< } Preliminaries

Majorant

An element

x

is called a majorant for a subset

Y

X

y \leq x

for every

y \in Y

Filtering Upward

We say that an order is filtering upward if every pair in

X

(and hence every finite subset of

X

) has a majorant.

Net

A net is a space

X

is a pair

(A, i)

where

A

is an upward-filtering ordered set and

i

is a map from

A \to X

If a Cauchy Sequence has a Convergent Subsequence, then the Original Converges There

Let

X, d

be a metric space, then if

(x_{n}) : N_{1} \to X

is cauchy, and there exists a subsequence

(x_{σ (n)})

that converges to

x

then

(x_{n}) \to x

as well.

Since

(x_{σ (n)}) \to x

then given

ϵ \in R^{+}

we have some

K

such that for any

k \geq K

we have

d (x_{σ (k)}, x) < \frac{ϵ}{2}

since

(x_{n})

is assumed cauchy then we obtain some

N^{'}

such that for all

n, m \geq N^{'}

we know that

d (x_{n}, x_{m}) < \frac{ϵ}{2}

, take

N = N^{'}

and let

n \geq N

since

σ

is strictly increasing then there is some

i \in N_{1}

such that

σ (i) \geq N

let

j : = max (i, K)

thus we have

d (x_{n}, x) \leq d (x_{n}, x_{σ (i)}) + d (x_{σ (i)}, x) < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ

as needed.

Normed Spaces

A bilinear form is a generlization of the idea of the dot product:

Bilinear Form

A bilinear form on a vector space

V

over a field

𝔽

is a map

H : V \times V \to 𝔽

such that

$H (v_{1} + v_{2}, w) = H (v_{1}, w) + H (v_{2}, w)$ , for all $v_{1}, v_{2}, w \in V$
$H (v, w_{1} + w_{2}) = H (v, w_{1}) + H (v, w_{2})$ , for all $v, w_{1}, w_{2} \in V$
$H (a v, w) = a H (v, w)$ , for all $v, w \in V, a \in 𝔽$
$H (v, a w) = a H (v, w)$ , for all $v, w \in V, a \in 𝔽$

A bilinear form $H$ is called symmetric if $H (v, w) = H (w, v)$ for all $v, w \in V$ .
A bilinear form $H$ is called skew-symmetric if $H (v, w) = - H (w, v)$ for all $v, w \in V$ .
A bilinear form $H$ is called non-degenerate if for all $v \in V$ , there exists $w \in V$ , such that $H (w, v) \neq 0$ .
A bilinear form $H$ defines a map $H^{#} : V \to V^{*}$ which takes $w$ to the linear map $v \mapsto H (v, w)$ . In other words, $H^{#} (w) (v) = H (v, w)$ .

Note that

H

is non-degenerate if and only if the map

H^{#} : V \to V^{*}

is injective. Since

V

and

V^{*}

are finite-dimensional vector spaces of the same dimension, this map is injective if and only if it is invertible.

Sesquilinear Form

Let

F

be a subfield of

C

and

U

and

V

be vector spacesover

F

, then a sesquilinear form is a function

u : U \times V \to C

such that

\forall α \in F, x_{1}, x_{2} \in U, y \in V

we have:

$u (α x_{1} + x_{2}, y) = α u (x_{1}, y) + u (x_{2}, y)$
$u (x, α y_{1} + y_{2}) = \bar{α} u (x, y_{1}) + u (x, y_{2})$

Note that if $F$ is a subfield of $R$ , then a sesquilinear form is a bilinear form.

Isomorphism Between Hilbert Spaces

Suppose that

H

and

K

are hilbert spaces then an isomorphism between them is a surjective linear operator

U : H \to K

such that

{⟨ U h, U k ⟩}_{K} = {⟨ h, k ⟩}_{H}

Isometry

A linear operator

T : V \to W

is an isometry if for all vectors

v_{1}, v_{2} \in V

we have that

{‖ T (v_{1}) - T (v_{2}) ‖}_{W} = {‖ v_{1} - v_{2} ‖}_{V}

Unitary Operator

We say that a linear operator

U : H \to K

between hilbert spaces is unitary whenever

U^{*} U = {id}_{H} and U U^{*} = {id}_{K}

Bounded Linear Operators on a Hilbert Space

Suppose that

H

is a hilbert space then we use the notation

B (H)

to denote all bounded linear operators

week 3

Unitary Operator Equivalences

Suppose that

U : H \to K

is a bounded linear operator between hilbert spaces, then the following are equivalent:

$U$ is a unitary operator
$U$ is a surjective and ${⟨ U h, U k ⟩}_{K} = {⟨ h, k ⟩}_{H}$
$U$ is a surjective isometry

Let's prove that $1 ⟹ 2$ so suppose that $U$ is unitary, so then we know that $U U^{*} = {id}_{K}$ which helps show that $U$ is surjective as given any $k \in K$ then we know that $U (U^{*} k) = k$ , but then additionally we know that $U^{*} U = id H$ so that then we note that $⟨ U x, U y ⟩ = ⟨ x, U^{*}, U ⟩ = ⟨ x, y ⟩$ as needed.

Now we prove that $2 ⟹ 3$ , but all we have to do is consider any $h \in H$ and thus we have that ${‖ U h ‖}^{2} = {⟨ U h, U h ⟩}_{K} = {⟨ h, h ⟩}_{H} = {‖ h ‖}^{2}$ and thus since the norm is non-negative then we can undo the square to obtain that ${‖ U h ‖}_{K} = {‖ h ‖}_{H}$ as needed.

Now we finish $3 ⟹ 1$ , we note that $⟨ U^{*} U x, x ⟩ = ⟨ U x, U x ⟩$ as the adjoint is an involution, but then $⟨ U x, U x ⟩ = {‖ U x ‖}^{2} = {‖ x ‖}^{2} = ⟨ x, x ⟩$ , thus we've shown that $⟨ U^{*} U x, x ⟩ = ⟨ x, x ⟩$ so we conclude that $U^{*} U = I$ which shows that $U$ is an invertible linear isometry with $U^{- 1} = U^{*}$ , in otherwords $U$ is unitary.

Unilateral Shift Operator

Let

H = ℓ^{2} (N)

then we define the unilateral shift operator as

S : H \to H

given by

S (x_{1}, x_{2}, x_{3}, x_{4}, \dots) = (0, x_{1}, x_{2}, x_{3}, \dots)

The Unilateral Shift Is an Isometry

As per title.

Consider the following

\begin{matrix} ‖ S (x) ‖ & = \sqrt{\sum_{n = 1}^{\infty} {| {(S x)}_{n} |}^{2}} \\ = \sqrt{\sum_{n = 2}^{\infty} 0 + {| x_{n - 1} |}^{2}} \\ = \sqrt{\sum_{n = 1}^{\infty} {| x_{n} |}^{2}} \\ = ‖ x ‖ \end{matrix}

The Unilateral Shift Is Not Unitary

As per title.

We show it's not unitary by showing that it is not surjective, which is the case as the element

(1, 0, 0, 0, \dots)

cannot be reached via

S

The Adjoint of the Unilateral Shift Is the Bilateral Shift

As per title.

https://math.stackexchange.com/questions/1043965/eigenvalue-of-a-unilateral-shift-operator/1044057#1044057 https://chatgpt.com/share/672160d4-7b14-8007-91cb-a5a8e41d359e Just note:

\begin{matrix} ⟨ S (x), y ⟩ & = ⟨ (0, x_{1}, x_{2}, \dots), (y_{1}, y_{2}, y_{3}, \dots) ⟩ \\ = \end{matrix}

First we take a look at

=

The Bilateral Shift Is Unitary

TODO: Add the content for the proposition here.

Closed Graph Theorem

Let

X, Y

be banach spaces and

T : X \to Y

a linear operator

Automorphism

Let

F

be a field, then an automorphism of

F

is a bijection from

F

to itself that preserves the operations of addition and multiplication.

Sesquilinear Form

A sesquilinear form on a vector space

V

over a field

F

is a map

⟨ \cdot, \cdot ⟩ : V \times V \to F

that is linear in the right argument and almost linear in the left, which is to say:

$⟨ v_{1}, c w_{1} ⟩ = c ⟨ v_{1}, w_{1} ⟩$
$⟨ v_{1}, w_{1} + w_{2} ⟩ = ⟨ v_{1}, w_{1} ⟩ + ⟨ v_{1}, w_{2} ⟩$
$⟨ c v_{1}, w_{1} ⟩ = \bar{c} ⟨ v_{1}, w_{1} ⟩$
$⟨ v_{1} + v_{2}, w_{1} ⟩ = ⟨ v_{1}, w_{1} ⟩ + ⟨ v_{2}, w_{1} ⟩$

Adjoint of a Sesquilinear Form

Given a sesquilinear form,

(\cdot, \cdot)

then we define the adjoint form as

{(x ∣ y)}^{*} = (\bar{y} ∣ \bar{x})

Self Adjoint Sesquilinear Form

We say that a sesquilinear form is self adjoint diff:

(x ∣ y) = {(x ∣ y)}^{*}

Hilbert Space

A hilbert space is a real or complex inner product space that is also a complete metric space with respect to the norm

‖ x ‖ = \sqrt{⟨ x, x ⟩}

. In other words the vector space under discussion is a Banach space.

Self Adjoint Matrix

A synonym for hermitian.

Convex Set

A convex set is a subset

C

of a vector space such that for any two points

x, y \in C

, the line segment connecting

x

and

y

is entirely contained within

C

. Formally, for all

λ \in [0, 1]

, the point

λ x + (1 - λ) y \in C

A Convex Subset of a Hilbert Space Has a Unique Closest Element to a Point in the Hilbert Space

C

is a closed, nonempty, convex subset of a hilbert space

H

, then for every

y \in H

there is a unique

x \in C

that minimizes the distance from

y

C

Let

C

be a subset as specified above, and let

y \in H

, finding the closest point in

C

y

is equivalent to finding the closest point in

D = C - y

0

, which is simpler to solve, so we prove this equivalent statement. Now set

p = inf ({‖ x ‖ : x \in D})

, , and note from the parallelogram law we have that for any

x, y \in D

‖ x - y ‖^{2} = 2 (‖ x ‖^{2} + ‖ y ‖^{2}) - ‖ x + y ‖^{2}

Note that since

\frac{1}{2} x + \frac{1}{2} y \in D

then we know that

‖ \frac{1}{2} x + \frac{1}{2} y ‖ \geq p

so that

‖ x + y ‖ \geq 2 p

in otherwords from the previous equality we obtain

‖ x - y ‖^{2} \leq 2 (‖ x ‖^{2} + ‖ y ‖^{2}) - 4 p^{2}

Now as

p

is the infimum ,then we can find a sequence

(x_{n}) : N_{1} \to D

such that

‖ x_{n} ‖ \to p

and more specifically

‖ x_{n} ‖^{2} \to p^{2}

, we'll show that

x_{n}

is cauchy, to do so let

ϵ \in R^{+}

then since

‖ x_{n} ‖^{2} \to p^{2}

then we obtain some

N^{'}

such that for any

k \geq N^{'}

we have

‖ x_{k} ‖ < p^{2} + ϵ

, now in our context we take

N = N^{'}

and let

n, m \geq N

therefore we have that:

‖ x_{n} - x_{m} ‖^{2} \leq 2 (‖ x_{n} ‖^{2} + ‖ y_{m} ‖^{2}) - 4 p^{2} \leq 2 ((p^{2} + ϵ) + (p^{2} + ϵ)) - 4 p^{2} = 4 ϵ

where we could then say

‖ x_{m} - x_{n} ‖ < 2 \sqrt{ϵ}

therefore

x_{n}

is cauchy, and thus

x_{n}

converges to some point

y \in H

, but since

D

is closed then also

y \in D

, moreover we know that

‖ y ‖ = {lim}_{n \to \infty} ‖ x_{n} ‖ = p

, thus we've shown that such a smallest element in

D

exists. To show it unique, then suppose there were two

u, v \in D

that were the smallest, but then revisiting our old inequality we have:

‖ u - v ‖ \leq 2 (‖ u ‖^{2} + ‖ v ‖^{2}) - 4 p^{2} = 2 (p^{2} + p^{2}) - 4 p^{2} = 0

so we deduce

u = v

Functional

Given a vector space

(V, F)

Spectrum of a Linear Operator

Spec (T) = {λ \in C : T - λ I is not invertible}

Co-kernel

The co-kernel of a linear opertor

T : V \to V

coker (T) = V ⧵ im (T)

Fredholm Operator

We say that a linear operator

T : X \to Y

is fredholm when

dim (ker (T)) < \infty

and

dim (Y ⧵ im (T)) < \infty

Finite Rank Linear Operator

A linear operator

T

on a hilbert space

H

has finite rank if

T (H)

is a finite dimensional subspace of

H

When you look at the spectrum of a finite rank operator their spectrum is always discrete and is only the eigen values that they have.

Compact Operators Are Limits of Finite Rank Operators

TODO: Add the content for the proposition here.

This means that the eigenvalues form a sequence that converges to 0, and this is what the spectrum of a finite rank operator looks like. If we have a compact self adjoint linear operator then we know that the eigenvalues all sit on $R$

Spectral Theorem for Compact Operators

Suppose that

T

is compact and self-adjoint operator on a seprable hilbert space

H

, then there is a sequence

{(λ_{n})}_{N_{1}}

of eigenvalues of

T

and an orthanormal basis

{(b_{n})}_{N_{1}}

H

such that

λ_{n} \to 0

and

T b_{n} = λ_{n} b_{n}

Self Adjoint Linear Operator

We say that

T = T^{*}

which means for all

x, y \in H

⟨ T x, y ⟩ = ⟨ x, T y ⟩

Real Spectrum Iff Self Adjoint

Spec (T) \subseteq R

iff

T

is self-adjoint

Positive Linear Operator

Spec (T) \subseteq [0, \infty)

and we say that

T \geq 0

The Square Root of a Linear Operator Exists If It Is Positive

proof diagonizliz T, and define

\sqrt{T}

\sqrt{T} b_{n} = \sqrt{λ_{n}} b_{n}

, show that the square roots commute because they square to T, and so you should be able to simulatenuously diagonilize them so the square of the eigens are equal so then the square roots are equal.

If T Is Self Adjoint Then T Is Bounded

TODO: Add the content for the theorem here. (Hellinger and Topis)

week 1

Nth Dirichlet Kernel

Let

T

be the unit circle and for each

n \in Z

let

χ_{n} : T \to C

be the function

χ_{n} (x) = e^{2 π i n x}

and for each

N \in N_{1}

let

D_{N} \in C (T)

be the n-th dirichlet kernel defined as

D_{N} (x) = \sum_{n = - N}^{N} χ_{n} (x)

The Dirichlet Kernel Is Real Valued and Integrates to 1

D_{N}

is real-valued, specifically:

D_{N} (x) = \frac{\sin ((N + \frac{1}{2}) 2 π x)}{\sin (π x)}

x \neq 0

, and

D_{N} (0) = 2 N + 1

. Moreover we have that

\int_{𝕋} D_{N} (x) d x = 1

When

x = 0

we have that

e^{2 π i n 0} = 1

therefore

\sum_{- N}^{N} χ_{n} (0) = 2 n + 1

where the plus one comes from

n = 0

, when we have that

x = 0

then we have the following

\begin{matrix} \sum_{- N}^{N} χ_{n} (0) & = \sum_{- N}^{N} {(e^{2 π i x})}^{n} \\ = \sum_{- N}^{N} {(e^{2 π i x})}^{n} \\ = {(e^{2 π i x})}^{- N} + \dots + {(e^{2 π i x})}^{0} + \dots + {(e^{2 π i x})}^{- N} \\ = e^{- 2 π i N x} (1 + \dots + {(e^{2 π i x})}^{2 N}) \end{matrix}

Now recall that

Bounded Linear Functional Integral

The linear functional

T_{n} : C (𝕋) \to ℝ

defined by

T_{n} f = \int_{𝕋} f (x) D_{n} (x) d x

is bounded, with

‖ T_{n} ‖ = \int_{𝕋} | D_{n} (x) | d x .

Lemma 4.7.

Embryionic Spectral Theorem

For any

p \in R [x]

and

T

which is adjoint and positive then if we know that for all

t \in [- ‖ T ‖, ‖ T ‖

we have

‖ p (T) ‖ \leq ‖ p (t) ‖

L2 Space

L^{2}

space constists of square integrable functions so that

L^{2}

is the collection of functions such that

\int {| f (x) |}^{2} d x < \infty

Multiplication Operator

Suppose that

f \in L^{2}

then we define the multiplication operator for

f

M_{f} : L^{2} \to L^{2}

such that

M_{f} (g) = f g

Trace-class Linear Operator

Let

ℋ

be a Hilbert space. A linear operator

A : ℋ \to ℋ

is called trace-class if its trace-class norm

‖ A ‖_{t c} = {sup}_{(v_{n}), (w_{n})} \sum_{n = 1}^{N} | ⟨ A v_{n}, w_{n} ⟩ |

is finite, where the supremum is taken over all integers

N ⩾ 0

and over any two finite lists of orthonormal vectors

(v_{1}, \dots, v_{N})

and

(w_{1}, \dots, w_{N})

of the same length

N

Algebra

(1.13). Definition (unitisation).

Consider $ℓ^{\infty} (X)$ (the collection of bounded functions from $S$ into $C$ with the sup norm, then it is a banach algebra, also it is unital with the constant 1 function, and moreover is commutative

$C_{b} (X)$ where $X$ is a non-empty topological space with the sup norm is a banach algebra also it is unital with the constant 1 function, and moreover is commutative

$C_{0} (X)$ the functions which vanish at infinity, where $X$ is a locally compact hausdorff space is also a banach algebra

in a measure space, then $L^{ω} (X, μ)$ the measure space is also a banach algebra

A = disk algebra, ${f \in C (\bar{D}) : f is holo on int (D)}$ where $D$ is the unit disk in $C$ is not a banach algebra. Note that this is not a $C^{*}$ algebra

$X$ is a NVS and $Y$ a baanch space, then $B (X, Y)$ the bounded functions? is a banach algebra

$M_{n} (C)$ the matrix algebras (n x n matrices with values in C), are banach algebras

$M_{n} (A)$ is a banach algebra whenever $A$ is.

Now a bunc of facts from tutotirla

If $A$ is a banach algebra and $I$ is a proper ideal in $A$ then $\bar{I}$ is also a proper ideal in $A$ implies that maximal ideals are closed.

I is a closed ideal in A implies A / Iis a banach algebra

Given a homomorphism $ϕ : A \to B$ then $ker (ϕ)$ is a closed ideal and $ϕ$ is unitl if $ϕ (1_{A}) = 1_{B}$

Suppose that $A$ is unital, and $p (z) \in C [x]$ then given any $a \in A$ we can construct a map ${ev}_{a} : C [z] \to A$ via $p (z) = p (a)$ is an algebra homomorphism.

Def: $a \in A$ , then $sp (a) = {λ \in C : a - λ \notin GL (A)}$ is closed, and you can show that $λ \in sp (a)$ then $‖ λ ‖ \leq ‖ a ‖$

If $A$ is a banach algebra, then we do Unitization, making $A^{'} = A \circplus C$ as a vector space, if you define multiplication component wise it turns out not to be unital, but instead if you do $(a, λ) \cdot (b, μ) = (a b + λ b + μ a, λ μ)$ this multiplication makes $A^{^{'}}$ into a uanital algebra with unit $(0, 1)$ and $‖ a + λ ‖ = max (‖ a ‖, | λ |)$

C Star Algebra

C^{*}

algebra

A

is a banach algebra over the complex numbers, equipped with an involution (the

*

operation) satisfying the following properties, first of the algebra:

$A$ is a vector space over $C$
$a (b + c) = a b + a c$
$(λ a) b = λ (a b)$

then of the involution, stating that there is a map

* : A \to A

called the involution satisfying

${(a^{*})}^{*} = a$
${(a + b)}^{*} = a^{*} + b^{*}$
${(λ a)}^{*} = \bar{λ} a^{*}$
${(a b)}^{*} = b^{*} a^{*}$

Interaction with the norm

$‖ a * a ‖ = ‖ a ‖^{2}$

Open Mapping Theorem

Let

B_{1}, B_{2}

be two banach space and let

T \in B (B_{1}, B_{2})

be a surjective linear operator, then

T

is a open map

Fredholm Operators Plus Finite Rank Operators Are Fredholm Operators

Fred (V) + FinRan (V) = Fred (V)

First step is to reduce to the special case of $index (T) = 0$ so that the dimension of the kernel and kernel have the same dimension and thus are isomorphic so there is a bijective funciton between them.

So now we reduce to the case where $T$ is invertible,

The Product of Linear Operators With Index Still Has Index

TODO: Add the content for the proposition here.

Spectral Theorem for Normal Operators

https://personal.math.ubc.ca/~feldman/m511/spectralReview.pdf https://math.dartmouth.edu/~dana/bookspapers/ln-spec-thm.pdf

A Bounded Linear Operator Has Non-empty Spectrum

Let

B

be a Banach space over

ℂ

and

T

a bounded linear operator on

B

the spectrum

sp (T)

T

is non-empty.

An Eigenvalue for a Normal Linear Operator Is an Eigenvalue for the Adjoint When Taking the Conjugate

T \in B (H)

is normal, and if

λ, v

is an eigen value vector pair, then

v

is an eigen vector for

T^{*}

with eigen value

\bar{λ}

Consider the eigenspace of

λ

, that is

ℰ_{λ} = {v \in ℋ : T v = λ v}

. Since

T

is normal then we have:

T T^{*} v = T^{*} T v = λ T^{*} v

, which shows that

T^{*} v \in ℰ_{λ}

. If

w \in ℰ_{λ}

, then

\begin{matrix} (T^{*} v - \bar{λ} v, w) & = (T^{*} v, w) - \bar{λ} (v, w) \\ = (v, T w) - \bar{λ} (v, w) \\ = 0 \end{matrix}

Since

T^{*} v - \bar{λ} v \in ℰ_{λ}

, we have

T^{*} v = \bar{λ} v

spectral theory of banach algebras

embryionic spectral theorem for any p in R[x] and T = T* (adjoint) >= 0, then || p (T) || <= sup |t| <= ||T| of |p(t)| recall that the linear operators with the operator norm is a complete space proof pn(T) is cauchy because pn - pm is small pn(T) - pm(T) is small so that is cauchy so the limit exists define f(T), f in C_r([-||T||, ||T||]) adding definition of dense and then proof in 2.1 in a hilbert space if you have a closed subspace then it always has a complementary subspace equilvalenlty thereis an orthoganal complement in linear algebra a a complementary subspace is one that is orgnaogonzl one and the union is the entire space, that's one of the the fundamental theorems in hilbert space theory any closed convex set in a hilbert space, and a point which is not in the space, then there is a unique closest point in the convex set to the point. (prove this one) if you have two inner products on a vector space and they turn them into a hilbert space and if they hav the given a hilbertspace then the number of elements in a ahilbert space basis is unique, so we prove that the number for here is the same as the number fo rhere, so the way you do it is that the humbe rof elements in a basis ais the same number of elements in a dense subset of the hilbert space, and you look at the smallest cardinal number in thedencse set and that's tthe number of elements in the basis, and so once you have the two orthanomral basis for the two bases given a hilbert space H and a bounded operators T = T* in B(H) bounded operator, polynomials are functions in the operator too, and there's something mysterious about them in the first instance, you have a varaible x in the first but then you can plug in an operator to a polynomial, so suppose that p in C[x] (complex coeffiecients), suppose that ||T|| <= 1, then what you do is that you say that ||p(t)|| <= eps for t in [-1, 1], then if we change t to T then we also know that ||p(T)|| <= eps (that's the spectral theorem, because once you know it's its fairly clear sailing) the tutorial suppose we have some field k and a vector space over k C* (* is an ivoluation) an algebra is a ring with addition, multiplication, and a norm gelfands theorem : If A is a commutative unital C*-algebra, then A =~ C(X) = {f: X -> C: f cts} for some compact hausdorff X. C0(R) = {f : R -> C : f cts and lim(fx) x -> oo = lim x -> -oo f(x) = 0 } modules are like vector spaces, and modules/k are precicely vector spaces / k, but modules don't always have a basis, but recall that a basis would be a subset B of M st B is lin indep (in a module this means given m1, m2 in B, then the solution to the equation r1m1 + r2m2 = 0 the same way we do in vector spaces), and forall m in M, we can find finitely many elements such that it is a linear comb of them. given a banach space over C, then the dimension is not aleph 0 consider l^oo (N) = {(x1, x2, ...) that are bounded}, the a basis could be e1, e2, ... and we would guess this is a basis for l^oo, but the problem is that we have an infinite sum, and so we need an infinite sum to exists and htus need converergence but we don't have a norm. so we have another idea which is the shauder basis, the {e_i}'s form a shauder basis for l^w, but this depends on the norm in the space, given a banach space over C it has an uncountable hamel basis. Exercise produce the dimension of X over C in a caoniacal way a banach space is a hilbert space iff the inner product satisfies the paralellogram law. From any ellipse you get an inner product, suppose it has major axis a and minor axis b then the inner product is given by ||(x, y)||^2 = (x/a)^2 + (y/b)^2 then the equation is given by <., .>: VxV -> R: symmetric biliear forms, is degenerate iff for every x, there is a y such that >0 Thm Riesz lemma ||<*, x>||_H* = ||x||_H add the https://www.math.mcgill.ca/jakobson/courses/ma667/mendelsontomberg-spectral.pdf https://www.math.uwo.ca/faculty/khalkhali/files/Fredholm.pdf https://math.stackexchange.com/questions/282140/the-inclusion-relation-sigmaab-subseteq-sigmaa-sigmab-is-not-true https://mathoverflow.net/questions/14246/spectra-of-sums-and-products-in-banach-algebras-was-spectrum-in-banach-algeb https://math.stackexchange.com/questions/4668576/infinity-norm-and-operator-norm-question https://www.math.ucdavis.edu/~anne/WQ2007/mat67-Ll-Spectral_Theorem.pdf https://math.libretexts.org/Bookshelves/Linear_Algebra/Book%3A_Linear_Algebra_(Schilling_Nachtergaele_and_Lankham)/11%3A_The_Spectral_Theorem_for_normal_linear_maps/11.01%3A_Self-adjoint_or_hermitian_operators http://www.math.ucdavis.edu/~anne/linear_algebra/mat67_course_notes.pdf https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2021/8fb8d5c170f1613151aca71de21027bc_MIT18_102s21_full_lec.pdf https://math.libretexts.org/Bookshelves/Linear_Algebra/Book%3A_Linear_Algebra_(Schilling_Nachtergaele_and_Lankham)/11%3A_The_Spectral_Theorem_for_normal_linear_maps/11.01%3A_Self-adjoint_or_hermitian_operators http://pfister.ee.duke.edu/courses/ecen601/notes_ch5.pdf https://math.stackexchange.com/questions/4050748/reference-for-wold-decomposition-theorem-operator-theory https://en.wikipedia.org/wiki/Direct_sum https://en.wikipedia.org/wiki/Orthogonal_complement https://arxiv.org/pdf/math/0701306#theorem.2.14.12 use this one for essay: https://tqft.net/web/teaching/current/Analysis3/LectureNotes/03.Compact.operators.pdf https://www.math.ucdavis.edu/~anne/WQ2007/mat67-Ll-Spectral_Theorem.pdf

🏗️ ΘρϵηΠατπ🚧 (under construction)

Normed Spaces

week 3

week 1

Algebra

Now a bunc of facts from tutotirla

Spectral Theorem for Normal Operators

spectral theory of banach algebras

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)