Optimal Gagliardo–Nirenberg interpolation inequality for rearrangement invariant spaces

Abstract

We prove the optimality of the Gagliardo–Nirenberg inequality:

$$\begin{aligned} \Vert \nabla u\Vert _{X}\lesssim \Vert \nabla ^2 u\Vert _Y^{1/2}\Vert u\Vert _Z^{1/2}, \end{aligned}$$

where Y and Z are rearrangement invariant Banach function spaces, and $X = Y^{1/2}Z^{1/2}$ is the Calderón–Lozanovskii space. By optimality, we mean that for a certain pair of spaces on the right-hand side, it is not possible to reduce the space on the left-hand side while remaining in the class of rearrangement invariant spaces. Our result establishes the optimality for Lorentz and Orlicz spaces, surpassing previous findings. Additionally, we discuss the significance of pointwise inequalities and present a counterexample that prohibits further improvements.

1 Introduction

The Sobolev–Gagliardo–Nirenberg interpolation inequality is typically an inequality in the form

$$\begin{aligned} \Vert \nabla ^j u\Vert _{X}\lesssim \Vert \nabla ^k u\Vert _Y^{\theta }\Vert u\Vert _Z^{1-\theta }, \end{aligned}$$

(1)

where $0\le j<k$ are the degrees of derivatives, $\theta \in [j/k, 1]$ and X, Y, Z are function spaces. The general form is the product of continued development. The first attempts at estimating an intermediate derivative were made at the beginning of the 20th century by Landau (see [24]) who proved inequality

$$\begin{aligned} \Vert u'\Vert _{L^\infty }\lesssim \Vert u''\Vert _{L^\infty }^{1/2}\Vert u\Vert _{L^\infty }^{1/2}. \end{aligned}$$

Later on, Kolmogoroff proved the result for higher-order derivatives [17]. In 1958 the problem was revisited by Nash (see [35]) and Ladyzenskaya who proved

$$\begin{aligned} \Vert u\Vert _{L^{4}}\lesssim \Vert \nabla u\Vert _{L^2}^{1/2}\Vert u\Vert _{L^2}^{1/2} \end{aligned}$$

in dimension two and also a similar type result in dimension three. Her research in [23] was motivated by the study of mathematical models for hydro-mechanics. Later, Stein in [44] improved Landau-Kolmogoroff’s result in one dimension as

$$\begin{aligned} \Vert u'\Vert _{L^p}\lesssim \Vert u''\Vert ^{1/2}_{L^p}\Vert u\Vert ^{1/2}_{L^p} \end{aligned}$$

for $p\in [1,\infty )$. The real brake-through came in 1959 as Gagliardo [11] and Nirenberg [37] independently introduced the general version of Gagliardo–Nirenberg interpolation inequality for Lebesgue spaces which took the form

$$\begin{aligned} \Vert \nabla ^j u\Vert _{L^p}\lesssim \Vert \nabla ^k u\Vert ^\theta _{L^r}\Vert u\Vert _{L^q}^{1-\theta }. \end{aligned}$$

(2)

The formula holds for parameters $0\le j<k$, $\theta =j/k$, and p, q, r satisfying $k/p=j/r+(k-j)/q$. Also a slightly more complicated setting for the case $\theta \in [j/k,1]$ is proven therein. The modern versions of the proof can be found in [10, 25].

Nowadays, the field is too wide for being covered without omitting a lot. The primary motivation is obtaining the estimate for the solution of PDE, see, for example, [2, 8, 38]. As the methods and the function spaces scales refine, the demand for a finer version of such an inequality appears. As for other applications, we may mention the chain rule for Sobolev spaces [34] or the study of the boundedness of bilinear multipliers [12, 43]. The problem is developed in several branches, and the term Gagliardo–Nirenberg inequality may be used for several slightly different settings.

We may split the field concerning (1) based on value of j. Either $j>0$ as done by Gagliardo or $j=0$ as done by Ladyzhenskaya. There are interesting modern results for $j>0$ arising from studying the inequality in the context of Orlicz spaces, see [16] or by relaxing $L^{\infty }$ space into BMO in [45]. In literature, the more dominant version is corresponding to Ladyzhenskaya’s result, where (1) is set for $j=0$ and typically $k=1$. This version is studied in [8, 33] and used in above mentioned [2, 38].

Generalizations have been made in several directions. Even in the original paper, Nirenberg extends the Lebesgue scale into negative p and claims the result for Hölder spaces. The other direction is to extend j and k from integer numbers into real, more precisely using the fractional Sobolev space. Such a setting may be found in [3, 4].

Yet another possible modification is the so-called nonlinear or strongly nonlinear Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla ^j u\Vert _{X}\lesssim F(\nabla ^k u,u), \end{aligned}$$

where F is a functional, such as in [15, 41]. These types may be beneficial in special cases of applications where standard multiplicative versions may not be enough.

To map the development, we outline the tools used in the formulation and proofs of the modern results. We already mentioned fractional Sobolev spaces, which not only create new settings but also help to prove older results with ease [3]. The problem was also investigated in the Lorentz spaces scale. Relaxations from the Lebesgue spaces to the Lorentz spaces scale were given for $j=0$ in [33] and, recently, in See also [21] for variable Lebesgue spaces setting and [3, 5, 6] for fractional Sobolev and Besov spaces setting.

2 Results

We start the paper by reformulating the version of the Gagliardo–Nirenberg inequality for rearrangement invariant Banach function spaces stated already in [9] (cf. [21]) in another form. We do so, since the statement from [9] allows only very limited analysis of optimality, as we explain in Sect. 4. The rougher versions of optimality may be found in [9] or even [37], where the optimality from the point of view of fundamental functions is given, covering for example the optimality on the scale of Lebesgue and Orlicz spaces. Some other special cases may be found elsewhere in the literature.

Theorem 1.1

(Gagliardo–Nirenberg inequality for r.i.B.f. spaces) Let Y, Z be rearrangement invariant Banach function spaces and let $X=Y^{j/k}Z^{1-j/k}$ be defined by the Calderón–Lozanovskii construction, where $1\le j< k$ and upper Boyd indices of both spaces Y and Z are smaller than 1. Then

$$\begin{aligned} \Vert \nabla ^j u\Vert _{X}\lesssim \Vert \nabla ^k u\Vert _Y^{\frac{j}{k}}\Vert u\Vert _Z^{1-\frac{j}{k}} \end{aligned}$$

(3)

holds for all $u\in W^{k,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)\cap Z$.

The use of the maximal operator to find a pointwise version of the Gagliardo–Nirenberg inequality was essential in achieving the result. Such a question was firstly undertaken by Kałamajska [13]. We use the following version of such a pointwise Gagliardo–Nirenberg inequality

$$\begin{aligned} |\nabla ^j u|\lesssim (M\nabla ^k u)^{\frac{j}{k}}(Mu)^{1-\frac{j}{k}},\ \ \mathrm{\ where\ }0<j\le k \end{aligned}$$

(4)

proved by Maz$'$ya and Shaposchnikova in [32].

It is worth mentioning here, that pointwise estimates have already been applied to prove more general forms of the Gagliardo–Nirenberg inequality, such as the Gagliardo–Nirenberg for Orlicz spaces considered by Kałamajska, Pietruska-Pałuba, and Krbec (see [14, 16]) or even for general Banach function spaces [9, 21].

Later, Lokharu focused on different types of maximal operators in [27, 28]. Notice also that early version of Theorem 2.1 appeared in the literature before [9] in [21], but without the focus on the optimality.

Having Theorem 2.1 in a suitable form, we are ready to discuss the optimality of it in the setting of rearrangement invariant Banach function spaces. By optimality we mean, that having fixed two spaces, say Y, Z, there is no space B such that it is smaller than X and (3) still holds.

It is already known that Theorem 2.1 is sharp in the scale of Lebesgue spaces. This follows from the so-called scaling argument, see [9, Thm. 1.1]. In fact, the scaling argument applied to the setting of rearrangement invariant Banach function spaces gives a necessary condition on the fundamental function of space X satisfying (3), as was proved in [9]. However, (except in the limit cases) there is a whole universe of rearrangement invariant Banach function spaces with the same fundamental function. Thus to decide whether the choice of $X=Y^{j/k}Z^{1-j/k}$ in (3) is optimal in the scale of rearrangement invariant Banach function spaces we need to introduce an essentially different and more sensitive approach, than the scaling argument. We fix our attention only on the most classical case of $j=1,k=2$.

Theorem 1.2

[Optimality of Gagliardo–Nirenberg inequality for r.i.B.f. spaces] Let Y, Z be r.i.B.f. spaces and $X=Y^{1/2}Z^{1/2}$. Assume that the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla u\Vert _{B}\lesssim \Vert \nabla ^2 u\Vert _Y^{\frac{1}{2}}\Vert u\Vert _Z^{\frac{1}{2}} \end{aligned}$$

holds for all $u\in W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)\cap Z$ and some rearrangement invariant Banach function space B.

• (i) If $Z\cap L^{\infty }\subset Y\cap L^{\infty }$, then $X\cap L^{\infty }\subset B\cap L^{\infty }$.

• (ii) If $Z\cap L^{1} \subset Y\cap L^{1}$, then $X\cap L^{1} \subset B\cap L^{1}$.

Remark 2.3

Notice that we do not need restrictions on upper Boyd indices of Y and Z, in contrast to Theorem 2.1.

When both assumptions of points (i) and (ii) are satisfied at the same time, we get the complete optimality of $X=Y^{j/k}Z^{1-j/k}$.

Corollary 1.4

Let Y, Z be r.i.B.f. spaces satisfying $Z\subset Y$. If the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla u\Vert _{B}\lesssim \Vert \nabla ^2 u\Vert _Y^{\frac{1}{2}}\Vert u\Vert _Z^{\frac{1}{2}} \end{aligned}$$

holds for all $u\in W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)\cap Z$ and some r.i.B.f. space B, then $Y^{1/2}Z^{1/2}\subset B$.

Concluding, we see that the choice of $X=Y^{1/2}Z^{1/2}$ is optimal among all r.i.B.f. spaces in the Gagliardo–Nirenberg inequality (3), provided that $Z\subset Y$. However, the assumption $Z\subset Y$ is quite restrictive and does not apply to the most classical r.i. spaces (Lebesgue spaces, Lorentz spaces, etc.), since usually there is no inclusion between such spaces over $\mathbb {R}_+$. It appears, however, that manoeuvring between points (i) and (ii) of Theorem 2.2 we can use it to give an almost complete answer to the question about the optimality of (3) among Lorentz spaces posted in [9, Remark 2.7].

Corollary 1.5

Let $1< R,Q< \infty$ and $1\le r,q\le \infty$. Then the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla u\Vert _{P,p}\lesssim \Vert \nabla ^2 u\Vert _{Q,q}^{\frac{1}{2}}\Vert u\Vert _{R,r}^{\frac{1}{2}} \end{aligned}$$

(5)

holds for all $u\in W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)\cap L^{R,r}$ with $P=(1/(2Q)+1/(2R))^{-1}$ and $p= (1/(2q)+1/(2r))^{-1}$.

On the other hand, if one of the following conditions holds

• (i) $R\not =Q$,

• (ii) $R=Q$ and $r<q$,

then the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla u\Vert _{P',p'}\lesssim \Vert \nabla ^2 u\Vert _{Q,q}^{\frac{1}{2}}\Vert u\Vert _{R,r}^{\frac{1}{2}} \end{aligned}$$

(6)

holds for all $u\in W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)\cap L^{R,r}$ if and only if $P'=P$ and $p'\ge p$.

Notice, that the corollary above does not cover the case of $Q=R$ and $r>q$, thus the following problem remains open.

Question 1.6

Is the Gagliardo–Nirenberg inequality (6) optimal when $1<Q=R$ and $r>q$, or is there any $p'<p{:=}(1/(2q)+1/(2r))^{-1}$ such that (6) holds with p replaced by $p'$ in that case, i.e. when $1<Q=R$ and $r>q$?

Our main results give also new information about the Gagliardo–Nirenberg inequality for Orlicz spaces.

Corollary 1.7

Let $\varphi ,\varphi _1,\varphi _2$ be Young functions such that $\varphi ^{-1}\approx \sqrt{\varphi _1^{-1}\varphi _2^{-1}}$ and upper Boyd indices of $L^{\varphi _1}$ and $L^{\varphi _2}$ are smaller than 1. Then the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla u\Vert _{\varphi }\lesssim \Vert \nabla ^2 u\Vert _{\varphi _1}^{\frac{1}{2}}\Vert u\Vert _{\varphi _2}^{\frac{1}{2}} \end{aligned}$$

(7)

holds for all $u\in W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)\cap L^{\varphi _2}$. On the other hand, if $L^{\varphi _2}\subset L^{\varphi _1}$ and the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla u\Vert _{B}\lesssim \Vert \nabla ^2 u\Vert _{\varphi _1}^{\frac{1}{2}}\Vert u\Vert _{\varphi _2}^{\frac{1}{2}} \end{aligned}$$

(8)

holds for all $u\in W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)\cap L^{\varphi _2}$ and some r.i.B.f. space B, then $L^{\varphi }\subset B$.

Notice that (7) has been already proved in [9] (cf. [16]). On the other hand, it was shown therein that $L^{\varphi }$ as above is optimal among all Orlicz spaces, while our result says that it is optimal also in the class of r.i.B.f. spaces, provided $L^{\varphi _2}\subset L^{\varphi _1}$.

3 Preliminaries

Given some positive-valued F(u), G(u) we write

$$\begin{aligned} F(u)\lesssim G(u) \end{aligned}$$

if there exists a constant $C>0$ (independent on u) such that

$$\begin{aligned} F(u)\le C G(u). \end{aligned}$$

If both $F(u)\lesssim G(u)$ and $G(u)\lesssim F(u)$, then we write

$$\begin{aligned} F(u)\approx G(u). \end{aligned}$$

3.1 Function spaces instruments

In our paper, we shall use the following standard notation. Given a Lebesgue-measurable set $A\subset \mathbb {R}^d$, we shall denote its Lebesgue measure by |A|. The symbol $L^0(A)$ denotes the space of all measurable functions over the given set A that are finite almost everywhere.

Let $X\subset L^0$ be a Banach space. We call X a Banach function space (B.f. space, for short) if it has the following properties:

• the ideal property, i.e. given $f\in L^0$ and $g\in X$ with $|f|\le g$ it holds that $\Vert f\Vert _X\le \Vert g\Vert _X$.

• the Fatou property, i.e. given $f\in L^0$ and an increasing sequence $(f_n)\in X$ such that $0\le f_n\rightarrow f$ a.e. and $\sup _n\Vert f_n\Vert _X<\infty$, there holds $f\in X$ and $\Vert f\Vert _X=\sup \limits _n\Vert f_n\Vert _X$.

Given a measurable function $f\in L^0(\mathbb {R}^d)$ and $\alpha \in \mathbb {R}$ we shall denote its level set shortly by

$$\begin{aligned} \{f>\alpha \}{:=}\{x\in \mathbb {R}^d: f(x)>\alpha \} \end{aligned}$$

and similarly in the case of $\{f\ge \alpha \}, \{f<\alpha \}, \{f\le \alpha \}$. Throughout the paper, a distribution function of f is denoted by

$$\begin{aligned} f_*(t){:=}\left| \{|f|>t\}\right| \ \textrm{for}\ t>0 \end{aligned}$$

and the non-increasing rearrangement by

$$\begin{aligned} f^*(t){:=}\inf _{s>0}\{f_*(s)\le t\}. \end{aligned}$$

A B.f. space X is called a rearrangement invariant B.f. space (we use the abbreviation r.i.B.f. space) if given two functions $f\in L^0$ and $g\in X$ which are equidistributed (i.e. $f_*=g_*$), it holds that $f\in X$ and $\Vert f\Vert _X=\Vert g\Vert _X$.

To simplify the notion, we make use of the Luxemburg representation theorem [1, Thm. 4.2, Ch. 2]. It says that for each r.i.B.f. space X over arbitrary $A\subset \mathbb {R}^d$ there is an r.i.B.f. space $\overline{X}$ over $(0,\infty )$ with the Lebesgue measure, such that for every measurable $f:A\rightarrow \mathbb {R}$ one has

$$\begin{aligned} \Vert f\Vert _X=\Vert f^*\Vert _{\overline{X}}. \end{aligned}$$

(9)

Vice versa, having r.i.B.f. spaces $\overline{X}$ defined over $(0,\infty )$ we can define a r.i.B.f. space X over $\mathbb {R}^d$ just by the same equality, i.e. (9) for $f\in L^0(\mathbb {R}^d)$.

Henceforth, speaking about a r.i.B.f. space X we always mean that X is defined over $(0,\infty )$ and, at the same time it defines a r.i.B.f. space over $\mathbb {R}^d$ by (9), but we will not differentiate between X and $\overline{X}$ (cf. [26, pp.114–115]).

Given a r.i.B.f. space X we define

$$\begin{aligned} X_{{{\,\textrm{loc}\,}}}{:=}\{u\in L^0: u^{*}\chi _{(0,1)}\in X\}. \end{aligned}$$

Moreover, we write

$$\begin{aligned} X{\mathop {\hookrightarrow }\limits ^{{{\,\textrm{loc}\,}}}}Y, \end{aligned}$$

if there exists a constant $C>0$ such that

$$\begin{aligned} \Vert f^*\chi _{(0,1)}\Vert _{Y}\le C\Vert f^*\chi _{(0,1)}\Vert _{X},\quad \forall f\in L^{0}. \end{aligned}$$

We also need the Hardy operator, which is defined for $u\in L^{1}_{{{\,\textrm{loc}\,}}}$ by the formula

$$\begin{aligned} Au(t){:=}\frac{1}{t}\int _0^t u(s)d s, \ \ t>0. \end{aligned}$$

Further, the symbol $u^{**}$ will be used for the Hardy-Littlewood maximal function of u defined by

$$\begin{aligned} u^{**}(t){:=}A(u^*)(t), \ \ t>0. \end{aligned}$$

It is known, that each r.i.B.f. space X satisfies the Hardy-Littlewood-Polya principle ( [1, Cor. 4.7, Ch. 2]) (or the majorant property, see [15]), which states that for $g\in X$ and $f\in L^{1}_{{{\,\textrm{loc}\,}}}$

$$\begin{aligned} if f^{**}(t)\le g^{**}(t) \text { for all } t>0 , then\ f\in X \mathrm{\ and\ }\Vert f\Vert _X\le \Vert g\Vert _X. \end{aligned}$$

(10)

The maximal function defined above is closely connected with the maximal operator which is defined for any function locally integrable on $\mathbb {R}^d$, by the formula

$$\begin{aligned} Mu(x){:=}\sup _{Q\ni x}\frac{1}{|Q|}\int _Q |u(y)|d y,\ \ x\in \mathbb {R}^d, \end{aligned}$$

where the supremum on the right-hand side is taken over all d–dimensional cubes Q containing x. The Riesz-Herz equivalence states that

$$\begin{aligned} (Mf)^*(t)\approx f^{**}(t)\quad \text { for all } t\in (0,\infty ), \end{aligned}$$

(11)

where the constant of equivalence depends only on the dimension d (see [1, Theorem 3.8]).

The dilation operator $D_s$ is defined for $s>0$ and $f\in L^0(0,\infty )$ (or $f\in L^0(\mathbb {R}^d)$) by the formula

$$\begin{aligned} D_sf(t){:=}f(st)\ \textrm{for} \ t>0\ [\textrm{or} \ t\in \mathbb {R}^d]. \end{aligned}$$

It is known ( [1, Prop. 5.11, Ch. 3]) that $D_s$ is bounded on each r.i.B.f. space X for each $s>0$. Moreover, for each r.i.B.f. space X on $(0,\infty )$ limits

$$\begin{aligned} \alpha _X=\lim _{s\rightarrow 0}\frac{\log \Vert D_{1/s}\Vert _{X\rightarrow X}}{\log s}, \quad \beta _X=\lim _{s\rightarrow \infty }\frac{\log \Vert D_{1/s}\Vert _{X\rightarrow X}}{\log s} \end{aligned}$$

exist. The numbers $\alpha _X$ and $\beta _X$ are called lower and upper Boyd indices of X, respectively. As in [1], we understand that Boyd of a r.i.B.f. space over $\mathbb {R}^d$ are just Boyd indices of its Luxemburg representant. Then for each r.i.B.f. space X, its Boyd indices belong to [0, 1] and $\alpha _X\le \beta _X$. For more information on Boyd indices and, in general, r.i.B.f. spaces we refer for instance to books [1, 22, 39].

The most significant examples of r.i.B.f. spaces except for the Lebesgue spaces are Lorentz and Orlicz spaces. Let us recall their definitions. For $1\le P,p<\infty$ the Lorentz space $L^{P,p}$ is given by the (quasi–) norm

$$\begin{aligned} \Vert f\Vert _{P,p}=\left( \int _0^{\infty }[t^{1/P}f^*(t)]^p\frac{dt}{t}\right) ^{1/p} \end{aligned}$$

or by

$$\begin{aligned} \Vert f\Vert _{P,\infty }=\sup _{0<t<\infty }t^{1/P}f^*(t) \end{aligned}$$

when $p=\infty$ and $1\le P<\infty$. When $P=\infty$ the space $L^{\infty ,p}$ is nontrivial only when also $p=\infty$ and then $L^{\infty ,\infty }=L^{\infty }$. In the case $1<P$ and $1\le p\le \infty$ the functional $\Vert \cdot \Vert _{P,p}$ is already equivalent to the norm, thus we will treat $L^{P,p}$ as a r.i.B.f. space. When $P=1$ we consider only $p=1$, since otherwise $L^{1,p}$ is no more a B.f. space.

A continuous, convex, increasing function $\varphi :[0,\infty )\rightarrow [0,\infty )$ satisfying $\varphi (0)=0$ is called a Young function. For a given Young function $\varphi$ we define the Orlicz space $L^{\varphi }$ by

$$\begin{aligned} L^{\varphi }=\left\{ f\in L^0:\int _0^{\infty }\varphi \left( \frac{|f(t)|}{\lambda }\right) dt <\infty \mathrm{\ for\ some \ }\lambda >0 \right\} \end{aligned}$$

with the Luxemburg norm

$$\begin{aligned} \Vert f\Vert _{\varphi }=\inf \left\{ \lambda >0:\int _0^{\infty }\varphi \left( \frac{|f(t)|}{\lambda }\right) dt \le 1\right\} . \end{aligned}$$

3.2 Derivatives

To denote the distributional derivative in one-dimensional case we use symbols $u',u'',u^{(k)}$. The symbols $\nabla ,\nabla ^k$ denote the distributional gradient, respectively the distributional gradient of order k. We set

$$\begin{aligned} \left| \nabla ^k u\right| {:=}\sum _{|\alpha |=k}\left| \frac{\partial ^\alpha u}{\partial x^\alpha }\right| . \end{aligned}$$

Given a r.i.B.f. space X we set

$$\begin{aligned} \Vert \nabla ^k u\Vert _X{:=}\Vert |\nabla ^k u|\Vert _X. \end{aligned}$$

We also define the maximal operator of the k-order gradient by

$$\begin{aligned} M(\nabla ^k u)(x){:=}\sup _{Q\ni x}\frac{1}{|Q|}\int _{Q}|\nabla ^k u|d x. \end{aligned}$$

The notion $W^{k,p}(A)$ will be used for the space of functions whose distributional derivatives up to degree k belong to $L^{p}(A)$, the sub-index “loc” is added if the distributional derivatives belong only to $L^{p}_{{{\,\textrm{loc}\,}}}(A)$.

3.3 Calderón–Lozanovskii construction and pointwise product spaces

Given $0<\theta <1$ and two B.f. spaces X, Y over the same measure space, the Calderón–Lozanovskii space $X^{\theta }Y^{1-\theta }$ is defined as

$$\begin{aligned} X^{\theta }Y^{1-\theta }=\{f\in L^0:|f|\le g^{\theta }h^{1-\theta } \mathrm{\ for \ some\ }0\le g\in X,0\le h\in Y \} \end{aligned}$$

and equipped with the norm

$$\begin{aligned} \Vert f\Vert _{X^{\theta }Y^{1-\theta }}=\inf \left\{ \max \left\{ \Vert g\Vert _X,\Vert h\Vert _Y\right\} :|f|\le g^{\theta }h^{1-\theta }, 0\le g\in X,0\le h\in Y\right\} . \end{aligned}$$

It is easily seen that one can replace “$\le$” by “$=$” in the above definitions. Moreover, the following inequality holds true for each $g\in X,h\in Y$ (see [30])

$$\begin{aligned} \Vert |g|^{\theta }|h|^{1-\theta }\Vert _{X^{\theta }Y^{1-\theta }}\le \Vert g\Vert _X^{\theta }\Vert h\Vert _Y^{1-\theta }. \end{aligned}$$

(12)

Given a B.f. space X and $\alpha >1$, we define the $\alpha$-convexification of X (respectively, $\alpha$-concavification when $0<\alpha <1$) by

$$\begin{aligned} X^\alpha {:=}\{f\in L^0:|f|^\alpha \in X\} \end{aligned}$$

with the (quasi–) norm given by

$$\begin{aligned} \Vert f\Vert _{X^{\alpha }}{:=}\left( \Vert |f|^\alpha \Vert _X\right) ^{1/\alpha }. \end{aligned}$$

It is easy to see that convexification is just a special case of the Calderón–Lozanovskii construction, namely

$$\begin{aligned} X^\alpha =X^{\frac{1}{\alpha }}(L^{\infty })^{1-\frac{1}{\alpha }}. \end{aligned}$$

In Sect. 5, we will once more need the constructions intimately connected with Calderón–Lozanovskii spaces. Given two B.f. spaces X, Y over the same measure space we define their pointwise product $X\odot Y$ by

$$\begin{aligned} X\odot Y=\{f\in L^0:|f|= gh \mathrm{\ for \ some\ } g\in X,h\in Y\}, \end{aligned}$$

equipped with the quasi-norm

$$\begin{aligned} \Vert f\Vert _{X\odot Y}=\inf \{ \Vert g\Vert _X\Vert h\Vert _Y:|f|= gh \mathrm{\ for \ some\ }g\in X,h\in Y\}. \end{aligned}$$

We will need the following identification from [19]

$$\begin{aligned} X^{\theta }Y^{1-\theta }=X^{\frac{1}{\theta }}\odot Y^{\frac{1}{1-\theta }}. \end{aligned}$$

(13)

4 New form of Gagliardo–Nirenberg inequality for r.i.B.f. spaces

We start this section with the proof of Theorem 2.1, which actually is an immediate consequence of (4).

Proof of Theorem 2.1

Let $u\in W^{k,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)$ and $1\le j<k$. We have by (4)

$$\begin{aligned} \begin{aligned} \Vert \nabla ^j u\Vert _X&\lesssim \Vert (M(\nabla ^k u))^{\frac{j}{k}} (Mu)^{1-\frac{j}{k}}\Vert _X. \end{aligned} \end{aligned}$$

Then by properties of the Calderon–Lozanovski construction

$$\begin{aligned} \begin{aligned} \Vert (M(\nabla ^k u))^{\frac{j}{k}} (Mu)^{1-\frac{j}{k}}\Vert _X \le \Vert M(\nabla ^{k}u)\Vert _{Y}^{\frac{j}{k}}\Vert Mu\Vert _{Z}^{\frac{k-j}{k}}. \end{aligned} \end{aligned}$$

Finally, we replace the maximal operators of functions with functions themselves in both norms on the right-hand side. The boundedness of the maximal operator in the considered spaces (based on the upper Boyd index being smaller than 1) gives

$$\begin{aligned} \Vert \nabla ^j u\Vert _X\lesssim \Vert (\nabla ^{k}u)\Vert _{Y}^{\frac{j}{k}}\Vert u\Vert _{Z}^{\frac{k-j}{k}}, \end{aligned}$$

as desired. $\square$

Let us note that without the assumption about the boundedness of the maximal operator, we cannot obtain the last estimate. Neither counterexamples nor the proof of the validity of the theorem without the assumption on the Boyd indices is known. Let us, however, notice, that the original proof of the Gagliardo–Nirenberg inequality for Lebesgue spaces allows also $L^1$ spaces (cf. [10, 11, 37]), the space where the maximal operator is not bounded. On the other hand, proofs of the inequality for Orlicz, Lorentz, or more general spaces (up to our knowledge) are based on the same kind of pointwise inequalities, while such an approach requires assumptions on Boyd indices (or generally on the boundedness of the maximal operator).

Question 1.8

Does Theorem 2.1 hold without the assumption on Boyd indices of Y and Z?

One of the possible ways of removing assumptions on Boyd indices would be a slight improvement of (4). In fact, according to [13] we can rewrite (actually, strengthen) (4) to

$$\begin{aligned} (\nabla ^j u)^{**}\lesssim (|\nabla ^k u|^{**})^{\frac{j}{k}}(|u|^{**})^{1-\frac{j}{k}}. \end{aligned}$$

Then, by the Riesz-Herz equivalence, we observe that it would be sufficient for our purpose if we could put the geometric mean of |u| and $|\nabla ^2 u|$ inside the double-star operator. Precisely, the question is if the following inequality holds

$$\begin{aligned} (\nabla ^j u)^{**}\lesssim (|\nabla ^k u|^{\frac{j}{k}}|u|^{1-\frac{j}{k}})^{**}? \end{aligned}$$

(14)

This inequality would then, by a simple use of the Hardy-Littlewood-Polya principle, imply that Theorem 2.1 holds with no restriction on Boyd indices. However, the counterexample below shows that one cannot hope for (14).

Example 4.2

Let $u_n:{{\,\mathrm{\mathbb {R}}\,}}\rightarrow {{\,\mathrm{\mathbb {R}}\,}}$ be a sequence of functions defined in the following way

$$\begin{aligned} u_n(0){:=}0, \quad u_n'(0)=0\quad u_n''(s){:=}n\left( \chi _{(0,\frac{1}{n})}-\chi _{(1-\frac{1}{n},1+\frac{1}{n})}+\chi _{(2-\frac{1}{n},2)}\right) . \end{aligned}$$

One easily verifies that all the functions are supported in interval [0, 2] and $u_n(2)=u_n'(2)=0$. Moreover, we estimate

$$\begin{aligned} (u_n')^*(s)=\chi _{(0,2-\frac{4}{n})}+\left( 1-n\left( s-\left( 2-\frac{4}{n}\right) \right) \right) \chi _{(2-\frac{4}{n},2)}, \end{aligned}$$

hence we have

$$\begin{aligned} 1\ge (u_n')^{**}(2)\ge \frac{1}{2}\int _0^2 \chi _{(0,2-\frac{4}{n})}=\frac{2-\frac{4}{n}}{2}=1-\frac{2}{n} \end{aligned}$$

On the other hand

$$\begin{aligned} \begin{aligned} (|u_n''|^{\frac{1}{2}}|u_n|^{\frac{1}{2}})^{**}(2)&=\frac{1}{2}\int _0^2|u_n''|^{\frac{1}{2}}|u_n|^{\frac{1}{2}}(s){{\,\textrm{d}\,}}s\\&\le \frac{1}{2}\left( \int _0^{\frac{1}{n}}\sqrt{n}{{\,\textrm{d}\,}}s+\int _{1-\frac{1}{n}}^{1+\frac{1}{n}}\sqrt{n}{{\,\textrm{d}\,}}s+\int _{2-\frac{1}{n}}^1\sqrt{n}\right) \\&=\frac{2}{\sqrt{n}} \end{aligned} \end{aligned}$$

One readily checks that

$$\begin{aligned} \lim _{n\rightarrow \infty } (u_n')^{**}(2)=1 \end{aligned}$$

while

$$\begin{aligned} \lim _{n\rightarrow \infty } (|u_n''|^{\frac{1}{2}}|u_n|^{\frac{1}{2}})^{**}(2)=0 \end{aligned}$$

Thus the inequality (14) cannot hold in general.

Let us finally explain why Theorem 1.2 of [9] required reformulation. For inequalities involving only two norms, like Sobolev or Poincare inequalities, the question about optimality is immediate: we fix one norm and ask how much the second one can be improved. For the Gagliardo–Nirenberg type inequalities the situation is more complicated. Namely, the question about optimality may be asked at least in three ways. Considering

$$\begin{aligned} \Vert \nabla ^j u\Vert _{X}\lesssim \Vert \nabla ^k u\Vert _{Y}^{\frac{j}{k}} \Vert u\Vert _{Z}^{1-\frac{j}{k}} \end{aligned}$$

(15)

we need to fix two spaces and ask for the optimality of the third one.

In our main results, we follow one possible approach: having two spaces Y, Z corresponding to a function and its higher derivative, roughly speaking we found or constructed an optimal space X corresponding to the middle derivative.

However, from the point of view of applications, another approach may be desirable. Namely, having spaces X and Y [or X and Z] fixed, we need to find the third space Z [or Y], possibly optimal, such that (15) holds. Such a point of view has already been presented in [9]. That approach, however, has some disadvantages.

To discuss it, we need to provide a new notation. Given a couple of r.i.B.f. spaces X, Y such that $Y{\mathop {\hookrightarrow }\limits ^{loc }} X$, we define the space M(X, Y) of pointwise multipliers from X to Y using the following formula

$$\begin{aligned} M(X,Y)=\{f\in L^0:fg\in Y\mathrm{\ for \ each \ }g\in X\} \end{aligned}$$

and equip it with the natural operator norm

$$\begin{aligned} \Vert f\Vert _{M(X,Y)}{:=}\sup _{\Vert g\Vert _X\le 1}\Vert fg\Vert _Y. \end{aligned}$$

With this norm M(X, Y) becomes an r.i.B.f. space. Moreover, the following general version of the Hölder inequality follows directly from the definition

$$\begin{aligned} \Vert fg\Vert _Y\le \Vert g\Vert _{X}\Vert f\Vert _{M(X,Y)}. \end{aligned}$$

Notice that M(X, Y) is a kind of generalization of the Köthe dual of X (i.e. $X'=M(X, L^1)$), but on the other hand, it may be regarded as a kind of point-wise quotient of the space Y by X. More information about the spaces of pointwise multipliers may be found in [18, 31].

Now we can state an alternative version of Theorem 2.1. Notice that point (i) is actually Theorem 1.2 of [9].

Theorem 1.10

Let $j,k\in \mathbb {N}$, $1\le j<k$.

• (i) If X, Y are r.i.B.f. spaces such that

  $$\begin{aligned} Y^{\frac{k}{j}}{\mathop {\hookrightarrow }\limits ^{loc }} X \end{aligned}$$

  and both Y and $M(Y^{k/j},X)^{1-j/k}$ have upper Boyd indices less than 1, then for all $u\in W^{k,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)$ it holds

  $$\begin{aligned} \Vert \nabla ^j u\Vert _{X}\lesssim \Vert \nabla ^k u\Vert _{Y}^{j/k} \Vert u\Vert _{M(Y^{k/j},X)^{1-j/k}}^{1-j/k}. \end{aligned}$$

  (16)

• (ii) If X, Z are r.i.B.f. spaces such that

  $$\begin{aligned} Z^{\frac{k}{k-j}}{\mathop {\hookrightarrow }\limits ^{loc }} X \end{aligned}$$

  and both Z and $M(Z^{k/(k-j)},X)^{j/k}$ have upper Boyd indices less than 1, then for all $u\in W^{k,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)$ it holds

  $$\begin{aligned} \Vert \nabla ^j u\Vert _{X}\lesssim \Vert \nabla ^k u\Vert _{M(Z^{k/(k-j)},X)^{j/k}}^{j/k} \Vert u\Vert _{Z}^{1-j/k}. \end{aligned}$$

  (17)

Proof

The proof is immediate once we know Theorem 2.1. Let us consider point (i). In fact, it is enough to see that

$$\begin{aligned} Y^{j/k}[M(Y^{k/j},X)^{1-j/k}]^{1-j/k}\subset X. \end{aligned}$$

We have by (13)

$$\begin{aligned}{} & {} Y^{j/k}[M(Y^{k/j},X)^{1-j/k}]^{\frac{k-j}{k}}=Y^{k/j}\odot [M(Y^{k/j},X)^{1-j/k}]^{\frac{k}{k-j}} \\{} & {} \quad =Y^{k/j}\odot M(Y^{k/j},X)\subset X. \end{aligned}$$

$\square$

Notice that the last inclusion above, i.e. $Y^{k/j}\odot M(Y^{k/j}, X)\subset X$, can not be replaced by equality. This is the reason that the formulation of Theorem 2.1 is more accurate than the claim of Theorem 4.3 above (thus also Theorem 1.2 of [9]). Let us explain it better. We already know that the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla ^j u\Vert _{X}\lesssim \Vert \nabla ^k u\Vert _{Y}^{\frac{j}{k}} \Vert u\Vert _{Z}^{1-\frac{j}{k}}. \end{aligned}$$

holds with $X=Y^{j/k}Z^{1-j/k}$ and is optimal in many cases (for more details see Sect. 5). So, to keep potential optimality, having spaces X, Y, we need to find the third space Z satisfying $X=Y^{j/k}Z^{1-j/k}$. However, such a space Z may not exist, and the best we can choose is the space

$$\begin{aligned} Z=M(Y^{k/j},X)^{1-j/k} \end{aligned}$$

appearing in the Theorem 4.3(i), which for sure gives only inclusion

$$\begin{aligned} Y^{j/k}[M(Y^{k/j},X)^{1-j/k}]^{1-j/k}\subset X, \end{aligned}$$

but need not give equality. Whether this inclusion becomes equality is exactly the problem of (Lozanovskii like) factorization.

Further, we say that Y factorizes X when

$$\begin{aligned} Y\odot M(Y, X)=X. \end{aligned}$$

The origins of factorization come from the Lozanovskii factorization theorem [29, Theorem 6] (which says that for each B.f. space X, there holds $X\odot X'=L^1$), while Pisier has considered factorization in terms of Calderon-Lozanovskii construction in [40] (cf. [36]). For a quite comprehensive study of factorization problems, see, for example, [19, 20, 42] and references therein.

In this language, our previous question becomes whether

$$\begin{aligned} Y^{k/j}\odot M(Y^{k/j},X)=X? \end{aligned}$$

This is, whether $Y^{k/j}$ factorizes X? If not, then one shouldn’t expect optimality of (16) (or (17), respectively).

Let us use an example to explain the situation. Consider $X=L^{2,\infty }$, $Y=L^{2,1}$ and $j=1,k=2$. Then $Y^2=L^{4,2}$. We have by [20, Theorem 4]

$$\begin{aligned} M(Y^{2},X)=M(L^{4,2},L^{2,\infty })=L^{4,\infty }. \end{aligned}$$

Thus

$$\begin{aligned} Z=M(Y^{2},X)^{1/2}=L^{2,\infty } \end{aligned}$$

and Theorem 4.3(i) gives

$$\begin{aligned} \Vert \nabla u\Vert _{L^{2,\infty }}\lesssim \Vert \nabla ^2 u\Vert _{L^{2,1}}^{1/2} \Vert u\Vert _{L^{2,\infty }}^{1/2}. \end{aligned}$$

On the other hand, we know from Theorem 2.1 or Corollary 2.5 that stronger inequalities hold, i.e.

$$\begin{aligned} \Vert \nabla u\Vert _{L^{2,2}}\lesssim \Vert \nabla ^2 u\Vert _{L^{2,1}}^{1/2} \Vert u\Vert _{L^{2,\infty }}^{1/2} \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla u\Vert _{L^{2,\infty }}\lesssim \Vert \nabla ^2 u\Vert _{L^{2,\infty }}^{1/2} \Vert u\Vert _{L^{2,\infty }}^{1/2} \end{aligned}$$

Concluding, we see that formulation of Theorem 4.3 may differ from that of Theorem 2.1 when respective spaces do not factorize each other. In consequence, only Theorem 2.1 has the potential to be optimal and, in fact, it is in many cases, as we will see in the next section.

5 Optimality

As announced in the introduction, we will show that in the most classical case of $k=2$ and $j=1$ of the Gagliardo–Nirenberg inequality which we just proved is optimal for a broad class of r.i.B.f. spaces. We have just shown that

$$\begin{aligned} \Vert \nabla u\Vert _{X}\lesssim \Vert \nabla ^2 u\Vert _{Y}^{\frac{1}{2}} \Vert u\Vert _{Z}^{\frac{1}{2}} \end{aligned}$$

(18)

holds with $X=Y^{1/2}Z^{1/2}$ under some assumptions on Y, Z. Now we study the following question of optimality. For fixed Y, Z, is the choice of X optimal among the r.i.B.f. spaces? More precisely, can X be replaced in (18) by strictly smaller r.i.B.f. space B (we mean that $B\subsetneq X$) such that the estimate is still valid for each $u\in W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)$? In this section, we show that in many cases of spaces Y, Z the choice $X=Y^{1/2}Z^{1/2}$ is optimal.

In the beginning, we need a few simple observations on Calderón–Lozanovskii construction.

Lemma 1.11

Let Y, Z be r.i.B.f. spaces.

• (i) If $Z\cap L^{\infty }\subset Y\cap L^{\infty }$ then for each $f=f^*\in Y^{1/2}Z^{1/2}$ of the form

  $$\begin{aligned} f=\sum _{k=1}^{\infty }c_{k}\chi _{[k-1,k)} \end{aligned}$$

  there are $g=g^*\in Y$ and $h=h^*\in Z$ of the same form, i.e.

  $$\begin{aligned} g=\sum _{k=1}^{\infty }a_{k}\chi _{[k-1,k)},\ \ h=\sum _{k=1}^{\infty }b_{k}\chi _{[k-1,k)} \end{aligned}$$

  such that

  $$\begin{aligned} f\le g^{1/2}h^{1/2}\ \text { and }\ h\le g. \end{aligned}$$

• (ii) If $Z\cap L^{1}\subset Y\cap L^{1}$, then for each $f=f^*\in Y^{1/2}Z^{1/2}$ such that ${{\,\textrm{supp}\,}}f\subset [0,1]$ there are $g=g^*\in Y$ and $h=h^*\in Z$ satisfying

  $$\begin{aligned} f\le g^{1/2}h^{1/2}\ \text { and }\ h\le g. \end{aligned}$$

Proof

(i) Let $f=f^*=\sum _{k=1}^{\infty }c_{k}\chi _{[k-1,k)}\in Y^{1/2}Z^{1/2}$. Then there are $u\in Y,v\in Z$ such that

$$\begin{aligned} f\le u^{1/2}v^{1/2}. \end{aligned}$$

By [22, p. 67] we have for each $t\ge 0$

$$\begin{aligned} f(t)=f^*(t)\le u^*(t/2)^{1/2}v^*(t/2)^{1/2}. \end{aligned}$$

Denoting $\eta {:=}D_{1/2}u^*,\gamma {:=}D_{1/2}v^*$ we get

$$\begin{aligned} f\le \eta ^{1/2}\gamma ^{1/2}, \end{aligned}$$

while $\eta \in Y,\gamma \in Z$ since Y, Z are r.i.B.f. spaces, as the dilation operator is bounded in every r.i.B.f. space.

Further, we define the averaging operator T (also known as conditional expectation operator) by the formula

$$\begin{aligned} T:w\mapsto \sum _{k=1}^{\infty } \left( \int _{k-1}^kw(t)\,dt \right) \chi _{[k-1,k)}. \end{aligned}$$

(19)

Since each r.i.B.f. space with the majorant property is an exact interpolation space for the couple $(L^{\infty },L^{1})$, the operator T is non-expanding mapping on such a space (see [1, 22]). In consequence,

$$\begin{aligned} r{:=}T\eta \in Y \ \mathrm{\ and\ }\ h{:=}T\gamma \in Z. \end{aligned}$$

Both r and h are of the form

$$\begin{aligned} r=r^*=\sum _{k=1}^{\infty }a_{k}\chi _{[k-1,k)},\ \ h=h^*=\sum _{k=1}^{\infty }b_{k}\chi _{[k-1,k)} \end{aligned}$$

where

$$\begin{aligned} a_{k}=\int _{k-1}^k\eta (t)dt \ \mathrm{\ and\ }\ b_k=\int _{k-1}^k\gamma (t)dt. \end{aligned}$$

We need to see that also $f\le r^{1/2}h^{1/2}$. Indeed, by the Hölder inequality we get immediately

$$\begin{aligned} \begin{aligned} c_{k}&=\int _{k-1}^kf(t) \,dt\le \int _{k-1}^k\eta (t)^{1/2}\gamma (t)^{1/2} \,dt \\&\le \left( \int _{k-1}^k\eta (t)\,dt\right) ^{1/2}\left( \int _{k-1}^k\gamma (t)\,dt\right) ^{1/2}=a_k^{1/2}b_k^{1/2}. \end{aligned} \end{aligned}$$

Consequently, it holds that $r\in Y\cap L^{\infty }$ and $h\in Z\cap L^{\infty }$, since they are of the special form. Therefore, applying the assumption $Z\cap L^{\infty }\subset Y\cap L^{\infty }$, we see that also

$$\begin{aligned} g{:=}\max \{r,h\}\in Y\cap L^{\infty }. \end{aligned}$$

Thus the functions g and h have all the properties declared in point (i) since the maximum of nonincreasing functions is also a nonincreasing function.

(ii) Let $Z\cap L^{1}\subset Y\cap L^{1}$ and choose $f^*=f\in Y^{1/2}Z^{1/2}$ such that ${{\,\textrm{supp}\,}}f\subset [0,1]$. Arguing as before, we conclude that there are $\eta =\eta ^*\in Y,\gamma =\gamma ^*\in Z$ such that

$$\begin{aligned} f\le \eta ^{1/2}\gamma ^{1/2}. \end{aligned}$$

Evidently, we can assume that also ${{\,\textrm{supp}\,}}\eta \subset [0,1]$ and ${{\,\textrm{supp}\,}}\gamma \subset [0,1]$. However, each r.i.B.f. space is contained in $L^1+L^{\infty }$, which implies that any element of a r.i.B.f. space whose support has finite measure is in $L^1$. Then $\eta \in Y\cap L^{1}$ and $\gamma \in Z\cap L^{1}$. Consequently, thanks to the assumption $Z\cap L^{1}\subset Y\cap L^{1}$, it is enough to take

$$\begin{aligned} g{:=}\max \{\eta ,\gamma \} \mathrm{\ and\ } h{:=}\gamma . \end{aligned}$$

$\square$

Lemma 1.12

For every $\alpha \ge 1$ there exists $u\in W^{2,\infty }(\mathbb {R})$, ${{\,\textrm{supp}\,}}u=[0,1]$, such that

• (i) $\bigl |\left\{ |u'|\ge \alpha \right\} \bigr |\ge 1/6$

• (i) $|u(t)|\le 1/3$ and $|u''(t)|\le 6\alpha ^2$.

Proof

On $(0,1/\alpha )$ we define a continuous piecewise affine $u'(t)$ as

$$\begin{aligned} u'(t){:=} {\left\{ \begin{array}{ll} 6\alpha ^2t &{}\text{ if } t\in [0,\frac{1}{6\alpha }]\\ \alpha &{}\text{ if } t\in [\frac{1}{6\alpha },\frac{2}{6\alpha }] \\ 3\alpha -6\alpha ^2t&{}\text{ if } t\in [\frac{2}{6\alpha },\frac{3}{4\alpha }] \\ -\alpha &{}\text{ if } t\in [\frac{4}{6\alpha },\frac{4}{5\alpha }] \\ -5\alpha +6\alpha ^2t &{}\text{ if } t\in [\frac{5}{6\alpha },\frac{1}{\alpha }]. \\ \end{array}\right. } \end{aligned}$$

and $u(t)=\int _{0}^tu'(s)\,ds$. This formula can be copied several times on intervals $[k/\alpha ,(k+1)/\alpha ]$ for $k\in \mathbb {N}, k<\alpha$ by $u(t){:=}u(t-k/\alpha )$. If $\alpha$ is integer number, this will cover whole [0, 1] interval, otherwise we extend by constant zero, as we illustrate by Fig. 1 for case $\alpha \in (2,3)$. Note that for $\alpha$ integer, we may get constant 1/3 in (i), but for $\alpha$ non-integer the achieved constant has to be lowered to 1/6. $\square$

Fig. 1

Sketch of the construction of $u'$ and u in Lemma 5.2

We are finally ready to prove the main theorem of the paper.

Proof of Theorem 2.2

(i) Assume that $Z\cap L^{\infty }\subset Y\cap L^{\infty }$ and suppose $X\cap L^{\infty }\not \subset B\cap L^{\infty }$. We will construct a function $\eta \in W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)$ such that

$$\begin{aligned} \Vert \nabla ^2\eta \Vert _Y^{\frac{1}{2}}\Vert \eta \Vert _Z^{\frac{1}{2}}< \infty , \end{aligned}$$

but

$$\begin{aligned} \Vert \nabla \eta \Vert _B=\infty . \end{aligned}$$

Firstly we will consider the case of $d=1$. Since $X\cap L^{\infty }\not \subset B\cap L^{\infty }$ there is $\tilde{f}=(\tilde{f})^*\in (X\cap L^{\infty })\backslash (B\cap L^{\infty })$ with $\Vert \tilde{f}\Vert _X=1$ and observe that $\tilde{f}$ can be chosen in the form of

$$\begin{aligned} \tilde{f}=\sum _{k=1}^{\infty }\tilde{c_k}\chi _{[k-1,k)}. \end{aligned}$$

On the other hand, $X=Y^{1/2}Z^{1/2}$, thus from Lemma 5.1 it follows that there are $g\in Y$, $h\in Z$ such that $\tilde{f}\le g^{1/2}h^{1/2}$,

$$\begin{aligned} g=g^*= & {} \sum _{k=1}^{\infty }a_k\chi _{[k-1,k)},\ \ \nonumber \\ h=h^*= & {} \sum _{k=1}^{\infty }b_k\chi _{[k-1,k)} \end{aligned}$$

(20)

and $h\le g$. Define

$$\begin{aligned} f=g^{1/2}h^{1/2}. \end{aligned}$$

Then $f=f^*$ is of the form

$$\begin{aligned} f=\sum _{k=1}^{\infty }c_k\chi _{[k-1,k)} \end{aligned}$$

and also $f\in X\backslash B$, since $0\le \tilde{f}\le f$. In particular, $\Vert f\Vert _B=\infty$.

By definition of f we have $c_k=\sqrt{a_kb_k}$ for each k, thus for each k

$$\begin{aligned} b_k\le c_k\le a_k, \end{aligned}$$

since $h\le g$. Now for each $k=1,2,...$ we define $u_k$ as u from Lemma 5.2 applied with $\alpha =c_k/b_k$. Further we define

$$\begin{aligned} \eta (t)=\sum _{k=1}^{\infty }b_ku_k(t-k+1). \end{aligned}$$

In consequence,

$$\begin{aligned} \eta '(t)=\sum _{k=1}^{\infty }b_ku'_k(t-k+1) \end{aligned}$$

From Lemma 5.2 it follows

$$\begin{aligned} \left| \left\{ |b_k u'_k(t)|\ge c_k\right\} \right| \ge \frac{1}{6}, \end{aligned}$$

which implies that for each $n\in \mathbb {N}$

$$\begin{aligned} (\eta ')^{*}\ge \sum _{k=1}^{n} c_k\chi _{[\frac{k-1}{6},\frac{k}{6})}=D_{6}\left( \sum _{k=1}^{n} c_k\chi _{[k-1,k)}\right) . \end{aligned}$$

(21)

Since dilation operators are bounded on r.i.B.f. spaces we get

$$\begin{aligned} \begin{aligned} \left\| D_{6}\left( \sum _{k=1}^{n} c_k\chi _{[k-1,k)}\right) \right\| _B&\ge \Vert D_{1/6}\Vert _{B\rightarrow B}^{-1}\left\| D_{1/6}D_6\left( \sum _{k=1}^{n} c_k\chi _{[k-1,k)}\right) \right\| _B \\&=\Vert D_{1/6}\Vert _{B\rightarrow B}^{-1}\left\| \sum _{k=1}^{n} c_k\chi _{[k-1,k)}\right\| _B. \end{aligned} \end{aligned}$$

(22)

The limit for $n\rightarrow \infty$ of the term on right-hand side is $\infty$ since $\sum _{k=1}^{n} c_k\chi _{[k-1,k)}\rightarrow f$ pointwise, while B has the Fatou property and $f\not \in B$. Together with estimates (21) and (22) it implies that

$$\begin{aligned} \Vert \eta '\Vert _B=\infty . \end{aligned}$$

Furthermore,

$$\begin{aligned} \eta ''(t)=\sum _{k=1}^{\infty }b_ku''_k(t-k+1) \end{aligned}$$

and the way we have chosen functions $u_k$ ensures that for each $k=1,2,...$ and each $t\in [k-1,k)$

$$\begin{aligned} b_k|u''_k(t-k+1)|\le 6b_k\left( \frac{c_k}{b_k}\right) ^2\le 6 a_k, \end{aligned}$$

while

$$\begin{aligned} u''_k(t-k+1)=0 \mathrm{\ for\ each \ }t\notin [k-1,k). \end{aligned}$$

Using this and (20) we get

$$\begin{aligned} |\eta ''|\le \sum _{k=1}^{\infty }a_k\chi _{[k-1,k)}=g. \end{aligned}$$

In consequence $\Vert \eta ''\Vert _Y<\Vert g\Vert _Y<\infty$. By (20) we have $\Vert \eta \Vert _Z\le \Vert h\Vert _Z<\infty$, since

$$\begin{aligned} \eta \le \sum _{k=1}^{\infty }b_k\chi _{[k-1,k)}=h. \end{aligned}$$

This finishes the proof of point (i) in the case of $d=1$.

In the case $d>1$ we proceed in the same fashion. Firstly we select f, g, h together with $b_k,c_k,a_k$ and define the functions $u_k$ exactly as above. Further, define $w'$ to be a piecewise affine continuous function with slots in points $(0,0),(1/4,1),(3/4,-1),(1,0)$. Precisely the formula is

$$\begin{aligned} w'(t){:=} {\left\{ \begin{array}{ll} 4t &{}\text{ if } t\in [0,\frac{1}{4}]\\ 2-4t &{}\text{ if } t\in [\frac{1}{4},\frac{3}{4}] \\ 4t-4&{}\text{ if } t\in [\frac{3}{4},1] \\ 0 &{}\text{ else. } \end{array}\right. } \end{aligned}$$

We set for $t>0$

$$\begin{aligned} w(t)=\int _0^tw'(s)\,ds. \end{aligned}$$

Then $w\in W^{2,\infty }(\mathbb {R})$ has the following properties:

• (a) ${{\,\textrm{supp}\,}}w= [0,1]$

• (b) $|w(t)|\le 1/4$, $|w'(t)|\le 1$ and $|w''(t)|\le 4$ for each $t\in [0,1]$

• (c) $|w(t)|\ge 1/8$ for each $t\in [1/4,3/4]$.

Finally, we are ready to define the desired function $\eta$ on $\mathbb {R}^d$. We put

$$\begin{aligned} \eta =\sum _{k=1}^{\infty }b_kv_k, \end{aligned}$$

where for each $k=1,2,3,...$

$$\begin{aligned} v_k(x_1,...,x_d)=u_k(x_1-k+1)w(x_2)w(x_3)...w(x_{d}). \end{aligned}$$

It remains to estimate the function $\eta$ and its derivatives analogously as in the previous part of the proof. By Lemma 5.2 and points (a), (b) above we have

$$\begin{aligned} \eta ^*\le h. \end{aligned}$$

Moreover, notice that $|u_k'|\le \frac{c_k}{b_k}$, therefore for each fixed $1<i\le d$

$$\begin{aligned} |b_ku_k'(x_1-k+1)w'(x_i)\Pi _{j=2,j\not =i}^{d}w(x_j)|\le b_k\frac{c_k}{b_k}\le a_k, \end{aligned}$$

while respective estimates for other second order derivatives appearing in $\nabla ^2\eta$ are immediate. In consequence,

$$\begin{aligned} (\nabla ^2\eta )^*\le 64^dd^2\sum _{k=1}^{\infty }a_k\chi _{[k-1,k)}=64^dd^2g. \end{aligned}$$

Finally, point (c) above implies that

$$\begin{aligned} |\nabla \eta |\ge \left| \frac{\partial \eta }{\partial x_1}\right| \ge \sum _{k=1}^{n}\frac{1}{8^d}c_k\chi _{C_k}, \end{aligned}$$

where $C_k=\left\{ |b_k u'_k|\ge c_k\right\} \times [1/4,3/4]^{d-1}$. Thus $|C_k|\ge \frac{1}{2^{d+2}}$ by definition of $u_k$’s. In consequence, for each $n\in \mathbb {N}$

$$\begin{aligned} |\nabla \eta |^{*}\ge \frac{1}{8^d}\sum _{k=1}^{n} c_k\chi _{[\frac{k-1}{2^{d+2}},\frac{k}{2^{d+2}})}=\frac{1}{8^d}D_{2^{d+2}}\left( \sum _{k=1}^{n} c_k\chi _{[k-1,k)}\right) \end{aligned}$$

and we can finish the proof as before.

(ii) Consider the case $Z\cap L^{1} \subset Y\cap L^{1}$ and suppose $X\cap L^{1}\not \subset B\cap L^{1}$. This time we will construct a sequence $(\eta _n)\subset W^{2,1}_{{{\,\textrm{loc}\,}}}(\mathbb {R}^d)$ such that

$$\begin{aligned} \frac{\Vert \nabla \eta _n\Vert _{B}}{\Vert \nabla ^2 \eta _n\Vert _{Y}^{\frac{1}{2}} \Vert \eta _n\Vert _{Z}^{\frac{1}{2}}}\rightarrow \infty . \end{aligned}$$

Once again we start with the case of $d=1$. It follows that there is $r = r^* \in (X\cap L^{1}) \backslash (B\cap L^{1})$. However, $r \chi _{[1,\infty )}\in L^{\infty }\cap L^1\subset B\cap L^{1}$ and consequently $r \chi _{[0,1]}\in (X\cap L^{1}) \backslash (B\cap L^{1})$. We define

$$\begin{aligned} {\tilde{f}}=r\chi _{[0,1]}. \end{aligned}$$

Then $\tilde{f}\in X=Y^{1/2}Z^{1/2}$, so by Lemma 5.1 there are $g^*=g\in Y$, $h^*=h\in Z$ such that $\tilde{f}\le g^{1/2}h^{1/2}$ and $h\le g$. Define for each n

$$\begin{aligned} g_n=g_n^*=\sum _{k=1}^{n}g\left( \frac{k}{n}\right) \chi _{[\frac{k-1}{n},\frac{k}{n}]}, \end{aligned}$$

$$\begin{aligned} h_n=h_n^*=\sum _{k=1}^{n}h\left( \frac{k}{n}\right) \chi _{[\frac{k-1}{n},\frac{k}{n}]}. \end{aligned}$$

Set

$$\begin{aligned} a_{n,k}{:=}g\left( \frac{k}{n}\right) , b_{n,k}{:=}h\left( \frac{k}{n}\right) and c_{n,k}{:=}\sqrt{a_{n,k}b_{n,k}}. \end{aligned}$$

Note that since $0\le h_n\le h$ and $0\le g_n\le g$ by the lattice property we obtain that

$$\begin{aligned} \Vert h_n\Vert _{Z}\le \Vert h\Vert _{Z} \mathrm{\ and\ } \Vert g_n\Vert _{Y}\le \Vert g\Vert _{Y}. \end{aligned}$$

Further, we put $f_n=g_n^{1/2}h_n^{1/2}$. Then $f_n$ is of the form

$$\begin{aligned} f_n=\sum _{k=1}^{n}c_{n,k}\chi _{[\frac{k-1}{n},\frac{k}{n}]}. \end{aligned}$$

The pointwise convergence $0\le f_n\rightarrow f= g^{1/2}h^{1/2}$ combined with the Fatou property of B yields

$$\begin{aligned} \left\| f_n\right\| _B\rightarrow \Vert f\Vert _B\ge \Vert \tilde{f}\Vert _B=\infty . \end{aligned}$$

Moreover, $c_{n,k}=\sqrt{a_{n,k}b_{n,k}}$ and

$$\begin{aligned} b_{n,k}\le c_{n,k}\le a_{n,k} \mathrm{\ for\ each \ }k=1,2,...,n. \end{aligned}$$

Now for each n and $k=1,2,...,n$ we select $u_{n,k}$ from Lemma 5.2 applied with $\alpha =c_{n,k}/b_{n,k}$. Finally we define

$$\begin{aligned} \eta _n(t)=\sum _{k=1}^{n}b_{n,k}u_{n,k}(nt-k+1). \end{aligned}$$

Then

$$\begin{aligned} \eta '_n(t)=n\sum _{k=1}^{n}b_{n,k}(u_{n,k})'(nt-k+1) \end{aligned}$$

and we have

$$\begin{aligned} \left| \left\{ t:b_{n,k} |u_{n,k}'(nt-k+1)|\ge c_{n,k}\right\} \right| \ge \frac{1}{6n} \end{aligned}$$

because by Lemma 5.2 (i)

$$\begin{aligned} \left| \left\{ t:b_{n,k} |u_{n,k}'(t)|\ge c_{n,k}\right\} \right| \ge \frac{1}{6}. \end{aligned}$$

Hence

$$\begin{aligned} (\eta '_n)^{*}(t)\ge n \sum _{k=1}^{n} c_{n,k}\chi _{[\frac{k-1}{6n},\frac{k}{6n})}=nD_{6}\left( \sum _{k=1}^{n} c_{n,k}\chi _{[\frac{k-1}{n},\frac{k}{n}]}\right) . \end{aligned}$$

We have

$$\begin{aligned} n\left\| \sum _{k=1}^{n} c_{n,k}\chi _{[\frac{k-1}{n},\frac{k}{n})}\right\| _B=n \left\| D_{1/6}D_6\left( \sum _{k=1}^{n} c_{n,k}\chi _{[\frac{k-1}{n},\frac{k}{n})}\right) \right\| _B\le \Vert D_{1/6}\Vert _{B\rightarrow B}\Vert \eta '_n\Vert _B, \end{aligned}$$

since dilations are bounded on B.

On the other hand, we have

$$\begin{aligned} |\eta _n(t)|\le \sum _{k=1}^{n}b_{n,k}\chi _{[\frac{k-1}{n},\frac{k}{n})}\le h_n \end{aligned}$$

and so $\Vert \eta _n\Vert _Z\le \Vert h_n\Vert _Z\le \Vert h\Vert _Z<\infty$. Moreover,

$$\begin{aligned} \eta ''_n(t)=n^2\sum _{k=1}^{n}b_{n,k}u_{n,k}''(nt-k+1) \end{aligned}$$

and, since

$$\begin{aligned} b_{n,k}|u''_{n,k}(nt-k+1)|\le b_{n,k}\left( \frac{c_{n,k}}{b_{n,k}}\right) ^2=a_{n,k}, \end{aligned}$$

we get

$$\begin{aligned} |\eta ''_n(t)|\le n^2\sum _{k=1}^{n}a_{n,k}\chi _{[\frac{k-1}{n},\frac{k}{n})}=n^2g_n \end{aligned}$$

and consequently $\Vert \eta ''_n\Vert _Y\le n^2\Vert g_n\Vert _Y\le n^2\Vert g\Vert _Y$. Finally, we have

$$\begin{aligned} \frac{\Vert \eta '_n\Vert _{B}}{\Vert \eta ''_n\Vert _{Y}^{\frac{1}{2}} \Vert \eta _n\Vert _{Z}^{\frac{1}{2}}}\ge \frac{n\Vert f_n\Vert _{B}}{(n^2\Vert g_n\Vert _{Y})^{\frac{1}{2}} \Vert h_n\Vert _{Z}^{\frac{1}{2}}}\rightarrow \infty \end{aligned}$$

and the proof of point (ii) is finished in the case of $d=1$. When $d>1$ the proof is a mixture of the above argument together with the idea used for the case $d>1$ in the proof of point (i). More precisely, we define

$$\begin{aligned} \eta _n=\sum _{k=1}^{n}b_{n,k}v_{n,k}, \end{aligned}$$

where

$$\begin{aligned} v_{n,k}(x_1,x_2,...,x_d)=u_{n,k}(nx_1-k+1)w(x_2)...w(x_d) \end{aligned}$$

and the rest of the proof repeats the same steps as before. $\square$

Assuming the spaces on the right-hand side of the Gagliardo–Nirenberg inequality satisfy assumption $Z\subset Y$, then Theorem 2.2 takes the following simplified form.

Corollary 1.13

Let Y, Z be r.i.B.f. spaces satisfying $Z\subset Y$. If the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla u\Vert _{B}\lesssim \Vert \nabla ^2 u\Vert _Y^{\frac{1}{2}}\Vert u\Vert _Z^{\frac{1}{2}} \end{aligned}$$

holds for all $u\in W^{2,0}(\mathbb {R}^d)$, then $Y^{1/2}Z^{1/2}\subset B$.

Proof

We need only to see that if $Z\subset Y$ and $Y^{1/2}Z^{1/2}\not \subset B$ then either assumptions of point (i) or point (ii) of Theorem 2.2 are satisfied. First of all, notice that $Z\subset Y$ implies both

$$\begin{aligned} Z\cap L^{\infty }\subset Y\cap L^{\infty }{\ \text {and}\ } Z\cap L^{1}\subset Y\cap L^{1}. \end{aligned}$$

Now let $0\le f\in Y^{1/2}Z^{1/2}\backslash B$ and define

$$\begin{aligned} f_a=f\chi _{\{f\ge 1\}}{\ \text {and}\ } f_b=f\chi _{\{f< 1\}}. \end{aligned}$$

Then either $f_b$ belongs to $(Y^{1/2}Z^{1/2})\backslash B$ or $f_a$ belongs to $(Y^{1/2}Z^{1/2})\backslash B$ (or both are valid). In case $f_b\in (Y^{1/2}Z^{1/2})\backslash B$ we apply Theorem 2.2 (i), since $f_b\in L^{\infty }$. Otherwise, we apply Theorem 2.2 (ii), since $f_a\in L^{1}$. $\square$

Let us apply previous thoughts on Orlicz spaces. As mentioned in the introduction, Kałamajska and Pietruska-Pałuba [16] studied the Gagliardo–Nirenberg inequality

$$\begin{aligned} \Vert \nabla u\Vert _{\varphi }\lesssim \Vert \nabla ^2 u\Vert _{\varphi _1}^{\frac{1}{2}}\Vert u\Vert _{\varphi _2}^{\frac{1}{2}}. \end{aligned}$$

They proved the choice of $L^{\varphi }$ for $\varphi$ satisfying

$$\begin{aligned} \varphi ^{-1}\approx \sqrt{\varphi _1^{-1}\varphi _2^{-1}} \end{aligned}$$

is valid when spaces satisfy some additional technical assumptions. Later the result was relaxed of some of these assumptions and the optimality of the choice $X=L^{\varphi }$ among all Orlicz spaces was given in [9]. Theorem 2.2 explains that it is already the optimal choice of space among all r.i.B.f. spaces, provided $L^{\varphi _2}\subset L^{\varphi _1}$.

Concluding, we see that the choice of $X=Y^{1/2}Z^{1/2}$ is optimal among all r.i.B.f. spaces in the Gagliardo–Nirenberg inequality (3), provided that $Z\subset Y$. However, the assumption $Z\subset Y$ is quite restrictive and does not apply to the most classical r.i.B.f. spaces on $\mathbb {R}_+$ (Lebesgue spaces, Lorentz spaces, etc.), since usually there is no inclusion between such spaces on $\mathbb {R}_+$. It appears, however, that maneuvering between points (i) and (ii) of Theorem 2.2 we can use it to give an almost complete answer to the question about the optimality of (3) among Lorentz spaces posted in [9] formulated as Corollary 2.5.

Proof of Corollary 2.5

The choice of $P=(1/(2Q)+1/(2R))^{-1}$ is the only possible. This is a consequence of the scaling argument (see [9, Theorem 1.1] for details) and of the shape of fundamental functions of Lorentz spaces. Thus we need only to explain that

$$\begin{aligned} \Vert \nabla u\Vert _{P,p'}\lesssim \Vert \nabla ^2 u\Vert _{Q,q}^{\frac{1}{2}}\Vert u\Vert _{R,r}^{\frac{1}{2}} \end{aligned}$$

does not hold when $p'<p{:=}(1/(2q)+1/(2r))^{-1}$ for $P=(1/(2Q)+1/(2R))^{-1}$. We will consider three cases. Firstly, if $R<Q$ then

$$\begin{aligned} L^{R,r}\cap L^{\infty }\subset L^{Q,q}\cap L^{\infty }. \end{aligned}$$

Thus we can apply point (i) of Theorem 2.2 with $B=L^{P,p'}$, since $L^{P,p'}\subsetneq L^{P,p}$.

In the second case, when $R>Q$, we have

$$\begin{aligned} L^{R,r}\cap L^{1}\subset L^{Q,q}\cap L^{1} \end{aligned}$$

and we apply point (ii) of Theorem 2.2.

Finally, when $R=Q$ and $r<q$, then assumptions of Corollary 5.3 are satisfied. $\square$