Algebra

Suppose that

A

is a vector space and a map

A \times A \to A

which is bilinear. An algebra is such a structure such that

(a b) c = a (b c)

Unital

An algebra is said to be unity if there exists an element

1 \in A

such that

1 a = a 1 = a

for any

a \in A

Unital Subalgebra

B

is called a unital subalgebra if

B

is a subalgebra of

A

and

1 \in B

Consider $A = C ([0, 1] \cup [2, 3])$ which is a unital algebra over pointwise operations, then if we set $B = {f \in A : f | [2, 3] = 0} \equiv C ([0, 1])$ which is not a unital subalgebra.

Subalgebra

A subalgebra of

A

is a subacpe of

B \subseteq A

such that for any

a, b \in B

such that

a b \in B

Normed Algebra

‖ \cdot ‖

is a norm on

A

and

‖ \cdot ‖

is submultiplicative, that is

‖ a b ‖ \leq ‖ a ‖ ‖ b ‖

Then

A

is a normed algebra

Unital Normed Algebra

A

is a normed algebra and

‖ 1 ‖ = 1

then

A

is said to be a unital normed algebra

Banach Algebra

A completed normed algebra is a Banach Algebra

Sup Norm

{‖ f ‖}_{\infty} = \sup_{x \in S} ‖ f (x) ‖

week 6

Continuous and Bounded Functions With the Sup-norm Form a Unital C* Algebra

Suppose that

X

is a compact hausdorff topological space, then

C_{b} (X) = {f : X \to ℂ : f is continuous and bounded}

is a unital

C^{*}

-algebra when equipped with the sup-norm

The set $C_{b} (X)$ consists of functions from $X$ to $ℂ$ that are continuous and bounded. It is closed under addition and scalar multiplication, if $f, g \in C_{b} (X)$ , then the function $(f + g) (x) = f (x) + g (x)$ is continuous (since the sum of continuous functions is continuous) and bounded (since the sum of two bounded functions is bounded). Similariy for any $α \in ℂ$ and $f \in C_{b} (X)$ , the function $(α f) (x) = α f (x)$ is also continuous and bounded. Therefore, $C_{b} (X)$ is a vector space over $ℂ$ .

We also need to show that $C_{b} (X)$ is closed under pointwise multiplication, so if $f, g \in C_{b} (X)$ , the function $(f \cdot g) (x) = f (x) g (x)$ is continuous (since the product of continuous functions is continuous) and bounded (the product of two bounded functions is bounded) so, $C_{b} (X)$ is an algebra over $ℂ$ .

We define the sup-norm $‖ f ‖_{\infty} = \sup_{x \in X} | f (x) |$ . We check the properties of a norm: For non-negativity, we see: $‖ f ‖_{\infty} \geq 0$ for all $f$ , and $‖ f ‖_{\infty} = 0$ if and only if $f = 0$ . Scalar multiplication also holds for $α \in ℂ$ and $f \in C_{b} (X)$ , $‖ α f ‖_{\infty} = \sup_{x \in X} | α f (x) | = | α | \sup_{x \in X} | f (x) | = | α | ‖ f ‖_{\infty} .$ as does the triangle inequality, given $f, g \in C_{b} (X)$ , $‖ f + g ‖_{\infty} = \sup_{x \in X} | f (x) + g (x) | \leq \sup_{x \in X} | f (x) | + \sup_{x \in X} | g (x) | = ‖ f ‖_{\infty} + ‖ g ‖_{\infty} .$

We must verify the C*-algebra condition, we need to show that: $‖ f^{*} f ‖_{\infty} = ‖ f ‖_{\infty}^{2} .$ Here, $f^{*}$ is defined as the complex conjugate of $f$ , so: $f^{*} f (x) = | f (x) |^{2} .$ Thus, $‖ f^{*} f ‖_{\infty} = \sup_{x \in X} | f (x) |^{2} = {(\sup_{x \in X} | f (x) |)}^{2} = ‖ f ‖_{\infty}^{2} .$

The constant function $1$ (where $1 (x) = 1$ for all $x \in X$ ) is in $C_{b} (X)$ and acts as a multiplicative identity: $f \cdot 1 = f for all f \in C_{b} (X) .$ This shows that our algebra is unital

Left Ideal

A left ideal in an algebra

A

is a vector subspace

I \subseteq A

such that if

a \in A

and

b \in I

then we have

a b \in I

The definition for right ideal is the same, but instead we have that $b a \in I$ . The definition for a general ideal is one such that $a b \in I$ and $b a \in I$ (which would happen automatically if the algebra was commutative.

Note that the sets ${0}$ and $A$ are ideals, and so we call them the trivial ideals, thus we call those ideals which are not ideals proper ones.

Maximal Ideal

A maximal ideal in

A

is a proper ideal in

A

that is not contained in any other proper ideal

Invertible Elements of an Algebra

Suppose that

A

is an algebra, then we define

Inv (A) = {a \in A : a is invertible}

Note that the above is a group under mulitplication.

Spectrum

σ (a) = {λ \in ℂ : λ 1_{A} - a \notin inv (A)}

Polynomials and Spectrum Commute

Let

a \in A

where

A

is a unital algebra. If

σ (a)

is non-empty and

p \in ℂ [z]

then we have

σ (p (a)) = p (σ (a))

We may suppose that

p

is non constant. Then if

w \in ℂ

then we have

r_{0}, \dots, r_{n} \in ℂ

with

r_{0} \neq 0

such that

p (z) - w = r_{0} (z - r_{1}) \dots (z - r_{n})

so that

p (a) = r_{0} (a - r_{1}) \dots (a - r_{n})

to conclude that

p (a) - w

is invertible iff

a - r_{1}, \dots, r_{n}

are invertible. So now

w \in σ (p (a))

iff

w = p (j)

for some

j \in σ (a)

so that

σ (p (a)) = p (σ (a))

as needed.

Newmann Series

Let

A

be a unital Banach algebra and let

a \in A

such that

‖ a ‖ < 1

, then

1 - a \in Inv (a)

and

{(1 - a)}^{- 1} = \sum_{n = 0}^{\infty} a^{n}

We leverge the geometric series and properties of the norm to see that $\sum_{n = 0}^{\infty} ‖ a^{n} ‖ = \sum_{n = 0}^{\infty} {‖ a ‖}^{n} = \frac{1}{1 - ‖ a ‖} < \infty$ thus we see that that the series $\sum_{n = 0} \to \infty a^{n}$ is convergent to some element $c \in A$ .

In the proof of the geometric series we noted that $(1 - a) (\sum_{i = 0}^{n} a^{i}) = 1 - a^{n + 1}$ then as $n \to \infty$ we note that this equality becomes $(1 - a) c = c (1 - a) = 1$ showing that $1 - a$ is invertible and its inverse is $c = \sum_{i = 0}^{\infty} a^{i}$ as needed.

The Spectrum of an Element From a Unital Banach Algebra Is Non-Empty

a \in A

where

A

is a unital banach algebra, then

σ (a) \neq \emptyset

Suppose for the sake of contradiction that $σ (a) = \emptyset$ . If $| λ | > 2 ‖ a ‖$ , then $‖ λ^{- 1} a ‖ < \frac{1}{2}$ so that , $1 - ‖ λ^{- 1} a ‖ \geq \frac{1}{2}$ . So we have: $\begin{matrix} ‖ {(1 - λ^{- 1} a)}^{- 1} - 1 ‖ & = ‖ \sum_{n = 1}^{\infty} {(λ^{- 1} a)}^{n} ‖ \\ \leq \frac{‖ λ^{- 1} a ‖}{1 - ‖ λ^{- 1} a ‖} \\ \leq 2 ‖ λ^{- 1} a ‖ \\ < 1 \end{matrix}$ To deduce that $‖ {(1 - λ^{- 1} a)}^{- 1} ‖ < 2$ so we have $‖ {(a - λ)}^{- 1} ‖ = ‖ λ^{- 1} {(1 - λ^{- 1} a)}^{- 1} ‖ < \frac{2}{| λ |} < {‖ a ‖}^{- 1}$ which is justified since $σ (a) = \emptyset$ so $a \neq 0$

Now note that the map $λ \mapsto {(a - λ)}^{- 1}$ is continuous and that it is bounded over the disk of radius $2 ‖ a ‖$ (as the disk is compact) so that we can conclude it must be bounded on all of $ℂ$ so there is some $R \in ℝ^{+}$ where we have $‖ {(a - λ)}^{- 1} ‖ \leq R$ for every $λ \in ℂ$

If $τ \in A^{*}$ (the dual space of $A$ ) then the function (the evaluation) $λ \mapsto τ {(a - λ)}^{- 1}$ is entire (because the dual space evaluation preseves the analyticity of ${(a - λ I)}^{- 1}$ ) and bounded by $R ‖ τ ‖$ .

Thus by Liouville's theorem we conclude that evaluation map is constant, and that in particular we have that $τ (a^{- 1}) = τ ({(a - 1)}^{- 1})$ for every $τ \in A^{*}$ we can say that $a^{- 1} = {(a - 1)}^{- 1}$ so that $a = a - 1$ which is a contradiction.

Gelfand

a

is an element of a unital banach algebra

A

then

σ (a) \neq \emptyset

TODO: Add the proof here.

Spectral Radius

Suppose that

a

is an element of a banach algebra

A

then its spectral radius is defined as

r (a) = \sup_{λ \in σ (a)} | λ |

Multiplicative Linear Functional

A multiplicative linear functional

τ

is a linear map

τ : A \to ℂ

that satisfies:

Multiplicativity: $τ (a b) = τ (a) τ (b)$ for all $a, b \in A$ .
Unital property: $τ (1) = 1$ , where $1$ is the multiplicative identity in $A$ .

Gelfand Space

In the context of a unital commutative Banach algebra

A

Δ (A)

denotes the Gelfand space of

A

defined as:

Δ (A) = {τ : A \to ℂ | τ is a nonzero multiplicative linear functional on A} .

Gelfand Transformation

Let

A

be an algebra. For

a \in A

, one defines

\hat{a} (τ) : = τ (a) (τ \in Δ (A))

The function

\begin{matrix} \hat{a} : Δ (A) & \to ℂ \\ τ & \mapsto \hat{a} (τ) . \end{matrix}

is called the Gelfand transform of

a

unitization

Unitization

Let

A

be an algebra. If

A

is unital, one defines

\tilde{A} : = A

. Assume now that

A

has no unit. One then defines

\tilde{A} : = ℂ \oplus︎ A

(direct sum of vector spaces). One imbeds

A

into

\tilde{A}

via

a \mapsto (0, a)

. One defines

e : = (1,0)

, so that

(λ, a) = λ e + a

for

λ \in ℂ

a \in A

. In order for

\tilde{A}

to become a unital algebra with

e

as unit and with multiplication extending the one in

A

, the multiplication in

\tilde{A}

must be defined by

(λ e + a) (μ e + b) = (λ μ) e + (λ b + μ a + a b) (λ, μ \in ℂ, a, b \in A)

and this definition indeed satisfies the requirements. One says that

\tilde{A}

is the unitisation of

A

. If

A

is a

*

-algebra, one makes

\tilde{A}

into a unital *-algebra by putting

{(λ e + a)}^{*} : = \overline{λ} e + a^{*} (λ \in ℂ, a \in A) .

Abstract Wiener's

Let

A

be a unital commutative banach algebra. An element

a

is not invertible in

A

iff

\hat{a}

vanishes at some

τ \in Δ (A)

TODO: Add the proof here.

Spectrum Is the Gelfand Image of Multiplicative Functionals

Let

A

be a commutative Banach algebra. For

a \in \tilde{A}

then we have

sp (a) = \hat{a} (Δ (\tilde{A}))

Note that the following statements are equivalent.

\begin{matrix} λ \in sp (a), \\ λ e - a is not invertible in \tilde{A}, \\ (\hat{λ e - a}) (\tilde{τ}) = 0 for some \tilde{τ} \in Δ (\tilde{A}), by Abstract Weiner’s \\ λ = \hat{a} (\tilde{τ}) for some \tilde{τ} \in Δ (\tilde{A}) . \end{matrix}

week 7

Spectral Sub Additivity and Multiplicativity

Let A be a commutative Banach algebra. For two elements

a, b

A

we have

\begin{matrix} sp (a + b) & \subset sp (a) + sp (b) \\ sp (a b) & \subset sp (a) sp (b) \end{matrix}

Suppose that $λ \in sp (a + b)$ then there exists $\tilde{τ} \in Δ (\tilde{A})$ such that $\tilde{τ} (a + b) = λ$ , but then $\tilde{τ}$ was linear so that we have $λ = \tilde{τ} (a) + \tilde{τ} (a)$ which is an element of $sp (a) + sp (b)$ by the same proposition, as needed.

For multiplication the same reasoning holds but instead of using the linearity we use the multiplicativity.