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B4a Banach spaces Michaelmas 2010 lectured by Bernd Kirchheim based on notes by CJK Batty
Lecture 3
Recall (X, ‖ · ‖) a normed space, then
- d(x, y) = ‖x − y‖ a metric on X
- converges xn → x ⇐⇒ ‖xn − x‖ → 0
- Y ⊂ X closed ⇔: (xn ∈ Y & xn → x ∈ X ⇒ x ∈ Y )
- T : X → X continuous ⇔: (xn → x ⇒ T xn → T x.
- (λ, x, y) → λx + y continuous from F × X × X → X 4.Corollary Let Y be a linear subspace of a normed vector space X. The closure Y of Y is also a linear subspace of X.
5.Example Consider sums A + B = {a + b : a ∈ A & b ∈ B} of sets.
- Let Y be a closed subset of a normed vector space X, and A be a compact subset of X. Then
Y + A := {y + a : y ∈ Y, a ∈ A}
is closed in X.
- Sums of closed subspaces are not necessarily closed Let X = ℓ^1 , with the subspaces Y = {y ∈ X : y 2 j = 0 for j ≥ 1 }, Z = {z ∈ X : z 2 j− 1 = jz 2 j all j ≥ 1 }.
Then Y and Z are closed linear subspaces of X and Y + Z is not closed. Idea: the point x = (0, 1 , 0 , 14 , 0 , 19 , 0 ,... ) can be approximated by
x(n)^ = (0, 1 , 0 , 14 , 0 , 19 , 0 ,... , 0 , (^) n^12 , 0 , 0 , 0 ,... ) = (− 1 , 0 , −^12 , 0 , −^13 , 0 ,... , − (^) n^1 , 0 , 0 , 0 , 0 ,... )
- (1, 1 , 12 , 14 , 13 , 19 , 14 ,... , (^1) n , (^) n^12 , 0 , 0 , 0 ,... ) ∈ Y + Z,
but is itself not in Y + Z. Remark Compare with Q3 on Sheet 1. We will show, that this situation can not occur in finite dimensions - however you can verify that the sum of the two closed set {x : x 1 x 2 = 1}, {x : x 1 = 0} in R^2 is not closed.
2 Completness - Banach spaces
We have seen already in Mods, that completeness is the most important property for doing analysis, and the reason why mathematicians can not happy live inside the rational numbers - where calculations are beautifully easy! There we formu- lated completness of the real numbers in terms of their order (existence of sup and inf), in several dimensions there is no natural order and we use Cauchy’s approach. 1.Definition A normed vector space (X, ‖ · ‖) is a Banach space if the associated metric is complete,
i.e. if every Cauchy sequence converges (in X), i.e. whenever (xn) is a sequence in X such that ‖xn − xm‖ → 0 as m, n → ∞, then there exists x ∈ X such that ‖xn − x‖ → 0 as n → ∞.
2.Example 1. Rm^ (or Cm) is complete in any norm. For ‖ · ‖∞ this is an easy excercise, compare with 2. below. For ‖ · ‖ 1 see Problem Sheet. For an arbitrary norm, this will follow from the “equivalences of norms” later in the course.
- Let Ω be any set and again consider Fb(Ω), the set of all x : Ω → F that are bounded, i.e. there is an R with |x(t)| < R for all t ∈ Ω, and with the supremums norm ‖x‖∞ = supΩ |x(t)|. This is a complete space.
An important example is ℓ∞^ = Fb(N), the space of all bounded sequences.
- Recall: if Ω is a topological space, then Cb(Ω) is defined as the space of bounded continuous functions on Ω. Based on 2. and our knowledge about uniform convergence it follows now easily that (Cb(Ω), ‖ · ‖∞) is complete. However, (C[0, 1], ‖ · ‖ 1 ) is not complete.
3.Definition Consider a series
1 xn, where^ xn^ ∈^ X. This series^ converges^ to^ x^ ∈^ X if: (^) ∥ ∥ ∥ ∥ ∥
∑^ m
n=
xn − x
→ 0 as m → ∞.
∑ xn is absolutely convergent if
‖xn‖ < ∞.
4.Theorem A normed vector space X is complete if and only if every absolutely convergent series in X is convergent to a limit in X.
Then B(x, ε) ⊃ B′(x, aε), B′(x, ε) ⊃ B(x, ε/b),
where B(x, ε) = {y ∈ X : ‖y − x‖ < ε}
etc. So
- the topologies coincide.
- The two norms have the same convergent sequences.
- They also have the same Cauchy sequences!!
- Thus if ‖ · ‖ is complete, so is any equivalent norm.
Remark Equivalence of norms is a stronger concept than equivalence of metrics! If two metrics ̺ ,σ are equivalent, i.e. induce the same topology, then in general the quotient ̺ (x, y)/σ(x, y) is not bounded away from zero (or infinity). For example, the metrics
̺ (x, y) = |x − y|, σ(x, y) = | arctan(x) − arctan(y)| on R
both induce the same (standart) topology. (R, σ(x, y)) is, however, not even complete. Indeed xn = n is a Cauchy sequence (since σ(xn, xm) < ε if n, m > tan(π− 2 ε)) which is not convergent - completness is not a topological property( not perserved under homeomorphisms)
