As an example, we spend most of the early chapters discussing the use of Schauder bases and fundamental sequences in theory. Included is an article by the editors [91] which gives a concise overview of the basic theory.
Bases and Basic Sequences
Schauder bases
There are several possible extensions of the basic concept to Banach spaces, but the following definition is the most useful. Let's emphasize that the order of the base is important; if we permute the elements of the base, the new set could very easily not be a base.
Examples: Fourier series
This is a classical result in Fourier analysis (a good reference is Katznelson [108]) which is equivalent to the statement that there is a continuous function f whose Fourier series does not converge uniformly. To show that it is not a basis, it is therefore sufficient to show that the sequence of operators (Tn)∞n=1 is not uniformly bounded.
Equivalence of bases and basic sequences
That is, (uk) satisfies Grunblum's condition (Proposition 1.1.9), therefore (uk) is a fundamental sequence with basis constant at most K. Here, as usual, Smden denotes the demte-partial sum operator with respect to the basis (en)∞n=1. We know that limn→∞Sr1xn= 0, therefore dern2> n1 is so. yk) is a block root sequence of the base (en).
Bases and basic sequences: discussion
If X is a separable Banach space, then X is isometrically embedded in C[0,1] (and therefore isometrically embedded in a monotone-based space). We are now ready to complete the proof of Fact 2 and, therefore, of the theorem.
Constructing basic sequences
Using Grunblum's condition (Proposition 1.1.9), we conclude that (x∗n)∞n=1 is a basic sequence with a basis constant at most 1. Now Theorem 1.5.2 brings about the existence of a basic sequence (xn) in the basis Swith a constant so close to 1 , as we wish.
The Eberlein-˘ Smulian Theorem
Not surprisingly, this has inspired more investigation and this is far from the end of the story. Later, Bourgain, Fremlin and Talagrand [16] proved similar results for subsets of the Baire class one functions on a compact metric space.
Problems
Mazur’s Weak Basis Theorem
This does not follow from basis series techniques, because it is no longer true that a cluster point of a basis series for pointwise convergence is necessarily zero. A function belongs to Baire class one if it is a pointwise limit of a set of continuous functions.
Krein-Milman-Rutman Theorem
In [74] the Eberlein-˘Smulian theorem is extended to bounded subsets of C(K) (Hausdorff compact Ka space) with the weak topology replaced by the pointwise convergence topology.
The Bounded Approximation Property
ThusX has (BAP) if and only if it is isomorphic to a complemented subspace of a space with a basis. We refer to [24] for a full discussion of the problems associated with the bounded approximation property.
The Classical Sequence Spaces
The isomorphic structure of the p -spaces and c 0
Therefore, to prove that T is compact, it is sufficient to show that T|BX is weak-to-norm continuous. Since the weak topology of
Proposition 2.2.1 gives an infinite dimensional subspace Z of Y such that Z is isomorphic vertex [respectively c0] and Z is complemented inp [respectively c0]. The problem of the complemented subspace. If X is a Banach space such that every closed subspace is complemented, then X is isomorphic to a Hilbert space.
The space 1
Every weakly Cauchy set (xn)∞n=1 in a Banach space X is norm bounded by the Uniform Boundedness principle. Statement 2.3.12. Any Banach space with the Schur property (in particular 1) is weakly sequentially complete.
Convergence of series
Let's look at a lemma first. n=1xn be a formal series in a Banach space X. Then the following are equivalent:. iii) There exists C >0 such that. In Chapter 10 we will see a similar and much more difficult result for Banach spaces that do not contain 1 (due to Rosenthal [197]), which when combined with the Bessaga-Pelczy´nski theorem gives a pair of books very elegant in Banach space theory.
Complementability of c 0
Of course, if E is a complemented subspace of a Banach space X, then X/E must be isomorphic to a subspace of X that is complementary to E.). We finish this chapter by noting that, in view of Theorem 2.5.8, it is natural to define a Banach space Y as separately injective if X is a separable Banach space, E a closed subspace of X is and T :E →Y a bounded operator then T can be extended to an operator ˜T :X.
Special Types of Bases
Unconditional bases
Boundedly-complete and shrinking bases
In the case that the basis (en)∞n=1 is monotone, equation (3.3) implies that X embeds isometrically into H∗. Note that the facts that the canonical basis of 1 does not shrink and that the canonical basis of c0 is not bounded-complete are now explained in the non-reflexivity of these spaces.
Nonreflexive spaces with unconditional bases
Using Theorem 2.4.10 we can derive a block basis sequence (xk) with respect to the canonical basis of c0 such that T|[xk] is an isomorphism on its range. Then, by Theorem 3.2.10, X is a dual space and so there exists a bounded projection of X∗∗ontoX (see the discussion after Proposition 2.5.2).
The James space J
Statement 3.4.2. The sequence(s)∞n=1 of standard unit vectors is a monotone basis for J in both norms · J and · 0. It is clear from the definition that J∗∗ coincides with ˜J, i.e. the space of sequences of bounded quadratic variation.
A litmus test for unconditional bases
Thus we have seen that having an unconditional basis is very special and cannot be relied upon to exist in most spaces. The Unconditional Basic Sequence Problem. Does every Banach space contain at least one unconditional basic sequence.
Orlicz sequence spaces
Show that the following conditions are equivalent:. i) Every base sequence in X is contracted;. ii) Every basic sequence in X is bounded complete;. This result is credited to Singer [206]; later Zippin [224] improved the result and replaced the base sequence with the base when X is known to have a base (see Problem 9.7).
Banach Spaces of Continuous Functions
Basic properties
If K is a compact Hausdorff topological space, then C(K)∗ is isometrically isomorphic to the space M(K) of all finite regular signed Borel measures on K with the norm µ=|µ|(K). To say the opposite, let's start by noting that if K is a metrizable compact Hausdorf space, then it is specifically separable.
A characterization of real C ( K )-spaces
Then A is isometrically isomorphic to the algebra C(K) of a compact Hausdorff space K if and only if. 4.1) On our way to the proof of Theorem 4.2.5, we need two preparatory Lemmas, which are based on the following simple deductions from the hypothesis. Let us observe that if we consider A = ∞ (with multiplication of two sequences defined coordinately), Theorem 4.2.5 gives that A=C(K) (isometric) for a compact Hausdorff space K.
Isometrically injective spaces
A Banach space X is isometrically injective if and only if it is isometrically isomorphic to an order-complete space C(K). Let us see that C(K) is order-complete, and then we will invoke Theorem 4.3.6 to deduce that C(K) is isometrically injective.
Spaces of continuous functions on uncountable compact metric spacescompact metric spaces
Then σ is continuous and one-to-one, hence a homeomorphism on σ(K). We repeatedly use the standard fact that a one-to-one continuous map from a compact space to a Hausdorf topological space is a homeomorphism on its range, since closed sets must map to compact, hence closed, sets.). If we combine Borsuk's theorem with Proposition 4.4.3, we see that an arbitrary C(K) with K an uncountably compact metric space (a) is isomorphic to a complemented subspace of C([0,1]N) and (b ) contains a complemented subspace isomorphic to C(∆), where ∆ = {0,1}N.
Spaces of continuous functions on countable compact metric spaces
This clearly contradicts the lower bound of −M on the set (an)∞n=1. The contradiction shows that K has a finite Cantor-Bendixson index. If K and L are countable compact metric spaces with different but finite Cantor-Bendixson indices, then K and L are not homeomorphic, but the spaces C(K) and C(L) are both isomorphic to c0.
Ransford’s proof of the Stone-Weierstrass Theorem [193]
Hint: Argue that if U :C(K)→ C(L) is any (on) isometry, then U∗ maps extreme points of the double ball to extreme points.].
De Branges’s proof of the Stone-Weierstrass Theorem [37]
4.6.(a) If K is extremely disconnected, show that for every bounded lower semi-continuous functionalf, the upper semi-continuous regularization. Show that if ϕ belongs to the predual, there is a regular Borel measure µonKso that µ(B) =ϕ(χB) for every Borel set.
The Amir-Cambern Theorem [4], [22]
In this case it is important to note that there exists only one such spaceL1(K,B, µ). More precisely, if µ is a non-atomic probability measure for K, then there is a bijectionσ: [0,1] → K such that both σ and σ−1 are Borel maps and.
Note that whenever (fn)∞n=1 is a sequence bounded above by an integrable function, then, in particular, (fn)∞n=1 is equi-integrable. Then there exists a subsequence(gn)∞n=1of(fn)∞n=1 and a sequence of disjoint measurable sets (An)∞n=1 such that ifBn= Ω\An then(gnχBn)∞n=1 is equivalent - integrable.
Using Hahn's decomposition theorem, there exists a Borel set Bn ⊂Un such that µn(Bn) =µ+n(Un)≥δ. By compactness, there exist N such that #N. i)⇒(iii) Let (Bn)∞n=1 be an arbitrary sequence of disjoint Borel sets in K and (µn)∞n=1 be an arbitrary sequence of measures in A.
The Dunford-Pettis property
By the Eberlein-˘Smulian theorem, the sequence (T∗yn∗)∞n=1 ⊂ T∗(BY∗) can be assumed weakly convergent to some x∗ in X∗.
The Dunford-Pettis Theorem)
- Weakly compact operators on C ( K )-spaces
- Subspaces of L 1 ( µ )-spaces and C ( K )-spaces
- Conditional expectations and the Haar basis
- Averaging in Banach spaces
- Properties of L 1
- Subspaces of L p
- The classical proof of Khintchine’s inequality
In Theorem 4.5.2, we showed by the technique of testing property (u) that C(K) embeds into a space with an unconditional basis if and only ifC(K)≈c0. Note that the normalization for Gaussians is slightly different, that is, the characteristic function of a normalized Gaussian would correspond to the case c = 1/2 in the previous definition.
Factorization Theory
Maurey-Nikishin factorization theorems
Thus we can resume our convention that µ is a probability measure. a)⇒(b) Since (Ω, h dµ) is a probability measure space andp < q, the Lp(hdµ) norm is less than the Lq(hdµ) norm and thus we have. Since Lr(µ) forr≥2 is type-2 space, we immediately get:. b) If a Banach space X is isomorphic to a closed subspace of both Lp(µ) for some1≤p <2 and Lr(µ)for some2< r <∞, then X is isomorphic to a Hilbert space.
Rearranging the last expression gives us an estimate. for some constant C, and by Kahane's inequality we deduce that X has cotype s. To obtain a lower estimate, we first use equation (7.13) in combination with Theorem 6.2.13 and Kahane's inequality to obtain.
Factoring through Hilbert spaces
In the proof of this result and others in the next chapter, we will make use of the following lemma. We can then define a complex inner product by "recalling" the complex structure of X and environment.
The Kwapie´ n-Maurey theorems for type-2 spaces
We have seen (Theorem 6.4.8) that ifp >2 any subspace of Lp that is isomorphic to a Hilbert space is necessarily complemented. Continued.) Suppose X is a Banach space of type p for approximately 1≤p <2. a) Show that for a suitable constant C we have the following estimate:
