Combinatorics in Banach space theory Lecture The next lemma considerably strengthens the assertion of Lemma.6(b). Lemma.9. For every Banach space X and any n N, either all the numbers n b n (X), c n (X) and d n (X) are equal to, or they are all less than. Proof. In view of Lemma.6(b), we shall only prove that c n = implies d n = and d n = implies n b n =, for every n N. First, assume c n = and fix any δ > 0. Then, there exist x,..., x n X such that n j= x j = n and r j (t)x j dt ( δ) n. (.) 0 j= To show that d n = we need to give an upper estimate for max j x j. Namely, we will show that max x j + (δn) /. (.) j n To this end, choose k 0 [n] such that x k0 = max j x j and apply Minkowski s inequality to get ( ) / ( ( n / x k0 = x k0 xk0 x j ) ) / ( ) /. + x j (.3) j= Moreover, by (.), we have ( xk0 x j ) j= i,j n j= j= ( xi x j ) ( = n x j x j j= ( = n x j ) n j= n ( δ) n 4δn, 0 j= r j (t)x j dt which, jointly with (.3), yields the claimed inequality (.). Combining this inequality with (.) we get ( / r j (t)x j dt) 0 j= j= δ n max + (δn) x / j, j n ) whence d n δ. + (δn) / δ 0
In the second part of the proof, we assume d n = and we want to show that n b n =. By the definition of d n, we have sup (x j ) n j= B X n ε j =± ε j x j = n. (.4) For any sequence (x j ) n j= B X pick a sequence (ˆε j ) n j= {, } n which minimises the norm of j ˆε jx j, that is, j ˆε jx j b n. Then, we have ε j =± j= ε n j x j = ˆε n j x j + n j= j= ε j =± (ε j ) (ˆε j ) ε j x j ( b n n + ( n )n ). Taking the supremum over all sequences (x j ) n j= B X, and making use of (.4), we obtain n n (b n + ( n )n ) which forces b n to be equal to n and completes the proof. Proof of Theorem.3. The equivalence (i) (ii) follows directly from Lemmas.4 and.5. Hence, both (i) and (ii) are in turn equivalent to (iii), in view of Lemma.9. Now, assuming (iii) choose k N so that σ := c k <. By the submultiplicativity of (c n ) n= (Lemma.6(d)), we have c k r σ r for each r N. For an arbitrary n N we may write n = k r + s with some r N 0 and 0 s < k r+ k r. Then, by Lemma.6(a), we have j= c n c k r+ kr + s +... k r + s k r+ k k r+ = r+ k k r + s r+ = k, k r whence c n k σ r+ for each n N, which implies (iv). We have proved so far that all conditions (i)-(iv) are pairwise equivalent. Now, assume that any of the conditions (i)-(iv) is satisfied (then, all of them are satisfied). We will show that if X contains l n s uniformly and λ is the uniformity constant, then b n λ n for every n N which will give a contradiction. So, suppose there is a sequence (E n ) n= of finite-dimensional subspaces of X such that for every n N there exists an isomorphism U n : l n E n such that U n = and Un λ. Fix any n N and let (ε j ) n j= {, } n be a sequence of signs that minimises the norm of n j= ε ju n (e j ). Then, n = j= ε j e j l n U n ε j U n (e j ) λ ε j U n (e j ) = λb n X X j= and so the proof of the implication (i) (v) has been completed. The implication (v) (vi) is trivial. Finally, to complete the proof we shall show that (vi) (i). Again, we prove it by contraposition. So, suppose that X is not B-convex. In light of Lemma.7, to guarantee l being finitely representable in X we shall merely show that X contains all l n s λ-uniformly, for every λ > (i.e. for every n N and λ > there exists an n- dimensional subspace E of X such that d BM (E, l n ) λ). j=
Fix any n N and λ >, and set δ = λ (0, ). Since X is not B-convex, there exist x,..., x n B X such that ε j x j n δ for every (ε j ) n j= {, } n. (.5) j= As shall be expected, the vectors x,..., x n span an almost isometric copy of l n inside X. In order to show this, we shall prove that the inequality λ a j a j x j a j (.6) j= j= holds true for arbitrary scalars a,..., a n. In fact, this would mean that the sequence (x j ) n j= is equivalent to the canonical basis (e j ) n j= of l n in the sense that there is an isomorphism T : span{x j } n j= l n satisfying T x j = e j for each j [n]. Moreover, the estimates (.6) would also indicate that T λ and T, thus d BM (span{x j } n j=, l n ) λ, as desired. After these explanations we may proceed to the proof of inequalities (.6). The second inequality in (.6) is obvious, since x j B X for each j [n]. For proving the first one, we may assume with no loss of generality that j a j =. Then, in view of (.5), we have ( n δ sgn(a j )x j = sgn(aj ) ( a j ) ) x j j= j= ( a j ) + a j x j = n + a j x j, j= which gives the first inequality in (.6). The proof is completed. Now, when we understand the essence of being B-convex pretty well, we shall quote the main result from Beck s paper [Bec6]. We will not prove it here, because it is beyond the main scope of our interest. However, it must be recalled, since it is a pioneering result which initiated the study of B-convex spaces. Recall that if (Ω, Σ, µ) is a measure space and X is a Banach space, then a map f : Ω X is called µ-measurable (or µ-strongly measurable), provided that there exists a sequence (g n ) n= of Σ-measurable step functions such that lim n f g n = 0 holds true µ-almost everywhere. The map f is Bochner integrable if and only if f dµ < ; Ω see [DU77, II.]. Now, if (Ω, Σ, P) is a probabilistic space, then any P-measurable function ξ : Ω X is called a random variable. If it is Bochner integrable, then the integral ξ dp Ω is called its expectation value and is denoted as Eξ. We also define the variation D ξ of the random variable ξ by D ξ = ξ Ω Eξ dp. We say that ξ is symmetric, if there exists a measure preserving map φ: Ω Ω such that ξ φ = ξ. A sequence (ξ n ) n= of X-valued random variables is called independent, whenever for every finite sequence By the classical Pettis measurability theorem, a function f : Ω X is µ-measurable if and only if it is essentially separably valued (i.e. for some µ-measure zero set E Σ the range f(ω \ E) is a separable subset of X) and weakly µ-measurable (i.e. for every x X the scalar function x f is Σ-measurable); see [DU77, II.]. j= j= j= 3
n <... < n k of natural numbers, and every choice of Borel sets B,..., B k X, we have k P(ξ n B... ξ nk B k ) = P(ξ nj B j ). The main results of Beck s paper [Bec6] may be summarised in the following theorem which, roughly speaking, asserts that the strong law of large numbers for X-valued random variables is valid if and only if X is B-convex. Theorem.0 (Beck, 96). Let (Ω, Σ, P) be a probabilistic space and X be a Banach space. If X is B-convex, then for every independent sequence (ξ n ) n= of Bochner integrable random variables ξ n : Ω X satisfying Eξ n = 0 for n N and sup n D ξ n < we have n j= ξ j n j= 0 P-almost everywhere. Moreover, if X fails to be B-convex, then there exists an independent sequence (ξ n ) n= of Bochner integrable, symmetric random variables ξ n : Ω X satisfying Eξ n = 0 and ξ n almost everywhere (n N) and such that ess sup Ω lim sup n n ξ j =. At the end of this section, we derive another infinite-dimensional counterpart of the Lyapunov convexity theorem for vector measures taking values in B-convex Banach spaces. The proof is essentially the same as the proof of the quantitative Theorem.6; just instead of Steinitz s lemma we use the basic property of B-convex spaces. Theorem. (V. Kadets, 99). If µ: Σ X is a σ-additive, non-atomic vector measure of bounded variation, defined on a σ-algebra Σ and taking values in a B-convex Banach space X, then the closure of µ(σ) is convex. Proof. Of course, it is enough to show that co(µ(σ)) = 0. This will follow from Lemma.8, if we prove that for every A Σ and ε > 0 there exists A ε A, A ε Σ, such that µ(a ε ) µ(a) ε. So, fix any A Σ and ε > 0. Let M = µ (Ω) be the total variation of µ. In view of Proposition., applied to the variation of µ, there exists a partition {A,..., A n } Π(A) such that µ (A j ) ε for each j [n] and n is the least natural number for which nε µ (A). Hence, (n )ε < µ (A) M, so n < ε M + < ε M for ε small enough. Now, repeating the calculation from the proof of Theorem.6, with the notation A J = j J A j for J [n], we obtain min J [n] µ(a J) µ(a) = min ε j µ(a j ) ε j =± b n max µ(a j) j n b nε < Mn b n. j= By the characterisation of B-convexity (see Theorem.3), the last expression goes to zero as n, thus the proof is completed. j= 4
Remark.. It is by no means straightforward to state what precisely the relation is between Theorem. and Uhl s Theorem.0. While it is easy to see that the latter does not follow from the former (for instance, l is a separable dual space which is not B-convex), it appears to be quite difficult to show that there exists any B-convex Banach space which is not reflexive (if it was not true, then Theorem. would be a direct consequence of Theorem.0). In 964, after discovering that B-convexity and reflexivity have much in common, James conjectured that every B-convex space is necessarily reflexive. Some partial confirmation of this conjecture was given in terms of (k, ε)-convexity. We say that a Banach space X is (k, ε)-convex (for some k and ε > 0), if sup (x j ) k j= B X min ε j =± k ε j x j ( ε)k. j= James [Jam64] showed that (, ε)-convexity implies reflexivity, for every ε > 0. In 973, Giesy confirmed James conjecture for Banach lattices ([Gi73]) and then he proved that for every k 3 and ε > 9 k every (k, ε)-convex Banach space must be reflexive. Finally, 4 in 974, James [Jam74] himself disproved his conjecture by constructing a non-reflexive Banach space X J such that for some λ > there is no isomorphism T from l 3 into X J satisfying λ x T x λ x for x l3 (hence, l is not finitely representable in X J and therefore X J is B-convex). He improved his construction later in [Jam78]. Consequently, Theorems.0 and. are incomparable. 3 B-convexity is a 3SP property In Section 8 we described the general framework for the three-space problem. Now, we turn our attention to the question whether B-convexity is a 3SP property, to which an affirmative answer, in the category of Banach spaces, was obtained by Giesy [Gie66]. Precisely saying, if X, Y and Z are all Banach spaces which form a short exact sequence 0 Y Z X 0 and if X and Y are B-convex, then Z also must be B-convex. Giesy s result was later expanded by Kalton [Kal78] who showed that it is not essential to assume that Z is a Banach space (which is, in fact, equivalent to assuming that Z is locally convex; see Proposition 3. below and recall the classical result, e.g. [Rud9, Theorem.39], which says that a linear topological space is normed iff it is locally bounded and locally convex). Instead, we may merely assume that Z is a locally bounded F -space (that is, a linear topological space having a bounded zero neighbourhood and being completely metrisable). Therefore, for the time being, we move to the world of linear topological spaces (not necessarily locally convex) and we aim to prove in this section the Kalton Giesy theorem, in particular to show how local convexity of Z follows from B-convexity of X and Y in the exact sequence above. Let us start with a simple result, due to Roelcke and Dierolf [RD8], which justifies our interest in locally bounded F -spaces. 5
Proposition 3. (Roelcke, Dierolf, 98). Let X be a linear topological space and Y be a closed subspace of X. (a) If both Y and X/Y are F -spaces, then so is X. (b) If both Y and X/Y are locally bounded, then so is X. Proof. (a): Recall that a linear topological space Z is metrisable if and only if it has a countable basis of zero neighbourhoods and in such a case there exists an invariant metric d on Z (i.e. d(x + z, y + z) = d(x, y) for all x, y, z Z) which is consistent with the given topology on Z (see [Rud9, Theorem.4]). So, there exists a decreasing sequence (V n ) n= of open subsets of X such that (Y V n ) n= is a basis of zero neighbourhoods in Y and (π(v n )) n= is a basis of zero neighbourhoods in X/Y, where π : X X/Y is the canonical map. Fix any zero neighbourhood U X; we are to prove that V m U for some m N. Pick any zero neighbourhood W X with W + W U. There exist m, n N with m > n such that Y V n W and π(v m ) π(w V n+ ), hence V m (W V n+ ) + Y. Fix any v V m ; v = w + y for some w W V n+ and y Y. Then and, consequently, y = v w Y [V m (W V n+ )] Y (V m V n+ ) V m (W V n+ ) [Y (V m V n+ )] (W V n+ ) + [Y (V n+ V n+ )] (W V n+ ) + (Y V n ) W + W U. Therefore, (V n ) n= is a (countable) basis of zero neighbourhoods in X. Now, assume that Y and X/Y are F -spaces and let d X be an invariant metric on X. Then, the formula d X/Y (π(x), π(y)) = inf{d X (x y, z): z Y } (3.) defines an invariant metric on X/Y which is consistent with the quotient topology on X/Y (see [Rud9, Theorem.4]). It follows that if (x n ) n= is a Cauchy sequence in X, then (π(x n )) n= is a Cauchy sequence in X/Y. Hence, there exists ξ X/Y such that lim n d X/Y (π(x n ), ξ) = 0. Choose any x 0 X with ξ = π(x 0 ). Then, in view of formula (3.), there exists a sequence (z n ) n= Y satisfying lim n d X (x n x 0, z n ) = 0. The inequality d X (z n, z m ) d X (x n x 0, z n ) + d X (x m x 0, z m ) + d X (x n, x m ) shows that (z n ) n= is a Cauchy sequence in Y and hence it is convergent to some z 0 Y. Then lim n d X (x n x 0, z 0 ) = 0, thus the sequence (x n ) n= converges to x 0 + z 0. (b): Assume that both Y and X/Y are locally bounded. Then, there is a zero neighbourhood U X such that both π(u) and Y U are bounded. Let V X be a balanced zero neighbourhood such that V + V U. We claim that V is bounded. Assume not and let (x n ) n= be a sequence in V such that ( n x n) n= is unbounded. Then n π(x n) 0 and hence n x n + y n 0 for some sequence (y n ) n= Y. Therefore, for n s large enough we have n x n + y n V, so y n V n V V + V U, which means that (y n) n= is bounded and forces the sequence ( n x n) n= to be bounded as well; a contradiction. 6
To prepare for the proof of the Kalton Giesy theorem we need to recall some material on geometric properties of locally bounded spaces. Definition 3.. Let X be a real or complex vector space. By a quasi-norm we mean any function X x x that satisfies the following three conditions: (i) x > 0 for every x X, x 0; (ii) λx = λ x for every x X and every scalar λ; (iii) x + y k ( x + y ) for all x, y X, where k is a constant independent of x and y. Any such constant is called the modulus of concavity of X. The class of quasi-normed spaces and the class of locally bounded spaces coincide. In fact, if X is quasi-normed by a quasi-norm, then the collection {εb X : ε > 0}, where B X = {x X : x }, is a base of bounded zero neighbourhoods for some linear topology on X. Conversely, if X is a locally bounded linear topological space, then the Minkowski functional of any bounded, balanced zero neighbourhood defines a quasi-norm on X which is consistent with the original topology. For quasi-normed spaces the notions of equivalent norms, quotient space, operator norm etc. are defined in a very the same way as for normed spaces. Definition 3.3. Let X be a quasi-normed space and 0 < p. We say that X is locally p-convex, provided that there exists a quasi-norm on X, equivalent to the original one, such that x + y p x p + y p for all x, y X. In such a case, the quasi-norm is called a p-norm. Let us now quote the fundamental result proved independently by Aoki and Rolewicz. The proof of the version presented below is explained in details in the hint to [Gra08, Exercise.4.6]. Theorem 3.4 (Aoki, 94 & Rolewicz, 957). Let be a quasi-norm on a vector space X with modulus of concavity k and let 0 < p be the solution of (k) p =. Then, there exists a p-norm on X which is equivalent to. More precisely, the quasi-norm satisfies the inequality x +... + x n p 4 ( x p +... x n p) for all x,..., x n X and the quasi-norm given by the formula { ( ) /p x = inf x j p : x,..., x n X and x = j= } x j j= is a p-norm equivalent to. Let X be a quasi-normed space and be a quasi-norm on X. Certainly, all the formulas defining the quantities: a n (X), b n (X), c n (X) and d n (X) (see Section ) make sense in this more general setting. We are ready to make the first essential step towards the proof of the Kalton Giesy theorem. Proposition 3.5 (Kalton, 978). A quasi-normed space X is locally convex if and only if sup n n a n (X) <. 7
Proof. The only if part is obvious. For the if part the only thing to be proved is that the convex hull of the unit ball {x X : x } is bounded. So, assume there is a constant C < such that a n Cn for each n N and fix arbitrary vectors x,..., x n X with x j (for j [n]) and arbitrary non-negative numbers α,..., α n with n j= α j =. For every j [n] and m N set k j,m = mα j. Plainly, lim m m k j,m = α j. For every m N we have k j,m x j a n j= k C k j,m j,m, hence j= j= m x j C. Letting m, and denoting k the modulus of concavity, we arrive at ( ) k j,m α j x j k m x j + ( α j k ) j,m x j m j= j= j= ( k C + k α j k ) j,m x j m k C, m so the result follows. k j,m j= j= 8