continuous homomorphism f : H → G. In other

words, G → G is a contravariant functor TopGrpdiscrete → TopGrpcpt. Moreover, G 7→ G defines a

natural isomorphism.

Remark 3.4.3. Thus, this gives a concrete realization of the abstract dual of the category of dis-crete commutative group as the category of compact topological groups. (Special case for discretecommutative group of the Pontryagin duality theorem).

We now consider the case when G is no longer discrete. Let G be a locally compact commutativetopological group.

Definition 3.4.4. For G commutative (locally compact, Hausdorff) topological group, we definethe dual group as G := ϕ : G→ T | ϕ continuous.

Proposition 3.4.5. Let G be a locally compact (Hausdorff) commutative topological group. Whenequipped with the topology of uniform convergence on compact sets, G is also a topological group.

Proof. ‖ϕψ − ϕ0ψ0‖∞−K (sup over compact set K, which are semi-norms). Well, we have that‖ϕψ − ϕ′ψ′‖ ≤ ‖ϕψ − ϕψ′‖ + ‖ϕψ′ − ϕ′ψ′‖ ≤ ‖ϕ‖‖ψ − ψ′‖ + ‖ψ′‖‖ϕ − ϕ′‖. Now, since ‖ϕ‖ =‖ψ‖ = 1, it is easy to check that nets converge. For the inverse, we use the trick ‖ϕ−1 − ψ−1|| =‖ϕ−1(ψ − ϕ)ψ−1‖.

Let G be a topological group (Hausdorff). Let π be an isometric action of G on a normed spaceV . How should we define π to be continuous? Well, Aut(V ) has the usual sup norm, and so wecan define π to be norm continuous if xλ → x in G, then ‖πxλ − πx‖ → 0. However, this is toostrong:

Example 3.4.6. Consider G = R acting on C∞(R) by translation. That is, for x ∈ R, we haveπx(f)(y) = f(y − x). Consider 1

n → 0. We see that π1/n 6→ π0 in End(C∞(R)) by consideringfunctions fn = “spike of height 1 on interval [−1/n, 1/n]” since ‖π1/n(fn) − π0(fn)‖ = 1 = ‖fn‖.Thus, this (rather natural) action is not norm continuous.

The more appropriate continuity condition to look as is the following:

Definition 3.4.7. Say that π is strongly continuous if for each v ∈ V , the function G → Vdefined by x 7→ πx(v) is continuous.

Proposition 3.4.8. To show that an action of π of G on V is strongly continuous, it suffices toshow that it is strongly continuous at e ∈ G.

Proof. ‖πxv− πx0v‖ = ‖πx0(πx−10 xv− v)‖ ≤ ‖πx0‖‖πx−1

0 xv− πev‖ and x→ x0 ⇐⇒ x−10 x→ e.

Proposition 3.4.9. Let π be an isometric action of G on V . If π is strongly continuous on a densesubspace W of V , then it is strongly continuous on V .

Proof. Given v ∈ V , and ε > 0, choose w ∈ W such that ‖v − w‖ < ε. Then ‖πxv − v‖ =‖πxv − πxw‖+ ‖πxw − w‖+ ‖v − w‖.

Example 3.4.10. To prove strong continuity of the action of R on on C∞(R), it suffices to showstrong continuity on Cc(R) at 0. We’ll do this in a more general setting next time.

22

3.5 Lecture 9/30/15 —

It is not always so simple to check strong continuity.

Definition 3.5.1. For G a topological group and M a topological space, by an action of Gon M we mean a group homomorphism α : G → AutTop(M) such that G ×M → M given by(x,m) 7→ αx(m) is jointly continuous.

Theorem 3.5.2. Let G be a locally compact (Hausdorff) group, and let α be an action of G ona locally compact (Hausdorff) space M . Define an action α of G on C∞(M) by (αx(f)(m)) =f(αx−1(m)). This action on C∞(M) is strongly continuous.

Proof. Suffices to show strong continuity on Cc(M), at eG. Suppose it is not strongly continuous,then ∃f ∈ Cc(M) such that αx(f) 6→ f as x → eG. In other words, ∃ a net xλ in G such thatxλ → eG, and ‖αxλ(f)−f‖∞ ≥ ε ∀λ for some ε > 0. Then, ∃mλ ∈M with |f(αx−1

λ(mλ))−f(mλ)| ≥

ε. Thus, either f(mλ) or f(αx−1λ

(mλ)) is nonzero. That is, letting K = supp(f), which is compact,

so either mλ or αx−1λ

(mλ) ∈ K. Since G is locally compact, we can assume that xλ ⊂ C, for a

compact neighborhood C of e with C−1 = C. Then, by joint continuity of α, we have αC(K) iscompact. Thus, mλ ∈ αC(K) for all λ. Take a subnet mΛ that converges to some m0 ∈ M . Byjoint continuity, αx−1

Λ(mΛ)→ m0. Since f is continuous, f(αx−1

Λ(mΛ))→ f(m0) →f(mΛ), so that

|f(αx−1Λ

(mΛ))− f(mΛ)| → 0, contradicting our condition on mλ.

Aside: Measure theory refresher

Measure, Borel Measure, Integration, σ-finite, finiteIntegration theorems: MCT, DCT, Fatou, and FubiniRadon measure, regular measure, locally finite, Riesz-Markov-KakutaniMutually singular, absolute continuityLebesgue-Radon-Nikodym

Definition 3.5.3. Let M be a locally compact space. By a positive Radon measure on M , wemean a positive linear functional µ on Cc(M). (positivity means: if f ≥ 0 then µ(f) ≥ 0).

Remark 3.5.4. These gives regular Borel measures on M . A Borel measure on M is a measureµ defined on the σ-ring generated by compact subsets of M , and having finite values on compactsubsets.

Definition 3.5.5. Let α be an action of G on M (locally compact), and let µ be a positive Radonmeasure of M . Then µ is α-invariant if

µ(αx(f)) = µ(f) ∀x ∈ G, ∀f ∈ Cc(M)

Remark 3.5.6. Now, given a positive Radon measure on M , we can form Lp(M,µ) (1 ≤ p < ∞).We do this by defining ‖f‖p := µ(|f |p)1/p for f ∈ Cc(M). By usual analysis, this is at least asemi-norm, and we take the abstract completion to be Lp(M,µ). Note that by construction Cc(M)is dense in Lp(M). To see this without the construction but from Borel measure, we need theregularity condition.

Theorem 3.5.7. If α is an action of G on M (locally compact), and if µ is an α-invariant positiveRadon measure on M , then we can define (extend) an isometric action α of G on Lp(M) by(αx(f))(m) := f(αx−1(m)) for f ∈ Cc(M) and for f ∈ Lp(M) by taking a limit. This action isstrongly continuous.

23

Proof. Need that if xλ → e and f ∈ Cc(M), then ‖αxλ(f) − f‖p → 0. Well, we know uniformconvergence but this does not imply convergence in the Lp norm. We can assume that xλ ∈ C withC compact neighborhood of e. All the αxλ(f) have support in αC(K) where K is the support off , and αC(K) has finite measure (since compact). On a finite measure space, and we have uniformconvergence implies convergence in Lp. So we have convergence we desired.

3.6 Lecture 10/2/15 — Haar measure and modular function

Notes by hand: TODO: Transfer here

3.7 Lecture 10/5/15 —

Let G be a locally compact group, and let dx be a choice of a left Haar measure. Then for Lp(G, dx),and the operator of left translation gives a strongly continuous action of G on these Banach spaces.

Definition 3.7.1. Let π be a strongly continuous isometric action of G on a Banach space V .The integrated form for f ∈ Cc(G) (or L1(G)) is the operator πf on V defined by πf (v) =∫f(x)πx(v)dx.

Remark 3.7.2. Weak integrals: for ϕ ∈ V ∗, can consider x 7→ ϕ(f(x)πx(V ))∗.

N.B. 3.7.3. Note that ‖πf (v)‖ ≤∫|f(x)|‖πx(v)‖dx = ‖f‖1‖v‖, and thus ‖πf‖ ≤ ‖f‖1. Moreover,

πf (πg(v)) = π(∫g(x)πx(v))

=∫f(y)πy(

∫g(x)πx(v))

=∫∫

f(y)g(x)πyx(v)dxdy=∫f(y)

∫(g(y−1x)πx(v)dxdy

=∫

(∫f(y)g(y−1x)dy)πx(v)dx

(since bounded operators commute with integrals?? and last equality by Fubini’s theorem). Fubini’sTheorem for compact support: consider spanf(x)g(y) : f ∈ Cc(X), g ∈ Cc(Y ), which is ansubalgebra of Cc(X × Y ) that is dense for ‖ ‖∞, but also dense for the inductive limit topology.Thus, we define (f ∗ g)(x) =

∫f(y)g(y−1x)dy, which really is λfg for λf left translation by f .

Definition 3.7.4. Define the convolution so that πfπg = πf∗g.

N.B. 3.7.5. Convolution is associative. ‖f∗g‖1 ≤∫∫|f(y)||g(y−1x)|dydx =

∫|f(y)|

∫|g(x)|dxdy =

‖f‖1‖g‖1. Thus, we see that L1(G) is a Banach algebra. Any strongly continuous isometric repre-sentation of G gives a representation of L1(G). One difficulty: the identity element should be δeG ,but if G is not discrete, the Haar measure of points is zero.

So let’s try to deal with Banach algebras A without 1. Well, we can always adjoin an identityelement. That is, A ⊕C where (a, α)(b, β) = (ab+aβ+αb, αβ). We denote this as A ∼. The norm isthat ‖(a, α)‖ = ‖a‖+ |α|. Now, if A = C∞(X) for locally compact, and add the constant function,but note that the norm above is really not what you get from compactification. Unfortunately, thisis the only thing we can do in general, but it might really gives us something weird. A is a 2-sidedideal in A ∼.

Approximate identity: in the case A = C∞(X), we can consider a net of constantly 1 oncompact set functions. That is, fK where fK ∈ [0, 1] and f |K ≡ 1. Given g ∈ C∞(X), we nowthat that limK fKg = g.

24

Definition 3.7.6. Let A be a normed algebra. By a left approximate identity for A , wemean a net eλ ⊂ A such that limλ eλa = a ∀a ∈ A . We likewise define right approximateidentity and 2-sided approximate identity. For any of these three notions, we say boundedif ‖eλ‖ ≤M ∀λ, ∃M . By an approximate identity of norm 1, we mean that ‖eλ‖ ≤ 1 ∀λ.

3.8 Lecture 10/7/15 —

Proposition 3.8.1. For any locally compact group G, the Banach algebra L1(G) has a 2-sidedapproximate identity of norm 1.

Proof. Let Λ be the collection of open neighborhoods of eG of compact closure, ordered by reverseinclusion. For each λ ∈ Λ, choose eλ ∈ Cc(G) such that 0 ≤ eλ, supp(eλ) ⊂ λ, ‖eλ‖1 =

∫f(x)dx =

1, and eλ(x−1) = eλ(x) for all x. We claim that eλ is an approximate identity.To check the claim, for f ∈ L1(G) (or suffices to say ∈ Cc(G)), (eλ ∗ f)(x) =

∫eλ(y)f(y−1x)dy

and so ((eλ ∗ f) − f)(x) =∫eλ(y)f(y−1x)dy −

∫eλ(y)f(x)dy =

∫eλ(y)(Lyf − f)dy(x). Since the

action of G on L1(G) by left-translation is strongly continuous, given any ε > 0, there exists aneighborhood λ0 of eG such that if y ∈ λ0 then ‖Lyf − f‖1 < ε. So, for any λ with λ ≥ λ0

(that is λ ⊂ λ0), we have that ‖eλf − f‖ = ‖∫eλ(y)(Lyf − f)dy‖ = ‖

∫λ eλ(y)(Lyf − f)dy‖ ≤∫

λ ey(y)dy · ‖Lyf − f‖1 < 1 · ε = ε.Likewise, we get that (f ∗eλ−f)(x) =

∫f(y)eλ(y−1x)dy−

∫eλ(y)f(x)dx =

∫f(xy)eλ(y−1)dy−∫

f(x)eλ(y) =∫

(f(xy) − f(x))eλ(y)dy =∫

(Ryf − f)(x)eλ(y), and the ‖‖1 of this is ≤ ‖Ryf =f‖1

∫λ eλ(y) (where y ∈ λ very close to eG, hence ‖Ryf − f‖1 ≤ ε by the fact that ∆ : G → R+ is

continuous).

Remark 3.8.2. This is often called the “approximate δ-function” at eG.

Definition 3.8.3. Let A be a normed algebra, and let (V, π) be a representation of A on V (anormed space). In other words, we have a homomorphism π : A → B(V ) with ‖π(a)‖ ≤ ‖a‖. Thenwe say that this representation is non-degenerate if spanπ(a)v : a ∈ A , v ∈ V is dense in V .

Remark 3.8.4. For any algebra, it is important to look at its left-regular representation of itself(representation by left multiplication).

Proposition 3.8.5. Let A be a normed algebra with approximate identity of norm 1 (or atleast bounded), and let (V, π) be a representation of A on a normed space V . Then (V, π) isnon-degenerate iff for any given approximate identity eλ one has π(eλ)v → v ∀v ∈ V .

Proof. If eλv → v for all v ∈ V , then clearly the representation is non-degenerate. Conversely,if a ∈ A and v ∈ V , then π(eλ)(π(a)v) = π(eλa)v → π(a)v (since eλ approximate identity).Note that this holds for any v of the form π(a)v, so if we have dense-ness of the span we have thecondition we wanted. (Written out: for any v ∈ V , and any ε > 0, choose v0 =

∑π(aj)vj with

‖v − v0‖ < ε/3, then ‖π(e)λ)v − v‖ ≤ ‖π(eλ)v − π(eλ)v0‖+ ‖π(eλ)v0 − v0‖+ ‖v0 − v‖).

Proposition 3.8.6. Let (V, π) be a strongly continuous action of G on V by isometries, and letπ also denote its integral form. Then this integrated form is a non-degenerate representation ofL1(G).

Proof. Let eλ be an approximate δ-function for L1(G), and let v ∈ V , then ‖π(eλ)v − v‖ =‖∫eλ(x)πxvdx −

∫eλ(v)vdx‖ ≤

∫eλ(x)‖πxv − v‖dx =

∫λ eλ(x)‖πxv − v‖, and now by strong

continuity, for λ small enough neighborhood of eG, we have ‖πxv − v‖ ≤ ε.

25

3.9 Lecture 10/9/15 —

Theorem 3.9.1. Any non-degenerate representation (V, π) of L1(G) comes as an integrated formfrom a strongly continuous action of G on V .

Remark 3.9.2. For G discrete, we had G → `1(G), so we just restricted the representation on `1(G)to G and we were done. We can’t do same thing here.

Before proving the theorem, let’s observe a statement that helps us in general: ‘Almost allnon-unital rings, algebras have a “Stone-Cech” compactification.’

N.B. 3.9.3. Let A = C∞(R) ⊂ Cb(R) ⊃ C∞((0, 1)). For a ring R, and and an (2-sided) idealI ⊂ R, we say that I is an essential ideal in R if for any r ∈ R, rI ≡ 0 ≡ Ir =⇒ r = 0 (i.e.AnnR(I) = 0). For example, C∞(R) is essential in Cb(R) but not C∞((0, 1)). Look for the largestunital ring or algebra that contains A as an essential ideal. If S is any ring containing A as anideal, then for s ∈ S, can define the operators Ls and Rs on A by Lsa = sa, Rsa = as. Callthe properties Ls(bc) = Ls(b)c left multiplier, and Rs(bc) = b(Rsc) right multiplier,and notebLsc = (Rsb)c.

Definition 3.9.4. A multiplier (double centralizer) of A is a pair (S, T ) where S and T areoperators on A , with S(ab) = (S(a))b, T (ab) = a(Tb), a(Sb) = (Ta)b.

There is a natural way of defining a + and · on the set of all multiplier so that the set forms aring / algebra. Denote this set (ring/algebra) by M(A ). Then there is a natural homomorphismA →M(A ) by a 7→ (La, Ra). The image is an ideal in M(A ), and in fact an essential ideal. Now,we check if A → M(A ) is injective: well, if a ∈ ker, then aA = 0 = A a (so the “almost” in thesentence means this doesn’t occur).

If A is a normed / Banach algebra, we require (S, T ) to be bounded operators (in fact, don’t

need this for Banach case) and ‖S‖ = ‖T‖, and so define ‖(S, T )‖ defn= ‖S‖ = ‖T‖. Then M(A )is a normed algebra (and Banach if A is). (1A , 1A ) is an identity element for M(A ). If A hasan approximate identity of norm 1, eλ, then we can use it to show that ‖(La, Ra)‖ = ‖a‖ viaLaeλ = aeλ → a. Moreover, if A is a ∗-normed algebra then we can define ∗ on M(A ).

Now, let G locally compact, not discrete, and consider L1(G). What is M(L1(G))?

Let S be a locally compact semigroup. Let µ, ν be finite signed Radon measures, and define (forf ∈ Cc(S)) (µ∗ν)f :=

∫f(xy)dµ(x)dν(y). Let M (S) be the vector space of signed Radon measures.

Then the convolution above gives a associative product on M (S). Define ‖µ‖ = total variation norm(For µ R-valued, write µ = µ+ − µ− disjoint. |µ| = µ+ + µ−, and define ‖µ‖ =

∫S 1d|µ| = |µ|(S)).

For this norm, M (S) forms a Banach algebra.

Let G be locally compact group. There is a homomorphism of L1(G) → M (G). given byf 7→ f(x)dx (where dx is the Haar measure). This is certainly injective. The surprise is that L1(G)is a 2-sided ideal in M (G). (in fact, essential ideal)

Theorem 3.9.5 (Weidel). M (G) 'M(L1(G)).

Proof of theorem 19.1? G not discrete, and let B(G) = L1(G) ⊕ `1(G) ⊂ M (G). Can define(δz ∗ f)(h) =

∫h(xy)dδz(x)f(y)dy =

∫h(zy)f(y)dy =

∫f(z−1y)h(y)dy = (Lzf)dy(h) for any

h ∈ Cc(G). From this, we see that L1(G) is a 2-sided ideal in B(G). Measures mutually singularand what?? B(G) is a unital Banach algebra (with norm ‖f +ϕ‖ = ‖f‖1 + ‖ϕ‖1 containing L1(G)as an essential ideal. G → B(G) with x 7→ δx.

26

So, the idea is that given a representation (V, π) of L1(G), extend it to B(G), and then restrictit to G. How to extend? Next time.

3.10 Lecture 10/12/15 —

We have been trying to prove the following:

Theorem 3.10.1. Let A be a normed algebra, and let I be an ideal (2-sided) in A , and assumethat I has a (2-sided) approximate identity eλ for itself. Then every non-degenerate boundedrepresentation (V, π) has a unique extension to (V, π) a bounded representation of A on V .

Proof. Uniqueness: Suppose an extension (V, π) of (V, π) exists. Then for any a ∈ A and d ∈ Iand v ∈ V , we have that πa(πd(v)) = πad(v) = πadv. This implies that π is determined already onelements of the form πd(v), and the non-degeneracy implies that elements of the form

∑finite πd(v)

is dense in A . And since πa need be continuous, πa is determined completely if it exists.Existence: first define π first on spanπIV by πa(

∑πdjvj) =

∑πadjvj . Now, is this well-

defined? Well, suppose we have∑πdjvj =

∑πd′jv

′j . Suffices to show that if

∑πdjvj = 0, then∑

πadjvj = 0. Well, if∑πdjvj = 0, then

πadjvj = lim∑

πa(eλdj)vj = lim∑

πaeλπdjvj = limπaeλ(∑

πdjvj) = 0

Now, πa is well-defined on 〈πIV 〉, and

‖πaeλ(∑

πdjvj)‖ ≤ ||πaeλ‖‖∑

πdjvj‖ ≤ ‖aeλ‖‖∑

...‖ ≤ ‖a‖‖eλ‖‖∑

...‖ ≤ ‖a‖‖∑

...‖

so πa on 〈πIV 〉 has a norm ≤ ‖a‖, and so πa extends to an operator on all of V with ‖πa‖ ≤ ‖a‖.One checks that π : A → End(V ) is indeed a (normed) algebra homormophism (and is of coursenon-degenerate).

N.B. 3.10.2. Let’s apply this to G locally compact and L1(G). Let (V, π) be a non-degeneraterepresentation of L1(G). Let A = L1(G) ⊕ `1(G). Note that I = L1(G) was shown to be 2-sidedideal with approximate identity. Thus, by the theorem, (V, π) extends uniquely to a representation(V, π) of A on V . Now, since G → `1(G) ⊂ A , we can restrict π to G to get a (isometric)representation π of G on V . Now, we need show that π on G is strongly continuous. Since action isisometric, it suffices to check on a dense subset, which we take to be 〈πL1(G)V 〉. So, given

∑πgjvj ,

for every x ∈ G, we see that πx(∑

gjvj) = πδx(

∑gjvj) =

∑πδx∗gjvj , but we saw that G→ L1(G)

by x 7→ δx ∗ gj was norm continuous, which implies that x 7→ πδx∗gj is norm continuous.

N.B. 3.10.3. Fact: Let (V, π) for V Banach space be a strongly continuous representation of Gon V by isometries. Let π′ be the integrated form of π, which is a non-degenerate representationof L1(G) on V . Form the extension π′ of π′ to L1(G)⊕ `1(G), and restrict it to G. This restrictionis actually (π, V ).

Theorem 3.10.4. There is a natural bijection between the strongly continuous representations ofG by isometries on Banach spaces and non-degenerate bounded representations of L1(G).

Let G be a locally compact commutative group, then L1(G) (values in C) is a commutativeBanach algebra. Let L1(G)∼ be L1(G) with the identity element adjoined. So, L1(G)∼ is acommutative Banach algebra with 1, and so we can consider its maximal ideal space (multiplicativelinear functionals). One such functional is sending L1(G) to 0, but all the others are 6≡ on L1(G),

27

mapping onto C. In this sense, we have a representation of L1(G) on V = C (obviously non-degenerate). Thus, this corresponds to a strongly continuous representation of G on V = C, i.e.corresponds to continuous homomorphism G→ C.

3.11 Lecture 10/14/15 —

Let G be locally compact commutative. Consider L1(G) and (L1(G)) ' G where G = Hom(G,T )

(continuous maps) . Last time, we showed that for L1(G) ' L1(G) ⊕ Cδe (adjoining the identity

element), we have L1(G) = ϕ : (L1(G)/L1(G))→ C∪G, where G as integrated representations ofL1(G). Giving this set the weak-∗ topology, we have compactness. Consider the Gelfand transformf 7→ f(ϕ) =

∫f(x)ϕ(x)dx for f ∈ L1(G), f ∈ C∞(G). This is called the Fourier transform.

Since f ∗ g = f g, the Fourier transform L1(G) → C∞(G) is a ring homomorphism. Moreover,if we define ∗ on L1(G) by f∗(x) = f(−x), then f∗(ϕ) =

∫f∗(x)ϕ(x)dx =

∫f(−x)ϕ(x)dx =∫

f(x)ϕ(−x)dx =∫f(x) · ϕ(x)dx = f(ϕ). In other words, f∗ = f . So, the image of L1(G)

in C∞(G) is closed under complex conjugation, separates points, and separates points from 0.Applying Stone-Weierstrass theorem, the image of L1(G) is dense in C∞(G).

Question: if f = 0, then must f ≡ 0?

Example 3.11.1. Let G = R, then G contains ϕs(t) = e2πist. So, f(x) =∫f(t)e2πistdt (the

original Fourier transform)

Let V be a finite dimensional vector space over R, and let G = V , forgetting scalar multipli-cation. For any ξ ∈ V ∗, ϕξ(v) = e2πi〈v,ξ〉 where 〈v, ξ〉 = ξ(v). Since V ∗ ' V (but not natural sochoose an inner product and isomorphism 〈−, v〉 ' v), so that ϕξ(v) = e2πiv·w

Given G = R, why is every continuous homomorphism ϕ : R→ T of the form ϕs(t) = eits for somes?

Proof. Well, we know that ϕ(0) = 1, and ϕ is continuous. Then we can choose a ∈ R close to 0such that A :=

∫ a0 ϕ(t)dt 6= 0. Then ϕ(s)A =

∫ a0 ϕ(t+ s)dt =

∫ a+ss ϕ(t)dt. Since A 6= 0, we thus see

that d/ds exists so that ϕ is differentiable, and ϕ′(s) = ϕ(a + s) − ϕ(s) = ϕ(s)(ϕ(a) − 1). Thus,ϕ′(s) = Cϕ(s) where C = A−1(ϕ(a)− 1). The solution for this is ϕ(s) = eCs = eits (since C needbe purely imaginary to be on the circle).

Need: on Rn = G, why does the weak-∗ topology agree with the usual topology of Rn? Well,f(s) =

∫f(t)eistdt is continuous for the usual topology in Rn, and f ∈ C∞(Rn) for usual topology.

Try f = χ[a,b]. (Riesz-Wieyl Lemma). Let Rn be the 1-point compactification of Rn with the usual

topology, and Rn′

be one with the weak-∗ topology. There is an obvious map Rn → Rn′

that iscontinuous:

Rn // Rn′

L1(G)

<<

28

3.12 Lecture 10/16/15 —

Recall: G locally compact commutative. Consider L1(G) and L1(G) = G = Homcts(G,T ). Somequestions we need ask in this case: Is G big? If f ∈ L1(G) and f ≡ 0, then is f ≡ 0? In the discretegroup case, we handled this by looking at left regular representations on the Hilbert space. Well,we have a left regular representations π on Lp(G) (1 ≤ p < ∞) that are stongly continuous, andhave the integrated forms as representations on L1(G).

Proposition 3.12.1. For each 1 ≤ p <∞, the representation π of L1(G) on Lp(G) is faithful.

Proof. For given p, and given f ∈ L1(G), for any ξ ∈ Cc(G) ⊂ Lp(G), we have that

(πfξ)(x) =

∫f(y)ξ(y−1x)dy = (f ∗ ξ)(x)

and we can view f ∗ ξ as an element in L1(G) or Lp(G) (doesn’t matter). Now, let ξ run throughan approximate identity for L1(G), so f 6≡ 0 almost everywhere then there’s a ξ such that f ∗ ξ 6≡ 0almost everywhere. Hence, πfξ 6= 0, so that πf = 0 (eh... not equal??).

Now, let’s look at the case p = 2. Let H be a Hilbert space, and let u be a unitary representationof G on H (strongly continuous), so it has integrated form. For f ∈ L1(G), the integrated formuf “ =

∫f(x)uxdx”, and (uf )∗ =

∫f(x)u∗xdx =

∫f(x)ux−1dx =

∫f(x−1)∆(x−1)uxdx. So define

f∗(x) := f(x−1)∆(x−1). Check that ‖f∗‖1 = ‖f‖1. So, L1(G) is a ∗-algebra, and for any unitaryrepresentation (u,H ), we have (uf )∗ = uf∗ .

Let λ be the left regular representation of L1(G) on L2(G) (which is faithful). The image ofL1(G) under λ will be a ∗-subalgebra of B(L2(G)). Its norm closure will be a C∗-algebra, denotedby C∗r (G). By faithfulness, λ is an injection of L1(G) onto a dense ∗-subalgebra of C∗r (G). IfG is commutative, then the L1(G) is commutative, and so C∗r (G) is commutative, and will nothave an identity element. Let C∗r (G)∼ be the C∗-subalgebra of B(H ) generated by C∗r (G) andIdH . Algebraically, C∗r (G) = C∗r (G) ⊕ C IdH (However, using ‖a + α1‖ = ‖a‖ + |α| is the wrongnorm: consider C∞(R)). Use the norm from B(H ). C∗r (G)∼ is a unital commutative C∗-algebra.Thus, by Little Gelfand-Naimark, C∗r (G) ' C(X) for some X compact. Considering the idealC∗r (G) ⊂ C∗r (G)∼, if we remove the multiplicative linear functional “∞” having C∗r (G) as kernel,we get C∗r (G) ' C∞(X \ ∞) = C∞(Y ) (Y locally compact space).

For any h ∈ C∞(Y ), with h 6≡ 0, there is y ∈ Y with h(y) 6= 0, we have ϕy, which is multiplicativelinear functional of evaluating at y (i.e. ϕy(h) 6= 0). Now, if h = λf with f 6≡ 0, so ϕy(λf ) 6= 0, ϕyrestricts to a multiplicative linear functional in λ(L1(G)), so gives an element of (L1(G))∨, as suchis a ϕ ∈ (L1(G))∨ with ϕ(f) 6= 0, i.e. f(ϕ) 6= 0. Now, viewing ϕ as in G, so we see ∃ϕ ∈ G withf(ϕ) 6= 0.

Remark 3.12.2. (Didn’t we do this already?). Give G = Homcts(G,T ) a topology of uniformconvergence on compact sets. G is then a topological group. Theorem: this topology agrees on Gwith the weak-∗ topology from viewing G ⊂ (L1(G))∨ = L∞(G).

If G and H are locally compact commutative groups, G, H are locally compact groups. By

precomposition, we have that G 7→ G is a contravariant functor. Moreover, consider G, G,G.

There is a canonical map G → G given by x 7→ ξx where ξx(ϕ) := ϕ(x) that is injective. The

Pontryagin duality theorem says that this map is in fact surjective as well (and furthermore, ahomeomorphism.

Thus, the abstract dual to the category of locally compact commutative groups has a concreterealization as itself, given by G 7→ G.

29

3.13 Lecture 10/19/15 —

Lemma 3.13.1. If G is compact commutative, then G is discrete. If ϕ ∈ G, then since G iscompact, then ϕ ∈ L1(G), so that we can consider

∫G ϕ(x)dx =

∫ϕ(x+ y)dx = ϕ(y)

∫ϕ(x)dx, so

if ϕ 6= 1 then∫ϕ(x)dx = 0.

Corollary 3.13.2. If ϕ,ψ ∈ G, and ϕ 6= ψ, then 〈ϕ,ψ〉L2(G) :=∫ϕ(x)ψ(x)dx = 0 unless ϕψ ≡ 1,

i.e. ψ = ϕ. If we view ϕ ∈ L1(G), then ϕ(ψ) =∫ϕ(x)ψ(x)dx = 0 unless ϕ = ψ. Thus, ϕ = cδϕ.

So, ϕ is open and closed in G.

Recall that the abstract dual of the category of compact abelian groups has concrete realizationas the category of discrete abelian groups (and vice versa).

What about non-commutative groups? How does Pontryagin duality work? What would thedual object be?

Well, in our case, for A a commutative Banach algebra, we had A → C∞(A ), which somehowhad a group structure. How? Well, for a commutative group G, the elements of G were actuallyHom(G,T ), and these elements “are” the one-dimensional unitary representations of G on Hilbertspaces. For any G, we can consider unitary representations of G on Hilbert spaces, and these willgive representations of L1(G). Given two representations (H , u), (K , v), can form H ⊗K (e.g.L2(X,µ) ⊗ L2(Y, ν) ' L2(X × Y, µ × ν)), and form the tensor product of representations u ⊗ vdefined as (u⊗ v)x(ξ ⊗ η) = uxξ ⊗ vxη, which gives a representation of L1(G) on H ⊗K .

If A is a finite dimensional k-algebra, and if M,N are two A -modules, in general there isno way to make M ⊗k N into a A -module. However, this is possible in this case. So what ishappening?

How to encode a (finite) group G in terms of an algebra. G as a set goes to an algebra C(G) :=

f : G→ C. OnG, we have a productG×G m→ G, which gives C(G)⊗CC(G) ' C(G×G)∆→C(G),

which is called the co-multiplication. Indeed, associative law holds (we call, co-associative). eis the identity element of G, then we get C(G) →C(e) = C, called the co-identity. (Note: forf ∈ C(G), (∆f)(x, y) = f(xy)).

Define a Hopf algebra to be an associative algebra equipped with an associative co-multiplication∆ such that ∆ : A → A ⊗ A is an algebra homomorphism, and has co-identity and co-inverse(some authors leave co-identity/inverse out).

Let A be a finite dimensional Hopf algebra. From the dual vector space A ∗. Then multiplica-tion in A goes to a co-multiplication, and the co-multiplication goes to multiplication. This is sortof a Pontryagin duality. If A = C(G) with G finite, then A ∗ = L1(G). We find that the multipli-cation on L1(G) from the co-multiplication on C(G) is exactly convolution. And co-multiplicationon L1(G) from the multiplication on C(G) is δx 7→ δx ⊗ δx. For any Hopf algebra A if M and N

are A -modules, then A∆→ A ⊗ A (aside: if M,N is A ,B-module, then M ⊗ N is an A ⊗B

module), and so M ⊗N becomes an A -module via ∆.Hopf algebra with enough structure are called quantum groups. There is a very good theory

of compact quantum groups using unital C∗-algebra, equipped with comultiplication, etc. 1985:Woronowing: discovered a quantum version of SU2

30

Chapter 4

Functional calculus

4.1 Lecture 10/21/15 — Holomorphic functional calculus

Today we study the topic of “holomorphic functional calculus.”

Let A be a Banach algebra over C (adjoining 1 if necessary). (Note that norm doesn’t affectthe spectrum, so that adjoining 1 is fine). Let a ∈ A , so σ(a) ⊂ C. Let O be an open subset of Cwith σ(a) ⊂ O. Let H (O) be defined as the algebra of C-valued holomorphic functions on O. Forf ∈H (O), want to define f(a). However, naively using power series can be tricky:

Example 4.1.1. Suppose σ(a) is not connected. Let let U t V be separation by two open sets.Defining f to be 1 on U and 0 on V , we see that f certainly cannot be expressed as a single powerseries.

We use the Cauchy integral formula:

Definition 4.1.2. Let A be a unital Banach algebra over C, a ∈ A , and σ(a) ⊂ O ⊂ C for Oopen. Choose a collection Ω (i.e. a 1-chain) of disjoint curves γ ⊂ O surrounding σ(a) such that

for any f ∈H (O), we have 12πi

∫γf(ζ)ζ−z dζ = f(z) if z ∈ σ(a) and 0 if z is “outside.” Then define

f(a) :=1

2πi

∫γ

f(ζ)

ζ − adζ =

1

2πi

∫γf(ζ)(ζ1− a)−1dζ

Remark 4.1.3. Since the integrand is continuous, we can just use Riemann integral. This definitionis independent of choice of γ (as long as it satisfies the condition stated above).

Theorem 4.1.4. f 7→ f(a) defined above is a (unital) algebra homomorphism from H (O)→ A .Moreover, note that ‖f(a)‖ ≤ 1

2π supγ |f(ζ)| supγ ‖(ζ−a)−1‖·(length of γ). So, if fn → f uniformlyon γ, then fn(a)→ f(a).

Proof. Let f, g ∈ H (O). WTS f(a)g(a) = (fg)(a) (linearity is clear from definition). Well, we

compute f(a)g(a) = 12π2

∫γ1

f(ζ1)ζ1−adζ1

∫γ2

g(ζ2)ζ2−adζ2. Choose γ1, γ2 so that γ2 is strictly inside γ1, then

f(a)g(a) = 12π2

∫∫f(ζ1)g(ζ2)((ζ1− a)−1(ζ2− a)−1))dζ1dζ2. And recalling that a−1− b−1 = a−1(b−

a)b−1, we have = 1(2π)2

∫∫f(ζ1)g(ζ2)) (ζ1−a)−1−(ζ2−a)−1

ζ2−ζ1 dζ1dζ2 = 1(2πi)2 [

∫γ1f(ζ1)

∫γ2

g(ζ2)(ζ1−a)−1

ζ2−ζ1 dζ2dζ1−∫γ2g(ζ2)

∫γ1

f(ζ1)(ζ2−a)−1

ζ2−ζ1 dζ1dζ2. The right term is (fg)(a) and the left term is zero (look where ζ1, ζ2

are inside or outside curves γ1, γ2 noting that γ2 is inside γ1). To check that the map is unital, in-crease γ to be centered at 0 with large radius contaiing σ(a): 1

2πi

∫γ

1ζ−adζ =

∫1ζ

11−a/ζ dζ = 1A .

31

Remark 4.1.5. Since f 7→ f(a) is an (unital) algebra homomorphism, we have f(z) = z impliesf(a) = a, and f(z) = zn implies f(a) = an. Thus, if f has a power series expression about O, it isalso valid on σ(a).

Theorem 4.1.6 (Spectral mapping theorem). For A , a, and f as before, we have σ(f(a)) =f(σ(a)).

Proof. If λ ∈ σ(a), then f(z) − f(λ) = (z − λ)g(z) for g holomorphic in O. So, f(a) − f(λ) =(a − λ)g(a), and so if f(a) − f(λ) has an inverse, then (a − λ) invertible, contradicting λ ∈ σ(a).Thus f(λ) ∈ σ(f(a)) (or, σ(f(a)) ⊃ f(σ(a))). If λ /∈ f(σ(a)), then f(z) − λ is not 0 on σ(a), so

1f(z)−λ is holomorphic on an open set O containing σ(a). Thus, (f(a) − λ)−1 is well-defined, i.e.

(f(a)− λ) is invertible, and so λ /∈ σ(f(a)).

Remark 4.1.7. Recall that `1(Z≥0) is the closed disk D. Now, L1(R≥0) with convolution is asubalgebra of L1(R) and is the set of bounded continuous homomorphisms of R≥0 → C given by

t 7→ e−tz = e−t(a+bi) = etae−tbi (so we need a ≥ 0). The Gelfand transform is f(z) =∫∞

0 f(t)e−tzdtwhere z ∈ Re(z) ≥ 0. This is the Laplace transform. Similarly, if f ∈ L1(R) supported on [a, b],

then f(z) =∫ ba f(t)eitzdt for z ∈ C is an entire function. Related to Palay-Wiener theory of growth

of entire functions.

32

Chapter 5

Operators on Hilbert spaces

5.1 Lecture 10/23/15 — Polarization identity and consequences

We now study operators on Hilbert spaces.

Proposition 5.1.1 (Polarization identity). Hilbert space over C has a sesquilinear inner product.Let B be any sesquilinear form (not necessarily positive). We have an identity:

B(v, w) =1

4

3∑k=0

ikB(v + ikw, v + ikw)

(in other words, a general sesquilinear form is determined by the quadratic forms).

Corollary 5.1.2. Let T ∈ B(H ). If 〈Tξ, ξ〉 = 0 for all ξ ∈H then T ≡ 0.

Proof. Use the polarization identity to B(ξ, η) := 〈Tξ, η〉.

Remark 5.1.3. Note that this fails over R (consider rotation by 90o degrees). So we assume ourHilbert spaces to be over C.

Corollary 5.1.4. If S, T ∈ B(H ) and 〈Sξ, ξ〉 = 〈Tξ, ξ〉 ∀ξ then S = T .

Proposition 5.1.5. 〈Tξ, ξ〉 ∈ R ∀ξ iff T = T ∗.

Proof. Immediate from 〈Tξ, ξ〉 = 〈ξ, T ∗ξ〉 = 〈ξ, T ξ〉 and the previous corollary.

Remark 5.1.6. Even if T = T ∗, T may not have any eigenvectors. For an example, considerH = L2([0, 1]) and let T be multiplication by t.

Proposition 5.1.7. For T ∈ End(H ), we have ker(T ∗) = Im(T )⊥

Proof. ξ ∈ ker(T ∗) iff 0 = 〈T ∗ξ, η〉 = 〈ξ, Tη〉 ∀η ∈H .

Proposition 5.1.8. T ∈ B(H ) is invertible iff ∃a, b ∈ R>0 such that ‖Tξ‖ ≥ a‖ξ‖ ∀ξ and‖T ∗ξ‖ ≥ b‖ξ‖ ∀ξ.

Proof. If T is invertible then a-statement holds, and T ∗ is invertible, so b-statement holds. Con-versely, if both hold, then T ∗ is injective, so Im(T ) is dense in H . Note that Im(T ) is closed, becauseif Tξn is a Cauchy sequence, then by a-statement ξn is a Cauchy sequence. Thus, Im(T ) = H .Since T was also injective, T is bijective linear map and so we can define T−1 (possibly unbounded)operator. But for any ξ ∈H , we have from a-statement that ‖ξ‖ ≥ a‖T−1ξ‖.

33

Corollary 5.1.9. If T is not invertible, then there is a sequence ξn with ‖ξn‖ = 1 ∀n such thatTξn → 0 or T ∗ξn → 0.

Corollary 5.1.10. If T −λI is not invertible, then either there is a sequence with all norm 1 witheither Tξn → λξn or T ∗ξn → λξn.

Corollary 5.1.11. If T = T ∗ then λ ∈ σ(T ) iff there is ξn with ‖ξn‖ = 1 such that (T −λI)ξn →0.

5.2 Lecture 10/26/15 — Normal and self-adjoint operators

Recall: we have T ∈ B(H ) for H Hilbert space. We had that T − λI is not invertible in B(H )iff there is a sequence ξn ⊂H with ‖ξn‖ = 1 and (T − λ)ξn → 0 or (T ∗ − λ)ξn → 0.

Definition 5.2.1. T is normal if TT ∗ = T ∗T .

Proposition 5.2.2. T is normal iff ‖Tξ‖ = ‖T ∗ξ‖ ∀ξ ∈H .

Proof. Write out ‖Tξ‖2 and ‖T ∗ξ‖2 as an inner product involving T ∗T and TT ∗, and then use thepolarization identity.

Corollary 5.2.3. Let T be normal (so T − λI is normal). Then T − λI is not invertible iff thereis a sequence ξn with ‖ξn‖ = 1 and (T − λI)ξn → 0.

Proposition 5.2.4. T is self-adjoint iff T is normal and σB(H )(T ) ⊂ R.

Proof. If T = T ∗, then and if λ ∈ σ(T ), there is ξn with ‖ξn‖ = 1 and (T − λ)ξn → 0, andhence 〈Tξn − λξn, ξn〉 = 〈Tξn, ξn〉 − λ‖ξn‖2 → 0, so 〈Tξn, ξn〉 → λ and 〈Tξn, ξn〉 ∈ R ∀n. Is theconverse actually true? For the converse, if 〈Tξ, ξ〉 = λ, then λ ∈ σ(T ) since for ‖ξ‖ = 1 then〈Tξ − λξ, ξ〉 = 0

Proposition 5.2.5. For T ∈ B(H ), the following are equivalent.

(a) T = T ∗ and σ(T ) ⊂ R≥0,

(b) 〈Tξ, ξ〉 ≥ 0 ∀ξ

(c) ∃S ∈ B(H ) with SS∗ = T . Moreover, one can choose S such that S = S∗, 〈Sξ, ξ〉 ≥ 0 ∀ξ,and S commutes with any operator R that commutes with T .

Proof. (b) =⇒ (a) and (c) =⇒ (b) is immediate. (a) =⇒ (c): Let A = (C∗-subalgebra of B(H )generated by T and I). Then A ' C(σ(T )). T 7→ f(t) = t, and f ≥ 0 on σ(T ) so +

√f ∈ C(σ(T )),

and let S be corresponding operator. So S = S∗, T = S2. S is a norm limit of polynomials in T ,so S commutes with every operator that commutes with T .

Definition 5.2.6. T is called positive if it satisfies any of (a), (b), (c) above.

Remark 5.2.7. Without the previous proposition, it is quite unclear just from condition (a) that ifT, T ′ are positive then so is T + T ′ (it is clear now from condition (b)).

We want to understand the structure of self-adjoint (or normal) operators acting on a Hilbertspace. For this purpose, consider T normal, we form the C∗-subalgebra A generated by T and I.Because T is normal, A is a unital C∗-algebra ' C(σ(T )). So, given a unital C∗-subalgebra ofB(H ), what is the structure of how it acts on H ?

34

Definition 5.2.8. Let A a unital C∗-subalgebra of B(H ). A (closed) subspace K of H is saidto be invariant under A if it is carried into itself by every operator in A .

Remark 5.2.9. Could make the same definition for any subset S of B(H ).

Lemma 5.2.10. Let S ⊂ B(H ), and assume that S ∗ = S (i.e. if T ∈ S then T ∗ ∈ S ). Thenif K is an S -invariant subspace, then so is K ⊥.

Proof. Let ξ ∈ K ⊥ and T ∈ S . We want Tξ ∈ K ⊥. Well, for any η ∈ K , we have 〈Tξ, η〉 =〈ξ, T ∗η〉 = 0 since T ∗ ∈ S and hence T ∗η ∈ K .

Consider the closed subalgebra generated by S and I, a C∗-algebra. We will continue nexttime by considering (for A a unital C∗-subalgebra of B(H )) K := A ξ, a A -invariant subspace,called the cyclic subspace, generated by ξ.

5.3 Lecture 10/28/15 — The GNS construction I

Definition 5.3.1. Let A be a unital ∗-algebra over C. A ∗-representation of A on a Hilbertspace H is a ∗-homomorphism π : A → B(H ) with π(1A ) = IdH . We say that a subspaceK ⊂H is π-invariant (or A -invariant if π is clear) if it is invariant under π(a) | a ∈ A .

Last time, we showed that K ⊥ is π-invariant if K is.

If 0 6= ξ ∈ H , then π(A )ξ is A -invariant, so consider K := π(A )ξ, a closed A -invariantsubspace. Then K ⊥ is a closed A -invariant subspace. Let 0 6= ξ′ ∈ K ⊥, and now considerπ(A )ξ′, which is a closed A -invariant subspace that is ⊥ to K . This process stops if H is finitedimensional. If infinite, it may not stop, but by invoking Zorn’s lemma, (note that C = A givesthe theorem about getting orthonormal basis of H ).

Proposition 5.3.2. Any ∗-representation of A on H can be expressed as the (possibly infinite)direct sum of A -cyclic subspaces.

Proof. Same as for obtaining an orthonormal basis.

Proposition 5.3.3. Conversely, let (Hλ, πλ) be a collection of ∗-representations of A . Thenforming

H :=⊕λ

Hλ := (ξλ)λ∈Λ : ξλ ∈Hλ and∑‖ξλ‖2Hλ

<∞

and noting that 〈ξ, η〉 :=∑〈ξλ, ηλ〉 is well-defined, the induced norm is complete, and so H is a

Hilbert space. Moreover,⊕

λ πλ naturally gives a map A → End(H ) (but may be unbounded!),but it is bounded if there exists Ma such that ‖πλ(a)‖ ≤Ma for all λ since then ‖(⊕πλ)(a)‖ ≤Ma.If A is a normed ∗-algebra, then usually we assume ‖πλ(a)‖ ≤ ‖a‖ ∀λ,∀a (in particular, π is normcontinuous).

Proof. The same as the proof that `2 is a Hilbert space.

This shows that to study a ∗-representaiton on a Hilbert space, it really suffices to considercyclic space. Thus, we want to study cyclic representations, i.e. ones that have a cyclic vector,denoted ξ0, such that H = π(A )ξ0

Given any ξ ∈H , define ϕξ(a) := 〈π(a)ξ, ξ〉. It is a linear functional, and |ϕξ(a)| ≤ ‖π(a)‖‖ξ‖2.Moreover, ϕξ(a

∗a) = 〈π(a∗a)ξ, ξ〉 = 〈π(a)ξ, π(a)ξ〉 ≥ 0.

35

Definition 5.3.4. Let A be a ∗-algebra. A linear function, ϕ, on A is said to be positive ifϕ(a∗a) ≥ 0 for all a ∈ A (“positive Radon measures” on A ). If also ϕξ(1A ) = 1, then we say ϕ isa state (“probability measures” on A ).

In the ϕξ(a) case, if ‖ξ‖ = 1, then ϕξ(1A ) = 1.

Let ϕ be a positive linear functional on A (a unital ∗-algebra), GNS-construction (Gelfand-Naimark-Segal). On A , define a pre-inner product by 〈a, b〉ϕ = ϕ(b∗a) (is a sesquilinear form).

Moreover, 〈a, a〉ϕ = ϕ(a∗a) ≥ 0. Need 〈a, b〉ϕ = 〈b, a〉ϕ: 0 ≤ ϕ(a + b, a + b) = ϕ(a∗a) + ϕ(b∗a) +ϕ(a∗b) + ϕ(b∗b), thus ϕ(b∗a) + ϕ(a∗b) ∈ R, and do the same computation for a+ ib.

Let Nϕ := a ∈ A : ϕ(a∗a) ≥ 0 =0? (instead of ≥ 0). Noting that Cauchy-Schwartz doesn’tneed positive definiteness, we have Nϕ = a : ϕ(ab) = 0∀b ∈ A . Nϕ is a linear subspace of A .Hence, we form A /Nϕ, and thus our pre-inner product becomes an (definite) inner product.

The A /Nϕ completed for the norm from inner product is thus a Hilbert space, denotedL2(A , ϕ). (Analogous to making L2([0, 1], µ)).

Definition 5.3.5. π, a ∗-representation of A on L2(A , ϕ) by defining π(a)b = ab (left-regularrepresentation). Note 〈π(a)b, c〉ϕ = ϕ(c∗(ab)) = ϕ((a∗c)∗b) = 〈b, π(a∗)c〉, and so π(a∗) = (π(a))∗.This is by we changed the order around when we defined 〈a, b〉ϕ.

From here, one can also see that this representation also drops down to A /Nϕ. (bounded?unbounded?)

5.4 Lecture 10/30/15 — The GNS construction II

Let A unital ∗-algebra, and µ is a positive linear functional on A . Then we have L2(A , µ) where〈a, b〉µ = µ(b∗a).

Example 5.4.1 (Non-example). Let A = polynomials over C, viewed as functions on R, withp∗ = p. Define µ(p) =

∫∞−∞ p(t)e

−t2dt. Let Lp be operator of pointwise multiplication by p where

deg p ≥ 1. This is an unbounded operator on L2(R, e−t2dt).

Theorem 5.4.2. Let A be a ∗-normed Banach ∗-algebra, and let µ be a positive linear functionalon A . Then µ is continuous for the norm of A , and ‖µ‖ = µ(1).

Proof. Let a ∈ A , a = a∗, and ‖a‖ ≤ 1. Consider f(z) = (1− z2)1/2, holomorphic on a open diskabout 0, with power series expansion

∑αnz

n about 0. Since f(z) has values in R for z ∈ R, wehave that all the αn’s are R. Let b = f(a), then b = b∗. And b∗b = b2 = (f(a))2 = 1 − a2. Thenµ(b∗b) ≥ 0 and so is µ(1) − µ(a), thus, µ(a) < µ(1), and applying to −a we have |µ(a)| < µ(1).Then for any a ∈ A with a∗ = a, we have |µ( a

‖a‖+ep)| < µ(1), i.e. |µ(a)| < µ(1)(‖a‖ + ε). Hence,

|µ(a)| ≤ µ(1)‖a‖. Thus, for any a, |µ(a)|2 = |µ(a1)|2 = |〈1, a〉µ|2 ≤ 〈1, 1〉)µ〈a, a〉µ = µ(1)µ(a∗a) ≤µ(1)µ(1)‖a∗a‖ ≤ µ(1)2‖a‖2.

Consider A /Nµ. Nµ is a left ideal in A so the left regular representation drops to representationof A on A /Nµ, i.e. if a ∈ A 4, b ∈ Nµ, then 〈λab, λb〉µ = 〈ab, ab〉µ = µ(b∗a∗ab) = 〈a∗ab, b〉 ≤〈a∗ab, a∗ab〉〈b, b〉µ = 0.

Theorem 5.4.3. Let A be a unital Banach ∗-algebra, and let µ be a positive linear functional onA , with L2(A , µ) the corresponding GNS Hilbert space. Let λ be the left regular representationof A on itself, and so on A /Nµ. Then for each a ∈ A , the operator λa is a bounded operator onA /Nµ for 〈−,−〉µ, and so extends to a bounded operator on L2(A , µ), and in fact, ‖λa‖ ≤ ‖a‖.

36

Proof. Let a, b ∈ A , 〈λab, λab〉µ = µ(b∗a∗ab). Define νb(a) := µ(b∗ab), which is a continuouspositive linear functional. Thus, we have νb(a

∗a) ≤ νb(1)‖a∗a‖ ≤ µ(b∗b)‖a‖2 = 〈b, b〉µ‖a‖. So,〈λab, λab〉µ ≤ 〈b, b, 〉µ‖a‖2. This all holds for A , so it drops down to A /Nµ. That is, ‖lama‖ ≤ ‖a‖on A /Nµ.

In L2(A , µ), let ξ0 “=” 1A +Nµ ∈ L2(A , µ). Then ξ0 is a cyclic vector for the representation ofA on L2(A , µ). The positive linear functional determined by ξ0 is a 7→ 〈Laξ0, ξ0〉µ = 〈a, 1〉µ = µ(a).

In this sense, the (continuous) positive linear functional corresponds to cyclic representationswith specific choice of the cyclic vector.

Proposition 5.4.4. Let A be a unital ∗-algebra, and let (H1, π1, ξ1) and (H2, π2, ξ2) be the cyclic∗-representations of A . Let µ1, µ2 be the corresponding positive linear functionals for ξ1, ξ2. Ifµ1 = µ2, then there is a unique unitary operator u : H1 →H2 (i.e. A -module map) that intertwinesπ1 and π2, i.e. u(π1(η)) = π2(u(η)) for η ∈H1 (uπ1 = π2u), such that uξ1 = ξ2.

Proof. Try to define u by u(π1(a)ξ1) = π2(a)ξ2. Check well-defined.

For uniqueness, if we have u, then for any a ∈ A , then we have u(π1(a)ξ1) = π2(a)(uξ1) =π2(a)ξ2. Thus, by cylic-ness, we have that u is determined on a dense set by this formula. Forexistence, we try to define u by u(π1(a)ξ1) = π2(a)ξ2. Let’s check that this is well-defined:〈u(π1(a)ξ1), u(π1(b)ξ1)〉 = 〈π2(a)ξ2, π2(b)ξ2〉 = 〈π2(b∗a)ξ2, ξ2〉 = µ2(b∗a) = µ1(b∗a) = 〈π1(b∗a)ξ1, ξ1〉 =〈π1(a)ξ1, π1(b)ξ1〉, and so if π1(a)ξ1 = π1(a′)ξ1 then π1(aa′)ξ2 = 0, for well-definedness suffices toshow that if π1(a)ξ1 = 0 then π2(a)ξ2 = 0. Thus, u is well-defined, and unitary follows from:u(π1(c)(π(a)ξ1)) = u(π1(ca)ξ1) = π2(ca)ξ2 = π2(c)π2(a)ξ2 = π2(c)u(π1(a)π1).

5.5 Lecture 11/2/05 —

Definition 5.5.1. Two representations (H , π) and (K , ρ) are unitarily equivalent if there is aunitary intertwining operator between them, i.e. u : H → K unitary with u π(a) = ρ(a) u.

Definition 5.5.2. A pointed cyclic ∗-representation of H is a representation (H , π) togetherwith a specified cyclic vector.

Remark 5.5.3. There is a bijection between pointed cyclic representation of A and positive linearfunctionals on A .

If ϕ is a state on A , then the cyclic vector for GNS representation has norm 1, and conversely,if ϕ(a∗a) ≥ 0, ϕ(1) = 1, the states form a convex subset of unit ball of A ∗ (tϕ+(1− t)ψ, 0 ≤ t ≤ 1)closed in the weak-∗ topology (and hence compact). Thus the state space S (A ) is a compact forweak-∗ topology. (By Krein-Milman theorem), we have that S (A ) is the closed convex hull of itsextreme points.

Definition 5.5.4. A ∗-representation (H , π) of A is said to be irreducible if it contains noproper π-invariant subspaces.

Example 5.5.5. If H is finite dimensional, it decomposes into irreducible (semi-simple-ness).However, let A = C([0, 1]), and H = L2([0, 1], µ) (work this out)

Theorem 5.5.6 (Math 208). The GNS representation for a given state is irreducible iff the stateis a extreme point of S (A ), which we call pure state.

Let G be a locally compact group. We have a left regular (faithful) representation of L1(G) onL2(G) (hence we have lots of states). Consider for example G = SL2(Z).

37

Theorem 5.5.7 (Schur’s Lemma). (H , π) is irreducible iff the commutant?T ∈ B(SH) : Tπ(a) =π(a)T = C IdH .

Proof. If (H , π) is not irreducible, then the projection operator on a proper π-invariant subspace isin the commutant. In the other direction, (π(A ))∗ is a closed unital ∗-subalgebra of B(H ) (closedin weak-∗, so is a von Neumann algebra). So if (π(A ))∗ is not C1 then ∃T ∈ (π(A ))∗ with T ∗ = Tand T /∈ C1 such that σ(T ) has more than one point, and so C∗(T, I) represented on H via πis C(σ(T )), which has more than one point, and hence can find g, f ∈ C(σ(T )) with f, g 6= 0 butfg = 0. Then π(f)H is a proper subspace with π(a)π(f)H = 0. Will finish next time.

5.6 Lecture 11/4/15 —

Recall Schur’s Lemma from last time.

Proposition 5.6.1. Let A be a commutative normed ∗-algebra. Then every irreducible ∗-representation is 1-dimensional.

Proof. If (H , π) is an irreducible ∗-representation, then π(A ) ⊂ (π(A ))∗ because A is commu-tative, and (π(A ))∗ = C IdH because π is irreducible, so π(A ) ⊂ C IdH . Then every subspaceproper subspace of H is π-invariant, so H must not have any proper subspace so dim H = 1.

this paragraph needs to be broken downNow, let A be commutative ∗-algebra, and let H , π) be some ∗-representation of A . So

π(A ) is a commutative ∗-subalgebra of B(H ). Let C be the norm-closure of π(A ), so C is acommutative C∗-subalgebra of B(H ). So, C is faithfully represented on H , so we can considera decomposition of H into cyclic subspaces Hλ for C . Choose a cyclic vector ξλ of length 1for each cyclic subspaces, and let πλ be the representation of C on Hλ by restriction. ThenH '

⊕λ Hλ. Each ξλ determines a state, say µλ, on C . Let X be the maximal ideal space of C ,

so C ' C(X), and each µλ is a probability Borel measure on X. Can form L2(X,µλ), and C(X) isrepresented on L2(X,µλ) by pointwise multiplication, call this representation ρλ. So, ρλ is also arepresentation of C via C ' C(X). For (Hλ, ρλ), we have a natural cyclic vector ηλ correspondingto 1 ∈ C(X). The vector state for ηλ is just µλ viewed as a positive linear functional. See that(Hλ, πλ, ξλ) ' (L2(X,µλ), ρλ, ηλ) via a natural unitary map. That is, (H , π) '

⊕λ(L2(X,µλ), ρλ).

This decomposition is generally not unique. Let B(X) denote the C∗-algebra of bounded complex-valued Borel functions on X. Notice that any f ∈ B(X) is measurable for every Borel measure onX. For each f ∈ B(X), we can define an operator Tλ on L2(X,µλ) by pointwise multiplication,and then set ρ(f) :=

⊕λ Tλ on H . This gives a ∗-representation of B(X) on H . If f ∈ B(X),

and if gn s a sequence of elements of C(X) with ‖gn‖ < K ∀n, and if f ≥ 0, and gn ↑ f pointwiseeverywhere, then for any ξλ, ηλ ∈ L2(X,µλ), we have 〈πλ(gn)ξλ, ηλ〉 =

∫gnξληλdµλ ↑

∫fξληλdµλ

by MCT. By definition, this means that πλ(gn)→ πλ(f) for the weak operator topology. It followsthat π(gn) → π(f). Similarly, for any f ∈ B(X), and if gn ⊂ C(X) with ‖gn‖ ≤ K ∀n, then〈π(gn)ξλ, ηλ〉 =

∫gnξληλdµλ → 〈ρ(f)ξλ, ηλ〉 (and hence again, π(gn)→ ρ(f), so this representation

does not depend on the cyclic decomposition.

Example 5.6.2. Let T ∈ B(H ), with T normal, let X = σ(T ) ⊂ C, and C = C∗(T, IdH ) ' C(X).By doing the whole cyclic decomposition thing, we get a ∗-homomorphism from B(X) → B(H )that extends the natural homomorphism C(X) → B(H ). In particular, for any f ∈ B(X), setf(T ) = ρ(f). this extends the continuous functional calculus to Borel functions (so this is calledthe Borel functional calculus).

38

Let C be a commutative C∗-subalgebra of B(H ). C ' C(X) ⊂ B(X). Well, by B(X) →B(H ), for any Borel set A ⊂ X consider χA ∈ B(X) and get an operator EA ∈ B(H ) (alsodenoted E(A). So, the map E maps the collection of Borel subsets of X to projection operators.Note that E(A) does not depend on the cyclic decomposition of H .

5.7 Lecture 11/6/15 —

Recall the map E : Borel sets → projections in B(H ). This map is an example of a projection-valued measure on X, with valued in B(H ).

Definition 5.7.1. A projection valued measure E on X, with values in B(H ), is a mapsatisfying

(a) E(∅) = 0, E(X) = IdH

(b) E(A ∩B) = E(A)E(B) (which =⇒ if A ∩B = ∅, then E(A)E(B) = 0, i.e. orthogonality)

(c) If Ajj∈N is a disjoint sequence of Borel sets, then E(⊕

j Aj) =∑E(Aj) where the conver-

gence is in the strong operator topology, i.e. for any ξ ∈ H (in other words, E(⊕Aj)ξ =∑

E(Aj)ξ ∀ξ ∈H .

Notation. We use the direct sum for disjoint union of sets.

Remark 5.7.2. For any ξ, η ∈ H , the map A → 〈E(A)ξ, η〉 is a C-valued Borel measure that isbounded by ≤ ‖ξ‖‖η‖. If A is a closed subset of X, then we can find a sequence fn ∈ C(X), withfn ↓ χA, then π(fn) ↓ E(A) in the strong operator topology

Conversely, given a projective valued measure of X to B(H ), we can use it to define a ∗-representation of B(X) (bounded Borel functions of X) on H such that π(χA) = E(A) for allχA ∈ B. Given f ∈ B(X), set π(f) =

∫f(λ)dE(λ).

Given ε > 0, find a finite partition Aj of X such that for x, y ∈ Aj we have |f(x)− f(y)| < ε.Choose xj ∈ Aj . Then ‖π(f)−

∑f(xj)E(Aj)‖ < ε.

If T is a normal operator on H , and let A = C∗(T, IdH ) ' C(ξ(T )) (normal implies commu-tative; note that T corresponds to f(z) = z). Then this determines a projection valued measure onσ(T ) such that T =

∫zdE(z). This is called the projective measure form of the spectral theorem for

normal operators. This is the generalization of the finite dimensional case where normal operatoradmits a orthonormal eigenvasis.

Now, if T = T ∗ (self-adjoint), so that σ(T ) ⊂ R, we set for t ∈ R, Et := E((−∞, t) ∩ σ(T )).Now, T =

∫tdE(t) in this case, and we get

∑tj(Etj+1 −Etj ) with tj+1 > tj , which converges (still

in strong operator topology) to T =∫tdE(t). This is the traditional form of the spectral theorem

for the self-adjoint operators.

39

Chapter 6

Compact and Fredholm operators

We now change our topic to compact operators.

Notation. Denote the (closed) unit ball of a normed space X by DX .

Definition 6.0.3. Let X,Y be Banach spaces (could be just normed spaces), and let T ∈ B(X,Y ).We say that T is compact if T (DX) has a compact closure in Y . Or equivalently, the image of abounded set of X is relative compact in Y .

N.B. 6.0.4. Recall that for metric spaces, complete + totally bounded ⇐⇒ compact. Hence, Tis compact iff T (DX) is totally bounded.

Notation. Denote the set of compact operators from X to Y by B0(X,Y ).

Example 6.0.5. T is of finite rank, i.e. the range of T is finite dimensional.

Example 6.0.6. Many integral operators are compact: (X,S, µ) measure space, and let K beC-valued measurable function on X × X, can define TK : Lp(X,µ) → Lp(X,µ) by (TKξ)(x) =∫K(x, y)ξ(y)dµ(y). This is the generalization of the finite dimensional case: Kv matrix indices in

X: (Tv)n =∑Knmvm.

Example 6.0.7. (X, d) a compact metric space, L (X) be the vector space of Lipschitz functionson X, then let ‖f‖L = ‖f‖∞ + L(f) (this makes L (X) into a Banach algebra). We also haveL (X) → C(X), and one can check that this map is a compact (follows from Arzela-Ascoli).

6.1 Lecture 11/9/15 —

Compact operators do occur in real life:

Example 6.1.1. C1(T) → C(T) is compact where C1(T) is given ‖f‖1 = ‖f‖∞+ ‖f ′‖∞. Anotherexample: for O ⊂ Rm, consider Hs(O)→ Ht(O). If s > t+m then the map is compact.

Proposition 6.1.2. Let X → Y and T ∈ B(X,Y ). Let B0(X,Y ) be the set of compact operators

in B(X,Y ). Then if XT→ Y

S→ Z with T ∈ B0(X,Y ) and S ∈ B(Y, Z), then ST ∈ B0(X,Z). Also,if instead T ∈ B(X,Y ) and S ∈ B0(Y, Z), then ST ∈ B0(X,Z)

Proof. Given ε, choose points y1, . . . , yn so the ε-balls about them cover T (DX). Then the ε‖S‖-balls about Syj ’s cover ST (DX). Likewise, if S is compact, then take since T (DX) is bounded inY , ST (DX) is totally bounded in Z.

40

Proposition 6.1.3. If S, T ∈ B0(X,Y ), then S + T ∈ B0(X,Y ) and αT ∈ B0(X,Y ).

Proof. Note that: If A,B ⊂ X, totally bounded, then A + B is totally bounded. For given ε > 0,choose y1, . . . , ym such that the ε/2-balls around them cover A, and find y′1, . . . , y

′n cover B by

ε/2-balls. Then for the ε-balls about the yi + y′j ’s cover A+B.

In other words, we have derived that:

N.B. 6.1.4. B0(X,Y ) is a B(Y ), B(X)-bimodule. Moreover, B0(X) is a two-sided ideal in B(X).

Proposition 6.1.5. B0(X,Y ) is a norm-closed subspace of B(X,Y ).

Proof. Let Tn ⊂ B0(X,Y ), and suppose Tn → T ∈ B(X,Y ). Given ε > 0, choose n0 such that‖T − Tn0‖ < ε/2. Let y1, . . . , ym be ε/2-dense in Tn0(DX). Then this same set is ε-dense inT (DX)) since if x ∈ DX , ∃y0 with ‖Tn0(x)− y0‖ ≤ ε/2, ‖Tx− Tn0(x)‖ ≤ ε/2

Integral operators: Let X be compact set, µ positive Borel measure. Let K ∈ C(X × X),and define (TKf)(x) :=

∫k(x, y)f(y)dµ(y). Claim: this is a compact operator (proof: by Stone-

Weierstrass, we can approximate K in norm by K ′(x, y) :=∑hj(x)kj(y) where hj , kj ∈ C(X),

then TK′f(x) ≡∑hj(x)

∫kj(y)f(y)dy, so K ′ is finite rank with range in C(X), so K ′ is compact.

‖TK − TK′‖ ≤ ‖K −K ′‖∞µ(X)).We can also look at TK : L2(X,µ)→ L2(X,µ). (TK′ − TK)f(x) =

∫(K ′(x, y)−K(x, y)f(y)dµ.

But then TKf(x) =∫K(x, y)fydµ(y), so then |Tkf |2(x) ≤ ‖y − K(x, y)‖2‖f‖2, so ‖TKf‖2 ≤∫

|K(x, y)|2dµ(x)dµ(y)‖f‖2 and so ‖TK‖ ≤ ‖K‖2. Hence, we conclude ‖TK′ − Tk‖ ≤ ‖K −K ′‖2.If G is a compact group, choose a Haar measure. f ∈ C(G), and let Lf : L2(G) be the

left regular representation on itself, then (Lfξ)(x) =∫f(y)ξ(y−1x)dy =

∫f((yx−1)−1)ξ(y)dy =∫

f(xy−1)ξ(y)dy = K(x, y). But for any f ∈ L1(G), can approximate it in L1(G) by elements ofC(G). Then Lf is approximated by Lg’s where g ∈ C(G). So, Lf is compact.

After some work, one can show that every irreducible representation of a compact group is finitedimensional.

Question. For general X,Y , is every compact operator approximated by finite rank operators? IfF (X,Y ) denote the finite rank operators, then is it dense in B0(X,Y )? Answer in general is no(1973, counterexample, but very hard). The strong operator topology on X: Tn → T for SOT ifTnξ → Tξ ∀ξ ∈ X, i.e. given by seminorms ‖T‖ξ = ‖T (ξ)‖ ∀ξ ∈ X. Fact: if Sα ⊂ B(Y ) is a netSα → S for SOT, then if ‖Sα‖ < K ∀α for some K, and T ∈ B0(X,Y ), then SαT → ST in norm.(False for switching this time).

6.2 Lecture 11/13/15 —

Proposition 6.2.1. Let T ∈ B0(X,Y ), and let Sλ be a net of operators in B(Y,Z) that convergesto S for strong operator topology, and assume that ‖Sα‖ ≤ K for some K. Then SαT convergesin norm to ST .

Proof. For any ε > 0, choose y1, . . . , yn ∈ Y such that ε-ball around these cover T (DX). Thenchoose λ0 ∈ Λ such that for λ ≥ λ0 we have ‖Sλyj − Syj‖ < ε ∀j = 1, . . . , n. Then, for anyx ∈ DX , there is a j0 such that ‖Tx − yj0‖ < ε. So, ‖SλTx − STx‖ = ‖(Sλ − S)(Tx)‖ ≤‖(Sλ − S)(Tx− yj0)‖+ ‖(Sλ − S)yj0‖ ≤ 2Kε+ ε

41

Remark 6.2.2. Note that this doesn’t work for TSλ → TS. Let S be the unilateral shift on `2(N).Let S∗ be its adjoint. Then (S∗)n → 0 for SOT, but it T = E1 projection on e1 = 1, thenE1(S∗)n 6→ 0 in norm.

Corollary 6.2.3. If a Banach space X has the property that there is a bounded net Sλ (i.e.‖Sλ‖ ≤ K ∃K) of finite rank operators that converges to IdX for SOT, then the set of finite rankoperators is dense (for norm) in B0(X).

Proof. Given T ∈ B0(X), T can be approximated by SαT in norm.

Example 6.2.4. For `p(N) (1 ≤ p <∞), and let Sn be the projection onto `2(1, . . . , n) ⊂ `p(N).More generally, for all `p spaces (1 ≤ p < ∞), every compact operator can be approximated innorm by finite rank operators. Hence, the same is true for any Hilbert space H .

Example 6.2.5. IfM is a compact Hausdorff space, and considerX = C(M). Let Λ be the directedPOSET of all finite open coverings of M ordered by refinement (i.e. λ1 ≤ λ2 if every open in λ2 iscontained in some open of λ1). For each λ ∈ Λ, let ϕλi be a partition of unity subordinate to λ.For each Uλj ∈ λ, choose a point mλ

j ∈ Uλj . Define Sλ ∈ B(C(M)) by (Sλf)(x) =∑

j f(mλj )ϕλj (x).

We claim that Sλf → f for each f , i.e. Sλ → IdX in SOT. (Well, from the definition it is clear that‖Sλ‖ = 1). So, given ε > 0, we can find a covering λ = Uλ1 , . . . , Uλkλ such that if m,n ∈ Uλj0 then|f(m) − f(n)| < ε. Similar ideas show that for Lp(X,M , µ) (1 ≤ p < ∞), again the finite rankoperators are dense in B0(Lp(X,M , µ)).

For H an infinite dimensional Hilbert space, B(H ) fails the approximation property.

Theorem 6.2.6 (Schauder). Let T ∈ B(X,Y ). then T ∈ B0(X,Y ) iff T ∗ ∈ B0(Y ∗, X∗).

Proof. Assume ‖T‖ ≤ 1. View the elements of Y ∗ as functions on DY . Then we claim that DY ∗ isa bounded equicontinuous family of functions on DY . Bounded is clear. For equicontinuous, notethat given ε > 0 and y1, y1 ∈ DY with ‖y1 − y2‖ < ε and ϕ ∈ DY ∗ , we have that ‖ϕ(y1)− ϕ(y2)‖ =

‖ϕ(y1 − y2)‖ ≤ ‖y1 − y2‖ < ε. Let A = T (DX) ⊂ DY is a compact subspace (T was compact). Sonow view the elements of Y ∗ as functions on A, then DY ∗ as sitting inside in C(A), this is againbounded and equicontinuous. Now applying Arzela-Ascoli, we have that DY ∗ as functions on A istotally bounded for ‖ · ‖∞−S , and thus given ε > 0, there are ϕ1, . . . , ϕn whose ε-balls for ‖ ‖∞−Scovers DY ∗ i.e. ∀ϕ ∈ DY ∗ , ∃ϕi such that |(ϕ−ϕj)(s)| ≤ ε for all s ∈ S and thus |(ϕ−ϕj)(Tx)| ≤ εfor all x ∈ DX , so |ϕ(Tx)−ϕj(Tx)| = |(T ∗ϕ)(x)− (T ∗ϕj)(x)| and hence ‖T ∗ϕ− T ∗ϕj‖ < ε. Thus,T ∗(DY ∗) is totally bounded and hence T ∗ is compact.

For the converse, if T ∗ is compact, then T ∗∗ is compact, but X → X∗∗ canonically, andT = T ∗∗|X , which is compact.

6.3 Lecture 11/16/15 —

Here is a good news about compact operators:

Theorem 6.3.1. Let T ∈ B0(H ) for H a Hilbert space, and assume that T is normal. Then ifλ ∈ σB(H )(T ) with λ 6= 0, then λ is an eigenvalue of T .

Proof. If λ ∈ σ(T ), then T − λ is not invertible, and is normal. Then there is a sequence ξnwith ‖ξn‖ = 1, such that (T − λ)ξn → 0. But Tξn ⊂ T (DH ) which has compact closure, andhence passing to a subsequence we have that Tξn → η for some η ∈ H . Now, since ‖η − λξn‖ ≤‖η − Tξn‖+ ‖Tξn − λξn‖, we have that ξn → 1

λη, and hence η is an λ-eigenvector.

42

Remark 6.3.2. We need λ 6= 0: Consider H = `2(N) and T = pointwise multiplication by f(n) = 1n .

Proposition 6.3.3. For any T ∈ B0(X) for X a normed space, if λ is an eigenvalue 6= 0, thenλ-eigensubspace Xλ is finite dimensional.

Proof. Note that T |Xλ = λ IdXλ is a compact operator.

Corollary 6.3.4. For T ∈ B0(H ) with T normal, then eigen-subspaces for any 2 distinct eigen-values are orthogonal. Thus, H is the orthogonal (possibly infinite) sum of the eigen-subspaces ofT .

Example 6.3.5. For K ∈ L2(X×X,S ×S , µ×µ), then TK on L2(X,S , µ) is a compact operator,and if K(x, y) = K(y, x), then TK is self-adjoint (so all eigenvalues are real)

Example 6.3.6. On L2([0, 1]) define the Voltera operator T by (Tf)(t) =∫ t

0 f(s)ds. This operator

is quasi-nilpotent (i.e. ‖Tn‖1/n → 0) so σ(T ) = 0.

We now move on to a different topic.

Definition 6.3.7. Let X be a normed space, and let V be a closed subspace. We say that V iscomplemented in X if there is a closed subspace W ⊂ X with V ∩W = 0 and V +W = X.

Let X be a Banach space, and V a closed subspace with complement W . On V ⊕W put norm‖(v, w)‖ = max‖v‖, ‖w‖) or ‖v‖+ ‖w‖, which makes V ⊕W into a Banach space. We have a mapV ⊕W → X taking (v, w) → v + w, which is bijective and continuous, and so the open mappingtheorem says the inverse map is continuous.

6.4 Lecture 11/18/15 —

Proposition 6.4.1. Let X be a Banach space, and suppose a closed subspace V has a complementW (so X = V ⊕W as vector spaces). Then, the projection of X on V along W is continuous.

Proof. On V ⊕W we put norm ‖(v, w)‖ = max‖v‖, ‖w‖) or ‖v‖+ ‖w‖, which makes V ⊕W intoa Banach space. We have a map V ⊕W → X given by (v, w) → v + w which is a continuousbijection. By open mapping theorem, this map is open, so its inverse is continuous. Now composethe inverse map X → V ⊕W with V ⊕W π→ V .

Proposition 6.4.2. Let X be normed space, and let V ⊂ X be closed, W ⊂ X be finite dimen-sional. Then V +W is closed in X.

Proof. Consider Xπ→ X/V , which is normed. Let W be the image of W in X/V . W is finite

dimensional, and hence complete, and thus W ⊂ X/V closed. Now, noting that π−1(W ) = W + V(and π is continuous), we have that V +W is closed.

Proposition 6.4.3. Let X be a Banach space, and let V ⊂ X be a finite dimensional, and letZ ⊂ X be closed with V ∩ Z = 0. Then V has a complement W in X such that W ⊃ Z.

Proof. Note that V + Z is closed by the previous proposition, and so a Banach space. Thus,V + Z

π→ V is continuous. If (v1, . . . , vn) is a basis for V , and denoting (ϕ1, . . . , ϕn) for the dualbasis, we can extend ϕj to V + Z by setting them = 0 on Z. By Hahn-Banach theorem, we canextend ϕj ’s to X as continuous linear functionals. Now, define P : X → V by P (x) :

∑nj=1 ϕj(x)vj ,

and note that P |V = IdV , P (Z) = 0, and P 2 = P . Set W := kerP .

43

Proposition 6.4.4. Let X be a vector space, and V ⊂ X such that X/V is finite dimensional.Then there is a W ⊂ X such that W ∩ V = 0 and V +W = X. (In particular, if X is normed,V closed, then W is closed since finite dimensional; i.e. W is a complement to V )

Proof. Let π : X → X/V . Let b1, . . . , bn be a basis for X/V . Choose vj ∈ X with π(vj) = bj foreach i. Now, set W = span(v1, . . . , vn).

Theorem 6.4.5. Let X be a Banach space, and let T ∈ B0(X). Then the range of I−T is closed.(Implies that for λ 6= 0, (λI − T ) has closed image).

Proof. Let N = ker(I − T ), i.e. 1-eigenspace for T . Since T compact, N is finite dimensional. So∃V ⊂ X closed and complement to N . Then Im(I − T ) = Im((I − T )|V ). The following claimfinishes the proof:

Claim: there exists δ > 0 such that ‖(I − T )v‖ ≥ δ‖v‖ ∀v ∈ V . FSOC, suppose ∃vj ⊂V, ‖vj‖ = 1 such that ‖(I − T )vj‖ → 0. Since T is compact, there is a subsequence denoted vjsuch that Tvj → x0 ∈ X. Then vj = Tvj + (I − T )vj so vj → x0, and thus x0 ∈ V so ‖x0‖ = 1.Then (I−T )x0 = lim(I−T )vj−0, but (I−T ) is injective on V , and hence x0 = 0, a contradiction.

Now to show that the range is closed, if there is a sequence vj with (I−T )vj → x0 ∈ X, thenby the claim, vj is a Cauchy sequence converging to v0 ∈ V , then x0 = (I − T )v0.

Remark 6.4.6. If X is normed, and V ( X closed, then ∀ε > 0 ∃x ∈ X such that ‖x‖ ≤ 1 + ε and‖x− v‖ ≥ 1 ∀v ∈ V . For the proof, consider X/V (normed) and choose x ∈ X/V with ‖x‖ = 1.

6.5 Lecture 11/20/15 —

Proposition 6.5.1. Let X be a Banach space, and T ∈ B0(X). Then Im(I − T ) has finitecodimension in X (i.e. dim(X/ Im(I − T )) <∞).

Proof. Let W = Im(I − T ), which is closed in X. If W = X then we are done. If W 6= X, thenchoose ε > 0. Then ∃x1 ∈ X such that ‖x1‖ ≤ 1 + ε and ‖x1 − w‖ ≥ 1 for all w ∈ W . SetW1 = span(W,x1), which is closed in X. If W1 6= X, then do the same to get W2 = span(W1, x2).If at any point Wn = X, then we are done. If Wn 6= X for all n then we have a sequence x1, x2, . . ..Then for n > m, Txn − Txm = xn + (T − I)xn − xm − (T − I)xm, and ‖Txn − Txm‖ ≥ 1. So allxn’s are in B1+ε(0) ⊂ X, contradicting T compact.

Definition 6.5.2. Let X,Y be vector spaces, and let T : X → Y be linear. We say that T isFredholm if dim(ker(T )) <∞ and dim(coker(T )) <∞. In this case, we define the index of T asindex(T ) := dim kerT − dim cokerT .

Example 6.5.3. If T ∈ B0(X), then (I − T ) is Fredholm.

Example 6.5.4. Let H = `2(Z+) with basis e1, e2, . . . (standard basis). Let S be the unilateralshift, Sej := ej+1. Note that ker(S) = 0 and Im(S) = span(e2, e3, . . .) (hence dim cokerS = 1).So, S is Fredholm with index(S) = −1. Also, S∗ which is S∗e1 = 0, Sej = ej−1, we have thatdim kerS∗ = 1 and dim cokerS∗ = 0 so that index(S∗) = 1.

If H ,K are Hilbert spaces and T ∈ B(H ,K ) (so T ∗ ∈ B(K ,H )), then kerT ∗ = (ImT )⊥.Thus, if ImT is closed, then dim cokerT = dim kerT ∗. Thus, if ImT is closed, T is Fredholm if(f)dim kerT <∞ and dim kerT ∗ <∞, in which case index dim kerT − dim kerT ∗.

Proposition 6.5.5. Let X,Y be Banach spaces, and let T : X → Y be Fredholm and bounded.Then Im(T ) is closed.

44

Proof. Since Y/TX is finite dimensional, Let (z1, . . . , zn) be the basis for it. Let π : Y → Y/TX.Now, choose y1, . . . , yn such that π(yj) = zj ∀j and let W = span(y1, . . . , yn), which is closedsubspace of Y . On X ⊕W , put norm ‖(x,w)‖ = max(‖x‖, ‖w‖) (so X ⊕W is a Banach space),and define ϕ : X ⊕W → Y by (x,w) 7→ Tx + w. Note that ϕ is bijective and continuous, soby open mapping theorem, its inverse is norm continuous, so that ϕ−1 : Y → X ⊕W given byy = Tx + w 7→ (Tx,w), and compose with projection onto W so that we have TX as the kernel,and hence TX is a closed subspace of Y .

Proposition 6.5.6. IfX,Y are finite dimensional, and if T : X → Y is linear. Then the index(T ) =dimX − dimY .

Proposition 6.5.7. If X,Y,X ′, Y ′ are vector spaces, and T, T ′ Fredholm linear maps, then (T ⊕T ′) : X ⊕X ′ → Y ⊕ Y ′ is Fredholm, and moreover, index(T ⊕ T ′) = index(T ) + index(T ′).

Example 6.5.8. For S the shift operator on `2(N), we thus have index(S ⊕ S∗) = 0.

6.6 Lecture 11/23/15 —

Theorem 6.6.1. Let X,Y, Z be vector spaces, and let XS→ Y

T→ Z with S, T Fredholm. ThenTS is Fredholm with index(TS) = index(T ) + index(S).

Proof. First, notice it is true for X,Y, Z finite dimensional. We now reduce the general case to thefinite dimensional case in several steps.

Step 1. Let X0 = ker(TS). Since kerS ⊂ kerTS, we have kerTS/ kerSS→ kerT , and since

kerS and kerT are finite dimensional, we have dim(kerTS) <∞.Step 2. Let X1 be a complementary subspace for X0 in X. Note that S is injective (in fact

isomorphism) on X1 to Y1 := S(X1). Now since X0 = kerTS, we have that T is also injective(hence an isomorphism) on Y1 to Z1 := T (Y1).

Step 3. Y1 has finite codimension in Y because Y = S(X) ⊕ f.d. = S(X1 ⊕ X0) + f.d. =S(X1) +S(X0) + f.d. and S(X0) is finite dimensional. Also, kerT ∩Y1 = 0 because T is injectiveon Y1.

Step 4. Choose Y0 a complement of Y1 in Y that contains kerT . Then T (Y0) ∩ T (Y1) = 0since if T (y0) = T (y1), then T (y0 − y1) = 0 and hence y0 − y1 ∈ kerT ⊂ Y0 so that y1 ∈ Y0

and thus y1 = 0. Also, since S(X0) = S(kerTS) ⊂ kerT , Z0 is f.d. because Z = T (Y ) + f.d. =T (Y1 + Y0) + f.d. = T (Y1) + T (Y0) + f.d. and T (Y0) is f.d.

Step 5. Choose a complement Z0 to Z1 with Z0 ⊃ S(X0). Now we have a diagram as follows:FIGUREAnd hence index(TS) = index(T0S0)+index(T1S1) = index(T0S0) = index(T0)+index(S0).

Theorem 6.6.2. Let T ∈ B(X,Y ). If ∃G1, G2 ∈ B(Y,X) such that G1T = IdX +K1 and TG2 =IdY +K2 where K1,K2 are compact operators, then T is Fredholm (as are G1 and G2).

N.B. 6.6.3. The pair (G1, G2) is called parametrix for T .

Proof. By earlier results we have IdX +K1 and IdY +K2 are Fredholm. kerT ⊂ ker(IdX +K1) whichis finite dimensional. Moreover, Im(T ) ⊃ Im(IdY +K) and hence has finite codimension. T is thusFredholm. That Im(G1) has finite codimension and G2 has finite dimensional kernel is likewiseeasy. Now, since G1 = G1 IdY = G1(TG2 −K2) = (G1T )G2 − G1K2 = (IdX +K1)G2 − G1K2 =G1 +K3.

45

Let F ∈ B(X,Y ) be Fredholm (where X,Y are Banach spaces), and let X0 = kerT and X1 bethe closed complement to X0 in X.

(end of class 11/23)Let T ∈ B(X,Y ) be Fredholm, X0 = kerT and let X1 its complementary closed in X. Y1, Y0

similar for Y . T an iso X1 → Y1. Let G = T−1 plus the 0 map on Y1, Y0. Note that G ∈ B(X,Y ).Claim: G a paremetricx for T .

6.7 Lecture 11/30/15 —

Recall: Let T ∈ B(X,Y ) be Fredholm, and X0 = kerT and X1 be a complement to X0. Let

Y1 = TX1 = Im(T ) and let Y0 be complement to Y1. Then we have XT→ Y decomposed into

X0 → Y0 (zero map) and X1 → Y1 (isomorphism) and by open mapping theorem T−1 is alsocontinuous so define G on Y by G on Y1 is T−1 and G on Y0 is 0.

Then GT is the projection of X on X1 along X0 which is IdX −projection of X on X0 along X1

and TG = IdY −projection of Y on Y0 along Y1. Thus we have:

Proposition 6.7.1. Let T ∈ B(X,Y ). Then TFAE:

1. T is Fredholm

2. There exists G ∈ B(Y,X) such that parametrix for T

3. There isG ∈ B(Y,X) withGT and TG satisfying what is above (GT = IdX = finite rank projection,etc.)

Remark 6.7.2. Well, GT = IdX +K for K compact, then let π : B(X) → B(X)/B0(X). Thenπ(GT ) = Id in B(X)/B0(X). And conversely. Notice that for T ∈ B(X), then T is Fredholm iffπ(T ) is invertible in B(X)/B0(X).

Proposition 6.7.3. Let T ∈ Fred(X,Y ). If index(T ) = 0, then

T = (an isomorphism from X to Y )− (finite rank)

Proof. Then dim(X0) = dim(Y0). So let T0 be an isomorphism of X0 with Y0, extended to be 0 onX1. Then T + T0 is an isomorphism of X onto Y .

Proposition 6.7.4. Fred(X,Y ) is open in B(X,Y ) (X,Y Banach spaces).

Proof. Let T ∈ Fred(X,Y ). Choose a parametrix G, so GT = IdX +K1, and TG = IdY +K2.Then once again let π : B(X) → B(X)/B0(X), so that π(GT ) = I, and let S ∈ B(X,Y ). Nowconsider GS. ‖π(GS)−π(GT )‖ = ‖π(G(S−T ))‖ Thus, if ‖π(GS)−I‖ < 1 then π(GS) is invertible,and ‖π(G(S − T ))‖ ≤ ‖G‖‖S − T‖ so need ‖S − T‖ < ‖G‖−1 so GS = invertible +K.

By a similar argument, the set of isomorphisms of X onto Y (if any) is an open subset ofB(X,Y ).

Theorem 6.7.5. The index is continuous on Fred(X,Y ).

46

Proof. Let T ∈ Fred(X,Y ). We show that index is locally constant. Decomposing as before,

the matrix for T is

[0 00 T22

]. Now let S ∈ Fred(X,Y ), has matrix

[S11 S12

S21 S22

](with respect

to the same decomposition for T ). S22 = proj of Y on Y1 along Y0 S|X1 . Then ‖S22 − T22‖ ≤‖proj of Y on Y1 along Y0‖‖S − T‖, and if this quantity is < 1 then S22 is an invertible operatorfrom X1 to Y1.

So, now assume S22 is an isomorphism of X1 to Y1. Claim: SX1∩Y0 = 0. Because for y1 ∈ X,

Sx1 =

[S12x1

S22x1

]6= 0 if x1 6= 0. Claim: SX1 is a complement to Y0. For y ∈ Y , write y =

[y0

y1

]and

∃x1 ∈ X with S22x1 = y1. So Sx1 =

[S21x1

S22x1

]=

[S21x1

y1

]so Sx1 +

[y0 − S21x1

0

]=

[y0

y1

]= y.

Now, let X2 = kerS. Clearly X2 ∩ X1 = 0. So there exists X3 with X = X1 ⊕ X2 ⊕ X3.Then Im(S) = SX1⊕SX3 and S is injective on X3. So dim(SX3) = dim(X3), and codim(ImS) =dim(Y/(Y1 ⊕ SX3)) = dim((Y/Y1)/(image of SX3 in it)) = dim(Y/Y1)− dim(SX3) = codim(T )−dim(X3). Then index(S) = dimX2 − dim cokerT − dimX3 = dimX2 + dimX3 − dim(cokerT ) =dim(X2 ⊕X3)− dim cokerT = dimX0 − dim cokerT = dim(kerT )− dim cokerT = index(T ).

6.8 Lecture 12/2/15 —

Definition 6.8.1. Given Banach space X, its projection constant is the smallest constant suchthat if X is isometrically embedded in any Banach Y , then there is a projection of Y onto X withnorm no biggr than this constant.

Example 6.8.2. Let R2 with Euclidean norm. Its projection constant is 4/π. Consider C(T ) andthe mapping (α, β)→ α sin t+ β cos t.

Proposition 6.8.3. Let T ∈ Fred(X,Y ), and K ∈ B0(X,Y ). Then T + K is Fredholm, andindex(T +K) = index(T ).

Proof. Let S be a parametrix for T , i.e. ST = IdX +K1 and TS = IdY +K2. Then S(T + K) =Idx +K1 + SK (latter two sum to compact operator) So S is a parametrix for T +K so T +K isFredholm. For t ∈ [0, 1], T + tK, norm continuous path through Fred(X,Y ) from T +K to T . Soby continuity of index, we have index(T +K) = index(T ).

Corollary 6.8.4. If K ∈ B0(X), then I +K has index 0.

Proposition 6.8.5. If T1, T2 ∈ Fred(X,Y ) and index(T1) = index(T2), then T2 = T1M +K whereM ∈ B(X) and M is invertible.

Proof. Let S be a parametrix for T1. Then index(S) = − index(T1) = index(T2), and 0 =index(ST2), so ST2 = M + K, M invertible on X. T1(ST2) = T1M + T1K and (T1S)T2 =(IdY +K2) = T2 + T2K2.

Proposition 6.8.6. If T ∈ Fred(X,Y ) and S is a parametrix for T , so ST = IdX +K1 andTS = IdY +K2, then index(S) = − index(T ).

Proof. From indexS + indexT = index(ST ) = 0.

47

Now, consider π : B(X) → B(X)/B0(X) := C . If T ∈ B(X) is Fredholm, then π(T ) isan invertible element in the quotient. The set of invertible elements of B(X)/B0(X) is open inB(X)/B0(X). A preimage of an invertible element B(X)/B0(X) is Fredholm, if β ∈ invtl(C ) andif the path t 7→ α+ tβ is in invtl(C ), and if T, S are preimages of α, β, so t 7→ T + tS is a path inFred(X), then indexT = indexS. Thus, the index is well-defined on invtl(C ) and is constant onpath components.

Remark 6.8.7. Let T ∈ Fred(X,Y ), and assume index(T ) = 0. The Fredholm alternative is either(i) the equation Tx = 0 has a unique solution, in which case the equation Tx = y has a (unique)solution for all y ∈ X, or (ii) the dimension of kerT is ≥ 1 in which case there are ϕ1, . . . , ϕn ∈ Y ∗such that Tx = y has a solution iff ϕj(y) = 0 ∀j = 1, . . . , n.

Proposition 6.8.8. Let K ∈ B0(X), for λσB(X)(K) with λ 6= 0, then K has a λ-eigenvector.

Proof. K−λI, Fredholm, with index 0. And since K−λI is not invertible, and since dim(ker(K−λI)) = codim(Im(K − λI)), so if the latter is zero then K − λI is invertible.

48