3)Banach and Hilbert

Question 1

True or false a normed space is also a metric space

Answer 1

a normed space (V, ∥ · ∥) is also a metric space
shown in the exercises

Question 2

Def 3.1 Banach Space and Hilbert Space

Answer 2

A normed space that is complete as
a metric space is called Banach space. An inner product space that is complete as a
metric space is called a Hilbert space

Question 3

hilbert space vs banach space vs metric spaces

Answer 3

Hilbert space ⊂ banach space ⊂ complete metric space
if i establish a result in banach follows to be true in metric etc

they also have a scaling and shift invariant property

Question 4

an example of a banach space in which V is dense

Answer 4

(V, ∥ · ∥) is a normed vector space
space of Cauchy sequences V` with the vector space structure. The Cauchy sequences converging to zero form a subspace V′_0

Define a norm on V`/V′_0 by identity y ∥[(xn)]∥ := limn→∞ ∥xn∥
resulting normed space is a Banach space in which V is dense

.

Question 5

THM 3.2 banach space for functions using a norm

Answer 5

(B(X), ∥ · ∥∞) is a Banach space

X is a set. Let B(X) be the space of bounded functions on X. On B(X) there is a natural norm ∥ · ∥∞ defined by
∥f∥∞ := supx∈X|f(x)|

This norm induces the metric d∞.

Question 6

PROOF (B(X), ∥ · ∥∞) is a Banach space (space of bounded functions on X)

Answer 6

STEP 1 Let f_n be a Cauchy sequence in B(X), i.e. for any ε > 0 we have
∥f_n − f_m∥∞ < ε
for large enough n and m. In particular this means that for any x ∈ X we have that
|f_n(x)−f_m(x)| < ε for sufficiently large n, m. Therefore, for any x ∈ X the sequence f_n(x)
is a Cauchy sequence of real numbers, and therefore converges to a value which we will
denote by
f(x) := lim f_n(x). We will complete the proof by showing that for such defined
function f we have:
1. f(x) is in (B(X), ∥ · ∥∞), that is f is bounded.
2. fn → f in the ∞-norm.
STEP 2
First we show that f(x) is bounded and thus in (B(X), ∥ · ∥∞). Since for some fixed ε > 0
and any x we have |fn(x) − fm(x)| < ε for sufficiently large n, m, we can pass to the limit
m → ∞ in the last inequality and obtain
|fn(x) − f(x)| < ε for all x. (3)
Then, the triangle inequality implies |f(x)| ≤ ∥fn∥∞ + ε for all x. Thus f is bounded and
∥f∥∞ ≤ ∥fn∥∞ + ε.
STEP 3
Finally, since for all ε > 0 we can find N such that for all n > N the inequality (3) is
satisfied, we conclude that fn → f in (B(X), ∥ · ∥∞).

Question 7

General scheme of completeness proof:

Answer 7

STEP 1For a general Cauchy sequence we build a “limit” in some point-wise sense.

STEP 2 At this stage it may not be clear either the constructed “limit” is at our space at all,
that shall be shown on the second step.

STEP 3 From the construction it does not follows that the “limit” is really the limit in the
metric of our space, that is done at the third step of the proof.

Question 8

DEF 3.4 Uniform Convergence)

Answer 8

We will refer to convergence in B(X) as uniform
convergence

space of bounded functions on x
Banach space under norm ∥ · ∥∞

Question 9

THM 3.6 ANOTHER BANACH SPACE with cb(X)

Answer 9

(Cb(X), ∥ · ∥∞) is a Banach space

In B(X) we have subspace
of bounded continuous functions , C_b(X). usually use ∥ · ∥∞ norm (unless otherwise stated)

Note that if X is compact then any continuous function on X
is bounded automatically, so in that case C(X) = C_b(X).

Question 10

Proof Thm 3.6. (C_b(X), ∥ · ∥∞) is a Banach space

Answer 10

Since a closed subspace of a Banach space is again a Banach space sufficient to check that C(X) is closed in B(X). ie check that the uniform limit of a sequence of continuous functions is continuous. using an ε/3-argument.

Suppose that
∥f_n −f∥∞ → 0.
We want to show that f is continuous at x. Let ε > 0. Now
choose n large such that
∥fn−f∥∞ < ε/3. In particular this means that |fn(x)−f(x)| < ε/3
for all x. Next choose δ > 0 such that d(x, y) < δ implies |fn(x) − fn(y)| < ε/3. Then, for
d(x, y) < δ we have
|f(x) − f(y)| = |f(x) − fn(x) + fn(x) − fn(y) + fn(y) − f(y)|
≤ |f(x) − fn(x)| + |fn(x) − fn(y)| + |fn(y) − f(y)| <
ε/3+ε/3+ε/3 = ε. showing f is continuous

diagram: Consider x values within B_δ(x) and show distance between uses ε/3 on the graph on the y

Question 11

Multi index notation

Answer 11

∂_k = (∂)/(∂x_k) partial deriv

For α ∈ N_0 ^m, so α = (α_1, . . . , α_m)

|α| := α_1 + . . . + α_m,
and
α! = α_1! · · · α_m!

define ∂^α :=∂_1 ^α_1…. ∂_m ^α_m
ie product of kth partial deriv wrt to the variable k for each

x^α = x_1^α1· · · x_m ^αm

Question 12

The degree N Taylor polynomial
T_N f of a C^N function f around a point x_0 is given by

Answer 12

(T_N f ) (x) =
Σ_{|α|≤N} of
(1/α! (∂^α f)(x_0) (x − x_0)^α

f(x_1+a_1, x_2+a_2)
f(x1,x2) + ∂/∂x_1 (f(x_1,x_2)a_1
+ ∂/∂x_2 (f(x_1,x_2)a_2
+ (1/2!)(∂²/∂x_1²)(f(x_1,x_2)*a_1^2+…..

Question 13

Any polynomial
p(x) = p(x1, x2, . . . , x_m) of the variables (x1, . . . , x_m) and degree N can be written as

Answer 13

p(x) = Σ_{|α|≤N} [(a_α)x^α],
with a_α ∈ R

Question 14

C^k -function

Answer 14

If k ∈ N_0, recall that a function f : U → R is said to be a C^k -function if all partial derivative
∂^α with |α| ≤ k exist and are continuous

Question 15

C^K (U)

Answer 15

he set of C^k
functions on an open set U

Question 16

C^k(Ū)

Answer 16

if U has compact closure we define C^k(Ū) as the set of functions
such that all partial derivative ∂^α f with |α| ≤ k exist and extend continuously to Ū

Question 17

Def 3.7(C^k -Norm)

Answer 17

For a bounded open set U the C^k-norm on C^k( Ū) is defined
by
∥f∥C^k =
Σ{|α|≤k} of
[∥∂^α f∥_∞

It is easy to check that this defines a norm and makes C^k(Ū) a normed vector space.

Question 18

THM 3.8 normed space is a banach space C^k

Answer 18

For any open and bounded set U⊂R^m the normed space
(C^k(U), ∥· ∥Ck
is a Banach space

proof: f_n cauchy, ∂^αf is for any |a| ≤ k a Cauchy sequence in C(U). (from ∥∂^αg∥∞ ≤ ∥g∥{C^k} , which then implies ∥∂^αf_n − ∂^αf_k∥∞ ≤ ∥fn − fk∥_{C^k} Hence, ∂^αf_n converges uniformly
for any |α| ≤ k. By a theorem in Analysis uniform convergence of the derivative implies
differentiability of the function and convergence of the derivative to the derivative of the
limit. To see this suppose that fn → f uniformly, and f′n → g uniformly. Now simply observe
that the integral fn(x) = fn(a) + Rx
∫[a,x] f′n (t)dt converges uniformly to f(x) = f(a) + Rx
∫[a,x] g(t)dt.
Since g is continuous, f is differentiable and f′ = g. Thereafter apply this to all derivatives
subsequently to conclude that all derivatives converge uniformly. This implies that sequence
converges in C^k (overlineU).

Question 19

C^k (K) can be defined

Answer 19

for any compact subset K ⊂ R^m simply as the restriction
of functions in C^k(U) of an open bounded neighbourhood. This definition generalises the
above one.

For k = 0 this is the Tietzsche extension theorem which states that any continuous function on a compact subset of R
m can be extended continuously.

Question 20

For example the space of continuous functions C(S^n)

Answer 20

For example the space of continuous functions C(S^n) in the sphere S^n is an example of a Banach space with the ∥ · ∥∞ as its natural norm

Question 21

Thm 3.11 Other Banach spaces

Answer 21

The spaces ℓ^p are Banach spaces.

Question 22

ℓ^p

Answer 22

Assume p ≥ 1. We denote by ℓ^p the space of sequences (a_n) with
Σ_n|an|^p < ∞.
On ℓ^p we will, unless stated otherwise, use the norm
∥(a_n)∥^p := (Σ_n|an|^p)^{1/p}

This defn. with p = 1 and p = 2 is a generalisation of norms ∥·∥1 and ∥·∥2

Also for p → ∞ we see that ∥(a_n)∥p → ∥(an)∥∞ for the norm from ∥f∥_∞:= sup_x∈X |f(x)|.

Question 23

PROOF THM 3.11 The spaces ℓ^p are Banach spaces

Answer 23

follow the three-step procedure outlined:
Step 1: ℓᵖ ⊂ ℓ^∞ and ∥x∥∞ ≤ ∥x∥ᵖ for any x ∈ ℓᵖ
Any cauchy seq (xₙ) of elements xₙ in ℓᵖ is also a cauchy seq in ℓ^∞ and hence converges in ℓ^∞ to some bounded sequence x.
.
Step2: Show x∈ ℓᵖ and sequence converges in ℓᵖ . Since x and each xₙ is a seq we use notation: j-th component of x x(j) so the ℓᵖ norm of x is . (Σ[j=1,∞]|x(j)|ᵖ ) ^(1/p) .

j-th component of xₙ is xₙ(j). Fix ε > 0 and then choose
N such that n, m > N implies ∥x_n − x_m∥_p < ε. Then for any K and m:

Σ_[j=1,..K] of xₙ(j) − xₘ(j)|ᵖ
≤ ∥xₙ − xₘ∥ᵖ < ϵᵖ
Let m → ∞ then Σ_[j=1,..K] of xₙ(j) − x(j)|ᵖ ≤ ϵᵖ
Let K → ∞ then Σ_[j=1,..∞] of xₙ(j) − x(j)|ᵖ ≤ ϵᵖ
Thus xₙ - x ∈ℓᵖ &b.c. ℓᵖ is a linear space then x = xₙ - (xₙ -x) is also in ℓᵖ
Step 3:
Finally, we saw above that for any ϵ > 0 there is N such that ∥xₙ − x∥ < ϵ for all n > N.
Thus xₙ → x.

Question 24

Multi index notation: When working with (x_1,x_2) in R^2 x^**α **

Answer 24

**α **= (α₁, α₂)

using notation x^α = x₁^{α₁}* x₂^{α₂}

e.g. if **α **= (α₁, α₂) = (5,7)
x^α = x₁^{5}* x₂^{7}

Question 25

Thm 3.12 sum banach space and sequence (useful obs about banach space)

Answer 25

Suppose that (V, ∥ · ∥) is a Banach space and suppose that (v_n) is a
sequence in V such that
Σ_k ∥vk∥ < ∞.
Then the sum Σ_k v_k converges in V .

proof: involves showing the sequence of partial sums is cauchy ( w_n = Σ_1,..n v_k using partial sums s_n=Σ_1,..n ∥vk∥ is cauchy and triangle inequality for d(w_n-w_m)

Question 26

A map is linear if

Answer 26

a map f between two linear spaces f : V → W is linear if
f(αv + βu) = αf(v) + βf(u), for all scalars α, β and all v, u ∈ V.

Such a linear map is often called an operator.

(E.g a linear map y=kx R to R

Question 27

Definition 3.13 (Bounded Linear Map).

(Operator norm)

Answer 27

.Let (V, ∥·∥V ) and (W, ∥·∥W ) be normed spaces.
Then a linear map (operator) f : V → W is called bounded if
∥f∥ := sup_{v∈V,∥v∥V =1} of [
∥f(v)∥_W] < ∞.

The value ∥f∥ is called norm of f.
(can be shown to meet norm conditions)

Question 28

Example linear map T:R^n to R^n

Answer 28

can be represented by matrix multiplied by (x_1,…x_n)

if infinite could consider infinite matrix

Question 29

We will consider infinite dimensional spaces: like spaces of sequences or spaces of functions

Answer 29

ℓ_2 : (x_1,x_2,….)
(sequences)

the norm for ℓ_2 : (x_1,x_2,…) is
∥x∥_2 ^2 = Σ_k (x_k)^2

sum of squares of the coefficients
where the norm is finite

Question 30

We will consider infinite dimensional spaces: linear operator for vector space C[0,1]

Answer 30

e.g linear operator
Tf = integral_[0,y] f(x).dx
is defined by integrals
T: C[0,1]^0 to C[0,1]

so maps one continuous funct to another cont funct on same space, we can verify linearity by using properties of the integrals (for sums and scalar multples. T(f+g) = Tf(y)+Tg(y)

Question 31

For a finite dimensonal space: consider closed ball of unit vectors
Definition 3.13 example (Bounded Linear Map).

Answer 31

If V is finite dimensional V=R^n
unit ball overline(B_1(0)) ={v ∈V : ∥v∥=1}
-is bounded and closed
-by Heine-Borel it is compact
- using function T take the norm
∥Tv∥ - is a continuous function on v
- a continuous function on a compact set is bounded and attains a maximal value
so
sup_{∥v∥=1} of ∥Tv∥ < ∞

(for infiite dim vector space a lot of operators are unbounded)

Question 32

d(x,y) as a norm and conditions used for Tv for ∥v∥=1

∥Tv∥
sup_{∥v∥=1}∥T( v)∥

Answer 32

d(x,y) = ∥x-y ∥
∥λx∥= |λ|∥x∥ property of norm linearity condition

∥Tv∥ on ∥ v∥=1 for 0 ≤λ ≤1

∥T (λ v)∥
by linearity = ∥λT( v)∥
homogenous =|λ| ∥T( v)∥
(if we are shrinking as making vector shorter then supremum will also maybe shrink proportionally)

sup_{∥v∥=1}∥T( v)∥ = sup_{∥v∥≤1} ∥T( v)∥
Here we take sup over larger set but all the values are either the same or shorter so they cant increase the sup so its equal

Question 33

for any v not equal to 0 unit norm defined

what is the norm of this unit vector?

Answer 33

u= (1/∥v∥)* v
By homogenuity of the norm:

∥ (1/∥v∥)* v ∥ = |(1/∥v∥)| ∥v∥
=(1/∥v∥) ∥v∥ = 1

∥u∥=1

Question 34

for u unit norm of v
what is ∥Tu∥

Answer 34

∥Tu∥ = ∥T (1/∥v∥)* v ∥ by homogeneity of the norm
= ( (1/∥v∥)* ∥Tv∥

sup_{∥u∥=1} ∥Tu∥ = sup_{v≠0} [ ∥Tv∥/∥v∥]

Question 35

In the previous exercises we computed the following identities for norm ∥f∥

Answer 35

for v In V
sup_{∥v∥_V=1} of ∥f(v)∥W
(over unit vectors)
=
sup{∥v∥_V ≤1} of ∥f(v)∥W
(adds shorter vectors)
=
sup{ v≠0} of [∥f(v)∥_W/ ∥v∥_V
(scaled down proportionally to unit vector)
better to know all three and use which is most suitable
in particular for the fraction one we will have…

(subindex tells us which subspace we are taking under the norms)

Question 36

EXERCISE
For any v ∈ V we have:
∥f(v)∥_W ≤

Answer 36

For any v ∈ V we have:
∥f(v)∥_W ≤∥v∥_V · ∥f∥.

Question 37

for any w not equal to 0 we have the following inequality using the bounded linear map

(in particular using sup_{ v≠0} of [∥f(v)∥_W/ ∥v∥_V)

Answer 37

∥Tw∥/∥w∥ ≤ ∥sup_ [∥Tv∥/ ∥v∥] = ∥T∥

for any vector w ≠0
∥Tw∥ ≤ ∥T∥* ∥w∥
this tells us bounded, function doesnt exceed this constant

Question 38

Thm 3.15 linear maps between normed space when are they continuous?

Answer 38

A linear map f : V → W between normed spaces is continuous if and
only if it is bounded.

(for general functs not connected but for linear maps continuity is the same as boundedness)
proof:
Operator T is continuous at every point of our vector space )T continuous at every x in V so T is continuous at 0 ( for all ε>0 ∃δ > 0 ∥x-y∥ < δ then ∥Tx-Ty)c<ε by linearity of operator ∥T(x-y)∥<ε, say ε = 1 ∃δ > 0 s.t. ∥v∥ < δ implies ∥T(v)∥ < 1. )
now conisder
∥v/δ ∥ < 1 so
∥T(v/δ) ∥ < (1/δ)
∥T∥≤ (1/δ) - bounded
(so continuity at 0 implies that operator is bounded)

show equiv boundedness to continuity:
we have ∥Tv∥≤ ∥T∥ ∥v∥ we need to show continuity.
e.g. take δ=ε/( ∥T∥+1)
∥Tx-Ty∥= ∥T(x-y)∥ ≤ ∥T(x-y)∥
≤ ∥T∥∥(x-y)∥ ≤ ∥T∥ ε/( ∥T∥+1)≤ ε

Question 39

thinking about vector spaces as a metric space…. linear maps being continuous and bounded?

Answer 39

Consider Unit ball B_1(0) and a vector v from centre to outside. We can scale to the size of the radius ∥λv∥ = 1
|λ|∥v∥ = 1 so ∥v∥= 1/|λ|
that means we can measure magnitude of any vector
if we consider d(x,y) = ∥x-y∥ then the unit ball defines the distance between any two vectors, in a general metric space this is not true! If we have the distance between x and y it wont tell us the distance for other points, but in vector spaces we use shift invariance to shift to origin then homogeneity to scale down

Question 41

Uniform continuity history

Answer 41

A continuous function on a closed and bounded interval is uniformly continuous. Peter Gustav Lejeune Dirichlet was the first to prove this and implicitly he used the existence of a finite subcover of a given open cover of a closed interval in his proof.[1]

Question 42

For banach spaces of functions we consider

Answer 42

(B(x),||.|| infinity)
whihc is a banach spave for bounded functions

METHOD to show its a BS we take any cauchy (f_n) 1) build a f pointwise often

2) show f is in our normed space

3)f_n converge to f in the sense of our soave

Question 43

Why does |f_n(x) -f(x)|<e give |f(x)| =<||f_n||_infinity +e

Answer 43

Given: |f_n(x) - f(x)| < e

By the triangle inequality: |f(x)| = |f(x) - f_n(x) + f_n(x)| ≤ |f(x) - f_n(x)| + |f_n(x)|

Now, substitute the given inequality: |f(x)| ≤ |f(x) - f_n(x)| + |f_n(x)| < e + |f_n(x)|

Since this holds for any x, we can take the supremum (infinity norm) on both sides: ||f||∞ ≤ ||f - f_n||∞ + ||f_n||∞ < e + ||f_n||∞

This leads to ||f||∞ < ||f_n||∞ + e, as desired.