## 1 Introduction

Let $$x=(x_1,\ldots ,x_n).$$ Denote by $$\hat{x}_k$$ the $$(n-1)$$-dimensional vector obtained from the n-tuple x by removal of its k-th coordinate. We shall write $$x=(x_k,\hat{x}_k).$$

Let $$\psi _1,\ldots ,\psi _n$$ be nonnegative measurable functions on $$\mathbb {R}^{n-1}~~~~(n\ge 2).$$ We consider the geometric mean

\begin{aligned} \mathcal {G}(x)=\left( \prod _{k=1}^n\psi _k(\hat{x}_k)\right) ^{1/n},\quad x\in \mathbb {R}^n. \end{aligned}

Gagliardo [12] proved the following theorem.

### Theorem 1.1

Assume that $$n\ge 2$$ and $$\psi _k\in L^1(\mathbb {R}^{n-1})$$    $$(k=1,\ldots ,n).$$ Then $$\mathcal {G}\in L^{n'}(\mathbb {R}^n)~~~(n'=n/(n-1))$$ and

\begin{aligned} ||\mathcal {G}||_{L^{n'}(\mathbb {R}^n)}\le \left( \prod _{k=1}^n||\psi _k||_{L^1(\mathbb {R}^{n-1})}\right) ^{1/n}. \end{aligned}
(1.1)

This theorem yields the embedding of the Sobolev space $$W_1^1(\mathbb {R}^n)$$ into $$L^{n'}(\mathbb {R}^n)$$ (see [12]). Later on, it was shown in [2] (see also [10, 11, 25]) that the space $$W_1^1(\mathbb {R}^n)$$ is embedded into the Lorentz space $$L^{n',1}(\mathbb {R}^n)$$ which is strictly smaller than $$L^{n'}(\mathbb {R}^n)$$. We emphasize that a similar refinement of Theorem 1.1 is not true; that is, the function $$\mathcal {G}$$ may not belong to $$L^{n',1}(\mathbb {R}^n)$$ (see [19, Remark 3.12]). However, set

\begin{aligned} \Psi (x)=\min _{1\le k\le n}\psi _k(\hat{x}_k), \quad x\in \mathbb {R}^n. \end{aligned}
(1.2)

Clearly, $$\Psi (x)\le \mathcal G(x).$$ The following theorem holds [11].

### Theorem 1.2

Assume that $$n\ge 2$$ and $$\psi _k\in L^1(\mathbb {R}^{n-1})$$ $$(\psi _k \ge 0,\,\,k=1,\ldots ,n).$$ Then $$\Psi \in L^{n',1}(\mathbb {R}^n)$$ and

\begin{aligned} ||\Psi ||_{L^{n',1}(\mathbb {R}^n)}\le \left( \prod _{k=1}^n||\psi _k||_{L^1(\mathbb {R}^{n-1})}\right) ^{1/n}. \end{aligned}
(1.3)

It is important to note that there are normalizing factors in the definitions of Lorentz norms given in Sect. 2.

Theorem 1.2 implies the embedding of $$W_1^1(\mathbb {R}^n)$$ into the Lorentz space $$L^{n',1}(\mathbb {R}^n).$$ Observe that this theorem was obtained by Fournier [11] in an equivalent form in terms of mixed norm spaces $$L^1_{\hat{x}_k}[L^\infty _{x_k}]$$ (see Sect. 3).

Different extensions of Theorem 1.2 and their applications have been studied in the works [1, 7, 17, 19, 23].

As in [1, 19], in this paper we consider Lorentz spaces defined in terms of iterated rearrangements. Let $$\Omega _n$$ be the collection of all permutations of the set $$\{1,\dots ,n\}$$. For each $$\sigma \in \Omega _n,$$ and $$0<p, r<\infty$$, we define a Lorentz space $$\mathcal L^{p,r}_{\sigma }(\mathbb R ^n)$$ (see Sect. 2). The relations between $$\mathcal L^{p,r}_{\sigma }$$-spaces and the classical Lorentz $$L^{p,r}$$-spaces are the following:

\begin{aligned} \mathcal L^{p,r}_\sigma \subset L^{p,r}\,~~ (r\le p),\quad L^{p,r}\subset \mathcal L^{p,r} _\sigma ~~ (p\le r), \end{aligned}
(1.4)

and for $$p\not =r$$ these embeddings are strict (see [26]).

We proved in [16] that Sobolev embeddings can be strengthened using the $$\mathcal L$$-norms. In particular, we have that for any $$f\in W_1^1(\mathbb {R}^n)$$

\begin{aligned} ||f||_{\mathcal L^{n',1}_\sigma }\le c \left( \prod _{k=1}^n||D_k f||_1\right) ^{1/n} \quad (n\ge 2,~~~~\sigma \in \Omega _n). \end{aligned}
(1.5)

By virtue of (1.4), (1.5) implies embedding $$W_1^1(\mathbb {R}^n)\subset L^{n',1}(\mathbb {R}^n).$$

In [1], there were proved estimates of generalized Lorentz norms in terms of mixed norm spaces $$L^1_{\hat{x}_k}[L^\infty _{x_k}]$$. These estimates provided a strengthening of Fournier’s estimates of classical Lorentz norms [11], but the constants were not optimal. Optimal constants were obtained in [19], where we proved the following extension of Theorem 1.2.

### Theorem 1.3

Assume that $$\psi _k\in L^1(\mathbb {R}^{n-1})$$ $$(\psi _k\ge 0,\,\,k=1,\ldots ,n).$$ Let $$\Psi$$ be defined by (1.2). Then $$\Psi \in \mathcal L^{n',1}_\sigma (\mathbb {R}^n)$$ and

\begin{aligned} ||\Psi ||_{\mathcal {L}^{n',1}_\sigma }\le \left( \prod _{k=1}^n||\psi _k||_{L^1(\mathbb {R}^{n-1})}\right) ^{1/n} \end{aligned}
(1.6)

for any $$\sigma \in \Omega _n$$.

Since $$||f||_{ L^{n',1}}\le ||f||_{\mathcal L^{n',1}_\sigma }$$ (see (2.5) below), inequality (1.6) gives a refinement of inequality (1.3). Inequality (1.6) cannot be improved; it becomes equality if $$\psi _k$$ are equal to the characteristic function of the unit cube $$[0,1]^{n-1}.$$

It follows from (1.6) that inequality (1.5) holds with $$c=1/2$$. This constant is optimal (see [19]).

We see that estimates of functions defined as minimum provide sharp embeddings of Sobolev spaces $$W_1^1(\mathbb {R}^n).$$ The main objective of this paper is to study another type of such functions. Let $$f_j$$ be nonnegative rearrangeable functions on $$\mathbb {R}$$. Set

\begin{aligned} f(x)=\min _{1\le j\le n}f_j(x_j), \,\,\, x=(x_1,\ldots ,x_n)\in \mathbb {R}^n. \end{aligned}
(1.7)

We observe that in the case, when the functions $$f_j$$ vanish outside $$\mathbb {R}_+$$ and decrease on $$\mathbb {R}_+$$, this definition and the expression for the rearrangement $$f^*(t)$$ (see Sect. 4) are closely related to the concept of average modulus of continuity (see [15, 18, p. 51]). This concept was useful in the study of embeddings of some anisotropic function classes.

We prove that

\begin{aligned} ||f||_{L^{n,1}}\le \left( \prod _{k=1}^n||f_k||_{L^1}\right) ^{1/n}. \end{aligned}
(1.8)

Further, we show that

\begin{aligned} \Psi ^*(t)\le f^*(t^{n-1})\quad \text{ for } \text{ all }\quad t>0 \end{aligned}
(1.9)

(where $$\Psi$$ is defined by (1.2) and $$f_j=\psi _j^*$$ in (1.7)). This implies that estimate (1.8) (with $$f_j=\psi _j^*$$) gives a strengthening of the estimate (1.3).

Our main results are related to generalized Lorentz norms (defined in terms of iterated rearrangements). A crucial role here is played by the following equality involving iterated rearrangements of the function (1.7): for any $$\alpha _j>0$$ and any $$\sigma \in \Omega _n$$

\begin{aligned} \int _{\mathbb {R}^n_+} \prod _{j=1}^n x_j^{\alpha _j-1}\mathcal R_\sigma f(x)\,dx=\left( \prod _{j=1}^n\alpha _j\right) ^{-1}\int _0^\infty \prod _{j=1}^n\lambda _{f_j}(y)^{\alpha _j}\,dy. \end{aligned}
(1.10)

First, this equality immediately implies that

\begin{aligned} ||f||_{\mathcal L^{p,1}_\sigma }=||f||_{L^{p,1}}\quad \text{ for } \text{ any }\quad \sigma \in \Omega _n\quad \text{ and } \text{ any } \quad p>0. \end{aligned}
(1.11)

The phenomenon exhibited in (1.11) is closely connected with an important geometric property of functions f defined by (1.7). Denote by $$\mathcal P_n$$ the class of all measurable nonnegative functions on $$\mathbb {R}^n$$ such that for any $$y>0$$ the Lebesgue set $$\{x\in \mathbb {R}^n: f(x)>y\}$$ is a cartesian product of measurable sets $$E_i(y)\subset \mathbb {R}.$$ It is obvious that any function f defined by (1.7) belongs to $$\mathcal P_n.$$ We prove that the inverse statement also is true, that is, any function $$f\in \mathcal P_n$$ can be represented as in (1.7). Thus, for any function $$f\in \mathcal P_n$$ its classical and generalized Lorentz norms coincide.

Another immediate consequence of (1.10) is the following inequality: if $$\sum _{k=1}^n\alpha _k=1$$ and $$f_j\in L^1(\mathbb {R})$$, then for any $$\sigma \in \Omega _n$$

\begin{aligned} \int _{\mathbb {R}^n} \prod _{j=1}^n x_j^{\alpha _j-1}\mathcal R_\sigma f(x)\,dx\le \left( \prod _{j=1}^n\alpha _j\right) ^{-1}\prod _{j=1}^n||f_j||_1^{\alpha _j}. \end{aligned}
(1.12)

Taking $$\alpha _j=1/n\,\, (j=1,\ldots ,n)$$ we obtain the inequality

\begin{aligned} \int _{\mathbb {R}^n} \prod _{j=1}^n x_j^{1/n-1}\mathcal R_\sigma f(x)\,dx\le n^n\left( \prod _{j=1}^n||f_j||_1\right) ^{1/n}. \end{aligned}
(1.13)

By (1.11), this inequality is the same as inequality (1.8). We show that inequality (1.13) (with $$f_j=\psi _j^*$$) is stronger than inequality (1.6).

Note also that (1.13) has an equivalent form in terms of mixed norm spaces $$L^1_{x_k}[L^\infty _{\hat{x}_k}]$$ (see Sect. 3).

We observe that our interest to these questions is inspired by their connections with embedding theorems and Loomis–Whitney type inequalities. Recall that Loomis–Whitney inequality [20] states that for any $$F_\sigma$$-set $$E\subset \mathbb {R}^n$$

\begin{aligned} {\text {mes}}_n E\le \prod _{k=1}^n ({\text {mes}}_{n-1}E_k)^{1/(n-1)}, \end{aligned}
(1.14)

where $$E_k$$ is the orthogonal projection of E onto the coordinate hyperplane $$x_k=0$$. Actually (1.14) can be immediately derived from inequality (1.1). Successively applying (1.14), one can obtain a more general inequality (see [14, Chapter 4, 4.4.2])

\begin{aligned} {\text {mes}}_n E\le \prod _{1\le i_1<\cdots < i_m\le n} ({\text {mes}}_{n-m}E_{i_1,\ldots ,i_m})^{1/\left( {\begin{array}{c}n-1\\ m\end{array}}\right) }, \end{aligned}
(1.15)

where $$E_{i_1,\ldots ,i_m}$$ is the projection of E onto $$(n-m)$$-dimensional coordinate plane $$x_{i_1}=\cdots =x_{i_m}=0$$ $$(1\le m\le n-1)$$. The case $$m=1$$ corresponds to the function (1.2). The case $$m=n-1$$ (in which (1.15) is obvious) corresponds to the function (1.7). Probably, it would be interesting to consider similar minimum-functions for any $$1<m<n-1.$$

The paper is organized as follows. In Sect. 2 we give the basic definitions of rearrangements and Lorentz spaces. In Sect. 3 we define mixed norm spaces and study some of their properties. In Sect. 4 we obtain some simple results concerning estimates of rearrangements and classical Lorentz norms of functions (1.2) and (1.7). Section 5 is devoted to the main results of this paper related to generalized Lorentz norms of the function (1.7). In Sect. 6 we show that $$\mathcal L_\sigma ^{n',1}$$-norms of the function (1.2) are majorized by $$\mathcal L_\sigma ^{n,1}$$-norms of the function (1.7) (where $$f_j=\psi _j^*$$).

## 2 Rearrangements and Lorentz spaces

A measurable and almost everywhere finite real-valued function f on $$\mathbb {R}^n$$ is said to be rearrangeable if

\begin{aligned} \lambda _f (y)= | \{x \in \mathbb {R}^n : |f(x)|>y \}| < \infty \quad \text {for each y>0}. \end{aligned}

A nonincreasing rearrangement of a rearrangeable function f defined on $$\mathbb {R}^n$$ is a nonnegative and nonincreasing function $$f^*$$ on $$\mathbb {R}_+ \equiv (0, + \infty )$$ which is equimeasurable with |f|, that is, $$\lambda _{f^*}=\lambda _f.$$ We shall assume in addition that the rearrangement $$f^*$$ is left-continuous on $$\mathbb {R}_+.$$ Under this condition it is defined uniquely by

\begin{aligned} f^*(t)=\inf \{y>0: \lambda _f (y)<t\}, \quad 0<t<\infty . \end{aligned}
(2.1)

Besides, we have the equality (see [9, p. 32])

\begin{aligned} f^*(t) = \sup _{|E|=t} \inf _{x \in E} |f(x)|,\quad 0<t<\infty . \end{aligned}
(2.2)

Let $$0<p,r<\infty .$$ A rearrangeable function f on $$\mathbb {R}^n$$ belongs to the Lorentz space $$L^{p,r}(\mathbb {R}^n)$$ if

\begin{aligned} \Vert f\Vert _{L^{p,r}}=\Vert f\Vert _{p,r} = \left( \frac{r}{p} \int _0^\infty \left( t^{1/p} f^*(t) \right) ^r\, \frac{dt}{t} \right) ^{1/r} < \infty . \end{aligned}

We emphasize that the latter integral is multiplied by r/p (as in the original definitions of Lorentz [21, 22]).

We have that $$||f||_{p,p}=||f||_p.$$ For a fixed p, the Lorentz spaces $$L^{p,r}$$ strictly increase as the secondary index r increases; that is, the strict embedding $$L^{p,r}\subset L^{p,s}~~~(r<s)$$ holds (see [5, Ch. 4]).

The Lorentz quasinorm can be given in an alternative form. Namely,

\begin{aligned} ||f||_{p,r}=\left( r\int _0^\infty y^{r-1}\lambda _f(y)^{r/p}\,dy\right) ^{1/r} \end{aligned}
(2.3)

(see [13, Proposition 1.4.9]).

Let f be a rearrangeable function on $$\mathbb {R}^n$$ and let $$1\le k \le n.$$ We fix $$\hat{x}_k\in \mathbb R^{n-1}$$ and consider the $$\hat{x}_k$$-section of the function f

\begin{aligned} f_{\hat{x}_k}(x_k)=f(x_k,\hat{x}_k), \quad x_k\in \mathbb R. \end{aligned}

For almost all $$\hat{x}_k\in \mathbb R^{n-1}$$ the function $$f_{\hat{x}_k}$$ is rearrangeable on $$\mathbb {R}$$. We set

\begin{aligned} \mathcal R_kf(u,\hat{x}_k)=f_{\widehat{x}_k}^*(u), \quad u\in \mathbb R_+. \end{aligned}

Stress that the k-th argument of the function $$\mathcal R_kf$$ is equal to u. The function $$\mathcal R_kf$$ is defined almost everywhere on $$\mathbb R_+\times \mathbb R^{n-1}$$; we call it the rearrangement of f with respect to the k-th variable. It is easy to show that $$\mathcal R_kf$$ is a measurable function equimeasurable with |f|. As above, let $$\Omega _n$$ be the collection of all permutations of the set $$\{1,\dots ,n\}$$. For each $$\sigma =\{k_1,\dots ,k_n\}\in \Omega _n$$ we call the function

\begin{aligned} \mathcal R_\sigma f(t)=\mathcal R_{k_n}\cdots \mathcal R_{k_1}f(t), \quad t\in \mathbb R_+^n, \end{aligned}

the $$\mathcal R_\sigma$$-rearrangement of f. Thus, we obtain $$\mathcal R_\sigma f$$ by “rearranging” f in non-increasing order successively with respect to the variables $$x_{k_1},\dots ,x_{k_n}$$. The rearrangement $$\mathcal R_\sigma f$$ is defined on $$\mathbb {R}_+^n.$$ It is nonnegative, nonincreasing in each variable, and equimeasurable with |f| (see [3, 4, 8, 16, 18, 19]).

Let $$0<p, r<\infty$$ and let $$\sigma \in \Omega _n\, (n\ge 2).$$ We denote by $$\mathcal L^{p,r}_{\sigma }(\mathbb R ^n)$$ the class of all rearrangeable functions f on $$\mathbb {R}^n$$ such that

\begin{aligned} \Vert f\Vert _{\mathcal L^{p,r}_{\sigma }}= ||f||_{p,r;\sigma }=\left( \left( \frac{r}{p}\right) ^n \int _{\mathbb R _+^n} \left( \prod _{k=1}^nt_k\right) ^{r/p-1}\mathcal R _\sigma f(t)^r\, dt\right) ^{1/r}<\infty \end{aligned}
(2.4)

(see [8]). The choice of a permutation $$\sigma$$ is essential.

The relations between $$L^{p,r}$$- and $$\mathcal L^{p,r}_{\sigma }$$-norms are described by embeddings (1.4). Moreover, the following proposition gives sharp constants in these relations (see [3, 19]).

### Proposition 2.1

Let f be a rearrangeable function on $$\mathbb {R}^n,~~n\ge 2.$$ Then for any $$\sigma \in \Omega _n$$

\begin{aligned} \Vert f\Vert _{p,r}\le \Vert f\Vert _{p,r;\sigma }\quad \text {if}\quad 0<r\le p<\infty \end{aligned}
(2.5)

and

\begin{aligned} \Vert f\Vert _{p,r;\sigma }\le \Vert f\Vert _{p,r}\quad \text {if}\quad 0<p\le r<\infty . \end{aligned}
(2.6)

These inequalities are optimal.

We stress again that for $$p\not =r$$ the norms $$||\cdot ||_{p,r}$$ and $$||\cdot ||_{p,r;\sigma }$$ may be essentially different. We consider the following simple example which will be also used in the sequel.

As usual, by $$\chi _E$$ we denote the characteristic function of a set E,  which is equal to 1 on E and 0 outside E.

### Example 2.2

Let $$N\ge 1$$. Set

\begin{aligned} g_N(x)= & {} \min \left( \frac{1}{x(\ln (e/x))^2}, \, N\right) , \,\, 0<x\le 1,\\ E_N= & {} \{(x,y):0<x\le 1, \, 0\le y\le g_N(x)\},\quad \text{ and } \quad f_N(x,y)=\chi _{E_N}(x,y). \end{aligned}

Then $$|E_N|\le 1$$ and

\begin{aligned} ||f_N||_{2,1}=\frac{1}{2}\int _0^{|E_N|} t^{-1/2}\,dt\le 1. \end{aligned}

On the other hand,

\begin{aligned} \begin{aligned}&\int _0^\infty \int _0^\infty (xy)^{-1/2}f_N(x,y)\,dx\,dy=\int _0^1 x^{-1/2}\,dx\int _0^{g_N(x)}y^{-1/2} \,dy\\&=2\int _0^1 x^{-1/2}g_N(x)^{1/2}\,dx\ge 2\int _{\alpha _n}^1\frac{1}{x\ln (e/x)}\,dx, \end{aligned} \end{aligned}

where $$\alpha _N$$ is determined by the equation $$\alpha _N(\ln (e/\alpha _N))^2=1/N$$ and $$\alpha _N\rightarrow 0$$ as $$N\rightarrow \infty .$$ Thus, $$||f_N||_{2,1;\sigma }\rightarrow \infty .$$

## 3 Mixed norm spaces

For a measurable set $$E\subset \mathbb {R}^d,$$ we denote by $${\text {mes}}_d E$$ the Lebesgue measure of E in $$\mathbb {R}^d.$$

Let f be a measurable function on $$\mathbb {R}^n$$ and let $$1\le k\le n.$$ By Fubini’s theorem, for almost all $$\hat{x}_k\in \mathbb {R}^{n-1}$$ the sections $$f_{\hat{x}_k}$$ are measurable functions on $$\mathbb {R}.$$ Moreover, the function

\begin{aligned} \psi _k(\hat{x}_k)=||f_{\hat{x}_k}||_{L^\infty (\mathbb {R})}={\text {ess ~sup}}_{x_k\in \mathbb {R}}|f(x_k,\hat{x}_k)| \end{aligned}
(3.1)

(defined a.e. on $$\mathbb {R}^{n-1}$$) is measurable. It suffices to prove the latter statement in the case when f is a bounded function with compact support. In this case we have

\begin{aligned} \psi _k(\hat{x}_k)=\lim _{\nu \rightarrow \infty } ||f_{\hat{x}_k}||_{L^\nu (\mathbb {R})}, \end{aligned}

and the functions $$\hat{x}_k\mapsto ||f_{\hat{x}_k}||_{L^\nu (\mathbb {R})}$$ are measurable by Fubini’s theorem.

Denote by $$\mathcal {M}_{dec}(\mathbb {R}^n_+)$$ the class of all nonnegative functions on $$\mathbb {R}^n$$ which vanish off $$\mathbb {R}^n_+$$ and are nonincreasing in each variable on $$\mathbb {R}^n_+$$.

Let $$f\in \mathcal {M}_{dec}(\mathbb {R}^n_+)$$. Assume that for almost all $$\hat{x}_k\in \mathbb {R}^{n-1}$$ the function f is bounded with respect to $$x_k.$$ Then

\begin{aligned} \psi _k(\hat{x}_k)=\lim _{x_k\rightarrow 0+} f(x_k,\hat{x}_k),\quad \hat{x}_k\in \mathbb {R}^{n-1}. \end{aligned}

Thus, in this case $$\psi _k$$ is the trace of f on the hyperplane $$x_k=0$$ in the sense of almost everywhere convergence (see [24, Chapter 6]).

Let

\begin{aligned} \mathcal {V}_k(\mathbb {R}^n)= L_{\hat{x}_k}^1(\mathbb {R}^{n-1})[L_{x_k}^\infty (\mathbb {R})],\quad 1\le k\le n, \end{aligned}

be the space of measurable functions on $$\mathbb {R}^n$$ with the finite mixed norm

\begin{aligned} \Vert f\Vert _{\mathcal {V}_k}=\Vert \psi _k\Vert _{L^1(\mathbb {R}^{n-1})}, \end{aligned}
(3.2)

where $$\psi _k$$ is defined by (3.1). As it was observed above, $$\psi _k$$ is measurable on $$\mathbb {R}^{n-1}$$, and thus this definition is correct. Denote also

\begin{aligned} \mathcal {V}(\mathbb {R}^n)=\bigcap _{k=1}^n \mathcal {V}_k(\mathbb {R}^n). \end{aligned}
(3.3)

An equivalent form of Theorem 1.2 proved by Fournier [11, Theorem 4.1] is the following.

### Theorem 3.1

If $$f\in \mathcal {V}(\mathbb {R}^n) ~~(n\ge 2),$$ then $$f\in L^{n',1}(\mathbb {R}^n)$$ and

\begin{aligned} ||f||_{L^{n',1}}\le \left( \prod _{k=1}^n||f||_{\mathcal V_k}\right) ^{1/n}. \end{aligned}
(3.4)

The constant is optimal.

Theorem 3.1 implies the embedding of $$W_1^1(\mathbb {R}^n)$$ into the Lorentz space $$L^{n',1}(\mathbb {R}^n).$$ Indeed, let $$f\in W_1^1(\mathbb {R}^n)$$ and let $$D_kf$$ denote the first order weak partial derivative of f with respect to $$x_k$$. Then

\begin{aligned} \Vert f\Vert _{\mathcal {V}_k}\le \frac{1}{2}\Vert D_kf\Vert _{L^1} \quad (k=1,\ldots ,n) \end{aligned}
(3.5)

(see [11, p. 57]). Thus, by (3.4),

\begin{aligned} \Vert f\Vert _{L^{n',1}}\le \frac{1}{2}\left( \prod _{k=1}^n\Vert D_kf\Vert _{L^1}\right) ^{1/n}. \end{aligned}
(3.6)

The following equivalent form of Theorem 1.3 was proved in [19].

### Theorem 3.2

If $$f\in \mathcal {V}(\mathbb {R}^n) ~~(n\ge 2),$$ then $$f\in \mathcal L^{n',1}_\sigma (\mathbb {R}^n)$$ and

\begin{aligned} ||f||_{\mathcal L^{n',1}_\sigma (\mathbb {R}^n)}\le \left( \prod _{k=1}^n||f||_{\mathcal V_k}\right) ^{1/n} \end{aligned}

for any $$\sigma \in \Omega _n.$$ The constant is optimal.

Now we introduce another type of mixed norm spaces (related to the traces on coordinate straight lines). Let

\begin{aligned} \mathcal {U}_k(\mathbb {R}^n)= L_{x_k}^1(\mathbb {R})[L_{\hat{x}_k}^\infty (\mathbb {R}^{n-1})],\quad 1\le k\le n, \end{aligned}
(3.7)

be the space of measurable functions on $$\mathbb {R}^n$$ with the finite mixed norm

\begin{aligned} ||f||_{\mathcal {U}_k}=\Vert f_k\Vert _{L^1(\mathbb {R})}, \quad \text {where}\quad f_k (x_k)={\text {ess ~sup}}_{\hat{x}_k\in \mathbb {R}^{n-1}}|f(x)|. \end{aligned}
(3.8)

Denote also

\begin{aligned} \mathcal {U}(\mathbb {R}^n)=\bigcap _{k=1}^n \mathcal {U}_k(\mathbb {R}^n). \end{aligned}

As above, it is easy to see that these definitions are correct.

Let $$f\in \mathcal {M}_{dec}(\mathbb {R}^n_+)$$. Assume that for almost all $$x_k\in \mathbb {R}$$ the function f is bounded with respect to $$\hat{x}_k.$$ Then

\begin{aligned} f_k(x_k)=\lim _{\hat{x}_k\rightarrow 0} f(x_k,\hat{x}_k),\quad x_k\in \mathbb {R}. \end{aligned}

Thus, in this case $$f_k$$ is the trace of f on the coordinate straight line $$\hat{x}_k=0$$ in the sense of almost everywhere convergence.

### Remark 3.3

Clearly, for $$n=2$$ the spaces $$\mathcal {U}$$ and $$\mathcal {V}$$ coincide. It is also easy to see that for $$n\ge 3$$ neither of the spaces $$\mathcal {U}$$ and $$\mathcal {V}$$ is contained in the other. Indeed, let $$n=3$$. The function

\begin{aligned} f(x,y,z)=\frac{1}{(x+y+z)^2}\chi _{[1,\infty )^3}(x,y,z) \end{aligned}

belongs to $$\mathcal {U}$$ but does not belong to $$\mathcal {V}.$$ On the other hand, the function

\begin{aligned} f(x,y,z)=\frac{1}{x+y+z}\chi _{(0,1)^3}(x,y,z) \end{aligned}

belongs to $$\mathcal {V}$$ but does not belong to $$\mathcal {U}.$$

By (3.5), $$\mathcal V_k$$-norms are estimated by $$L^1$$-norms of the first order weak partial derivatives. Similarly, $$\mathcal {U}_k$$-norms can be estimated by $$L^1$$-norms of the pure (non-mixed) partial derivatives of the order $$n-1$$.

Assume that for a function $$f\in L^1(\mathbb {R}^m)$$ all pure weak partial derivatives

\begin{aligned} D_k^m f=\frac{\partial ^m f}{\partial x_k^m}\quad \,\,(k=1,\ldots ,m) \end{aligned}

of the order m exist and belong to $$L^1(\mathbb {R}^m).$$ Sobolev’s theorem asserts that in this case the function f can be modified on a set of measure zero so as to become uniformly continuous and bounded on $$\mathbb {R}^m$$ and

\begin{aligned} ||f||_\infty \le c\sum _{k=1}^m||D_k^mf||_1 \end{aligned}
(3.9)

(see [6, §10]).

Assume that a function $$f\in L^1(\mathbb {R}^n)$$ $$(n\ge 2)$$ has all pure weak partial derivatives of the order $$n-1.$$ Then for any $$1\le j\le n$$ and almost every fixed $$x_j\in \mathbb {R}$$ the function $$f_{x_j}(\hat{x}_j)=f(x)$$ satisfies the conditions of Sobolev’s theorem for $$m=n-1$$ and therefore by (3.9)

\begin{aligned} {\text {ess ~sup}}_{\hat{x}_j\in \mathbb {R}^{n-1}}|f(x)|\le c\sum _{k\not =j}\int _{\mathbb {R}^{n-1}}|D_k^{n-1}f(x_j,u)|du. \end{aligned}

This implies that

\begin{aligned} ||f||_{\mathcal {U}_j}\le c\sum _{k\not =j}||D_k^{n-1}f||_1\quad (j=1,\ldots ,n). \end{aligned}

## 4 Embeddings into classical Lorentz spaces

We begin with the following simple theorem.

### Theorem 4.1

Let $$f_j$$ $$(j=1,\ldots ,n)$$ be nonnegative rearrangeable functions on $$\mathbb {R}$$ and let

\begin{aligned} f(x)=\min _{1\le j\le n}f_j(x_j), \,\,\, x=(x_1,\ldots ,x_n)\in \mathbb {R}^n. \end{aligned}
(4.1)

Then

\begin{aligned} \lambda _f(y)=\prod _{j=1}^n\lambda _{f_j}(y) \quad \text{ for } \text{ any } \quad y>0. \end{aligned}
(4.2)

If, in addition, $$f_j\in L^1(\mathbb {R})$$, then $$f\in L^{n,1}(\mathbb {R}^n)$$ and

\begin{aligned} ||f||_{n,1}\le \left( \prod _{j=1}^n||f_j||_1\right) ^{1/n}. \end{aligned}
(4.3)

The constant in (4.3) is optimal.

### Proof

Let $$y>0.$$ Then $$f(x)>y$$ if and only if $$f_j(x_j)>y$$ for all $$j=1,\ldots ,n.$$ That is,

\begin{aligned} \{x\in \mathbb {R}^n: f(x)>y\}=\prod _{j=1}^n \{x_j\in \mathbb {R}:f_j(x_j)>y\}. \end{aligned}
(4.4)

This implies (4.2).

Now, applying (2.3), (4.2), and using Hölder’s inequality, we get

\begin{aligned} \begin{aligned}&||f||_{n,1}=\int _0^\infty \lambda _f(y)^{1/n}\,dy=\int _0^\infty \left( \prod _{j=1}^n\lambda _{f_j}(y)\right) ^{1/n}\,dy\\&\quad \le \left( \prod _{j=1}^n\int _0^\infty \lambda _{f_j}(y)\,dy\right) ^{1/n}= \left( \prod _{j=1}^n||f_j||_1\right) ^{1/n}. \end{aligned} \end{aligned}

If $$f_j(u)=\chi _{[0,1]}(u),$$ then $$f(x)=\chi _{[0,1]^n}(x),$$ $$f^*(t)=\chi _{[0,1]}(t),$$ and we have equality in (4.3). Thus, the constant is optimal. $$\square$$

Similarly to Theorems 3.1, 4.1 can be expressed in terms of mixed norms defined by (3.8). Namely, the following theorem is equivalent to Theorem 4.1.

### Theorem 4.2

If $$f\in \mathcal {U}(\mathbb {R}^n) ~~(n\ge 2),$$ then $$f\in L^{n,1}(\mathbb {R}^n)$$ and

\begin{aligned} ||f||_{L^{n,1}(\mathbb {R}^n)}\le \left( \prod _{k=1}^n||f||_{\mathcal {U}_k}\right) ^{1/n}. \end{aligned}
(4.5)

To show the equivalence, we first assume that $$f\in \mathcal {U}(\mathbb {R}^n)$$. Let $$f_k$$ be defined by (3.8). Then

\begin{aligned} |f(x)|\le \min _{1\le k\le n}f_k(x_k)\quad \text{ a.e. } \text{ on }\quad \mathbb {R}^n. \end{aligned}

Since $$||f||_{\mathcal {U}_k}=||f_k||_1,$$ (4.5) follows immediately from (4.3). Conversely, assume that $$f_j\in L^1(\mathbb {R})$$ $$(j=1,\ldots ,n)$$ are nonnegative functions on $$\mathbb {R}$$ and f is defined by (4.1). Then

\begin{aligned} {\text {ess ~sup}}_{\hat{x}_k\in \mathbb {R}^{n-1}}f(x)\le f_k(x_k)\quad \text{ and }\quad ||f||_{\mathcal {U}_k}\le ||f_k||_1 \,\, (k=1,\ldots ,n). \end{aligned}

Thus, (4.5) implies (4.3).

We observe that without loss of generality we can assume that in (4.1) $$f_j$$ are nonincreasing functions on $$\mathbb {R}_+$$ with $$f_j(+\infty )=0$$ that vanish off $$\mathbb {R}_+$$ $$(j=1,\ldots ,n).$$ Indeed, let f be defined by (4.1). Let $$\sigma \in \Omega _n$$ (recall that by $$\Omega _n$$ we denote the collection of all permutations of the set $$\{1,\dots ,n\}$$). We consider the iterated rearrangement $$\mathcal R_\sigma f(t),\,\,t\in \mathbb {R}^n_+.$$ We have

\begin{aligned} \mathcal R_\sigma f(t)=\min _{1\le j\le n}f_j^*(t_j), \quad t=(t_1,\ldots ,t_n)\in \mathbb {R}^n_+ \end{aligned}
(4.6)

for any $$\sigma \in \Omega _n.$$ To show this, it is sufficient to use the following observation: if $$g(x)\ge 0\,\,(x\in \mathbb {R}), \,\,a\ge 0,$$ and $$h(x)=\min (g(x),a),$$ then $$h^*(t)=\min (g^*(t), a).$$

It is easy to obtain the rearrangement of the function $$\Phi =\mathcal R_\sigma f$$ in an explicit form.

### Theorem 4.3

Let $$\varphi _j$$ be nonnegative nonincreasing and left-continuous functions on $$\mathbb {R}_+$$ with $$\varphi _j(+\infty )=0$$ $$(j=1,\ldots ,n)$$. Set

\begin{aligned} \Phi (x)=\min _{1\le j\le n}\varphi _j(x_j), \,\,\, x=(x_1,\ldots ,x_n)\in \mathbb {R}_+^n. \end{aligned}
(4.7)

Then

\begin{aligned} \Phi ^*(t)=\sup \{\Phi (u):\prod _{j=1}^n u_j=t,\,\,\,u_j>0\} \end{aligned}
(4.8)

for any $$t>0.$$

### Proof

Fix $$t>0.$$ Assume that

\begin{aligned} \prod _{j=1}^n u_j=t, \quad u_j>0. \end{aligned}

Since the function $$\Phi$$ decreases in each variable $$x_j\in \mathbb {R}_+,$$ we have $$\Phi (x)\ge \Phi (u)$$ for any $$x\in (0,u_1]\times \cdots \times (0,u_n]$$, that is, for any x from a set of measure t. By (2.2), this implies that $$\Phi ^*(t)\ge \Phi (u).$$ Denoting

\begin{aligned} \omega (t)=\sup \{\Phi (u):\prod _{j=1}^n u_j=t,\,\,\,u_j>0\}, \end{aligned}

we have that $$\Phi ^*(t)\ge \omega (t)$$. Let $$\varepsilon >0$$. Denote $$y=\Phi ^*(t)-\varepsilon$$. Then $$\lambda _\Phi (y)\ge t.$$ Thus, by (4.2),

\begin{aligned} \prod _{j=1}^n\lambda _{\varphi _j}(y)\ge t. \end{aligned}

Therefore, there exist $$\tilde{u}_j\le \lambda _{\varphi _j}(y)$$ $$(j=1,\ldots ,n)$$ such that

\begin{aligned} \prod _{j=1}^n \tilde{u}_j=t. \end{aligned}
(4.9)

Since the functions $$\varphi _j$$ are nonincreasing and left-continuous, we have

\begin{aligned} \varphi _j(\tilde{u}_j)\ge \varphi _j(\lambda _{\varphi _j}(y))\ge y,\quad j=1,\ldots ,n. \end{aligned}

This implies that $$\Phi (\tilde{u})\ge y$$ $$(\tilde{u}=(\tilde{u}_1,\ldots ,\tilde{u}_n))$$ and, by (4.9), we obtain that $$\omega (t)\ge y=\Phi ^*(t)-\varepsilon$$. It follows that $$\omega (t)\ge \Phi ^*(t)$$. Thus, $$\Phi ^*(t)=\omega (t).$$ $$\square$$

### Remark 4.4

Assume, in addition, that all functions $$\varphi _j$$ are continuous and strictly decreasing on $$\mathbb {R}_+$$. Then there exist functions $$u_j$$ on $$\mathbb {R}_+$$ such that for any $$t>0$$

\begin{aligned} \Phi ^*(t)=\varphi _j(u_j(t)),\quad j=1,\ldots ,n, \end{aligned}
(4.10)

and

\begin{aligned} \prod _{j=1}^n u_j(t)=t. \end{aligned}
(4.11)

Indeed, taking into account that $$\Phi ^*(t)<\varphi _j(+0)$$ for any $$t>0$$ and any $$j=1,\ldots ,n$$, define

\begin{aligned} u_j(t)=\varphi _j^{-1}(\Phi ^*(t)). \end{aligned}

Then we have (4.10). Further, $$(\Phi ^*)^{-1}(y)=\lambda _\Phi (y)$$ and $$\varphi _j^{-1}(y)=\lambda _{\varphi _j}(y).$$ Thus, by (4.2),

\begin{aligned} \prod _{j=1}^n\varphi _j^{-1}(y)=(\Phi ^*)^{-1}(y). \end{aligned}

Setting $$y=\Phi ^*(t)$$, we obtain that

\begin{aligned} \prod _{j=1}^n u_j(t)=\prod _{j=1}^n \varphi _j^{-1}(\Phi ^*(t))=(\Phi ^*)^{-1}(\Phi ^*(t))=t, \end{aligned}

and (4.11) holds.

Now we show that inequality (4.3) implies Fournier’s inequality (1.3). First, we have the following theorem.

### Theorem 4.5

Let $$\psi _j$$ $$(j=1,\ldots ,n,\,\,n\ge 2)$$ be nonnegative rearrangeable functions on $$\mathbb {R}^{n-1}.$$ Set

\begin{aligned} \Psi (x)=\min _{1\le j\le n} \psi _j(\hat{x}_j),\quad x=(x_1,\ldots ,x_n)\in \mathbb {R}^n. \end{aligned}
(4.12)

Let $$\varphi _j=\psi ^*_j$$ and

\begin{aligned} \Phi (u)=\min _{1\le j\le n}\varphi _j(u_j), \,\,\, u\in \mathbb {R}_+^n. \end{aligned}
(4.13)

Then

\begin{aligned} \Psi ^*(t)\le \Phi ^*(t^{n-1})\quad \text{ for } \text{ all }\quad t>0 \end{aligned}
(4.14)

and

\begin{aligned} ||\Psi ||_{n',1}\le ||\Phi ||_{n,1}. \end{aligned}
(4.15)

### Proof

Let $$t>0$$ and let $$E\subset \mathbb {R}^n$$ be an $$F_\sigma$$-set with $${\text {mes}}_nE=t.$$ Let $$E_j$$ be the projection of E onto the hyperplane $$x_j=0 \,\,(j=1,\ldots ,n).$$ Then for any $$x\in E$$ and any $$j=1,\ldots ,n$$

\begin{aligned} \inf _{\hat{x}_j\in E_j} \psi _j(\hat{x}_j)\ge \inf _{x\in E} \Psi (x). \end{aligned}
(4.16)

Set

\begin{aligned} u_j={\text {mes}}_{n-1} E_j\quad \text{ and }\quad \mu =\prod _{j=1}^n u_j. \end{aligned}

By (2.2) and (4.16),

\begin{aligned} \varphi _j(u_j)=\psi ^*_j(u_j)\ge \inf _{x\in E} \Psi (x) \quad (j=1,\ldots ,n). \end{aligned}

Further, the functions $$\varphi _j$$ are left-continuous because they are defined as nonincreasing rearrangements (see (2.1)). Therefore, by Theorem 4.3, for all $$t>0$$ we have equality (4.8). Thus,

\begin{aligned} \Phi ^*(\mu )\ge \inf _{x\in E} \Psi (x). \end{aligned}
(4.17)

By the Loomis–Whitney inequality (1.14),

\begin{aligned} \mu = \prod _{j=1}^n {\text {mes}}_{n-1} E_j\ge ({\text {mes}}_n E)^{n-1}=t^{n-1}. \end{aligned}

Hence, by (4.17),

\begin{aligned} \Phi ^*(t^{n-1})\ge \inf _{x\in E} \Psi (x). \end{aligned}

Here $$E\subset \mathbb {R}^n$$ is an arbitrary $$F_\sigma$$-set with $${\text {mes}}_n E=t.$$ Thus, by (2.2), the latter inequality implies (4.14).

Now, applying inequality (4.14), we obtain

\begin{aligned}&\int _0^\infty t^{1/n'-1}\Psi ^*(t)\,dt\le \int _0^\infty t^{1/n'-1}\Phi ^*(t^{n-1})\,dt\\&\quad =\frac{1}{n-1}\int _0^\infty u^{1/n-1}\Phi ^*(u)\,du. \end{aligned}

This implies inequality (4.15). $$\square$$

Inequalities (4.15) and (4.3) yield the following strengthening of inequality (1.3).

### Corollary 4.6

Let $$\psi _j\in L^1(\mathbb {R}^{n-1})$$ $$(j=1,\ldots ,n,\,\,n\ge 2)$$ be nonnegative functions. Set

\begin{aligned} \Psi (x)=\min _{1\le j\le n} \psi _j(\hat{x}_j),\quad x=(x_1,\ldots ,x_n)\in \mathbb {R}^n \end{aligned}

and

\begin{aligned} \Phi (u)=\min _{1\le j\le n}\psi ^*_j(u_j), \,\,\, u\in \mathbb {R}_+^n. \end{aligned}

Then

\begin{aligned} ||\Psi ||_{n',1}\le ||\Phi ||_{n,1}\le \left( \prod _{k=1}^n||\psi _j||_{L^1}\right) ^{1/n}. \end{aligned}
(4.18)

.

In comparison with (1.3), inequality (4.18) contains the intermediate term $$||\Phi ||_{n,1}.$$ We shall see that this term can be essentially greater that $$||\Psi ||_{n',1}.$$

### Remark 4.7

First, we observe that for $$n=2$$ we have $$\Psi ^*(t)=\Phi ^*(t)$$ for all $$t>0.$$ Indeed, let $$t>0$$. Assume that $$u_1,u_2\in \mathbb {R}_+, \,\,u_1u_2=t.$$ There exist $$F_\sigma$$-sets $$E_1,E_2\subset \mathbb {R}$$ such that

\begin{aligned} {\text {mes}}_1 E_j=u_j\quad \text{ and }\quad \varphi _j(u_j)=\psi ^*_j(u_j)=\inf _{x_j\in E_j}\psi _j(x_j)\quad (j=1,2). \end{aligned}

Let $$E=E_1\times E_2.$$ Then $${\text {mes}}_2E=t$$ and

\begin{aligned} \min (\varphi _1(u_1),\varphi _2(u_2))=\inf _{(x_1,x_2)\in E} \Psi (x_1,x_2)\le \Psi ^*(t). \end{aligned}

By (4.8), this implies that $$\Phi ^*(t)\le \Psi ^*(t)$$, and we obtain that $$\Phi ^*(t)=\Psi ^*(t).$$

Similar arguments fail for $$n\ge 3$$. Indeed, let $$n=3$$ and assume that $$0<\varepsilon <1.$$ Let $$I=[0,\varepsilon ]\times [0,1/\varepsilon ].$$ Set $$\psi _j=\chi _I \,\,\,(j=1,2,3)$$. Then

\begin{aligned} \begin{aligned} \Psi (x_1,x_2,x_3)&=\min (\psi _1(x_2,x_3), \psi _2(x_1,x_3), \psi _3(x_1,x_2))\\&=\chi _{P_\varepsilon }(x_1,x_2,x_3), \end{aligned} \end{aligned}

where $$P_\varepsilon =[0,\varepsilon ]\times [0,\varepsilon ]\times [0,1/\varepsilon ].$$ Thus, $$\Psi ^*(t)=\chi _{[0,\varepsilon ]}(t).$$ On the other hand, $$\psi _j^*(u)=\chi _{[0,1]}(u)$$ $$(j=1,2,3)$$, and

\begin{aligned} \Phi (u_1,u_2,u_3)=\chi _{[0,1]^3}(u_1,u_2,u_3),\quad \Phi ^*(t)=\chi _{[0,1]}(t). \end{aligned}

We have $$||\Phi ||_{3,1}=1$$ and $$||\Psi ||_{3/2,1}=\varepsilon ^{2/3}.$$ Thus, for $$n\ge 3$$ inequality (4.3) (with $$f_j=\psi _j^*$$) is essentially stronger than (1.3).

## 5 Estimates of generalized Lorentz norms

In this section we obtain our main results.

As above, by $$\Omega _n$$ we denote the collection of all permutations of the set $$\{1,\ldots ,n\}.$$

First, we prove the following theorem.

### Theorem 5.1

Let $$f_j\in L^1(\mathbb {R})$$ $$(j=1,\ldots ,n)$$ be nonnegative rearrangeable functions on $$\mathbb {R}$$ and let

\begin{aligned} f(x)=\min _{1\le j\le n}f_j(x_j), \,\,\, x=(x_1,\ldots ,x_n)\in \mathbb {R}^n. \end{aligned}
(5.1)

Let $$\alpha _j>0\,\,(j=1,\ldots ,n).$$ Then for any $$\sigma \in \Omega _n$$

\begin{aligned} \int _{\mathbb {R}^n_+} \prod _{j=1}^n x_j^{\alpha _j-1}\mathcal R_\sigma f(x)\,dx=\left( \prod _{j=1}^n\alpha _j\right) ^{-1}\int _0^\infty \prod _{j=1}^n\lambda _{f_j}(y)^{\alpha _j}\,dy. \end{aligned}
(5.2)

### Proof

Set

\begin{aligned} \varphi _j=f_j^*\quad \text{ and }\quad \Phi (x)=\min _{1\le j\le n} \varphi _j(x_j),\quad x\in \mathbb {R}^n_+. \end{aligned}
(5.3)

By (4.6), for any $$\sigma \in \Omega _n$$

\begin{aligned} \mathcal R_\sigma f(x)dx=\Phi (x)\quad \text{ for } \text{ all }\quad x\in \mathbb {R}^n_+. \end{aligned}
(5.4)

Thus, (5.2) is equivalent to the equality

\begin{aligned} \int _{\mathbb {R}^n_+} \prod _{j=1}^n x_j^{\alpha _j-1}\Phi (x)\,dx=\left( \prod _{j=1}^n\alpha _j\right) ^{-1}\int _0^\infty \prod _{j=1}^n\lambda _{\varphi _j}(y)^{\alpha _j}\,dy. \end{aligned}
(5.5)

We may assume that all functions $$\varphi _j$$ are continuously differentiable, $$\varphi _j(+0)=+\infty ,$$ and

\begin{aligned} \varphi _j(t)>0, \,\,\varphi _j'(t)<0\quad \text{ for } \text{ all }\quad t>0\quad \text{ and } \text{ all } \quad j=1,\ldots ,n. \end{aligned}

Then we have $$\varphi _j(\mathbb {R}_+)=\mathbb {R}_+$$ $$(j=1,\ldots ,n).$$ Denote

\begin{aligned} \eta _j(y)=\varphi _j^{-1}(y)=\lambda _{\varphi _j}(y),\quad y>0, \quad j=1,\ldots ,n. \end{aligned}

Let $$A_j$$ be the set of all $$x\in \mathbb {R}_+^n$$ such that $$\Phi (x)=\varphi _j(x_j),\,\,j=1,\ldots ,n$$. Then

\begin{aligned} \bigcup _{j=1}^n A_j=\mathbb {R}_+^n\quad \text{ and }\quad {\text {mes}}_n (A_i\cap A_j)=0\quad \text{ if }\quad i\not =j. \end{aligned}

Indeed, the first equality is obvious. Further, let $$i\not = j.$$ Then $$x\in A_i\cap A_j$$ if and only if $$x_j=\eta _j(\varphi _i(x_i)).$$ Thus, the projection of the set $$A_i\cap A_j$$ onto the 2-dimensional plane $$(x_i,x_j)$$ has the 2-dimensional measure zero and therefore $${\text {mes}}_n(A_i\cap A_j)=0$$.

Denote $$\gamma _{j,k}(x_k)=\eta _j(\varphi _k(x_k)), \,\, x_k\in \mathbb {R}_+.$$ For a fixed $$1\le j\le n$$ we have

\begin{aligned} \begin{aligned} Q_j&\equiv \int _{A_j}\prod _{k=1}^n x_k^{\alpha _k-1}\Phi (x)\,dx\\&=\int _{\mathbb {R}_+^{n-1}}\prod _{k\not =j} x_k^{\alpha _k-1}\int _{\sigma _j(\hat{x}_j)}^\infty x_j^{\alpha _j-1}\varphi _j(x_j)\,dx_j \,d\hat{x}_j, \end{aligned} \end{aligned}

where $$\sigma _j(\hat{x}_j)=\max _{k\not = j} \gamma _{j,k}(x_k).$$ Indeed, $$x\in A_j$$ if and only if $$\hat{x}_j\in \mathbb {R}_+^{n-1}$$ and $$x_j\ge \gamma _{j,k}(x_k)$$ for any $$k\not =j.$$ Thus, by Fubini’s theorem, we have

\begin{aligned} \begin{aligned} Q_j&=\int _0^\infty x_j^{\alpha _j-1}\varphi _j(x_j)\prod _{k\not = j}\int _0^{\gamma _{k,j}(x_j)} x_k^{\alpha _k-1}\,dx_k\, dx_j\\&=\left( \prod _{k\not = j}\alpha _k\right) ^{-1}\int _0^\infty x_j^{\alpha _j-1}\varphi _j(x_j)\prod _{k\not = j} \eta _k(\varphi _j(x_j))^{\alpha _k}\,dx_j. \end{aligned} \end{aligned}

Change of variable $$y=\varphi _j(x_j)$$ gives

\begin{aligned} \begin{aligned} Q_j&=-\left( \prod _{k\not = j}\alpha _k\right) ^{-1}\int _0^\infty \eta _j(y)^{\alpha _j-1}\prod _{k\not = j} \eta _k(y)^{\alpha _k}y\,d\eta _j(y)\\&=-\left( \prod _{k\not = j}\alpha _k\right) ^{-1}\int _0^\infty \prod _{k=1}^n\eta _k(y)^{\alpha _k}y\,d(\ln \eta _j(y))\\&=-\left( \prod _{k=1}^n\alpha _k\right) ^{-1}\int _0^\infty \prod _{k=1}^n\eta _k(y)^{\alpha _k}y\,d(\ln [\eta _j(y)^{\alpha _j}]). \end{aligned} \end{aligned}

From here, setting

\begin{aligned} c_n=\left( \prod _{k=1}^n\alpha _k\right) ^{-1}\quad \text{ and }\quad \zeta (y)=\prod _{k=1}^n\eta _k(y)^{\alpha _k}, \end{aligned}

we obtain

\begin{aligned} \begin{aligned}&\int _0^\infty \prod _{k=1}^n x_k^{\alpha _k-1}\Phi (x)\,dx=\sum _{j=1}^n Q_j\\&\quad = -c_n\int _0^\infty \zeta (y)y\,d\left( \sum _{j=1}^n \ln [\eta _j(y)^{\alpha _j}]\right) \\&\quad =-c_n\int _0^\infty \zeta (y)y\,d(\ln \zeta (y))\\&\quad =-c_n\int _0^\infty y\,d\zeta (y)=c_n\int _0^\infty \zeta (y)\,dy. \end{aligned} \end{aligned}

Since $$\eta _k(y)=\lambda _{\varphi _k}(y),$$ this implies (5.5). $$\square$$

### Corollary 5.2

\begin{aligned} \sum _{k=1}^n\alpha _k=1. \end{aligned}

Then for any $$\sigma \in \Omega _n$$

\begin{aligned} \int _{\mathbb {R}^n_+} \prod _{k=1}^n x_k^{\alpha _k-1}\mathcal R_\sigma f(x)\,dx\le \left( \prod _{j=1}^n\alpha _j\right) ^{-1}\prod _{j=1}^n||f_j||_1^{\alpha _j}. \end{aligned}
(5.6)

Indeed, applying Hölder’s inequality at the right-hand side of (5.2), we obtain

\begin{aligned} \int _{\mathbb {R}^n_+} \prod _{j=1}^n\lambda _{f_j}(y)^{\alpha _j}\,dy\le \prod _{j=1}^n\left( \int _0^\infty \lambda _{f_j}(y)\,dy\right) ^{\alpha _j}=\prod _{j=1}^n\left( \int _0^\infty f_j(t)\,dt\right) ^{\alpha _j}. \end{aligned}

In the case $$f_j(t)=\chi _{[0,1]}(t)\,\,\,(j=1,\ldots ,n)$$ we have equality in (5.6). This shows that the constant in (5.6) is optimal.

Now we suppose that, in Theorem 5.1, $$\alpha _1=\cdots =\alpha _n=1/p.$$ Then equality (5.2) assumes the form

\begin{aligned} \int _{\mathbb {R}^n_+} \prod _{j=1}^n x_j^{1/p-1}\mathcal R_\sigma f(x)\,dx=p^{n}\int _0^\infty \prod _{j=1}^n\lambda _{f_j}(y)^{1/p}\,dy. \end{aligned}
(5.7)

By (5.3), (4.2), and (5.4),

\begin{aligned} \prod _{j=1}^n\lambda _{f_j}(y)=\lambda _\Phi (y)=\lambda _f(y). \end{aligned}

Hence, using (2.3), we obtain that

\begin{aligned} \frac{1}{p^{n}}\int _{\mathbb {R}^n_+} \prod _{j=1}^n x_j^{1/p-1}\mathcal R_\sigma f(x)\,dx=\int _0^\infty \lambda _f(y)^{1/p}\,dy=||f||_{p,1} \end{aligned}
(5.8)

for any $$\sigma \in \Omega _n$$. Thus, we have the following statement.

### Theorem 5.3

Let $$f_j\in L^1(\mathbb {R})$$ $$(j=1,\ldots ,n)$$ be nonnegative functions on $$\mathbb {R}$$ and let

\begin{aligned} f(x)=\min _{1\le j\le n}f_j(x_j), \,\,\, x=(x_1,\ldots ,x_n)\in \mathbb {R}^n. \end{aligned}
(5.9)

Then for any $$p>0$$

\begin{aligned} \Vert f\Vert _{\mathcal L^{p,1}_{\sigma }}=||f||_{p,1}\quad \text{ for } \text{ any }\quad \sigma \in \Omega _n. \end{aligned}
(5.10)

We have emphasized in Sect. 2that, in general, the spaces $$\mathcal L_{p,1;\sigma }(\mathbb {R}^n)$$ are strictly smaller than $$L_{p,1}(\mathbb {R}^n)$$ (see Example 2.2). We shall now analyse the situation that we have in Theorem 5.3.

First, denote by $$\mathcal Q_n$$ the class of all measurable functions on $$\mathbb {R}^n$$ such that for any $$y>0$$ the set

\begin{aligned} E(y)= \{x\in \mathbb {R}^n: |f(x)|>y\} \end{aligned}
(5.11)

is essentially a cube in $$\mathbb {R}^n$$ (that is, for any $$y>0$$ there exists a cube $$Q\subset \mathbb {R}^n$$ with edges parallel to coordinate axes which differs from $$E_f(y)$$ by a set of measure 0). It was shown in [19] that for any function $$f\in \mathcal Q_n$$ $$(n\ge 2)$$

\begin{aligned} \Vert f\Vert _{\mathcal L^{n',1}_{\sigma }}=||f||_{n',1}\quad \text{ for } \text{ any }\quad \sigma \in \Omega _n. \end{aligned}
(5.12)

We shall show that equality (5.12) is true for a much wider class of functions.

Denote by $$\mathcal P_n$$ the class of all measurable functions on $$\mathbb {R}^n$$ such that for any $$y>0$$ the set (5.11) is a cartesian product of measurable sets $$E_i(y)\subset \mathbb {R},$$

\begin{aligned} E(y)=E_1(y)\times \cdots \times E_n(y). \end{aligned}
(5.13)

By virtue of equality (4.4), any function f defined by (4.1) belongs to $$\mathcal P_n.$$ We shall prove the inverse statement.

### Proposition 5.4

Let $$f\in \mathcal P_n\,\, (n\ge 2)$$ be a nonnegative rearrangeable function on $$\mathbb {R}^n.$$ Set

\begin{aligned} f_k(x_k)={\text {ess ~sup}}_{\hat{x}_k\in \mathbb {R}^{n-1}} f(x), \,\, k=1,\ldots ,n. \end{aligned}

Then

\begin{aligned} f(x)=\min _{1\le k\le n} f_k(x_k)\quad \text{ for } \text{ almost } \text{ all }\quad x\in \mathbb {R}^n. \end{aligned}
(5.14)

### Proof

First, we have for all $$x\in \mathbb {R}^n$$

\begin{aligned} f(x)\le \min _{1\le k\le n} f_k(x_k). \end{aligned}
(5.15)

Fix $$x\in \mathbb {R}^n$$ and denote $$y=f(x).$$ Let

\begin{aligned} E(y)=\{z\in \mathbb {R}^n: f(z)>y\}. \end{aligned}

Then E(y) is a cartesian product (5.13), where $$E_i(y)\subset \mathbb {R}.$$ Since $$x\not \in E(y),$$ there exists $$j\in \{1,\ldots ,n\}$$ such that $$x_j\not \in E_j(y).$$ Whatever be $$u\in \mathbb {R}^{n-1}$$, we have $$(x_j, u)\not \in E_y$$ and therefore

\begin{aligned} f_j(x_j)={\text {ess ~sup}}_{u\in \mathbb {R}^{n-1}} f(x_j,u)\le y. \end{aligned}

Thus, $$f_j(x_j)\le f(x).$$ Taking into account (5.15), we obtain (5.14). $$\square$$

Applying Proposition 5.4 and Theorem 5.3, we get the following result.

### Theorem 5.5

Let $$f\in \mathcal P_n\,\, (n\ge 2)$$ be a rearrangeable function on $$\mathbb {R}^n.$$ Then for any $$p>0$$

\begin{aligned} \Vert f\Vert _{\mathcal L^{p,1}_{\sigma }}=||f||_{p,1}\quad \text{ for } \text{ any }\quad \sigma \in \Omega _n. \end{aligned}
(5.16)

### Remark 5.6

We observe that the coincidence of the classical and generalized Lorentz norms which holds for any function f defined by (5.9), may not hold for the function $$\Psi$$ defined by (4.12). Let $$N\ge 1$$ and let $$g_N$$ be the function defined in Example 2.2. Set

\begin{aligned} \psi _1(y,z)= & {} {\left\{ \begin{array}{ll} 2&{} \text {if }0\le y\le N, \, 0\le z\le 1,\\ 0&{} \text {otherwise}, \end{array}\right. }\\ \psi _2(x,z)= & {} {\left\{ \begin{array}{ll} 2&{} \text {if }0\le x\le 1, \, 0\le z\le 1,\\ 0&{} \text {otherwise}, \end{array}\right. }\\ \psi _3(x,y)= & {} {\left\{ \begin{array}{ll} 1&{} \text {if }0\le x\le 1, \, 0\le y\le g_N(x),\\ 0&{} \text {otherwise}. \end{array}\right. } \end{aligned}

Let

\begin{aligned} \Psi _N(x,y,z)=\min (\psi _1(y,z), \psi _2(x,z),\psi _3(x,y)). \end{aligned}

Then $$\Psi _N=\chi _{E_N}$$, where

\begin{aligned} E_N=\{(x,y,z): 0\le x\le 1, 0\le y\le g_N(x), 0\le z\le 1\}. \end{aligned}

We have $$|E_N|\le 1$$ and $$||\Psi _N||_{2,1}\le 1.$$ At the same time,

\begin{aligned} \iiint _{E_N}(xyz)^{-1/2} dxdydz\ge c\ln \ln N\quad (c>0). \end{aligned}

Thus, $$||\Psi _N||_{2,1;\sigma }\rightarrow \infty$$ as $$N\rightarrow \infty .$$

Finally, we return to Corollary 5.2. Taking $$\alpha _k=1/n \,\,(k=1,\ldots ,n),$$ we obtain

### Theorem 5.7

Let $$f_k\in L^1(\mathbb {R})$$ $$(k=1,\ldots ,n)$$ be nonnegative functions on $$\mathbb {R}$$ and let f be defined by (4.1). Then for any $$\sigma \in \Omega _n$$

\begin{aligned} ||f||_{\mathcal {L}^{n,1}_\sigma }\le \left( \prod _{j=1}^n||f_j||_1\right) ^{1/n}, \end{aligned}
(5.17)

and the constant is optimal.

Similarly to Theorems 4.1, 5.7 can be expressed in terms of mixed norms defined by (3.8). Namely, the following theorem is equivalent to Theorem 5.7.

### Theorem 5.8

Assume that $$f\in \mathcal {U}(\mathbb {R}^{n})$$ $$(n\ge 2).$$ Then $$f\in \mathcal L^{n,1}_\sigma (\mathbb {R}^n)$$ and

\begin{aligned} ||f||_{\mathcal {L}^{n,1}_\sigma }\le \left( \prod _{k=1}^n||f||_{\mathcal {U}_k}\right) ^{1/n} \end{aligned}
(5.18)

for any $$\sigma \in \Omega _n$$.

The equivalence follows by standard arguments. First, assume that $$f\in \mathcal {U}(\mathbb {R}^{n})$$. We have

\begin{aligned} |f(x)|\le {\text {ess ~sup}}_{\hat{x}_k\in \mathbb {R}^{n-1}}|f(x_k,\hat{x}_k)|\quad (k=1,\ldots ,n) \end{aligned}

for almost all $$x\in \mathbb {R}^n$$. Denote the right-hand side of the above inequality by $$f_k(x_k)$$ and set

\begin{aligned} \tilde{f}(x)= \min _{1\le k\le n}f_k(x_k). \end{aligned}

We have $$||f||_{\mathcal {U}_k}=||f_k||_1$$ and $$|f(x)|\le \tilde{f}(x).$$ Thus, (5.18) follows immediately from Theorem 5.7.

Conversely, under the conditions of Theorem 5.7,

\begin{aligned} {\text {ess ~sup}}_{\hat{x}_k\in \mathbb {R}^{n-1}}f(x_k,\hat{x}_k)\le f_k(x_k)\quad (k=1,\ldots ,n) \end{aligned}

and therefore $$||f||_{\mathcal {U}_k}\le ||f_k||_1$$. Thus, (5.17) follows from Theorem 5.8.

However, there is a difference between inequalities (5.17) and (5.18). Namely, by Theorem 5.3, the left-hand side of (5.17) coincides with $$||f||_{n,1}$$. At the same time, the condition $$f\in \mathcal {U}(\mathbb {R}^{n})$$ doesn’t imply such coincidence, and the norm $$||f||_{\mathcal {L}^{n,1}_\sigma }$$ at the left-hand side of (5.18) may be much greater than $$||f||_{n,1}$$ (see Example 2.2).

## 6 Comparison of estimates (1.13) and (1.6)

In Sect. 4we obtained inequality (4.15) between classical Lorentz norms of functions (4.12) and (4.13). In this section we prove a similar inequality between generalized Lorentz norms of these functions. For this, we apply a special bijection of $$\mathbb {R}_+^n$$ onto $$\mathbb {R}_+^n$$.

Assume that $$u\in \mathbb {R}^m_+,\,\, m\ge 2.$$ Then

\begin{aligned} \prod _{k=1}^m\prod _{j\not =k}u_j=\left( \prod _{j=1}^m u_j\right) ^{m-1}. \end{aligned}
(6.1)

In what follows we denote

\begin{aligned} \pi _m(u)=\prod _{k=1}^m u_k \quad \text{ for } \text{ any } \quad u\in \mathbb {R}^m_+. \end{aligned}
(6.2)

### Lemma 6.1

Let $$n\ge 2$$ and let the mapping $$y=\Lambda (x)$$ be defined by

\begin{aligned} y_j=\pi _{n-1}(\hat{x}_j), \quad j=1,\ldots ,n,\quad x\in \mathbb {R}^n_+. \end{aligned}
(6.3)

Then $$\Lambda$$ is a differentiable bijective mapping from $$\mathbb {R}^n_+$$ onto $$\mathbb {R}^n_+$$ with the jacobian

\begin{aligned} |J\Lambda (x)|=(n-1)\pi _n(x)^{n-2}, \quad x\in \mathbb {R}_+^n. \end{aligned}
(6.4)

### Proof

First we show that $$\Lambda$$ is injective. For $$n=2$$ it is obvious. Let $$n\ge 3.$$ Assume that there exist $$x',x''\in \mathbb {R}^n_+$$ such that

\begin{aligned} x'_1>x''_1\quad \text{ but } \quad y'=y'',\quad \text{ where } \quad y'=\Lambda (x'), \, y''=\Lambda (x''). \end{aligned}
(6.5)

By our assumption,

\begin{aligned} x_1'\prod _{j\not =1,k}x_j'=x_1''\prod _{j\not =1,k}x_j''\quad \text{ for } \text{ all }\quad k=2,\ldots ,n. \end{aligned}

Thus

\begin{aligned} \prod _{j\not =1,k}x_j'<\prod _{j\not =1,k}x_j''\quad \text{ for } \text{ all }\quad k=2,\ldots ,n. \end{aligned}

Multiplying these inequalities and using (6.1), we obtain that

\begin{aligned} \left( \prod _{k=2}^n x_k'\right) ^{n-2}<\left( \prod _{k=2}^n x_k''\right) ^{n-2}\quad \text{ and } \text{ thus }\quad y'_1<y_1'', \end{aligned}

Next, we show that $$\Lambda$$ is surjective. Let $$y\in \mathbb {R}_+^n.$$ Set

\begin{aligned} x_j=\pi _n(y)^{1/(n-1)}/y_j, \quad j=1,\ldots ,n. \end{aligned}

Then

\begin{aligned} \pi _n(x)= \pi _n(y)^{n/(n-1)}/\pi _n(y)=\pi _n(y)^{1/(n-1)}. \end{aligned}

Thus,

\begin{aligned} y_j=\pi _n(x)/x_j=\pi _{n-1}(\hat{x}_j) \,\, (j=1,\ldots ,n),\quad \text{ and }\quad y=\Lambda (x). \end{aligned}

It remains to prove equality (6.4). Fix $$x\in \mathbb {R}_+^n$$ and consider the Jacobi matrix of $$\Lambda$$ at the point x. The i-th row of this matrix is formed by the partial derivatives of the function

\begin{aligned} x\mapsto \pi _{n-1}(\hat{x}_i), \quad x\in \mathbb {R}^n. \end{aligned}

Assume that $$x_k\not =0$$ for all $$k=1,\ldots ,n.$$ The j-th element of the i-th row is equal to

\begin{aligned} \prod _{k\not =i,j}x_k=\frac{\pi _n(x)}{x_ix_j}\quad \text{ if }\quad i\not =j, \end{aligned}

and it is equal to 0 if $$i=j.$$ It is easily verified that the determinant of this matrix is equal to $$\pi _n(x)^{n-2}\det (\gamma _{ij}),$$ where

\begin{aligned} \gamma _{ij}={\left\{ \begin{array}{ll} 0&{} \text {if }i=j,\\ 1&{} \text {if }i\not =j. \end{array}\right. } \end{aligned}

Adding to the first row of the matrix $$(\gamma _{ij})$$ all other rows, we obtain the matrix $$(\tilde{\gamma }_{ij}),$$ where $$\tilde{\gamma }_{1j}=n-1$$ and $$\tilde{\gamma }_{ij}= \gamma _{ij}$$ for $$i\ge 2$$ $$\,(j=1,\ldots ,n).$$ Next, from all rows of the matrix $$(\tilde{\gamma }_{ij})$$ beginning from the second, we subtract its first row divided by $$n-1$$. We get a triangular matrix with the diagonal $$n-1, -1, \ldots ,-1.$$ Thus, $$\det (\gamma _{ij})=(-1)^{n-1}(n-1)$$, and we obtain equality (6.4).

$$\square$$

As above, we denote by $$\mathcal {M}_{dec}(\mathbb {R}^m_+)$$ the class of all nonnegative functions on $$\mathbb {R}^m$$ which vanish off $$\mathbb {R}^m_+$$ and are nonincreasing in each variable on $$\mathbb {R}^m_+$$. If $$f\in \mathcal {M}_{dec}(\mathbb {R}^m_+)$$, then

\begin{aligned} f(u)\le f^*(\pi _m(u)),\quad u\in \mathbb {R}^m_+. \end{aligned}
(6.6)

Indeed,

\begin{aligned} f(v)\ge f(u)\quad \text{ for } \text{ any }\quad v\in E_u=(0,u_1]\times \cdots \times (0,u_m], \end{aligned}

and $${\text {mes}}_m E_u=\pi _m(u).$$ Applying (2.2), we obtain (6.6).

### Theorem 6.2

Assume that $$n\ge 2$$ and $$\psi _j\in L^1(\mathbb {R}^{n-1})$$ $$(j=1,\ldots ,n)$$. Set

\begin{aligned} \Psi (x)=\min _{1\le j\le n}\psi _j(\hat{x}_j), \quad x\in \mathbb {R}^n. \end{aligned}

Let $$\varphi _j(t)=\psi _j^*(t),\,\,t\in \mathbb {R}_+$$, and

\begin{aligned} \Phi (x)=\min _{1\le j\le n}\varphi _j(x_j), \quad x\in \mathbb {R}_+^n. \end{aligned}

Then for any $$\sigma \in \Omega _n$$

\begin{aligned} \int _{\mathbb {R}^n_+} \pi _n(x)^{-1/n}\mathcal R_\sigma \Psi (x)\,dx \le \frac{1}{n-1} \int _{\mathbb {R}^n_+}\pi _n(x)^{-1/n'}\Phi (x)\,dx. \end{aligned}
(6.7)

### Proof

Let $$\sigma \in \Omega _n$$ and let $$\sigma _k$$ be obtained from $$\sigma$$ by removal of k. Then for any $$1\le k\le n$$

\begin{aligned} \mathcal R_\sigma \Psi (u)\le \widetilde{\psi }_k(\hat{u}_k), \quad \text{ where }\quad \widetilde{\psi }_k(\hat{u}_k)=\mathcal R_{\sigma _k}\psi _k(\hat{u}_k). \end{aligned}

Set

\begin{aligned} \widetilde{\Psi }(u)=\min _{1\le k\le n}\widetilde{\psi }_k(\hat{u}_k). \end{aligned}

Then $$\mathcal R_\sigma \Psi (u)\le \widetilde{\Psi }(u), \,\,u\in \mathbb {R}^n_+,$$ and $$\widetilde{\psi }_k^*(t)=\psi _k^*(t),\,\, t>0.$$ Since $$\widetilde{\psi }_k\in \mathcal {M}_{dec}(\mathbb {R}^{n-1}_+)$$ and $$\widetilde{\psi }_k^*=\varphi _k,$$ we have by (6.6)

\begin{aligned} \widetilde{\psi }_k(\hat{x}_k)\le \varphi _k(\pi _{n-1}(\hat{x}_k)),\quad k=1,\ldots ,n. \end{aligned}
(6.8)

We consider the mapping $$y=\Lambda (x)$$ from $$\mathbb {R}_+^n$$ to $$\mathbb {R}_+^n$$ defined in Lemma 6.1,

\begin{aligned} y_j=\pi _{n-1}(\hat{x}_j), \quad j=1,\ldots ,n,\quad x\in \mathbb {R}^n_+. \end{aligned}
(6.9)

By Lemma 6.1, $$\Lambda$$ is bijective and $$|J\Lambda (x)|=(n-1)\pi _{n}(x)^{n-2}$$. For $$y=\Lambda (x)$$ we have $$\pi _n(y)=\pi _n(x)^{n-1}$$ (see (6.1)). Thus, the inverse mapping $$G=\Lambda ^{-1}$$ satisfies the equality

\begin{aligned} |JG(y)|=\frac{1}{|J\Lambda (x)|}=\frac{1}{(n-1)\pi _n(y)^{(n-2)/(n-1)}}, \quad y=\Lambda (x). \end{aligned}
(6.10)

Applying inequality (6.8), performing the change of variables (6.9), and using equality (6.10), we get

\begin{aligned} \begin{aligned}&\int _{\mathbb {R}^n_+}\pi _n(x)^{-1/n}\widetilde{\Psi }(x)\,dx \le \int _{\mathbb {R}^n_+}\pi _n(x)^{-1/n}\min _{1\le j\le n}\varphi _j(\pi _{n-1}(\hat{x}_j))\,dx\\&\quad =\int _{\mathbb {R}^n_+}\pi _n(y)^{-1/(n(n-1))}\Phi (y)|JG(y)|\,dy =\frac{1}{n-1}\int _{\mathbb {R}^n_+}\pi _n(y)^{1/n-1}\Phi (y)\,dy. \end{aligned} \end{aligned}

Since $$\mathcal R_\sigma \Psi (x)\le \widetilde{\Psi }(x),$$ this yields (6.7). $$\square$$

### Remark 6.3

The example given above in Remark 4.7 shows that inequality (6.7) cannot be reverted, even by inserting an arbitrarily small constant to the right-hand side. Namely, in this example

\begin{aligned} \iiint _{\mathbb {R}^3_+} (x_1x_2x_3)^{-1/3}\Psi (x_1,x_2,x_3)\,dx_1\,dx_2\,dx_3=27\varepsilon ^{2/3}/8 \end{aligned}

and

\begin{aligned} \iiint _{\mathbb {R}^3_+} (x_1x_2x_3)^{-2/3}\Phi (x_1,x_2,x_3)\,dx_1\,dx_2\,dx_3=27. \end{aligned}

Theorem 6.2 and Remark 6.3 show that inequality (1.13) (with $$f_j=\psi _j^*$$) is stronger than inequality (1.6).

However, we observe that inequality (6.6) which was used in the proof of Theorem 6.2 is rather rough. Apparently, the constant in inequality (6.7) may be improved. Namely, Theorems 1.3 and 5.7 suggest that the optimal constant should be $$(n-1)^{-n}$$.