1 Introduction

The first appearence of generalized grand Lebesgue spaces—a class of Banach function spaces, see e.g. the monograph by Bennett and Sharpley [4]—is in the final remark in [5], where the factor against the \(L^{p-\varepsilon }\) norm has been further generalized from a power into an increasing function. The growing interest in literature for this class of spaces has been motivated either by their utility in the theory of PDEs (see e.g. [1, 8, 11]), either in Function Spaces theory (see [6, 9, 10, 16,17,18, 21]). We refer to [7] for a study of these spaces, to [13] for a survey, to [12] for a recent characterization of its norm in term of the decreasing rearrangement. The increasing interest on these spaces led to “maximize” the generalization of the original norm

$$\begin{aligned} \Vert f\Vert _{L^{p)}(0,1)}=\sup _{0<\varepsilon <p-1}\left( \varepsilon \int _0^1|f|^{p-\varepsilon }dx\right) ^\frac{1}{p-\varepsilon }, \end{aligned}$$

due to Iwaniec and Sbordone in 1992 (see [14]), into

$$\begin{aligned} \Vert f\Vert _{L^{p),\delta }(0,1)}= \mathrm{ess}\sup _{0<\varepsilon <p-1}\left( \delta (\varepsilon )\int _0^1|f|^{p-\varepsilon }dx\right) ^\frac{1}{p-\varepsilon }, \end{aligned}$$

where \(\delta \) is a nonnegative measurable bounded function on \(]0,p-1]\) (see [7, Theorem 2.1] for details; see [2, 3, 15, 19] for a generalization also with respect to \(\varepsilon \)). Even without a deep knowledge of the literature on these spaces, already the expression of the norm \(\Vert \cdot \Vert _{L^{p),\delta }(0,1)}\) suggests clearly that the most natural assumption on \(\delta \) is the property to be nondecreasing of the function

$$\begin{aligned} {\widehat{\delta }}(\varepsilon ):=\delta (\varepsilon )^{\frac{1}{p-\varepsilon }}\, , \qquad \varepsilon \in ]0,p-1]\, . \end{aligned}$$

In [12, 3.8] it has been observed that if \({\widehat{\delta }}\) is nondecreasing, then \(\delta \) is nondecreasing, and that the viceversa does not hold. Hence it is not a surprise that the main results in [7, 12] have the “strong” assumption on the monotonicity of \({\widehat{\delta }}\) and not on the monotonicity of \(\delta \).

The \(\Delta _2\) condition is a notion familiar for researchers working in Orlicz spaces (see e.g. [20]). When the so-called \(\Delta _2\) condition near the origin (we write \(\delta \in \Delta _2\) if \(\delta (2\varepsilon )\le c\delta (\varepsilon )\) for \(\varepsilon \) small, for some \(c>1\)) plays a role, again from [12, 3.8] we know that under the “weak” assumption that \(\delta \) is nondecreasing, \(\delta \in \Delta _2\Leftrightarrow {\widehat{\delta }}\in \Delta _2\).

The considerations above show that

$$\begin{aligned} {\widehat{\delta }} \,\, {nondecreasing} \qquad \text {and} \qquad {\widehat{\delta }}\in \Delta _2 \end{aligned}$$
(1.1)

is a stronger assumption with respect to

$$\begin{aligned} \delta \,\, {nondecreasing} \qquad \text {and} \qquad \delta \in \Delta _2 \end{aligned}$$
(1.2)

and the viceversa does not hold, a simple example being \(\delta (\varepsilon )\equiv 1/2\).

The novelty of this paper is that, for the theory built on generalized grand Lebesgue spaces, the weaker assumption (1.2) is sufficient for all statements containing (1.1) as hypothesis: in fact, essentially, we prove (see Theorem 1 for the precise statement) that if (1.2) holds, then \({\widehat{\delta }}\) can be replaced, up to equivalence, by its nondecreasing envelope. We may therefore sharpen a number of results (see the last Sect. 4). In next Sect. 2 we state and prove Theorem 1, and in Sect. 3 we show that the result fails without the assumption \(\Delta _2\).

2 The Main Result

Theorem 1

Let \(1<p<\infty \), and let \(\delta :]0,p-1]\rightarrow ]0,\infty [\) be continuous, nondecreasing, and satisfying the \(\Delta _2\) condition near the origin, i.e.

$$\begin{aligned} \exists c>1\,, \,\, \exists \varepsilon _0\in ]0,p-1]\quad \text {such that} \quad \delta (2\varepsilon )\le c\delta (\varepsilon )\qquad \forall \, 0<\varepsilon <\varepsilon _0\, . \end{aligned}$$
(2.1)

The function

$$\begin{aligned} {\bar{\delta }}(\varepsilon ):=\left[ \sup _{0<\zeta <\varepsilon }\delta (\zeta )^{\frac{1}{p-\zeta }}\right] ^{p-\varepsilon }, \qquad \varepsilon \in ]0,p-1] \end{aligned}$$
(2.2)

is such that for some \(M>1\)

$$\begin{aligned} \delta (\varepsilon )\le {\bar{\delta }}(\varepsilon )\le M \delta (\varepsilon ) \qquad \forall \, 0<\varepsilon \le p-1. \end{aligned}$$
(2.3)

Proof

The left wing inequality in (2.3) is an immediate consequence of the continuity of \(\delta \): in fact,

$$\begin{aligned} {\bar{\delta }}(\varepsilon )=\left[ \sup _{0<\zeta <\varepsilon }\delta (\zeta )^{\frac{1}{p-\zeta }}\right] ^{p-\varepsilon }\ge \left[ \delta (\varepsilon )^{\frac{1}{p-\varepsilon }}\right] ^{p-\varepsilon }=\delta (\varepsilon ), \end{aligned}$$

therefore the proof consists of showing the right wing inequality.

We begin observing that \({\bar{\delta }}\) is not affected, up to equivalence, multiplying \(\delta \) by a positive constant k: assuming, without loss of generality, \(k>1\), we have

$$\begin{aligned} {\bar{\delta }}(\varepsilon )&\le \overline{k\delta }(\varepsilon )=\left\{ \sup _{0<\zeta<\varepsilon }[k\delta (\zeta )]^{\frac{1}{p-\zeta }}\right\} ^{p-\varepsilon }\\&=\left[ \sup _{0<\zeta<\varepsilon }k^{\frac{1}{p-\zeta }}\delta (\zeta )^{\frac{1}{p-\zeta }}\right] ^{p-\varepsilon } \le \left[ \sup _{0<\zeta<\varepsilon }k\delta (\zeta )^{\frac{1}{p-\zeta }}\right] ^{p-\varepsilon }\\&\le k^p\left[ \sup _{0<\zeta<\varepsilon }\delta (\zeta )^{\frac{1}{p-\zeta }}\right] ^{p-\varepsilon }= k^p{\bar{\delta }}(\varepsilon )\qquad \forall \, 0<\varepsilon \le p-1\, . \end{aligned}$$

As a consequence, dividing \(\delta \) by \(2\delta (p-1)\), we may assume without loss of generality that \(\delta (\varepsilon )\in ]0,1]\) for every \(\varepsilon \in ]0,p-1]\) and, since \(\delta \) is nondecreasing, that \(\delta _0:=\delta (0+)\le 1/2\). If \(\delta _0>0\), then for some \(\varepsilon _1\in ]0,p-1]\) we have

$$\begin{aligned} \delta _0\le \delta (\zeta )<2\delta _0\qquad \forall \, 0<\zeta <\varepsilon _1\, , \end{aligned}$$

hence

$$\begin{aligned} {\bar{\delta }}(\varepsilon )= & {} \left[ \sup _{0<\zeta<\varepsilon }\delta (\zeta )^{\frac{1}{p-\zeta }}\right] ^{p-\varepsilon } \le \left[ \sup _{0<\zeta<\varepsilon } (2\delta _0)^{\frac{1}{p-\zeta }}\right] ^{p-\varepsilon }\\= & {} \sup _{0<\zeta<\varepsilon } (2\delta _0)^{\frac{p-\varepsilon }{p-\zeta }} \le (2\delta _0)^{\frac{p-\varepsilon _1}{p}} \\= & {} (2\delta _0)^{\frac{-\varepsilon _1}{p}} \cdot (2\delta _0)=2(2\delta _0)^{\frac{-\varepsilon _1}{p}}\delta _0\le 2(2\delta _0)^{\frac{-\varepsilon _1}{p}}\delta (\varepsilon ) \qquad \forall \, 0<\varepsilon <\varepsilon _1\, ; \end{aligned}$$

on the other hand,

$$\begin{aligned} {\bar{\delta }}(\varepsilon )=\frac{{\bar{\delta }}(\varepsilon )}{\delta (\varepsilon )}\delta (\varepsilon )\le \frac{1}{\delta (\varepsilon _1)}\delta (\varepsilon ) \qquad \forall \, \varepsilon _1<\varepsilon \le p-1\, . \end{aligned}$$
(2.4)

The two relations above show that in the case \(\delta _0>0\), then (2.3) holds with \(M=\max \{2(2\delta _0)^{\frac{-\varepsilon _1}{p}}, 1/\delta (\varepsilon _1)\}\). In the following we may therefore assume that \(\delta (0+)=0\).

For every \(\varepsilon \in ]0,\min \{1,p-1\}[\) let \(n=n(\varepsilon )\in {{\mathbb {N}}}\) be such that

$$\begin{aligned} 2^{-n}\le \varepsilon <2^{-n+1}. \end{aligned}$$
(2.5)

By continuity of \(\delta \), \(\delta (0+)=0\) and, on the other hand, we recall that \(\delta (\zeta )>0\) for \(\zeta \in ]0,p-1]\); therefore for every \(\varepsilon \in ]0,\min \{1,p-1\}[\) there exists \(\zeta _\varepsilon \in ]0,\varepsilon ]\) such that

$$\begin{aligned} \sup _{0<\zeta <\varepsilon }\delta (\zeta )^{\frac{1}{p-\zeta }}=\delta (\zeta _\varepsilon )^{\frac{1}{p-\zeta _\varepsilon }}; \end{aligned}$$
(2.6)

let \(m=m(\varepsilon )\in {{\mathbb {N}}}\), \(m\ge n\), be such that \(2^{-m}\le \zeta _\varepsilon <2^{-m+1}\). Hence for every \(\varepsilon \in ]0,\min \{1,p-1\}[\) we have the existence of integers n, m such that \(m\ge n\), both depending on \(\varepsilon \), for which

where the last inequality is due to:

$$\begin{aligned} \zeta _\varepsilon<2^{-m+1}\Rightarrow & {} \delta (\zeta _\varepsilon )\le \delta (2^{-m+1})\\ \varepsilon <2^{-n+1}\Rightarrow & {} p-\varepsilon >p-2^{-n+1}\\ \zeta _\varepsilon \ge 2^{-m}\Rightarrow & {} p-\zeta _\varepsilon \le p-2^{-m}\\ \varepsilon \ge 2^{-n}\Rightarrow & {} \delta (\varepsilon )\ge \delta (2^{-n}). \end{aligned}$$

We go on with the estimate as follows:

$$\begin{aligned}&\mathbf{A}=\frac{\delta (2^{-m+1})^{\frac{p-2^{-n+1}}{p-2^{-m}}}}{\delta (2^{-n})}\\&\quad =\frac{\delta (2^{-m+1})}{\delta (2^{-n})}\delta (2^{-m+1})^{\frac{p-2^{-n+1}}{p-2^{-m}}-1}\\&\quad =\frac{\delta (2^{-m+1})}{\delta (2^{-n})}\delta (2^{-m+1})^{\frac{2^{-m}-2^{-n+1}}{p-2^{-m}}}\\&\quad =\frac{\delta (2^{-m+1})}{\delta (2^{-n})}\delta (2^{-m+1})^{\frac{2^{-m}}{p-2^{-m}}}\delta (2^{-m+1})^{-\frac{2^{-n+1}}{p-2^{-m}}}\le \frac{\delta (2^{-m+1})^{1-\frac{2^{-n+1}}{p-2^{-m}}}}{\delta (2^{-n})}:=\mathbf{B}\, , \end{aligned}$$

where the last inequality is due to the fact that \(0<\delta (2^{-m+1})<1\) and \(\frac{2^{-m}}{p-2^{-m}}>0\).

Now fix any \(n_0\in {{\mathbb {N}}}\) such that

$$\begin{aligned} n_0>1-\log _2\left( \min \{\varepsilon _0, p-1\}\right) ; \end{aligned}$$
(2.7)

note that it depends only on p and \(\delta \) and that from (2.7) we have

$$\begin{aligned} -n_0+1<\log _2\left( \min \{\varepsilon _0, p-1\}\right) \, , \end{aligned}$$

from which

$$\begin{aligned} \varepsilon _2:=2^{-n_0+1}<\min \{\varepsilon _0, p-1\}\, , \end{aligned}$$
(2.8)

where \(\varepsilon _0\) is from assumption (2.1).

Consider for the moment \(\varepsilon \) in the interval \(]0,\varepsilon _2[\), so that from (2.5)

$$\begin{aligned} 2^{-n}\le \varepsilon <2^{-n_0+1}\, ; \end{aligned}$$

this means that \(-n<-n_0+1\), i.e. \(n\ge n_0\), hence, from (2.8), we have

$$\begin{aligned} 2^{-n}\le \varepsilon<2^{-n+1}\le 2^{-n_0+1}<\min \{\varepsilon _0, p-1\}. \end{aligned}$$
(2.9)

This chain of inequalities gives us the possibility to establish that the exponent in the numerator of B is positive: in fact

$$\begin{aligned} 1-\frac{2^{-n+1}}{p-2^{-m}}>1-\frac{p-1}{p-1}=0\, . \end{aligned}$$

Hence, taking into account that \(m\ge n\) and that \(\delta \) is nondecreasing,

$$\begin{aligned} \mathbf{B}=\frac{\delta (2^{-m+1})^{1-\frac{2^{-n+1}}{p-2^{-m}}}}{\delta (2^{-n})}\le \frac{\delta (2^{-n+1})^{1-\frac{2^{-n+1}}{p-2^{-m}}}}{\delta (2^{-n})}:=\mathbf{C}\, . \end{aligned}$$

By (2.9) and (2.1)

$$\begin{aligned} \delta (2^{-n+1})=\delta (2\cdot 2^{-n})\le c\delta (2^{-n}) \end{aligned}$$
(2.10)

and therefore we may estimate C as follows:

$$\begin{aligned}&\mathbf{C}=\frac{\delta (2^{-n+1})^{1-\frac{2^{-n+1}}{p-2^{-m}}}}{\delta (2^{-n})}\le \frac{\delta (2^{-n+1})^{1-\frac{2^{-n+1}}{p-2^{-m}}}}{c^{-1}\delta (2^{-n+1})}\\&\quad = c\delta (2^{-n+1})^{-\frac{2^{-n+1}}{p-2^{-m}}}\\&\quad = c\left[ \frac{1}{\delta (2^{-n+1})}\right] ^{\frac{2^{-n+1}}{p-2^{-m}}}:=\mathbf{D}\, . \end{aligned}$$

Observe that, since \(m\ge n\), we have \(2^{-m}\le 2^{-n}\) and therefore, by (2.9), \(2^{-m}\le p-1\), from which \(p-2^{-m}\ge 1\). On the other hand, by (2.8), iterating the argument in (2.10), we have

$$\begin{aligned} \delta (2^{-n_0+1})\le & {} c\delta (2^{-n_0})\le c^2\delta (2^{-n_0-1})\le \cdots \le c^k\delta (2^{-n_0-k+1})\\\le & {} \cdots \le c^{n-n_0}\delta (2^{-n+1})\,. \end{aligned}$$

We may therefore estimate

$$\begin{aligned}&\mathbf{D}=c\left[ \frac{1}{\delta (2^{-n+1})}\right] ^{\frac{2^{-n+1}}{p-2^{-m}}}\\&\quad \le c\left[ \frac{1}{c^{n_0-n}\delta (2^{-n_0+1})}\right] ^{2^{-n+1}}\\&\quad = c\left[ \frac{1}{c^{n_0}\delta (2^{-n_0+1})}\right] ^{2^{-n+1}}c^{n2^{-n+1}}\rightarrow c\, . \end{aligned}$$

Overall, we obtained that D is smaller than a term of a bounded sequence depending only on \(\delta \), c (from (2.1)), and \(n_0\) (which in turn depends again on (2.1) and p), hence \({\bar{\delta }}(\varepsilon )/\delta (\varepsilon )\) is bounded for \(0<\varepsilon <\varepsilon _2\), where \(\varepsilon _2\) depends on \(n_0\), too. Finally, we conclude arguing as in (2.4):

$$\begin{aligned} {\bar{\delta }}(\varepsilon )=\frac{{\bar{\delta }}(\varepsilon )}{\delta (\varepsilon )}\delta (\varepsilon )\le \frac{1}{\delta (\varepsilon _2)}\delta (\varepsilon ) \qquad \forall \, \varepsilon _2<\varepsilon \le p-1. \end{aligned}$$

\(\square \)

3 A Counterexample

The assumption \(\delta \in \Delta _2\) in Theorem 1 cannot be dropped. We are going to exhibit an example of function \(\delta \) continuous and nondecreasing, such that \({\bar{\delta }}\not \approx \delta \) (i.e., as usual, positive constants \(c_1,c_2\) such that \(c_1\delta (\varepsilon )\le {\bar{\delta }}(\varepsilon )\le c_2\delta (\varepsilon )\) for every \(\varepsilon \) small cannot exist).

Example 1

Let \(1<p<\infty \), and let \(\delta :]0,p-1]\rightarrow ]0,\infty [\) be defined, close to the origin, as follows:

$$\begin{aligned} \delta (\varepsilon ):={\left\{ \begin{array}{ll}\exp (-5^n)\,\,\,\,\,\qquad \qquad \text {if} \,\,\varepsilon \in \left[ a_n,b_n\right] \\ \text {affine and continuous in} \left[ b_{n+1},a_n\right] \end{array}\right. }\,\, ,\qquad n\in {{\mathbb {N}}}\quad \text {large} \end{aligned}$$

where

$$\begin{aligned} a_n=2^{-n}\, , \qquad b_n=2^{-n}+\frac{p-2^{-n}}{4^n}\, . \end{aligned}$$

We stress that the definition is well posed, because

$$\begin{aligned}&b_{n+1}<a_n\, \Leftrightarrow \, 2^{-n-1}\\&\quad +\frac{p-2^{-n-1}}{4^{n+1}}<2^{-n}\, \Leftrightarrow \, p-2^{-n-1}<4^{n+1}(2^{-n}-2^{-n-1})=2^{n+1}\, , \end{aligned}$$

which is true for n large. The function \(\delta \) is continuous by definition and clearly nondecreasing (because the sequence \(\left( \exp (-5^n)\right) _{n\in {{\mathbb {N}}}}\) is decreasing).

Since \(a_n<b_n\) and \(\delta (a_n)=\delta (b_n)\), we have

$$\begin{aligned} \frac{{\bar{\delta }}(b_n)}{\delta (b_n)}&=\frac{1}{\delta (b_n)} \left[ \sup _{0<\zeta <b_n}\delta (\zeta )^{\frac{1}{p-\zeta }}\right] ^{p-b_n}\ge \frac{1}{\delta (b_n)}\left[ \delta (a_n)^{\frac{1}{p-a_n}}\right] ^{p-b_n}\\&=\delta (a_n)^{\frac{p-b_n}{p-a_n}-1} =[\exp (-5^n)]^{-\frac{1}{4^n}}\rightarrow \infty \, , \end{aligned}$$

and therefore \({\bar{\delta }}\not \approx \delta \).

We observe also that \({\widehat{\delta }}\) cannot be nondecreasing, otherwise it would be \({\bar{\delta }}=\delta \). The lost monotonicity of \({\widehat{\delta }}\) could be verified also directly, because of the constant behavior of \(\delta \) in the intervals \(\left[ a_n,b_n\right] \).

Finally, we observe that \(\delta \notin \Delta _2\): this is a direct consequence of our Theorem 1, but it can be verified also directly. In fact, writing explicitly the values of \(a_n\) and \(b_{n+1}\), it is immediate to realize that \(a_n<2b_{n+1}\), hence

$$\begin{aligned} \frac{\delta (2b_{n+1})}{\delta (b_{n+1})}\ge \frac{\delta (a_n)}{\delta (b_{n+1})}= \frac{\exp (-5^n)}{\exp (-5^{n+1})}=\exp (4\cdot 5^n)\uparrow \infty \, . \end{aligned}$$

4 Applications

4.1 On the Definition of Generalized Grand Lebesgue Spaces

After [7], the interesting class of spaces \(L^{p),\delta }\) is for functions \(\delta \) defined pointwise, therefore the norm can be written with \(\sup \) instead of \( \mathrm{ess}\sup _{}\quad \,\,\,\,\):

$$\begin{aligned} \Vert f\Vert _{L^{p),\delta }(0,1)}= & {} \sup _{0<\varepsilon<p-1}\left( \delta (\varepsilon )\int _0^1|f|^{p-\varepsilon }dx\right) ^\frac{1}{p-\varepsilon }\nonumber \\= & {} \sup _{0<\varepsilon <p-1}{\widehat{\delta }}(\varepsilon )\left( \int _0^1|f|^{p-\varepsilon }dx\right) ^\frac{1}{p-\varepsilon }. \end{aligned}$$
(4.1)

Again after [7], the whole class of spaces is covered by the assumption \({\widehat{\delta }}\) nondecreasing, and this is in agreement with the heart of these spaces, which is based on the monotonicity of the Lebesgue norm \(\Vert \cdot \Vert _{p-\varepsilon }\), consequence of the Hölder’s inequality (see [13] for details). With this assumption in order, one has that \(\delta \) itself is nondecreasing, too.

If one replaces the assumption “\({\widehat{\delta }}\) nondecreasing” with “\(\delta \) nondecreasing”, then one gets functions \({\widehat{\delta }}\) which are not nondecreasing (see Example 1): in principle, it gives a wider range of spaces. However, in [7] it is shown that replacing \(\delta \) by \({\bar{\delta }}\) one has an equivalent norm, and the corresponding \(\widehat{{\bar{\delta }}}\) is nondecreasing. The reader must be aware of the fact that in this case the norms are equivalent, but \(\delta \) and \({\bar{\delta }}\) are not necessarily equivalent. After our Theorem 1, they are equivalent if and only if \(\delta \in \Delta _2\). This essentially confirms that “\({\widehat{\delta }}\) nondecreasing” is the best option for the consideration of the whole class of spaces.

The point is that several results using techniques from Interpolation-Extrapolation theory involve the \(\Delta _2\) condition. With the addition of this assumption, after our Theorem 1, we know that the replacement of \(\delta \) by \({\bar{\delta }}\) gives not only the equivalence of the norms, but also the equivalence \({\bar{\delta }}\approx \delta \).

The above considerations lead to state the following

Theorem 2

If \(1<p<\infty \) and \(\delta ,\delta _1:]0,p-1]\rightarrow ]0,\infty [\) are continuous and nondecreasing, then

$$\begin{aligned} \delta \approx \delta _1\quad \Rightarrow \quad L^{p),\delta }=L^{p),\delta _1}\, , \end{aligned}$$

and the viceversa does not hold. Moreover,

$$\begin{aligned} {\bar{\delta }}\approx \delta \quad \Rightarrow \quad L^{p),{\bar{\delta }}}=L^{p),\delta }\, , \end{aligned}$$

and the viceversa does not hold. If, moreover, \(\delta \in \Delta _2\), then

$$\begin{aligned} {\bar{\delta }}\approx \delta \quad \Leftrightarrow \quad L^{p),{\bar{\delta }}}=L^{p),\delta }\, . \end{aligned}$$

We stress that the last sentence in Theorem 2 comes from the fact that when \(\delta \) is nondecreasing and \(\Delta _2\), by Theorem 1 we know that \({\bar{\delta }}\approx \delta \) without the use of \(L^{p),{\bar{\delta }}}=L^{p),\delta }\).

4.2 On the Fundamental Function of Generalized Grand Lebesgue Spaces

Let \(1<p<\infty \), and let \(\delta :]0,p-1]\rightarrow ]0,1]\) be continuous. From [12, 3.7] and [7, Proposition 2.3] we know that if \(\delta \) is such that \({\widehat{\delta }}(\cdot ):=\delta (\cdot )^{\frac{1}{p-\cdot }}\) is nondecreasing and \(\Delta _2\) near the origin, then, setting

$$\begin{aligned} A(t):=t^p\delta \left( \frac{p-1}{\log (e+t)}\right) \, , \qquad t\ge 0\, , \end{aligned}$$

and denoting by \(\varphi _X\) the fundamental function of a rearrangement invariant Banach function space X, the following holds:

(i):

\(\delta (t)\approx {\bar{\delta }}(t)\)  ,    t small

(ii):

\({\mathbb \varphi }_{L^{A}}(t)\approx t^{\frac{1}{p}}\delta \left( \frac{1}{\log (e+1/t)}\right) ^{\frac{1}{p}}\)  ,    t small

(iii):

\({\mathbb \varphi }_{L^{A}}(t)\approx t^{\frac{1}{p}}{\bar{\delta }}\left( \frac{1}{\log (e+1/t)}\right) ^{\frac{1}{p}}\)  ,    t small

(iv):

\(L^{p),\delta }(0,1)=L^{p),{\bar{\delta }}}(0,1)\)

(v):

\({\mathbb \varphi }_{L^{p),\delta }}\approx t^{\frac{1}{p}}\delta \left( \frac{1}{\log (e+1/t)}\right) ^{\frac{1}{p}}\)  ,    t small

Here there are the easy proofs or references: (i) since \({\widehat{\delta }}\) is nondecreasing, we have \(\delta ={\bar{\delta }}\), hence trivially \(\delta \approx {\bar{\delta }}\); (ii) comes from direct computation, taking into account that from the classical theory \({\mathbb \varphi }_{L^{A}}(t)=1/A^{-1}(1/t)\); (iii) is obvious from (i); (iv) was proved in [7, Proposition 2.3]; (v) comes from the fact that by [12, 3.8] \({\widehat{\delta }}\) is nondecreasing implies that \(\delta \) is nondecreasing, and therefore, since \({\widehat{\delta }}\in \Delta _2\), again by [12, 3.8], also \(\delta \in \Delta _2\); hence by [12, 3.5] we have \({\mathbb \varphi }_{L^{p),\delta }}\approx t^{\frac{1}{p}}{\widehat{\delta }}\left( \frac{1}{\log (e+1/t)}\right) \) and this implies \({\mathbb \varphi }_{L^{p),\delta }}\approx t^{\frac{1}{p}}\delta \left( \frac{1}{\log (e+1/t)}\right) ^{\frac{1}{p}}\).

The arguments above use heavily the assumptions on \({\widehat{\delta }}\), even if this function does not appear explicitly in (i)-(v). After our Theorem 1, all the statements (i)-(v) are still true in the weaker assumption that \(\delta \) is nondecreasing and \(\Delta _2\). Note that such weaker assumption allows for instance \(\delta \equiv 1/2\) (corresponding to \(A(t)=(1/2)t^p\)), which is a case not included in the original assumption, implicit in [12, 3.7], that \(\delta \) must be such that \({\widehat{\delta }}(\cdot ):=\delta (\cdot )^{\frac{1}{p-\cdot }}\) is nondecreasing.

If one drops the assumption \(\Delta _2\) and assumes just \(\delta \) nondecreasing, then Example 1 shows that (i), (iii) and (v) in general fail, while (ii) and (iv) remain true (because they have been proved without the \(\Delta _2\) assumption).

We stress that the problem of a complete characterization of the behavior of the fundamental function in generalized grand Lebesgue spaces remains still open.

Remark 1

Theorem 1 shows that the sentences in [12, 3.7] are correct. The equivalences stated therein use implicitly the stronger assumption that \(\delta \) (called \(\varphi \) in [12]) must be such that \({\widehat{\delta }}(\cdot ):=\delta (\cdot )^{\frac{1}{p-\cdot }}\) is nondecreasing, however, their correct (omitted and detailed) justification is a consequence of our main result.

4.3 The Extension of the Validity of a Sharp Blow-up Estimate

Let \(1<p<\infty \), and let \(\psi :]0,p-1]\rightarrow ]0,\infty [\) be continuous and nondecreasing. In this assumption we know from [12] that if, moreover, \(\psi \in \Delta _2\), then

$$\begin{aligned} \sup _{0<\varepsilon<p-1} \psi (\varepsilon ) \Vert f\Vert _{L^{p-\varepsilon }(0,1)} \approx \sup _{0<t<1}\psi \left( \frac{p-1}{1-\log t}\right) \Vert f^*\Vert _{L^{p}(t,1)}\, . \end{aligned}$$
(4.2)

With respect to the notation in (4.1), this means that

$$\begin{aligned} \sup _{0<\varepsilon<p-1} {\widehat{\delta }}(\varepsilon ) \Vert f\Vert _{L^{p-\varepsilon }(0,1)} \approx \sup _{0<t<1}{\widehat{\delta }}\left( \frac{p-1}{1-\log t}\right) \Vert f^*\Vert _{L^{p}(t,1)}\, . \end{aligned}$$
(4.3)

The assumptions on \(\psi \), in terms of \(\delta \), entrain—as we already observed above—that \(\delta \) is nondecreasing and \(\Delta _2\), but the viceversa does not hold. After our Theorem 1, we can assert that (4.3) holds also assuming just \(\delta \) nondecreasing and \(\Delta _2\). In fact, we know that \({\bar{\delta }}\approx \delta \) and therefore also \({\bar{\delta }}\in \Delta _2\) and \({\bar{\delta }}(\cdot )^\frac{1}{p-\cdot }\in \Delta _2\). Moreover, since by definition of \({\bar{\delta }}\) we have that \({\bar{\delta }}(\cdot )^\frac{1}{p-\cdot }\) is nondecreasing,

$$\begin{aligned}&\sup _{0<\varepsilon<p-1} {\widehat{\delta }}(\varepsilon ) \Vert f\Vert _{L^{p-\varepsilon }(0,1)}\\&\quad =\sup _{0<\varepsilon<p-1}\left( \delta (\varepsilon )\int _0^1|f|^{p-\varepsilon }dx\right) ^\frac{1}{p-\varepsilon }\approx \sup _{0<\varepsilon<p-1}\left( {\bar{\delta }}(\varepsilon )\int _0^1|f|^{p-\varepsilon }dx\right) ^\frac{1}{p-\varepsilon }\\&\quad =\sup _{0<\varepsilon<p-1}{\bar{\delta }}(\varepsilon )^\frac{1}{p-\varepsilon }\left( \int _0^1|f|^{p-\varepsilon }dx\right) ^\frac{1}{p-\varepsilon }\approx \sup _{0<t<1}{\bar{\delta }}\left( \frac{p-1}{1-\log t}\right) ^\frac{1}{p-\frac{p-1}{1-\log t}}\Vert f^*\Vert _{L^{p}(t,1)}\\&\quad \approx \sup _{0<t<1}\delta \left( \frac{p-1}{1-\log t}\right) ^\frac{1}{p-\frac{p-1}{1-\log t}}\Vert f^*\Vert _{L^{p}(t,1)} =\sup _{0<t<1}{\widehat{\delta }}\left( \frac{p-1}{1-\log t}\right) \Vert f^*\Vert _{L^{p}(t,1)}\, . \end{aligned}$$

An extension of (4.2) to the case where the \(\Delta _2\) assumption is dropped is still an open problem.