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Albrecht PIETSCH

Biberweg 7 07749 Jena — Germany [email protected]

Received: June 10, 2008 Accepted: September 30, 2008

ABSTRACT

We study the Banach envelope of the quasi-Banach  space l1,∞ consisting of all x=(ξ ) s (x)=O 1 s (x) k with n n , where n denotes the non-increasing rearrangement of x=(ξk). The situation turns out to be much more complicated ◦ than that in the well-known case  of the separable subspace l1,∞, whose members s (x)=o 1 are characterized by n n .

Key words: Banach envelope, Marcinkiewicz space l1,∞, weak l1-space. 2000 Subject Classification: 46A16, 46B45.

1. Notation and terminology

We use standard notation and terminology of theory; see [21], for example. Throughout, X and Y are real quasi-normed linear spaces. The symbol ||| · ||| denotes quasi-norms, while ·is reserved for norms. By an embedding we a one-to-one continuous linear map J : X → Y . Let N := {1, 2,... }. The cardinality of a set S is denoted by |S|.

2. spaces

th The n approximation number of a bounded real sequence x=(ξk) is defined by   sn(x):=inf sup |ξk| : |F|

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sn(x) := inf{t>0:|S(x, t)| t}.  S x S x, t {k ξ  } x ξ We refer to ( ):= t>0 ( )= : k =0 as the support of =( k). A sequence x is called finite if it has a finite support. The finite sequences form a linear space, denoted by c00. A rearrangement of a sequence x=(ξk) is obtained by permutating its coordinates ξk, and adding or deleting 0’s. In the case of a null sequence, sn(x) is the non- increasing rearrangement of |x|=(|ξk|). The Banach space of all bounded real sequences x=(ξk) equipped with the

x|l∞ := sup |ξk| 1≤k<∞

is denoted by l∞, and c0 stands for the closed linear subspace of all null sequences. Let 0

1/p 1/p |||x|lp,∞||| := sup n sn(x) = sup t |S(x, t)| 1≤n<∞ t>0 is finite. It easily turns out that ||| · |lp,∞||| is a quasi-norm, which satisfies the quasi-   |||x + y|lp,∞||| ≤ cp |||x|lp,∞||| + |||y|lp,∞||| .

1/p In passing, we note that cp =2 is the best possible constant. Indeed, substituting x ,...,1 ,..., 1 , , ,... =( 1 n1/p (2n−1)1/p 0 0 ) and y 1 ,...,1 ,..., , , ,... , =((2n−1)1/p n1/p 1 0 0 )

|||x|l ||| |||y|l ||| s x y 2 −1 1/p ≤ c we may infer from p,∞ = p,∞ =1 and 2n−1( + )= n1/p that (2 n ) p. ◦ The little Marcinkiewicz space or little weak lp space, denoted by lp,∞, is defined ◦ to be the closed hull of c00 in lp,∞. The members of lp,∞ are characterized by the 1/p condition that lim n sn(x)=0. n→∞ In analogy to an observation of Hunt [9, pp. 259–260], it turns out that lp,∞ fails to be normable for 0

From now on, we concentrate on the case p=1.

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The Sargent space m1,∞ is the collection of all x=(ξk) for which s x ... s x x|m 1( )+ + n( ) 1,∞ := sup 1 1 ≤n<∞ ... 1 1 + + n is finite. Note that m1,∞ is a Banach space. Using the abbreviation n n 1 1 μn(x):= sk(x) Ln := , Ln with k k=1 k=1 we get x|m1,∞ = sup μn(x). 1≤n<∞ m ◦ c m The little Sargent space 1,∞ is defined to be the closed hull of 00 in 1,∞, while m • l m m ◦ 1,∞ denotes the closed hull of 1,∞ in 1,∞. The members of 1,∞ are characterized by lim μn(x)=0. We have the strict inclusions n→∞ l ◦ ⊂ l ⊂ m • ,l◦ ⊂ m ◦ ⊂ m • , m • ⊂ m . 1,∞ 1,∞ 1,∞ 1,∞ 1,∞ 1,∞ and 1,∞ 1,∞

1 ∈ l \l ◦ 1 ∈ m• \m ◦ a(λ) α(λ) <λ≤ Obviously ( k ) 1,∞ 1,∞ and ( k ) 1,∞ 1,∞. Define =( k ) with 0 1 by (λ) λ −m2 (m−1)2 m2 αk := m 2 if 2

The linear space b1,∞ consists of all x=(ξk) that admit a representation ∞ ∞ x= xk (coordinatewise), where x1,x2,... ∈l1,∞ and |||xk|l1,∞||| <∞. k=1 k=1

Obviously, l1,∞ ⊆ b1,∞. This inclusion is strict. Indeed, a result of Markus, m ◦ ⊆b Mityagin, and Ariño mentioned at the beginning of section 4 implies that 1,∞ 1,∞. a(λ) ∈ b \ l <λ< x |m ≤ |||x |l ||| Therefore 1,∞ 1,∞ for 0 1. It follows from k 1,∞ k 1,∞ and ∞ |||x |l ||| <∞ ∞ x m b k=1 k 1,∞ that k=1 k converges in 1,∞. Hence 1,∞ is a dense subset m• of 1,∞.

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We can easily check that b1,∞ becomes a Banach space under the norm ∞ x|b1,∞ := inf |||xk|l1,∞|||, k=1 the infimum being taken over all representations described above. Observe that the closed unit balls

Bl :={x∈l1,∞ : |||x|l1,∞||| ≤ 1} and Bm :={x∈m1,∞ : x|m1,∞≤1} have the same extreme points xex. The collection of these extreme points, which are ex 1 ex characterized by sn(x )= n for n=1, 2,... , is denoted by B . Since m1,∞ can be identified with the dual of l∞,1, we know from the classical ∗ ex Kre˘ın-Milman theorem that Bm is the weakly closed convex hull of B . Recall that the extreme points of the closed unit ball B∞ :={x∈l∞ : x|l∞≤1} ex have the form x∞ =(±1). The σ-convex hull of a bounded subset B of a Banach space is defined by  ∞ ∞  σ-conv(B):= λkxk : x1,x2,... ∈B, λ1,λ2,... ≥0, λk =1 . k=1 k=1 In the following, we need a trivial case of the Choquet theorem; see [2, § 27] and [19, chap. 3].

Lemma. The closed unit ball B∞ coincides with the σ-convex hull of its extreme points.

Proof. For every sequence x=(ξk)∈B∞, let

+1 if ξk ≥0, a =(αk), where αk := −1 if ξk <0.

Then 2x − a|l∞≤1. Therefore, beginning with x1 := x, we inductively find a1, x2 :=2x1 − a1,a2,x3 :=2x2 − a2,a3, ... such that

xi|l∞≤1 and xi+1 := 2xi − ai. This construction yields n ∞  −i −n  −i x = 2 ai +2 xn+1 = 2 ai. i=1 i=1 ex Using the preceding lemma and the formula Bl = B∞ ·B , we can easily show ex that the norm of b1,∞ is just the Minkowski functional of the σ-convex hull of B .In particular, it follows that b1,∞ consists of all sequences that admit a representation ∞ ∞ ex ex 1 x= ξkxk (coordinatewise), where sn(xk )= n and |ξk|<∞. k=1 k=1

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Historical remarks. The definition of Lp,∞[0, 1] goes back to the famous interpolation theorem of Marcinkiewicz [14, p. 1272], who considered functions U such that A[mesE(|U(x)| >A)]1/p is uniformly bounded for all A>0. spaces that are non-atomic forerunners of l∞,1 and m1,∞ were invented by Lorentz [13, pp. 418–420]. Later on, Sargent [23, p. 162; 24, pp. 64–65] and Garling [5, p. 96–99] introduced l m m ◦ the sequence spaces ∞,1, 1,∞, and 1,∞. Almost simultaneously, the Soviet school presented a third approach that was created for treating non-self-adjoint operators on . The ideals Sω, SΩ, S (◦) l m m ◦ and Ω are the operator-theoretical counterparts of ∞,1, 1,∞, and 1,∞; see [7, pp. 139–150] as well as the original papers of Gokhberg/Kre˘ın [6], Matsaev [16], Markus [15], Mityagin [17], and Russu [22], written in the early 1960’s. Curiously enough, the three lines of development took place without reference to each other. Thus it is impossible to decide who was the first inventor of the sequence spaces under consideration. In any case, all of them could be called Lorentz spaces.

3. The Banach envelope of a quasi-normed linear space

The following definition is made in the spirit of category theory. The objects are the real quasi-normed linear spaces and, since we are interested in the isometric theory, the morphisms are the contractions, |||T : X → Y ||| ≤ 1. The Banach envelope of a quasi-normed linear space X is a Banach space Xban together with a contraction E from X into Xban (not necessarily one-to-one) such that for every contraction T from X into an arbitrary Banach space Y there exists a unique ‘extension’ T ban: Xban H HH T ban 6 HH E H HH HHj XYT -

Note that Xban is unique up to an ; see [25, pp. 170–171]. The Banach envelope Xban can be obtained as the closed hull of K(X) in X∗∗, where K denotes the canonical map from X into its bidual.

Another construction goes as follows. Letting  n n  max π (x):=inf |||xk||| : x1,...,xn ∈X, xk = x, n=1, 2,... k=1 k=1 yields a semi-norm on X that coincides with the Minkowski functional of the convex hull of BX :={x∈X : |||x||| ≤ 1},

max π (x):=inf{ ≥ 0:x ∈  conv(BX )}.

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Obviously, πmax is the largest semi-norm π on X for which π(x) ≤ |||x|||. Passing to the quotient space X/N with N :={x∈X : πmax(x)=0}, we obtain a normed linear space, whose completion is the required Banach envelope. The underlying norm is denoted by ·|Xban. Since Lp[0, 1] with 0

The rest of this section is concerned with a ‘philosophical’ question. Let J : X → Y be an embedding of a normed linear space X into a Banach space Y , which means that X can be viewed as a linear subspace of Y . If the unique continuous extension JÜ: Xf → Y is one-to-one as well, then we may say that the character of the members of X is preserved in passing to the completion Xf. For instance, we think of the case that the members of Y are sequences, functions, or operators. Unfortunately, as shown by a trivial example, JÜ need not be one-to-one. Note that ∞ ∞ 1/2 2 2 x := ξk + |ξk| k=1 k=1

is a norm on c00, the linear space of all finite sequences x =(ξk). Let J be the embedding of c00 into l∞, and consider the sequences

n-times  x 1 ,..., 1 , ,... . n :=( n n 0 ) Then   1 1 1 xm − xn = m − n for m

the (abstract!) element x˜ ∈ ec00 generated by the Cauchy sequence (xn) is different Ü from o, while Jx˜ = o. This means that the complete hull ec00 cannot be canonically identified with a .

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The most famous example occurs in the theory of operator ideals and tensor products. Let F(X)=X∗ ⊗ X denote the ideal of finite rank operators on a Banach space X, and define the norm

n ◦  ∗ ν (T ):=inf xkxk, k=1 the infimum being taken over all finite representations

n  ∗ Tx = x, xk xk for x∈X, k=1 x∗,...,x∗ ∈ X∗ x ,...,x ∈ X F X where 1 n and 1 n . The completion of ( ) with respect to ν◦ can be identified with the projective tensor product X∗⊗ÒX. We know from Grothendieck [8, chap. I, p. 165] that the continuous extension JÜof the embedding J : F(X) → L(X) is one-to-one if and only if X has the ; see also [21, p. 280]. Another necessary and sufficient condition requires that ν◦ coincides with the nuclear norm ∞  ∗ ν(T ):=inf xkxk. k=1

In the latter case the infimum ranges over all infinite representations

∞  ∗ Tx = x, xk xk for x∈X, k=1

x∗,x∗,...∈X∗ x ,x ,...∈X where 1 2 and 1 2 .

4. Old and new results, open problems

For 0

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It is hoped that the diagram J ∗∗ l ∗∗ m - m ∗∗ 1,∞ 1,∞ HYH ¨¨* H ¨ 6 H J ban ¨ 6 H m• ¨ l ban - m • 1,∞ 1,∞ H J ban J ¨* H b ¨  HHj ¨ @ @ b ,∞ 1 1 PP @  PP  PP @  PP  P @@R  PPq Jm - l1,∞ m1,∞ l ban helps to understand the complications that occur in the case of 1,∞. Almost all arrows denote canonical embeddings. Possible exceptions could be J ban l ban → b ,Jban l ban → m• , J ∗∗ l∗∗ → m∗∗ . b : 1,∞ 1,∞ m• : 1,∞ 1,∞ and m : 1,∞ 1,∞ These maps are the continuous extensions and the bidual of the canonical embeddings

• J l → b ,J• l → m , J l → m , b : 1,∞ 1,∞ m : 1,∞ 1,∞ and m : 1,∞ 1,∞ respectively. ·|l ban ·|m As shown in section 6, the norms 1,∞ and 1,∞ are not equivalent on ban l1,∞. Therefore Jm• fails to be an . This implies that at least one of the conjecture below must be true. Conjecture 1. ·|l ban ·|b l The norms 1,∞ and 1,∞ are not equivalent on 1,∞.In J ban l ban → b other words, the map b : 1,∞ 1,∞ is onto but not one-to-one.

Conjecture 2. The norms ·|b1,∞ and ·|m1,∞ are not equivalent on l1,∞.In J b → m• other words, the map : 1,∞ 1,∞ is one-to-one but not onto. Recall that n ∞ x|l ban |||x |l ||| x|b |||x |l |||, 1,∞ := inf k 1,∞ and 1,∞ := inf k 1,∞ k=1 k=1 where the infima range over all finite and infinite representations n ∞ x= xk (n=1, 2,...) and x= xk (coordinatewise), k=1 k=1 respectively. Comparing these definitions with those of ν◦(T ) and ν(T ) shows that Conjecture 1 is not unimaginable. In summary, we notice a surprising (formal?) analogy: lban ↔ X∗⊗ÒX 1,∞ projective tensor product, b1,∞ ↔ N(X) ideal of nuclear operators, m1,∞ ↔ I(X) ideal of operators.

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5. The Kalton-Sukochev norm      ban  ban Explicit expressions for the norms ·|L1,∞[0, 1] and ·|L1,∞[0, ∞) were discovered by Cwikel/Fefferman [3, p. 150] and [4, p. 277]; see also Kupka/Peck [12, p. 237]. Unfortunately, their approach was strictly limited to the non-atomic case. This gap could be closed only recently by Kalton/Sukochev [11]. Following the latter authors, for every bounded sequence x=(ξk) and m=1, 2,..., we define the finite or infinite quantity n sk(x) β x k=ma+1 , m( ) := sup n ma

On the other hand, if βm(x) is finite for some m, then x ∈ l1,∞. Indeed, by A.1, we get 2ma 2ma 1 mas2ma(x) ≤ sk(x) ≤ βm(x) k ≤ (log 2m) βm(x) for a=1, 2,... . k=ma+1 k=a+1 Moreover, n n 1 nsn(x) ≤ sk(x) ≤ βm(x) k ≤ (1 + log 2m) βm(x) for n ≤ 2m. k=1 k=1 Thus |||x|l1,∞||| = sup ksk(x) ≤ (4 log 2m) βm(x). 1≤k<∞

The results presented in the remaining part of this section are essentially borrowed from Kalton/Sukochev [11]. For the convenience of the reader, I have added the elementary proofs. n  n−a  Sublemma. sk(x) = inf sh(x − u):|S(u)|≤a for x∈c0, k=a+1 h=1 a=0, 1, 2,... and n=1, 2,... such that a

Proof. If |S(u)|≤a, then sa+1(u)=0. Hence n n−a n−a   n−a sk(x)= sa+h(x) ≤ sa+1(u)+sh(x − u) = sh(x − u), k=a+1 h=1 h=1 h=1 which implies that n  n−a  sk(x) ≤ inf sh(x − u):|S(u)|≤a . k=a+1 h=1

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In order to get equality, we may assume that |ξ1|≥|ξ2|≥... . Putting

u := (ξ1,...,ξa, 0 ,..., 0 ,...) yields x−u =(0,..., 0 ,ξa+1,...,ξn,...).

Thus sh(x−u)=|ξa+h|=sa+h(x). Lemma. If a, b=0, 1, 2,... and n=1, 2,... such that a+b

Proof. By the preceding sublemma, we may choose u, v ∈c00 such that n n−a sk(x)= sh(x−u) and |S(u)|≤a, k=a+1 h=1 n n−b sk(y)= sh(y−v) and |S(v)|≤b. k=b+1 h=1

Then |S(u+v)|≤|S(u)|+|S(v)|≤a+b and, therefore, n n−a−b n−a−b n−a−b sk(x+y) ≤ sh(x+y−u−v) ≤ sh(x−u)+ sh(y−v) k=a+b+1 h=1 h=1 h=1 n−a n−b n n ≤ sh(x−u)+ sh(y−v)= sk(x)+ sk(y). h=1 h=1 k=a+1 k=b+1

Proposition 1. β(x):= lim βm(x) defines a norm on l ,∞. m→∞ 1 Proof. We conclude from n n n sk(x + y) ≤ sk(x)+ sk(y) for x, y ∈ c0 k=2ma+1 k=ma+1 k=ma+1 that β x y ≤ β x β y x, y ∈ c , 2m( + ) m( )+ m( ) for 0 which implies the triangle inequality, β(x+y) ≤ β(x)+β(y).

Proposition 2. β(x) ≥x|m1,∞ for x∈l1,∞.

Proof. Letting a=0 in the definition of βm(x), we obtain n sk(x) β x ≥ k=1 x|m . m( ) sup n = 1,∞ 1≤n<∞ 1 k k=1

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Proposition 3. β x x|m x∈l ◦ ( )= 1,∞ for 1,∞.

Proof. Given any finite sequence x, we let m0 := |S(x)|. Then it follows from s x m0+1( )=0 that n sk(x)=0 whenever n>ma≥m0. k=ma+1 n Thus sk(x) β x k=1 x|m m ≥ m , m( ) = sup n = 1,∞ if 0 1≤n<∞ 1 k k=1 x∈c β x x|m which proves the required equality for 00. Since ( ) and 1,∞ are continuous |||·|l ||| x∈l ◦ with respect to quasi-norm 1,∞ , the result extents to all 1,∞. A theorem of Kalton-Sukochev says that β x x|l ban x∈l . ( )= 1,∞ for 1,∞ The proof of this fundamental achievement is rather cumbersome. Luckily, we need only its elementary part: Proposition 4. β x ≤x|l ban x∈l ( ) 1,∞ for 1,∞.

Proof. In view of β(x) ≤ |||x|l1,∞||| for x ∈ l1,∞, the inequality follows from the fact ·|l ban l |||·|l ||| that 1,∞ is the greatest norm on 1,∞ majorized by 1,∞ .

6. Tricky sequences

The following theorem, which is the main result of this paper, belongs to the ‘frustrating’ part of Banach space theory. Theorem. ·|m ·|l ban l The norms 1,∞ and 1,∞ fail to be equivalent on 1,∞. x|m ≤x|l ban x ∈ l Proof. Obviously, 1,∞ 1,∞ for 1,∞. To see that there holds no x|l ban≤cx|m converse estimate 1,∞ 1,∞ , a family of ‘tricky’ sequences will be constructed. The symbols A.1, ... ,A.4 refer to some elementary facts from calculus that are summarized in the appendix.

(A) Given natural numbers i1,i2,... such that h 2 ih ≤ih+1, (1) n :=0 h we define 0 ,  m nh := (2 −1)im for h=1, 2,..., m=1 and p nh,p := nh−1 +(2−1)ih for p=0,...,h. (2)

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Then n n

It follows by induction that h nh ≤ 2 ih −i1. (3)

h−1 Indeed, the case h =1 is trivial, since n1 = i1. Assume that nh−1 ≤ 2 ih−1 −i1 is true for some h≥2. Using (1), we obtain

h h−1 h h h nh = nh−1 +(2 −1)ih ≤ 2 ih−1 − i1 +(2 −1)ih ≤ ih +(2 −1)ih − i1 =2 ih − i1.

Combining (1) and (3) yields nh ≤ ih+1 − i1. (4) In view of (2), we get p nh,p ≤ 2 ih − i1. (5)

(B) Let

Nh,p :={n : nh,p−1

p−1 h Note that |Nh,p|=2 ih and |Nh|=(2 −1)ih. Consequently, the size of the sets Nh and Nh,p increases rapidly:

h+1 |Nh+1| > 2 |Nh| and |Nh,p+1| =2|Nh,p|.

(C) Define the sequence x=(ξn) by ξ 1 1 n∈N . n := p− = for h,p 2 1ih |Nh,p|

The condition (1) guarantees that (ξn) does not increase:

ξ 1 ξ 1 . nh = h−1 is greater than nh+1 = 2 ih ih+1

Thus sn(x)=ξn. We conclude from (5) that

ns x ≤ n 1 < n ∈ N . n( ) h,p p− 2 for h,p 2 1ih

Hence x∈l1,∞.

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(D) In the next step, the norm β(x) will be estimated from below. Obviously,  ξk =1. k∈Nh,p

q h Because of (2) and (3), we have (2 −1)ih ≤nh,q and nh <2 ih. Therefore

h 2ih nh h  sk(x) ≥ sk(x)= ξk = h − q. q k=(2 −1)ih+1 k=nh,q +1 p=q+1 k∈Nh,p

Moreover, by A.1,

h  hi 2ih 2 h 1 1 h k ≤ t dt = log 2 ih − log ih = h log 2. k=ih+1 ih

h Hence 2ih sk(x) q k=(2 −1)ih+1 h−q β q x ≥ ≥ . 2 −1( ) h h 2ih log 2 1 k k=ih+1 If h:=qr, then r− β q x ≥ 1 . 2 −1( ) r log 2

Letting q →∞yields β x ≥ r−1 . ( ) r log 2 Consequently, r →∞gives β x ≥ 1 . ( ) log 2 (6)

(E) To estimate x|m1,∞ from above, we use the quantity n λ x 1 s x n( ):= 1+log n k( ) k=1 instead of n n 1 1 μn(x):= sk(x) Ln := . Ln with k k=1 k=1

By A.1,

x|m1,∞log := sup λn(x) 1≤n<∞

x|m ≤x|m ≤ 3 x|m defines an equivalent norm: 1,∞ log 1,∞ 2 1,∞ log.

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(F) For the special indexes nh,weget h  ξ k 1 m k∈N h(h +1) λ x =1 m 2 . nh ( )= = 1 + log nh 1 + log nh

Thus, given ε with 0<ε≤1, we can successively find i1,i2,... such that

hi ≤ i λ x ≤ 1 ε. 2 h h+1 and nh ( ) log 2

λ x ≤ 1 ε n In the rest of this section, it will be shown that n( ) log 2 for all ’s.

(G) As an intermediate step, we treat the indexes nh,p with p=1,...,h−1 and h≥2. In this case,

h−1  p  ξk + ξk 1 m=1 k∈Nm q=1 k∈Nh,q (h − 1)h + p λ x 2 . nh,p ( )= = 1 + log nh,p 1 + log nh,p Let ah + log[t + bh] Fh(t):= , 1 + log t where a 1 h− h − i b i − n . h := 2 ( 1) log 2 log h and h := h h−1 p We deduce from (1) that nh,p +bh =2 ih. Therefore   a n b 1 h− h p , h + log[ h,p + h]= 2 ( 1) + log 2 which in turn yields λ x 1 F n . nh,p ( )= log 2 h( h,p) In particular,

1 1 λn x Fh nh− λn x Fh nh . h−1 ( )= log 2 ( 1) and h ( )= log 2 ( )

We infer from (4) that bh = ih −nh−1 ≥ i1 ≥ 1. i F n ≤ ε ≤ t n According to the choice of the h’s, we have h( h−1) 1. Thus, with 0 = h−1, all conditions posed in A.2 are satisfied, and by A.4 we get

λ x 1 F n ≤ 1 {F n ,F n }≤ 1 ε. nh,p ( )= log 2 h( h,p) log 2 max h( h−1) h( h) log 2

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(H) What remains is to consider those λn(x)’s whose index n differs from the nh,p’s. First of all, we observe that 1 n 1 1 λn(x)= < = λn (x) ≤ ε for 1≤n

From now on, let nh,p−1

1 N h(h+1)N Remarks. Let x( ) be the sequence associated with ih := 2 2 , where N is any h h natural number. Then, it follows from (2 −1)ih

The following diagram indicates the behavior of λn(x(2)): λ x(2) → 1 . ... 6 n( ) log 4 =07213 0.6 n s 4 n q s3 r q qn 0.5 n q q r 4,3 2 r q q n s r q , n q c q 4 2 q 3,2 n n q , s1 q r 4 1 q c q n 0.4 r 3,1 q q n q q 2,1 c s q c local maxima local minima logarithmic scale n 0.25 s - 1 4   196 28868 15757508 (2) The sequence λn(x ) has one local minimum on every interval Nh,1. Curiously enough, beginning with h=6, there is also one local minimum on every interval Nh,2. However,  this phenomenon appears only for N =2. In other words, if N > 2,then N λn(x( )) has no other local maxima than those attained at the nh’s.

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7. Appendix: Some elementary facts from calculus

A.1 We have  b b b−1 1 1 1 k < t dt < k . a+1 a a n 1 The asymptotic behavior of the quantity Ln := k is described by k=1 n 2 n < 1 n ≤ 1 ≤ n. 3 (1 + log ) n + log k 1 + log k=1 A.2 Let a + log[t + b] F (t):= for t ≥ 1. 1 + log t

If b>0, t0 ≥ 1, and F (t0) ≤ 1, then either F (t) increases on [t0, ∞) or there exists t >t F t t ,t t , ∞ some point min 0 such that ( ) decreases on [ 0 min] and increases on [ min ). Proof. If t≥1, then the sign of t(1 + log t) − (t + b)(a + log[t + b]) F (t)= t(t + b)(1 + log t)2 is determined by that of the nominator N(t):=t(1 + log t) − (t + b)(a + log[t + b]). Regarding N(t) as a function on (0, ∞),wehave N t − a t − t b − a − b N t 1 − 1 > . ( )=1 + log log[ + ]=1 log[1 + t ] and ( )= t t+b 0

The assumption F (t0)≤1 implies that a+log[t0 + b]≤1+log t0. Hence a<1. Let s>0 denote the unique zero of the increasing function N (t). Then N(t) is increasing on

[s, ∞). In view of F (t0)≤1, we obtain log[1 + b ] ≤ 1 − a = log[1 + b ]. t0 s

Thus s ≤ t0, and N(t) is increasing on [t0, ∞).IfN(t0) ≥ 0, it follows that F (t) is t , ∞ N t < t >t increasing on [ 0 ). In the case that ( 0) 0, there exists a point min 0 at N t − F t t ,t which ( ) changes from to + , which means that ( ) decreases on [ 0 min] and increases on [tmin, ∞).

A.3 Let c+t G(t):= for t ≥ 1. 1+log t  If c ≥ 0, then the function F (t) decreases on [1,tmin ] and increases on [ tmin, ∞ , where t ≥1 is the unique solution of the equation t log t=c. min  A.4 H t ,t t , ∞ If a function decreases on [ 0 min ] and increases on [ min , then H t ≤ {H t ,H t } t ≤ t

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