Optimal Cwikel–Solomyak Estimates

Abstract

We obtain the optimal version of Cwikel-type estimates for the uniform operator norm which implies the optimality of M.Z. Solomyak’s results (Proc. Lond. Math. Soc. 71(1):53–75, 1995) within the classes of Orlicz/Lorentz spaces. Our methods are based on finding the distributional version of the Sobolev inequality.

1 Introduction

This text is inspired by result of Solomyak [23, 24] establishing (for $d\in 2\mathbb {N}$) that the (symmetrized) Cwikel–Solomyak operator

$$\begin{aligned} (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}} \end{aligned}$$

belongs to the weak trace class ideal $\mathcal {L}_{1,\infty }$ when $f\in L\log L(\mathbb {T}^d).$ Here, $\mathbb {T}^d$ is d-dimensional torus, $M_f$ is an (unbounded) multiplication operator by f on $L_2(\mathbb {T}^d)$ and $\Delta _{\mathbb {T}^d}$ is the torical Laplacian.

The space $L\log L$ is simultaneously characterised as the Orlicz space $L_M$ with $M(t)=t\log (e+t)$ and the Lorentz space $\Lambda _{\phi }$ with $\phi (t)=t\log \big (\frac{e}{t}\big )$ (see e.g. Theorem 5.1 in [5]). In this paper we show that the result of Solomyak is optimal in the classes of Orlicz and Lorentz spaces. The principal result of this paper(which implies this optimality result) asserts that the Marcinkiewicz space $\texttt{M}_{\psi }$ with $\psi (t)=\frac{1}{\log \big (\frac{e}{t}\big )}$ is the largest possible symmetric function space such that the symmetrized Cwikel–Solomyak operator is bounded. We combine our principal result with that of Solomyak to obtain new estimates in Schatten–Lorentz ideals in Corollary 25.

In the course of the proof of our principal result, we have made two notable contributions to the theory of Sobolev Inequality and Sobolev Embedding Theorem.

Firstly, we establish a distributional version of Sobolev inequality written in terms of Hardy–Littlewood submajorization (Theorem 3). We demonstrate that our version of this inequality is optimal in the sense that it cannot be improved in terms of the distribution function (Proposition 6).

Secondly, we show that our version of Sobolev inequality yields the optimal Sobolev embedding theorem (Proposition 7 and Corollary 14). These results enhance earlier results due to Hansson [12], Brezis and Wainger [7], and Cwikel and Pustylnik [9].

In the following section, we will establish any necessary notation and preliminaries. Technical details and the relation to prior work on Cwikel–Solomyak-type estimates are discussed in Sect. 3. Then in Sect. 4 we prove a distributional form of the Sobolev-type inequality, followed by an optimal Sobolev embedding theorem in Sect. 5. Finally, in Sect. 6 we use these results to prove our optimal Cwikel–Solomyak type estimates.

2 Preliminaries

2.1 Symmetric Function Spaces

For detailed information about decreasing rearrangements and symmetric spaces briefly discussed below, we refer the reader to the standard textbook [14] (see also [15, 16]). Let $(\Omega ,\nu )$ be a measure space. Let $S(\Omega ,\nu )$ be the collection of all $\nu$-measurable functions on $\Omega$ such that, for some $n\in \mathbb {N},$ the function $|f|\chi _{\{|f|>n\}}$ is supported on a set of finite measure. For every $f\in S(\Omega ,\nu )$ one can define its decreasing rearrangement, $\mu (f)$, see e.g. [16, Sect. 2.3] (and also [14, Sect. II.2], [15, pp. 116–117]). This is a positive decreasing function on $\mathbb {R}_+$ equimeasurable with |f|. Note that the notation is somewhat unconventional. In the literature it is common to denote the decreasing rearrangement of |f| by $f^{*}.$ Similarly, we denote $f^{*}(s)$ by $\mu (s,f).$

Let $E(\Omega ,\nu )\subset S(\Omega ,\nu )$ and let $\Vert \cdot \Vert _E$ be Banach norm on $E(\Omega ,\nu )$ such that

• (1) if $f\in E(\Omega ,\nu )$ and $g\in S(\Omega ,\nu )$ be such $|g|\le |f|,$ then $g\in E(\Omega ,\nu )$ and $\Vert g\Vert _E\le \Vert f\Vert _E;$

• (2) if $f\in E(\Omega ,\nu )$ and $g\in S(\Omega ,\nu )$ be such $\mu (g)=\mu (f),$ then $g\in E(\Omega ,\nu )$ and $\Vert g\Vert _E=\Vert f\Vert _E.$

We say that $(E(\Omega ,\nu ),\Vert \cdot \Vert _E)$ (or simply E) is a symmetric Banach function space (symmetric space, for brevity). The most well-known examples of symmetric spaces are the Lebesgue $L_p$-spaces, ($L_p(\Omega ,\nu ), \Vert \cdot \Vert _p)$, for $1\le p\le \infty .$ That such spaces are symmetric is easily verified.

If $\Omega =\mathbb {R}_+,$ then the function

$$\begin{aligned} t\rightarrow \Vert \chi _{(0,t)}\Vert _E,\quad t>0, \end{aligned}$$

is called the fundamental function of E. This may also be extended to the case where $\Omega$ is an interval or an arbitrary $\sigma$-finite measure space (see [14, Chap. II, Sect. 8]). The concrete examples of measure spaces $(\Omega , \nu )$ considered in this paper are d-dimensional tori $\mathbb {T}^d$ (equipped with the respective Haar measures), $\mathbb {R}_+$ and $\mathbb {R}^d$ (equipped with Lebesgue measures), their measurable subsets and compact d-dimensional Riemannian manifolds (X, g).

In this paper, we focus on the following concrete examples of symmetric spaces: $L_p$-spaces, Orlicz spaces, Lorentz spaces and Marcinkiewicz spaces.

Given an Orlicz function M on $\mathbb {R}_+$ (that is continuous convex increasing function M satisfying $M(0)=0$ and $M(t)\rightarrow \infty$ as $t\rightarrow \infty$) the Orlicz space $L_M(\Omega ,\nu )$ is defined by setting

$$\begin{aligned} L_M(\Omega ,\nu )=\Big \{f\in S(\Omega ,\nu ):\ M \left( \frac{|f|}{\lambda } \right) \in L_1(\Omega ,\nu ) \text{ for } \text{ some } \lambda >0\Big \}. \end{aligned}$$

We equip it with a Banach norm

$$\begin{aligned} \Vert f\Vert _{L_M}=\inf \Big \{\lambda >0:\ \Big \Vert M\Big (\frac{|f|}{\lambda }\Big )\Big \Vert _1\le 1\Big \}. \end{aligned}$$

We refer the reader to [13,14,15] for further information about Orlicz spaces.

Given a concave increasing function $\psi ,$ the Lorentz space $\Lambda _{\psi }$ is defined by setting

$$\begin{aligned} \Lambda _{\psi }(\Omega ,\nu )=\Big \{f\in S(\Omega ,\nu ):\ \int _0^{\infty }\mu (s,f)d\psi (s)<\infty \Big \}. \end{aligned}$$

The Marcinkiewicz space $\texttt{M}_{\psi }$ is then defined by setting

$$\begin{aligned} \texttt{M}_{\psi }(\Omega ,\nu )=\Big \{f\in S(\Omega ,\nu ):\ \sup _{t>0}\frac{1}{\psi (t)}\int _0^t\mu (s,f)ds<\infty \Big \}. \end{aligned}$$

Setting $\psi (t)=t^{\frac{1}{2}}$, the spaces $L_{2,1}$ and $L_{2,\infty }$ are then the Lorentz and Marcinkiewicz spaces corresponding to $\psi (t)$ (see e.g. [15]).

All the examples of symmetric spaces listed above are closed with respect to the Hardy–Littlewood submajorization; a pre-order (denoted $\prec \prec$) defined on equivalence classes of functions in $L_1+L_{\infty }$ by $y\prec \prec x$ if and only if

$$\begin{aligned} \int _0^t\mu (s,y)ds\le \int _0^t\mu (s,x)ds,\quad t>0. \end{aligned}$$

Let us fix, throughout the text, the two concave functions

$$\begin{aligned} \psi (t)=\frac{1}{\log \big (\frac{e}{t}\big )},\quad \textrm{and}\quad \phi (t)=t\log \Big (\frac{e}{t}\Big ), \quad t\in (0,1) \end{aligned}$$

and consider Orlicz functions

$$\begin{aligned} M(t)=t\log (e+t),\quad \textrm{and}\quad G(t)=e^t-1,\quad t>0. \end{aligned}$$

Throughout this text, the interplay between the Lorentz spaces $\Lambda _{\psi }(0,1)$ and $\Lambda _{\phi }(0,1)$, the Orlicz spaces $L_M(0,1)$ and $L_G(0,1)$, and the Marcinkiewicz spaces $\texttt{M}_{\psi }(0,1)$ and $\texttt{M}_{\phi }(0,1)$ plays a fundamental role. The space $\texttt{M}_{\psi }$ (respectively, the space $\Lambda _{\phi }$) is the largest (respectively, the smallest) symmetric Banach function space with the fundamental function $\phi .$

First of all, we observe that the spaces $\texttt{M}_{\phi }(0,1)$ and $L_G(0,1)$ coincide (the corresponding norms are equivalent). Indeed, the fundamental functions of those spaces coincide, hence $L_G(0,1)\subset \texttt{M}_{\phi }(0,1).$ On the other hand, $\phi '\in L_G(0,1)$ and, therefore, $\texttt{M}_{\phi }(0,1)\subset L_G(0,1).$ This fact was first observed in [4].

By invoking Köthe duality (see [14] and [15]), we infer that the spaces $\Lambda _\phi (0,1)$ and $L_M(0,1)$ also coincide and (the corresponding norms are equivalent).

In the statement of our main result, we use the 2-convexification of the symmetric space. Given a symmetric space $E(\Omega ,\nu ),$ we set

$$\begin{aligned} E^{(2)}(\Omega ,\nu )= & {} \{f\in S(\Omega ,\nu ):\ |f|^2\in E(\Omega ,\nu )\}, \\ \Vert f\Vert _{E^{(2)}}= & {} \big \Vert |f|^2\big \Vert _E^{\frac{1}{2}},\quad f\in E^{(2)}(\Omega ,\nu ). \end{aligned}$$

More details on the 2-convexification of a function space may be found in the book [15].

We also need a definition of dilation operator $\sigma _u,$ $u>0,$ which acts on $S(\mathbb {R})$ by the formula

$$\begin{aligned} (\sigma _uf)(t)=f\left( \frac{t}{u}\right) ,\quad f\in S(\mathbb {R}). \end{aligned}$$

We frequently use the mapping $r_d:\mathbb {R}^d\rightarrow \mathbb {R}_+$ defined by the formula

$$\begin{aligned} r_d(t)=|t|^d,\quad t\in \mathbb {R}^d. \end{aligned}$$

3 Connection with Earlier Results

3.1 Connection with Cwikel–Solomyak Type Estimates

Let $d\in \mathbb {N}$ and let f be a measurable function on d-dimensional torus $\mathbb {T}^d$ and let $M_f$ be an (unbounded) multiplication operator by f on $L_2(\mathbb {T}^d).$ Let $\Delta _{\mathbb {T}^d}$ be the torical Laplacian. In Proposition 19 we show that (symmetrized) Cwikel–Solomyak operator

$$\begin{aligned} (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}} \end{aligned}$$

is bounded for $f\in \texttt{M}_{\psi }(\mathbb {T}^d).$ We denote the $*$-algebra of all bounded linear operators on the Hilbert space $L_2(\mathbb {T}^d)$ by $\mathcal {L}_{\infty }.$

Estimates of the symmetrized Cwikel–Solomyak operator in the weak-trace ideal $\mathcal {L}_{1,\infty }$ as well as in the ideal $\mathcal {M}_{1,\infty }$ are important in Non-commutative Geometry and in Mathematical Physics. For the background concerning weak trace ideals $\mathcal {L}_{p,\infty }$, $1\le p<\infty$ and Marcinkiewicz ideal $\mathcal {M}_{1,\infty }$ we refer to [16] (see also Sect. 6.3 below).

Estimates of the symmetrized Cwikel–Solomyak operator in $\mathcal {L}_{1,\infty }$ (on the torus) for even d appeared in the foundational papers [23, 24]. The estimate there was given in terms of the Orlicz norm $\Vert \cdot \Vert _{L_M}$ (which is equivalent to $\Vert \cdot \Vert _{\Lambda _{\phi }}$). Recently, symmetrized Cwikel–Solomyak estimates in the ideal $\mathcal {M}_{1,\infty }$ were established (on the Euclidean space) in [17]. The estimate was given in terms of the Lorentz norm

$$\begin{aligned} f\rightarrow \Vert \mu (f)\chi _{(0,1)}\Vert _{\Lambda _{\phi }(0,1)}+\Vert f\Vert _{L_1(\mathbb {R}^d)}. \end{aligned}$$

The surprising fact that $\Lambda _\phi (0,1)=L_M(0,1)$ demonstrates the convergence of those totally unrelated approaches.

The results in this paper complement the results cited in the preceding paragraph. Indeed, Cwikel–Solomyak operator belongs to $\mathcal {L}_{\infty }$ for $f\in \texttt{M}_{\psi }$ and to $\mathcal {L}_{1,\infty }$ for $f\in \Lambda _{\phi }.$ This opens an avenue for a given function f to determine (using interpolation methods) the least ideal to which the corresponding Cwikel–Solomyak operator belongs.

3.2 Connection with Sobolev-Type Estimates

The following Sobolev inequality is customarily credited to [12] and [7]. Actually, none of those papers contain a complete proof or even a clear-cut statement. The proof is available in [9]. For a notion of Sobolev space on $\Omega \subset \mathbb {R}^d$ we refer the reader to Chap. 7 in [2].

Theorem 1

Let d be even and let $\Omega$ be a bounded domain in $\mathbb {R}^d$ (conditions apply). There exists a constant $c_{\Omega }$ such that

$$\begin{aligned} \Vert u\Vert _{\Lambda _{\psi }^{(2)}}\le c_{\Omega }\Vert u\Vert _{W^{\frac{d}{2},2}},\quad u\in W^{\frac{d}{2},2}(\Omega ). \end{aligned}$$

This result is optimal due to the following theorem (proved in [9]).

Theorem 2

Let d be even and let $\Omega$ be a bounded domain in $\mathbb {R}^d$ (conditions apply). If X be a Banach symmetric function space on $\Omega$ such that

$$\begin{aligned} \Vert u\Vert _X\le c_{\Omega }\Vert u\Vert _{W^{\frac{d}{2},2}},\quad u\in W^{\frac{d}{2},2}(\Omega ) \end{aligned}$$

for some constant $c_{\Omega },$ then $\Lambda _{\psi }^{(2)}\subset X.$

In the next sections, we prove the assertions which substantially strengthen theorems above and extend them to arbitrary dimensions $d\ge 1.$ In the course of the proof, we also recast Sobolev inequality using distribution functions. Those results are very much inspired by [9], however, our technique is a substantial improvement of that in [9]. We refer the reader to [18] and [19] for further development of the inequalities in [9].

4 Distributional Sobolev-Type Inequality

In this section, we introduce Hardy-type operator T (see e.g. [14]) and employ it as a technical tool for our distributional version of Sobolev inequality. In the second subsection, we show that our result is optimal.

4.1 Distributional Sobolev-Type Inequality

Let $T:L_2(0,1)\rightarrow L_2(0,1)$ be the operator defined by the formula

$$\begin{aligned} (Tx)(t)=t^{-\frac{1}{2}}\int _0^tx(s)ds+\int _t^1\frac{x(s)ds}{s^{\frac{1}{2}}},\quad x\in L_2(0,1). \end{aligned}$$

The boundedness of $T:L_2(0,1)\rightarrow L_2(0,1)$ is guaranteed by the fact that T is actually a Hilbert–Schmidt operator. Indeed, its integral kernel is given by the formula

$$\begin{aligned} (t,s)\rightarrow t^{-\frac{1}{2}}\chi _{\{s<t\}}+s^{-\frac{1}{2}}\chi _{\{t<s\}},\quad 0<s,t<1, \end{aligned}$$

which is, obviously, square-integrable.

The gist of (the critical case of the) Sobolev inequality on $\mathbb {R}^d$ may be informally understood as the boundedness of the operator $(1-\Delta )^{-\frac{d}{4}}$ from $L_2(\mathbb {R}^d)$ into a symmetric Banach function space on $\mathbb {R}^d$ which is sufficiently close to $L_{\infty }(\mathbb {R}^d).$ We suggest to view this result as a statement concerning distribution functions of elements $(1-\Delta )^{-\frac{d}{4}}x$ and $T(\mu (x)\chi _{(0,1)})$ where $x\in L_2(\mathbb {R}^d).$

Our distributional version of Sobolev inequality is as follows:

Theorem 3

Let $d\in \mathbb {N}.$ For every $x\in L_2(\mathbb {R}^d),$ we have

$$\begin{aligned} \mu \big ((1-\Delta )^{-\frac{d}{4}}x\big )\chi _{(0,1)}\prec \prec c_dT(\mu (x)\chi _{(0,1)}). \end{aligned}$$

Here, $\prec \prec$ denotes the Hardy–Littlewood submajorization.

We need the following lemma stated in terms of convolutions.

Lemma 4

Let $x\in L_2(\mathbb {R}^d)$ and $g\in (L_{2,\infty }\cap L_1)(\mathbb {R}^d).$ We have

$$\begin{aligned} \mu (x*g)\chi _{(0,1)}\prec \prec 4\Vert g\Vert _{L_{2,\infty }\cap L_1}T(\mu (x)\chi _{(0,1)}). \end{aligned}$$

Proof

Lemma 1.5 in [20] states that

$$\begin{aligned} \int _0^t\mu (s,x*g)ds\le \int _0^t\mu (s,x)ds\cdot \int _0^t\mu (s,g)ds+t\int _t^{\infty }\mu (s,x)\mu (s,g)ds. \end{aligned}$$

Obviously,

$$\begin{aligned} \int _0^t\mu (s,x)ds\le \Vert g\Vert _{2,\infty }\cdot \int _0^ts^{-\frac{1}{2}}ds=2t^{\frac{1}{2}}\Vert g\Vert _{2,\infty }\le 2\Vert g\Vert _{L_{2,\infty }\cap L_1}\cdot t^{\frac{1}{2}}. \end{aligned}$$

For $t\in (0,1),$ we have

$$\begin{aligned} t\int _t^{\infty }\mu (s,x)\mu (s,g)ds&=t\int _t^1\mu (s,x)\mu (s,g)ds+t\int _1^{\infty }\mu (s,x)\mu (s,g)ds\\&\le t\Vert g\Vert _{2,\infty }\int _t^1\mu (s,x)\frac{ds}{s^{\frac{1}{2}}}+t\mu (1,x)\Vert g\Vert _1\\&\le \Vert g\Vert _{L_{2,\infty }\cap L_1}\cdot \Big (t\int _t^1\mu (s,x)\frac{ds}{s^{\frac{1}{2}}}+t\mu (1,x)\Big ). \end{aligned}$$

We, therefore, have

$$\begin{aligned} \int _0^t\mu (s,x*g)ds\le 2\Vert g\Vert _{L_{2,\infty }\cap L_1}\cdot F(t),\quad t\in (0,1), \end{aligned}$$

(1)

where

$$\begin{aligned} F(t)=t^{\frac{1}{2}}\int _0^t\mu (s,x)ds+t\int _t^1\mu (s,x)\frac{ds}{s^{\frac{1}{2}}}+t\mu (1,x),\quad t>0. \end{aligned}$$

Computing the derivative, we obtain

$$\begin{aligned} F'(t)&=\frac{1}{2}t^{-\frac{1}{2}}\int _0^t\mu (s,x)ds+\int _t^1\mu (s,x)\frac{ds}{s^{\frac{1}{2}}}+\mu (1,x)\\&\le (T\mu (x)\chi _{(0,1)})(t)+\mu (1,x). \end{aligned}$$

Next,

$$\begin{aligned} (T\mu (x)\chi _{(0,1)})(t)&\ge t^{-\frac{1}{2}}\int _0^t\mu (1,x)ds+\int _t^1\frac{\mu (1,x)ds}{s^{\frac{1}{2}}}\\&=\mu (1,x)\cdot (2-t^{\frac{1}{2}})\ge \mu (1,x),\qquad t\in (0,1). \end{aligned}$$

Thus,

$$\begin{aligned} F'(t)\le 2(T\mu (x)\chi _{(0,1)})(t),\quad t\in (0,1). \end{aligned}$$

It is now immediate that

$$\begin{aligned} F(t)\le 2\int _0^t\Big ((T\mu (x)\chi _{(0,1)})(s)\Big )ds,\quad t\in (0,1). \end{aligned}$$

Combining this equation with (1), we obtain

$$\begin{aligned} \int _0^t\mu (s,x*g)ds\le 4\Vert g\Vert _{L_{2,\infty }\cap L_1}\int _0^t\Big ((T\mu (x)\chi _{(0,1)})(s)\Big )ds,\quad t\in (0,1). \end{aligned}$$

This is exactly the required assertion. $\square$

Proof of Theorem 3

We rewrite $(1-\Delta )^{-\frac{d}{4}}$ as a convolution operator. Namely,

$$\begin{aligned} (1-\Delta )^{-\frac{d}{4}}x=x*g,\quad x\in L_2(\mathbb {R}^d), \end{aligned}$$

where g is the Fourier transform of the function

$$\begin{aligned} t\rightarrow (1+|t|^2)^{-\frac{d}{4}},\quad t\in \mathbb {R}^d. \end{aligned}$$

Precise expression for the function g involves Macdonald function $K_{\frac{d}{4}}$ and is given in [3] (see formulae (2.7) and (2.10) there) as follows

$$\begin{aligned} g(t)=c_d|t|^{-\frac{d}{4}}K_{\frac{d}{4}}(|t|),\quad t\in \mathbb {R}^d. \end{aligned}$$

Let $\mathbb {B}^d$ be the unit ball in $\mathbb {R}^d.$ If $\frac{d}{4}$ is not integer, then it follows from formulae (9.6.2) and (9.6.10) in [1] that $g\chi _{\mathbb {B}^d}\in L_{2,\infty }(\mathbb {B}^d)\subset L_1(\mathbb {B}^d).$ If $\frac{d}{4}$ is integer, then formula (9.6.11) in [1] yields that $g\chi _{\mathbb {B}^d}\in L_{2,\infty }(\mathbb {B}^d)\subset L_1(\mathbb {B}^d).$ Also, $g\chi _{\mathbb {R}^d\backslash \mathbb {B}^d}\in (L_1\cap L_{\infty })(\mathbb {R}^d\backslash \mathbb {B}^d)$ by formula (9.7.2) in [1]. Thus, $g\in (L_{2,\infty }\cap L_1)(\mathbb {R}^d).$ The assertion follows from Lemma 4. $\square$

4.2 Optimality of the Distributional Sobolev-Type Inequality

We start with the following useful observation.

Lemma 5

For $x\in S(0,\infty ),$ we have

$$\begin{aligned} \mu (x\circ r_d)=\sigma _{\omega _d}\mu (x),\quad r_d(t)=|t|^d,\quad t\in \mathbb {R}^d, \end{aligned}$$

where $\omega _d$ is the volume of the unit ball in $\mathbb {R}^d.$

Proof

Indeed, for every interval (a, b),  we have

$$\begin{aligned} m(r_d^{-1}(a,b))&=m(\{t\in \mathbb {R}^d:\ a<|t|^d<b\})\\&=m(\{t\in \mathbb {R}^d:\ |t|<b^{\frac{1}{d}}\})-m(\{t\in \mathbb {R}^d:\ |t|<a^{\frac{1}{d}}\})\\&=\omega _d\cdot (b-a). \end{aligned}$$

Hence, for every measurable set $A\subset (0,\infty ),$ we have

$$\begin{aligned} m(r_d^{-1}(A))=\omega _dm(A). \end{aligned}$$

Set $\gamma _d(t)=\omega _d|t|^d,$ $t\in \mathbb {R}^d.$ In other words, the mapping $\gamma _d:\mathbb {R}^d\rightarrow \mathbb {R}$ preserves a measure. Thus

$$\begin{aligned} x\circ r_d=\sigma _{\omega _d}x\circ \gamma _d \text{ and, } \text{ hence, } \mu (x\circ r_d)=\mu (\sigma _{\omega _d}x\circ \gamma _d)=\mu (\sigma _{\omega _d}x)=\sigma _{\omega _d}\mu (x). \end{aligned}$$

$\square$

The following proposition shows (with the help of Lemma 5) that the result of Theorem 3 is optimal.

Proposition 6

There exists a strictly positive constant $c_d'$ depending only on d such that

$$\begin{aligned} (1-\Delta )^{-\frac{d}{4}}(x\circ r_d)\ge c_d'(Tx)\circ r_d \end{aligned}$$

for every $0\le x\in L_2(0,\infty )$ supported in the interval (0, 1).

Proof

As in the proof of Theorem 3 above, we rewrite $(1-\Delta )^{-\frac{d}{4}}$ as a convolution operator. Namely,

$$\begin{aligned} (1-\Delta )^{-\frac{d}{4}}z=z*g,\quad z\in L_2(\mathbb {R}^d) \end{aligned}$$

where g is given by the formula

$$\begin{aligned} g(t)=c_d|t|^{-\frac{d}{4}}K_{\frac{d}{4}}(|t|),\quad t\in \mathbb {R}^d. \end{aligned}$$

By formula (9.6.23) in [1], this is a strictly positive function.

Let $\mathbb {B}^d$ be the unit ball in $\mathbb {R}^d.$ Since g is strictly positive, it follows from the formulae (9.6.2) and (9.6.10) in [1] (when $d\ne 0\textrm{mod}4$) or from the formula (9.6.11) in [1] (when $d=0\textrm{mod}4$) that

$$\begin{aligned} g(t)\ge c_d'|t|^{-\frac{d}{2}},\quad 0<|t|<2. \end{aligned}$$

Since $x\ge 0,$ it follows that

$$\begin{aligned} \big ((1-\Delta )^{-\frac{d}{4}}(x\circ r_d)\big )(t)=((x\circ r_d)*g)(t)\ge c_d'\int _{|t-s|<2}|t-s|^{-\frac{d}{2}}x(|s|^d)ds. \end{aligned}$$

When $|t|<1,$ we have

$$\begin{aligned} \Big \{s\in \mathbb {R}^d:\quad |s|<1\Big \}\subset \Big \{s\in \mathbb {R}^d: |t-s|<2\Big \}. \end{aligned}$$

Thus,

$$\begin{aligned} \big ((1-\Delta )^{-\frac{d}{4}}(x\circ r_d)\big )(t)\ge c_d'\int _{|s|<1}|t-s|^{-\frac{d}{2}}x(|s|^d)ds,\quad |t|<1. \end{aligned}$$

Obviously,

$$\begin{aligned} |t-s|\le |t|+|s|\le 2\max \{|t|,|s|\} \end{aligned}$$

and, therefore,

$$\begin{aligned} |t-s|^{-\frac{d}{2}}\ge 2^{-\frac{d}{2}}\min \{|t|^{-\frac{d}{2}},|s|^{-\frac{d}{2}}\}. \end{aligned}$$

It follows that

$$\begin{aligned} \big ((1-\Delta )^{-\frac{d}{4}}(x\circ r_d)\big )(t)\ge 2^{-\frac{d}{2}}c_d'\int _{|s|<1}\min \{|t|^{-\frac{d}{2}},|s|^{-\frac{d}{2}}\}x(|s|^d)ds,\quad |t|<1. \end{aligned}$$

Passing to spherical coordinates, we obtain

$$\begin{aligned} \int _{|s|<1}\min \{|t|^{-\frac{d}{2}},|s|^{-\frac{d}{2}}\}x(|s|^d)ds=\int _0^1\min \{|t|^{-\frac{d}{2}},r^{-\frac{d}{2}}\}r^{d-1}x(r^d)dr\cdot \int _{\mathbb {S}^{d-1}}ds. \end{aligned}$$

Making the substitution $r^d=u,$ we write

$$\begin{aligned} \int _0^1\min \{|t|^{-\frac{d}{2}},r^{-\frac{d}{2}}\}r^{d-1}x(r^d)dr&=\frac{1}{d}\int _0^1\min \{|t|^{-\frac{d}{2}},u^{-\frac{1}{2}}\}x(u)du\\&=\frac{1}{d}\Big (\int _0^{|t|^d}|t|^{-\frac{d}{2}}x(u)du+\int _{|t|^d}^1u^{-\frac{1}{2}}x(u)du\Big )\\&=\frac{1}{d}(Tx)(|t|^d). \end{aligned}$$

Combining the last two equations, we complete the proof. $\square$

5 Sobolev Embedding Theorem in Arbitrary Dimension

In this section, we extend Theorems 1 (see Proposition 7) and 2 (see Corollary 14) to Euclidean spaces of arbitrary dimension. We provide new proofs of both theorems in their original setting (for both even and odd d).

In what follows, we extend $\psi$ to $(0,\infty )$ by setting

$$\begin{aligned} \psi (t)= {\left\{ \begin{array}{ll} \frac{1}{\log (\frac{e}{t})},&{} t\in (0,1)\\ t,&{} t\in [1,\infty ). \end{array}\right. } \end{aligned}$$

So-defined $\psi$ is a concave increasing function on $(0,\infty ).$

Below we consider the space $\Lambda _{\psi }$ and its 2-convexification on $\mathbb {R}^d.$

Proposition 7

Let $d\in \mathbb {N}.$ For every $x\in L_2(\mathbb {R}^d),$ we have

$$\begin{aligned} \Vert (1-\Delta )^{-\frac{d}{4}}x\Vert _{\Lambda _{\psi }^{(2)}}\le c_d\Vert x\Vert _2. \end{aligned}$$

The proof of Proposition 7 requires some preparation.

With some effort, the crucial lemma below can be inferred from the proof of Theorem 5.1 in [10]. We provide a direct proof. For the definition of real interpolation method employed below we refer the reader to [6] (see also [15]).

Lemma 8

We have

$$\begin{aligned}{}[\texttt{M}_{\phi }(0,1),L_{\infty }(0,1)]_{\frac{1}{2},2}\subset \Lambda _{\psi }^{(2)}(0,1). \end{aligned}$$

Proof

Let $t\in (0,1)$ and set $s=\psi ^{-1}(t).$

Let $x\in (\texttt{M}_{\phi }+L_{\infty })(0,1)$ and let $x=f+g,$ where $f\in \texttt{M}_{\phi }(0,1)$ and where $g\in L_{\infty }(0,1).$ Set

$$\begin{aligned} A=|x|^{-1}([\mu (s,x),\infty )),\ B=|f|^{-1}\Big (\Big [\frac{1}{2}\mu (s,x),\infty \Big )\Big ),\ C=|g|^{-1}\Big (\Big [\frac{1}{2}\mu (s,x),\infty \Big )\Big ). \end{aligned}$$

Note that $m(A)\ge s.$

We have

$$\begin{aligned} |f|(u)<\frac{1}{2}\mu (s,x),\quad |g|(u)<\frac{1}{2}\mu (s,x),\quad u\in B^c\cap C^c. \end{aligned}$$

Thus,

$$\begin{aligned} |x|(u)\le |f|(u)+|g|(u)<\mu (s,x),\quad u\in B^c\cap C^c. \end{aligned}$$

So, $u\in B^c\cap C^c$ implies that $u\in A^c.$ That is, $B^c\cap C^c\subset A^c$ or, equivalently, $A\subset B\cup C.$

If $m(B)\ge \frac{1}{2} m(A),$ then

$$\begin{aligned} \Vert f\Vert _{M_{\phi }}+t\Vert g\Vert _{L_{\infty }}&\ge \Vert f\chi _B\Vert _{M_{\phi }} \ge \frac{1}{2}\mu (s,x) \Vert \chi _B\Vert _{M_{\phi }} \\&= \frac{1}{2}\mu (s,x) \psi (m(B)) \ge \frac{1}{4}\mu (s,x)\psi (m(A)) \ge \frac{1}{4} t\mu (s,x). \end{aligned}$$

If $m(C)\ge \frac{1}{2} m(A),$ then

$$\begin{aligned} \Vert f\Vert _{M_{\phi }}+t\Vert g\Vert _{L_{\infty }} \ge t\Vert g\chi _C\Vert _{L_{\infty }} \ge \frac{1}{2}t\mu (s,x). \end{aligned}$$

In either case, we have

$$\begin{aligned} K(t,x,\texttt{M}_{\phi },L_{\infty })\ge \frac{1}{4}t\mu (s,x). \end{aligned}$$

Thus,

$$\begin{aligned} \Vert x\Vert _{[\texttt{M}_{\phi },L_{\infty }]_{\frac{1}{2},2}}^2&=\int _0^1 \Big (\frac{1}{t}K(t,x,\texttt{M}_{\phi },L_{\infty })\Big )^2 dt \\&\ge \frac{1}{16}\int _0^1 \mu ^2(s,x)dt {\mathop {=}\limits ^{t=\psi (s)}}\frac{1}{16} \int _0^1\mu ^2(s,x)d\psi (s). \end{aligned}$$

$\square$

The following lemma shows that the receptacle of the operator T is strictly smaller than the space suggested by the Moser–Trudinger inequality.

Lemma 9

We have $T:L_2(0,1)\rightarrow \Lambda _{\psi }^{(2)}(0,1).$

Proof

Let $x\in L_{2,1}(0,1).$ It is immediate that

$$\begin{aligned} |(Tx)(t)|&\le t^{-\frac{1}{2}}\int _0^t|x(s)|ds+\int _t^1|x(s)|\frac{ds}{s^{\frac{1}{2}}}\\&\le t^{-\frac{1}{2}}\int _0^t\mu (s,x)ds+\int _0^1|x(s)|\frac{ds}{s^{\frac{1}{2}}}\\&\le t^{-\frac{1}{2}}\int _0^t\mu (s,x)ds+\int _0^1\mu (s,x)\frac{ds}{s^{\frac{1}{2}}}\\&=t^{-\frac{1}{2}}\int _0^t\mu (s,x)ds+2\Vert x\Vert _{2,1}\\&\le t^{-\frac{1}{2}}\Vert x\Vert _{2,\infty }\int _0^t\frac{ds}{s^{\frac{1}{2}}}+2\Vert x\Vert _{2,1}\\&=2\Vert x\Vert _{2,\infty }+2\Vert x\Vert _{2,1}\\&\le c_{\textrm{abs}}\Vert x\Vert _{2,1}. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert T\Vert _{L_{2,1}\rightarrow L_{\infty }}\le c_{\textrm{abs}}. \end{aligned}$$

Let $x\in L_{2,\infty }(0,1).$ It is immediate that

$$\begin{aligned} |(Tx)(t)|&\le \int _0^1|x(s)|\cdot \min \{t^{-\frac{1}{2}},s^{-\frac{1}{2}}\}ds\\&\le \int _0^1\mu (s,x)\cdot \min \{t^{-\frac{1}{2}},s^{-\frac{1}{2}}\}ds\\&\le \Vert x\Vert _{2,\infty }\int _0^1s^{-\frac{1}{2}}\cdot \min \{t^{-\frac{1}{2}},s^{-\frac{1}{2}}\}ds. \end{aligned}$$

Obviously,

$$\begin{aligned} \int _0^1s^{-\frac{1}{2}}\cdot \min \{t^{-\frac{1}{2}},s^{-\frac{1}{2}}\}ds=t^{-\frac{1}{2}}\int _0^t\frac{ds}{s^{\frac{1}{2}}}+\int _t^1\frac{ds}{s}=\log \Big (\frac{e^2}{t}\Big ). \end{aligned}$$

That is,

$$\begin{aligned} |(Tx)(t)|\le \Vert x\Vert _{2,\infty }\log \Big (\frac{e^2}{t}\Big ),\quad t\in (0,1). \end{aligned}$$

Since the function $t\rightarrow \log \big (\frac{e^2}{t}\big ),$ $t\in (0,1),$ falls into $\texttt{M}_{\phi }(0,1),$ it follows that

$$\begin{aligned} \Vert T\Vert _{L_{2,\infty }\rightarrow \texttt{M}_{\phi }}\le c_{\textrm{abs}}. \end{aligned}$$

By real interpolation, we have

$$\begin{aligned} T:[L_{2,\infty },L_{2,1}]_{\frac{1}{2},2}\rightarrow [\texttt{M}_{\phi },L_{\infty }]_{\frac{1}{2},2} \end{aligned}$$

is a bounded mapping. By Lemma 8, we have

$$\begin{aligned}{}[L_{2,\infty },L_{2,1}]_{\frac{1}{2},2}=L_2,\quad [\texttt{M}_{\phi },L_{\infty }]_{\frac{1}{2},2}{\mathop {\subset }\limits ^{L.8}}\Lambda _{\psi }^{(2)}. \end{aligned}$$

Thus, $T:L_2(0,1)\rightarrow \Lambda _{\psi }^{(2)}(0,1)$ is a bounded mapping. $\square$

Proof of Proposition 7

By Theorem 3 and Lemma 9, we have

$$\begin{aligned} \Big \Vert \mu \big ((1-\Delta )^{-\frac{d}{4}}x\big )\chi _{(0,1)}\Big \Vert _{\Lambda _{\psi }^{(2)}}\le c_d\Vert x\Vert _2, \quad x\in L_2(\mathbb {R}^d). \end{aligned}$$

(2)

On the other hand, we trivially have

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}x\Big \Vert _2\le \Vert x\Vert _2,\quad x\in L_2(\mathbb {R}^d). \end{aligned}$$

(3)

For every measurable function f (e.g. on $\mathbb {R}^d$) we have

$$\begin{aligned} \Vert f\Vert _{\Lambda _{\psi }^{(2)}}&=\Big (\int _0^{\infty }\mu ^2(s,f)d\psi (s)\Big )^{\frac{1}{2}}\\&=\Big (\int _0^1\mu ^2(s,f)d\psi (s)+\int _1^{\infty }\mu ^2(s,f)ds\Big )^{\frac{1}{2}}\\&\le \Big (\Vert \mu (f)\chi _{(0,1)}\Vert _{\Lambda _{\psi }^{(2)}}^2+\Vert f\Vert _2^2\Big )^{\frac{1}{2}}. \end{aligned}$$

Applying the latter inequality to $f=(1-\Delta )^{-\frac{d}{4}}x$ and using (2) and (3) we obtain

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}x\Big \Vert _{\Lambda _{\psi }^{(2)}}\le (c_d^2+1)^{\frac{1}{2}}\Vert x\Vert _2, \quad x\in L_2(\mathbb {R}^d). \end{aligned}$$

$\square$

The next assertion should be compared with Theorem 4 in [9] (it is proved in the companion paper [10] and constitutes the key part of the proof of Theorem 5.7 in that paper). Note that our T is different from that in [9]. This difference is the reason why our proof is so much simpler.

Theorem 10

For every $z=\mu (z)\in \Lambda _{\psi }^{(2)}(0,1),$ there exists $x=\mu (x)\in L_2(0,1)$ such that

$$\begin{aligned} \mu (z)\le Tx \text{ and } \Vert x\Vert _2\le c_{\textrm{abs}}\Vert z\Vert _{\Lambda _{\psi }^{(2)}}. \end{aligned}$$

The technical part of the proof of Theorem 10 is concentrated in the next lemma.

Lemma 11

Let $z\in \Lambda _{\psi }^{(2)}(0,1)$ and let

$$\begin{aligned} y(t)=\sup _{0<s<t}\psi (s)\mu (s,z),\quad 0<t<1. \end{aligned}$$

We have

$$\begin{aligned} \Vert y\Vert _{L_2((0,1),\frac{dt}{t})}\le 3^{\frac{1}{2}}\Vert z\Vert _{\Lambda _{\psi }^{(2)}}. \end{aligned}$$

Proof

Recall that the Cesaro operator $C:L_2(0,1)\rightarrow L_2(0,1)$ is defined by the formula

$$\begin{aligned} (Cx)(t)=\frac{1}{t}\int _0^tx(s)ds,\quad t\in (0,1),\quad x\in L_2(0,1). \end{aligned}$$

We claim that (here C is the classical Cesaro operator)

$$\begin{aligned} \int _0^1\sup _{0<s<t}\psi ^2(s)(C\mu (w))(s)\frac{dt}{t}\le 3\Vert w\Vert _{\Lambda _{\psi }},\quad w\in \Lambda _{\psi }. \end{aligned}$$

The functional on the left hand side is normal, subadditive and positively homogeneous. By Lemma II.5.2 in [14], it suffices to prove the inequality for the indicator functions.

Let $w=\chi _A$ and let $m(A)=u.$ Obviously, $\mu (w)=\chi _{(0,u)}$ and

$$\begin{aligned} (C\mu (w))(s)=\frac{\min \{u,s\}}{s},\quad 0<s<1. \end{aligned}$$

Thus,

$$\begin{aligned}&\quad \int _0^1\sup _{0<s<t}\psi ^2(s)(C\mu (w))(s)\frac{dt}{t}\\&=\int _0^u\sup _{0<s<t}\psi ^2(s)\frac{dt}{t}+ \int _u^1\max \Big \{\sup _{0<s<u}\psi ^2(s),\sup _{u<s<t}\psi ^2(s)\frac{u}{s}\Big \}\frac{dt}{t}\\&=\int _0^u\psi ^2(t)\frac{dt}{t}+\int _u^1\max \Big \{\psi ^2(u),u\sup _{u<s<t}s^{-1}\psi ^2(s)\Big \}\frac{dt}{t}. \end{aligned}$$

The function $s\rightarrow s^{-1}\psi ^2(s),$ $s\in (0,1),$ decreases on the interval $(0,e^{-1})$ and increases on the interval $(e^{-1},1).$ Thus,

$$\begin{aligned} \sup _{u<s<t}s^{-1}\psi ^2(s)= {\left\{ \begin{array}{ll} u^{-1}\psi ^2(u),&{} t\in (0,e^{-1})\\ t^{-1}\psi ^2(t),&{} u\in (e^{-1},1)\\ \frac{e}{4},&{} u<e^{-1}<t \end{array}\right. } \end{aligned}$$

For $u\in (e^{-1},1),$ we have

$$\begin{aligned} \int _0^1\sup _{0<s<t}\psi ^2(s)(C\mu (w))(s)\frac{dt}{t}&=\int _0^u\psi ^2(t)\frac{dt}{t}+\int _u^1\max \Big \{\psi ^2(u),ut^{-1}\psi ^2(t)\Big \}\frac{dt}{t}\\&=\int _0^u\psi ^2(t)\frac{dt}{t}+\int _u^1\psi ^2(u)\frac{dt}{t}\\&=\psi (u)+\psi ^2(u)\log (\frac{1}{u})\\&\le 2\psi (u). \end{aligned}$$

For $u\in (0,e^{-1}),$ we have

$$\begin{aligned}&\quad \int _0^1\sup _{0<s<t}\psi ^2(s)(C\mu (w))(s)\frac{dt}{t}\\&=\int _0^u\psi ^2(t)\frac{dt}{t}+\int _u^{e^{-1}}\psi ^2(u)\frac{dt}{t}+\int _{e^{-1}}^1\max \Big \{\psi ^2(u),\frac{e}{4}u\Big \}\frac{dt}{t}\\&=\psi (u)+\psi ^2(u)\log (\frac{1}{eu})+\max \Big \{\psi ^2(u),\frac{e}{4}u\Big \}\\&\le 3\psi (u). \end{aligned}$$

This yields the claim.

Let $w=\mu ^2(z)\in \Lambda _{\psi }.$ Since $\mu (w)\le C\mu (w),$ it follows from the preceding paragraph that

$$\begin{aligned} \int _0^1\sup _{0<s<t}\psi ^2(s)\mu (s,w)\frac{dt}{t}\le \int _0^1\sup _{0<s<t}\psi ^2(s)(C\mu (w))(s)\frac{dt}{t}\le 3\Vert w\Vert _{\Lambda _{\psi }}. \end{aligned}$$

Noting that

$$\begin{aligned} \sup _{0<s<t}\psi ^2(s)\mu (s,w)=y^2(t),\quad \Vert w\Vert _{\Lambda _{\psi }}=\Vert z\Vert _{\Lambda _{\psi }^{(2)}}^2, \end{aligned}$$

we complete the proof. $\square$

The following lemma affords a very substantial simplification and streamlining of the arguments employed in Theorem 5.7 in [10] (see also Theorem 4 in [9]).

Lemma 12

Let $y\in L_2\big ((0,1),\frac{dt}{t}\big )$ and let $x(t)=t^{-\frac{1}{2}}y(t),$ $0<t<1.$ If y is positive and increasing, then

$$\begin{aligned} \psi (t)\cdot (Tx)(t)\ge \frac{1}{4e}y\Big (\frac{t}{2}\Big ),\quad 0<t<1. \end{aligned}$$

Proof

Suppose $t\in (0,e^{-1}).$ We have

$$\begin{aligned} \psi (t)\cdot (Tx)(t)\ge \psi (t)\int _t^1\frac{y(s)ds}{s}\ge y(t)\cdot \psi (t)\int _t^1\frac{ds}{s}\ge \frac{1}{2} y(t)\ge \frac{1}{4e}y\Big (\frac{t}{2}\Big ). \end{aligned}$$

Suppose $t\in (e^{-1},1).$ We have

$$\begin{aligned} \psi (t)\cdot (Tx)(t)\ge \frac{1}{2}(Tx)(t)\ge \frac{1}{2t^{\frac{1}{2}}}\int _0^ty(s)ds\ge \frac{1}{2}\int _{\frac{t}{2}}^ty(s)ds\ge \frac{1}{4e}y\Big (\frac{t}{2}\Big ). \end{aligned}$$

$\square$

Lemma 13

For every $0\le x\in L_2(0,1),$ we have $Tx\le T\mu (x).$

Proof

Fix $t>0$ and denote

$$\begin{aligned} f_t(s)=\min \{t^{-\frac{1}{2}},s^{-\frac{1}{2}}\},\quad s\in (0,1). \end{aligned}$$

We have (see e.g. inequality 2.24 in Chap. II of [14])

$$\begin{aligned} (Tx)(t)=\int _0^1x(s)f_t(s)ds\le \int _0^1\mu (s,x)\mu (s,f_t)ds. \end{aligned}$$

Since $f_t=\mu (f_t),$ it follows that

$$\begin{aligned} (Tx)(t)\le \int _0^1\mu (s,x)f_t(s)ds=(T\mu (x))(t). \end{aligned}$$

$\square$

Proof of Theorem 10

Let $z=\mu (z)\in \Lambda _{\psi }^{(2)}(0,1).$ Let y be as in Lemma 11 and let x be as in Lemma 12. It follows from Lemma 11 that $\Vert y\Vert _2\le 3^{\frac{1}{2}}\Vert z\Vert _{\Lambda _{\psi }^{(2)}}.$ Obviously, y is positive and increasing. It follows from Lemma 12 that

$$\begin{aligned} \psi (t)\cdot (Tx)(t)\ge \frac{1}{4e}y\Big (\frac{t}{2}\Big )\ge \frac{1}{4e}\psi \Big (\frac{t}{2}\Big )\mu \Big (\frac{t}{2},z\Big )\ge \frac{1}{8e}\psi (t)\mu (t,z),\quad 0<t<1. \end{aligned}$$

Thus,

$$\begin{aligned} T\mu (x){\mathop {\ge }\limits ^{L.1313}} Tx\ge \frac{1}{8e}\mu (z). \end{aligned}$$

$\square$

The next corollary shows that the result of Proposition 7 is optimal in the class of symmetric function spaces.

Corollary 14

Let $E(\mathbb {R}^d)$ be a symmetric Banach function space on $\mathbb {R}^d.$ If

$$\begin{aligned} (1-\Delta )^{-\frac{d}{4}}:L_2(\mathbb {R}^d)\rightarrow E(\mathbb {R}^d), \end{aligned}$$

then $\Lambda _{\psi }^{(2)}(\mathbb {R}^d)\subset (E+L_{\infty })(\mathbb {R}^d).$

Proof

Clearly, $E(\mathbb {R}^d)\subset (E+L_{\infty })(\mathbb {R}^d).$ Thus,

$$\begin{aligned} (1-\Delta )^{-\frac{d}{4}}:L_2(\mathbb {R}^d)\rightarrow (E+L_{\infty })(\mathbb {R}^d). \end{aligned}$$

Hence, we may assume without loss of generality that $E+L_{\infty }=E$ so that $L_{\infty }\subset E.$

Take $z=\mu (z)\in \Lambda _{\psi }^{(2)}(0,1)$ and, using Theorem 10, find $x=\mu (x)\in L_2(0,1)$ such that $\mu (z)\le Tx.$ Extend x to $(0,\infty )$ by setting $x=0$ outside (0, 1). By Proposition 6, we have

$$\begin{aligned} (1-\Delta )^{-\frac{d}{4}}(x\circ r_d)\ge c_d'(Tx)\circ r_d \end{aligned}$$

Since $x\in L_2(0,\infty ),$ it follows that $x\circ r_d\in L_2(\mathbb {R}^d).$ By assumption, we have $(1-\Delta )^{-\frac{d}{4}}(x\circ r_d)\in E(\mathbb {R}^d).$ Thus, $(Tx)\circ r_d\in E(\mathbb {R}^d)$ and $Tx\in E(0,1).$ Since $\mu (z)\le Tx,$ it follows that $z\in E(0,1).$ Thus, $\Lambda _{\psi }^{(2)}(0,1)\subset E(0,1).$ $\square$

6 Cwikel–Solomyak Estimate in $\mathcal {L}_{\infty }$

We consider operators

$$\begin{aligned} (1-\Delta )^{-\frac{d}{4}}M_f(1-\Delta )^{-\frac{d}{4}} \text{ and } (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}} \end{aligned}$$

which act, respectively, on $L_2(\mathbb {R}^d)$ and $L_2(\mathbb {T}^d)$ and evaluate their uniform norms. We show that the maximal (symmetric Banach function) space E such that the operators above are bounded for every $f\in E$ is the Marcinkiewicz space $\texttt{M}_{\psi }.$ For Euclidean space, this follows from Proposition 15 and Theorem 16 below. For torus, this follows from Proposition 19 and Theorem 20 below.

6.1 Estimates for Euclidean Space

Recall that $\psi$ is extended to a concave increasing function on $(0,\infty )$ in Sect. 5.

Proposition 15

Let $d\in \mathbb {N}.$ We have

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_f(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }\le c_d\Vert f\Vert _{\texttt{M}_{\psi }},\quad f\in \texttt{M}_{\psi }(\mathbb {R}^d). \end{aligned}$$

Proof

Without loss of generality, f is real-valued and positive. Recall that, for positive operator $A:L_2(\mathbb {R}^d)\rightarrow L_2(\mathbb {R}^d),$ we have

$$\begin{aligned} \Vert A\Vert _{\infty }=\sup _{\Vert x\Vert _2\le 1}\langle Ax,x\rangle . \end{aligned}$$

Thus,

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_f(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }&=\sup _{\Vert x\Vert _2\le 1}\Big |\langle (1-\Delta )^{-\frac{d}{4}}M_f(1-\Delta )^{-\frac{d}{4}}x,x\rangle \Big |\\&=\sup _{\Vert x\Vert _2\le 1}\Big |\langle f\cdot (1-\Delta )^{-\frac{d}{4}}x,(1-\Delta )^{-\frac{d}{4}}x\rangle \Big |\\&=\sup _{\Vert x\Vert _2\le 1}\Big \Vert f\cdot \Big |(1-\Delta )^{-\frac{d}{4}}x\Big |^2\Big \Vert _1. \end{aligned}$$

By Hölder inequality, we have

$$\begin{aligned} \Big \Vert f\cdot \Big |(1-\Delta )^{-\frac{d}{4}}x\Big |^2\Big \Vert _1\le \Vert f\Vert _{\texttt{M}_{\psi }}\Big \Vert (1-\Delta )^{-\frac{d}{4}}x\Big \Vert _{\Lambda _{\psi }^{(2)}}^2. \end{aligned}$$

Thus,

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_f(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }\le \Vert f\Vert _{\texttt{M}_{\psi }}\sup _{\Vert x\Vert _2\le 1}\Big \Vert (1-\Delta )^{-\frac{d}{4}}x\Big \Vert _{\Lambda _{\psi }^{(2)}}^2. \end{aligned}$$

The assertion follows now from Proposition 7. $\square$

The proof of the converse inequality will be based on Proposition 6 and Theorem 10. Recall the notation $r_d$ from Lemma 5.

Theorem 16

Let $d\in \mathbb {N}.$ Let $f=\mu (f)\in \texttt{M}_{\psi }(0,\infty ).$ We have

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }\ge c_d\Vert f\Vert _{\texttt{M}_{\psi }}. \end{aligned}$$

Proof

We have

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }&=\sup _{\Vert \xi \Vert _2\le 1}\Big |\langle (1-\Delta )^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta )^{-\frac{d}{4}}\xi ,\xi \rangle \Big |\\&=\sup _{\Vert \xi \Vert _2\le 1}\Big |\langle (f\circ r_d)\cdot (1-\Delta )^{-\frac{d}{4}}\xi ,(1-\Delta )^{-\frac{d}{4}}\xi \rangle \Big |\\&=\sup _{\Vert \xi \Vert _2\le 1}\Big \Vert (f\circ r_d)\cdot \Big |(1-\Delta )^{-\frac{d}{4}}\xi \Big |^2\Big \Vert _1. \end{aligned}$$

Let us now restrict the supremum to the radial functions $\xi .$ That is, let $\xi =x\circ r_d,$ where $x\in L_2(0,\infty )$ and $\Vert x\Vert _2\le \omega _d^{-\frac{1}{2}}$ (see Lemma 5). We have

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty } \ge \sup _{\Vert x\Vert _2\le \omega _d^{-\frac{1}{2}}}\Big \Vert (f\circ r_d)\cdot \Big |(1-\Delta )^{-\frac{d}{4}}(x\circ r_d)\Big |^2\Big \Vert _1. \end{aligned}$$

Let us further assume that $x=\mu (x)$ is supported on the interval (0, 1). By Proposition 6 we have

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }&\ge \sup _{\begin{array}{c} x=\mu (x)\\ \Vert x\Vert _2\le \omega _d^{-\frac{1}{2}}\\ x=0 \text{ on } (1,\infty ) \end{array}}\Big \Vert (f\circ r_d)\cdot \Big |(1-\Delta )^{-\frac{d}{4}}(x\circ r_d)\Big |^2\Big \Vert _1\\&\ge c_d\cdot \sup _{\begin{array}{c} x=\mu (x)\\ \Vert x\Vert _2\le \omega _d^{-\frac{1}{2}}\\ x=0 \text{ on } (1,\infty ) \end{array}}\Big \Vert (f\circ r_d)\cdot \Big |Tx\circ r_d\Big |^2\Big \Vert _1\\&=c_d\omega _d\cdot \sup _{\begin{array}{c} x=\mu (x)\\ \Vert x\Vert _2\le \omega _d^{-\frac{1}{2}}\\ x=0 \text{ on } (1,\infty ) \end{array}}\big \Vert f\cdot |Tx|^2\big \Vert _1. \end{aligned}$$

By Theorem 10, we have

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }&\ge c_d\omega _d\cdot \sup _{\begin{array}{c} z=\mu (z)\in \Lambda _{\psi }^{(2)}(0,1)\\ \Vert z\Vert _{\Lambda _{\psi ^{(2)}}}\le c_{\textrm{abs}}^{-1}\omega _d^{-\frac{1}{2}} \end{array}}\big \Vert f\cdot z^2\big \Vert _1\\&=c_dc_{\textrm{abs}}^{-2}\cdot \sup _{\begin{array}{c} z=\mu (z)\in \Lambda _{\psi }(0,1)\\ \Vert z\Vert _{\Lambda _{\psi }}\le 1 \end{array}}\big \Vert f\cdot z\big \Vert _1. \end{aligned}$$

By Theorem II.5.2 in [14] we have

$$\begin{aligned} \sup _{\begin{array}{c} z=\mu (z)\in \Lambda _{\psi }(0,1)\\ \Vert z\Vert _{\Lambda _{\psi }}\le 1 \end{array}}\big \Vert f\cdot z\big \Vert _1=\Vert f\chi _{(0,1)}\Vert _{\texttt{M}_{\psi }}. \end{aligned}$$

Since $f=\mu (f)$ and since $\psi$ is linear on $(1,\infty ),$ it follows that

$$\begin{aligned} \Vert f\chi _{(0,1)}\Vert _{\texttt{M}_{\psi }}=\Vert \mu (f)\chi _{(0,1)}\Vert _{\texttt{M}_{\psi }}=\Vert f\Vert _{\texttt{M}_{\psi }}. \end{aligned}$$

Combining the last three equations, we complete the proof. $\square$

6.2 Estimates for the Torus

The following lemma is taken from [25] (see Lemmas 4.5 and 4.6 there).

Lemma 17

Let h be a measurable function on $[-1,1]^d.$ We have

$$\begin{aligned} M_h(1-\Delta )^{-\frac{d}{2}}M_h\Big |_{L_2([-1,1]^d)}=M_ha(\nabla _{\mathbb {T}^d})M_h\Big |_{L_2([-1,1]^d)}, \end{aligned}$$

where

$$\begin{aligned} a(n)=(1+|n|^2)^{-\frac{d}{2}}+b(n),\quad b(n)=O\Big ((1+|n|^2)^{-\frac{d+1}{2}}\Big ),\quad n\in \mathbb {Z}^d. \end{aligned}$$

Lemma 18

Let $h\in L_1(\mathbb {T}^d).$ We have

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}M_h(1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}\Big \Vert _{\infty }\le c_d\Vert h\Vert _1. \end{aligned}$$

Proof

Without loss of generality, h is real-valued and positive. We have

$$\begin{aligned}&~\quad \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}M_h(1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}\Big \Vert _{\infty }\\&=\sup _{\Vert x\Vert _2\le 1}\Big |\langle (1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}M_h(1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}x,x\rangle \Big |\\&=\sup _{\Vert x\Vert _2\le 1}\Big |\langle h\cdot (1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}x,(1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}x\rangle \Big |\\&=\sup _{\Vert x\Vert _2\le 1}\Big \Vert h\cdot \Big |(1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}x\Big |^2\Big \Vert _1\le \sup _{\Vert x\Vert _2\le 1}\Vert h\Vert _1\Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}x\Big \Vert _{\infty }^2. \end{aligned}$$

Recall the post-critical Sobolev inequality (see e.g. Theorem 7.57 (c) in [2]):

$$\begin{aligned} W^{\frac{d+1}{2},2}(\mathbb {T}^d)\subset W^{\frac{d+1}{2},2}([-\pi ,\pi ]^d)\subset C([-\pi ,\pi ]^d)\subset L_{\infty }(\mathbb {T}^d). \end{aligned}$$

Therefore,

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}\Big \Vert _{L_2(\mathbb {T}^d)\rightarrow L_{\infty }(\mathbb {T}^d)}\le c_d. \end{aligned}$$

Combining these estimates, we complete the proof. $\square$

It is of crucial importance that the estimate in the preceding lemma is given in terms of $\Vert h\Vert _1$ rather than $\Vert h\Vert _{\texttt{M}_{\psi }}.$

The following proposition is a version of Proposition 15 for $\mathbb {T}^d.$ A direct computation yields $\psi '\notin L_M(0,1),$ $M(t)=t\log (e+t),$ $t>0.$ By Theorem II.5.7 in [14], Marcinkiewicz space $\texttt{M}_{\psi }(0,1)$ contains (and, therefore, strictly contains) the Orlicz space $L_M(0,1).$ The result below delivers the sharp estimate of the symmetrized Cwikel–Solomyak operator

$$\begin{aligned} (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}. \end{aligned}$$

The sharpness of the estimate will be demonstrated below in Theorem 20.

Proposition 19

Let $d\in \mathbb {N}.$ Let $f\in \texttt{M}_{\psi }(\mathbb {T}^d).$ We have

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\le c_d\Vert f\Vert _{\texttt{M}_{\psi }}. \end{aligned}$$

Proof

Without loss of generality, f is real-valued, positive and supported on $[-1,1]^d.$

In this proof, we frequently use the following property: let $H_0\subset H$ be a Hilbert subspace and let $A:H\rightarrow H_0.$ If A vanishes on the orthogonal complement of $H_0,$ then $\Vert A\Vert _{\infty }=\Vert A|_{H_0}\Vert _{\infty }.$

Firstly, note that

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_f(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }&=\Big \Vert M_{f^{\frac{1}{2}}}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}}\Big \Vert _{\infty }\\&=\Big \Vert M_{f^{\frac{1}{2}}}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}}\Big |_{L_2([-1,1]^d)}\Big \Vert _{\infty }. \end{aligned}$$

By Lemma 17, we have

$$\begin{aligned}&~\quad M_{f^{\frac{1}{2}}}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}}\Big |_{L_2([-1,1]^d)}\\&=M_{f^{\frac{1}{2}}}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{2}}M_{f^{\frac{1}{2}}}\Big |_{L_2([-1,1]^d)}+M_{f^{\frac{1}{2}}}b(\nabla _{\mathbb {T}^d}) M_{f^{\frac{1}{2}}}\Big |_{L_2([-1,1]^d)}. \end{aligned}$$

By triangle inequality, we have

$$\begin{aligned}&~\quad \Big \Vert M_{f^{\frac{1}{2}}}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{2}}M_{f^{\frac{1}{2}}}\Big |_{L_2([-1,1]^d)}\Big \Vert _{\infty }\\&\le \Big \Vert M_{f^{\frac{1}{2}}}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}}\Big |_{L_2([-1,1]^d)}\Big \Vert _{\infty }+ \Big \Vert M_{f^{\frac{1}{2}}}b(\nabla _{\mathbb {T}^d})M_{f^{\frac{1}{2}}}\Big |_{L_2([-1,1]^d)}\Big \Vert _{\infty }\\&=\Big \Vert M_{f^{\frac{1}{2}}}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}}\Big \Vert _{\infty } +\Big \Vert M_{f^{\frac{1}{2}}}b(\nabla _{\mathbb {T}^d})M_{f^{\frac{1}{2}}}\Big \Vert _{\infty }\\&\le \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_f(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }+c_d\Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}\Big \Vert _{\infty }. \end{aligned}$$

The first summand is estimated in Proposition 15. The second summand is estimated in Lemma 18. Combining these estimates, we complete the proof. $\square$

The following theorem is a version of Theorem 16 for $\mathbb {T}^d.$ Here, f is a measurable function on (0, 1) and $f\circ r_d$ is a measurable function on $\mathbb {B}^d$ extended to $[-\pi ,\pi ]^d$ by setting $f=0$ outside of $\mathbb {B}^d$ and identified with a measurable function on $\mathbb {T}^d.$

Theorem 20

Let $d\in \mathbb {N}.$ Let $f=\mu (f)\in \texttt{M}_{\psi }(0,1).$ We have

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\ge c_d\Vert f\Vert _{\texttt{M}_{\psi }}. \end{aligned}$$

Proof

Firstly, note that

$$\begin{aligned} \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }&=\Big \Vert M_{f^{\frac{1}{2}}\circ r_d}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}\circ r_d}\Big \Vert _{\infty }\\&=\Big \Vert M_{f^{\frac{1}{2}}\circ r_d}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}\circ r_d}\Big |_{L_2([-1,1]^d)}\Big \Vert _{\infty }. \end{aligned}$$

By Lemma 17, we have

$$\begin{aligned}&~\quad M_{f^{\frac{1}{2}}\circ r_d}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}\circ r_d}\Big |_{L_2([-1,1]^d)}\\&=M_{f^{\frac{1}{2}}\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{2}}M_{f^{\frac{1}{2}}\circ r_d}\Big |_{L_2([-1,1]^d)}+M_{f^{\frac{1}{2}}\circ r_d}b(\nabla _{\mathbb {T}^d}) M_{f^{\frac{1}{2}}\circ r_d}\Big |_{L_2([-1,1]^d)}. \end{aligned}$$

By triangle inequality, we have

$$\begin{aligned}&~\quad \Big \Vert M_{f^{\frac{1}{2}}\circ r_d}(1-\Delta )^{-\frac{d}{2}}M_{f^{\frac{1}{2}}\circ r_d}\Big |_{L_2([-1,1]^d)}\Big \Vert _{\infty }\\&\le \Big \Vert M_{f^{\frac{1}{2}}\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{2}}M_{f^{\frac{1}{2}}\circ r_d}\Big |_{L_2([-1,1]^d)}\Big \Vert _{\infty } +\Big \Vert M_{f^{\frac{1}{2}}\circ r_d}b(\mathbb {T}^d)M_{f^{\frac{1}{2}}\circ r_d}\Big |_{L_2([-1,1]^d)}\Big \Vert _{\infty }\\&=\Big \Vert M_{f^{\frac{1}{2}}\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{2}}M_{f^{\frac{1}{2}}\circ r_d}\Big \Vert _{\infty }+\Big \Vert M_{f^{\frac{1}{2}}\circ r_d}b(\mathbb {T}^d)M_{f^{\frac{1}{2}}\circ r_d}\Big \Vert _{\infty }. \end{aligned}$$

Recall that

$$\begin{aligned} |b(n)|\le c_d(1+|n|^2)^{-\frac{d+1}{2}},\quad n\in \mathbb {Z}^d. \end{aligned}$$

Thus,

$$\begin{aligned}&~\quad \Big \Vert (1-\Delta )^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta )^{-\frac{d}{4}}\Big \Vert _{\infty }\\&\le \Big \Vert M_{f^{\frac{1}{2}}\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{2}}M_{f^{\frac{1}{2}}\circ r_d}\Big \Vert _{\infty }+\Big \Vert M_{f^{\frac{1}{2}}\circ r_d}b(\mathbb {T}^d)M_{f^{\frac{1}{2}}\circ r_d}\Big \Vert _{\infty }\\&\le \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }+c_d\Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d+1}{4}}\Big \Vert _{\infty }. \end{aligned}$$

By Theorem 16 and Lemma 18, we have

$$\begin{aligned} c_d'\Vert f\Vert _{\texttt{M}_{\psi }}\le \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }+c_d''\Vert f\Vert _1. \end{aligned}$$

In other words, we have

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\ge c_d'\Vert f\Vert _{\texttt{M}_{\psi }}-c_d''\Vert f\Vert _1. \end{aligned}$$

(4)

On the other hand, we have

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }&\ge \langle (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}1,1\rangle \\&=\langle (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}1,1\rangle \\&=\langle M_{f\circ r_d}1,(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}1\rangle \\&=\langle M_{f\circ r_d}1,1\rangle . \end{aligned}$$

Thus,

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\ge \Vert f\Vert _1. \end{aligned}$$

(5)

Combining (4) and (5), we obtain

$$\begin{aligned} (1+c_d'')\Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{f\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\ge c_d'\Vert f\Vert _{\texttt{M}_{\psi }}. \end{aligned}$$

This completes the proof. $\square$

6.3 Optimality of Cwikel–Solomyak Estimates Within the Classes of Orlicz and Lorentz Spaces

The following material is standard; for more details we refer the reader to [16, 21]. Let H be a complex separable infinite dimensional Hilbert space, and let B(H) denote the set of all bounded operators on H, and let K(H) denote the ideal of compact operators on H. Given $T\in K(H),$ the sequence of singular values $\mu (T) = \{\mu (k,T)\}_{k=0}^\infty$ is defined as:

$$\begin{aligned} \mu (k,T) = \inf \{\Vert T-R\Vert _{\infty }:\quad \textrm{rank}(R) \le k\}. \end{aligned}$$

Let $p \in (0,\infty ).$ The weak Schatten class $\mathcal {L}_{p,\infty }$ is the set of all operators $T\in K(H)$ such that $\mu (T)$ is in the weak $L_p$-space $l_{p,\infty }$, with quasi-norm:

$$\begin{aligned} \Vert T\Vert _{p,\infty } = \sup _{k\ge 0} (k+1)^{\frac{1}{p}}\mu (k,T) < \infty . \end{aligned}$$

Obviously, $\mathcal {L}_{p,\infty }$ is an ideal in B(H).

Our next corollary follows from Theorem 20 and provides one of the two key ingredients in the proof of one of our main results.

Corollary 21

Let $E(\mathbb {T}^d)$ be a symmetric Banach function space on $\mathbb {T}^d.$ Suppose that

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\le c_{d,E}\Vert f\Vert _E,\quad f\in E(\mathbb {T}^d). \end{aligned}$$

It follows that $E\subset \texttt{M}_{\psi }.$

Proof

Take $h=\mu (h)\in E(0,1)$ and let $f=h\circ r_d\in E(\mathbb {T}^d)$ (by Lemma 5). We have that

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_{h\circ r_d}(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\le c_{d,E}\omega _d\Vert h\Vert _E,\quad h=\mu (h)\in E(0,1). \end{aligned}$$

By Theorem 20, it follows that

$$\begin{aligned} c_d\Vert h\Vert _{\texttt{M}_{\psi }}\le c_{d,E}\omega _d\Vert h\Vert _E,\quad h=\mu (h)\in E(0,1). \end{aligned}$$

This completes the proof. $\square$

The next lemma demonstrates efficiency of general theory of symmetric function spaces in the study of Cwikel–Solomyak estimates.

Lemma 22

If $L_N(0,1)$ is an Orlicz space such that $L_N(0,1)\subset \texttt{M}_{\psi }(0,1),$ then $L_N(0,1)\subset L_M(0,1),$ where $M(t)=t\log (e+t),$ $t>0.$

Proof

We have

$$\begin{aligned} \Vert \chi _{(0,t)}\Vert _{\texttt{M}_{\psi }}\le c_N\Vert \chi _{(0,t)}\Vert _{L_N},\quad 0<t<1. \end{aligned}$$

Thus,

$$\begin{aligned} t\log (\frac{e}{t})\le \frac{c_N}{N^{-1}(\frac{1}{t})},\quad t\in (0,1). \end{aligned}$$

Setting $u=t^{-1},$ we write

$$\begin{aligned} u^{-1}\log (eu)\le \frac{c_N}{N^{-1}(u)},\quad u>1. \end{aligned}$$

Setting $v=N^{-1}(u),$ we write

$$\begin{aligned} N(v)^{-1}\log (eN(v))\le \frac{c_N}{v},\quad v>N^{-1}(1). \end{aligned}$$

Equivalently,

$$\begin{aligned} N(v)\ge c_N^{-1}v\log (eN(v)),\quad v>N^{-1}(1). \end{aligned}$$

By convexity,

$$\begin{aligned} N(v)\ge \frac{v}{N^{-1}(1)},\quad v>N^{-1}(1). \end{aligned}$$

Thus,

$$\begin{aligned} N(v)\ge c_N^{-1}v\log (\frac{ev}{N^{-1}(1)}),\quad v>N^{-1}(1). \end{aligned}$$

Thus,

$$\begin{aligned} N(v)\ geq c_N'M(v),\quad v>c_N''. \end{aligned}$$

$\square$

Lemma 23

If $\Lambda _{\theta }(0,1)$ is a Lorentz space such that $\Lambda _{\theta }(0,1)\subset \texttt{M}_{\psi }(0,1),$ then $\Lambda _{\theta }(0,1)\subset \Lambda _{\phi }(0,1).$

Proof

By assumption, there exists a constant $c_{\theta }$ such that

$$\begin{aligned} \Vert x\Vert _{\texttt{M}_{\psi }}\le c_{\theta }\Vert x\Vert _{\Lambda _{\theta }},\quad x\in \Lambda _{\theta }(0,1). \end{aligned}$$

Setting $x=\chi _{(0,t)},$ we obtain

$$\begin{aligned} \phi (t)=\Vert \chi _{(0,t)}\Vert _{\texttt{M}_{\psi }}\le c_{\theta }\Vert \chi _{(0,t)}\Vert _{\Lambda _{\theta }}=c_{\theta }\theta (t),\quad t\in (0,1). \end{aligned}$$

Thus, $\Lambda _{\theta }(0,1)\subset \Lambda _{\phi }(0,1).$ $\square$

The following corollary is one of our main results. It demonstrates that Cwikel inequality proved by Solomyak (for even dimensions d) cannot be improved within the classes of Orlicz and Lorentz spaces. Observe that a version of Solomyak inequality for an arbitrary dimension d (in more general setting of Riemannian manifolds) is established in [26].

It is interesting to compare the result of the following corollary with Theorem 9.4 in [22] which is proved under an artificial condition on the Orlicz function N. In contrast, the following result holds for an arbitrary Orlicz function.

Corollary 24

If $L_N$ is an Orlicz space such that

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\le c_{d,N}\Vert f\Vert _{L_N},\quad f\in L_N(\mathbb {T}^d), \end{aligned}$$

then $L_N(0,1)\subset L_M(0,1).$

If $\Lambda _{\theta }$ is a Lorentz space such that

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{\infty }\le c_{d,\theta }\Vert f\Vert _{\Lambda _{\theta }},\quad f\in \Lambda _{\theta }(\mathbb {T}^d), \end{aligned}$$

then $\Lambda _{\theta }(0,1)\subset L_M(0,1).$

Proof

By Corollary 21, $L_N\subset \texttt{M}_{\psi }$ (respectively, $\Lambda _{\theta }(0,1)\subset \texttt{M}_{\psi }(0,1)$). By Lemma 22 (respectively, by Lemma 23), $L_N(0,1)\subset L_M(0,1)$ (respectively, $\Lambda _{\theta }(0,1)\subset \Lambda _{\psi }(0,1)$). $\square$

6.4 Cwikel–Solomyak Estimates in Schatten–Lorentz Ideals

Proposition 19 yields the sharp uniform norm estimate for the Cwikel–Solomyak operator, whereas Solomyak [23, 24] proved the $\Vert \cdot \Vert _{1,\infty }$-quasi-norm estimate for it. The next natural step is to apply real interpolation (see e.g. [6, 15]) to obtain Lorentz (p, q)-quasi-norm estimates for the Cwikel operator. Recall that the Lorentz (p, q)-quasi-norm is defined by the formula

$$\begin{aligned} \Vert T\Vert _{p,q}=\Big (\sum _{k\ge 0}\mu (k,T)^q(k+1)^{\frac{q}{p}-1}\Big )^{\frac{1}{q}}. \end{aligned}$$

For discussion of function spaces $L_{p,q}(\Omega , \nu )$ and their noncommutative counterparts, we refer the reader to [15, pp. 142–144, pp. 228–229] and [11, 16, 21] respectively.

Corollary 25

Let $1<p<\infty$ and $1\le q\le \infty .$ We have

$$\begin{aligned} \Big \Vert (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}\Big \Vert _{p,q}\le c_{p,q,d}\Vert f\Vert _{[\Lambda _{\phi },\texttt{M}_{\psi }]_{1-\frac{1}{p},q}},\quad f\in [\Lambda _{\phi },\texttt{M}_{\psi }]_{1-\frac{1}{p},q}(\mathbb {T}^d). \end{aligned}$$

Proof

Let $A:\texttt{M}_{\psi }(\mathbb {T}^d)\rightarrow \mathcal {L}_{\infty }$ be a bounded operator defined by the setting

$$\begin{aligned} A:f\rightarrow (1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}M_f(1-\Delta _{\mathbb {T}^d})^{-\frac{d}{4}}. \end{aligned}$$

We have that $A:\Lambda _{\psi }(\mathbb {T}^d)\rightarrow \mathcal {L}_{1,\infty }$ is bounded. By real interpolation, we have

$$\begin{aligned} A:[\Lambda _{\phi },\texttt{M}_{\psi }]_{1-\frac{1}{p},q}(\mathbb {T}^d)\rightarrow [\mathcal {L}_{1,\infty },\mathcal {L}_{\infty }]_{1-\frac{1}{p},q}. \end{aligned}$$

Combining Proposition 2.g.20 in [15] and Theorem 3.2 in [11], we obtain

$$\begin{aligned}{}[\mathcal {L}_{1,\infty },\mathcal {L}_{\infty }]_{1-\frac{1}{p},q}=\mathcal {L}_{p,q}. \end{aligned}$$

This completes the proof. $\square$

Let us note that the spaces $[\Lambda _{\phi },\texttt{M}_{\psi }]_{1-\frac{1}{p},q}(\mathbb {T}^d)$ are different for different pairs (p, q)—see e.g. Theorem 4.6.24 in [8].