On the full range of Zippin and inclusion indices of rearrangement-invariant spaces

Abstract

Let X be a rearrangement-invariant space on [0, 1]. It is known that its Zippin indices $\underline{\beta }{}_X,\overline{\beta }{}_X$ and its inclusion indices $\gamma _X,\delta _X$ are related as follows: $0\le \underline{\beta }{}_X\le 1/\gamma _X \le 1/\delta _X\le \overline{\beta }{}_X\le 1$. We show that given $\underline{\beta },\overline{\beta }\in [0,1]$ and $\gamma ,\delta \in [1,\infty ]$ satisfying $\underline{\beta }\le 1/\gamma \le 1/\delta \le \overline{\beta }$, there exists a rearrangement-invariant space X such that $\underline{\beta }{}_X=\underline{\beta }$, $\overline{\beta }{}_X=\overline{\beta }$ and $\gamma _X=\gamma$, $\delta _X=\delta$.

1 Introduction

Indices associated to quasiconcave functions are an important tool for studying rearrangement-invariant (r.i. in short) spaces and the operators acting on them. An uppermost example are the Boyd indices of a r.i. space X, $\underline{\alpha }{}_X$ and $\overline{\alpha }{}_X$, which in general satisfy $0\le \underline{\alpha }{}_X\le \overline{\alpha }{}_X\le 1$ and characterize the boundedness of the Hilbert transform acting on X, i.e., when $0< \underline{\alpha }{}_X\le \overline{\alpha }{}_X<1$; see [2, Ch. 3, Section 5]. Related, and simpler, indices are the Zippin indices, $\underline{\beta }{}_X$ and $\overline{\beta }{}_X$ (see bellow for the definition), which satisfy

$$\begin{aligned} 0\le \underline{\alpha }{}_X\le \underline{\beta }{}_X \le \overline{\beta }{}_X\le \overline{\alpha }{}_X\le 1. \end{aligned}$$

R.i. spaces X satisfying $\underline{\alpha }{}_X= \underline{\beta }{}_X$ and $\overline{\beta }{}_X= \overline{\alpha }{}_X$ are known as spaces of fundamental type. The class of r.i. spaces of fundamental type include most of the classical r.i. spaces; see [5].

The study of the fine spectra of the finite Hilbert transform acting on a r.i. space X over $(-1,1)$ depends in a relevant way on the following inclusion indices

$$\begin{aligned} \gamma _X:=\inf \left\{ p\in [1,\infty ]: L^p\hookrightarrow X\right\} , \quad \delta _X:=\sup \left\{ p\in [1,\infty ]: X\hookrightarrow L^p\right\} . \end{aligned}$$

In particular, the condition that the Boyd indices, the Zippin indices, and the inverse of the inclusion indices all coincide allows giving a full description of the fine spectra, see [4, Theorem 7.2]. The Zippin indices and the inverse of the inclusion indices satisfy

$$\begin{aligned} 0\le \underline{\beta }{}_X \le 1/\gamma _X \le 1/\delta _X \le \overline{\beta }{}_X\le 1, \end{aligned}$$

(1)

see [4, Lemma 6.1(a)].

The inclusion indices appear in the study by Hernández and Rodríguez-Salinas of Orlicz spaces having a sublattice lattice isomorphic to $L^p$, see [9, p. 185], [10, p. 192], [11, p. 11]. García del Amo, Hernández, Sánchez, and Semenov have used them to study disjoint strict singularity of inclusions between r.i. spaces [8, p. 249]. The incluson indices have been specifically studied by Fernández-Cabrera [6, 7]; Fernández-Cabrera, Cobos, Hernández and Sánchez [7]; Cobos, Fernández-Cabrera, Manzano and Martínez [3].

The aim of this paper is to discuss the distribution of values in inequalities (1). Hernández and Rodríguez-Salinas in [9, Theorem A] proved that given a triple $\alpha ,\beta ,p$ satisfying $0<\alpha<p\le \beta <\infty$ there exist an Orlicz space having the Zippin indices $1/\beta$ and $1/\alpha$ and a sublattice which is lattice isomorphic to $L^p$. Further, they proved in [10, Theorem 1] that for a triple $\alpha ,\beta ,\gamma$ satisfying $0<\alpha \le \gamma \le \beta <\infty$, there exists an Orlicz space with the upper inclusion index $\gamma$ and the Zippin indices $1/\beta$ and $1/\alpha$. This implies that the second inequality in (1) may be strict. By a duality argument based on formulae (2) and (4) below, this fact yields that the penultimate inequality in (1) also may be strict.

In this regard we establish the following result.

Theorem 1

Given $\underline{\beta },\overline{\beta }\in [0,1]$ and $\gamma ,\delta \in [1,\infty ]$ satisfying

$$\begin{aligned} \underline{\beta }\le 1/\gamma \le 1/\delta \le \overline{\beta }, \end{aligned}$$

there exists a quasiconcave function $\varphi :[0,1] \rightarrow [0,\infty )$ such that for every r.i. space X with the fundamental function equivalent to $\varphi$ one has

$$\begin{aligned} \underline{\beta }{}_X=\underline{\beta }, \quad \overline{\beta }{}_X=\overline{\beta }, \quad \gamma _X=\gamma , \quad \delta _X=\delta . \end{aligned}$$

Note that, by [12, Ch. II, Theorem 4.2], a function $\varphi :[0,1]\rightarrow [0,\infty )$ is a fundamental function of an r.i. space if and only if it is quasiconcave. If $\varphi$ is quasiconcave, then the Marcinkiewicz space $M_\varphi$ is an r.i. space whose fundamental function coincides with $\varphi$. Further, let $\widetilde{\varphi }$ be the least concave majorant of the quasiconcave function $\varphi$. Then the fundamental function of the Lorentz space $\Lambda _{\widetilde{\varphi }}$ is equal to $\widetilde{\varphi }$ and $\widetilde{\varphi }/2\le \varphi \le \widetilde{\varphi }$ (see Sect. 2 below).

A consequence of the above theorem is the following.

Corollary 2

If $p\in (1,\infty )$, then there exists an r.i. space X such that for every $\varepsilon \in (0,p-1)$,

$$\begin{aligned} L^{p+\varepsilon ,\infty } \hookrightarrow X \hookrightarrow L^{p-\varepsilon ,1}, \end{aligned}$$

and its Zippin indices are trivial, that is, $\underline{\beta }{}_X=0$ and $\overline{\beta }{}_X=1$.

The paper is organised as follows. In Sect. 2, we collect necessary definitions. Section 3 contains the proofs of Theorem 1 and its Corollary 2.

2 Preliminaries

Let $L^0$ be the space of all equivalence classes of complex-valued Lebesgue measurable functions on [0, 1] and let m denote the Lebesgue measure on [0, 1]. The distribution function of $f\in L^0$ is defined by

$$\begin{aligned} d_f(y):=m\{t\in [0,1]:|f(t)|>y\}, \quad y>0. \end{aligned}$$

Functions $f,g\in L^0$ are called equimeasurable if $d_f=d_g$. The nonincreasing rearrangement of f is given by

$$\begin{aligned} f^*(t)=\sup \{y>0:d_f(y)>t\}=\inf \{y>0:d_f(y)\le t\}. \end{aligned}$$

Following Semenov [15], a Banach subspace X of $L^0$ is called a symmetric space if

• (a) for any $g\in L^0$ equimeasurable to $f\in X$, one has $g\in X$ and $\Vert g\Vert _X=\Vert f\Vert _X$;

• (b) for every $g\in L^0$ and $f\in X$ the inequality $|f|\le |g|$ a.e. implies that $g\in X$ and $\Vert g\Vert _X\le \Vert f\Vert _X$.

Lebesgue spaces $L^p$ with $p\in [1,\infty ]$, Orlicz spaces $L^\Phi$, and Lorentz spaces $L^{p,q}$ (see below) are the most widely used examples of symmetric spaces.

If X is symmetric, then the function $\varphi _X(t):=\Vert \chi _E\Vert _X$, where $E\subset [0,1]$ is a measurable set with $m(E)=t$, is well defined and is called the fundamental function of X.

The associate space $X'$ of X consists of all functions $g\in L^0$ satisfying

$$\begin{aligned} \int _0^1 |f(x)g(x)|\,dx<\infty \end{aligned}$$

for all $f\in X$. It is equipped with the norm

$$\begin{aligned} \Vert g\Vert _{X'}:= \sup \left\{ \int _0^1 |f(x)g(x)|\,dx: \Vert f\Vert _X\le 1\right\} . \end{aligned}$$

Semenov proved [15, Theorem 2] (see also [12, Ch. II, Theorem 4.1]), that if X is symmetric, then $L^\infty \hookrightarrow X\hookrightarrow L^1$, where $\hookrightarrow$ denotes a continuous embedding. A symmetric space X is said to have the Fatou property if for every sequence $\{f_n\}$ in X such that $0\le f_n\uparrow f$ a.e., one has either $f\in X$ and $\Vert f_n\Vert _X\uparrow \Vert f\Vert _X$, or $\Vert f_n\Vert _X\uparrow \infty$. Symmetric spaces with the Fatou property are usually called rearrangement-invariant Banach function spaces (or, shortly, r.i. spaces), see [2, Ch. 1–2].

A function $\varphi :[0,1]\rightarrow [0,\infty )$ is said to be quasiconcave if $\varphi (t)=0$ precisely when $t=0$, the function $\varphi (t)$ is increasing and the function $\varphi (t)/t$ is decreasing on (0, 1]. Following [14, Section 2], for a quasiconcave, and hence measurable, function $\varphi :[0,1]\rightarrow [0,\infty )$, one can define its dilation function

$$\begin{aligned} M_\varphi (x):=\sup _{s\in (0,\min \{1,1/x\}]}\frac{\varphi (xs)}{\varphi (s)}. \end{aligned}$$

It follows from [12, Ch. II, Theorem 1.3] (see also [14, Theorem 1.2]) that

$$\begin{aligned} \underline{\alpha }(\varphi )&:= \sup _{0<x<1} \frac{\log M_\varphi (x)}{\log x} = \lim _{x\rightarrow 0^+} \frac{\log M_\varphi (x)}{\log x}, \\ \overline{\alpha }(\varphi )&:= \inf _{x>1} \frac{\log M_\varphi (x)}{\log x} = \lim _{x\rightarrow \infty } \frac{\log M_\varphi (x)}{\log x}. \end{aligned}$$

If X is an r.i. space, then its fundamental function $\varphi _X$ is quasiconcave (see, e.g., [15, Theorem 1], [2, Ch. 2, Corollary 5.3]). The numbers

$$\begin{aligned} \underline{\beta }{}_X:=\underline{\alpha }(\varphi _X), \quad \overline{\beta }{}_X:=\overline{\alpha }(\varphi _X) \end{aligned}$$

are called the Zippin (fundamental) indices of the r.i. space X (see [14, p. 27], [16], and also [2, Ch. 3, Exercise 14]). It is well known that

$$\begin{aligned} 0\le \underline{\beta }{}_X\le \overline{\beta }{}_X\le 1, \quad \underline{\beta }{}_X+\overline{\beta }{}_{X'}=1, \quad \underline{\beta }{}_{X'}+\overline{\beta }{}_{X}=1 \end{aligned}$$

(2)

(see, e.g., [14, formulae (4.14)–(4.15)]).

The inclusion indices $\gamma _X$ and $\delta _X$ can be expressed as follows:

$$\begin{aligned} \delta _X=\liminf _{t\rightarrow 0^+}\frac{\log t}{\log \varphi _X(t)}, \quad \gamma _X=\limsup _{t\rightarrow 0^+}\frac{\log t}{\log \varphi _X(t)} \end{aligned}$$

(3)

(see [8, p. 249] or [3, Theorems 1.1$-$1.2]). Since the fundamental functions of X and $X'$ satisfy $\varphi _X(t)\varphi _{X'}(t)=1$ (see [2, Ch. 2, Theorem 5.2]), the following relation between the inclusion indices of X and $X'$ hold:

$$\begin{aligned} \frac{1}{\gamma _X}+\frac{1}{\delta _{X'}}=1, \quad \frac{1}{\gamma _{X'}}+\frac{1}{\delta _X}=1. \end{aligned}$$

(4)

It follows from (3) that the middle inequality in (1) becomes the equality if and only if the limit

$$\begin{aligned} \lim _{t\rightarrow 0^+}\frac{\log t}{\log \varphi _X(t)} \end{aligned}$$

exists (see, e.g., [6, p. 669] or [3, Corollary 1.3]).

Suppose $1<p<\infty$ and $1\le q\le \infty$. The Lorentz spaces $L^{p,q}$ consist of all measurable functions $f:[0,1]\rightarrow \mathbb {C}$ such that $\Vert f\Vert _{(p,q)}<\infty$, where

$$\begin{aligned} \Vert f\Vert _{(p,q)}:= \left\{ \begin{array}{ll} \displaystyle \left( \int _0^1 \left\{ t^{1/p}f^{**}(t)\right\} ^q\frac{dt}{t}\right) ^{1/q}, &{} 1\le q<\infty , \\ \displaystyle \sup _{0<t\le 1}\left\{ t^{1/p} f^{**}(t)\right\} , &{} q=\infty , \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} f^{**}(t)=\frac{1}{t}\int _0^t f^*(y)\,dy. \end{aligned}$$

The spaces $L^{p,\infty }$ are frequently called weak $L^p$-spaces or Marcinkiewicz spaces. It follows from [2, Ch. 4, Theorem 4.6] that $L^{p,q}$ are rearrangement-invariant Banach function spaces. In view of [2, Ch. 2, Theorem 5.13], the Lorentz space $L^{p,1}$ and the Marcinkiewicz space $L^{p,\infty }$ are respectively the smallest and the largest of all r.i. spaces having the same fundamental function as $L^p$.

As usual, two functions $\phi ,\psi :[0,1]\rightarrow [0,\infty )$ are said to be equivalent if there exist constants $c,C\in (0,\infty )$ such that

$$\begin{aligned} c\phi (t)\le \psi (t)\le C\phi (t), \quad t\in [0,1]. \end{aligned}$$

For a quasiconcave function $\varphi :[0,1]\rightarrow [0,\infty )$, let $M_\varphi$ be the Marcinkiewicz space consisting of measurable functions $f:[0,1]\rightarrow \mathbb {C}$ satisfying

$$\begin{aligned} \Vert f\Vert _{M_\varphi }:=\sup _{0<t\le 1}\big \{\varphi (t)f^{**}(t)\big \}<\infty . \end{aligned}$$

Then $M_\varphi$ is an r.i. space whose fundamental function coincides with $\varphi$ (see, e.g., [2, Ch. 2, Proposition 5.8] or [12, formula (4.28)]).

For each quasiconcave function $\varphi :[0,1]\rightarrow [0,\infty )$, its least concave majorant $\widetilde{\varphi }$ satisfies

$$\begin{aligned} \widetilde{\varphi }(t)/2\le \varphi (t)\le \widetilde{\varphi }(t), \quad t\in [0,1] \end{aligned}$$

(see, e.g., [2, Ch. 2, Proposition 5.10]). The Lorentz space $\Lambda _{\widetilde{\varphi }}$ consists of all measurable functions $f:[0,1]\rightarrow \mathbb {C}$ such that

$$\begin{aligned} \Vert f\Vert _{\Lambda _{\widetilde{\varphi }}}:= \int _0^1 f^*(t)\,d\widetilde{\varphi }(t)<\infty . \end{aligned}$$

It is well known that $\Lambda _{\widetilde{\varphi }}$ is an r.i. space whose fundamental function is $\widetilde{\varphi }$ (see, e.g., [2, Ch. 2, Theorem 5.13] or [12, formula (4.28)]).

3 Proofs

Inequalities (1) were proved in [4, Lemma 6.1(a)] under the assumption that the Boyd indices of X are non-trivial. For completeness, we include a proof of them.

3.1 Proof of inequalities (1)

Inequality $1/\gamma _X\le 1/\delta _X$ follows immediately from equalities (3).

It follows from [13, Lemma 4.2] that if $\overline{\beta }{}_X<1/p$, then $X\hookrightarrow L^p$. Hence

$$\begin{aligned} 1/\overline{\beta }{}_X = \sup \{p\in [1,\infty ]:\overline{\beta }{}_X<1/p\} \le \sup \{p\in [1,\infty ]: X\hookrightarrow L^p\}=\delta _X, \end{aligned}$$

which implies that $1/\delta _X\le \overline{\beta }{}_X$.

Let $p':=p/(p-1)$. It follows from (2) that $\overline{\beta }{}_{X'}<1/p'$ if and only if $\underline{\beta }{}_X>1/p$. By [2, Ch. 1, Proposition 2.10], $X'\hookrightarrow L^{p'}$ if and only if $L^p\hookrightarrow X$. So, if $\underline{\beta }{}_X>1/p$, then $L^p\hookrightarrow X$. Hence

$$\begin{aligned} 1/\underline{\beta }{}_X = \inf \{p\in [1,\infty ]:\underline{\beta }{}_X>1/p\} \ge \inf \{p\in [1,\infty ]:L^p\hookrightarrow X\} = \gamma _X. \end{aligned}$$

Therefore, $\underline{\beta }{}_X\le 1/\gamma _X$.

$\square$

3.2 The case of coinciding inclusion indices

In this subsection, we will prove Theorem 1 in the case where the inclusion indices coincide. The proof will follow from the theorem below.

Theorem 3

Let $p \in [1, \infty ]$ and $\underline{\beta }, \overline{\beta } \in [0, 1]$ be such that $\underline{\beta } \le 1/p \le \overline{\beta }$, and let $\rho : [0, 1] \rightarrow [0, 1]$ be an increasing continuous function such that $\rho (t)=0$ precisely when $t=0$. Then there exists a quasiconcave function $\varphi : [0, 1] \rightarrow [0, \infty )$ such that

$$\begin{aligned} t^{1/p} \rho (t) \le \varphi (t) \le t^{1/p}/ \rho (t) , \quad t \in (0, 1], \end{aligned}$$

(5)

and

$$\begin{aligned} \underline{\alpha }(\varphi ) = \underline{\beta }, \quad \overline{\alpha }(\varphi ) = \overline{\beta } . \end{aligned}$$

(6)

Proof

Our construction is inspired by the construction presented in [1, p. 261], and it works as follows. It requires three sequences $\{a_k\}_{k \in \mathbb {N}}$, $\{b_k\}_{k \in \mathbb {N}}$, $\{c_k\}_{k \in \mathbb {N}}$, whose terms satisfy

$$\begin{aligned} 0< a_{k+1}<c_k< b_k < a_k \le 1, \quad k \in \mathbb {N}, \end{aligned}$$

and a continuous function $\varphi : [0, 1] \rightarrow [0, \infty )$ satisfying

$$\begin{aligned}&\frac{\varphi (t)}{t^{\underline{\beta }}} \ \text{ is } \text{ constant } \text{ on } \left[ b_k, a_k\right) , \nonumber \\&\frac{\varphi (t)}{t^{\overline{\beta }}} \ \text{ is } \text{ constant } \text{ on } \left[ c_k, b_k\right) , \nonumber \\&\frac{\varphi (t)}{t^{\frac{1}{p}}} \ \text{ is } \text{ constant } \text{ on } \left[ a_{k + 1}, c_k\right) . \end{aligned}$$

(7)

The terms $b_k$ should be defined so that the intervals $\left[ b_k, a_k\right)$ are large enough for the first condition in (7) to ensure that the first equality in (6) holds. Similarly, the terms $c_k$ should be defined so that the intervals $\left[ c_k, b_k\right)$ are large enough for the second condition in (7) to ensure that the second equality in (6) holds. Finally, the sequence $\{a_k\}_{k \in \mathbb {N}}$ should converge to 0 sufficiently rapidly so that, via the third condition in (7), it is guaranteed that $\varphi (t)/t^{\frac{1}{p}}$ is constant “most of the time", which leads to (5).

Following the above, for any sequence $\{a_k\}_{k \in \mathbb {N}}$ with the terms in (0, 1], set

$$\begin{aligned} b_k:=\frac{a_k}{k}, \quad c_k:=\frac{b_k}{k}=\frac{a_k}{k^2}, \quad k\in \mathbb {N}. \end{aligned}$$

Choose a sequence $\{a_k\}$ so that

$$\begin{aligned} a_{k+1}<c_k, \quad \rho (a_k) \le k^{-k(k + 1)\left( \frac{1}{p} - \underline{\beta } \right) }, \quad \rho (b_k) \le k^{-k(k + 1)\left( \overline{\beta } - \frac{1}{p}\right) } , \end{aligned}$$

(8)

and set $a_0:= 1$. Let

$$\begin{aligned} \varphi (t):= t^{\frac{1}{p}}, \quad t \in [a_1, 1]. \end{aligned}$$

Suppose we have defined $\varphi$ on $[a_k, 1]$ with $k\ge 1$ in such a way that

$$\begin{aligned} t^{\frac{1}{p}} \rho (t)^{1 - \frac{1}{k}} \le \varphi (t) \le \frac{t^{\frac{1}{p}}}{\rho (t)^{1 - \frac{1}{k}}}\,, \quad t \in [a_k,a_{k-1}). \end{aligned}$$

(9)

Let us define $\varphi$ on $[a_{k + 1}, a_k)$. We start with

$$\begin{aligned} \varphi (t) := \varphi (a_k) \left( \frac{t}{a_k}\right) ^{\underline{\beta }}\,, \quad t \in \left[ b_k, a_k\right) . \end{aligned}$$

(10)

It follows from (9) with $t = a_k$ and the second inequality in (8) that for all $t\in [b_k,a_k)$,

$$\begin{aligned} \varphi (t)&\le \frac{a_k^{\frac{1}{p}}}{\rho (a_k)^{1 - \frac{1}{k}}} \left( \frac{t}{a_k}\right) ^{\underline{\beta }} = \frac{a_k^{\frac{1}{p}}}{\rho (a_k)^{1 - \frac{1}{k}}} \left( \frac{t}{a_k}\right) ^{\underline{\beta } - \frac{1}{p}} \left( \frac{t}{a_k}\right) ^{\frac{1}{p}} \\&\le \frac{a_k^{\frac{1}{p}}}{\rho (a_k)^{1 - \frac{1}{k}}}\, \left( \frac{b_k}{a_k}\right) ^{\underline{\beta } - \frac{1}{p}} \left( \frac{t}{a_k}\right) ^{\frac{1}{p}} = \frac{k^{\frac{1}{p} - \underline{\beta }}}{\rho (a_k)^{1 - \frac{1}{k}}}\, t^{\frac{1}{p}} \le \frac{\rho (a_k)^{-\frac{1}{k(k + 1)}}}{\rho (a_k)^{1 - \frac{1}{k}}}\, t^{\frac{1}{p}} \\&= \frac{t^{\frac{1}{p}}}{\rho (a_k)^{1 - \frac{1}{k + 1}}} \le \frac{t^{\frac{1}{p}}}{\rho (t)^{1 - \frac{1}{k + 1}}}. \end{aligned}$$

On the other hand, it follows from (9) with $t = a_k$ and (10) that for $t\in [b_k,a_k)$,

$$\begin{aligned} \varphi (t) \ge a_k^{\frac{1}{p}} \rho (a_k)^{1 - \frac{1}{k}} \left( \frac{t}{a_k}\right) ^{\underline{\beta }} \ge a_k^{\frac{1}{p}} \rho (a_k)^{1 - \frac{1}{k}} \left( \frac{t}{a_k}\right) ^{\frac{1}{p}} = \rho (a_k)^{1 - \frac{1}{k}} t^{\frac{1}{p}} \ge t^{\frac{1}{p}} \rho (t)^{1 - \frac{1}{k}}. \end{aligned}$$

So,

$$\begin{aligned} t^{\frac{1}{p}} \rho (t)^{1 - \frac{1}{k}} \le \varphi (t) \le \frac{t^{\frac{1}{p}}}{\rho (t)^{1 - \frac{1}{k + 1}}}\,, \quad t\in [b_k,a_k). \end{aligned}$$

(11)

Now, take

$$\begin{aligned} \varphi (t) := \varphi (b_k) \left( \frac{t}{b_k}\right) ^{\overline{\beta }}\,, \quad t \in \left[ c_k, b_k\right) . \end{aligned}$$

(12)

It follows from (11) with $t = b_k$ and the third inequality in (8) that for all $t\in [c_k,b_k)$,

$$\begin{aligned} \varphi (t)&\ge b_k^{\frac{1}{p}} \rho (b_k)^{1 - \frac{1}{k}} \left( \frac{t}{b_k}\right) ^{\overline{\beta }} = b_k^{\frac{1}{p}} \rho (b_k)^{1 - \frac{1}{k}} \left( \frac{t}{b_k}\right) ^{\overline{\beta } - \frac{1}{p}} \left( \frac{t}{b_k}\right) ^{\frac{1}{p}} \\&\ge b_k^{\frac{1}{p}} \rho (b_k)^{1 - \frac{1}{k}} \left( \frac{c_k}{b_k}\right) ^{\overline{\beta } - \frac{1}{p}} \left( \frac{t}{b_k}\right) ^{\frac{1}{p}} = \rho (b_k)^{1 - \frac{1}{k}} k^{\frac{1}{p} - \overline{\beta }} t^{\frac{1}{p}} \\&\ge \rho (b_k)^{1 - \frac{1}{k}} \rho (b_k)^{\frac{1}{k(k + 1)}} t^{\frac{1}{p}} = \rho (b_k)^{1 - \frac{1}{k + 1}} t^{\frac{1}{p}} \ge \rho (t)^{1 - \frac{1}{k + 1}}\, t^{\frac{1}{p}}. \end{aligned}$$

On the other hand, it follows from (11) with $t = b_k$ and (12) that for all $t\in [c_k,b_k)$,

$$\begin{aligned} \varphi (t) \le \frac{b_k^{\frac{1}{p}}}{\rho (b_k)^{1 - \frac{1}{k + 1}}} \left( \frac{t}{b_k}\right) ^{\overline{\beta }} \le \frac{b_k^{\frac{1}{p}}}{\rho (b_k)^{1 - \frac{1}{k + 1}}} \left( \frac{t}{b_k}\right) ^{\frac{1}{p}} = \frac{t^{\frac{1}{p}}}{\rho (b_k)^{1 - \frac{1}{k + 1}}} \le \frac{t^{\frac{1}{p}}}{\rho (t)^{1 - \frac{1}{k + 1}}}. \end{aligned}$$

So, combining the above two inequalities with inequality (11), we arrive at

$$\begin{aligned} t^{\frac{1}{p}} \rho (t)^{1 - \frac{1}{k + 1}} \le \varphi (t) \le \frac{t^{\frac{1}{p}}}{\rho (t)^{1 - \frac{1}{k + 1}}}\,, \quad t\in [c_k,a_k). \end{aligned}$$

(13)

Finally, let

$$\begin{aligned} \varphi (t) := \varphi (c_k) \left( \frac{t}{c_k}\right) ^{\frac{1}{p}}\,, \quad t \in \left[ a_{k + 1}, c_k\right) . \end{aligned}$$

(14)

Then it follows from (13) with $t = c_k$ that for all $t \in \left[ a_{k + 1}, c_k\right)$,

$$\begin{aligned}&\varphi (t) \le \frac{c_k^{\frac{1}{p}}}{\rho (c_k)^{1 - \frac{1}{k + 1}}} \left( \frac{t}{c_k}\right) ^{\frac{1}{p}} = \frac{t^{\frac{1}{p}}}{\rho (c_k)^{1 - \frac{1}{k + 1}}} \le \frac{t^{\frac{1}{p}}}{\rho (t)^{1 - \frac{1}{k + 1}}}\,, \\&\varphi (t) \ge c_k^{\frac{1}{p}} \rho (c_k)^{1 - \frac{1}{k + 1}} \left( \frac{t}{c_k}\right) ^{\frac{1}{p}} = t^{\frac{1}{p}} \rho (c_k)^{1 - \frac{1}{k + 1}} \ge t^{\frac{1}{p}} \rho (t)^{1 - \frac{1}{k + 1}}. \end{aligned}$$

So, the above two inequalities and (13) imply that

$$\begin{aligned} t^{\frac{1}{p}} \rho (t)^{1 - \frac{1}{k + 1}} \le \varphi (t) \le \frac{t^{\frac{1}{p}}}{\rho (t)^{1 - \frac{1}{k + 1}}}, \quad t \in [a_{k+1},a_k) \end{aligned}$$

(cf. (9)). Since $\rho (t)\in (0,1]$, this inequality implies (5).

For $k\in \mathbb {N}$, let

$$\begin{aligned} I_{1, k}:= [b_k, a_k], \quad I_{2, k}:= [c_k, b_k], \quad I_{3, k}:= [a_{k + 1}, c_k] \end{aligned}$$

and

$$\begin{aligned} \beta _1:= \underline{\beta }, \quad \beta _2:= \overline{\beta }, \quad \beta _3 = \frac{1}{p}. \end{aligned}$$

The above inductive argument produces a continuous function on (0, 1] satisfying (5) and such that $\varphi (t) = t^{1/p}$ for $t \in [a_1, 1]$,

$$\begin{aligned} \varphi (t) = \textrm{const}_{j, k}\, t^{\beta _j} , \quad t\in I_{j, k} , \quad j = 1, 2, 3, \quad k \in \mathbb {N} . \end{aligned}$$

(15)

Set $\varphi (0) = 0$. It is clear that $\varphi$ is quasiconcave. Since $\log \varphi$ is continuous and piecewise smooth, it follows from (15) that if $x > 1$ and $0 < s \le 1/x$, then

$$\begin{aligned} \log \varphi (xs) - \log \varphi (s) = \int _s^{xs} (\log \varphi (t))'\, dt \le \int _s^{xs}\overline{\beta }\, \frac{dt}{t} = \overline{\beta }\, \log {x}, \end{aligned}$$

and hence

$$\begin{aligned} \frac{\varphi (xs)}{\varphi (s)} \le x^{\overline{\beta }}. \end{aligned}$$

(16)

Similarly, if $x, s \in (0, 1)$, then

$$\begin{aligned} \log \varphi (s) - \log \varphi (xs) = \int _{xs}^{s} (\log \varphi (t))'\, dt \ge \int _{xs}^{s} \underline{\beta }\,\frac{dt}{t} = -\underline{\beta }\, \log {x}, \end{aligned}$$

and

$$\begin{aligned} \frac{\varphi (xs)}{\varphi (s)} \le x^{\underline{\beta }}\,. \end{aligned}$$

(17)

Take any $x > 1$ and choose $k > x$, $k \in \mathbb {N}$. Then $xc_k\in [c_k,b_k)$ and it follows from (12) that

$$\begin{aligned} M_\varphi (x) = \sup _{0 < s \le 1/x} \frac{\varphi (xs)}{\varphi (s)} \ge \frac{\varphi (xc_k)}{\varphi (c_k)} = x^{\overline{\beta }}. \end{aligned}$$

Hence $M_\varphi (x) = x^{\overline{\beta }}$ (see (16)), and

$$\begin{aligned} \overline{\alpha }(\varphi ) = \inf _{x > 1} \frac{\log {M_\varphi (x)}}{\log {x}} = \overline{\beta }. \end{aligned}$$

Similarly, take any $x < 1$ and choose $k \in \mathbb {N}$ such that $1/k < x$. Then $xa_k\in [b_k,a_k)$ and it follows from (10) that

$$\begin{aligned} M_\varphi (x) = \sup _{0 < s \le 1} \frac{\varphi (xs)}{\varphi (s)} \ge \frac{\varphi (xa_k)}{\varphi (a_k)} = x^{\underline{\beta }}. \end{aligned}$$

Hence $M_\varphi (x) = x^{\underline{\beta }}$ (see (17)), and

$$\begin{aligned} \underline{\alpha }(\varphi ) = \sup _{0< x < 1} \frac{\log {M_\varphi (x)}}{\log {x}} = \underline{\beta }, \end{aligned}$$

which completes the proof of (6). $\square$

Corollary 4

Given $p\in [1,\infty ]$ and $\underline{\beta },\overline{\beta }$ such that $0\le \underline{\beta }\le 1/p\le \overline{\beta }\le 1$, there exists a quasiconcave function $\varphi :[0,1]\rightarrow [0,\infty )$ such that for every r.i. space X with the fundamental function equivalent to $\varphi$, its inclusion indices satisfy $\gamma _X=\delta _X=p$ and its Zippin indices satisfy $\underline{\beta }{}_X=\underline{\beta }$ and $\overline{\beta }=\overline{\beta }{}_X$.

Proof

Take $\rho (t)=1/\log (e - 1 + 1/t)$, $t\in (0, 1]$, and $\rho (0) = 0$. It follows from Theorem 3 that there exists a quasiconcave function $\varphi :[0,1]\rightarrow [0,\infty )$ such that

$$\begin{aligned} t^{1/p}\frac{1}{\log (e - 1 + 1/t)}\le \varphi (t)\le t^{1/p}\log (e - 1 + 1/t), \quad t\in (0,1], \end{aligned}$$

(18)

and (6) holds.

Let X be an r.i. space such that $\varphi _X$ is equivalent to $\varphi$. Then there exist constants $c,C\in (0,\infty )$ such that

$$\begin{aligned} c\varphi (t)\le \varphi _X(t)\le C\varphi (t), \quad t\in [0,1]. \end{aligned}$$

Hence

$$\begin{aligned} \delta _X&= \liminf _{t\rightarrow 0^+}\frac{\log t}{\log \varphi _X(t)} = \liminf _{t\rightarrow 0^+}\frac{\log t}{\log \varphi (t)}, \\ \gamma _X&= \limsup _{t\rightarrow 0^+}\frac{\log t}{\log \varphi _X(t)} = \limsup _{t\rightarrow 0^+}\frac{\log t}{\log \varphi (t)}. \end{aligned}$$

Moreover, in view of (6), the Zippin indices of X satisfy

$$\begin{aligned} \underline{\beta }{}_X=\underline{\beta }, \quad \overline{\beta }{}_X=\overline{\beta }. \end{aligned}$$

It follows from (18), for $t\in (0,1]$, that

$$\begin{aligned} \frac{1}{p}\log t - \log \log (e - 1 + 1/t) \le \log \varphi (t) \le \frac{1}{p}\log t+\log \log (e - 1 + 1/t). \end{aligned}$$

Dividing by $\log t < 0$, we get

$$\begin{aligned} \frac{1}{p}+\frac{\log \log (e - 1 + 1/t)}{\log t} \le \frac{\log \varphi (t)}{\log t} \le \frac{1}{p} - \frac{\log \log (e - 1 + 1/t)}{\log t}. \end{aligned}$$

Since

$$\begin{aligned} \lim _{t\rightarrow 0^+}\frac{\log \log (e - 1 + 1/t)}{\log t}&= \lim _{t\rightarrow 0^+}\frac{1}{1/t}\frac{-1/t^2}{(e - 1 + 1/t)\log (e - 1 + 1/t)} \nonumber \\&= \lim _{t\rightarrow 0^+}\frac{-1}{((e - 1)t + 1)\log (e - 1 + 1/t)}=0, \end{aligned}$$

(19)

we arrive at

$$\begin{aligned} \delta _X=\gamma _X=p \end{aligned}$$

(see (3)), which completes the proof. $\square$

3.3 The case of distinct inclusion indices

In this subsection, we prove Theorem 1 in the case where the inclusion indices are distinct. The proof will follow from the theorem below.

Theorem 5

Let $\underline{\beta }, \overline{\beta } \in [0, 1]$ and $\gamma , \delta \in [1, \infty ]$ be such that

$$\begin{aligned} \underline{\beta }\le 1/\gamma < 1/\delta \le \overline{\beta }. \end{aligned}$$

(20)

Then there exists a quasiconcave function $\psi : [0, 1] \rightarrow [0, \infty )$ such that

$$\begin{aligned} \underline{\alpha }(\psi )=\underline{\beta }, \quad \overline{\alpha }(\psi ) = \overline{\beta }, \quad \limsup _{t\rightarrow 0^+}\frac{\log t}{\log \psi (t)}= \gamma , \quad \liminf _{t\rightarrow 0^+}\frac{\log t}{\log \psi (t)}= \delta . \end{aligned}$$

(21)

Proof

We first outline the proof. We will choose p and $\rho (t)$ such that $1/\gamma< 1/p < 1/\delta$ and $\rho (t)$ decays to 0 sufficiently slowly as $t \rightarrow 0$. Then Theorem 3 produces a quasiconcave function $\varphi$ such that

$$\begin{aligned} \frac{t^{\frac{1}{\delta }}}{\rho (t)} \le \varphi (t) \le t^{\frac{1}{\gamma }} \rho (t), \quad t \in (0, 1], \end{aligned}$$

and the first two equalities in (21) (with $\varphi$ in place of $\psi$) are satisfied while the last two are not.

We need to modify $\varphi$ on a part of [0, 1] to get the third and the fourth equalities in (21). To achieve this, we construct inductively a subsequence $\{a_{k(j)}\}_{j \in \mathbb {N}}$ of the sequence $\{a_{k}\}_{k \in \mathbb {N}}$ from the proof of Theorem 3. Take $\tau _j\in (0,a_{k(j)}]$ and set $\psi (t):= \textrm{const}\, t^{\underline{\beta }}$, $t \in [\tau _j, a_{k(j)}]$, where the constant is such that $\psi (a_{k(j)}) = \varphi (a_{k(j)})$. We can choose $\tau _j$ in such a way that

$$\begin{aligned} \psi (\tau _j) = \tau _j^{\frac{1}{\gamma }} \rho (\tau _j). \end{aligned}$$

(22)

This is because $t^{\frac{1}{\gamma }} \rho (t)/t^{\underline{\beta }} \rightarrow 0$ as $t \rightarrow 0^+$. Next, take $\upsilon _j\in (0,\tau _j]$ and set $\psi (t):= \textrm{const}\, t^{\overline{\beta }}$, $t \in [\upsilon _j, \tau _j)$, where the constant is such that $\psi$ is continuous at $\tau _j$. We can choose $\upsilon _j$ in such a way that

$$\begin{aligned} \psi (\upsilon _j) = \frac{\upsilon _j^{\frac{1}{\delta }}}{\rho (\upsilon _j)}\,. \end{aligned}$$

(23)

This is because

$$\begin{aligned} \frac{t^{\overline{\beta }}}{t^{\frac{1}{\delta }}/\rho (t)} \rightarrow 0 \quad \text{ as }\quad t \rightarrow 0^+. \end{aligned}$$

Finally, take $\varsigma _j\in (0,\upsilon _j]$ and set $\psi (t):= \textrm{const}\, t^{\underline{\beta }}$, $t \in [\varsigma _j, \upsilon _j)$, where the constant is such that $\psi$ is continuous at $\upsilon _j$. We can choose $\varsigma _j$ in such a way that

$$\begin{aligned} \psi (\varsigma _j) = \varphi (\varsigma _j). \end{aligned}$$

This is because $0 \le \varphi (t)/t^{\underline{\beta }} \le t^{\frac{1}{\gamma }} \rho (t)/t^{\underline{\beta }} \rightarrow 0$ as $t \rightarrow 0^+$. We then choose $k(j + 1)$ in such a way that $a_{k(j + 1) - 1} \le \varsigma _j$ and keep

$$\begin{aligned} \psi (t) = \varphi (t)\text { on } [a_{k(j + 1)}, \varsigma _j] \supseteq [a_{k(j + 1)}, a_{k(j + 1) - 1}]. \end{aligned}$$

The last equality ensures that the first two equalities in (21) hold, while (22) and (23) allow one to prove the last two equalities in (21). Details of the above argument are given below.

Let

$$\begin{aligned} \frac{1}{p}:= \frac{1}{2}\left( \frac{1}{\delta } + \frac{1}{\gamma }\right) , \quad \kappa := \frac{1}{4}\left( \frac{1}{\delta } - \frac{1}{\gamma }\right) = \frac{1}{2}\left( \frac{1}{\delta } - \frac{1}{p}\right) = \frac{1}{2}\left( \frac{1}{p} - \frac{1}{\gamma }\right) > 0, \end{aligned}$$

and

$$\begin{aligned} \rho (t) := \frac{1}{\log ^\kappa (e - 1 + 1/t)}\,, \quad t \in (0, 1] , \quad \rho (0) := 0. \end{aligned}$$

(24)

Integrating the inequality

$$\begin{aligned} \frac{1}{s^2(e - 1 + 1/s)}< \frac{1}{s^2}, \quad 0 < s \le 1 \end{aligned}$$

between $t \in (0, 1]$ and 1, one gets $\log (e - 1 + 1/t) - 1 \le 1/t - 1$. Therefore

$$\begin{aligned} \frac{1}{\log (e - 1 + 1/t)} \ge t, \end{aligned}$$

which implies that

$$\begin{aligned} 1\ge \rho ^2(t) \ge t^{2\kappa } = t^{\frac{1}{\delta }-\frac{1}{p}} = t^{\frac{1}{p}-\frac{1}{\gamma }}, \quad t\in (0,1]. \end{aligned}$$

Hence

$$\begin{aligned} \frac{t^{\frac{1}{\delta }}}{\rho (t)} \le t^{\frac{1}{p}} \rho (t) \le \frac{t^{\frac{1}{p}}}{\rho (t)} \le t^{\frac{1}{\gamma }} \rho (t), \quad t \in (0, 1]. \end{aligned}$$

Let $\varphi :[0,1]\rightarrow [0,\infty )$ be the quasiconcave function constructed in the proof of Theorem 3, which satisfies

$$\begin{aligned} \frac{t^{\frac{1}{\delta }}}{\rho (t)} \le t^{\frac{1}{p}} \rho (t) \le \varphi (t) \le \frac{t^{\frac{1}{p}}}{\rho (t)} \le t^{\frac{1}{\gamma }} \rho (t), \quad t \in (0, 1]. \end{aligned}$$

(25)

Below, we also use the notation $a_k$ from the proof of Theorem 3. Note that the sequence $\{a_k\}$ is decreasing and $a_k\rightarrow 0$ as $k\rightarrow \infty$ (see the first inequality in (8)).

Let

$$\begin{aligned} \psi (t):= \varphi (t) = t^{\frac{1}{p}}, \quad t \in [a_1, 1], \end{aligned}$$

and $k(1):=1$. In the next step, $j = 1$, but it is convenient for future reference to write it up for a general $j \in \mathbb {N}$.

It follows from the last two inequalities in (25) with $t=a_{k(j)}$ that

$$\begin{aligned} \varphi (a_{k(j)}) \le \frac{a_{k(j)}^{\frac{1}{p}}}{\rho (a_{k(j)})} \le a_{k(j)}^{\frac{1}{\gamma }}\rho (a_{k(j)}), \quad j\in \mathbb {N}, \end{aligned}$$

which implies that

$$\begin{aligned} 1\le \frac{a_{k(j)}^{\frac{1}{\gamma }}\rho (a_{k(j)})}{\varphi (a_{k(j)}) \left( \displaystyle \frac{a_{k(j)}}{a_{k(j)}}\right) ^{\underline{\beta }}}, \quad j\in \mathbb {N}. \end{aligned}$$

(26)

Since $1/\gamma -\underline{\beta }\ge 0$ in view of (20), we have

$$\begin{aligned} t^{\frac{1}{\gamma }-\underline{\beta }}\rho (t)\rightarrow 0 \quad \text{ as }\quad t\rightarrow 0^+. \end{aligned}$$

(27)

Then

$$\begin{aligned} \frac{t^{\frac{1}{\gamma }}\rho (t)}{\varphi (a_{k(j)}) \left( \displaystyle \frac{t}{a_{k(j)}}\right) ^{\underline{\beta }}} \rightarrow 0 \quad \text{ as }\quad t \rightarrow 0^+, \quad j\in \mathbb {N}. \end{aligned}$$

(28)

It follows from (26) and (28) that there exists $\tau _j\in (0,a_{k(j)}]$ such that

$$\begin{aligned} \frac{\tau _j^{\frac{1}{\gamma }}\rho (\tau _j)}{\varphi (a_{k(j)}) \left( \displaystyle \frac{\tau _j}{a_{k(j)}}\right) ^{\underline{\beta }}} =1, \quad 1\le \frac{t^{\frac{1}{\gamma }}\rho (t)}{\varphi (a_{k(j)}) \left( \displaystyle \frac{t}{a_{k(j)}}\right) ^{\underline{\beta }}}, \quad t\in [\tau _j,a_{k(j)}]. \end{aligned}$$

Hence

$$\begin{aligned} \varphi (a_{k(j)}) \left( \frac{\tau _j}{a_{k(j)}}\right) ^{\underline{\beta }} = \tau _j^{\frac{1}{\gamma }}\rho (\tau _j), \quad \varphi (a_{k(j)}) \left( \frac{t}{a_{k(j)}}\right) ^{\underline{\beta }} \le t^{\frac{1}{\gamma }}\rho (t), \ t \in [\tau _j, a_{k(j)}] . \end{aligned}$$

(29)

Let

$$\begin{aligned} \psi (t) := \varphi (a_{k(j)}) \left( \frac{t}{a_{k(j)}}\right) ^{\underline{\beta }} , \quad t \in [\tau _j, a_{k(j)}) . \end{aligned}$$

(30)

Then $\psi$ is continuous at $a_{k(j)}$. It follows from (17) with $s = a_{k(j)}$ and $x = t/a_{k(j)}$ that

$$\begin{aligned} \frac{\varphi (t)}{\varphi (a_{k(j)})} \le \left( \frac{t}{a_{k(j)}}\right) ^{\underline{\beta }}, \quad t\in [\tau _j,a_{k(j)}). \end{aligned}$$

Hence $\varphi (t) \le \psi (t)$ for $t \in [\tau _j, a_{k(j)}]$. Combining inequalities (25) with the above inequality, the inequality in (29) and definition (30), we get

$$\begin{aligned} \frac{t^{\frac{1}{\delta }}}{\rho (t)} \le \psi (t) \le t^{\frac{1}{\gamma }}\rho (t), \quad t \in [\tau _j, a_{k(j)}]. \end{aligned}$$

(31)

It follows from the first inequality in (31) with $t=\tau _j$ that

$$\begin{aligned} 1\le \frac{\psi (\tau _j) \left( \displaystyle \frac{\tau _j}{\tau _j}\right) ^{\overline{\beta }}}{\tau _j^{\frac{1}{\delta }}/\rho (\tau _j)}, \quad j\in \mathbb {N}. \end{aligned}$$

(32)

Since $\overline{\beta }-1/\delta \ge 0$ in view of (20), we have

$$\begin{aligned} t^{\overline{\beta }-\frac{1}{\delta }}\rho (t)\rightarrow 0 \quad \text{ as }\quad t\rightarrow 0^+. \end{aligned}$$

Then

$$\begin{aligned} \frac{\psi (\tau _j) \left( \displaystyle \frac{t}{\tau _j}\right) ^{\overline{\beta }}}{t^{\frac{1}{\delta }}/\rho (t)} \rightarrow 0 \quad \text{ as }\quad t\rightarrow 0^+, \quad j\in \mathbb {N}. \end{aligned}$$

(33)

It follows from (32) and (33) that there exists $\upsilon _j\in (0,\tau _j]$ such that

$$\begin{aligned} \frac{\psi (\tau _j) \displaystyle \left( \frac{\upsilon _j}{\tau _j}\right) ^{\overline{\beta }}}{\upsilon _j^{\frac{1}{\delta }}/\rho (\upsilon _j)} =1, \quad 1\le \frac{\psi (\tau _j) \displaystyle \left( \frac{t}{\tau _j}\right) ^{\overline{\beta }}}{t^{\frac{1}{\delta }}/\rho (t)}, \quad t\in [\upsilon _j,\tau _j]. \end{aligned}$$

Hence

$$\begin{aligned} \psi (\tau _j)\left( \frac{\upsilon _j}{\tau _j}\right) ^{\overline{\beta }} = \frac{\upsilon _j^{\frac{1}{\delta }}}{\rho (\upsilon _j)}, \quad \frac{t^{\frac{1}{\delta }}}{\rho (t)} \le \psi (\tau _j)\left( \frac{t}{\tau _j}\right) ^{\overline{\beta }}, \quad t \in [\upsilon _j, \tau _j]. \end{aligned}$$

(34)

Let

$$\begin{aligned} \psi (t) := \psi (\tau _j)\left( \frac{t}{\tau _j}\right) ^{\overline{\beta }} , \quad t \in [\upsilon _j, \tau _j) . \end{aligned}$$

(35)

Then $\psi$ is continuous at $\tau _j$. It follows from the above definition and the second inequality in (31) with $t = \tau _j$ that

$$\begin{aligned} \psi (t) \le \tau _j^{\frac{1}{\gamma }}\rho (\tau _j) \left( \frac{t}{\tau _j}\right) ^{\overline{\beta }} \le \tau _j^{\frac{1}{\gamma }}\rho (\tau _j) \left( \frac{t}{\tau _j}\right) ^{\frac{1}{\gamma }} = \rho (\tau _j) t^{\frac{1}{\gamma }} \le t^{\frac{1}{\gamma }}, \quad t \in [\upsilon _j, \tau _j]. \end{aligned}$$

Combining inequalities (31) with the above inequality, the inequality in (34) and definition (35), we get

$$\begin{aligned} \frac{t^{\frac{1}{\delta }}}{\rho (t)} \le \psi (t) \le t^{\frac{1}{\gamma }} , \quad t \in [\upsilon _j, a_{k(j)}]. \end{aligned}$$

(36)

It follows from definition (35), the equality in (34) and the first two inequalities in (25) that

$$\begin{aligned} \psi (\upsilon _j) = \frac{\upsilon _j^{\frac{1}{\delta }}}{\rho (\upsilon _j)} \le \upsilon _j^{\frac{1}{p}}\rho (\upsilon _j) \le \varphi (\upsilon _j), \end{aligned}$$

which implies that

$$\begin{aligned} 1\le \frac{\varphi (\upsilon _j)}{\psi (\upsilon _j) \displaystyle \left( \frac{\upsilon _j}{\upsilon _j}\right) ^{\underline{\beta }}}, \quad j\in \mathbb {N}. \end{aligned}$$

(37)

The last two inequalities in (25) and the asymptotic relation in (27) imply that

$$\begin{aligned} 0< \frac{\varphi (t)}{\psi (\upsilon _j) \displaystyle \left( \frac{t}{\upsilon _j}\right) ^{\underline{\beta }}} \le \frac{t^{\frac{1}{\gamma }} \rho (t)}{\psi (\upsilon _j) \displaystyle \left( \frac{t}{\upsilon _j}\right) ^{\underline{\beta }}} \rightarrow 0 \quad \text{ as }\quad t \rightarrow 0^+, \quad j\in \mathbb {N}. \end{aligned}$$

(38)

It follows from (37) and (38) that there exists $\varsigma _j\in (0,\upsilon _j]$ such that

$$\begin{aligned} \frac{\varphi (\varsigma _j)}{\psi (\upsilon _j) \displaystyle \left( \frac{\varsigma _j}{\upsilon _j}\right) ^{\underline{\beta }}} =1, \quad 1\le \frac{\varphi (t)}{\psi (\upsilon _j) \displaystyle \left( \frac{t}{\upsilon _j}\right) ^{\underline{\beta }}}, \quad t\in [\varsigma _j,\upsilon _j]. \end{aligned}$$

Hence

$$\begin{aligned} \psi (\upsilon _j)\left( \frac{\varsigma _j}{\upsilon _j}\right) ^{\underline{\beta }} = \varphi (\varsigma _j), \quad \psi (\upsilon _j)\left( \frac{t}{\upsilon _j}\right) ^{\underline{\beta }} \le \varphi (t) , \quad t \in [\varsigma _j, \upsilon _j] . \end{aligned}$$

(39)

Let

$$\begin{aligned} \psi (t) := \psi (\upsilon _j)\left( \frac{t}{\upsilon _j}\right) ^{\underline{\beta }} , \quad t \in [\varsigma _j, \upsilon _j). \end{aligned}$$

(40)

Then $\psi$ is continuous at $\upsilon _j$. It follows from the first inequality in (36) with $t = \upsilon _j$ that

$$\begin{aligned} \psi (t) \ge \frac{\upsilon _j^{\frac{1}{\delta }}}{\rho (\upsilon _j)}\left( \frac{t}{\upsilon _j}\right) ^{\underline{\beta }} \ge \frac{\upsilon _j^{\frac{1}{\delta }}}{\rho (\upsilon _j)}\left( \frac{t}{\upsilon _j}\right) ^{\frac{1}{\delta }} = \frac{t^{\frac{1}{\delta }}}{\rho (\upsilon _j)} \ge t^{\frac{1}{\delta }}, \quad t\in [\varsigma _j,\upsilon _j]. \end{aligned}$$

The above inequality, definition (40), inequality in (39), and the last two inequalities in (25) imply that

$$\begin{aligned} t^{\frac{1}{\delta }} \le \psi (t) \le \varphi (t) \le t^{\frac{1}{\gamma }}\rho (t), \quad t\in [\varsigma _j,\upsilon _j]. \end{aligned}$$

Combining these inequalities with inequalities (36), we get

$$\begin{aligned} t^{\frac{1}{\delta }} \le \psi (t) \le t^{\frac{1}{\gamma }} , \quad t \in [\varsigma _j, a_{k(j)}]. \end{aligned}$$

(41)

Let

$$\begin{aligned} k(j + 1):= 1 + \min \left\{ k \in \mathbb {N}: \ a_k \le \varsigma _j\right\} \end{aligned}$$

and

$$\begin{aligned} \psi (t) := \varphi (t), \quad t \in [a_{k(j + 1)}, \varsigma _j). \end{aligned}$$

(42)

It follows from the equality in (39) and definition (40) that $\psi (\varsigma _j) = \varphi (\varsigma _j)$. Hence $\psi$ is continuous at $\varsigma _j$. Inequalities (25), (41), and definition (42) imply that

$$\begin{aligned} t^{\frac{1}{\delta }} \le \psi (t) \le t^{\frac{1}{\gamma }}, \quad t \in [a_{k(j + 1)}, a_{k(j)}]. \end{aligned}$$

Above, we had $j = 1$. Repeating the same procedure for $j = 2, 3, \dots$, we get a continuous function $\psi : (0, 1] \rightarrow (0, \infty )$ such that

$$\begin{aligned} \psi (t) = \varphi (t), \quad t \in [a_{k(j)}, a_{k(j) - 1}] , \quad j \in \mathbb {N} \end{aligned}$$

(43)

(note that $[a_{k(j)},a_{k(j)-1}]\subset [a_{k(j)},\varsigma _{j-1}]$), and

$$\begin{aligned} t^{\frac{1}{\delta }} \le \psi (t) \le t^{\frac{1}{\gamma }}, \quad t \in (0, 1]. \end{aligned}$$

(44)

It follows from definition (30) and the equality in (29) that

$$\begin{aligned} \psi (\tau _j) = \tau _j^{1/\gamma }\rho (\tau _j), \quad j\in \mathbb {N}. \end{aligned}$$

(45)

Analogously, it follows from definition (35) and the equality in (34) that

$$\begin{aligned} \psi (\upsilon _j) = \frac{\upsilon _j^{1/\delta }}{\rho (\upsilon _j)}\,, \quad j \in \mathbb {N}. \end{aligned}$$

(46)

Finally, there is a partition $\cup _{l \in \mathbb {N}} [\eta _{l + 1}, \eta _l) = (0, 1)$ such that

$$\begin{aligned} \psi (t) = \textrm{const}_l\, t^{\beta _l} , \quad t \in [\eta _{l + 1}, \eta _l), \quad \beta _l \in \left\{ \underline{\beta },\, \overline{\beta }, 1/p\right\} ,\, \quad l \in \mathbb {N} . \end{aligned}$$

(47)

Set $\psi (0) = 0$. It follows from (47) and the continuity of $\psi$ that $\psi$ is quasiconcave and (16) and (17) remain true with $\psi$ in place of $\varphi$. Hence

$$\begin{aligned} M_\psi (x)&= \sup _{0<s\le 1/x}\frac{\psi (xs)}{\psi (s)} \le x^{\overline{\beta }}, \quad x\in (1,\infty ),\end{aligned}$$

(48)

$$\begin{aligned} M_\psi (x)&= \sup _{0<s\le 1}\frac{\psi (xs)}{\psi (s)} \le x^{\underline{\beta }}, \quad x\in (0,1). \end{aligned}$$

(49)

Let the sequences $b_k=a_k/k$ and $c_k=b_k/k$ be as in the proof of Theorem 3. Take any $x>1$ and choose $j\in \mathbb {N}$ so that $k(j) - 1>x$. Then $xc_{k(j) - 1}\in [c_{k(j) - 1},b_{k(j) - 1})$ and it follows from (43) and (12) that

$$\begin{aligned} M_\psi (x) = \sup _{0<s\le 1/x}\frac{\psi (xs)}{\psi (s)} \ge \frac{\psi (xc_{k(j) - 1})}{\psi (c_{k(j) - 1})} =x^{\overline{\beta }}. \end{aligned}$$

Hence $M_\psi (x)=x^{\overline{\beta }}$ (see (48)), and

$$\begin{aligned} \overline{\alpha }(\psi ) = \inf _{x>1}\frac{\log M_\psi (x)}{\log x}=\overline{\beta }. \end{aligned}$$

Similarly, take any $x<1$ and choose $j\in \mathbb {N}$ so that $1/(k(j) - 1)<x$. Then $xa_{k(j) - 1}\in [b_{k(j) - 1},a_{k(j) - 1})$ and it follows from (43) and (10) that

$$\begin{aligned} M_\psi (x) = \sup _{0<s\le 1}\frac{\psi (xs)}{\psi (s)} \ge \frac{\psi (xa_{k(j) - 1})}{\psi (a_{k(j) - 1})} = x^{\underline{\beta }}. \end{aligned}$$

Hence $M_\psi (x)=x^{\underline{\beta }}$ (see (49)), and

$$\begin{aligned} \underline{\alpha }(\psi ) = \sup _{0<x<1}\frac{\log M_\psi (x)}{\log x}=\underline{\beta }. \end{aligned}$$

So, the first two equalities in (21) hold.

Estimates (44) imply

$$\begin{aligned} \limsup _{t\rightarrow 0^+}\frac{\log t}{\log \psi (t)} \le \gamma , \quad \liminf _{t\rightarrow 0^+}\frac{\log t}{\log \psi (t)} \ge \delta . \end{aligned}$$

(50)

Since $0<a_{k(j+1)}<\upsilon _j\le \tau _j\le a_{k(j)}$ for all $j\in \mathbb {N}$ and $a_k\rightarrow 0$ as $k\rightarrow \infty$, we conclude that $\upsilon _j\rightarrow 0$ and $\tau _j\rightarrow 0$ as $j\rightarrow \infty$. Finally, it follows from (45)–(46), (24), and (19) that

$$\begin{aligned} \limsup _{t\rightarrow 0^+} \frac{\log t}{\log \psi (t)}&\ge \lim _{j \rightarrow \infty } \frac{\log \tau _j}{\log \psi (\tau _j)} = \lim _{j \rightarrow \infty } \frac{\log \tau _j}{\frac{1}{\gamma }\log \tau _j + \log \rho (\tau _j)} \\&= \lim _{j \rightarrow \infty } \frac{1}{\frac{1}{\gamma } - \kappa \frac{\log \log (e - 1 + 1/\tau _j)}{\log \tau _j}} = \gamma , \\ \liminf _{t\rightarrow 0^+}\frac{\log t}{\log \psi (t)}&\le \lim _{j \rightarrow \infty } \frac{\log \upsilon _j}{\log \psi (\upsilon _j)} = \lim _{j \rightarrow \infty } \frac{\log \upsilon _j}{\frac{1}{\delta }\log \upsilon _j + \log \rho (\upsilon _j)} \\&= \lim _{j \rightarrow \infty } \frac{1}{\frac{1}{\delta } - \kappa \frac{\log \log (e - 1 + 1/\upsilon _j)}{\log \upsilon _j}} = \delta . \end{aligned}$$

Combining the above inequalities with (50), we arrive at

$$\begin{aligned} \limsup _{t\rightarrow 0^+}\frac{\log t}{\log \psi (t)} = \gamma , \quad \liminf _{t\rightarrow 0^+}\frac{\log t}{\log \psi (t)} = \delta , \end{aligned}$$

which completes the proof of the last two equalities in (21). $\square$

Corollary 6

Let $\underline{\beta }, \overline{\beta } \in [0, 1]$ and $\gamma , \delta \in [1, \infty ]$ satisfy $\underline{\beta }\le 1/\gamma < 1/\delta \le \overline{\beta }$. Then there exists a quasiconcave function $\psi :[0,1]\rightarrow [0,\infty )$ such that for every r.i. space X with the fundamental function equivalent to $\psi$ one has

$$\begin{aligned} \underline{\beta }{}_X=\underline{\beta }, \quad \overline{\beta }{}_X = \overline{\beta }, \quad \gamma _X= \gamma , \quad \delta _X = \delta . \end{aligned}$$

This result follows from Theorem 5 and (3).

Theorem 1 immediately follows form Corollaries 4 and 6.

3.4 Inclusion indices in terms of Lorentz and Marcinkiewicz spaces

The following fact was observed in the case of r.i. spaces with nontrivial Boyd indices in [4, formulae (3.5)–(3.6) and (6.1)–(6.2)].

Lemma 7

Let X be an r.i. space. If $1<\delta _X\le \gamma _X<\infty$, then

$$\begin{aligned} \gamma _X = \inf \left\{ p\in (1,\infty ): L^{p,\infty }\hookrightarrow X\right\} , \quad \delta _X = \sup \left\{ p\in (1,\infty ):X\hookrightarrow L^{p,1}\right\} . \end{aligned}$$

Proof

Since $1<\delta _X\le \gamma _X<\infty$, we have

$$\begin{aligned} \gamma _X = \inf \left\{ p\in (1,\infty ): L^p\hookrightarrow X\right\} , \quad \delta _X = \sup \left\{ p\in (1,\infty ):X\hookrightarrow L^p\right\} . \end{aligned}$$

Set

$$\begin{aligned} \gamma _X^\infty := \inf \left\{ p\in (1,\infty ): L^{p,\infty }\hookrightarrow X\right\} , \quad \delta _X^1:= \sup \left\{ p\in (1,\infty ):X\hookrightarrow L^{p,1}\right\} . \end{aligned}$$

Since $L^{p,1}\hookrightarrow L^p\hookrightarrow L^{p,\infty }$ for every $p\in (1,\infty )$, we have

$$\begin{aligned} \gamma _X\le \gamma _X^\infty , \quad \delta _X^1\le \delta _X. \end{aligned}$$

If $\gamma _X<\gamma _X^\infty$, then there exist $p_1,p_2$ such that $\gamma _X<p_1<p_2<\gamma _X^\infty$. It follows from the definitions of $\gamma _X$ and $\gamma _X^\infty$ that $L^{p_1}\hookrightarrow X$ and $L^{p_2,\infty }\not \hookrightarrow X$. Since $L^{p_2,\infty }\hookrightarrow L^{p_1}$ (see, e.g., [2, p. 217]), the absence of inclusion of $L^{p_2,\infty }$ into X is impossible. Then $\gamma _X\ge \gamma _X^\infty$. The proof of $\delta _X^1\ge \delta _X$ is similar. Thus $\gamma _X=\gamma _X^\infty$ and $\delta _X=\delta _X^1$. $\square$

3.5 Proof of Corollary 2

It follows from Corollary 4 with $\underline{\beta }=0$ and $\overline{\beta }=1$ that there exists an r.i. space X such that its inclusion indices are $\gamma _X=\delta _X=p$ and its Zippin indices are trivial, that is, $\underline{\beta }{}_X=0$ and $\overline{\beta }{}_X=1$. Since $\gamma _X=\delta _X=p$, in view of Lemma 7, for every $\varepsilon \in (0,p-1)$, one has $L^{p+\varepsilon ,\infty } \hookrightarrow X \hookrightarrow L^{p-\varepsilon ,1}$, which completes the proof. $\square$