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B4a Banach spaces Michaelmas 2010 lectured by Bernd Kirchheim based on notes by CJK Batty
Lectures 9,10,11 and 12
Recall ‖T x‖ ≤ ‖T ‖ · ‖x‖ ∀x ∈ X
where
T : X → Y linear and ‖T ‖ = sup{‖T x‖ : ‖x‖ ≤ 1 } < ∞ (T ∈ B(X, Y )).
6.Theorem The linear space of bounded operators Let X, Y be normed vector spaces and for T 1 , T 2 ∈ B(X, Y ), α 1 , α 2 ∈ F and for x ∈ X, define
(α 1 T 1 + α 2 T 2 )x = α 1 T 1 x + α 2 T 2 x.
Then α 1 T 1 + α 2 T 2 ∈ B(X, Y ) and B(X, Y ) is a normed vector space with respect to the norm[ T 7 → ‖T ‖ and a Banach Space if Y is a Banach Space. You can (try to) imagine B(X, Y ) as rectangular matricies, perhaps “infinitely big”
]
A way to simplify the proof essentially is to notice that the restriction operator Φ : T ∈ B(X, Y ) → T∣∣B 1 (X)^ ∈ Fb(B 1 (X))
is an isometry (but not onto) with respect to the norms ‖ · ‖ from Definition 3 and the supremum norm ‖ · ‖∞. 7.Proposition Let X, Y, Z be normed vector spaces, let T ∈ B(X, Y ), S ∈ B(Y, Z) and define ST for all x ∈ X by (ST )x = S(T x) (usual composition). Then ST ∈ B(X, Z) and
‖ST ‖ ≤ ‖S‖ ‖T ‖
Remark Let X be a normed vector space and let B(X) = B(X, X). Then the multiplication (S, T ) 7 → ST, (B(X) × B(X) → B(X)) is bilinear and associative, and if I ∈ B(X), I(x) = x ∀ x ∈ B(X), then T I = IT = T for all T ∈ B(X) and ‖I‖ = 1.
(i) This multiplication is not commutative (or rather, not commutative iff dimX > 1).
(ii) If X is m-dimensional, then B(X) ∼= Mm(F)
8.Theorem Unique extension of bounded operators Let X be a normed vector space, let Z be a dense subspace of X, i.e. Z = X. Let Y be a Banach Space and let T ∈ B(Z, Y ). Then there exists a unique T˜ ∈ B(X, Y ) such that T˜ |Z = T. Moreover, ‖ T˜ ‖ = ‖T ‖.
Remark The extension operator T → T˜ therefore gives an isometric isomorphism between B(Z, Y ) and B(X, Y ) and in the sequel we will often identify B(Z, Y ) and B(X, Y ) (if Z = X).
6 The dual space and dual operators
1.Definition Let X be a normed vector space over F. A linear operator f : X → F is said to be a linear functional (or a linear form), and if f ∈ B(X, C) is said to be a bounded linear functional (Then f is contin- uous by Theorem 5.4). The Banach space of bounded linear functionals is denoted by X′^ = B(X, F ) and called the dual space of X.
Remark
(i) X∗^ is sometimes used for X′
(ii) As in B(X, Y ), there are 4 expressions for the norm on X′, all equal to ‖f ‖X′ = sup{|f (x)| : ‖x‖ ≤ 1 }.
(iii) Let X be a normed vector space and let Z be a dense subspace. Then Z′^ = X′^ in light of Theorem 5.8 and the Remark after.
2.Theorem Let f be a linear functional on the normed vector space X. Then ker(f ) is closed if and only if f is bounded.
Remark For a general linear operator T : X → Y we have ker(T ) = T −^1 ({ 0 }), so ker(T ) is closed if T is bounded (i.e. continuous). The converse does not hold as the following example shows. Let X ⊂ ℓ^1 be the span of the set of all canonical unit vectors ek = (δkj )∞ j=1. This means X is the space of all sequences x = (αj )j eventually
One can show that there is not even any bounded linear J : ℓ^1 onto (ℓ∞)′, in fact the later space is not separable. It is, however, even for our J chosen above not obvious how to construct even a single a linear functional f ∈ (ℓ∞)′^ \ J(ℓ^1 ). We might for instance consider
f ((αj )) = lim j
αj for (αj ) ∈ c,
where c is the closed subspace of ℓ∞^ consisting of all convergent sequence. Then f is a continous linear functional on c and if there would be a y = (βj ) ∈ ℓ^1 such that f ((αj )) = Jy((αj )) for all (αj ) ∈ c then βk = f (ek) = 0 for all k implies y = 0 and Jy(x) = 0 for all x ∈ c ⊂ ℓ∞. But f ((1, 1 , 1 ,.. .)) = 1 6 = 0! So f would a a candidate to be in (ℓ∞)′^ \J(ℓ^1 ). For this to be mathematically correct, f has to be defined on all of ℓ∞. The existence of such a linear and bounded extension, is a highly non-trivial question - and ensured by the famous Hahn-Banach Theorem. It will be the topic of the next lecture. Summary The above approach worked in the Example 3 and 4 to show that J : Y → X′^ isometric isomorphism since
a) We have a bilinear form B : Y × X → F (here B(y, x) =
r xryr) which satisfies a Hoelder type inequality (see Section 1)
‖y‖ = sup{|B(y, x)| : ‖x‖ ≤ 1 }.
This allows to carry out steps (i)–(v).
b) the space X whose dual we want to classify is separable, i.e.
X = span({ek, k ∈ N}),
for a “basis sequence” - this is crucial in step (vi). Nevertheless, we also need to find the right basis sequence, which allows us to show that our candidate y (to satisfy Jy = f ) is indeed in Y - this “right choice” depends on the spaces X, Y at hand, but in sequence spaces the canonical basis often works.
To illustrate point b), we just say that despite the problems discussed above, we can easily classify c′^ ∼= ℓ^1. We use a Jˆ which takes the additional basis element e∞ = (1, 1 , 1 ,.. .) into account, so we can choose
Jˆ(y) : x ∈ c → y 1 lim r
xr +
∑^ ∞
r=
yr+1xr if y ∈ ℓ^1.
Verify now (i)–(vi) on your own. We briefly mention more classification results for duals.
- (Rn, ‖ · ‖ 1 )′^ ∼= (Rn, ‖ · ‖∞);
- (Rn, ‖ · ‖∞)′^ ∼= (Rn, ‖ · ‖ 1 );
- (Rn, ‖ · ‖ 2 )′^ ∼= (Rn, ‖ · ‖ 2 );
- (ℓ^2 )′^ ∼= ℓ^2 ;
- L^2 (R)′^ ∼= L^2 (R);
- L^1 (R)′^ ∼= L∞(R);
- C[0, 1]′^ is a space of “measures”;
- (ℓ∞)′^ is very unpleasant, but it contains ℓ^1 (see remark above).
In cases 1–4, the isomorphism is given by: (Jy)(x) =
( xryr^ (and^ J−^1 φ^ = φ(er)
r; in cases 5,6, it is given by: (Jg)(f^ ) =^
f g.
5.Theorem Hahn-Banach Let X be a normed vector space, let M be a subspace of X (wlog M 6 = { 0 }, X) and let f ∈ M ′. Then there exists g ∈ X′^ such that g|M = f and ‖g‖X′^ = ‖f ‖M ′^.
We will prove this theorem in the next Section (and only for separable spaces). Let us, however, already now look at some applications.
6.Corollary Let X be a normed vector space and let x 0 ∈ X. Then there exists g 0 ∈ X′^ such that
‖g 0 ‖X′^ = 1, g 0 (x 0 ) = ‖x 0 ‖.
Therefore,
‖x 0 ‖ = max
|g(x 0 )| : g ∈ X′, ‖g‖X′^ = 1
for all x 0 ∈ X.
This describes that trick for all problems involving Hahn-Banach Theo- rem: do whatever you need to do on a subspace on which it is easily doable,
Remarks
(i) For any subsets M 1 , M 2 ⊆ X, if M 1 ⊆ M 2 then M 2 ◦ ⊆ M 1 ◦.
(ii) For N a closed subspace of X′, it is not true in general that (N◦)◦^ = N. For example, if X = l^1 and N = c 0 in l∞^ = X′^ then N◦ = { 0 } and thus (N◦)◦^ = l∞^6 = N. (The proof goes wrong in the construction of g 0 , as H-BT works only “upwards” - giving objects in the dual).
13.Example For n ≥ 1 let sn = nen − (n + 1)en+1 where en = (δn,j )j is the n-th canonical unit vector and Y = span({sn : n ≥ 1 }) their linear span. Then Y is a subspace of c 0 and l^2 , but Y is dense in c 0 and not dense in l^2.
As in Linear algebra we can also consider the original vectors as func- tionals acting on the dual space, and hence as elements of the bi-dual.
14.Theorem Let x be an element of the normed vector space X and let xˆ : X′^ → F be defined for all f ∈ X′^ by
xˆ(f ) = f (x).
Then ˆx ∈ X′′^ and ‖ˆx‖X′′ = ‖x‖X and the mapping x 7 → xˆ is a linear isometry (and hence 1-1) from X onto a subspace Xˆ of X′′^ which is closed if and only if X is a Banach Space.
Remarks
(i) If X is a normed vector space then the isometric mapping x 7 → xˆ takes X onto Xˆ and embeds X in X′′. Since X′′^ = B(X′, F) is complete, we see that Xˆ is a completion of X with i : x → xˆ.
(iii) A very important question is if it true in general that for X a Banach Space, Xˆ = X′′?
15.Examples
(i) If X is finite dimensional, then Xˆ = X′′.
(ii) If (^1) p + (^1) q = 1 then (ℓp)′^ ∼= ℓq, so (ℓp)′′^ ∼= ℓp.
(iii) (c 0 )′^ ∼= ℓ^1 , (ℓ^1 )′^ ∼= ℓ∞, so ˆc 0 $ ℓ∞^ since ˆc 0 is separable.
16.Definition If Xˆ = X′′, then X is said to be reflexive.
17.Theorem Let X be a Banach space, then X is reflexive if and only if X′^ is reflexive.
