Unknown Book Proceedings Series Volume 00, 1997

Random Rearrangements and Operators

Stephen Montgomery-Smith and Evgueni Semenov

Dedicated to Professor Selim Krein on his 80th birthday

0. Introduction

Let n be a positive integer, x = (x_{ij})_{1≤i,j≤n}, and S_{n} the group of permutations of {1, 2, …, n}. Denote by x_{1}^{∗}, x_{2}^{∗}, …, x_{n^{2}}^{∗} the rearrangement of |x_{ij}| in decreasing order. S. Kwapien and C. Schütt proved the following results.

Theorem A [KS1]. We have that

\frac{1}{2n} ∑_{k=1}^{n} x_{k}^{∗} ≤ \frac{1}{n!} ∑_{π∈S_{n}} max_{1≤i≤n} |x_{i,π(i)}| ≤ \frac{1}{n} ∑_{k=1}^{n} x_{k}^{∗}.

Theorem B [Sc]. If 1 ≤ p ≤ q < ∞, then

\frac{1}{10} ((\frac{1}{n} ∑_{k=1}^{n}(x_{k}^{∗})^{p})^{1/p} + (\frac{1}{n} ∑_{k=n+1}^{n^{2}}(x_{k}^{∗})^{q})^{1/q})

≤ (\frac{1}{n!} ∑_{π∈S_{n}} (∑_{i=1}^{n} |x_{i,π(i)}|^{q})^{p/q})^{1/p}

≤ (\frac{1}{n} ∑_{k=1}^{n}(x_{k}^{∗})^{p})^{1/p} + (\frac{1}{n} ∑_{k=n+1}^{n^{2}}(x_{k}^{∗})^{q})^{1/q} .

There are two ways to generalize these results. They were presented in [S1],

[S2] and [M2]. This article is devoted to the development of these methods. Let 1 ≤ q ≤ ∞, and ℓ be a one-to-one correspondence of S_{n} to {1, 2, …, n!}.

We define the quasi-linear operator T_{q} as follows:

T_{q}x(t) = (∑_{i=1}^{n} |x_{i,π(i)}|^{q})^{1/q} , t ∈ [\frac{ℓ(π) − 1}{n!}, \frac{ℓ(π)}{n!}) ,

\begin{matrix}c \\ ◯ \end{matrix}1997 American Mathematical Society

1991 Mathematics Subject Classification. 46E30, 46B45, 46B70.

with the usual modification for q = ∞. The operator T_{q} acts from the set of n × n matrices into the step functions. Clearly it depends upon the choice of ℓ. However, if E is a rearrangement invariant space (see Section 1 for the definitions), then ∥T_{q}x∥_{E} does not depend upon ℓ. We define the operator

Ux(t) = ∑_{k=1}^{n} x_{k}^{∗}χ_{((k−1)/n,k/n)}(t)

on the set of n × n matrices x = (x_{ij}), where χ_{e} is the characteristic function of e ⊂ [0, 1].

(We would like to mention that there is another modification of this construction. If E is an r.i. space, we can construct a space Ẽ on the group S_{n} equipped with the Haar measure. In this case it is not necessary to introduce the function ℓ. However, some additional difficulties appear.)

The inequality

\frac{1}{12} (∥U_{x}∥_{E} + (\frac{1}{n} ∑_{k=n+1}^{n^{2}} x_{k}^{∗q})^{1/q}) ≤ ∥T_{q}x∥_{E}  (1)

was proved [S1] for any matrix x, rearrangement invariant space E, and 1 ≤ q < ∞. A more exact estimate is valid for q = ∞:

\frac{1}{2} ∥Ux∥_{E} ≤ ∥T_{∞}x∥_{E} ≤ ∥Ux∥_{E} .  (2)

The inverse inequality

∥T_{q}x∥_{E} ≤ C (∥Ux∥_{E} + (\frac{1}{n} ∑_{k=n+1}^{n^{2}} x_{k}^{∗q})^{1/q})  (3)

was established under the additional assumptions α_{E} > 0, where α_{E} is the lower Boyd index of an r.i. space E, and C does not depend upon x or n. It is evident that (3) fails for E = L_{∞} and 1 ≤ q < ∞.

In this article we generalize these results. In Chapter 4, we give a complete criterion for Lorentz spaces for which (3) holds. In Chapter 5, we consider other r.i. spaces for which (3) holds, and give some interpolation results. All this is based on Chapter 3, where it is shown that one can reduce (3) to the special case of diagonal matrices.

In Chapter 7, we consider a completely different generalization, where in effect E is L_{1}, but T_{q} is replaced by something analogous to T_{X}, where X is a symmetric sequence space. For the analogous result to (3), one needs the concept of a D^{∗}- convex space. For this reason, in Chapter 6, we develop the theory of D and D^{∗}-convex spaces, building on earlier work of Kalton [K]. In Chapter 8 we develop this idea further, and classify which Lorentz spaces are D or D^{∗}-convex.

Some results from this article were announced in [S2] and [M2].

1. Preliminaries

If x(t) is a measurable function on [0, 1], we denote by x^{∗}(t) the decreasing rearrangement of |x(t)|. A Banach space E on [0, 1] is said to be rearrangement invariant (r.i.) if y ∈ E and x^{∗} ≤ y^{∗} implies that x ∈ E and ∥x∥_{E} ≤ ∥y∥_{E}.

The embeddings L_{∞} ⊂ E ⊂ L_{1} are true for every r.i. space E. In fact, 1 ∈ E. Without loss of generality (except in Section 7 and parts of Section 6), we may assume that

∥1∥_{E} = 1.  (4)

We write x ≺ y if

∫_{0}^{τ} x^{∗}(t) dt ≤ ∫_{0}^{τ} y^{∗}(t) dt

for each τ ∈ [0, 1]. If E is separable or isometric to the conjugate of some separable r.i. space, that x ≺ y implies ∥x∥_{E} ≤ ∥y∥_{E}. Denote

E′ = {x : x ∈ L_{1}, ∫_{0}^{1} xy dt < ∞ ∀y ∈ E} ,

and equip it with the norm

∥x∥_{E′} = sup_{∥y∥_{E}≤1} ∫_{0}^{1} xy dt.

Then E′ is an r.i. space. The embedding E ⊂ E′′ is isometric. In fact ∥x∥_{E} = ∥x∥_{E′′} for all x ∈ L_{∞}. The function ∥χ_{(0,s)}∥_{E} is called the fundamental function of E.

Throughout this paper we will assume that all r.i. spaces are either maximal or minimal in the sense of Lindenstrauss and Tzafriri [LT] (or one may restrict to the case that the r.i. space is separable orisometric to the dual of a separable r.i. space). Thus for x ∈ E with x ≥ 0, we have that lim_{t→∞} ∥min{x, t}∥_{E} = ∥x∥_{E}.

Let φ(t) be an increasing function from [0, 1] to [0, 1], with φ(0) = 0 and φ(1) = 1, and continuous on (0, 1]. Let 1 ≤ r < ∞. The Lorentz space Λ_{r}(φ) consists of those functions on [0, 1] for which the functional

∥x∥_{Λ_{r}(φ)} = (∫_{0}^{1}(x^{∗}(t))^{r} d(φ(t))^{r})^{1/r}

is finite.

Let us set Φ_{r} to be the collection of those φ satisfying the above conditions, and also that φ(t)^{r} is concave. Then we see that if φ is in Φ_{r}, then Λ_{r}(φ) satisfies the triangle inequality, and hence is an r.i. space.

Also, for r > 1, it is also known that this is equivalent to a norm satisfying the triangle inequality if and only if there exist c > 0 and ϵ > 0 such that φ(ts) ≥ c^{−1}t^{1−ϵ}φ(s) for 0 ≤ s, t ≤ 1 (see [Sa]).

In the earlier sections, we shall primarily be interested in the case when r = 1, and so we will write Λ(φ) for Λ_{1}(φ), and Φ for Φ_{1}.

If φ ∈ Φ is continuous at 0, then (Λ(φ))^{∗} is equal to the Marcinkiewicz space, M(φ), where M(φ) is the space of functions on [0, 1] for which the functional

∥x∥_{M(φ)} = sup_{0<s≤1} \frac{∫_{0}^{s} x^{∗}(t) dt}{φ(s)}

is finite. This space is also an r.i. space. The spaces (Λ(φ))′ and M(φ) coincide for every φ ∈ Φ.

If φ ∈ Φ, then φ(t) and t/φ(t) increase on [0, 1]. A function having these properties is said to be quasi-concave. If ψ is quasi-concave, then there exists φ ∈ Φ such that \frac{1}{2}φ ≤ ψ ≤ φ. Indeed, φ may be chosen as the concave majorant of ψ.

Let M(t) be a concave even function on (−∞, ∞), M(0) = 0. Then we define another r.i. space, the Orlicz space L_{M}, to consist of all functions on [0, 1] for which the functional

∥x∥_{L_{M}} = inf \{λ : λ > 0, ∫_{0}^{1} M (\frac{|x(t)|}{λ}) dt ≤ 1\}

is finite.

We shall use Peetre’s K-method in this article. Therefore we present the main definitions. Let (E_{0}, E_{1}) be a compatible pair of Banach spaces. The K-functional is defined for each x ∈ E_{0} + E_{1} and t > 0 by

K(t, x) = K(t, x, E_{0}, E_{1}) = inf(∥x_{0}∥_{E_{0}} + t ∥x_{1}∥_{E_{1}}),

where the infimum is taken over all representations x = x_{0} + x_{1} with x_{0} ∈ E_{0} and x_{1} ∈ E_{1}. If 0 < θ < 1 and 1 ≤ p ≤ ∞, then the space (E_{0}, E_{1})_{θ,p} consists of all x ∈ E_{0} + E_{1} for which the functional

∥x∥_{θ,p} = (∫_{0}^{∞}(t^{−θ}K(t, x))^{p} \frac{dt}{t})^{1/p}

is finite. The space (E_{0}, E_{1})_{θ,p} is an interpolation space with respect to E_{0}, E_{1}. All the above mentioned properties of r.i. spaces and the K-method can be found in [BS], [BK], [KPS], [LT].

2. Hardy-Littlewood semiordering

The semiordering ≺ can be applied to vectors. The definition is completely analogous. In this section, we shall establish some preliminary statements about this semiordering.

If x, y ∈ ℝ^{n}, x, y ≥ 0,

X(t) = ∑_{k=1}^{n} x_{k}χ_{((k−1)/n,k/n)}, Y (t) = ∑_{k=1}^{n} y_{k}χ_{((k−1)/n,k/n)},

then the correlations x ≺ y and X ≺ Y are equivalent.

In general, if a, b, c ∈ ℝ^{n}, a, b, c ≥ 0, then it does not follow from a ≺ b that a + c ≺ b + c.

Given x ∈ ℝ^{n}, I ⊆ {1, 2, …, n}, denote

(xχ_{I})_{k} = \{ \begin{matrix}x_{k} & if k ∈ I \\ 0 & if k ∉ I, 1 ≤ k ≤ n. \end{matrix}

Lemma 1. Let {1, 2, …, n} = I_{1} ∪ I_{2} ∪ ⋯ ∪ I_{s}, where I_{1}, I_{2}, …, I_{s} are disjoint subsets. If xχ_{I_{k}} ≺ yχ_{I_{k}} for all k = 1, 2, …, s, then x ≺ y.

The proof is obvious. This statement remains true if we consider xχ_{I_{k}} and yχ_{I_{k}} as elements of ℝ^{|I_{k}|}.

Let u = (u_{i,j})_{1≤i,j,≤n}, u ≥ 0, u_{11} > 0, u_{i,2} = 0 for i = 1, 2, …, n. Put

v_{ij} = \{ \begin{matrix}0 & if i = j = 1 \\ u_{11} & if i = 1, j = 2 \\ u_{ij} & otherwise. \end{matrix}

Given an n × n matrix x, consider the vector

(T x) = (∑_{k=1}^{n} x_{k,π(k)}, π ∈ S_{n})

as an element of ℝ^{n!}.

Lemma 2. T u ≺ T v.

Proof. Denote

S_{0} = {π ∈ S_{n} : π(1) ≠ 1, π(2) ≠ 1}
S(1, j) = {π ∈ S_{n} : π(1) = 1, π(2) = j}
S(j, 1) = {π ∈ S_{n} : π(1) = j, π(2) = 1},

where j = 2, 3, …, n. It is evident that

S_{n} = S_{0} ∪ ⋃_{j=2}^{n}(S(1, j) ∪ S(j, 1))  (5)

is a disjoint decomposition of S_{n}. Clearly

(T u)χ_{S_{0}} = (T v)χ_{S_{0}}.  (6)

Put

(T̃x)(π) = x_{1,π(1)} + x_{2,π(2)}.

If μ ∈ S(1, j) for some j ∈ {2, 3, …, n}, then there exists a unique ν ∈ S(j, 1) such that μ(k) = ν(k) for k = 3, 4, …, n. Therefore,

\begin{aligned}u_{k,μ(k)} & = v_{k,ν(k)} \\ u_{k,ν(k)} & = v_{k,μ(k)} \end{aligned}  (7)

for every k = 3, …, n. For such μ, ν we have

((T̃u)(μ), (T̃u)(ν)) ≺ ((T̃v)(μ), (T̃v)(ν)).  (8)

From (6) and (7) we get

((T u)(μ), (T u)(ν)) ≺ ((T v)(μ), (T v)(ν)).

Given j ∈ {3, 4, …, n}, there exist μ_{i} ∈ S(1, j) and ν_{i} ∈ S(j, 1) such that

S(1, j) ∪ S(j, 1) = ⋃_{i=1}^{(n−2)!}{μ_{i}, ν_{i}}

and

((T u)(μ_{i}), (T u)(ν_{i})) ≺ ((T v)(μ_{i}), (T v)(ν_{i}))

for each i = 1, 2, …, (n − 2)!. By Lemma 1

(T u)χ_{S(1,j)} ≺ (T v)χ_{S(j,1)}  (9)

for every j = 2, 3, …, n. Taking into account (5), (6), and (9), and applying Lemma 1 again, we get that T u ≺ T v. □

Denote by Q_{n} the set of (n × n) matrices (x_{ij}) such that x_{ij} = 0 or 1 for every 1 ≤ i, j ≤ n and

|{(i, j) : x_{ij} = 1}| = n.

The identity matrix is denoted by I_{n}.

Lemma 3. If x ∈ Q_{n}, then T x ≺ T I_{n}.

Proof. Clearly T x and T I_{n} are equidistributed for every permutation matrix x. If x is not a permutation matrix, then there exists a pair of columns or rows such that the first one contains two or more 1’s, and the second one contains no 1’s. Without loss of generality we may assume that x_{11} = x_{21} = 1, and that x_{i,2} = 0 for each i = 1, 2, …, n. Put

y_{ij} = \{ \begin{matrix}0 & if i = j = 1 \\ 1 & if i = 1, j = 2 \\ x_{ij} & otherwise. \end{matrix}

By Lemma 2, T x ≺ T y. If y is a permutation matrix, then the lemma is proved. If y is not a permutation matrix, we can use this construction again. We obtain a permutation matrix after less than n iterations. □

Lemma 4. Let x be a (n × n) matrix such that

∑_{i,j=1}^{n} x_{ij} ≤ n, 0 ≤ x_{ij} ≤ 1.  (10)

Then

T x ≺ T I_{n}.  (11)

Proof. Denote by P_{n} the set of (n × n) matrices satisfying condition (10). It is evident that the set of extremal points of P_{n} coincides with Q_{n}. Given j = 1, 2, …, n, consider the functional

f_{j}(x) = max ∑_{π∈R} ∑_{k=1}^{n} x_{k,π(k)},

where the maximum is taken over all subsets R ⊂ S_{n} with |R| = j. The functional f_{j} is convex. Therefore

max_{x∈P_{n}} f_{j}(x) = max_{x∈ext P_{n}} f_{j}(x).

If x ∈ P_{n}, we can find y ∈ Q_{n} such that f_{j}(x) ≤ f_{j}(y). By Lemma 3, f_{j}(y) ≤ f_{j}(I_{n}), and consequently, f_{j}(x) ≤ f_{j}(I_{n}). Thus (11) is proved. □

Let x, f ∈ L_{1}, and x, f ≥ 0. It is well known ([KPS], II.2.2) that x ≺ f implies x^{α} ≺ f^{α} for α ≥ 1. This is not true in general if 0 < α < 1. However, under some additional assumptions on f the implication x ≺ f ⇒ x^{α} ≺ C f^{α} is true. Denote by O_{f} the set of x ≥ 0 such that x ≺ f.

Lemma 5. Let f ≥ 0, f ∈ L_{1}, 0 < α < 1, C > 1. Then
x^{α} ≺ C f^{α} ∀x ∈ O_{f}  (12)

if and only if

τ^{1−α} (∫_{0}^{τ} f(t) dt)^{α} ≤ C ∫_{0}^{τ} f(t)^{α} dt ∀τ ∈ [0, 1].  (13)

Proof. We may assume that f = f^{∗}. Given τ ∈ (0, 1], consider the function

x_{τ} (t) = χ_{(0,τ)}(t)\frac{1}{τ} ∫_{0}^{τ} f(s) ds.

Since x_{τ} ∈ O_{f}, then (12) implies

∫_{0}^{τ} x_{τ} (t)^{α} dt = (\frac{1}{τ} ∫_{0}^{τ} f(s) ds)^{α} τ ≤ C ∫_{0}^{τ} f(t)^{α} dt.

It is equivalent to (13).

Let inequality (13) be valid. By Hölder’s inequality

∫_{0}^{τ} x^{∗}(t)^{α} dt ≤ (∫_{0}^{τ} x^{∗}(t) dt)^{α} τ^{1−α} ≤ (∫_{0}^{τ} f(t) dt)^{α} τ^{1−α} ≤ C ∫_{0}^{τ} f(t)^{α} dt.

□

Lemma 6. The function f(t) = T_{1}I_{n}(t) satisfies (12) for α > 0, where C = 6.

Proof. There exists a sequence 1 > τ_{1} ≥ τ_{2} ≥ ⋯ ≥ τ_{n+1} = 0 such that

f^{∗}(t) = ∑_{j=1}^{n} jχ_{[τ_{j+1},τ_{j}]}(t).

The function f is closely connected with the classical coincidence problem. It is well known (see [W], 4.9, 10) that

s_{j} := τ_{j} − τ_{j+1} = \frac{1}{j!} ∑_{k=0}^{n−j} \frac{(−1)^{k}}{k!} .

Denote

q_{j} = ∑_{k=0}^{n−j} \frac{(−1)^{k}}{k!} .

Then q_{n−1} = 0, q_{n} = 1 and \frac{1}{3} ≤ q_{j} ≤ \frac{1}{2} for j = 1, 2, …, n − 2. Hence

τ_{j} = ∑_{i=j}^{n} s_{i} ≤ 3s_{j}  (14)

8 STEPHEN MONTGOMERY-SMITH AND EVGUENI SEMENOV

for j ≠ n − 1. Since

js_{j} = \frac{1}{(j − 1)!} ∑_{k=0}^{n−j} \frac{(−1)^{k}}{k!} ,

we get

∑_{i=j}^{n} is_{i} ≤ 3js_{j}  (15)

for j ≠ n − 1. Using (14) and (15), we have

τ^{1−α} (∑_{i=j}^{n} is_{i})^{α} ≤ (3s_{j})^{1−α}(3js_{j})^{α} = 3j^{α}s_{j} < 3 ∑_{i=j}^{n} i^{α}s_{i}

for 1 ≤ j ≤ n. Consequently, the inequality

τ^{1−α} (∫_{0}^{τ} f^{∗}(t) dt)^{α} ≤ 3 ∫_{0}^{τ} f^{∗}(t)^{α} dt

is proved for τ = τ_{j}, 1 ≤ j ≤ n.

If τ ∈ [0, τ_{1}], we can find 1 ≤ j ≤ n and λ ∈ [0, 1] such that

τ = \{ \begin{matrix}τ_{j+1} + λs_{j} & if τ > τ_{n−2} or τ ≤ τ_{n} \\ τ_{n} + λs_{n−2} & if τ_{n} < τ ≤ τ_{n−2}. \end{matrix}

It is sufficient to consider only the case τ > τ_{n−2}. By (14) and (15),

( ∑_{i=j+1}^{n} s_{i} + λs_{j})^{1−α} ( ∑_{i=j+1}^{k} is_{i} + λjs_{j})^{α} ≤ (3s_{j+1} + λs_{j})^{1−α}(3(j + 1)s_{j+1} + λjs_{j})^{α}

≤ 3(j + 1)^{α}(s_{j+1} + λs_{j})

≤ 6j^{α}(s_{j+1} + λs_{j})

≤ 6 ( ∑_{i=j+1}^{n} i^{α}s_{i} + λj^{α}s_{j}) .

The obtained inequality shows that

τ^{1−α} (∫_{0}^{τ} f^{∗}(t) dt)^{α} ≤ 6 ∫_{0}^{τ} f^{∗}(t)^{α} dt.

□

3. Reduction to diagonal matrices Given an integer n, denote by D_{n} the set of diagonal matrices. It is evident

that if x ∈ D_{n}, then x_{k}^{∗} = 0 for n < k ≤ n^{2}.

Theorem 7. Let C, q > 1, and let E be an r.i. space. If ∥T_{q}y∥_{E} ≤ C ∥Uy∥_{E} for any y ∈ D_{n}, then

∥T_{q}x∥_{E} ≤ 7C (∥Ux∥_{E} + (\frac{1}{n} ∑_{k=n+1}^{n^{2}} x_{k}^{∗q})^{1/q})

for any (n × n) matrix x.

Proof. Let

∥Ux∥_{E} ≤ 1  (16)
\frac{1}{n} ∑_{k=n+1}^{n^{2}} x_{k}^{∗q} ≤ 1.  (17)

We can find matrices y = (y_{ij}) and z = (z_{ij}) such that x = y + z, |supp y| ≤ n, and Ux = Uy. Denote by w a diagonal matrix such that Uw = Uy. By Lemma 2,

T_{1}y ≺ T_{1}w.

Hence,

∥T_{1}y∥_{E} ≤ ∥T_{1}w∥_{E} ≤ C ∥Uw∥_{E} = C ∥Ux∥_{E}

and

∥T_{q}y∥_{E} ≤ ∥T_{1}y∥_{E} ≤ C ∥Ux∥_{E} ≤ C.  (18)

If x_{n+1}^{∗} > 1, then

min_{0≤t≤1} Ux(t) = x_{n}^{∗} ≥ x_{n+1}^{∗} > 1.

By (4),

∥Ux∥_{E} > ∥1∥_{E} = 1.

The obtained inequality contradicts (16). Consequently, for every 1 ≤ i, j ≤ n

|z_{ij}| ≤ x_{n+1}^{∗} ≤ 1.

Denoting |x_{ij}|^{q} by v_{ij}, we get

0 ≤ v_{ij} ≤ 1,  1 ≤ i, j, ≤ n
∑_{i,j=1}^{n} v_{ij} ≤ n
∥T_{q}z∥_{E} = ∥(T_{1}v)^{1/q}∥_{E} .

By Lemma 4,

T_{1}v ≺ T_{1}I_{n}.

Applying Lemmas 5 and 6, we have

(T_{1}v)^{1/q} ≺ 6(T_{1}I_{n})^{1/q}.

We have mentioned in Section 1 that this inequality implies

∥(T_{1}v)^{1/q}∥_{E} ≤ 6 ∥(T_{1}I_{n})^{1/q}∥_{E} .

Since

∥(T_{1}I_{n})^{1/q}∥_{E} = ∥T_{q}I_{n}∥_{E} ≤ C ∥UI_{n}∥_{E} = C,

then

∥T_{q}z∥_{E} ≤ 6C.

Using the obtained inequality and (18), we get

∥T_{q}x∥_{E} ≤ ∥T_{q}y∥_{E} + ∥T_{q}z∥_{E} ≤ 7C.
□

Theorem 7 and (1) lead to the following.

Corollary 8. Let E be an r.i. space, and 1 ≤ q < ∞. The equivalence

∥T_{q}x∥_{E} ≈ ∥Ux∥_{E} + (\frac{1}{n} ∑_{k=n+1}^{n^{2}} x_{k}^{∗q})^{1/q} ,  (19)

where the equivalence constants depend neither upon the matrix x nor on n, takes place if and only if the estimate

∥T_{q}y∥_{E} ≤ C ∥Uy∥_{E}

is valid for every diagonal matrix y.

Denote by ∥T_{q}∥_{E} the least C in the last inequality, and by F_{q} the set of r.i. spaces satisfying condition (19). Given an r.i. space E, denote by ω(E) the set of q ∈ [1, ∞] such that equivalence (19) takes place, and put

τ(E) = inf ω(E).

The monotonicity of the function q ↦ ∥T_{q}x∥_{E}, (21), and Corollary 8 imply that w(E) = [τ(E), ∞] or ω(E) = (τ(E), ∞]. Some examples show that both of these possibilities may be realized.

4. Lorentz spaces

Given 0 < j ≤ k < n, denote

I_{n,k} = diag(\underbrace{1, 1, …, 1}_{k}, \underbrace{0, 0, …, 0}_{n−k})

and put

μ_{n,k,j} = mes{t ∈ [0, 1] : T_{1}I_{n,k}(t) = j}.

Lemma 9. If 0 < s < 1, then

\frac{s^{j}}{ej!} ≤ sup μ_{n,k,j} ≤ \frac{s^{j}}{j!} ,  (20)

where the supremum is taken over (n, k) such that k ≤ ns.

Proof. We have

\begin{aligned}μ_{n,k,j} & ≤ \frac{C_{k}^{j}(n − j)!}{n!} = \frac{k!(n − j)!}{j!(k − j)!n!} \\ & = \frac{k(k − 1) … (k − j + 1)}{j!n(n − 1) … (n − j + 1)} ≤ \frac{1}{j!} (\frac{k}{n})^{j} ≤ \frac{s^{j}}{j!} . \end{aligned}  (21)

Following ([W], 4.9.B), denote B_{n,k,j} = n!μ_{n,k,j}. It is known that

B_{n,k,j} = \frac{j + 1}{k + 1}B_{n+1,k+1,j+1}.

Therefore,

μ_{n,k,j} = \frac{1}{n!}B_{n,k,j} = \frac{k}{n!j} B_{n−1,k−1,j−1} = \frac{k}{nj} μ_{n−1,k−1,j−1}.

Hence

μ_{n,k,j} = \frac{k(k − 1) … (k − j + 1)}{j!n(n − 1) … (n − j + 1)}μ_{n−j,k−j,0}

= \frac{k(k − 1) … (k − j + 1)}{j!n(n − 1) … (n − j + 1)} (1 − \frac{1}{n − j} )^{k−j} .

The last part tends to s^{j}e^{−s}/j! if k = [ns] and n tends to infinity. This proves the left part of (20). □

The above proved statement allows us to solve completely the problem on the validity of equivalence (19) in the class of Lorentz spaces. Recall that we denote by Φ the set of increasing concave functions on [0, 1] with φ(0) = 0 and φ(1) = 1.

Theorem 10. Let φ ∈ Φ, 1 ≤ q < ∞. The equivalence

∥T_{q}x∥_{Λ(φ)} ≈ ∥Ux∥_{Λ(φ)} + (\frac{1}{n} ∑_{k=n+1}^{n^{2}} x_{k}^{q})^{1/q}  (22)

takes place if and only if

Γ_{φ,q} := sup_{0<t≤1} \frac{1}{φ(t)} ∑_{j=1}^{∞} j^{(1/q)−1}φ(t^{j}/j!) < ∞.  (23)

Moreover,

∥T_{q}∥_{Λ(φ)} ≤ Γ_{φ,q} ≤ qe∥T_{q}∥_{Λ(φ)}.

Proof. We use the notation of Lemma 9. If (22) is fulfilled, then there exists a constant C > 0 such that

∥T_{q}I_{n,k}∥_{Λ(φ)} ≤ C∥UI_{n,k}∥_{Λ(φ)} = Cφ(k/n).

Since

∥T_{q}I_{n,k}∥_{Λ(φ)} = ∥∑_{j=1}^{k}(j^{1/q} − (j − 1)^{1/q})χ_{(0,μ_{n,k,j})}∥_{Λ(φ)}

= ∑_{j=1}^{k}(j^{1/q} − (j − 1)^{1/q})φ(μ_{n,k,j}), (24)

then it follows by Lemma 9 that for each t ∈ (0, 1) and for each integer m that we have

∑_{j=1}^{m}(j^{1/q} − (j − 1)^{1/q})φ(t^{j}/ej!) ≤ Cφ(t).

Hence,

∑_{j=1}^{∞} j^{(1/q)−1}φ(t^{j}/j!) ≤ Cqeφ(t).

This proves the first part of the theorem.

Now suppose that, for every t ∈ (0, 1],

∑_{j=1}^{∞} t^{(1/q)−1}φ(t^{j}/j!) ≤ Cφ(t).

Applying the obvious inequality

j^{1/q} − (j − 1)^{1/q} ≤ j^{(1/q)−1}

and (24), we get

∥T_{q}I_{n,k}∥_{Λ(φ)} = ∑_{j=1}^{k}(j^{1/q} − (j − 1)^{1/q}φ(μ_{n,k,j})

≤ ∑_{j=1}^{k} j^{(1/q)−1}φ(1/j!(k/n)^{j})

≤ Cφ(k/n)

= C∥UI_{n,k}∥_{Λ(φ)}.  (25)

All Lorentz spaces Λ(φ) have the following property ([KPS], II.5.2). If a convex functional is uniformly bounded on the set of characteristic functions, then it is uniformly bounded on the set of step functions. We apply this property to matrices. Then (25) implies that

∥T_{q}y∥_{Λ(φ)} ≤ C∥Uy∥_{Λ(φ)}

for each y ∈ D. By Corollary 8, (22) is valid. □

In other words, Λ(φ) ∈ F_{q} if and only if Γ_{φ,q} < ∞. We mention that

∥Ux∥_{Λ(φ)} = ∑_{k=1}^{n} x_{k}(φ(k/n) − φ((k − 1)/n)).

Let us study condition (23) in detail.

Lemma 11. Let 0 < α ≤ 1, a, q ≥ 1, and φ ∈ Φ with φ(t) ≤ at^{α} for every t ∈ [0, 1]. Then Λ(φ) ∈ F_{q} and Γ_{φ,q} ≤ 5a/α.

Proof. Since Γ_{φ,q} ≤ Γ_{φ,1}, we shall estimate only Γ_{φ,1}. Given s ∈ (0, a^{−1/α}), we construct the function

φ_{s}(t) = \{ \begin{matrix}at^{α} & if 0 ≤ t ≤ s, \\ as^{α} & if s ≤ t ≤ as^{α}, \\ t & if as^{α} ≤ t ≤ 1. \end{matrix}

The set of the quasi-concave functions φ_{s} possesses the following property. If φ ∈ Φ and t_{1} ∈ (0, 1), then we can find s ∈ (0, a^{−1/α}) such that φ_{s}(t) ≤ φ(t) for t ∈ [t_{1}, 1], and φ_{s}(t) ≥ φ(t) for t ∈ [0, t_{1}]. Therefore it is sufficient to obtain the needed estimate only for the function φ_{s} and t = as^{α}. Put

N = max \{j : \frac{(as^{α})^{j}}{j!} ≥ s\} = max\{j : (j!)^{1/j} ≤ as^{α−1/j}\}.

Then

∑_{j=1}^{∞} φ_{s} (\frac{(as^{α})^{j}}{j!} ) = Nas^{α} + a ∑_{j=N+1}^{∞} (\frac{(as^{α})^{j}}{j!} )^{α} ≤ as^{α} (N + a ∑_{j=1}^{∞}(j!)^{−α}) .

Since (j/3)^{j} ≤ j!, it follows that

N ≤ max{j : j ≤ 3as^{α−1/j}} ≤ max(3a, 1/α).

Hence,

Γ_{φ,1} ≤ max(3a, 1/α) + a ∑_{j=1}^{∞}(j!)^{−α} ≤ 3a/α + a ∑_{j=1}^{∞} 2^{−jα} ≤ 5a/α.

□

The assumption φ(1) = 1 is essential in Lemma 11. Indeed, if we let ψ_{ϵ}(t) = min(\sqrt{t}, ϵ), then Γ_{ψ_{ϵ},1} tends to ∞ when ϵ tends to 0.

5. Interpolation Spaces

Theorem 10 may be extended on a wider class of r.i. spaces. Given numbers α ∈ (0, 1], a ≥ 1, denote by Φ(a, α) the set of functions φ ∈ Φ such that φ(t) ≤ at^{α} for each t ∈ [0, 1]. Let E be an r.i. space. Given y ∈ E′, ∥y∥_{E′} = 1, we put

ψ_{y}(t) = \frac{∫_{0}^{t} y^{∗}(s)ds + t}{∥y∥_{L_{1}} + 1}

and

∥x∥_{1} = sup_{∥y∥_{E′} =1} ∥x∥_{Λ(ψ_{y})}.

We mention that ψ_{y} ∈ Φ.

Lemma 12.

i) The norms ∥ · ∥_{E′′} and ∥ · ∥_{1} are equivalent and

\frac{1}{2}∥x∥_{E} ≤ ∥x∥_{1} ≤ 2∥x∥_{E}  for all x ∈ L_{∞}.  (26)

ii) If L_{p} ⊂ E for some p < ∞ and

∥x∥_{E} ≤ a∥x∥_{L_{p}}  for all x ∈ L_{p},  (27)
then ψ_{y} ∈ Φ(a + 1, 1/p) for every y ∈ E′ with ∥y∥_{E′} = 1.

Proof. i) By the Hardy-Littlewood theorem on rearrangements (see [KPS], II.2.2.17) it follows that for x ∈ E′′ we have

∥x∥_{E′′} = sup_{∥y∥_{E′} =1} ∫_{0}^{1} x(t)y(t)dt = sup_{∥y∥_{E′} =1} ∫_{0}^{1} x^{∗}(t)y^{∗}(t)dt = sup_{∥y∥_{E′} =1} ∥x∥_{Λ(φ_{y})},

where φ_{y}(t) = ∫_{0}^{t} y^{∗}(s)ds. Since

∥y∥_{L_{1}} + 1 ≤ ∥y∥_{E′} + 1 = 2,

then ψ_{y} ≥ \frac{1}{2}φ_{y} and

∥x∥_{1} ≥ \frac{1}{2}∥x∥_{E′′}  for all x ∈ E′′.  (28)

On the other hand,

∥x∥_{1} ≤ ∥x∥_{E′′} + ∥x∥_{L_{1}} ≤ 2∥x∥_{E′′}  for all x ∈ E′′.  (29)

The norms ∥ · ∥_{E} and ∥x∥_{E′′} coincide on L_{∞} ([BS], 1.2.7). Therefore (28) and (29) imply (26).

ii) If \frac{1}{p} + \frac{1}{p′} = 1, then E′ ⊂ L_{p′} and

∥y∥_{L_{p′}} ≤ a∥y∥_{E′} = a.

By Hölder’s inequality,

ψ_{y}(t) ≤ ∫_{0}^{t} y^{∗}(s)ds + t ≤ ∥y∥_{E′}t^{1/p} + t ≤ at^{1/p} + t ≤ (a + 1)t^{1/p}.
□

Theorem 13. Let E be an r.i. space, E ⊃ L_{p} for some p < ∞, and 1 ≤ q < ∞. Then (19) is fulfilled, that is, E ∈ F_{q}.

Proof. There is a constant a > 1 such that (27) is valid. By Lemma 12 (ii), ψ_{y} ∈ Φ(a + 1, 1/p) for every y ∈ E′ with ∥y∥_{E′} = 1. By Lemma 11, we know that

Γ_{ψ_{y},q} ≤ 5(a + 1)p.

Applying the second part of Theorem 10, we get that

∥T_{q}x∥_{Λ(ψ_{y})} ≤ 5(a + 1)p∥Ux∥_{Λ(ψ_{y})}

for each x ∈ D, and y ∈ E′ with ∥y∥_{E′} = 1. Hence,

∥T_{q}x∥_{1} ≤ 5(a + 1)p∥Ux∥_{1}.

This and (26) imply that

∥T_{q}x∥_{E} ≤ 20(a + 1)p∥Ux∥_{E}.

By Corollary 8, it follows that (19) is fulfilled. □

We mention that the conditions

1) E ⊃ L_{p} for some p < ∞;

2) t^{−α} ∈ E for some α > 0 are equivalent.

Let E_{1}, E_{2} be r.i. spaces, 1 ≤ q < ∞, E_{1} ⊃ E_{2} and E_{2} ∈ F_{q}. Does it follow that E_{1} ∈ F_{q}? Theorem 13 shows that the answer to this question is positive if E_{2} ⊃ L_{p} for some p < ∞. In general, the answer is negative. We now show this.

Theorem 14. Let φ ∈ Φ and 1 ≤ q < ∞. The following conditions are equivalent:

i) if ψ ∈ Φ and ψ ≤ φ, then Γ_{ψ,q} < ∞; ii) there exists a constant C > 0 such that if ψ ∈ Φ and ψ ≤ φ, then Γ_{ψ,q} ≤ C. iii) there exist numbers α ∈ (0, 1] and a ≥ 1 such that φ(t) ≤ at^{α} for every

t ∈ [0, 1].

Proof. The implication (iii) ⇒ (ii) was proved in Lemma 11. The implication (ii) ⇒ (i) is trivial. Therefore, we need prove only the implication (i) ⇒ (iii). Let

φ ∉ ⋃_{0<α≤1≤a} Φ(a, α),

so that

sup_{0<t≤1} φ(t)t^{−1/n} = ∞  n = 1, 2, …  (30)

Using (30) we can find a sequence t_{n} ↓ 0 such that t_{1} = 1 and

φ(t_{n}) ≥ \frac{nφ(t_{n−1})}{t_{n−1}}t_{n}^{1/n}

for every n = 2, 3 …. Then

(\frac{t_{n−1}}{φ(t_{n−1})}φ(t_{n}))^{n} ≥ n^{n}t_{n} > n!t_{n}.  (31)

If we put

s_{n} = \frac{t_{n−1}}{φ(t_{n−1})}φ(t_{n}),

then t_{n} < s_{n} < t_{n−1} for every n = 2, 3, …. Define

ψ(t) = \{ \begin{matrix}φ(t_{n}) & if t_{n} ≤ t ≤ s_{n}, \\ tφ(t_{n−1})/t_{n−1} & if s_{n} ≤ t ≤ t_{n−1}, n = 2, 3, …, \\ 0 & if t = 0. \end{matrix}

The function ψ(t) is quasi-concave on [0, 1], and ψ ≤ φ. By (31) we see that

ψ(s_{n}^{n}/n!) = ψ(t_{n}) = φ(t_{n}) = ψ(s_{n}).

For every integer n we get

Γ_{ψ,q} ≥ \frac{1}{ψ(s_{n})} ∑_{k=1}^{n} k^{(1/q)−1}ψ(s_{n}^{k}/k!) = ∑_{k=1}^{n} k^{(1/q)−1} > ∑_{k=1}^{n} k^{−1}.

Hence, Γ_{ψ,q} = ∞. Denote by ν(t) the concave majorant of ψ(t). Then ν ∈ Φ, ν ≤ φ and Γ_{ν,q} = ∞. □

Theorem 13 is practically an interpolation theorem. It shows that if E is an r.i. space, E = E′′ and E ⊃ L_{p} for some p < ∞, then E is an interpolation space with respect to the set {Λ(φ), Γ_{φ,1} < ∞}. One can prove that the assumption E = E′′ may be replaced with the separability of E. Using the K-method, we can obtain another sufficient condition for E ∈ F_{q}.

Theorem 15. Let φ_{0}, φ_{1} ∈ Φ and 0 < γ, θ < 1, and suppose that the function φ_{0}^{γ}(t)/φ_{1}(t) increases on (0, 1]. Then

(Λ(φ_{0}), Λ(φ_{1}))_{θ,∞} ≈ M(φ̃_{θ}),  (32)

where

φ̃_{θ}(t) = \frac{t}{φ_{θ}(t)} = \frac{t}{φ_{0}^{1−θ}(t)φ_{1}^{θ}(t)}.

Theorem 15 is a special case of a more general result which is contained in [S3]. Using some results on the stability of the interpolation functions [A], one can obtain a similar statement.

Lemma 16. Let E_{0}, E_{1} ∈ F_{q} be r.i. spaces, where q ∈ [1, ∞), and suppose that E_{2} is an interpolation space with respect to E_{0}, E_{1}. Then E_{2} ∈ F_{q}.

Proof. Given an integer n, we consider the operator

B_{n}x = diag (n ∫_{(k−1)/n}^{k/n} x(s) ds, 1 ≤ k ≤ n) .

The operator B_{n} acts from L_{1} into the set of diagonal matrices. The operator UB_{n} is an averaging operator, and UB_{n}x is the conditional expectation of x with respect to the set of intervals \{(\frac{k−1}{n}, \frac{k}{n}), 1 ≤ k ≤ n\}. By Theorem 2.a.4 [LT], it follows that

∥UB_{n}∥_{E} = 1.

Corollary 8 shows that E ∈ F_{q} if and only if

sup_{n} ∥T_{q}B_{n}∥_{E} < ∞.

This proves the Lemma. □

Lemma 17. Suppose that 1 ≤ q < ∞, that φ ∈ Φ satisfies condition (23), that 1 < μ < λ, and that φ^{λ} ∈ Φ. Then M(φ̃^{μ}) ∈ F_{q}.

Proof. Applying (23) and Jensen’s inequality, we get that

∑_{j=1}^{∞} j^{(1/q)−1}φ^{λ}(t^{j}/j!) ≤ (∑_{j=1}^{∞} j^{(1/q)−1}φ(t^{j}/j!))^{λ} ≤ C^{λ}φ^{λ}(t)

for every t ∈ (0, 1]. It means that φ^{λ} satisfies condition (23) with the constant C^{λ}. By Theorem 15, M(φ̃^{μ}) is an interpolation space with respect to Λ(φ) and Λ(φ^{λ}). The required statements now follow from Lemma 16. □

Consider the following example. Given p > 0, we put

φ_{p}(t) = (\frac{log(1 + 1/t)}{log 2})^{−1/p} .  (33)

If p ≥ 1, then φ_{p} ∈ Φ. If p ∈ (0, 1), then φ_{p}(t) is concave in a sufficiently small neighborhood of the origin. Consequently, φ_{p} is concave up to equivalence.

Lemma 18. Let p > 0 and 1 ≤ q < ∞. Then Γ_{φ_{p},q} < ∞ if p < q. If φ ∈ Φ and φ ≥ Cφ_{q} for some C > 0, then Γ_{φ,q} = ∞.

Proof. Let p < q and 0 < t ≤ 1. We have that

\frac{1}{φ_{p}(t)} ∑_{j=1}^{∞} j^{(1/q)−1}φ_{p}(t^{j}/j!) ≤ 1 + ∑_{j=2}^{∞} j^{(1/q)−1} (\frac{log(1 + 1/t)}{log(1 + j!/t^{j})})^{1/p}

≤ 1 + ∑_{j=2}^{∞} j^{(1/q)−1} (\frac{1}{j − 1})^{1/p} < ∞.

Therefore Γ_{φ_{p},q} < ∞. If φ ≥ Cφ_{q} and 0 < t ≤ 1, then

∑_{j=2}^{∞} j^{(1/q)−1}φ(t^{j}/j!) ≥ C(log 2)^{−1/q} ∑_{j=2}^{∞} j^{(1/q)−1} log^{−1/q}(1 + j!/t^{j})

≥ C ∑_{j=2}^{∞} j^{(1/q)−1}(j log(j/t))^{−1/q}

= C ∑_{j=2}^{∞} j^{−1}(log(j/t))^{−1/q} = ∞.

□

Let us consider the Orlicz space exp L_{p}. It is generated by the function

M_{p}(u) = e^{|u|^{p}} − 1.

If p ≥ 1, then M_{p}(u) is convex and the fundamental function of exp L_{p} is equal to (log(1 + 1/t))^{−1/p}. If 0 < p < 1, then M_{p}(u) is convex for sufficiently large u.

Theorem 19. Let 1 ≤ q < ∞. If p < q, then exp L_{p} ∈ F_{q}. If p > q, then exp L_{p} ∉ F_{q}.

Proof. Let p ≥ 1. By [Lo], the spaces exp L_{p} and M(φ̃_{p}) coincide, where φ_{p} is defined by (33). Applying Lemma 18, we get that exp L_{p} ∈ F_{q} if p < q. The first part of the theorem is proved.

If E is an r.i. space and E ∈ F_{q}, then

sup_{n} ∥T_{q}I_{n}∥_{E} < ∞.

In fact, let E be an Orlicz space L_{M}. Lemma 9 shows that

∑_{j=1}^{∞} \frac{e^{ϵ^{p}j^{p/q}} − 1}{j!} < ∞

for some ϵ > 0. This series diverges for any p > q and ϵ > 0. Hence, p ≤ q.

So, the theorem has been proved for p ≥ 1. If 0 < p < 1, we can change M_{p}(u) for a convex equivalent function. Therefore the theorem is valid for every p > 0. □

6. D and D^{∗}-convex Spaces

The notion of D-convexity was introduced by Kalton [K] (Section 5). Indeed, much of the proof of this section is inspired by his proof of Lemma 5.5.

Given a function x on [0, 1], we will define its distribution function d_{x}(t) = mes({|f| > t}). Thus the decreasing rearrangement x^{∗}(t) is essentially the inverse function of d_{x}. Given functions x_{1}, x_{2}, …, x_{n} on [0, 1], we define their dilated disjoint sum to be the function on [0, 1]:

C(x_{1}, …, x_{n})(t) = |x_{k}(nt − k + 1)| ((k − 1)/n < t < k/n).

Thus

d_{C(x_{1},...,x_{n})}(t) = \frac{1}{n} ∑_{k=1}^{n} d_{x_{k}}(t).

We will say that an r.i. space E is D-convex if there is a constant c > 0 such that

∥C(x_{1}, …, x_{n})∥_{E} ≤ c sup_{1≤k≤n} ∥x_{k}∥_{E} ,

and that E is D^{∗}-convex if there is a constant c > 0 such that

∥C(x_{1}, …, x_{n})∥_{E} ≥ c^{−1} inf_{1≤k≤n} ∥x_{k}∥_{E} .

There is another way to define these notions. Let us consider the vector space V of right continuous functions from [0, ∞) to ℝ of bounded variation. Define the subsets

B_{c}^{≤} = {d_{x} : ∥x∥_{E} ≤ c},
B_{c}^{≥} = {d_{x} : ∥x∥_{E} ≥ c},
B_{c}^{=} = {d_{x} : ∥x∥_{E} = c}.

Then E is D-convex if and only if there exists a constant c > 0 such that conv B_{1}^{≤} is contained in B_{c}^{≤}, and E is D^{∗}-convex if and only if there exists a constant c > 0 such that conv B_{1}^{≥} is contained in B_{c}^{≥}.

Note that Lorentz spaces as defined in Section 1 are all D^{∗}-convex, Marcinkiewicz spaces are all D-convex, and Orlicz spaces are both D and D^{∗}-convex. An easy argument shows that if E is D-convex, then E′ is D^{∗}-convex, and it follows from Corollary 24 below that if E is D^{∗}-convex, then E′ is D-convex.

Suppose that M : [0, ∞) → [0, ∞) is increasing. We will say that M is p-convex if M(t^{1/p}) is convex, and we will say that M is q-concave if −M(t^{1/q}) is convex. By convention, we will say that M is always ∞-concave. We have the following result.

Lemma 20. Suppose that M : [0, ∞) → [0, ∞) is such that there exist 1 ≤ p < q ≤ ∞ and a constant c > 0 such that for all 0 < s < 1 that

c^{−1}s^{q}M(t) ≤ M(st) ≤ cs^{p}M(t),

(where we shall suppose that the first inequality is missing if q = ∞). Then there exists an increasing, p-convex, q-concave function M_{1} such that there exists a constant c_{1} > 0 with c_{1}^{−1}M ≤ M_{1} ≤ c_{1}M.

Proof. Let M_{2}(t) = sup_{s<1} M(st)/s^{p}, and let M_{3}(t) = inf_{s<1} M(st)/s^{q} (M_{3} = M_{2} if q = ∞). From now on, if q = ∞, we shall suppose that any inequality involving q is automatically true. Then c^{−1}M ≤ M_{3} ≤ cM, and

s^{q}M_{3}(t) ≤ M_{3}(st) ≤ s^{p}M_{3}(t),

that is, M_{3}(t^{1/p})/t is an increasing function, and M_{3}(t^{1/q})/t is a decreasing function. Now set

M_{1}(t) = ∫_{0}^{t} \frac{M_{3}(s)}{s} ds

= ∫_{0}^{t^{p}} \frac{M_{3}(s^{1/p})}{ps} ds

= ∫_{0}^{t^{q}} \frac{M_{3}(s^{1/q})}{qs} ds,

where the last equality holds only if q < ∞. Then M_{1} is p-convex and q-concave. Further, M_{1} ≤ M_{3}/p, and

M_{1}(t) ≥ ∫_{t^{p}/2}^{t^{p}} \frac{M_{3}(s^{1/p})}{ps} ds ≥ 1/(2p)M_{3}(t/2^{1/p}) ≥ 1/(4p)M_{3}(t).

□

If 1 ≤ p < q ≤ ∞, we say that E is an interpolation space for (L_{p}, L_{q}) if there is a constant c > 0 such that whenever T : L_{p} ∩ L_{q} → L_{p} ∩ L_{q} is a linear operator, such that ∥T ∥_{L_{p}→L_{p}} ≤ 1 and ∥T ∥_{L_{q}→L_{q}} ≤ 1, then ∥T ∥_{E→E} ≤ c.

The following result is an immediate consequence of results in [HM] and Lemma 20 (see also [AC]).

Theorem C. Suppose that E is an interpolation space for (L_{p}, L_{q}). Then there is a constant c > 0 such that whenever ∥x∥_{L_{M}} ≤ ∥y∥_{L_{M}} for all increasing p-convex and q-concave functions M : [0, ∞) → [0, ∞) and if y ∈ E, then x ∈ E and ∥x∥_{E} ≤ c ∥y∥_{E}.

Now let us state the main results of this section.

Theorem 21. Suppose that E is a D-convex interpolation space for (L_{p}, L_{q}), where 1 ≤ p < q ≤ ∞. Then there exists a constant c > 0 such that for every x ∈ E with ∥x∥_{E} = 1, there exists an increasing, p-convex, q-concave function M : [0, ∞) → [0, ∞) such that ∫ M(|x|) ds ≥ c^{−1}, and ∫ M(|y|) ds ≤ c whenever ∥y∥ ≤ c^{−1}.

Thus there exists a family of increasing, p-convex, q-concave functions M_{α} : [0, ∞) → [0, ∞) (α ∈ A) such that ∥·∥_{E} is equivalent to sup_{α∈A} ∥·∥_{L_{Mα}}.

Theorem 22. Suppose that E is a D^{∗}-convex interpolation space for (L_{p}, L_{q}), where 1 ≤ p < q ≤ ∞. Then there exists a constant c > 0 such that for every x ∈ E with ∥x∥_{E} = 1, there exists an increasing, p-convex, q-concave function M : [0, ∞) → [0, ∞) such that ∫ M(|x|) ds ≤ c, and ∫ M(|y|) ds ≥ c^{−1} whenever ∥y∥ ≥ c.

Thus there exists a family of increasing, p-convex, q-concave functions M_{α} : [0, ∞) → [0, ∞) (α ∈ A) such that ∥·∥_{E} is equivalent to inf_{α∈A} ∥·∥_{L_{Mα}}.

Theorem 23. Suppose that E is D-convex and D^{∗}-convex. Then there exists an increasing function M : [0, ∞) → [0, ∞) such that E is equivalent to L_{M}.

Corollary 24. Suppose that E is an interpolation space for (L_{p}, L_{q}), where 1 ≤ p < q ≤ ∞. If E is D-convex, then E is p-convex, and if q < ∞ then there is a constant c > 0 such that given functions x_{1}, x_{2}, …, x_{n} on [0, 1]

∥C(x_{1}, …, x_{n})∥_{E} ≤ c (\frac{1}{n} ∑_{k=1}^{n} ∥x_{k}∥_{E}^{q})^{1/q} .

If E is D^{∗}-convex, then E is q-concave, and there is a constant c > 0 such that given functions x_{1}, x_{2}, …, x_{n} on [0, 1]

∥C(x_{1}, …, x_{n})∥_{E} ≥ c^{−1} (\frac{1}{n} ∑_{k=1}^{n} ∥x_{k}∥_{E}^{p})^{1/p} .

Proof. Let us provide the proof of the stated inequality in the case that E is D-convex. The other results have almost identical proofs.

From Theorem 21, we see that it is sufficient to show that if M : [0, ∞) → [0, ∞) is increasing, convex, and q-concave (with q < ∞), then given functions x_{1}, x_{2}, …, x_{n} on [0, 1], we have that

∥C(x_{1}, …, x_{n})∥_{L_{M}} ≤ (\frac{1}{n} ∑_{k=1}^{n} ∥x_{k}∥_{L_{M}}^{q} )^{1/q} .

Let us suppose that the left hand side is bounded below by 1. Thus

∫ M(|C(x_{1}, …, x_{n})|) ds = \frac{1}{n} ∑_{k=1}^{n} ∫ M(|x_{k}|) ds ≥ 1.

Thus there exists a sequence c_{k} ≥ 0 with ∑_{k=1}^{n} c_{k} = n such that

∫ M(|x_{k}|) ds ≥ c_{k}.

Since M is q-concave, it follows that

∫ M(|x_{k}| /c_{k}^{1/q}) ds ≥ 1,

that is, ∥x_{k}∥_{L_{M}} ≥ c_{k}^{1/q}. The result follows. □

Let us now proceed with the proofs of the main theorems.

Lemma 25. If E ≠ L_{∞}, then for each ϵ > 0, there exists a strictly increasing function N : [0, ∞) → [0, ∞) with N(0) = 0, such that if ∫_{0}^{1} N(x^{∗}(t)) dt ≤ 1, then ∥x∥_{E} < ϵ.

Proof. Let k ∈ E \ L_{∞} such that k^{∗}(1) ≤ 1, ∥k∥_{E} < ϵ, and k^{∗} is strictly decreasing. Define

N(t) = \{ \begin{matrix}1/(k^{∗})^{−1}(t) & if t ≥ 1 \\ t/(k^{∗})^{−1}(1) & if t ≤ 1. \end{matrix}

Now suppose that ∥f∥_{L_{N}} ≤ 1. Then

∫_{0}^{1} N(f^{∗}(t)) dt ≤ 1,

which implies that tN(f^{∗}(t)) ≤ 1, that is f^{∗}(t) ≤ k^{∗}(t). Therefore ∥f∥_{E} ≤ ∥k∥_{E} < ϵ. □

For any L > 0, we will write

V_{L} = {f ∈ V : f(t) = 0 for t > L}.

Note that d_{x} ∈ V_{L} if and only if ∥x∥_{∞} ≤ L. Notice also that V_{L} has a predual, C([0, L]), defined by the pairing

⟨f, N⟩ = − ∫_{[0,L]} N(s)df(s).

Notice that if ∥x∥_{∞} ≤ L, then

⟨d_{x}, N⟩ = ∫_{0}^{1} N(|x(s)|) ds.

Lemma 26. Suppose that E ≠ L_{∞}, and that L > 0. Then the set B_{c}^{=} ∩ V_{L} is weak* compact in V_{L}.

Proof. It is clear that B_{c}^{=} is a bounded set in V, and hence it is sufficient to show that B_{c}^{≤} ∩ V_{L} is weak* closed in V_{L}. Suppose that ∥x_{n}∥_{E} = c, that d_{x_{n}} ∈ V_{L}, and that d_{x_{n}} → g weak*. Then it is easy to see that g is decreasing with g(0) ≤ 1, and that d_{x_{n}} → g pointwise except possibly at discontinuities of g. Therefore g = d_{y} for some y ∈ V_{L}, and x_{n}^{∗} → y^{∗} pointwise except possibly at points of discontinuity of y^{∗}, of which there are only countably many. Hence by Lebesgue’s Dominated Convergence Theorem, it follows that for any continuous function N that ∫_{0}^{1} N(x_{n}^{∗} − y^{∗}) dt → 0, and hence by Lemma 1, it follows that ∥x_{n}^{∗} − y^{∗}∥_{E} → 0. □

Now, if 1 ≤ p < q ≤ ∞, and L > 0, we define the subset C_{p,q,L} of V_{L} to be the set of all those f ∈ V_{L} such that

− ∫_{[0,L]} N df ≥ 0

for all increasing N : [0, ∞) → [0, ∞) that are p-convex and q-concave. Notice that C_{p,q,L} is weak* closed in V_{L} for all L > 0.

Proof of Theorem 21. We may suppose that E ≠ L_{∞}. Since we have that ∥min{|x| , t}∥_{E} → ∥x∥_{E} as t → ∞, we may suppose without loss of generality that x ∈ L_{∞}. By a further slight approximation, we may suppose that x^{∗} is strictly decreasing and x^{∗}(1) = 0, that is, we may suppose that d_{x} is absolutely continuous.

By the definition of D-convexity, and Theorem C, we know that there exists a constant c > 0 such that for all L > 0 we have that conv(B_{c^{−1}}^{=}) does not intersect with {d_{x}} + C_{p,q,L}. Hence, by the Hahn-Banach Theorem, for each L > ∥x∥_{∞}, there exists M_{L} ∈ C_{0}([0, L]) such that for some constant S = ±1

∫_{0}^{1} M_{L}(|y|) ds = − ∫_{0}^{L} M_{L} d(d_{y}) ≤ S for d_{y} ∈ B_{c^{−1}}^{≤} ∩ V_{L},

and

− ∫_{0}^{L} M_{L} d(d_{cx} + f) ≥ S for f ∈ C_{p,q,L}.

Hence

∫_{0}^{1} M_{L}(|x|) ds = − ∫_{0}^{L} M_{L}d(d_{x}) ≥ S.

Further, since C_{p,q,L} is a cone, it follows that

− ∫_{0}^{L} M_{L} df ≥ 0 for f ∈ C_{p,q,L}.

For 0 ≤ a < b ≤ L, consider the functions

f_{1}(s) = \{ \begin{matrix}1 & if a ≤ s ≤ b \\ 0 & otherwise, \end{matrix}

f_{2}(s) = \{ \begin{matrix}−1 & if a^{1/p} ≤ s ≤ ((a + b)/2)^{1/p} \\ 1 & if ((a + b)/2)^{1/p} ≤ s ≤ b^{1/p} \\ 0 & otherwise, \end{matrix}

f_{3}(s) = \{ \begin{matrix}1 & if a^{1/q} ≤ s ≤ ((a + b)/2)^{1/q} \\ −1 & if ((a + b)/2)^{1/q} ≤ s ≤ b^{1/q} \\ 0 & otherwise. \end{matrix}

It is easily seen that f_{1}, f_{2} and f_{3} are in C_{p,q,L}, and hence it may be seen that M_{L} is positive, increasing, p-convex, and q-concave on [0, L]. Hence S = 1.

By Lemma 25, there exists N : [0, ∞) → [0, ∞) with N(0) = 0 and that is strictly increasing, and such that if ∫_{0}^{1} N(|z|) ds ≤ 1, then ∥z∥_{E} ≤ 1. Hence, if z with ∥z∥_{∞} ≤ L, and if ∫_{0}^{1} N(|z|) ds ≤ 1, then ∫ M_{L}(|z|) ds ≤ 1. Hence M_{L} ≤ N.

Notice that (M_{L}/N) is a bounded sequence in L_{∞}([0, ∞)). Let M/N be a weak* limit point of this sequence. Since N(s)\frac{d}{ds}(d_{z}(s)) is in L_{1}([0, ∞)) whenever z ∈ L_{N}, and d_{x} is absolutely continuous, it is easy to see that M satisfies the requirements of Theorem 1. (Initially one would have to restrict to those y ∈ L_{∞}, but an application of Lebesgue’s Monotone Convergence Theorem will deal with this.) □

Proof of Theorem 22. Following the first part of the proof of Theorem 21 with only minor modifications, we can show the following. If x ∈ L_{∞} with ∥x∥_{∞} = L, then there exists an increasing, p-convex, q-concave function M_{1} : [0, L] → [0, ∞) such that

∫ M_{1}(|x|) ds ≤ c,

and whenever ∥y∥_{∞} ≤ L with ∥y∥_{E} ≥ c, then

∫ M_{1}(|y|) ds ≥ 1.

If q = ∞, consider the function N generated by Lemma 25 in the case when ϵ = 1. Notice that if we set

N_{1}(t) = \{ \begin{matrix}N(t) & if t ≥ 1 \\ N(1)t^{p} & if t < 1, \end{matrix}

and

N_{2}(t) = sup_{s<1} N_{1}(st)/s^{p},

then N_{2} still satisfies the conclusion of Lemma 25. Furthermore, if we set

M(t) = \{ \begin{matrix}M_{1}(t) & if t ≤ L \\ (M_{1}(L)/N_{2}(L))N_{2}(t) & if t > L, \end{matrix}

then we see that M satisfies the hypotheses of Lemma 20. Also, if ∥y∥_{E} ≥ 2c, with y ≥ 0, then either ∥min{y, L}∥_{E} ≥ c, in which case ∫ M(min{y, L}) ds = ∫ M_{1}(min{y, L}) ds ≥ c^{−1}, or ∥yI_{y>L}∥_{E} ≥ c, in which case ∫ M(yI_{y>L}) ds = (M_{1}(L)/N_{2}(L)) ∫ N_{2}(yI_{y>L}) ds ≥ c^{−1}. In either case, ∫ M(y) ds ≥ c^{−1}.

If q < ∞, it is an easy matter to see that L_{q} ⊆ E, and that there exists a constant c_{1} > 0 such that ∥z∥_{E} ≤ c_{1} ∥z∥_{q} for all z ∈ L_{q}. Set

M(t) = \{ \begin{matrix}M_{1}(t) & if t ≤ L \\ M_{1}(L)(t/L)^{q} & if t > L. \end{matrix}

Thus M satisfies the hypotheses of Lemma 20. Furthermore, if ∥y∥_{E} ≥ 2c, then ∫ M(|y|) ds ≥ c^{−1}, by the same argument as in the case when q = ∞.

Now let us consider the case for general x. Note that E ⊆ L_{p}, and that there is a constant c_{2} > 0 such that ∥z∥_{p} ≤ c_{2} ∥z∥_{E} for all z ∈ E. Without loss of generality, x ≥ 0. Write x = x_{1} + x_{2}, where x_{1} and x_{2} have disjoint support, x_{1} ∈ L_{∞}, ∥x_{1}∥_{E} ≥ 1/2, and ∥x_{2}∥_{p} ≤ 1.

Let M_{1} be the function described by Theorem 22 for x_{1}, and let M(t) = min{M_{1}(t), t^{p}}. It is clear that M satisfies the hypotheses of Lemma 20, and also that ∫ M(x) ds ≤ c. Now suppose that ∥y∥ ≥ 4c. We may suppose that y ≥ 0. Write y = y_{1} + y_{2}, where y_{1} = yI_{M_{1}(y)≤y^{p}}. Then either ∥y_{1}∥_{E} ≥ 2c, in which case ∫ M(y) ds ≥ ∫ y_{1}^{p} ds ≥ 2c_{2}^{−1}, or ∥y_{2}∥_{E} ≥ 2c, in which case ∫ M(y) ds ≥ ∫ M_{1}(y_{2}) ds ≥ c^{−1}. □

We will leave the proof of Theorem 23 to the reader, as it follows the ideas of the previous proofs.

We also leave with a problem that was given to the first named author by Carsten Schütt. If E is D^{∗}-convex, does there exist an appropriately measurable family of increasing, convex functions M_{α} : [0, ∞) → [0, ∞) (α ∈ A), where A is a measurable space with measure μ, such that ∥·∥_{E} is equivalent to ∫ ∥·∥_{M_{α}} dμ(α)?

7. Another generalization of Theorem B

In this section we will consider another generalization of Theorem B. Suppose that X is a symmetric sequence space on sequences x = (x_{i})_{1≤i≤n} ∈ ℝ^{n}. Let us suppose that ∥(1, 0, …, 0)∥_{X} = 1. Then we define its associated r.i. space, E_{X} by the following formula:

∥x∥_{E_{X}} = ∥(\frac{1}{n} ∫_{(i−1)/n}^{i/n} x^{∗}(s) ds)_{1≤i≤n}∥_{X} .

Let us show that E_{X} really does satisfy the triangle inequality. It is clear that ∥x^{∗} + y^{∗}∥_{E_{X}} ≤ ∥x∥_{E_{X}} + ∥y∥_{E_{X}}. It is also easy to see that if x ≺ y, then ∥x∥_{E_{X}} ≤ ∥y∥_{E_{X}}. Since x + y ≺ x^{∗} + y^{∗}, we are done.

To save space, if A and B are two quantities depending upon certain parameters, we will write A ≈ B if there exists a constant c > 0, independent of the parameters, such that c^{−1}A ≤ B ≤ cA. If t is a real number, we will write [t] for the greatest integer less than t.

Theorem 27. There exists a constant c > 0 such that if (x_{i,j})_{1≤i,j≤n} is an n × n matrix, then

\frac{1}{n!} ∑_{π∈S_{n}} ∥(x_{i,π(i)})_{1≤i≤n}∥_{X} ≥ c^{−1} (\frac{1}{n} ∑_{k=1}^{n} x_{k}^{∗} + ∥(x_{kn}^{∗})_{1≤k≤n})∥_{X}) .

Furthermore, if the associated r.i. space is D^{∗}-convex, then there exists a constant c > 0 such that if (x_{i,j})_{1≤i,j≤n} is an n × n matrix, then

\frac{1}{n!} ∑_{π∈S_{n}} ∥(x_{i,π(i)})_{1≤i≤n}∥_{X} ≤ c (\frac{1}{n} ∑_{k=1}^{n} x_{k}^{∗} + ∥(x_{kn}^{∗})_{1≤k≤n})∥_{X}) .

We do not know whether the condition that the associated space be D^{∗}-convex is necessary in order for the second inequality to hold. In order to show this result, we will use the following result due to Kwapień and Schütt.

Theorem D [KS2]. There exist a constant c > 0 such that for any n × n × n array y = (y_{i,j,k})_{1≤i,j,k≤n}, we have that

\frac{1}{(n!)^{2}} ∑_{π,σ∈S_{n}} max_{1≤i≤n} |y_{i,π(i),σ(i)}| ≈ \frac{1}{n^{2}} ∑_{k=1}^{n^{2}} y_{k}^{∗}.

Proof of Theorem 27. Let us first consider the case when X_{m} is the symmetric sequence space given by

∥z∥_{X_{m}} = ∑_{k=1}^{m} z_{k}^{∗}.

Suppose that given z, one forms the array

y_{i,j} = \{ \begin{matrix}z_{i} & if j ≤ n/m \\ 0 & otherwise. \end{matrix}

Then by Theorem A, it may be seen that

\frac{1}{n!} ∑_{π∈S_{n}} max_{1≤i≤n} |y_{i,π(i)}| ≈ \frac{1}{n} ∑_{k=1}^{n} y_{k}^{∗}

≈ \frac{[n/m]}{n} ∑_{k=1}^{n/[n/m]} z_{k}^{∗}

≈ \frac{1}{m} ∥z∥_{X_{m}} .

Now, given x as in the hypothesis of the theorem, form the following array:

y_{i,j,k} = \{ \begin{matrix}x_{i,j} & if k ≤ n/m \\ 0 & otherwise. \end{matrix}

In that case

\frac{1}{n!} ∑_{π∈S_{n}} ∥(x_{i,π(i)})_{1≤i≤n}∥_{X_{m}} ≈ \frac{m}{(n!)^{2}} ∑_{π,σ∈S_{n}} max_{1≤i≤n} |y_{i,π(i),σ(i)}|

≈ \frac{m}{n^{2}} ∑_{k=1}^{n^{2}} y_{k}^{∗}

≈ \frac{m[n/m]}{n^{2}} ∑_{k=1}^{n^{2}/[n/m]} x_{k}^{∗}

≈ \frac{1}{n} ∑_{k=1}^{nm} x_{k}^{∗}

≈ \frac{1}{n} ∑_{k=1}^{n} x_{k}^{∗} + ∥(x_{kn}^{∗})_{1≤k≤n}∥_{X_{m}} .

Now let us consider more general symmetric sequence spaces X. We know that

∥z∥_{X} = sup_{∥w∥_{X^{∗}} ≤1} ∑_{k=1}^{n} z_{k}^{∗}w_{k}^{∗}

= sup_{∥w∥_{X^{∗}} ≤1} ∑_{m=1}^{n}(w_{m}^{∗} − w_{m+1}^{∗}) ∥z∥_{X_{m}} ,

where by convention w_{n+1}^{∗} = 0. From this, we immediately see that for some constant c > 0

\frac{1}{n!} ∑_{π∈S_{n}} ∥(x_{i,π(i)})_{1≤i≤n}∥_{X} ≥ c^{−1} sup_{∥w∥_{X^{∗}} ≤1} ∑_{m=1}^{n}(w_{m}^{∗} − w_{m+1}^{∗}) (\frac{1}{n} ∑_{k=1}^{n} x_{k}^{∗} + ∥(x_{kn}^{∗})_{1≤k≤n}∥_{X_{m}})

≥ c^{−1} (\frac{1}{n} ∑_{k=1}^{n} x_{k}^{∗} + ∥(x_{kn}^{∗})_{1≤k≤n}∥_{X}) .

since w_{1}^{∗} ≤ 1 whenever ∥w∥_{X^{∗}} ≤ 1.

Now let us show the second inequality when E_{X} is D^{∗}-convex. Let us consider the following functions:

z_{π}(t) = x_{i,π(i)} t ∈ [(i − 1)/n, i/n),
w(t) = x_{k}^{∗} t ∈ [(k − 1)/n^{2}, k/n^{2}).

It is an easy matter to see that w^{∗} = C(z_{π} : π ∈ S_{n})^{∗}. Hence, by Corollary 24, we see that for some constant c > 0 depending only on X

\frac{1}{n!} ∑_{π∈S_{n}} ∥z_{π}∥_{E_{X}} ≤ c ∥w∥_{E_{X}} .

The result now follows after we notice that

∥x∥_{E_{X}} ≈ \frac{1}{n} ∫_{0}^{1/n} x^{∗}(s) ds + ∥(x^{∗}(k/n))_{1≤k≤n}∥_{X} .

□

8. D and D^{∗}-convex Lorentz Spaces

Although the results in this section are primarily concerned with Lorentz spaces, in order to prove our results, we will need a wider class of spaces, known as Orlicz-Lorentz spaces. If M, N : [0, ∞) → [0, ∞) are strictly increasing bijections, then we define the space L_{M,N} to be the set of those measurable functions x : [0, 1] → ℝ such that

∥x∥_{L_{M,N}} = ∥x^{∗} ∘ M̃ ∘ Ñ^{−1}∥_{L_{N}} ,

where M̃ = 1/M(1/t), and ∘ denotes function composition. It is not clear what are necessary and sufficient conditions for L_{M,N} to have an equivalent norm that satisfies the triangle inequality, but this will not be relevant to our discussion. It is clear that L_{M} = L_{M,M}, and that Λ_{r}(φ) = L_{φ̃^{−1},t^{r}} with equality of norms.

Following [M1], we say that an increasing bijection M : [0, ∞) → [0, ∞) is almost convex if there are numbers a > 1, b > 1, and a positive integer p such that for all positive integers m, the cardinality of the set of integers n such that we do not have M(a^{n+m}) ≥ a^{m−p}M(a^{n}) is less than b^{m}. It is clear that this notion also can be made to make sense if M is only a bijection from [0, 1] → [0, 1], or a bijection from [1, ∞) → [1, ∞), by stating that the inequality is true whenever it is undefined.

The following result is essentially Theorem 4.2 from [M1]. The results from [M1] are concerned with function spaces on ℝ rather than [0, 1], but the change is not too hard to do.

Theorem E. Let M, N_{1}, N_{2} be increasing bijections [0, ∞) → [0, ∞) that map 1 to 1, such that one of N_{1} or N_{2} is convex and q-concave for some q < ∞. Then the following are equivalent.

i) For some c > 0 we have that ∥x∥_{L_{M,N_{1}}} ≤ c ∥x∥_{L_{M,N_{2}}} for all measurable

x : [0, 1] → ℝ.

ii) N_{1} ∘ N_{2}^{−1} restricted to [1, ∞) is almost convex.

Theorem 28. If φ : [0, 1] → [0, 1] is an increasing bijection, and 1 ≤ r < ∞, such that Λ_{r}(φ) is equivalent to a norm, then Λ_{r}(φ) is D-convex if and only if (φ̃(t))^{r} is almost convex, and D^{∗}-convex if and only if φ̃^{−1}(t^{1/r}) is almost convex.

Proof. Suppose that Λ_{r}(φ) is D-convex. Define

M(t) = \{ \begin{matrix}t & if 0 ≤ t < 1 \\ φ̃^{−1}(t) & if t ≥ 1. \end{matrix}

It is clear that ∥I_{[0,t]}∥_{Λ_{r}(φ)} = ∥I_{[0,t]}∥_{L_{M}} = φ(t) for 0 ≤ t ≤ 1, and that Λ_{r}(φ) = L_{M,t^{r}}.

We will show that there is a constant c_{1} > 0 such that ∥x∥_{Λ_{r}(φ)} ≤ c_{1} ∥x∥_{L_{M}}. Then the result will follow easily from Theorem E.

For, by Theorem 21, there exists a constant c_{2} > 0 so that the following holds. Suppose that ∥x∥_{Λ_{r}(φ)} = 1. Then there exists an increasing convex bijection N : [0, ∞) → [0, ∞) such that ∥x∥_{L_{N}} ≥ c_{2}^{−1}, but that in general ∥y∥_{L_{N}} ≤ ∥y∥_{Λ_{r}(φ)}. By considering y = I_{[0,t]}, we see that N(t) ≤ M(t) for t ≥ 1. Now, for any λ < c_{2}^{−1}, we have that

∫_{0}^{1} N(|x| /λ) ds > 1.

Since N is convex, for any λ < c_{2}^{−1}/2, we have that

∫_{0}^{1} N(|x| /λ) ds > 2.

Further, for any λ > 0

∫_{|x|≤λ} N(|x| /λ) ds ≤ N(1) ≤ 1.

Hence for λ < c_{2}^{−1}/2, we have that

∫_{0}^{1} M(|x| /λ) ds ≥ ∫_{|x|≥λ} M(|x| /λ) ds ≥ ∫_{|x|≥λ} N(|x| /λ) ds > 1,

and hence ∥x∥_{L_{M}} ≥ c_{2}^{−1}/2.

The case when Λ_{r}(φ) is D^{∗}-convex is almost identical. □

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Department of Mathematics, University of Missouri, Columbia, MO 65211

    E-mail address: stephen@math.missouri.edu

Department of Mathematics, Voronezh State University, 394693 Voronezh, Russia

    E-mail address: root@mathd.vucnit.voronezh.su