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B4a Banach spaces Michaelmas 2010 lectured by Bernd Kirchheim based on notes by CJK Batty
Lectures 13 and 14
The last still missing concept known from the treatment of duality in Linear Algebra is the adjoint map (operator). 18.Theorem Let X, Y be normed normed vector spaces and let T ∈ B(X, Y ). Fix g ∈ Y ′, then gT ∈ X′^ and hence we can define T ′^ : Y ′^ → X′ by T ′g = gT , i.e. (T ′g)(x) = g(T (x)).
Then T ′^ ∈ B(Y ′, X′) and ‖T ′‖B(Y ′,X′) = ‖T ‖B (X, Y ).
Moreover, if Z is also a normed vector space and S ∈ B(Y, Z) then for S′^ ∈ B(Z′, Y ′) and (ST )′^ ∈ B(Z′, X′) introduced as above, we have
(ST )′^ = T ′S′.
19.Definition For T ∈ B(X, Y ), T ′^ ∈ B(Y ′, X′) as defined above is said to be the (Banach Space) adjoint of T.
Remark So with hindsight, the very abstract first part of the proof of Theorem 17 just makes the formal argument X ∼= X′′^ ⇒ X′^ ∼= (X′′)′ rigorous by showing that for the isomorphismˆ: X → X′′^ the adjoint is still an isomorphism from X′′′^ onto X′^ and equals (ˆ: X′^ → X′′′)−^1. 20.Example Let X be lp, p ∈ [1, ∞), or c 0 , Y = lq, q ∈ [1, ∞). or Y = c 0. Then X′^ = lp ′ , Y = lq ′ where p′^ = p/(p − 1) if p ∈ (1, ∞), p′^ = ∞ if p = 1 and p = 1 if X = c 0 , similarly for q′. Let a T ∈ B(X, Y ) be given by
T ((αj )) =
j=
tk,j αj )∞ k=1.
Then T ′^ ∈ B(lq ′ , lp ′ ) is given by
T ((βj )) =
j=
tj,kβj )∞ k=1.
Formally spoken, the matrix representation of T ′^ is the transposed of the matrix representation of T.
In particular, if
T ((α 1 , α 2 , α 3 , α 4 ,.. .)) = (0, α 1 , α 2 , α 3 ,.. .),
the right shift operator, then tk,j = δk,j+1. Its adjoint
T ′((α 1 , α 2 , α 3 , α 4 ,.. .)) = (α 2 , α 3 , α 4 , α 5 ,.. .)
is the left shift operator, with t′ k,j = δj,k+1 = δk,j− 1. Notice, although T ∈ B(X) is always an isometry, we have T ′(e 1 ) = 0 so T ′^ is not an isometry (but still ‖T ′‖ = 1). This is related to the fact that T is not onto - but still injective, a truely infinite dimensional phenomenon. 21.Practising excercise Let X, Y be Banach spaces and S ∈ B(Y ′, X′). Show there is T ∈ B(X, Y ) such that T ′^ = S iff S′( Xˆ) ⊂ Yˆ. Now, we consider S : (co)′^ → (l^2 )′^ defined by
S : f ∈ c 0 →
∑^ ∞
k=
f (ek)e′ 1 ∈ (l^2 )′, where e′ k(ej ) = δk,j.
Show that S is well-defined and in B((co)′, (l^2 )) but that there is no T ∈ B(l^2 , c 0 ) with S = T ′. (Any solutions upon request marked by BK.)
7 Hahn-Banach Theorem and convexity
The Hahn-Banach Theorem 6.5 was the central tool in the last section, here we are going to prove it1)^ rigorously. It turns out, however, that the natural setting for the idea of Hahn and Banach is the notion of (control by) a convex function rather then a norm. Therefore, we introduce/recall: 1.Definition Let X be a real linear space.
(i) a subset M ⊂ X is said to be convex if
∀x, y ∈ M ∀λ ∈ (0, 1)λx + (1 − λ)y ∈ M, i.e. [x, y] ⊂ M.
(ii) if M ⊂ X is a convex, then a function f : M → R is convex if
∀x, y ∈ M ∀λ ∈ (0, 1) : f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y). 1)up to a rather technical “size” condition on the space X, the heart of the matter - Propostion 2 below - is, however, non-trivial and surprising already in two dimensions.
Remarks a) Clearly, there is in general non-uniqueness of the required linear exten- sion G ≤ F. We can avoid this by defining a canonical HB-extension from Y to Y ⊕ span({x 0 }).
b) For future reference we notice the following (crucial) criterion when two sets A,B in R can be separated in the sense of being in in two disjoint halflines. Our proof essentially shows/uses that this is possible iff
- no a ∈ A is between b 1 < b 2 from B and no b ∈ B is between two a 1 < a 2 , iff
- Aco^ ∩ Bco^ = ∅. We will see that the last (trivially looking) condition is also in any normed space equivalent to A,B being in two disjoint half-spaces. It turns out that this separation property is equivalent to the HB-Theorem, so all what our proof (calulation) does is to extend to simple but pow- erful geometric observation to any dimension. 3.Corollary Let X be a real linear space, Y its subspace and (xi)∞ i=1 a sequence in X. If F : X → R is convex, g : Y → R linear and g(y) ≤ F (y) for all y ∈ Y then there is
G : X˜ = span(Y ∪ {xi, i ≥ 1 }) → R linear, such that
G|Y = g and G(x) ≤ F (x)∀x ∈ X.˜
Idea: Set Yn = span(Y ∪ {xi, 1 ≤ i ≤ n}), then X˜ =
n Yn. By induction ∀n uniquely ∃gn : Yn → R linear s.t. g|Yk is the canonical HB-extension of g|Yk− 1 if 1 ≤ k ≤ n. Then gn = gm|Yn if n < m and G = ∪ngn does what is needed.
- Lemma Let (X, ‖ · ‖) be a real normed space and X˜ a dense subspace of X. Suppose g : X˜ → R linear and F : X → R satisfy g ≤ F on X˜. If
∀t ∈ R : {x : F (x) < t} is open in X (e.g. if F continuous),
then g ∈ X˜′^ and its unique continuous extension G ∈ X′^ (see Theorem 5.8) satisfies G ≤ F on X.
5.Corollary The Hahn-Banach Theorem is true for every separable real normed space (X, ‖ · ‖):
Y subspace of X, g ∈ Y ′^ ⇒ ∃G ∈ X′^ : G|Y = g and ‖G‖X′^ = ‖g‖Y ′^.
