Orlicz spaces associated to a quasi-Banach function space: applications to vector measures and interpolation

Abstract

The Orlicz spaces $X^{\varPhi }$ associated to a quasi-Banach function space X are defined by replacing the role of the space $L^1$ by X in the classical construction of Orlicz spaces. Given a vector measure m,  we can apply this construction to the spaces $L^1_w(m),$ $L^1(m)$ and $L^1(\Vert m\Vert )$ of integrable functions (in the weak, strong and Choquet sense, respectively) in order to obtain the known Orlicz spaces $L^{\varPhi }_w(m)$ and $L^{\varPhi }(m)$ and the new ones $L^{\varPhi }(\Vert m\Vert ).$ Therefore, we are providing a framework where dealing with different kind of Orlicz spaces in a unified way. Some applications to complex interpolation are also given.

1 Introduction

The Banach lattice $L^1(m)$ of integrable functions with respect to a vector measure m (defined on a $\sigma$-algebra of sets and with values in a Banach space) has been systematically studied during the last 30 years and it has proved to be a efficient tool to describe the optimal domain of operators between Banach function spaces (see [18] and the references therein). The Orlicz spaces $L^{\varPhi }(m)$ and $L^{\varPhi }_w(m)$ associated to m were introduced in [8] and they have recently shown in [5] their utility in order to characterize compactness in $L^1(m).$

On the other hand, the quasi-Banach lattice $L^1(\Vert m\Vert )$ of integrable functions (in the Choquet sense) with respect to the semivariation of m was introduced in [9]. Some properties of this space and their corresponding $L^p(\Vert m\Vert )$ with $p>1$ have been obtained, but in order to achieve compactness results in $L^1(\Vert m\Vert )$ we would need to dispose of certain Orlicz spaces related to $L^1(\Vert m\Vert ).$

In [10] some generalized Orlicz spaces $X_{\varPhi }$ have been obtained by replacing the role of the space $L^1$ by a Banach function space X in the classical construction of Orlicz spaces. Moreover, the spaces X they consider are allways supposed to possess the $\sigma$-Fatou property. However, these Orlicz spaces do not cover our situation since:

• the space $L^1(\Vert m\Vert )$ is only a quasi-Banach function space, and

• in most of the time $L^1(m)$ lacks the $\sigma$-Fatou property.

Thus, the purpose of this work is to provide a construction of certain Orlicz spaces $X^{\varPhi }$ valid for the case of X being an arbitrary quasi-Banach function space (in general without the $\sigma$-Fatou property), with the underlying idea that it can be applied simultaneously to the spaces $L^1(\Vert m\Vert )$ and $L^1(m)$ among others. In a subsequent paper [6] we shall employ these Orlicz spaces $L^{1}(\Vert m\Vert )^{\varPhi }$ and their main properties here derived in order to study compactness in $L^1(\Vert m\Vert ).$

The organization of the paper goes as follows: Section 2 contains the preliminaries which we will need later. Section 3 contains a discussion of completeness in the quasi-normed context without any additional hypothesis on $\sigma$-Fatou property. Section 4 is devoted to introduce the Orlicz spaces $X^{\varPhi }$ associated to a quasi-Banach function space X and obtain their main properties. In Sect. 5, we show that the construction of the previous section allows to capture the Orlicz spaces associated to a vector measure and we take advantage of its generality to introduce the Orlicz spaces associated to its semivariation. Finally, in Sect. 6 we present some applications of this theory to compute their complex interpolation spaces.

2 Preliminaries

Throughout this paper, we shall always assume that $\varOmega$ is a nonempty set, $\varSigma$ is a $\sigma$-algebra of subsets of $\varOmega ,$ $\mu$ is a finite positive measure defined on $\varSigma$ and $L^0(\mu )$ is the space of ($\mu$-a.e. equivalence classes of) measurable functions $f:\varOmega \rightarrow {\mathbb {R}}$ equipped with the topology of convergence in measure.

Recall that a quasi-normed space is any real vector space X equipped with a quasi-norm, that is, a function $\Vert \cdot \Vert _{X} : X\rightarrow [0,\infty )$ which satisfies the following axioms:

• (Q1) $\Vert x\Vert _{X}=0$ if and only if $x=0.$

• (Q2) $\Vert \alpha x\Vert _{X} = |\alpha | \Vert x\Vert _{X},$ for $\alpha \in {\mathbb {R}}$ and $x\in X.$

• (Q3) There exists $K\ge 1$ such that $\left\| x_1+x_2\right\| _{X} \le K\left( \left\| x_1\right\| _{X} + \left\| x_2\right\| _{X}\right) ,$ for all $x_1,x_2\in X.$

The constant K in (Q3) is called a quasi-triangle constant of X. Of course if we can take $K=1,$ then $\Vert \cdot \Vert _{X}$ is a norm and X is a normed space. A quasi-normed function space over $\mu$ is any quasi-normed space X satisfying the following properties:

• (a) X is an ideal in $L^0(\mu )$ and a quasi-normed lattice with respect to the $\mu$-a.e. order, that is, if $f\in L^0(\mu ),$ $g\in X$ and $|f|\le |g|$ $\mu$-a.e., then $f\in X$ and $\Vert f\Vert _{X} \le \Vert g\Vert _{X}.$

• (b) The characteristic function of $\varOmega ,$ $\chi _{\varOmega },$ belongs to X.

If, in addition, the quasi-norm $\Vert \cdot \Vert _{X}$ happens to be a norm, then X is called a normed function space. Note that, with this definition, any quasi-normed function space over $\mu$ is continuously embedded into $L^0(\mu ),$ as it is proved in [18, Proposition 2.2].

Remark 1

Many of the results that we will present in this paper are true if we assume that the measure space $\left( \varOmega ,\varSigma ,\mu \right)$ is $\sigma$-finite. In this case, the previous condition (b) must be replaced by

• (b’) The characteristic functions $\chi _{A}$ belong to X for all $A\in \varSigma$ such that $\mu (A)<\infty .$

Nevertheless we prefer to present the results in the finite case for clarity and simplicity in the proofs.

We say that a quasi-normed function space X has the $\sigma$-Fatou property if for any positive increasing sequence $(f_n)_n$ in X with $\displaystyle \sup _{n}\Vert f_n\Vert _{X}<\infty$ and converging pointwise $\mu$-a.e. to a function f,  then $f\in X$ and $\displaystyle \Vert f\Vert _{X} = \sup _{n}\Vert f_n\Vert _{X}.$ And a quasi-normed function space X is said to be $\sigma$-order continuous if for any positive increasing sequence $(f_n)_n$ in X converging pointwise $\mu$-a.e. to a function $f\in X,$ then $\left\| f-f_n\right\| _{X}\rightarrow 0.$

A complete quasi-normed function space is called a quasi-Banach function space (briefly q-B.f.s.). If, in addition, the quasi-norm happens to be a norm, then X is called a Banach function space (briefly B.f.s.). It is known that if a quasi-normed function space has the $\sigma$-Fatou property, then it is complete and hence a q-B.f.s. (see [18, Proposition 2.35]) and that inclusions between q-B.f.s. are automatically continuous (see [18, Lemma 2.7]).

Given a countably additive vector measure $m:\varSigma \rightarrow Y$ with values in a real Banach space Y,  there are several ways of constructing q-B.f.s. of integrable functions. Let us recall them briefly. The semivariation of m is the finite subadditive set function defined on $\varSigma$ by

$$\begin{aligned} \Vert m\Vert (A):=\sup \left\{ \left| \langle m, y^*\rangle \right| (A) : y^*\in B_{Y^*} \right\} , \end{aligned}$$

where $\left| \langle m, y^*\rangle \right|$ denotes the variation of the scalar measure $\langle m, y^*\rangle :\varSigma \rightarrow {\mathbb {R}}$ given by $\langle m, y^*\rangle (A):= \langle m(A), y^*\rangle$ for all $A\in \varSigma ,$ and $B_{Y^*}$ is the unit ball of $Y^*,$ the dual of Y. A set $A\in \varSigma$ is called m-null if $\Vert m\Vert (A)=0.$ A measure $\mu := \left| \langle m, y^*\rangle \right| ,$ where $y^*\in B_{Y^*},$ that is equivalent to m (in the sense that $\Vert m\Vert (A)\rightarrow 0$ if and only if $\mu (A)\rightarrow 0$) is called a Rybakov control measure for m. Such a measure always exists (see [7, Theorem 2, p.268]). Let $L^0(m)$ be the space of (m-a.e. equivalence classes of) measurable functions $f:\varOmega \rightarrow {\mathbb {R}}.$ Thus, $L^0(m)$ and $L^0(\mu )$ are just the same whenever $\mu$ is a Rybakov control measure for m.

A measurable function $f:\varOmega \rightarrow {\mathbb {R}}$ is called weakly integrable (with respect to m) if f is integrable with respect to $|\langle m, y^*\rangle |$ for all $y^*\in Y^*.$ A weakly integrable function f is said to be integrable (with respect to m) if, for each $A\in \varSigma$ there exists an element (necessarily unique) $\displaystyle \int _A f \, dm \in Y,$ satisfying

$$\begin{aligned} \left\langle \int _A f \, dm, y^* \right\rangle = \int _A f \, d\langle m, y^*\rangle , \quad y^*\in Y^*. \end{aligned}$$

Given a measurable function $f:\varOmega \rightarrow {\mathbb {R}},$ we shall also consider its distribution function (with respect to the semivariation of the vector measure m)

$$\begin{aligned} \Vert m\Vert _{f}: t\in [0,\infty )\rightarrow \Vert m\Vert _f(t):=\Vert m\Vert \left( \left[ |f|>t\right] \right) \in [0,\infty ), \end{aligned}$$

where $\left[ |f|>t\right] := \left\{ w\in \varOmega : |f(w)|>t \right\} .$ This distribution function is bounded, non-increasing and right-continuous.

Let $L^1_w(m)$ be the space of all (m-a.e. equivalence classes of) weakly integrable functions, $L^1(m)$ the space of all (m-a.e equivalence classes of) integrable functions and $L^1(\Vert m\Vert )$ the space of all (m-a.e. equivalence classes of) measurable functions f such that its distribution function $\Vert m\Vert _{f}$ is Lebesgue integrable in $(0,\infty ).$ Letting $\mu$ be any Rybakov control measure for m,  we have that $L^1_w(m)$ becomes a B.f.s. over $\mu$ with the $\sigma$-Fatou property when endowed with the norm

$$\begin{aligned} \Vert f\Vert _{L^1_w(m)} := \sup \left\{ \int _{\varOmega } |f| \, d|\langle m, y^*\rangle | : y^*\in B_{Y^*} \right\} . \end{aligned}$$

Moreover, $L^1(m)$ is a closed $\sigma$-order continuous ideal of $L^1_w(m).$ In fact, it is the closure of ${\mathscr {S}}(\varSigma ),$ the space of simple functions supported on $\varSigma .$ Thus, $L^1(m)$ is a $\sigma$-order continuous B.f.s. over $\mu$ endowed with same norm (see [18, Theorem 3.7] and [18, p.138])). It is worth noting that space $L^1(m)$ does not generally have the $\sigma$-Fatou property. In fact, if $L^1(m) \ne L^1_w(m),$ then $L^1(m)$ does not have the $\sigma$-Fatou property. See [2] for details.

On the other hand, $L^1(\Vert m\Vert )$ equipped with the quasi-norm

$$\begin{aligned} \Vert f\Vert _{L^1(\Vert m\Vert )}:= \int _{0}^{\infty } \Vert m\Vert _{f}(t)\, dt. \end{aligned}$$

is a q-B.f.s. over $\mu$ with the $\sigma$-Fatou property (see [4, Proposition 3.1]) and it is also $\sigma$-order continuous (see [4, Proposition 3.6]). We will denote by $L^{\infty }(m)$ the B.f.s. of all (m-a.e. equivalence classes of) essentially bounded functions equipped with the essential sup-norm.

3 Completeness of quasi-normed lattices

In this section we present several characterizations of completeness which will be needed later. We begin by recalling one of them valid for general quasi-normed spaces (see [10, Theorem 1.1]).

Theorem 1

Let X be a quasi-normed space with a quasi-triangle constant K. The following conditions are equivalent:

• (i) X is complete.

• (ii) For every sequence $(x_n)_n\subseteq X$ such that $\displaystyle \sum _{n=1}^{\infty }K^n\Vert x_n\Vert _{X}<\infty$ we have $\displaystyle \sum _{n=1}^{\infty } x_n \in X.$ In this case, the inequality $\displaystyle \left\| \sum _{n=1}^{\infty } x_n \right\| _{X} \le K \sum _{n=1}^{\infty } K^n \Vert x_n\Vert _{X}$ holds.

The next result is a version of Amemiya’s Theorem ( [10, Theorem 2, p.290]) for quasi-normed lattices.

Theorem 2

Let X be a quasi-normed lattice. The following conditions are equivalent:

• (i) X is complete.

• (ii) For any positive increasing Cauchy sequence $(x_n)_n$ in X there exists $\displaystyle \sup _{n} x_n \in X.$

Proof

(i) $\Rightarrow$ (ii) is evident because the limit of increasing convergent sequences in a quasi-normed lattice is always its supremum.

(ii) $\Rightarrow$ (i) Let $(x_n)_n$ be a positive increasing Cauchy sequence in X. It is sufficient to prove that $(x_n)_n$ is convergent in X for X being complete (see, for example [1, Theorem 16.1]). By hypothesis, there exists $x:=\displaystyle \sup _{n} x_n \in X.$ We have to prove that $(x_n)_n$ converges to x and for this it is enough the convergence of a subsequence of $(x_n)_n.$ So, let us take a subsequence of $(x_n)_n,$ that we still denote by $(x_n)_n,$ such that $\Vert x_{n+1}-x_n\Vert _{X}\le \displaystyle \frac{1}{K^n n^3},$ for all $n\in {\mathbb {N}}$ where K is a quasi-triangle constant of X. Thus, the sequence $y_n:=\displaystyle \sum _{i=1}^{n} i (x_{i+1}-x_i)$ is positive, increasing and Cauchy. Indeed, given $m>n,$ we have

$$\begin{aligned} \Vert y_m-y_n\Vert _{X} \le \sum _{i=n+1}^{m} i K^{i-n} \Vert x_{i+1}-x_{i}\Vert _{X} \le \frac{1}{K^n}\sum_{i=n+1}^{m}\frac{1}{i^2}. \end{aligned}$$

Applying (ii) again, we deduce that there exists $y:=\displaystyle \sup _{n} y_n\in X.$ Moreover, given $n\in {\mathbb {N}},$ we have

$$\begin{aligned} n(x-x_n)= \,\,&{} n \left( \sup _{m>n} x_{m+1} - x_n\right) \ = \ n \sup _{m>n} (x_{m+1}-x_n)=\,\, &{} n \sup _{m>n} \sum _{i=n}^{m} (x_{i+1}-x_i) \ \le \ \sup _{m>n} y_n \ = \ y. \end{aligned}$$

Therefore, $0\le x-x_n \le \displaystyle \frac{1}{n}y$ and hence $\Vert x-x_n\Vert _{X}\le \displaystyle \frac{1}{n}\Vert y\Vert _{X}\rightarrow 0.$ $\square$

Applying Theorem 2 to the sequence of partial sums of a given sequence, we see that completeness in quasi-normed lattices can still be characterized by a Riesz-Fischer type property.

Corollary 1

Let X be a quasi-normed lattice with a quasi-triangle constant K. The following conditions are equivalent:

• (i) X is complete.

• (ii) For every positive sequence $(x_n)_n\subseteq X$ such that $\displaystyle \sum _{n=1}^{\infty }K^n\Vert x_n\Vert _{X}<\infty$ there exists $\displaystyle \sup _{n} \sum _{i=1}^{n} x_i \in X.$

4 Orlicz spaces $X^{\varPhi }$

In this section we introduce the Orlicz spaces $X^{\varPhi }$ associated to a quasi-Banach function space X and a Young function $\varPhi$ and obtain their main properties.

Recall that a Young function is any function $\varPhi :[0,\infty )\rightarrow [0,\infty )$ which is strictly increasing, continuous, convex, $\varPhi (0)=0$ and $\displaystyle \lim _{t\rightarrow \infty }\varPhi (t)=\infty .$ A Young function $\varPhi$ satisfies the following useful inequalities (which we shall use without explicit mention) for all $t\ge 0$:

$$\begin{aligned} \left\{ \begin{array}{ccl} \varPhi (\alpha t) \le \alpha \, \varPhi (t) & \quad {\text{ if }} & 0\le \alpha \le 1, \\ \varPhi (\alpha t) \ge \alpha \, \varPhi (t) &\quad {\text{ if }} & \alpha \ge 1. \end{array} \right. \end{aligned}$$

In particular, from the second of the previous inequalities it follows that for all $t_0>0$ there exists $C>0$ such that $\varPhi (t)\ge C t$ for all $t\ge t_0.$ For a given $t_0>0,$ just take $\displaystyle C:=\frac{\varPhi (t_0)}{t_0}>0$ and observe that $\displaystyle \varPhi (t) = \varPhi \left( t_0\frac{t}{t_0}\right) \ge \frac{t}{t_0}\varPhi (t_0) = C t$ for all $t\ge t_0.$

Moreover, it is easy to prove using the convexity of $\varPhi$ that

$$\begin{aligned} \varPhi \left( \sum _{n=1}^{N} t_n \right) \le \sum _{n=1}^{N} \frac{1}{2^n \alpha ^n} \varPhi (2^n \alpha ^n t_n) \end{aligned}$$

(1)

for all $N\in {\mathbb {N}},$ $\alpha \ge 1$ and $t_1,\dots ,t_N\ge 0.$

A Young function $\varPhi$ has the $\varDelta _2$-property, written $\varPhi \in \varDelta _2,$ if there exists a constant $C>1$ such that $\varPhi (2t)\le C\varPhi (t)$ for all $t\ge 0.$ Equivalently, $\varPhi \in \varDelta _2$ if for any $c>1$ there exists $C>1$ such that $\varPhi (ct)\le C\varPhi (t),$ for all $t\ge 0.$

Definition 1

Let $\varPhi$ be a Young function. Given a quasi-normed function space X over $\mu ,$ the corresponding (generalized) Orlicz class $\widetilde{X}^{\varPhi }$ is defined as the following set of ($\mu$-a.e. equivalence classes of) measurable functions:

$$\begin{aligned} \widetilde{X}^{\varPhi } := \left\{ f\in L^0(\mu ) : \varPhi (|f|)\in X \right\} . \end{aligned}$$

Proposition 1

Let $\varPhi$ be a Young function and X be a quasi-normed function space over $\mu .$ Then, $\widetilde{X}^{\varPhi }$ is a solid convex set in $L^0(\mu ).$ Moreover, $\widetilde{X}^{\varPhi } \subseteq X.$

Proof

Let $f,g\in \widetilde{X}^{\varPhi }$ and $0\le \alpha \le 1.$ According to the convexity and monotonicity properties of $\varPhi$ we have $\varPhi (|\alpha f + (1-\alpha )g|) \le \alpha \varPhi (|f|) + (1-\alpha )\varPhi (|g|) \in X.$ The ideal property of X yields $\varPhi (|\alpha f + (1-\alpha )g|)\in X$ which means that $\begin{aligned} \alpha f + (1-\alpha )g \in \widetilde{X}^{\varPhi } \end{aligned}$ and proves the convexity of $\widetilde{X}^{\varPhi }.$ Clearly, $\widetilde{X}^{\varPhi }$ is solid, since $|h|\le |f|$ implies that $\varPhi (|h|)\le \varPhi (|f|)\in X,$ for any $h\in L^0(\mu ).$ Moreover, since $\varPhi$ is a convex function, there exists $C>0$ such that $\varPhi (t)\ge C t,$ for all $t > 1.$ Thus, for all $f\in \widetilde{X}^{\varPhi },$

$$\begin{aligned} |f| = |f|\chi _{\left[ |f|>1\right] } + |f|\chi _{\left[ |f|\le 1\right] } \le \frac{1}{C}\varPhi \left( |f|\chi _{\left[ |f| > 1\right] }\right) + \chi _{\varOmega } \le \frac{1}{C}\varPhi \left( |f|\right) + \chi _{\varOmega } \in X, \end{aligned}$$

which gives $f\in X.$ $\square$

Definition 2

Let $\varPhi$ be a Young function. Given a quasi-normed function space X over $\mu ,$ the corresponding (generalized) Orlicz space $X^{\varPhi }$ is defined as the following set of ($\mu$-a.e. equivalence classes of) measurable functions:

$$\begin{aligned} X^{\varPhi }:= \left\{ f\in L^0(\mu ) : \exists \, c >0 : \frac{|f|}{c}\in \widetilde{X}^{\varPhi } \right\} . \end{aligned}$$

Proposition 2

Let $\varPhi$ be a Young function and X be a quasi-normed function space over $\mu .$ Then, $X^{\varPhi }$ is a linear space, an ideal in $L^0(\mu )$ and $\widetilde{X}^{\varPhi } \subseteq X^{\varPhi }\subseteq X.$

Proof

Let $f,g\in X^{\varPhi }$ and $\alpha \in {\mathbb {R}}.$ Then, there exist $c_1, c_2>0$ such that $\displaystyle \frac{|f|}{c_1}, \displaystyle \frac{|g|}{c_2}\in \widetilde{X}^{\varPhi }.$ Setting $c:=\max \{c_1,c_2\}$ and using the convexity of $\widetilde{X}^{\varPhi }$ we have

$$\begin{aligned} \frac{|f+g|}{2c} \le \frac{1}{2}\frac{|f|}{c} + \frac{1}{2}\frac{|g|}{c} \le \frac{1}{2}\frac{|f|}{c_1} + \frac{1}{2}\frac{|g|}{c_2} \in \widetilde{X}^{\varPhi } \end{aligned}$$

and hence $\displaystyle \frac{|f+g|}{2c} \in \widetilde{X}^{\varPhi }$ since $\widetilde{X}^{\varPhi }$ is solid, which proves that $f+g\in X^{\varPhi }.$ Note that this also implies that $nf\in X^{\varPhi }$ for any $n\in {\mathbb {N}}.$ Taking $n_0\in {\mathbb {N}}$ such that $|\alpha |\le n_0,$ it follows that there exists $c_0>0$ such that $\displaystyle \frac{|\alpha f|}{c_0}\le \frac{n_0|f|}{c_0} \in \widetilde{X}^{\varPhi },$ which yields $\displaystyle \frac{|\alpha f|}{c_0}\in \widetilde{X}^{\varPhi }$ and so $\alpha f\in X^{\varPhi }.$

It is evident that $\widetilde{X}^{\varPhi } \subseteq X^{\varPhi }$ and $X^{\varPhi }$ inherits the ideal property from $\widetilde{X}^{\varPhi },$ since $|h|\le |f|$ implies that $\displaystyle \frac{|h|}{c_1} \le \frac{|f|}{c_1}\in \widetilde{X}^{\varPhi }$ for any $h\in L^0(\mu ).$ Moreover, taking into account Proposition 1, we have $\displaystyle \frac{|f|}{c_1}\in \widetilde{X}^{\varPhi } \subseteq X$ and so $f\in X$ which proves that $X^{\varPhi } \subseteq X.$ $\square$

Definition 3

Let $\varPhi$ be a Young function and X be a quasi-normed function space over $\mu .$ Given $f\in X^{\varPhi },$ we define

$$\begin{aligned} \Vert f\Vert _{X^{\varPhi }} := \inf \left\{ k>0 : \frac{|f|}{k}\in \widetilde{X}^{\varPhi } {\text{ with }} \left\| \varPhi \left( \frac{|f|}{k}\right) \right\| _{X}\le 1 \right\} . \end{aligned}$$

The functional $\Vert \cdot \Vert _{X^{\varPhi }}$ in $X^{\varPhi }$ is called the Luxemburg quasi-norm.

Proposition 3

Let $\varPhi$ be a Young function and X be a quasi-normed function space (respectively, normed function space) over $\mu .$ Then, $\Vert \cdot \Vert _{X^{\varPhi }}$ is a quasi-norm (respectively, norm) in $X^{\varPhi }.$ Moreover, $X^{\varPhi }$ equipped with the Luxemburg quasi-norm, is a quasi-normed (respectively, normed) function space over $\mu .$

Proof

First, note that $\Vert \cdot \Vert _{X^{\varPhi }}:X^{\varPhi }\rightarrow [0,\infty ).$ Given $f\in X^{\varPhi },$ there exists $c>0$ such that $\varPhi \displaystyle \left( \frac{|f|}{c}\right) \in X.$ Let $M:=\displaystyle \left\| \varPhi \left( \frac{|f|}{c}\right) \right\| _X<\infty .$ On the one hand, if $M\le 1$ then $\Vert f\Vert _{X^{\varPhi }}\le c < \infty .$ On the other hand, if $M>1$ then $\displaystyle \varPhi \left( \frac{|f|}{Mc}\right) \le \frac{1}{M}\varPhi \left( \frac{|f|}{c}\right) \in X$ and so $\displaystyle \left\| \varPhi \left( \frac{|f|}{Mc}\right) \right\| _X \le \frac{1}{M} \left\| \varPhi \left( \frac{|f|}{c}\right) \right\| =1,$ which implies that $\Vert f\Vert _{X^{\varPhi }}\le Mc < \infty .$

If $f=0,$ then $\displaystyle \left\| \varPhi \left( \frac{|f|}{c}\right) \right\| _X=0\le 1$ for all $c>0$ and so $\Vert f\Vert _{X^{\varPhi }}=0.$ Now, suppose that $\Vert f\Vert _{X^{\varPhi }}=0$ and that $\mu \left( \left[ f\ne 0\right] \right) >0,$ that is, $\displaystyle \left\| \varPhi \left( \frac{|f|}{c}\right) \right\| _X\le 1$ for all $c>0$ and there exist $\varepsilon >0$ and $A\in \varSigma$ such that $\mu (A)>0$ and $|f|\chi _A \ge \varepsilon \chi _A.$ Given $c>0,$ we have $\displaystyle \varPhi \left( \frac{\varepsilon }{c}\right) \chi _A \le \varPhi \left( \frac{|f|\chi _A}{c}\right) \le \varPhi \left( \frac{|f|}{c}\right) .$ Therefore,

$$\begin{aligned} \left\| \varPhi \left( \frac{|f|}{c}\right) \right\| _X \ge \left\| \varPhi \left( \frac{\varepsilon }{c}\right) \chi _A \right\| _X = \varPhi \left( \frac{\varepsilon }{c}\right) \Vert \chi _A\Vert _X \end{aligned}$$

and keeping in mind that $\displaystyle \lim _{t\rightarrow \infty }\varPhi (t)=\infty ,$ we can take $c>0$ such that

$$\begin{aligned} \varPhi \left( \frac{\varepsilon }{c}\right) \Vert \chi _A\Vert _X > 1 \end{aligned}$$

which yields a contradiction.

On the other hand, given $f\in X^{\varPhi }$ and $\lambda \in {\mathbb {R}},$ it is clear that

$$\begin{aligned} \Vert \lambda f\Vert _{X^{\varPhi }}=\,\, & {} \inf \left\{ k>0 : \left\| \varPhi \left( \frac{|\lambda f|}{k}\right) \right\| _{X}\le 1 \right\} =\, \inf \left\{ k>0 : \left\| \varPhi \left( \frac{|f|}{\frac{k}{|\lambda |}}\right) \right\| _{X}\le 1 \right\} \\=\,\, & {} |\lambda | \inf \left\{ \frac{k}{|\lambda |}>0 : \left\| \varPhi \left( \frac{|f|}{\frac{k}{|\lambda |}}\right) \right\| _{X}\le 1 \right\} \ = \ |\lambda | \Vert f\Vert _{X^{\varPhi }}. \end{aligned}$$

Now, let $f, g\in X^{\varPhi }$ and take $K\ge 1$ as in (Q3). Given $a, b>0$ such that $\displaystyle \left\| \varPhi \left( \frac{|f|}{a}\right) \right\| _{X}\le 1$ and $\displaystyle \left\| \varPhi \left( \frac{|g|}{b}\right) \right\| _{X}\le 1,$ we have

$$\begin{aligned} \varPhi \left( \frac{|f+g|}{K(a+b)}\right)\le & {} \frac{1}{K} \ \varPhi \left( \frac{|f+g|}{a+b}\right) \ \le \ \frac{1}{K} \ \varPhi \left( \frac{a}{(a+b)}\frac{|f|}{a} + \frac{b}{(a+b)}\frac{|g|}{b}\right) \\\le & {} \frac{1}{K} \frac{a}{(a+b)} \varPhi \left( \frac{|f|}{a}\right) + \frac{1}{K} \frac{b}{(a+b)} \varPhi \left( \frac{|g|}{b}\right). \end{aligned}$$

Hence, $\displaystyle \left\| \varPhi \left( \frac{|f+g|}{K(a+b)}\right) \right\| _{X} \le \frac{a}{(a+b)} \left\| \varPhi \left( \frac{|f|}{a}\right) \right\| _{X} + \frac{b}{(a+b)} \left\| \varPhi \left( \frac{|g|}{b}\right) \right\| _{X} \le 1$ which implies that $\Vert f+g\Vert _{X^{\varPhi }} \le K(a+b).$ By the arbitrariness of a and b we deduce that $\Vert f+g\Vert _{X^{\varPhi }} \le K(\Vert f\Vert _{X^{\varPhi }}+\Vert g\Vert _{X^{\varPhi }}).$

Thus, we have proved that $\Vert \cdot \Vert _{X^{\varPhi }}$ is a quasi-norm in $X^{\varPhi }$ with the same quasi-triangle constant as the one of the quasi-norm of X. Moreover, we have already proved that $X^{\varPhi }$ equipped with the Luxemburg quasi-norm is a quasi-normed space and an ideal in $L^0(\mu ).$ It is also clear that the Luxemburg quasi-norm is a lattice quasi-norm: $|f|\le |g|$ implies that $\displaystyle \varPhi \left( \frac{|f|}{k}\right) \le \varPhi \left( \frac{|g|}{k}\right)$ for all $k>0$ and this guarantees that $\Vert f\Vert _{X^{\varPhi }} \le \Vert g\Vert _{X^{\varPhi }}.$ In addition, $\chi _{\varOmega }\in X^{\varPhi },$ since $\displaystyle \varPhi \left( \frac{|\chi _{\varOmega }|}{c}\right) = \varPhi \left( \frac{1}{c}\right) \chi _{\varOmega }\in X,$ for all $c>0,$ and hence $X^{\varPhi }$ is in fact a quasi-normed function space. $\square$

Remark 2

The inclusion of $X^{\varPhi }\subseteq X$ is continuous provided X and $X^{\varPhi }$ be q-B.f.s. We will see in Theorem 3 that the completeness is transferred from X to $X^{\varPhi }.$

Once we have checked that $X^{\varPhi }$ is quasi-normed function space, it is immediate that $L^{\infty }(\mu )$ is contained in $X^{\varPhi }$ and this inclusion is continuous with norm $\Vert \chi _{\varOmega }\Vert _{X^{\varPhi }}.$ The next result establishes the relation between the norm of this inclusion and the norm $\Vert \chi _{\varOmega }\Vert _{X}$ of the continuous inclusion of $L^{\infty }(\mu )$ into X.

Lemma 1

Let $\varPhi$ be a Young function and X be a quasi-normed function space over $\mu .$

• (i) For all $A\in \varSigma$ with $\mu (A)>0,$ $\Vert \chi _A\Vert _{X^{\varPhi }} = \displaystyle \frac{1}{\varPhi ^{-1}\left( \frac{1}{\Vert \chi _A\Vert _X} \right) }.$

• (ii) For all $f\in L^{\infty }(\mu ),$ $\Vert f\Vert _{X^{\varPhi }} \le \displaystyle \frac{\Vert f\Vert _{L^{\infty }(\mu )}}{\varPhi ^{-1}\left( \frac{1}{\Vert \chi _{\varOmega }\Vert _X} \right) }.$

Proof

(i) Write $\alpha :=\displaystyle \frac{1}{\varPhi ^{-1}\left( \frac{1}{\Vert \chi _A\Vert _X} \right) }.$ On the one hand,

$$\begin{aligned} \left\| \varPhi \left( \frac{|\chi _A|}{\alpha }\right) \right\| _{X} = \varPhi \left( \frac{1}{\alpha }\right) \Vert \chi _A\Vert _{X} = \varPhi \left( \varPhi ^{-1}\left( \frac{1}{\Vert \chi _A\Vert _X} \right) \right) \Vert \chi _A\Vert _{X} = 1, \end{aligned}$$

and so $\Vert \chi _A\Vert _{X^{\varPhi }}\le \alpha .$ On the other hand, given $k>0$ such that $\displaystyle \frac{\chi _A}{k}\in \widetilde{X}^{\varPhi }$ with $\displaystyle \left\| \varPhi \left( \frac{\chi _A}{k}\right) \right\| _{X}\le 1,$ we have $\displaystyle \varPhi \left( \frac{1}{k}\right) \Vert \chi _A\Vert _{X}\le 1,$ that is, $\displaystyle \varPhi \left( \frac{1}{k}\right) \le \frac{1}{\Vert \chi _A\Vert _{X}}$ or, equivalently, $\displaystyle \frac{1}{k} \le \varPhi ^{-1}\left( \frac{1}{\Vert \chi _A\Vert _{X}}\right) ,$ which finally leads to $\alpha \le k$ and so $\alpha \le \Vert \chi _A\Vert _{X^{\varPhi }}.$

(ii) Since $|f|\le \Vert f\Vert _{L^{\infty }(\mu )}\chi _{\varOmega },$ for any $f\in L^{\infty }(\mu ),$ we have $\begin{aligned} \Vert f\Vert _{X^{\varPhi }} \le \Vert f\Vert _{L^{\infty }(\mu )} \Vert \chi _{\varOmega }\Vert _{X^{\varPhi }} \end{aligned}$ and the result follows applying (i) to $\chi _{\varOmega }.$ $\square$

The following two results explore the close relationship between the quantities $\Vert f\Vert _{X^{\varPhi }}$ and $\Vert \varPhi (|f|)\Vert _X.$ This entails interesting consequences on boundedness in $X^{\varPhi },$ allowing us to obtain a sufficient condition and a necessary condition for it.

Lemma 2

Let $\varPhi$ be a Young function, X be a quasi-normed function space over $\mu$ and $H\subset L^0(\mu ).$

• (i) If $f\in \widetilde{X}^{\varPhi },$ then $\Vert f\Vert _{X^{\varPhi }} \le \max \{1, \Vert \varPhi (|f|)\Vert _X \}.$

• (ii) If $\{ \varPhi (|h|) : h\in H \}$ is bounded in X,  then H is bounded in $X^{\varPhi }.$

Proof

(i) On the one hand, $\Vert \varPhi (|f|)\Vert _X\le 1$ directly implies that

$$\begin{aligned} \Vert f\Vert _{X^{\varPhi }} \le 1 = \max \{1, \Vert \varPhi (|f|)\Vert _X \}. \end{aligned}$$

On the other hand, if $\Vert \varPhi (|f|)\Vert _X\ge 1,$ then $\begin{aligned} \displaystyle \varPhi \left( \frac{|f|}{\Vert \varPhi (|f|)\Vert _X}\right) \le \displaystyle \frac{1}{\Vert \varPhi (|f|)\Vert _X} \varPhi (|f|)\in X \end{aligned}$ and hence $\displaystyle \varPhi \left( \frac{|f|}{\Vert \varPhi (|f|)\Vert _X}\right) \in X$ with $\displaystyle \left\| \varPhi \left( \frac{|f|}{\Vert \varPhi (|f|)\Vert _X}\right) \right\| _{X} \le 1.$ This also leads to $\Vert f\Vert _{X^{\varPhi }} \le \Vert \varPhi (|f|)\Vert _X = \max \{1, \Vert \varPhi (|f|)\Vert _X \}.$

(ii) If $\Vert \varPhi (|h|)\Vert _X \le M<\infty ,$ for all $h\in H,$ according to (i) we have that $\Vert h\Vert _{X^{\varPhi }} \le \max \{1, \Vert \varPhi (|h|)\Vert _X \}\le \max \{1,M\}<\infty ,$ for all $h\in H.$ $\square$

Lemma 3

Let $\varPhi$ be a Young function, X be a quasi-normed function space over $\mu$ and $f\in X^{\varPhi }.$

• (i) If $\Vert f\Vert _{X^{\varPhi }}<1,$ then $f\in \widetilde{X}^{\varPhi }$ with $\Vert \varPhi (|f|)\Vert _X \le \Vert f\Vert _{X^{\varPhi }}.$

• (ii) If $\Vert f\Vert _{X^{\varPhi }}>1$ and $f\in \widetilde{X}^{\varPhi },$ then $\Vert \varPhi (|f|)\Vert _X \ge \Vert f\Vert _{X^{\varPhi }}.$

• (iii) If $H\subseteq X^{\varPhi }$ is bounded, then there exists a Young function $\Psi$ such that the set $\{ \Psi (|h|) : h\in H \}$ is bounded in X.

Proof

(i) Given $0<k<1$ such that $\displaystyle \frac{|f|}{k}\in \widetilde{X}^{\varPhi }$ with $\displaystyle \left\| \varPhi \left( \frac{|f|}{k}\right) \right\| _{X}\le 1,$ we have

$$\begin{aligned} \varPhi (|f|) = \varPhi \left( k\frac{|f|}{k}\right) \le k \ \varPhi \left( \frac{|f|}{k}\right) \in X. \end{aligned}$$

Therefore, $\varPhi (|f|)\in X$ with $\Vert \varPhi (|f|)\Vert _X \le k \displaystyle \left\| \varPhi \left( \frac{|f|}{k}\right) \right\| _{X}\le k$ and keeping in mind that $\Vert f\Vert _{X^{\varPhi }}<1,$ we obtain

$$\begin{aligned} \Vert \varPhi (|f|)\Vert _X \le \inf \left\{ 0<k<1 : \frac{|f|}{k}\in \widetilde{X}^{\varPhi } {\text{ with }} \left\| \varPhi \left( \frac{|f|}{k}\right) \right\| _{X}\le 1 \right\} = \Vert f\Vert _{X^{\varPhi }}. \end{aligned}$$

(ii) Let $0<\varepsilon <\Vert f\Vert _{X^{\varPhi }}-1$ and observe that $\displaystyle \left\| \varPhi \left( \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}-\varepsilon }\right) \right\| _{X} > 1.$ Thus,

$$\begin{aligned} \Vert \varPhi (|f|)\Vert _X=\,\, & {} \left\| \varPhi \left( (\Vert f\Vert _{X^{\varPhi }}-\varepsilon )\frac{|f|}{\Vert f\Vert _{X^{\varPhi }}-\varepsilon }\right) \right\| _X\\\ge\,\, & {} (\Vert f\Vert _{X^{\varPhi }}-\varepsilon ) \left\| \varPhi \left( \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}-\varepsilon }\right) \right\| _{X} \ge \Vert f\Vert _{X^{\varPhi }}-\varepsilon , \end{aligned}$$

and letting $\varepsilon \rightarrow 0,$ it follows that $\Vert \varPhi (|f|)\Vert _X \ge \Vert f\Vert _{X^{\varPhi }}.$

(iii) Take $M>0$ such that $\Vert h\Vert _{X^{\varPhi }}<M,$ for all $h\in H.$ Since $\displaystyle \left\| \frac{h}{M}\right\| _{X^{\varPhi }}< 1,$ for all $h\in H,$ (i) guarantees that $\displaystyle \varPhi \left( \frac{|h|}{M}\right) \in X$ with $\displaystyle \left\| \varPhi \left( \frac{|h|}{M}\right) \right\| _{X}\le \left\| \frac{h}{M}\right\| _{X^{\varPhi }}<1,$ for all $h\in H.$ Defining $\Psi (t):=\varPhi \displaystyle \left( \frac{t}{M}\right) ,$ for all $t\ge 0,$ we produce a Young function such that $\{ \Psi (|h|) : h\in H \}$ is bounded in X. $\square$

We are now in a position to establish the remarkable fact that Orlicz spaces $X^{\varPhi }$ are always complete for any q-B.f.s. X. It is worth pointing out that standard proofs in the Banach setting require the $\sigma$-Fatou property of X to obtain the $\sigma$-Fatou property of $X^{\varPhi }$ (see the next Theorem 4) and as a byproduct, the completeness of this last space. However, as we have said before, there are many complete spaces without the $\sigma$-Fatou property, to which it is not possible to apply Theorem 4. Herein lies the importance of the result that we will show next about completeness of $X^\varPhi .$

Theorem 3

Let $\varPhi$ a Young function and X be a q-B.f.s. over $\mu .$ Then, $X^{\varPhi }$ is complete (and hence it is a q-B.f.s. over $\mu$).

Proof

Let $(h_n)_n$ be a positive increasing Cauchy sequence in $X^{\varPhi }$ and take $K\ge 1$ as in (Q3). Then, we can choose a subsequence of $(h_n)_n,$ that we denote by $(f_n)_n,$ such that $\Vert f_{n+1}-f_{n}\Vert _{X^{\varPhi }} < \displaystyle \frac{1}{2^{2n} K^{2n}},$ for all $n\in {\mathbb {N}}.$ Thus,

$$\begin{aligned} \left\| 2^{n} K^{n} (f_{n+1}-f_{n}) \right\| _{X^{\varPhi }}< \displaystyle \frac{1}{2^{n} K^{n}}< 1 \end{aligned}$$

for all $n\in {\mathbb {N}},$ and by Lemma 3 it follows that

$$\begin{aligned} \left\| \varPhi \left( 2^{n} K^{n} \left( f_{n+1}-f_{n}\right) \right) \right\| _{X} \le \left\| 2^{n} K^{n} \left( f_{n+1}-f_{n}\right) \right\| _{X^{\varPhi }} < \frac{1}{2^{n} K^{n}}, \quad n\in {\mathbb {N}}, \end{aligned}$$

which proves that $\displaystyle \sum _{n=1}^{\infty } K^n \left\| \varPhi \left( 2^{n} K^{n} \left( f_{n+1}-f_{n}\right) \right) \right\| _{X} \le \sum _{n=1}^{\infty } \frac{1}{2^{n}} < \infty .$ The completeness of X ensures that the function $f:=\displaystyle \sum _{n=1}^{\infty } \varPhi \left( 2^{n} K^{n} \left( f_{n+1}-f_{n}\right) \right) \in X,$ by Theorem 1. Note that $f\in L^0(\mu )$ and the convergence of that series is also $\mu$-a.e, since X is continuously included in $L^0(\mu ).$ Given $N\in {\mathbb {N}},$ let $g_N:=\displaystyle \sum _{n=1}^{N} (f_{n+1}-f_{n})$ and denote by $g:=\displaystyle \sup _{N} g_N$ pointwise $\mu$-a.e. Applying (1) with $\alpha :=K,$ it follows that for all $N\in {\mathbb {N}},$

$$\begin{aligned} \varPhi (g_N)=\,\, & {} \varPhi \left( \sum _{n=1}^{N} (f_{n+1}-f_{n}) \right) \le \sum _{n=1}^{N} \frac{1}{2^n K^n} \varPhi \left( 2^n K^n \left( f_{n+1} -f_{n}\right) \right) \\\le & {} \sum _{n=1}^{N} \varPhi (2^{n} K^{n} (f_{n+1}-f_{n}) ) \le f \end{aligned}$$

Therefore, $0\le g_N\le \varPhi ^{-1}(f)\in L^0(\mu )$ for all $N\in {\mathbb {N}}$ and so $g\in L^0(\mu )$ with $0\le g\le \varPhi ^{-1}(f) \in X^{\varPhi },$ which guarantees that $g\in X^{\varPhi }.$ But

$$\begin{aligned} f_{N+1} =\sum _{n=1}^{N} (f_{n+1}-f_{n}) + f_{1} = g_{N} + f_1 \end{aligned}$$

for all $N\in {\mathbb {N}}$ and so there also exists $\displaystyle \sup _{n} f_n = g + f_1 \in X^{\varPhi }.$ Since $(f_n)_n$ is a subsequence of the original increasing sequence $(h_n)_n,$ the supremum of the whole sequence must exists and be the same as the supremum of the subsequence. By applying Amemiya’s Theorem 2 we conclude that $X^{\varPhi }$ is complete. $\square$

If the q-B.f.s. X has the $\sigma$-Fatou property, then we can improve a little more our knowledge about $X^{\varPhi }$ as the following proposition makes evident.

Theorem 4

Let $\varPhi$ be a Young function and X be a q-B.f.s. over $\mu$ with the $\sigma$-Fatou property.

• (i) If $0\ne f\in X^{\varPhi }$ then $\displaystyle \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}\in \widetilde{X}^{\varPhi }$ with $\displaystyle \left\| \varPhi \left( \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}\right) \right\| _{X}\le 1.$

• (ii) If $f\in X^{\varPhi }$ with $\Vert f\Vert _{X^{\varPhi }}\le 1$ then $f\in \widetilde{X}^{\varPhi }$ with $\Vert \varPhi (|f|)\Vert _X \le \Vert f\Vert _{X^{\varPhi }}.$

• (iii) $X^{\varPhi }$ also has the $\sigma$-Fatou property.

Proof

(i) Take a sequence $(k_n)_n$ such that $k_n \downarrow \Vert f\Vert _{X^{\varPhi }}$ and $\displaystyle \left\| \varPhi \left( \frac{|f|}{k_n}\right) \right\| _{X}\le 1,$ for all $n\in {\mathbb {N}}.$ Then, $\displaystyle \frac{|f|}{k_n}\uparrow \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}$ and so $\displaystyle \varPhi \left( \frac{|f|}{k_n}\right) \uparrow \varPhi \left( \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}\right) ,$ since $\varPhi$ is continuous and increasing. The $\sigma$-Fatou property of X guarantees that $\displaystyle \varPhi \left( \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}\right) \in X$ and

$$\displaystyle \left\| \varPhi \left( \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}\right) \right\| _{X} = \sup _{n} \left\| \varPhi \left( \frac{|f|}{k_n}\right) \right\| _X \le 1.$$

(ii) According to (i) and the inequality

$$\begin{aligned} \varPhi (|f|) = \varPhi \left( \Vert f\Vert _{X^{\varPhi }}\frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}\right) \le \Vert f\Vert _{X^{\varPhi }} \ \varPhi \left( \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}\right) \end{aligned}$$

we deduce that $\varPhi (|f|)\in X$ and $\displaystyle \Vert \varPhi (|f|)\Vert _X \le \Vert f\Vert _{X^{\varPhi }} \left\| \varPhi \left( \frac{|f|}{\Vert f\Vert _{X^{\varPhi }}}\right) \right\| _{X}\le \Vert f\Vert _{X^{\varPhi }}.$

(iii) Let $(f_n)_n$ in $X^{\varPhi }$ with $0\le f_n\uparrow f$ $\mu$-a.e. and $M:=\displaystyle \sup _n \Vert f_n\Vert _{X^{\varPhi }}<\infty .$ Then, $\displaystyle \varPhi \left( \frac{f_n}{M}\right) \uparrow \varPhi \left( \frac{f}{M}\right)$ $\mu$-a.e. and $\displaystyle \left\| \frac{f_n}{M}\right\| _{X^{\varPhi }}\le 1$ for all $n\in {\mathbb {N}}.$ Applying (ii), we deduce that $\displaystyle \varPhi \left( \frac{f_n}{M}\right) \in X$ with $\displaystyle \left\| \varPhi \left( \frac{f_n}{M}\right) \right\| _{X}\le 1$ for all $n\in {\mathbb {N}}$ and using the $\sigma$-Fatou property of X,  it follows that $\displaystyle \varPhi \left( \frac{f}{M}\right) \in X$ with $\begin{aligned} \left\| \varPhi \left( \frac{f}{M}\right) \right\| _{X} = \sup _n \left\| \varPhi \left( \frac{f_n}{M}\right) \right\| _{X}\le 1. \end{aligned}$ This implies that $f\in X^{\varPhi }$ with $\Vert f\Vert _{X^{\varPhi }}\le M$ and we also have $M\le \Vert f\Vert _{X^{\varPhi }},$ since $f_n\le f\in X^{\varPhi }.$ Thus, $\Vert f\Vert _{X^{\varPhi }}=M,$ which proves that $X^{\varPhi }$ has the $\sigma$-Fatou property. $\square$

The relation between the Orlicz class and its corresponding Orlicz space is greatly simplified when the Young function has the $\varDelta _2$-property. In addition, this has far-reaching consequences on convergence in $X^{\varPhi }$ as we state in the next result.

Theorem 5

Let X be a quasi-normed function space over $\mu$ and $\varPhi \in \varDelta _2.$

• (i) The Orlicz space and the Orlicz class coincide: $X^{\varPhi } = \widetilde{X}^{\varPhi }.$

• (ii) $\Vert f_n\Vert _{X^{\varPhi }}\rightarrow 0$ if and only if $\Vert \varPhi (|f_n|)\Vert _{X}\rightarrow 0,$ for all $(f_n)_n\subseteq X^{\varPhi }.$

• (iii) If X is $\sigma$-order continuous, then $X^{\varPhi }$ is also $\sigma$-order continuous.

Proof

(i) Given $f\in X^{\varPhi },$ there exists $c>0$ such that $\displaystyle \varPhi \left( \frac{|f|}{c}\right) \in X.$ If $c\le 1,$ then

$$\begin{aligned} \varPhi (|f|) = \varPhi \left( c \ \frac{|f|}{c}\right) \le c \ \varPhi \left( \frac{|f|}{c}\right) \in X, \end{aligned}$$

and if $c>1,$ then there exist $C>1$ such that $\varPhi (ct)\le C\varPhi (t)$ for all $t\ge 0$ by the $\varDelta _2$-property of $\varPhi .$ Therefore, $\displaystyle \varPhi (|f|) = \varPhi \left( c \ \frac{|f|}{c}\right) \le C \ \varPhi \left( \frac{|f|}{c}\right) \in X.$ In any case, it follows that $\varPhi (|f|)\in X,$ which means that $f\in \widetilde{X}^{\varPhi }.$

(ii) If $\Vert f_n\Vert _{X^{\varPhi }}\rightarrow 0,$ then $\Vert \varPhi (|f_n|)\Vert _{X}\rightarrow 0$ as a consequence of Lemma 3 (i). Suppose now that $\Vert f_n\Vert _{X^{\varPhi }}$ does not converges to 0. Then, there exists $\varepsilon >0$ and a subsequence $(f_{n_k})_k$ of $(f_n)_n$ such that $\Vert f_{n_k}\Vert _{X^{\varPhi }}>\varepsilon$ for all $k\in {\mathbb {N}}.$ We can assume that $\varepsilon <1$ and that $(f_{n_k})_k$ is the whole $(f_n)_n$ without loss of generality. Since $\varPhi \in \varDelta _2$ and $\displaystyle \frac{1}{\varepsilon }>1,$ there exist $C>1$ such that $\displaystyle \varPhi \left( \frac{|f_n|}{\varepsilon }\right) \le C \varPhi (|f_n|).$ By (i), we deduce that $\displaystyle \varPhi \left( \frac{|f_n|}{\varepsilon }\right) \in X$ and hence $\displaystyle \left\| \varPhi \left( \frac{|f_n|}{\varepsilon }\right) \right\| _{X} > 1.$ Thus, $\begin{aligned} \Vert \varPhi (|f_n|)\Vert _{X} \ge \displaystyle \frac{1}{C} \displaystyle \left\| \varPhi \left( \frac{|f_n|}{\varepsilon }\right) \right\| _{X}> \frac{1}{C} >0, \end{aligned}$ which means that $\Vert \varPhi (|f_n|)\Vert _{X}$ does not converges to 0.

(iii) Let $(f_n)_n$ and f in $X^{\varPhi }$ such that $0\le f_n\uparrow f$ $\mu$-a.e. Then, $\varPhi \left( f-f_n\right) \downarrow 0$ $\mu$-a.e. Since X is $\sigma$-order continuous, it follows that $\Vert \varPhi \left( f-f_n\right) \Vert _{X}\rightarrow 0$ and by (ii) this implies that $\Vert f-f_n\Vert _{X^{\varPhi }}\rightarrow 0,$ which gives the $\sigma$-order continuity of $X^{\varPhi }.$ $\square$

5 Application: Orlicz spaces associated to a vector measure

First of all observe that classical Orlicz spaces $L^{\varPhi }(\mu )$ with respect to a positive finite measure $\mu$ are obtained applying the construction $X^{\varPhi }$ of section 4 to the B.f.s. $X=L^1(\mu ),$ that is, $L^{\varPhi }(\mu ) = L^1(\mu )^{\varPhi }$ equipped with the norm $\Vert \cdot \Vert _{L^{\varPhi }(\mu )} := \Vert \cdot \Vert _{L^1(\mu )^{\varPhi }}.$ Using these classical Orlicz spaces, the Orlicz spaces $L^{\varPhi }_w(m)$ and $L^{\varPhi }(m)$ with respect to a vector measure $m:\varSigma \rightarrow Y$ were introduced in [8] in the following way:

$$\begin{aligned} L^{\varPhi }_w(m) := \left\{ f\in L^0(m) : f\in L^{\varPhi }(|\langle m, y^*\rangle |), \ \forall \, y^*\in Y^* \right\} , \end{aligned}$$

equipped with the norm

$$\begin{aligned} \Vert f\Vert _{L^{\varPhi }_w(m)} := \sup \left\{ \Vert f\Vert _{L^{\varPhi }(|\langle m, y^*\rangle |)} : y^*\in B_{Y^*}\right\} , \end{aligned}$$

and $L^{\varPhi }(m)$ is the closure of simple functions ${\mathscr {S}}(\varSigma )$ in $L^{\varPhi }_w(m).$ The next result establishes that these Orlicz spaces $L^{\varPhi }_w(m)$ and $L^{\varPhi }(m)$ can be obtained as generalized Orlicz spaces $X^{\varPhi }$ by taking X to be $L^1_w(m)$ and $L^1(m),$ respectively.

Proposition 4

Let $\varPhi$ be a Young function and $m:\varSigma \rightarrow Y$ a vector measure.

• (i) $L^{\varPhi }_w(m) = L^1_w(m)^{\varPhi }$ and $\Vert f\Vert _{L^{\varPhi }_w(m)} = \Vert f\Vert _{L^1_w(m)^{\varPhi }},$ for all $f\in L^{\varPhi }_w(m).$

• (ii) $L^{\varPhi }(m) \subseteq L^1(m)^{\varPhi }$ and if $\varPhi \in \varDelta _2,$ then $L^{\varPhi }(m) = L^1(m)^{\varPhi }.$

Proof

(i) Suppose that $f\in L^1_w(m)^{\varPhi }$ and let $k>0$ such that $\displaystyle \varPhi \left( \frac{|f|}{k}\right) \in L^1_w(m)$ with $\displaystyle \left\| \varPhi \left( \frac{|f|}{k}\right) \right\| _{L^1_w(m)}\le 1.$ Given $y^*\in B_{Y^*}$ we have $\displaystyle \varPhi \left( \frac{|f|}{k}\right) \in L^1(|\langle m, y^*\rangle |)$ with

$$\displaystyle \left\| \varPhi \left( \frac{|f|}{k}\right) \right\| _{L^1(|\langle m, y^*\rangle |)} \le \displaystyle \left\| \varPhi \left( \frac{|f|}{k}\right) \right\| _{L^1_w(m)}\le 1.$$

This implies that $f\in L^{\varPhi }(|\langle m, y^*\rangle |)$ with $\Vert f\Vert _{L^{\varPhi }(|\langle m, y^*\rangle |)}\le k.$ Hence, $f\in L^{\varPhi }_w(m)$ with $\Vert f\Vert _{L^{\varPhi }_w(m)} \le \Vert f\Vert _{L^1_w(m)^{\varPhi }}.$

Reciprocally, suppose now that $f\in L^{\varPhi }_w(m),$ write $M:=\Vert f\Vert _{L^{\varPhi }_w(m)}$ and let $y^*\in B_{Y^*}.$ Since $f\in L^{\varPhi }(|\langle m, y^*\rangle |)$ and $\Vert f\Vert _{L^{\varPhi }(|\langle m, y^*\rangle |)} \le M,$ we have that $\displaystyle \frac{f}{M}\in L^{\varPhi }(|\langle m, y^*\rangle |)$ with $\displaystyle \left\| \frac{f}{M}\right\| _{L^{\varPhi }(|\langle m, y^*\rangle |)} \le 1.$ Applying Theorem 4 (ii) to the space $X=L^{1}(|\langle m, y^*\rangle |),$ it follows that $\displaystyle \varPhi \left( \frac{|f|}{M}\right) \in L^1(|\langle m, y^*\rangle |)$ with $\begin{aligned} \left\| \varPhi \left( \frac{|f|}{M}\right) \right\| _{L^1(|\langle m, y^*\rangle |)} \le \displaystyle \left\| \frac{f}{M}\right\| _{L^{\varPhi }(|\langle m, y^*\rangle |)} \le 1. \end{aligned}$ Then, the arbitrariness of $y^*\in B_{Y^*}$ guarantees that $\displaystyle \varPhi \left( \frac{|f|}{M}\right) \in L^1_w(m)$ with $\displaystyle \left\| \varPhi \left( \frac{|f|}{M}\right) \right\| _{L^1_w(m)} \le 1$ and hence $f\in L^1_w(m)^{\varPhi }$ with $\Vert f\Vert _{L^1_w(m)^{\varPhi }} \le M.$

(ii) Since $L^1(m)^{\varPhi }$ is a B.f.s., simple functions ${\mathscr {S}}(\varSigma ) \subseteq L^1(m)^{\varPhi }$ and $L^1(m)^{\varPhi }$ is a closed subspace of $L^1_w(m)^{\varPhi }.$ Thus, taking in account (i), we deduce that $L^{\varPhi }(m) \subseteq L^1(m)^{\varPhi }.$ If in addition $\varPhi \in \varDelta _2,$ we have

$$\begin{aligned} L^1(m)^{\varPhi } = \{ f\in L^0(m) : \varPhi (|f|)\in L^1(m) \} = L^{\varPhi }(m), \end{aligned}$$

where the first equality is due to Theorem 5 (i) applied to $X=L^1(m)$ and the second one can be found in [8, Proposition 4.4]. $\square$

The Orlicz spaces $L^{\varPhi }(m)$ have been recently employed in [5] to locate the compact subsets of $L^{1}(m).$ Motivated by the idea of studying compactness in $L^{1}(\Vert m\Vert )$ (see [6] for details), we introduce the Orlicz spaces $L^{\varPhi }(\Vert m\Vert )$ as the Orlicz spaces $X^{\varPhi }$ associated to the q-B.f.s. $X = L^1(\Vert m\Vert ).$ For further reference, we collect together all the information that our general theory provide about these new Orlicz spaces.

Definition 4

Let $\varPhi$ be a Young function and $m:\varSigma \rightarrow Y$ a vector measure. We define the Orlicz spaces associated to the semivariation of m as $L^{\varPhi }(\Vert m\Vert ):=L^1(\Vert m\Vert )^{\varPhi }$ equipped with $\Vert f\Vert _{L^{\varPhi }(\Vert m\Vert )}:= \Vert f\Vert _{L^1(\Vert m\Vert )^{\varPhi }},$ for all $f\in L^{\varPhi }(\Vert m\Vert ).$

Corollary 2

Let $\varPhi$ be a Young function, $m:\varSigma \rightarrow Y$ a vector measure and $\mu$ any Rybakov control measure for m. Then,

• (i) $L^{\varPhi }(\Vert m\Vert )$ is a q-B.f.s. over $\mu$ with the $\sigma$-Fatou property.

• (ii) If $\varPhi \in \varDelta _2,$ then $L^{\varPhi }(\Vert m\Vert )$ is $\sigma$-order continuous.

• (iii) $L^{\varPhi }(\Vert m\Vert ) \subseteq L^1(\Vert m\Vert )$ with continuous inclusion.

Proof

Apply Theorems 3, 4 and 5 to the q-B.f.s $X=L^1(\Vert m\Vert ).$ See also Proposition 2 and Remark 2. $\square$

Corollary 3

Let $\varPhi$ be a Young function, $m:\varSigma \rightarrow Y$ a vector measure, $f\in L^{\varPhi }(\Vert m\Vert )$ and $H\subseteq L^0(m).$

• (i) If $\varPhi (|f|)\in L^1(\Vert m\Vert ),$ then $\Vert f\Vert _{L^{\varPhi }(\Vert m\Vert )} \le \max \{1, \Vert \varPhi (|f|)\Vert _{L^1(\Vert m\Vert )} \}.$

• (ii) If $\Vert f\Vert _{L^{\varPhi }(\Vert m\Vert )}\le 1,$ then $\varPhi (|f|)\in L^1(\Vert m\Vert )$ and $\Vert \varPhi (|f|)\Vert _{L^1(\Vert m\Vert )} \le \Vert f\Vert _{L^{\varPhi }(\Vert m\Vert )}.$

• (iii) If $\Vert f\Vert _{L^{\varPhi }(\Vert m\Vert )}>1$ and $\varPhi (|f|)\in L^1(\Vert m\Vert ),$ then $\Vert \varPhi (|f|)\Vert _{L^1(\Vert m\Vert )} \ge \Vert f\Vert _{L^{\varPhi }(\Vert m\Vert )}.$

• (iv) If $\{\varPhi (|h|) : h\in H \}$ is bounded in $L^{1}(\Vert m\Vert ),$ then H is bounded in $L^{\varPhi }(\Vert m\Vert ).$

• (v) If H is bounded in $L^{\varPhi }(\Vert m\Vert ),$ then there exists a Young function $\Psi$ such that $\{\Psi (|h|) : h\in H \}$ is bounded in $L^{1}(\Vert m\Vert ).$

Proof

Particularize Lemmas 2 and 3 to $X=L^1(\Vert m\Vert ).$ Note that, in fact, we can use (ii) of Theorem 4. $\square$

Corollary 4

Let $\varPhi \in \varDelta _2,$ $m:\varSigma \rightarrow Y$ a vector measure and $(f_n)_n\subseteq L^{\varPhi }(\Vert m\Vert ).$

• (i) $L^{\varPhi }(\Vert m\Vert ) = \{ f\in L^0(m) : \varPhi (|f|)\in L^1(\Vert m\Vert ) \}.$

• (ii) $\Vert f_n\Vert _{L^{\varPhi }(\Vert m\Vert )}\rightarrow 0$ if and only if $\Vert \varPhi (|f_n|)\Vert _{L^1(\Vert m\Vert )}\rightarrow 0.$

Proof

Apply Theorem 5 to the space $X=L^1(\Vert m\Vert ).$ $\square$

6 Application: interpolation of Orlicz spaces

In this section all the q-B.f.s. will be supposed to be complex. This means that $L^0(\mu )$ will be assumed to be in fact the space of all ($\mu$-a.e. equivalence classes of) ${\mathbb {C}}$-valued measurable functions on $\varOmega .$ Recall that a complex q-B.f.s X over $\mu$ is the complexification of the real q-B.f.s. $X_{{\mathbb {R}}}:=X\cap L^0_{{\mathbb {R}}}(\mu ),$ where $L^0_{{\mathbb {R}}}(\mu )$ is the space of all ($\mu$-a.e. equivalence classes of) ${\mathbb {R}}$-valued measurable functions on $\varOmega$ (see [18, p.24] for more details) and this allows to extend all the real q-B.f.s. defined above to complex q-B.f.s. following a standard argument.

The complex method of interpolation, $[X_0,X_1]_{\theta }$ with $0<\theta <1,$ for pairs $(X_0,X_1)$ of quasi-Banach spaces was introduced in [10] as a natural extension of Calderón’s original definition for Banach spaces. It relies on a theory of analytic functions with values in quasi-Banach spaces which was developed in [10] and [10]. It is important to note that there is no analogue of the Maximum Modulus Principle for general quasi-Banach spaces, but there is a wide subclass of quasi-Banach spaces called analytically convex (A-convex) in which that principle does hold. For a q-B.f.s. X it can be proved that analytical convexity is equivalent to lattice convexity (L-convexity), i.e., there exists $0<\varepsilon <1$ so that if $f\in X$ and $0\le f_i\le f,$ $i=1,\dots ,n,$ satisfy $\displaystyle \frac{f_1+\cdots + f_n}{n}\ge (1-\varepsilon )f,$ then $\displaystyle \max _{1\le i \le n}\Vert f_i\Vert _{X} \ge \varepsilon \Vert f\Vert _{X}$ (see [10, Theorem 4.4]). This is also equivalent to X be s-convex for some $s>0$ (see [10, Theorem 2.2]). We recall that X is called s-convex if there exists $C\ge 1$ such that

$$\begin{aligned} \left\| \left( \sum _{k=1}^{n}|f_k|^s \right) ^{\frac{1}{s}} \right\| _{X} \le C \left( \sum _{k=1}^{n} \Vert f_k\Vert _{X}^{s} \right) ^{\frac{1}{s}} \end{aligned}$$

for all $n\in {\mathbb {N}}$ and $f_1, \dots , f_n\in X.$ Observe that, X is s-convex if and only if its s-th power $X_{[s]}$ is 1-convex, where the s-th power $X_{[s]}$ of a q-B.f.s. X over $\mu$ (for any $0<s<\infty$) is the q-B.f.s. $X_{[s]}:=\displaystyle \left\{ f\in L^0(\mu ) : |f|^{\frac{1}{s}}\in X \right\}$ equipped with the quasi-norm $\Vert f\Vert _{X_{[s]}} = \displaystyle \left\| |f|^{\frac{1}{s}} \right\| _{X}^{s},$ for all $f\in X_{[s]}$ (see [18, Proposition 2.22]).

The following result provides a condition under which the L-convexity of X can be transferred to its Orlicz space $X^{\varPhi }.$ When X possesses the $\sigma$-Fatou property, this can be derived from [10, Proposition 3.3], but we make apparent that this property can be dropped. Recall that a function $\psi$ on the semiaxis $[0,\infty )$ is said to be quasiconcave if $\psi (0)=0,$ $\psi (t)$ is positive and increasing for $t>0$ and $\displaystyle \frac{\psi (t)}{t}$ is decreasing for $t>0.$ Observe that a quasiconcave function $\psi$ satisfies the following inequalities for all $t\ge 0$:

$$\begin{aligned} \left\{ \begin{array}{ccl} \psi (\alpha t) \ge \alpha \, \psi (t) & {\text{ if }} & 0\le \alpha \le 1, \\ \psi (\alpha t) \le \alpha \, \psi (t) & {\text{ if }} & \alpha \ge 1. \end{array} \right. \end{aligned}$$

Theorem 6

If X is an L-convex q-B.f.s. and $\varPhi \in \varDelta _2,$ then $X^{\varPhi }$ is L-convex.

Proof

Since $\varPhi \in \varDelta _2,$ there exists $s>1$ such that $\varPhi (2t)\le s\varPhi (t)$ for all $t\ge 0.$ From the inequality

$$\begin{aligned} t \varPhi '(t)\le \int _{t}^{2t} \varPhi '(u)\, du \le \int _{0}^{2t} \varPhi '(u)\, du = \varPhi (2t)\le s\varPhi (t), \ t>0 \end{aligned}$$

it is easy to check that $\displaystyle \frac{\varPhi (t)}{t^s}$ is decreasing and then $\displaystyle \frac{\varPhi \left( t^{\frac{1}{s}}\right) }{t}$ so is. Therefore, the function $\psi (t):=\varPhi \left( t^{\frac{1}{s}}\right)$ is quasiconcave. Take $0<\delta <1$ such that $(1-\delta )^s=1-\varepsilon ,$ where $\varepsilon$ is the constant from the L-convexity of X. Let $f\in X^{\varPhi }$ and $0\le f_i\le f,$ $i=1,\dots ,n$ satisfying $\displaystyle \frac{f_1+\cdots + f_n}{n}\ge (1-\delta )f.$ We can also assume that $\Vert f\Vert _{X^{\varPhi }}=1$ without loss of generality. Note that this implies $\Vert \varPhi (f)\Vert _{X} \ge 1.$ If we suppose, on the contrary, that $\Vert \varPhi (f)\Vert _{X}<1$ and we take $0<k<1$ such that $\Vert \varPhi (f)\Vert _{X}< k^s < 1,$ then

$$\begin{aligned} \left\| \varPhi \left( \frac{f}{k}\right) \right\| _X = \left\| \psi \left( \frac{f^s}{k^s} \right) \right\| _{X} \le \frac{1}{k^s} \Vert \psi (f^s) \Vert _{X} = \frac{1}{k^s} \Vert \varPhi (f) \Vert _{X} <1, \end{aligned}$$

and therefore $\Vert f\Vert _{X^{\varPhi }}<k<1.$ Moreover, we have $0\le \varPhi (f_i)\le \varPhi (f)\in X$ and

$$\begin{aligned} \frac{\varPhi (f_1)+\cdots +\varPhi (f_n)}{n}\ge\,\, & {} \varPhi \left( \frac{f_1+\cdots + f_n}{n}\right) \ \ge \ \varPhi ( (1-\delta )f )\\\ge\,\, & {} (1-\delta )^s \psi (f^s) \ = \ (1-\delta )^s \varPhi (f) \ = \ (1-\varepsilon ) \varPhi (f). \end{aligned}$$

Thus, the L-convexity of X implies that $\displaystyle \max _{1\le i \le n}\Vert \varPhi (f_i)\Vert _{X} \ge \varepsilon \Vert \varPhi (f)\Vert _{X} \ge \varepsilon$ and hence $\displaystyle \max _{1\le i \le n}\Vert f_i\Vert _{X^{\varPhi }} \ge \varepsilon >\delta$ by (i) of Lemma 3. $\square$

The Calderón product $X_0^{1-\theta }X_1^{\theta }$ of two q-B.f.s. $X_0$ and $X_1$ over $\mu$ is the q-B.f.s. of all functions $f\in L^0(\mu )$ such that there exist $f_0\in B_{X_0},$ $f_1\in B_{X_1}$ and $\lambda >0$ for which

$$\begin{aligned} |f(w)| \le \lambda |f_0(w)|^{1-\theta } |f_1(w)|^{\theta }, \quad w\in \varOmega \ \ (\mu {\text{-a.e. }}) \end{aligned}$$

(2)

endowed with the quasi-norm $\Vert f\Vert _{X_0^{1-\theta }X_1^{\theta }}= \inf \lambda ,$ where the infimum is taken over all $\lambda$ satisfying (2). The complex method gives the result predicted by the Calderón product for nice pairs of q-B.f.s. (see [10, Theorem 3.4]).

Theorem 7

Let $\varOmega$ be a Polish space and let $\mu$ be a finite Borel measure on $\varOmega .$ Let $X_0,$ $X_1$ be a pair of $\sigma$-order continuous L-convex q-B.f.s. over $\mu .$ Then $X_0+X_1$ is L-convex and $[X_0,X_1]_{\theta } = X_0^{1-\theta }X_1^{\theta }$ with equivalence of quasi-norms.

On the other hand, it is easy to compute the Calderón product of two Orlicz spaces associated to the same q-B.f.s:

Proposition 5

Let X be a q-B.f.s. over $\mu ,$ $\varPhi _0,$ $\varPhi _1$ Young functions, $0< \theta <1$ and $\varPhi$ such that $\varPhi ^{-1} := (\varPhi _{0}^{-1})^{1-\theta } (\varPhi _{1}^{-1})^{\theta }.$ Then $\left( X^{\varPhi _0}\right) ^{1-\theta }\left( X^{\varPhi _1}\right) ^{\theta } = X^{\varPhi }.$

Proof

Given $f\in X^{\varPhi },$ there exists $c>0$ such that $h:=\displaystyle \varPhi \left( \frac{|f|}{c}\right) \in X$ and hence $f_0:=\varPhi _0^{-1}(h)\in X^{\varPhi _0}$ and $f_1:=\varPhi _1^{-1}(h)\in X^{\varPhi _1}.$ Taking $\alpha :=\max \{ \Vert f_0\Vert _{X^{\varPhi _0}}, \Vert f_1\Vert _{X^{\varPhi _1}} \},$ it follows that

$$\begin{aligned} |f|=\,\, & {} c\, \varPhi ^{-1}(h) \ = \ c \, (\varPhi _0^{-1}(h))^{1-\theta } (\varPhi _1^{-1}(h))^{\theta } \ = \ c |f_0|^{1-\theta } |f_1|^{\theta }\le\,\, & {} c\alpha \left( \frac{f_0}{\alpha }\right) ^{1-\theta } \left( \frac{f_1}{\alpha }\right) ^{\theta }, \end{aligned}$$

which yields $f\in \left( X^{\varPhi _0}\right) ^{1-\theta }\left( X^{\varPhi _1}\right) ^{\theta }.$

Conversely, if $f\in \left( X^{\varPhi _0}\right) ^{1-\theta }\left( X^{\varPhi _1}\right) ^{\theta },$ then there exist $\lambda >0,$ $f_0\in X^{\varPhi _0}$ and $f_1\in X^{\varPhi _1}$ such that $|f|\le \lambda |f_0|^{1-\theta } |f_1|^{\theta }.$ This implies the existence of $c>0$ such that $h_0 := \displaystyle \varPhi _0\left( \frac{|f_0|}{c}\right) \in X$ and $h_1 := \displaystyle \varPhi _1\left( \frac{|f_1|}{c}\right) \in X.$ Thus, taking $h:=h_0+h_1\in X,$ we deduce that

$$\begin{aligned} |f|\le\,\, & {} \lambda |f_0|^{1-\theta } |f_1|^{\theta } = \lambda c \left( \frac{|f_0|}{c}\right) ^{1-\theta } \left( \frac{|f_1|}{c}\right) ^{\theta } = \lambda c (\varPhi _0^{-1}(h_0))^{1-\theta } (\varPhi _1^{-1}(h_1))^{\theta }\\\le\,\, & {} \lambda c (\varPhi _0^{-1}(h))^{1-\theta } (\varPhi _1^{-1}(h))^{\theta } = \lambda \varPhi ^{-1}(h) \in X^{\varPhi }, \end{aligned}$$

and hence $f\in X^{\varPhi }.$ $\square$

Combining the three previous results, we obtain conditions under which the complex method applied to Orlicz spaces associated to a q-B.f.s. over $\mu$ keeps on producing an Orlicz space associated to the same q-B.f.s.

Corollary 5

Let $\varOmega$ be a Polish space and let $\mu$ be a finite Borel measure on $\varOmega .$ Let X be an L-convex, $\sigma$-order continuous q-B.f.s. over $\mu ,$ $\varPhi _0, \varPhi _1\in \varDelta _2,$ $0<\theta <1$ and $\varPhi$ such that $\varPhi ^{-1} := (\varPhi _{0}^{-1})^{1-\theta } (\varPhi _{1}^{-1})^{\theta }.$ Then, $[X^{\varPhi _0},X^{\varPhi _1}]_{\theta } = X^{\varPhi }.$

Proof

According to Theorems 5 and 6, the hypotheses guarantee that $X^{\varPhi _0}$ and $X^{\varPhi _1}$ are L-convex, $\sigma$-order continuous q-B.f.s. Therefore, the result follows by applying Theorem 7 and Proposition 5. $\square$

Let us denote $L^s(\Vert m\Vert ):=L^1(\Vert m\Vert )_{\left[ \frac{1}{s}\right] },$ for $0<s<\infty$ and $m:\varSigma \rightarrow Y$ a vector measure. In [4, Proposition 4.1] we proved that if $s>1,$ then $L^s(\Vert m\Vert )$ is r-convex for every $r < s.$ In fact, this is true for all $0<s<\infty$ because if $0<s\le 1$ and $r<s,$ then $\displaystyle \frac{s}{r}>1$ and hence $L^{\frac{s}{r}}(\Vert m\Vert )$ is 1-convex, that is $L^s(\Vert m\Vert )_{[r]}$ is 1-convex, which is equivalent to $L^s(\Vert m\Vert )$ be r-convex. This means that $L^s(\Vert m\Vert )$ is L-convex for all $0<s<\infty .$ In particular, $L^1(\Vert m\Vert )$ is L-convex and we can apply Corollary 5 to it.

Corollary 6

Let $\varOmega$ be a Polish space and let $\mu$ be a Borel measure which is a Rybakov control measure for m. Let $\varPhi _0, \varPhi _1\in \varDelta _2,$ $0<\theta <1$ and $\varPhi$ such that $\varPhi ^{-1} := (\varPhi _{0}^{-1})^{1-\theta } (\varPhi _{1}^{-1})^{\theta }.$ Then, $[L^{\varPhi _0}(\Vert m\Vert ),L^{\varPhi _1}(\Vert m\Vert )]_{\theta } = L^{\varPhi }(\Vert m\Vert ).$

For a similar result about complex interpolation of Orlicz type spaces $L^{\Phi}(m)$ and $L_w^{\Phi}(m)$ see [3, Corollary 4.2 and Theorem 4.5].

Note that, for $p>1,$ $\displaystyle \frac{1}{p}$-th powers are an special case of Orlicz spaces, since $X_{\left[ \frac{1}{p}\right] } = X^{\varPhi _{[p]}},$ where $\varPhi _{[p]}(t)=t^p.$ If we particularize the previous Corollary to these powers, then we obtain the interpolation result below for $L^p(\Vert m\Vert )$ spaces. In fact, this result is valid for all $0<p_0, p_1<\infty$ due to the fact that the Calderón product commutes with powers for all indices.

Corollary 7

Let $\varOmega$ be a Polish space and let $\mu$ be a Borel measure which is a Rybakov control measure for m. Let $0<\theta <1$ and $0<p_0, p_1<\infty .$ Then $[L^{p_0}(\Vert m\Vert ),L^{p_1}(\Vert m\Vert )]_{\theta } = L^{p}(\Vert m\Vert ),$ where $\displaystyle \frac{1}{p} = \frac{1-\theta }{p_0} + \frac{\theta }{p_1}.$