Algebra i analiz St. Petersburg Math. J. Tom. 17 (2005), vyp. 4 Vol. 17 (2006), No. 4, Pages 683–698 S 1061-0022(06)00925-3 Article electronically published on May 3, 2006
GENERATING BOREL SETS BY BALLS
E. RISS
Abstract. It is proved that an arbitrary infinite-dimensional Banach space with basis admits an equivalent norm such that any Borel set can be obtained from balls by taking complements and countable disjoint unions. For reflexive spaces, the new norm can be chosen arbitrarily close to the initial norm.
Introduction Let X be a separable Banach space. The following question will be discussed: is it possible to obtain an arbitrary Borel set from balls by using the operations of taking complements and at most countable disjoint unions? We start with some definitions (see, e.g., [1] or [2]).
Definition 1. A nonempty family D of subsets of a set A is called a Dynkin system if D is closed under taking complements and at most countable disjoint unions.1
Definition 2. A nonempty family M of subsets of a set A is called a monotone system if M is closed under taking at most countable disjoint unions and countable monotone limits (i.e., increasing unions and decreasing intersections). It is easy to show that any Dynkin system is a monotone system; see [1]. Consider the minimal Dynkin system D(X) that contains all balls in X. All elements of D(X) are Borel sets. Therefore, our main question can be reformulated as follows: is it true that D(X) coincides with the Borel σ-algebra B(X)ofthespaceX? A similar question is meaningful for the minimal monotone system M(X)thatcontains all balls. Clearly, if M(X)=B(X), then D(X)=B(X). Furthermore, we can refine the question, using only balls belonging to some fixed class (for example, only “small” balls, i.e., those with radius not exceeding 1, or, by contrast, only “large” balls with radius at least 1). In what follows, the balls we consider are assumed to be open. A detailed survey of the history and results concerning the problem under discussion, including proofs, can be found in the book [1] and the paper [2]. We say a few words about history. For the first time, the question concerning the coincidence of the minimal Dynkin system generated by balls and the Borel σ-algebra for separable Banach spaces was raised at the 3rd Conference on Topology and Measure (1980), in connection with the problem of coincidence of two finite Borel measures taking the same values at all balls of the space. Later on, these two questions were separated from each other. For finite-dimensional spaces, the problem of generating the Borel sets by small balls was solved completely (in the affirmative).
2000 Mathematics Subject Classification. Primary 46B20, 46B25, 28A05. Key words and phrases. Dynkin system, monotone system, small balls, large balls, Borel sets. 1Such systems were considered and studied much earlier by W. Sierpi´nski and some other mathe- maticians after him. The author thanks Professor Z. Lipecki for this remark.
c 2006 American Mathematical Society 683
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Theorem A. Let X be a finite-dimensional Banach space, and let C be a family of balls such that for every x ∈ X and every r0 > 0 there exists r ∈ (0,r0) such that B(x, r) ∈ C (B(x, r) is the open ball centered at x and of radius r). Then the minimal monotone system containing C coincides with the Borel σ-algebra of X. (This remarkable theorem is Corollary 8.4 in [1].) In [1], Theorem A was obtained as a consequence of a general statement whose proof occupies several pages and refers the reader to some strong abstract covering theorems. The history of this result is rather rich in events. In 1988, Olej˘cek [3] proved that in Rn the minimal Dynkin system containing all balls coincides with the Borel σ-algebra for n = 2, and in 1995, in [4], he proved that the same is true for n = 3. Later, Jackson and Mauldin in [5], and also Zelen´y in [6] independently extended this result to Rn with an arbitrary n. In fact, the proof given by Jackson and Mauldin works in the case of an arbitrary finite-dimensional space (not only Euclidean), and M. Zelen´yusedonly monotone systems. For infinite-dimensional Banach spaces the situation is completely different. In [2] it was proved that the minimal Dynkin system containing all balls in the Hilbert space l2 does not coincide with the Borel σ-algebra.
Theorem B. There exists a Borel subset of l2 that is not generated by balls (with the help of taking complements and at most countable disjoint unions). Nevertheless, in some infinite-dimensional spaces the situation is the same as in finite- dimensional spaces. For instance, in c0 the small balls generate the Borel sets, and the same is true for the large balls. (Apparently, this fact, together with a simple proof, should be regarded as well known.) The following natural questions arise: (1) What are the infinite-dimensional Banach spaces, besides c0, in which the balls generate all Borel sets? (2) Is it true that every separable Banach space admits an equivalent norm for which the balls generate the Borel sets? In other words, is the question we consider purely topological, or is the geometry of the unit ball also important? As a result of misunderstanding, in the paper [2] some strong statements in this direction were announced and ascribed to the present author. Now some weaker versions of those statements are proved indeed; the proofs are presented below. The paper is organized as follows. In §1 we give an example of a “nice” space (in the sense of question (1)): the minimal Dynkin system containing the large balls in C[0, 1] coincides with the Borel σ-algebra of C[0, 1] (Proposition 1). In §3 we present a simple construction allowing us to “correct” slightly the unit ball B in the spaces lp (1 ≤ p<∞). As a result, we obtain an equivalent ball B0 such that the new large balls (the sets RB0 + x, R ≥ 1, x ∈ lp) generate the Borel sets. The construction is described in the proof of Proposition 2 and admits much freedom, which motivates seeking generalizations. Such generalizations can be found in §§4, 5 (unfortunately, the transparentness of the initial construction is lost there). Theorem 1 (in §4) says that if a Banach space possesses a Schauder basis, then this space admits an equivalent norm whose large balls generate the Borel sets. Unfortunately, the norm obtained in Theorem 1 may fail to be close to the original norm (our correction of the ball is far from being delicate). In §5 we try to refine Theorem 1. In Theorem 2 we prove that in the case of a reflexive Banach space with basis, the “nice” ball of Theorem 1 can be chosen as close to the original ball B as we wish.
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Figure 1.
Figure 2.
The author wants to express her sincere gratitude to Professor J. Ti˘ser for useful discussions and to the referee for advice that helped to improve the structure of the paper.
§1. The large balls of C[0, 1] generate all Borel sets Proposition 1. Let D be the minimal Dynkin system of subsets of C[0, 1] containing all balls with radius r ≥ 1.ThenD coincides with the Borel σ-algebra.
Proof. Step 1. We fix a point t0 ∈ (0, 1) and a number ε>0, ε
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Figure 3.
we see that
A(q1,q2,t)=A¯(q1,t) \ A(q2,t) ∈ D. Thus,
(1) {x ∈ C[0, 1] : q1 ≤ x(t) ≤ q2}∈D.
Step 2. In (1), we can take any finite set {t1,...,tn}⊂(0, 1) instead of one point t:
(2) {x ∈ C[0, 1] : x(ti) ∈ [ai,bi],i=1,...,n}∈D
for any t1,t2,...,tn ∈ (0, 1) and any a1,...,an,b1,...,bn ∈ R. We explain how to pass from one point to two points. We fix t1, t2 ∈ (0, 1), a1 ≤ b1, a2 ≤ b2 (if a1 >b1 or a2 >b2, then the set in question is empty, and ∅ ∈ D). As the centers of balls we take broken lines with two suitably high peaks at t1 and t2 (see def Figure 3). As before, we see that the set E1 = {x ∈ C[0, 1] : x(t1) >a1, x(t2) >a2} belongs to D. Repeating the arguments of Step 1, we see that
E1 = {x ∈ C[0, 1] : x(t1) ≥ a1,x(t2) ≥ a2}∈D
and { ∈ ≥ }∈ E1 = x C[0, 1] : x(t1) >b1,x(t2) a2 D. def { ∈ ≤ ≤ ≥ } \ ∈ Then E2 = x C([0, 1] : a1 x(t1) b1,x(t2) a2 = E1 E1 D. { ≤ ≤ ≥ 1 }∈ Next, x : a1 x(t1) b1, x(t2) b2 + n D, and this sequence is monotone increasing, so that its union
E3 = {x : a1 ≤ x(t1) ≤ b1,x(t2) >b2}
belongs to D. Taking the difference of the sets E2 and E3, we conclude that
{x : x(t1) ∈ [a1,b1],x(t2) ∈ [a2,b2]}∈D,
as required. Similar arguments give (2) by induction.
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k n − Step 3. For fixed n, we put tk = 2n , k =1, 2,...,2 1. Then (2) takes the following form: k (3) x ∈ C[0, 1] : a ≤ x ≤ b ,k=1,...,2n − 1 ∈ D k 2n k ∈ N ≤ n − ∈ for all n and all ak, bk with ak bk, k =1,..., 2 1. Suppose f, g C[0, 1], ≤ ∈ N k k n − f g.Forn ,in(3)wetakeak = f n , bk = g n , k =1, 2,...,2 1. Let 2 2 ∈ k ≤ k ≤ k n − Cn = x C[0, 1] : f n x n g n ,k=1,...,2 1 . 2 2 2 ⊃ ⊃ ··· ∈ ∈ N ∞ Then C1 C2 ,andCn D, n . Therefore, the set C = n=1 Cn belongs to D.ButC = {x ∈ C[0, 1] : f ≤ x ≤ g}. The result of the first 3 steps is the following: (4) {x ∈ C[0, 1] : f ≤ x ≤ g}∈D for all f,g ∈ C[0, 1] with f ≤ g. So, all order intervals (and, in particular, all balls with radius less than 1) belong to D.
Step 4. Let B(x, r) be the closed ball with center x ∈ C[0, 1] and radius r>0. Observe ∈ n that for any x1, x2,...,xn C[0, 1] and any r1,...,rn > 0theset k=1 B(xk,rk)is equal to {x ∈ C[0, 1] : f ≤ x ≤ g},where f =max{x − r ,x − r ,...,x − r },g=min{x + r ,x + r ,...,x + r } 1 1 2 2 n n 1 1 2 2 n n n ∈ (in particular, it can be empty). Therefore, k=1 B(xk,rk) D,by(4). Thus, the intersections of finite families of closed balls in C[0, 1] belong to D.
Step 5. Observe that if the finite intersections of balls belong to D, then the same is n ∈ ∈ true for the finite unions of balls: k=1 B(xk,rk) D for all xi X, ri > 0. (For two balls we have B1 ∪ B2 = B1 ∪ (B2 \ (B1 ∩ B2)) ∈ D. In the general case the argument is similar: the inclusion-exclusion formula works.) To finish the proof of Proposition 1, now it suffices to check the following statement, which will be used repeatedly in the sequel. Lemma 1. If (X, ρ) is a separable metric space and M is the minimal monotone system in X containing all closed balls and all unions of finite families of balls, then the system M coincides with the Borel σ-algebra of (X, ρ).
Proof. First, if a monotone system contains the finite unions of balls, then it also contains ∞ ∞ N the countable unions: n=1 Bn = N=1 n=1 Bn is the limit of a monotone increasing sequence of subsets of M. By the Lindel¨of theorem, any open set in the separable space (X, ρ) can be represented as the union of a finite or countable family of closed balls. Thus, the system M contains all open sets. To complete the proof of Lemma 1, we need to show that any monotone system containing all open sets contains all Borel sets. This fact can be found, e.g., in [1, Subsection 8.4]. For completeness, we present a proof, for which the author thanks the referee. Containing all open sets, the system M contains all Gδ-sets and, therefore, all sets that are Gδ and Fσ simultaneously. Since such sets form an algebra, the minimal monotone system containing them is the Borel σ-algebra. This completes the proof of Lemma 1 and, with it, of Proposition 1. Remark. The above arguments can be repeated almost word for word to obtain a similar property for the space c0: the large balls in c0 also generate all Borel sets.
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§2. Some auxiliary notions and facts
{ }∞ def Suppose that X is a Banach space with a normalized basis en 1 .WesetXn = span{e1,...,en},denotebyPn the canonical projection onto Xn, and put Ln = span{en+1,en+2,...} =KerPn. −1 ⊂ Definition 3. A set of the form C = Pn (A)=A + Ln,whereA Xn is a Borel set, is called a cylinder with base A.IfA is a cone, then the cylinder with base A is called a wedge. Let Mc be the minimal monotone system containing all cylinders.
Lemma 2. The system Mc coincides with the Borel σ-algebra of X.
Proof. By Lemma 1, it suffices to show that Mc contains all finite unions of closed { }∞ balls. Since the situation is purely topological, we may assume that the basis en 1 is monotone. For simplicity, we consider only two balls B1 = B(x, r1)andB2 = B(y, r2). Let ∪ −1 ∪ F = B1 B2 and Fn = Pn (Pn(B1 B2)). We must show that F ∈ Mc. It is clear that the family {Fn} is monotone decreasing ∞ and that all cylinders Fn belong to Mc. Therefore, Fn ∈ Mc. We shall check that ∞ n=1 F = n=1 Fn. ∈ ∈ ∈ Fix z =(z1,z2,...) F , k N. The definition of Fk implies that z Fk.Con- ∈ ∞ ∈ versely, suppose z =(z1,z2,...) k=1 Fk. We want to show that z F .Ifnot, then z/∈ B1 and z/∈ B2, i.e., z − x >r1 and z − y >r2.Forx =(x1,x2,...) we have (z1 − x1,z2 − x2,...) = r1 + ε for some ε>0. Then, for some k,wehave (z1 − x1,...,zk − xk, 0, 0,...) >r1. Using the monotonicity of the basis, we see that
(z1 − x1,...,zk − xk,αk+1 − xk+1,...) >r1
for all αk+1,αk+2,.... This means that for arbitrary αk+1,αk+2,... the point (z1,z2,..., zk,αk+1,αk+2,...)doesnotbelongtoB1. Similarly, the condition z/∈ B2 implies that for some m ∈ N any point of the form (z1,z2,...,zm,βm+1,βm+2,...) lies off B2. Assuming ≥ −1 ∪ ∈ that m k, we see that the points of Pm (Pmz)donotbelongtoB1 B2, i.e., z/Fm. This contradiction finishes the proof of Lemma 2. { −1 ⊂ } Remark. Let Cn denote the class Pn (A):A Xn, A is Borel of all cylinders with base of dimension not exceeding n.Let{nk} be a monotone sequence of integers tending ⊃ ≤ to infinity. Since Cnk Cm for m nk, the minimal monotone system containing all sets
of all Cnk coincides with Mc. Thus, this new system coincides with the Borel σ-algebra. So, the cylinders generate all Borel sets. Now we shall show that, instead of cylinders, it suffices to use only wedges of a special kind.
Definition 4. Let {y1,...,yn} be a linearly independent system of points in Xn.The cone with vertex 0 spanned by co{y1,...,yn}, and also all shifts of this cone, will be called nice cones in Xn. A wedge is called a nice wedge of dimension n if its base is a nice cone in Xn. Lemma 3. Let K2, K3,... be nice cones in the spaces X2,X3,..., and let K2, K3,... be wedges with bases K2, K3,..., respectively. Then the minimal Dynkin system that contains the wedges K2, K3,... and their shifts Kn + zn, zn ∈ Xn, coincides with the Borel σ-algebra. Proof. It suffices to show that, for each n =2, 3,..., if a Dynkin system D in X contains all shifts of the cone K ,thenD (∗) n n contains all Borel sets of Xn.
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Suppose the cone Kn is generated by a system {y1,...,yn}. We regard this system as a basis of Xn,andtheconeKn as the positive hyperoctant P = {(x1,...,xn):xk ≥ 0,k =1,...,n}. We show that all shifts of the cubes
Lr = {(x1,x2,...,xn):0≤ xk ≤ r} are contained in D. Indeed, taking the union of the monotone increasing system of hyperoctants Pk = 1 { ≥ 1 ≥ ≥ } P + k y1 = (x1,...,xn):x1 k ,x2 0,...,xn 0 , we see that the set (1) P = {(x1,...,xn):x1 > 0,x2 ≥ 0,...,xn ≥ 0} (1) belongs to D. Therefore, the set P \ (P + ry1)={(x1,...,xn):0≤ x1 ≤ r, x2 ≥ 0, ...,xn ≥ 0} belongs to D. Repeating these arguments for the other coordinates, we obtain Lr ∈ D. Obviously, all shifts of the cube Lr (the sets Lr + z,z ∈ Xn)alsobelongtoD.Butweknowthatin ∞ the finite-dimensional space ln every Dynkin system containing all small balls contains all Borel sets. The lemma is proved. So, in order to generate the Borel sets starting with balls, it suffices to generate some nice wedge of dimension n for every n; moreover, it suffices to do this for n running through some sequence (nk),nk →∞. Precisely this will be done in the proof of Proposition 2 and Theorems 1 and 2 below. Namely, the original ball will be modified so that the new ball acquires the form of a nice wedge in small neighborhoods of certain points of Xn. Then we can “blow” the new ball up so as to get an entire nice wedge. More precisely, to generate a nice wedge of dimension n, we require that the new ball B0 satisfy the following conditions (Kn):
There exists a vector xn ∈ Xn, xn = 1, such that − ⊂ − (Kn) 1) B 0( kxn,k) B0( (k +1)xn,k+1)fork =1, 2,...; ∞ − 2) k=1 B0( kxn,k) is a nice wedge with base in Xn.
§3. A “good” norm close to the standard norm in lp
Proposition 2. Let B = B(0, 1) be the unit ball of lp,1≤ p<∞, and let ε be a positive number. There exists an equivalent norm with unit ball B0 = B0(0, 1) such that 1) dist(B,B0) <ε(in the Hausdorff metric); 2) the minimal Dynkin system D containing all balls B0(x, R),x∈ lp, R ≥ 1, coincides with the Borel σ-algebra of lp. Remark. Proposition 2 is a special case of Theorem 1 (to be proved in §4). However, here we give a sketch of the proof, because this proof is quite simple and clear, follows the same leading idea, and is not overloaded with many technical details that arise unavoidably in the proof of Theorem 1.
Proof. The construction of the unit ball B0 can be described as follows. We fix ε>0, assuming that ε is sufficiently small. The ball B0 will be obtained as the monotone intersection of some intermediate balls Bn, n =2, 3,..., which will be constructed by ∞ ⊃ ⊃ ⊃··· { }∞ induction, B0 = n=2 Bn, B B2 B3 . For the standard basis ek k=1 of lp,we denote Xn =span{e1,...,en}.
Base of induction (n =2). In the two-dimensional space X2 we draw the line parallel to X1 and passing through the point (1 − ε)e2. This line intersects the oval B ∩ X2 − ε at points M and N (see Figure 4). Let A be the point 1 2 e2,letK2 be the angle ∠ ∞ MAN,andletK2 be the cylinder with the base K2. ∞ ∩ − ∞ ∩ We put B2 = K2 ( K2 ) B.
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Figure 4.
Step of induction. Suppose that the balls B2, B3,...,Bn−1 have already been con- structed, B ⊃ B2 ⊃ B3 ⊃···⊃Bn−1. In the space Xn we take the hyperplane parallel to Xn and passing through the point (1 − ε)en.LetS denote the intersection of this hyperplane and ∂Bn−1. We choose n points y1,...,yn on S in such a way that, for some δn > 0withδn <ε, { − − }n 1) the system (1 ε + δn)en yk k=1 is linearly independent; 2) the ball B(0, 1 − ε) is contained in the cone with the vertex (1 − ε + δn)en spanned by the system {y1,...,yn}. Observe that in the space l2 the points y1,...,yn can be taken at the vertices of a regular simplex. For the present, let us agree that the existence of such a system {y1,...,yn} is obvious. Below, in the proof of Theorem 1, this fact will be proved. Let Kn denote the cone with the vertex (1 − ε + δn)en spanned by the system { } ∞ y1,...,yn . The wedge with the base Kn is denoted by Kn . We put def ∞ ∩ − ∞ ∩ Bn = Kn ( Kn ) Bn−1.
Observe that the ball Bn is obtained from Bn−1 by “grinding” in some neighborhood Un of the point en.Ifε is sufficiently small, then the neighborhoods Un and Um are disjoint for m = n. Thus, the “grindings” at different steps do not affect each other. ∞ So,wehaveconstructedtheballB0 = n=2 Bn. Obviously, by construction we have B0 ⊂ B and dist(B0,B) <ε. Now we verify the main property of B0 (property (Kn)for ∈ N ≥ − ≥ − ε n =2, 3,...). We fix n , n 2, and put xn =(1 ε+δn)en for n 3, x2 = 1 2 e2. Denoting the new norm by · 0,wehave xn 0 =1.
Lemma 4. For the point xn and the ball B0, conditions (Kn) are fulfilled: − ⊂ − 1) B0( kxn,k) B0( (k +1)xn,k+1),k =1, 2,..., and ∞ − ∞ − − 2) k=1 B0( kxn,k)=(Kn xn) is a nice wedge with the base Kn xn. ∈ N k+1 Proof. 1) We fix k and consider the homothety with center 0 and coefficient k . The image of B0(−kxn,k) under this homothety is the ball B0(−(k +1)xn,k+ 1). Since
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Figure 5.
0 ∈ B0(−kxn,k)andthesetB0(−kxn,k) is convex, we have
B0(−(k +1)xn,k+1)⊃ B0(−kxn,k). ∞ − 2) Observe that k=1 B0( kxn,k) is the cone with vertex 0 spanned by the ball B0(−xn, 1). In some neighborhood U of the point zero, the ball B0(−xn, 1) coincides ∞ − − ∩ ∞ − ∩ with the wedge Kn xn, because B0( xn, 1) U =(Kn xn) U. Therefore, ∞ − ∞ − B0( kxn,k)=Kn xn. k=1 { − }∞ By Lemma 4, the limit of the monotone increasing sequence B0( kxn, k) k=1 of ∞ − balls, which belongs to the Dynkin system D, coincides with the wedge Kn xn.Thus, for every n ∈ N \{1} we have a nice wedge of dimension n lying in D. Now, to finish the proof of Proposition 2, it remains to apply Lemma 3.
§4. The existence of a “good” equivalent norm in a space with basis Theorem 1. Let (X, · ) be an infinite-dimensional Banach space with a normalized { }∞ · basis en 1 .ThenX admits an equivalent norm 0 with unit ball B0 such that the minimal Dynkin system D containing the large balls RB0 + x (x ∈ X, R ≥ 1) coincides with the Borel σ-algebra of (X, · ). { }∞ · If the basis en 1 is monotone, then for every ε>0 the norm 0 can be chosen to be eε-equivalent to the norm · . { }∞ · Proof. Let en 1 be a monotone basis, let B be the unit ball in (X, ), and let ε>0. We are going to generalize the construction described in the proof of Proposition 2. As ⊃ ⊃ ⊃ ··· before, we construct balls B2, B3,... such that B B2 B3 by induction and ∞ { } put B0 = n=2 Bn.Asalways,Xn =span e1,...,en . Base of induction (n = 2). In the two-dimensional space X2, we consider the ball 2 B = B ∩ X2. By [7, Corollary 5.18], every weakly compact set in a Banach space is the closure of the convex hull of its strictly exposed points. (Recall that a point e2 on the unit sphere ∗ is said to be strictly exposed for the unit ball if there exists a functional x2 such that
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∗ ∗ ∗ x2 =1and1=x2(e2) >x2(y) for all y with y =1,y = e2. Moreover, it is required { ∈ ∗ − }−−−→ that diam x B : x2(x) > 1 δ 0.) δ→0 2 In ∂B we can find a strictly exposed point e2, e2 = 1, such that e2 = αe1 + βe2 with β>1 − ε. (Such a point does exist. Indeed, assuming that every strictly exposed point has a coefficient β not exceeding 1 − ε in the representation e = αe1 + βe2,wesee that the point e2 admits no approximation by convex combinations of strictly exposed points.) ∗ Let x2 be a functional occurring in the definition of a strictly exposed point: 1 = ∗ ∗ ∗ − ε ∈ 2 − 8 x2 = x2(e2) >x2(y) for all y ∂B , y = e2. We find δ,0<δ<1 e , such that ∗ − ∈ 2 − the inequality x2(x) > 1 δ for x B implies that x e2 <ε. Consider the slice ∗ { ∈ 2 ∗ − } ∈ A2 = A(e2,x2,δ)= x B : x2(x) > 1 δ (see Figure 5). Then for all x A2 we have − { ∗ − } 2 x e2 <ε. The intersection of the line x : x2(x)=1 δ and the unit sphere S · − δ consists of points a and b.LetK2 denote the cone with the vertex e2 1 spanned ∞ 2 by the segment [a, b], and let K2 be the wedge with the base K2. As before, we put ∞ ∩ − ∞ ∩ B2 = K2 ( K2 ) B.
Step of induction. Suppose the balls B2, B3,...,Bn−1 have already been constructed, ⊃ ⊃ ⊃ ··· ⊃ n B B2 B3 Bn−1. We consider the intersection Bn−1 of the ball Bn−1 and n the space Xn. On the sphere of the n-dimensional ball Bn−1, we find a strictly exposed − ∗ ∗ point en =(α1,...,αn)with1 ε<αn;letxn be a functional, xn = 1, that attains − ε n ≤ − 2n+1 its strict maximum on Bn−1 at the point en. Next, we take δ>0, δ 1 e ,such ∗ − ∈ n − that the inequality xn(x) > 1 δ for x Bn−1 implies that x en <ε.Asabove,the ∗ diameter of the slice An = A(en,xn,δ)islessthanε. Thenextstepistochooseanumberσ>0withσ<δand points y1,...,yn in the n { ∗ − } intersection of the sphere Sn−1 and the hyperplane x : xn(x)=1 δ in such a way that { − − }n (5) the points (1 δ + σ)en yk k=1 are linearly independent, and − { } the cone Kn with the vertex (1 δ + σ)en spanned by co y1,...,yn (6) −ε/2n+1 · n includes the ball e Bn−1.
The following two lemmas show that this can be done indeed.
Lemma 5. Let Xn be an n-dimensional Banach space, let B = B(0, 1) be its unit ball, ∈ ∈ ∗ let e ∂B be a strictly exposed point of B,andletf Xn be a functional that attains its strict maximum on B at the point e.WeputYn =Kerf and F =(Yn +(1− δ)e) ∩ ∂B,whereδ ∈ (0, 1). Suppose that y1,...,yn ∈ F and (1 − δ)e ∈ int co{y1,...,yn}. Then there exists σ>0 such that the cone with the vertex (1 − δ + σ)e spanned by co{y1,...,yn−1} includes the ball (1 − δ)B.
Proof. Since (1 − δ)e ∈ int co{y1,...,yn},forsomer0 > 0 the intersection of the Eu- clidean ball B2((1−δ)e, r0) with the hyperplane (1−δ)e+Yn is included in co{y1,...,yn}. Observe that Γ = Yn +(1− δ)e is the support plane to the ball B(0, 1 − δ)atthepoint (1 − δ)e, and the point (1 − δ)e is a strictly exposed point of that ball. Therefore, Γ intersects the ball B(0, 1 − δ)atauniquepoint.Thus,forsomeα>0, the distance between the finite-dimensional Euclidean sphere ∂B2((1 − δ)e, r0) ∩ ((1 − δ)e + Yn)and ∂B(0, 1 − δ)isatleastα. As σ, we take any number in the interval (0,α). Then, obviously, the cone with the vertex (1−δ+σ)e spanned by B2((1−δ)e, r0)∩((1−δ)e+Yn) includes the ball B(0, 1−δ).
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Figure 6.
Figure 7.
Thereby, this ball is included in the cone with the vertex (1 − δ + σ)e and spanned by co{y1,...,yn−1} (Figure 6).
Lemma 6. Let Xn be an n-dimensional Banach space, let Y ⊂ X be its (n − 1)- dimensional subspace, and let e ∈ Xn \Y .IfF is a closed, bounded, and convex subset of Y + e, where e ∈ int F (in the topology of Y + e), then there exist points y1,...,yn ∈ ∂F such that the system {y1,...,yn} is linearly independent and e ∈ int co{y1,...,yn}.
Proof. We use induction on the dimension n.Ifn =2,thenF =[a, b]. We put y1 = a, y2 = b. To pass from n to n +1,firstwetakeapoint˜y1 on the boundary of F and strictly separate it from e by a functional f.LetKerf = Z,andletZ ∩ F = Fn.ThenFn is a bounded convex closed set (Figure 7). We draw the line l passing through the pointsy ˜1 and e and put def l ∩ Fn = en,l∩ ∂F = {y˜1,y1}.
We have constructed the point y1. The required linearly independent points y2,...,yn+1 are constructed on ∂Fn with the help of the induction hypothesis in such a way that en ∈ int co{y2,..., yn+1}. Then, obviously, e ∈ int co{y1,...,yn+1}, and the system {y1,...,yn+1} is linearly independent. The lemma is proved. Lemmas 5 and 6 allow us to find a number σ>0andpointsy1,...,yn satisfying conditions (5) and (6). Indeed, we only need to check that if a system γ = {y1,...,yn}
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is linearly independent and e ∈ co{y1,...,yn}, then for every α = 0 the system γ = { − − − } y1 (1 + α)e, y2 (1 + α)e,...,yn (1 + α)e is also linearly independent. Suppose n ≥ n e = i=1 αiyi, αi 0, i=1 α i = 1. If the system γ is linearly dependent, then for n | | n − some numbers β1,...,β n with i=1 β i =0wehave i=1( yi (1 + α) e) βi =0.Then n n n n n n i=1 βi yi =(1+ α) i=1 β ie =(1+ α) j=1 βj i=1 αiyi = i=1(1+α) j=1 βj αiyi, n − n whence i=1 βi (1+α) j=1 β j αi yi = 0. Since the system γ is linearly independent, n this means that βi =(1+α ) j=1 βj α i for i =1,...,n , i.e., βi = cαi with c = n n n n · (1 + α) j=1 βj.Butthen i=1 βi = c i=1 αi = c = i=1 βi (1 + α). We see n n that either α = 0, which is impossible, or i=1 βi = 0, which yields i=1 βiyi =0, contradicting the linear independence of γ. So, we have chosen a number σ>0andpointsy1,...,yn such that conditions (5) and (6) are fulfilled. − { } As before, let Kn be the cone with vertex (1 δ +σ)en and spanned by co y1,...,yn , ∞ ∞ ∞ and let K be the wedge with the base Kn. We put Bn = K ∩(−K )∩Bn−1. Finally, n ∞ n n we form B0 = k=2 Bk; this completes the construction of the desired unit ball. −ε All the required properties of B0 except for the inclusion B0 ⊃ e B can be checked precisely as in the proof of Proposition 2. (Observe that if ε is sufficiently small, then n−1 the slice An does not intersect the ball Bn−1 . Therefore, the transformations of the ball at the nth step do not affect the preceding transformations, which were made in Xn−1.) − − ε ⊃ ε n ⊃ 2n+1 n We check that B0 e B. By Lemma 5, we have Bn e Bn−1. We show that also − ε (7) Bn ⊃ e 2n+1 Bn−1. − ε Let x ∈ e 2n+1 Bn−1, x =(x1,x2,...,xn−1,xn,...). Then x ∈ Bn−1.Byconstruction, for the norm · n−1 with the unit ball Bn−1 we always have
(z1,z2,...,zn,...) n−1 ≥ (z1,...,zn, 0, 0,...) n−1. − ε ∈ 2n−1 n ⊂ n ∈ ∈ This implies that (x1,...,xn) e Bn−1 Bn .Thus,x Bn−1 and (x1,...,xn) n ∈ Bn . By construction, this means that x Bn.Nowwehave
ε ε ε − − − n Bn ⊃ e 2n+1 Bn−1 ⊃ e 2n+1 · e 2 Bn−2 ⊃··· ε ε ε ε ε ε ε − − n − − − n − − ⊃ e 2n+1 · e 2 ···e 24 B2 ⊃ e 2n+1 · e 2 ···e 24 · e 23 B − ε ·(1+ 1 +··· ) − ε −ε ⊃ e 8 2 B = e 4 B ⊃ e B, as required. Theorem 1 is proved.
§5.The“good”normsaredenseinthecaseofareflexivespacewithbasis Theorem 2. Let (X, · ) be a reflexive infinite-dimensional Banach space with a nor- { }∞ malized basis en n=1.LetB = B(0, 1) be the unit ball of X, and let ε be a positive number. Then X admits an equivalent norm · 0 with unit ball B0 such that 1) dist(B0,B) <ε(in the Hausdorff metric); 2) the minimal Dynkin system D containing all large balls RB0 + x (x ∈ X, R ≥ 1) coincides with the Borel σ-algebra. Thus, for every reflexive Banach space with basis, the family of “good ” balls is dense (in the Hausdorff metric) in the space of convex bounded symmetric sets. Proof. We start with the construction of small spherical segments (slices of the form B ∩ Γ, where Γ is a half-space), in which the subsequent “grinding” of the ball B will be made.
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Step 1. Since the space X is reflexive, its ball B is weakly compact and, thus, is the (1) (1) ∈ closure of the convex hull of its strictly exposed points. Let x1 =(t1 ,t2 ,...) B be a strictly exposed point, and let f1, f1 = 1, be a functional attaining its strict maximum ε ∈ on B at the point x1. We find a number δ1 < 4 such that, for x B, ε (1 ) f (x) > 1 − δ =⇒ x − x < . 1 1 1 1 8 { }∞ Being a basis of a reflexive space X, the system en n=1 is shrinking (this means that | −−−−→ ∈ ∗ { }∞ f span{en+1,en+2,... } 0 for all f X ). Let M1 be the basis constant of en n=1, n→∞ and let N1 be such that | ε f1 span{e ,... } < 2 , N1+1 4(1+M1+M ε) (21) 1 (0,...,0,t(1) ,t(1) ,...) < ε . N1+1 N1+2 8 { }∞ Now we modify the basis en n=1 somewhat: we want to replace the vectors e1,...,eN1 with vectors e ,...,e in such a way that x ∈ span{e ,...,e } and 1 N1 1 1 N1
N1 ε (3 ) e − e < . 1 i i 4 i=1
We explain how to do this. Denoting (t(1),...,t(1))=¯x ,(0,...,0,t(1) )=˜x , 1 N1 1 N1+1,... 1 ε ⇒ we have x1 =¯x1 +˜x1.Using(21), we get x˜1 < 8 .Ifε is small, then x1 =1 = (1) (1) sgn t(1) ≥ − ε ⇒| | ··· | |≥ − ε ≥ 1 i x¯1 1 = t + + t 1 .Pute = ei + (1) (1) x˜1.Then 8 1 N1 8 2 i |t |+···+|t | 1 N1
N1 N1 N1 (1) (1) 1 · (1) ti ei = ti ei + x˜1 ti sgn ti =¯x1 +˜x1 = x1, |t(1)| + ···+ |t(1)| i=1 i=1 1 N1 i=1 so that x ∈ span{e ,...,e },and 1 1 N1
N1 − · 1 ε · ε ei ei = x˜1 < 2= . |t(1)| + ···+ |t(1)| 8 4 i=1 1 N1 Thus, both required conditions are satisfied.
{ }∞ Step 2. We recall the Kre˘ın–Milman–Rutman theorem (see [8, p. 246]). If en 1 is a { }∞ normalized basis of X with basis constant K and xn 1 is a sequence in X satisfying ∞ − 1 { }∞ { }∞ n=1 en xn < 2K ,then xn 1 is a basis of X equivalent to en 1 . So, if ε>0 is sufficiently small, then the system {e ,...,e , e ,e } is a 1 N1 N1+1 N1+2,... basis of X. In what follows we shall work with this latter basis. Let M2 be its basis 1 constant. Then dist(eN +1, span{e ,...,e }) ≥ .SincethepointeN +1 lies in 1 1 N1 2M2 1 the closure of the convex hull of the strictly exposed points of B, there exists a strictly ∞ exposed point x =(t(2),t(2),...) (i.e., x = N1 t(2)e + t(2)e ) such that 2 1 2 2 i=1 i i i=N1+1 i i
∗ { } ≥ 1 ( 2)dist(x2, span e1,...,eN1 ) . 4M2
Let f2 be a functional of norm 1 that attains its strict maximum on B at the point x2. ε ∈ This functional gives us arbitrarily small slices: for some δ2 with 0 <δ2 < 2 ,forx B we have 1 ε ε (12) f2(x) > 1 − δ2 =⇒ x − x2 < min − , . 4M2 4 8
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Let N2 be a number⎧ for which ⎨ | ε f2 span{eN +1,...,} < 2 , 2 4(1+M1+ M1 ε) (22) ⎩ (0,...,0,t(2) ,...) < min ε , 1 . N2+1 16·4M2 8M2 As above, we modify the vectors {e ,...,e } so as to get vectors {e ,...,e } N1+1 N2 N1+1 N2 such that x ∈ span{e ,...,e } and 2 1 N2 N2 ε (3 ) e − e < . 2 i i 8 i=N1+1 (2) (2) (2) 1 1 1 (By (∗2)and(22), we now have |t | + |t | + ···+ |t |≥ − = .) N1+1 N1+2 N2 4M2 8M2 8M2 Conditions (11)and(12) imply that the slice {x ∈ B : f2(x) > 1 − δ2} does not intersect the slice {x ∈ B : f1(x) > 1 − δ1} constructed at Step 1. Indeed, suppose x belongs to the latter slice and y belongs to the former. Then
x − y ≥ x2 − x1 − x1 − x − y2 − y ∗ 1 ε ε ( 2) 1 1 ε ε ε ≥ x2 − x1 − − − ≥ − + − = > 0, 4M2 4 8 4M2 4M2 4 8 8 whence x = y. We continue the process described above. At the nth step, we deal with a basis {e ,...,e ,e ,...} with basis constant M . We find a strictly exposed point 1 Nn−1 Nn−1+1 n (n) (n) xn =(t1 ,t2 ,...) satisfying 1 ∗ − { } ≥ ( n 1)dist(xn, span e1,...,eNn−1 ) , 4Mn ε the corresponding functional fn,andnumbersNn and δn < such that, for x ∈ B, 8 1 ε ε (1n) fn(x) > 1 − δn =⇒ x − xn < min − , , 4Mn 4 8 and ⎧ ⎨ | ε fn span{e ,e ,... } < 2 , Nn+1 Nn+2 4(1+ M1+M ε) (2 ) 1 n ⎩ (n) ε 1 (0,...,0,t ...) < min n+2 , . Nn+1, 4Mn·2 8Mn { } Then we modify the part eNn−1+1,...,eNn of the original basis so as to get a system {e ,...,e } such that x ∈ span{e ,...,e } and Nn−1+1 Nn n 1 Nn Nn ε (3 ) e − e < n i i 2n+1 i=Nn−1+1
(this is done precisely as at Steps 1 and 2, by using inequalities (∗n)and(2n)). { }∞ By the Kre˘ın–Milman–Rutman theorem, the resulting system en n=1 is a basis of X equivalent to {en}. Moreover, obviously, all intermediate basis constants Mn are bounded from above by some number M.(Forsmallε, this allows us to keep the inequalities (1n) satisfied.) We also note that span{e ,e ,...} =span{e ,e ,...} Nk+1 Nk+2 Nk+1 Nk+2 for every k. So,wehavepreparedslicesAn = {x ∈ B : fn(x) > 1 − δn}.AsatStep2,itis easy to check that these slices are pairwise disjoint. Now we start “grinding” the ball B within An to obtain the required ball B0 (we shall “grind” the part of B in An to get a part of a nice wedge with the vertex xn). As always, the ball B0 will be obtained
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as the intersection of balls Bn, n =1, 2, 3,..., constructed by induction. We denote { } ∩ k span e1,...,ek by Xk and Bn Xk by Bn. We construct the ball B . Recall that M is the basis constant of the basis {e ,...,e , 1 2 1 N1 e }.InthespaceX , which contains the point x , we consider two slices N1+1,... N1 1
N1 { ∈ N1 − } A1 = x B : f1(x) > 1 δ1 , δ1 A˜N1 = x ∈ BN1 : f (x) > 1 − . 1 1 2 δ Let Γ = {x ∈ X : f (x)=1− δ } and Γ = {x ∈ X : f (x)=1− 1 }. 1 N1 1 1 1 N1 1 2 N1 The intersection of the hyperplane Γ1 and the ball B is a bounded closed set F . As in the proof of Theorem 1, we choose a number σ1 > 0 and linearly independent { } − points y1,...,yN1 in F so that the cone K1 with the vertex (1 σ1)x1 spanned by { } − ε N1 co y1,...,yN1 contains the ball 1 4 B . A natural modification of the proof of Lemma 5 allows us to ensure the following additional condition:
N1 (9) ∂K1 ∩ Γ1 lies outside of the ball (3 + M1)B . ∞ As always, we denote by K1 the wedge with the base K1: ∞ { ∈ } K1 = (z1,z2,...):(z1,...,zN1 , 0,...) K1 (here the coordinates correspond to the basis {e ,...,e ,e ,...}). 1 N1 N1+1 We show that ε (10) K∞ ⊃ 1 − B 1 2 and ∞ ⊃ \ (11) K1 B0 A1. ∞ To check (10), we take z ∈ 1 − ε B, z = N1 z e + z e , and denotez ¯ = 2 i=1 i i i=N1+1 i i ∞ N1 N1 ∈ ∞ i=1 ziei,˜z = i=N +1 ziei,¯z = i=1 ziei. We need to verify that z K1 , i.e., 1 z¯ ∈ K1. Assuming that this is not true, we consider two cases: 1) f1(¯z ) ≥ 1 − δ1 andz ¯ ∈/ K1; 2) f1(¯z ) < 1 − δ1 andz ¯ ∈/ K1. Inthefirstcasewehave ε f (z)=f (¯z)+f (˜z) ≥ 1 − δ + f (˜z) ≥ 1 − + f (˜z) 1 1 1 1 1 4 1 (2 ) ≥1 − ε − ε · 1 2 z˜ 4 4(1 + M1 + M1 ε) ε ε ε ≥ 1 − − =1− , 4 4 2 because z˜ ≤ z + z¯ ≤ z + z¯ + z¯ − z¯
N1 ≤ − z + M1 z +2M1 z¯ ei ei i=1 ε ≤ z (1 + M )+2M · z¯ 1 1 2 ≤ · · · 2 z (1 + M1)+M1ε M1 z = z (1 + M1 + M1 ε) ≤ 2 1+M1 + M1 ε. ≤ − ε This contradicts the inequality f(z1) z < 1 2 . Thus, case 1) is impossible.
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In case 2), condition (9) implies thatz ¯ ∈/ (3 + M )BN1 , i.e., z¯ ≥3+M .Butthen 1 1 ε z ≥ z¯ − z˜ ≥3+M − (1 + M + M 2ε) 1 − > 2 − M 2ε>1 1 1 1 2 1 ≤ · 2 for sufficiently small ε (recall that above it was shown that z˜ z (1 + M1 + M1 ε)). Since we have a contradiction again, case 2) is also impossible. This proves (10). The inclusion (11) was actually verified when we analyzed case 2). As usual, we put ∞ ∩ − ∞ ∩ B1 = K1 ( K1 ) B. N2 Now, we perform similar constructions in the ball B1 . The inclusion (11) means that all modifications of the ball were made only within the slice A1. Since the slice A2 appearing at the second step does not intersect A1, the “grinding” to be made when we construct B2 is independent of a similar process occurring in the construction of B1. Also, observe \ − ε that all modifications of the ball B are made only inside the spherical layer B 1 2 B. ⊃ ⊃ ··· ∞ Continuing the construction, we obtain balls B1 B2 and put B0 = n=1 Bn. Obviously, B0 ⊃ (1 − ε)B. It remains to show that the balls {RB0 + x : x ∈ X, R ≥ 1} − generate the Borel σ-algebra. Applying Lemma 4 to each point (1 σn)xn in XN ,we ∞ n see that all shifts of the nice wedge Kn belong to the Dynkin system D. By the proof of Lemma 3 (see (∗)), for all k and all Borel sets A ⊂ X the set Nk
Nk ∞ Nk def ∈ ∈ A = x X : x = αnen + αnen, αnen A
n=1 n=Nk+1 n=1 belongs to D. But Nk ∞ Nk ∈ ∈ A = x X : x = αnen + αnen, αn,en A .
n=1 n=Nk+1 n=1 { }∞ Thus, all the sets A are cylinders relative to the basis en n=1. By the remark after Lemma 2, D coincides with the Borel σ-algebra. References
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Russian State Pedagogical University, Mo˘ıka 48, St. Petersburg 191186, Russia Received 11/APR/2005
Translated by THE AUTHOR
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