3)Banach and Hilbert Flashcards
(43 cards)
True or false a normed space is also a metric space
a normed space (V, ∥ · ∥) is also a metric space
shown in the exercises
Def 3.1 Banach Space and Hilbert Space
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
hilbert space vs banach space vs metric spaces
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
an example of a banach space in which V is dense
(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
.
THM 3.2 banach space for functions using a norm
(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∞.
PROOF (B(X), ∥ · ∥∞) is a Banach space (space of bounded functions on X)
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), ∥ · ∥∞).
General scheme of completeness proof:
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.
DEF 3.4 Uniform Convergence)
We will refer to convergence in B(X) as uniform
convergence
space of bounded functions on x
Banach space under norm ∥ · ∥∞
THM 3.6 ANOTHER BANACH SPACE with cb(X)
(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).
Proof Thm 3.6. (C_b(X), ∥ · ∥∞) is a Banach space
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
Multi index notation
∂_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
The degree N Taylor polynomial
T_N f of a C^N function f around a point x_0 is given by
(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+…..
Any polynomial
p(x) = p(x1, x2, . . . , x_m) of the variables (x1, . . . , x_m) and degree N can be written as
p(x) = Σ_{|α|≤N} [(a_α)x^α],
with a_α ∈ R
C^k -function
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
C^K (U)
he set of C^k
functions on an open set U
C^k(Ū)
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
Def 3.7(C^k -Norm)
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.
THM 3.8 normed space is a banach space C^k
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).
C^k (K) can be defined
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.
For example the space of continuous functions C(S^n)
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
Thm 3.11 Other Banach spaces
The spaces ℓ^p are Banach spaces.
ℓ^p
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)|.
PROOF THM 3.11 The spaces ℓ^p are Banach spaces
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.
Multi index notation: When working with (x_1,x_2) in R^2 x^**α **
**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}
