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B4a Banach spaces Michaelmas 2010 lectured by Bernd Kirchheim based on notes by CJK Batty
Lectures 5 and 6
Recall If X is a normed vector space which is not complete, then it has a (essentialy) unique completion. This means that there is a complete normed vector space X˜ and an linear isometry
i : X → X˜ s.t. ‖i(x)‖ (^) X˜ = ‖x‖X , i(λx + y) = λi(x) + i(y) ∀λ, x, y
such that i(X) = X˜. Moreover, X˜ is unique up to norm-preserving bijection. (ˆi : X → Xˆ any other completion, then i◦ˆi−^1 has a norm preserving bijective extension I : Xˆ → X˜, i.e. I is an isometric linear isomorphism.) In general, if X is a Banach space and Y a linear subspace, then the closure Y¯ in X is a completion of Y. 7.Definition Two norms ‖ · ‖, ‖ · ‖′^ on the same space X are said to be equivalent if there are constants a > 0, b such that
a‖x‖ ≤ ‖x‖′^ ≤ b‖x‖ for all x ∈ 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)
3 Finite-dimensional normed vector spaces
Recall that if V is a finite-dimensional vector space over F, then V is linearly isomorphic to Fm. Indeed, if {v 1 ,... , vm} is a basis of V , then the map:
λ 1 v 1 + · · · + λmvm 7 → (λ 1 ,... , λm)
is a linear isomorphism. We shall see that, if V has a norm, then this map is also a homeomorphism, so any two normed vector spaces of the same finite dimension have the same linear and topological structure. Consider the norms ‖ · ‖ 1 , ‖ · ‖ 2 , and ‖ · ‖∞, on Fm. Then
‖x‖∞ = max j
|xj | ≤
j
|xj |^2
= ‖x‖ 2 ≤
j
|xj | = ‖x‖ 1 ≤ m‖x‖∞.
As mentioned before, such inequalities imply that the topologies on Fm^ in- duced by the three norms are the same, i.e. each is the Euclidean topology, since the three families of ε-balls nest inside each other. They also have the same Cauchy sequences. Thus if ‖ · ‖ is complete, so is any equivalent norm. We will show that all norms on a fixed finite-dimensional normed space are equivalent and complete! 1.Proposition Let (X, ‖ · ‖) be a normed vector space of dimension m, with basis x 1 ,... , xm. Then there exist constants a > 0, b such that
a (|λ 1 | + · · · + |λm|) ≤ ‖λ 1 x 1 + · · · + λmxm‖ ≤ b (|λ 1 | + · · · + |λm|)
for all scalars λj.
4 Separability
Recall for (X, d) a metric space and M ⊂ X we define the distance of a point x ∈ X to the set M as
dist(M, x) = distM (x) = inf{d(x, y) : y ∈ M }. Then |distM (x) − distM (y)| ≤ d(x, y) for all x, y ∈ X, so distM (·) is lipschitz continuous, and dist(M, x) = dist( M , x¯ ). Moreover,
x ∈ M¯ = {y : exist (yn) ⊂ Y s.t. yn → y} ⇔ dist(M, x) = 0.
- Definition A metric space x is said to be separable if there exists a countable set D such that D = X (X has a countable dense subset). A normed space X is separable if X is separable in the induced metric.
- Examples (i) X finite-dimensional.
(ii) X = c 0 = {(αj ) : αj → 0 as j → ∞ }, ‖(αj )‖∞ = sup|αj | is separable.
(iii) X = l∞^ = {(αj ) : sup|αj | exists}, ‖(αj )‖∞ = sup|αj | is not separable.
(iv) The space c od all convergent sequences with the ‖ · ‖-norm is... - see Sheet 2 Q3.
- Theorem Let X be a separable metric space and let Y be a subset of X, then the metric subspace Y is separable.
- Theorem Let X be a normed vector space and let S be a count- able subset of X such that linS = X (linS is the smallest subspace of X containing S), then X is separable.
- Corollary Let X be a normed vector space for which there exists a
countable set S such that linS = X. Then X is separable.
- Examples (i) Ca,b = {f : R → F : f (x) = 0 ∀x /∈ [a, b]} with the supremum norm is separable.
(ii)
n=1 C[−n,n]^ is dense in (L
(^1) (R), ‖ · ‖ 1 ), and hence L (^1) (R) is separable.
5 Linear Operators
- Definition Let X, Y be normed vector spaces. A mapping T : X → Y such that for all x 1 , x 2 ∈ X and α 1 , α 2 ∈ C
T (α 1 x 1 + α 2 x 2 ) = α 1 T x 1 + α 2 T x 2
is said to be a Linear Operator from X to Y.
The linear operator T is s.t.b. bounded if there exists M > 0 such that for all x ∈ X ‖T x‖Y ≤ M ‖x‖X
Let B(X, Y ) denote the set of bounded linear operators from X to Y (we write B(X) = B(X, Y ) if X = Y ).
- Lemma Let X, Y be normed vector spaces, let T ∈ B(X, Y ) and let
N 1 (T ) = inf {M : ‖T x‖ ≤ M ‖x‖ ∀x ∈ X}
N 2 (T ) = sup{
‖T x‖ ‖x‖
: x ∈ X x 6 = 0}
N 3 (T ) = sup{‖T x‖ : x ∈ X 1 } (X 1 = {x ∈ X : ‖x‖ ≤ 1 }) N 4 (T ) = sup{‖T x‖ : x ∈ X ‖x‖ = 1}
Then N 1 (T ) = N 2 (T ) = N 3 (T ) = N 4 (T ) and for N 1 (T ) the inf is attained. Notice, that if x ∈ X, x 6 = 0 then (^) ‖xx‖ ∈ X 1 , ‖ (^) ‖xx‖ ‖ = 1
- Definition For T ∈ B(X, Y ) as above, the common value of Nj , j = 1 , 2 , 3 , 4 is said to be the norm of T and is denoted ‖T ‖.
- Theorem For X, Y normed vector spaces and a linear operator T from X to Y , the following are equivalent ( )
(i) T is bounded
(ii) T is continuous at all points x 0 ∈ X
(iii) T is continuous at 0
