Positive $L^p$-Bounded Dunkl-Type Generalized Translation Operator and Its Applications

Abstract

We prove that the spherical mean value of the Dunkl-type generalized translation operator $\tau ^y$ is a positive $L^p$-bounded generalized translation operator $T^t$. As applications, we prove the Young inequality for a convolution defined by $T^t$, the $L^p$-boundedness of $\tau ^y$ on radial functions for $p>2$, the $L^p$-boundedness of the Riesz potential for the Dunkl transform, and direct and inverse theorems of approximation theory in $L^p$-spaces with the Dunkl weight.

1 Introduction

During the last three decades, many important elements of harmonic analysis with Dunkl weight on $\mathbb {R}^d$ and ${\mathbb {S}}^{d-1}$ were proved; see, e.g., the papers by Dunkl [14,15,16], Rösler [40,41,42,43], de Jeu [24, 25], Trimèche [52, 53], Xu [54, 55], and the recent works [1, 11, 12, 19, 20].

Yet there are still several gaps in our knowledge of Dunkl harmonic analysis. In particular, Young’s convolution inequality, several important polynomial inequalities, and basic approximation estimates are not established in the general case. One of the main reasons is the lack of tools related to the translation operator. Needless to say, the standard translation operator $f\mapsto f(\cdot +y)$ plays a crucial role both in classical approximation theory and harmonic analysis, in particular, by introducing several smoothness characteristics of f. In Dunkl analysis, its analogue is the generalized translation operator $\tau ^y$ defined by Rösler [40]. Unfortunately, the $L^p$-boundedness of $\tau ^y$ is not obtained in general.

To overcome this difficulty, the spherical mean value of the translation operator $\tau ^y$ was introduced in [28] and was studied in [42], where, in particular, its positivity was shown. Our main goal in this paper is to prove that this operator is a positive $L^p$-bounded operator $T^t$, which may be considered as a generalized translation operator. It is worth mentioning that this operator can be applied to problems where it is essential to deal with radial multipliers. This is because by virtue of $T^t$ we can define the convolution operator that coincides with the known convolution introduced by Thangavelu and Xu in [48] using the operator $\tau ^y$.

For this convolution, we prove the Young inequality and, subsequently, an $L^p$-boundedness of the operator $\tau ^y$ on radial functions for $p>2$. For $1\le p\le 2$ it was proved in [48].

Let us mention here two applications of the operator $T^t$. The first one is the Riesz potential defined in [49], where its boundedness properties were obtained for the reflection group $\mathbb {Z}_2^d$. For the general case see [21]. Using the $L^p$-boundedness of the operator $T^t$ allows us to give a different simple proof, which follows ideas of Thangavelu and Xu [49]. Another application is basic inequalities of approximation theory in the weighted $L^p$ spaces. With the help of the operator $T^t$ one can define moduli of smoothness, which are equivalent to the K-functionals, and prove the direct and inverse approximation theorems. For the reflection group $\mathbb {Z}_2^d$, basic approximation inequalities were studied in [11, 12].

The paper is organized as follows. In the next section, we give some basic notation and facts of Dunkl harmonic analysis. In Sect. 3, we study the operator $T^t$, define a convolution operator, and prove the Young inequality. As a consequence, we obtain an $L^p$-boundedness of the operator $\tau ^y$ on radial functions. The weighted Riesz potential is studied in Sect. 4. Section 5 consists of a study of interrelation between several classes of entire functions. We also obtain multidimensional weighted analogues of Plancherel–Polya–Boas inequalities, which are of their own interest. In Sect. 6, we introduce moduli of smoothness and the K-functional, associated to the Dunkl weight, and prove equivalence between them as well as the Jackson inequality. Section 7 consists of weighted analogues of Nikol’skiǐ, Bernstein, and Boas inequalities for entire functions of exponential type. In Sect. 8, we obtain that moduli of smoothness are equivalent to the realization of the K-functional. We conclude with Sect. 9, where we prove the inverse theorems in $L^p$-spaces with the Dunkl weight.

2 Notation

In this section, we recall the basic notation and results of Dunkl harmonic analysis, see, e.g., [43].

Throughout the paper, $\langle x,y\rangle$ denotes the standard Euclidean scalar product in d-dimensional Euclidean space $\mathbb {R}^{d}$, $d\in \mathbb {N}$, equipped with a norm $|x|=\sqrt{\langle x,x\rangle }$. For $r>0$ we write $B_r=\{x\in \mathbb {R}^d:|x|\le r\}$. Define the following function spaces:

• $C(\mathbb {R}^d)$ the space of continuous functions,

• $C_b(\mathbb {R}^d)$ the space of bounded continuous functions with the norm $\Vert f\Vert _{\infty }=\sup _{\mathbb {R}^d}|f|$,

• $C_0(\mathbb {R}^d)$ the space of continuous functions that vanish at infinity,

• $C^{\infty }(\mathbb {R}^d)$ the space of infinitely differentiable functions,

• $\mathcal {S}(\mathbb {R}^d)$ the Schwartz space,

• $\mathcal {S}'(\mathbb {R}^d)$ the space of tempered distributions,

• $X(\mathbb {R}_+)$ the space of even functions from $X(\mathbb {R})$, where X is one of the spaces above,

• $X_\mathrm {rad}(\mathbb {R}^d)$ the subspace of $X(\mathbb {R}^d)$ consisting of radial functions $f(x)=f_{0}(|x|)$.

Let a finite subset $R\subset \mathbb {R}^{d}\setminus \{0\}$ be a root system; $R_{+}$ a positive subsystem of R; $G(R)\subset O(d)$ the finite reflection group, generated by reflections $\{\sigma _{a}:a\in R\}$, where $\sigma _{a}$ is a reflection with respect to hyperplane $\langle a,x\rangle =0$; $k:R\rightarrow \mathbb {R}_{+}$ a G-invariant multiplicity function. Recall that a finite subset $R\subset \mathbb {R}^{d}\setminus \{0\}$ is called a root system if

$$\begin{aligned} R\cap \mathbb {R}a=\{a, -a\}\quad \text {and}\quad \sigma _{a}R=R\quad \text {for all}\ a\in R. \end{aligned}$$

Let

$$\begin{aligned} v_{k}(x)=\prod _{a\in R_{+}}|\langle a,x\rangle |^{2k(a)} \end{aligned}$$

be the Dunkl weight,

$$\begin{aligned} c_{k}^{-1}=\int _{\mathbb {R}^{d}}e^{-|x|^{2}/2}v_{k}(x)\,\mathrm{d}x,\quad \mathrm{d}\mu _{k}(x)=c_{k}v_{k}(x)\,\mathrm{d}x, \end{aligned}$$

and $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $0<p<\infty$, be the space of complex-valued Lebesgue measurable functions f for which

$$\begin{aligned} \Vert f\Vert _{p,\mathrm{d}\mu _{k}}=\left( \int _{\mathbb {R}^{d}}|f|^{p}\,\mathrm{d}\mu _{k}\right) ^{1/p}<\infty . \end{aligned}$$

We also assume that $L^{\infty }\equiv C_b$ and $\Vert f\Vert _{\infty ,\mathrm{d}\mu _{k}}=\Vert f\Vert _{\infty }$.

Example

If the root system R is $\{\pm e_1,\dots ,\pm e_d\}$, where $\{e_1,\dots ,e_d\}$ is an orthonormal basis of $\mathbb {R}^{d}$, then $v_{k}(x)=\prod _{j=1}^d|x_j|^{2k_j}$, $k_{j}\ge 0$, $G=\mathbb {Z}_2^d$.

Let

$$\begin{aligned} D_{j}f(x)=\frac{\partial f(x)}{\partial x_{j}}+ \sum _{a\in R_{+}}k(a)\langle a,e_{j}\rangle \,\frac{f(x)-f(\sigma _{a}x)}{\langle a,x\rangle },\quad j=1,\dots ,d, \end{aligned}$$

be differential-differences Dunkl operators and $\Delta _k=\sum _{j=1}^dD_j^2$ be the Dunkl Laplacian. The Dunkl kernel $e_{k}(x, y)=E_{k}(x, iy)$ is a unique solution of the system

$$\begin{aligned} D_{j}f(x)=iy_{j}f(x),\quad j=1,\dots ,d,\quad f(0)=1, \end{aligned}$$

and it plays the role of a generalized exponential function. Its properties are similar to those of the classical exponential function $e^{i\langle x, y\rangle }$. Several basic properties follow from an integral representation [41]:

$$\begin{aligned} e_k(x,y)=\int _{\mathbb {R}^d}e^{i\langle \xi ,y\rangle }\,\mathrm{d}\mu _x^k(\xi ), \end{aligned}$$

where $\mu _x^k$ is a probability Borel measure, whose support is contained in

$$\begin{aligned} \mathrm {co}{}(\{gx:g\in G(R)\}), \end{aligned}$$

the convex hull of the G-orbit of x in $\mathbb {R}^d$. In particular, $|e_k(x,y)|\le 1$.

For $f\in L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, the Dunkl transform is defined by the equality

$$\begin{aligned} \mathcal {F}_{k}(f)(y)=\int _{\mathbb {R}^{d}}f(x)\overline{e_{k}(x, y)}\,\mathrm{d}\mu _{k}(x). \end{aligned}$$

For $k\equiv 0$, $\mathcal {F}_{0}$ is the classical Fourier transform $\mathcal {F}$. We also note that $\mathcal {F}_k(e^{-|\,\cdot \,|^2/2})(y)=e^{-|y|^2/2}$ and $\mathcal {F}_{k}^{-1}(f)(x)=\mathcal {F}_{k}(f)(-x)$. Let

$$\begin{aligned} \mathcal {A}_k=\left\{ f\in L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})\cap C_0(\mathbb {R}^d):\mathcal {F}_{k}(f)\in L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})\right\} . \end{aligned}$$

(2.1)

Let us now list several basic properties of the Dunkl transform.

Proposition 2.1

• (1) For $f\in L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $\mathcal {F}_{k}(f)\in C_0(\mathbb {R}^d)$.

• (2) If $f\in \mathcal {A}_k$, we have the pointwise inversion formula

  $$\begin{aligned} f(x)=\int _{\mathbb {R}^{d}}\mathcal {F}_{k}(f)(y)e_{k}(x, y)\,\mathrm{d}\mu _{k}(y). \end{aligned}$$

• (3) The Dunkl transform leaves the Schwartz space $\mathcal {S}(\mathbb {R}^d)$ invariant.

• (4) The Dunkl transform extends to a unitary operator in $L^{2}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$.

Let $\lambda \ge -1/2$ and $J_{\lambda }(t)$ be the classical Bessel function of degree $\lambda$ and

$$\begin{aligned} j_{\lambda }(t)=2^{\lambda }\Gamma (\lambda +1)t^{-\lambda }J_{\lambda }(t) \end{aligned}$$

be the normalized Bessel function. Set

$$\begin{aligned} b_{\lambda }^{-1}=\int _{0}^{\infty }e^{-t^{2}/2}t^{2\lambda +1}\, \mathrm{d}t=2^{\lambda }\Gamma (\lambda +1),\quad \mathrm{d}\nu _{\lambda }(t)=b_{\lambda }t^{2\lambda +1}\,\mathrm{d}t,\quad t\in \mathbb {R}_{+}. \end{aligned}$$

The norm in $L^{p}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda })$, $1\le p<\infty$, is given by

$$\begin{aligned} \Vert f\Vert _{p,\mathrm{d}\nu _{\lambda }}=\left( \int _{\mathbb {R}_{+}}|f(t)|^{p}\,\mathrm{d}\nu _{\lambda }(t)\right) ^{1/p}. \end{aligned}$$

Define $\Vert f\Vert _{\infty }={\text {ess\,sup}}_{t\in \mathbb {R}_{+}}|f(t)|$.

The Hankel transform is defined as follows:

$$\begin{aligned} \mathcal {H}_{\lambda }(f)(r)=\int _{\mathbb {R}_{+}}f(t)j_{\lambda }(rt)\,\mathrm{d}\nu _{\lambda }(t),\quad r\in \mathbb {R}_{+}. \end{aligned}$$

It is a unitary operator in $L^{2}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda })$ and $\mathcal {H}_{\lambda }^{-1}=\mathcal {H}_{\lambda }$ [2, Chap. 7].

Note that if $\lambda =d/2-1$, the Hankel transform is a restriction of the Fourier transform on radial functions, and if $\lambda =\lambda _k=d/2-1+\sum _{a\in R_+}k(a)$, of the Dunkl transform.

Let $\mathbb {S}^{d-1}=\{x'\in \mathbb {R}^d:|x'|=1\}$ be the Euclidean sphere and $\mathrm{d}\sigma _k(x')=a_kv_k(x')\,\mathrm{d}x'$ be the probability measure on $\mathbb {S}^{d-1}$. We have

$$\begin{aligned} \int _{\mathbb {R}^{d}}f(x)\,\mathrm{d}\mu _{k}(x)= \int _{0}^{\infty }\int _{\mathbb {S}^{d-1}}f(tx')\,\mathrm{d}\sigma _k(x')\, \mathrm{d}\nu _{\lambda _k}(t). \end{aligned}$$

(2.2)

We need the following partial case of the Funk–Hecke formula [55]

$$\begin{aligned} \int _{\mathbb {S}^{d-1}}e_{k}(x, ty')\,\mathrm{d}\sigma _k(y')=j_{\lambda _k}(t|x|). \end{aligned}$$

(2.3)

Throughout the paper, we will assume that $A\lesssim B$ means that $A\le C B$ with a constant C depending only on nonessential parameters.

3 Generalized Translation Operators and Convolutions

Let $y\in \mathbb {R}^d$ be given. Rösler [40] defined a generalized translation operator $\tau ^y$ in $L^{2}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ by the equation

$$\begin{aligned} \mathcal {F}_k(\tau ^yf)(z)=e_k(y, z)\mathcal {F}_k(f)(z). \end{aligned}$$

Since $|e_k(y, z)|\le 1$, we have $\Vert \tau ^y\Vert _{2\rightarrow 2}\le 1$. If $f\in \mathcal {A}_k$ (recall that $\mathcal {A}_k$ is given by (2.1)), then, for any $x, y\in \mathbb {R}^d$,

$$\begin{aligned} \tau ^yf(x)=\int _{\mathbb {R}^d}e_k(y, z)e_k(x, z)\mathcal {F}_k(f)(z)\,\mathrm{d}\mu _k(z). \end{aligned}$$

(3.1)

Note that $\mathcal {S}(\mathbb {R}^d)\subset \mathcal {A}_k\subset L^{2}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$. Trimèche [53] extended the operator $\tau ^y$ on $C^{\infty }(\mathbb {R}^d)$.

The explicit expression of $\tau ^yf$ is known only in the case of the reflection group $\mathbb {Z}_2^d$. In particular, in this case $\tau ^yf$ is not a positive operator [39]. Note that in the case of symmetric group $S_d$, the operator $\tau ^yf$ is also not positive [48].

It remains an open question whether $\tau ^yf$ is an $L^p$ bounded operator on $\mathcal {S}(\mathbb {R}^d)$ for $p\ne 2$. It is known [39, 48] only for $G=\mathbb {Z}_2^d$. Note that a positive answer would follow from the $L^1$-boundedness.

Let

$$\begin{aligned} \lambda _k=d/2-1+\sum _{a\in R_+}k(a). \end{aligned}$$

We have $\lambda _k\ge -1/2$ and, moreover, $\lambda _k=-1/2$ only if $d=1$ and $k\equiv 0$. In what follows, we assume that $\lambda _k>-1/2$.

Define another generalized translation operator $T^t:L^{2}(\mathbb {R}^{d},\mathrm{d}\mu _{k})\rightarrow L^{2}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $t\in \mathbb {R}$, by the relation

$$\begin{aligned} \mathcal {F}_k(T^tf)(y)=j_{\lambda _k}(t|y|)\mathcal {F}_k(f)(y). \end{aligned}$$

Since $|j_{\lambda _k}(t)|\le 1$, it is a bounded operator such that $\Vert T^t\Vert _{2\rightarrow 2}\le 1$ and

$$\begin{aligned} T^tf(x)=\int _{\mathbb {R}^d}j_{\lambda _k}(t|y|)e_k(x,y)\mathcal {F}_k(f)(y)\,\mathrm{d}\mu _k(y). \end{aligned}$$

This gives $T^t=T^{-t}$. If $f\in \mathcal {A}_k$, then from (2.3) and (3.1) we have (pointwise)

$$\begin{aligned} T^tf(x)=\int _{\mathbb {R}^d}j_{\lambda _k}(t|y|)e_k(x, y)\mathcal {F}_k(f)(y)\,\mathrm{d}\mu _k(y)=\int _{\mathbb {S}^{d-1}}\tau ^{ty'}f(x)\, \mathrm{d}\sigma _k(y'). \end{aligned}$$

(3.2)

Note that the operator $T^t$ is self-adjoint. Indeed, if $f, g\in \mathcal {A}_k$, then

$$\begin{aligned} \int _{\mathbb {R}^d}T^tf(x)\,g(x)\,\mathrm{d}\mu _k(x)&=\int _{\mathbb {R}^d} \int _{\mathbb {R}^d}j_{\lambda _k}(t|y|)e_k(x, y)\mathcal {F}_k(f)(y)\,\mathrm{d}\mu _k(y)\,g(x)\,\mathrm{d}\mu _k(x)\\&=\int _{\mathbb {R}^d}j_{\lambda _k}(t|y|)\mathcal {F}_k(f)(y) \mathcal {F}_k(g)(-y)\,\mathrm{d}\mu _k(y)\\&=\int _{\mathbb {R}^d}j_{\lambda _k}(t|y|)\mathcal {F}_k(g)(y) \mathcal {F}_k(f)(-y)\,\mathrm{d}\mu _k(y)\\&=\int _{\mathbb {R}^d}f(x)\,T^tg(x)\,\mathrm{d}\mu _k(x). \end{aligned}$$

Rösler [42] proved that the spherical mean (with respect to the Dunkl weight) of the operator $\tau ^y$, i.e., $\int _{\mathbb {S}^{d-1}}\tau ^{ty'}f(x)\,\mathrm{d}\sigma _k(y')$, is a positive operator on $C^{\infty }(\mathbb {R}^d)$ and obtained its integral representation. This implies that $T^t$ is a positive operator on $C^{\infty }(\mathbb {R}^d)$ and, moreover, for any $t\in \mathbb {R}$, $x\in \mathbb {R}^d$,

$$\begin{aligned} T^tf(x)=\int _{\mathbb {R}^d}f(z)\,\mathrm{d}\sigma _{x,t}^k(z), \end{aligned}$$

(3.3)

where $\sigma _{x,t}^k$ is a probability Borel measure,

$$\begin{aligned} {\text {supp}}\sigma _{x,t}^k\subset \bigcup \limits _{g\in G}\{z\in \mathbb {R}^d:|z-gx|\le t\} \end{aligned}$$

(3.4)

and the mapping $(x, t)\rightarrow \sigma _{x,t}^k$ is continuous with respect to the weak topology on probability measures.

The representation (3.3) gives a natural extension of the operator $T^t$ on $C_b(\mathbb {R}^d)$; namely, for $f\in C_b(\mathbb {R}^d)$ we define $T^tf(x)\in C(\mathbb {R}\times \mathbb {R}^d)$ by (3.3), and, moreover, the estimate $\Vert T^tf\Vert _{\infty }\le \Vert f\Vert _{\infty }$ holds.

Note that for $k\equiv 0$, $T^t$ is the usual spherical mean

$$\begin{aligned} T^tf(x)=S^tf(x)=\int _{\mathbb {S}^{d-1}}f(x+ty')\,\mathrm{d}\sigma _0(y'). \end{aligned}$$

(3.5)

Theorem 3.1

If $1\le p\le \infty$, then, for any $t\in \mathbb {R}$ and $f\in \mathcal {S}(\mathbb {R}^d)$,

$$\begin{aligned} \Vert T^tf\Vert _{p,\mathrm{d}\mu _k}\le \Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(3.6)

Remark 3.2

• (i) The inequality $\Vert T^tf\Vert _{p,\mathrm{d}\mu _k}\le c\Vert f\Vert _{p,\mathrm{d}\mu _k}$ was proved in [48] for $G=\mathbb {Z}_2^d$.

• (ii) $\mathcal {S}(\mathbb {R}^d)$ is dense in $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $1\le p<\infty$, so for any $t\in \mathbb {R}_+$ the operator $T^t$ can be defined on $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ and estimate (3.6) holds.

• (iii) If $d=1$, $v_k(x)=|x|^{2\lambda +1}$, $\lambda >-1/2$, inequality (3.6) was proved in [7]. In this case the integral representation of $T^t$ is of the form

  $$\begin{aligned} T^tf(x)=\frac{c_{\lambda }}{2}\int _0^{\pi }\{f(A)(1+B)+f(-A)(1-B)\} \sin ^{2\lambda }\varphi \,\mathrm{d}\varphi , \end{aligned}$$

  where, for $(x,t)\ne (0,0)$,

  $$\begin{aligned} c_{\lambda }=\frac{\Gamma (\lambda +1)}{\sqrt{\pi }\Gamma (\lambda +1/2)},\quad A=\sqrt{x^2+t^2-2xt\cos \varphi },\quad B=\frac{x-t\cos \varphi }{A}. \end{aligned}$$

  (3.7)

  If $\lambda =-1/2$, i.e., $k\equiv 0$, then $T^tf(x)=\frac{1}{2} (f(x+t)+f(x-t))$.

Proof

Let $t\in \mathbb {R}_+$ be given and the operator $T^t$ be defined on $\mathcal {S}(\mathbb {R}^d)$ by (3.3). Using (3.2), we have

$$\begin{aligned} \sup \{\Vert T^tf\Vert _{2}:f\in \mathcal {S}(\mathbb {R}^d),\, \Vert f\Vert _{2}\le 1\}\le 1, \end{aligned}$$

and $T^t$ can be extended to the space $L^{2}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ with preservation of norm; moreover, this extension coincides with (3.2). Furthermore, (3.3) yields

$$\begin{aligned} \sup \{\Vert T^tf\Vert _{\infty }:f\in \mathcal {S}(\mathbb {R}^d),\, \Vert f\Vert _{\infty }\le 1\}\le 1. \end{aligned}$$

(3.8)

Since the operator $T^t$ is self-adjoint, by (3.8),

$$\begin{aligned}&\sup \{\Vert T^tf\Vert _{1,\mathrm{d}\mu _k}:f\in \mathcal {S}(\mathbb {R}^d),\, \Vert f\Vert _{1,\mathrm{d}\mu _k}\le 1\}\\&\quad =\sup \left\{ \int _{\mathbb {R}^d}T^tf\,g\,\mathrm{d}\mu _k:f, g\in \mathcal {S}(\mathbb {R}^d),\ \Vert f\Vert _{1,\mathrm{d}\mu _k}\le 1,\ \Vert g\Vert _{\infty }\le 1\right\} \\&\quad =\sup \left\{ \int _{\mathbb {R}^d}f\,T^tg\,\mathrm{d}\mu _k:f, g\in \mathcal {S}(\mathbb {R}^d),\ \Vert f\Vert _{1,\mathrm{d}\mu _k}\le 1,\ \Vert g\Vert _{\infty }\le 1\right\} \\&\quad =\sup \{\Vert T^tg\Vert _{\infty }:g\in \mathcal {S}(\mathbb {R}^d),\ \Vert g\Vert _{\infty }\le 1\}\le 1. \end{aligned}$$

Hence, $T^t$ can be extended to $L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ with preservation of the norm such that this extension coincides with (3.2) on $L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})\cap L^{2}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$.

By the Riesz–Thorin interpolation theorem we obtain

$$\begin{aligned} \sup \{\Vert T^tf\Vert _{p,\mathrm{d}\mu _k}:f\in \mathcal {S}(\mathbb {R}^d),\ \Vert f\Vert _{p,\mathrm{d}\mu _k}\le 1\}\le 1,\quad 1\le p\le 2. \end{aligned}$$

Let $2<p<\infty$, $1/p+1/p'=1.$ As for $p=1$, we get

$$\begin{aligned}&\sup \{\Vert T^tf\Vert _{p,\mathrm{d}\mu _k}:f\in \mathcal {S}(\mathbb {R}^d),\, \Vert f\Vert _{p,\mathrm{d}\mu _k}\le 1\}\\&\quad =\sup \{\Vert T^tg\Vert _{p',\mathrm{d}\mu _k}:g\in \mathcal {S}(\mathbb {R}^d),\, \Vert g\Vert _{p',\mathrm{d}\mu _k}\le 1\}\le 1. \end{aligned}$$

$\square$

For any $f_0\in L^{p}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda })$, $1\le p\le \infty$, $\lambda >-1/2$, let us define the Gegenbauer-type translation operator (see, e.g., [34, 35])

$$\begin{aligned} R^tf_0(r)=c_\lambda \int _{0}^{\pi }f_0\left( \sqrt{r^2+t^2-2rt\cos \varphi }\right) \sin ^{2\lambda }\varphi \,\mathrm{d}\varphi , \end{aligned}$$

where $c_\lambda$ is defined by (3.7). We have that $\Vert R^t\Vert _{p\rightarrow p}\le 1$ and $\mathcal {H}_{\lambda }(R^tf_0)(r)=j_\lambda (tr)\mathcal {H}_{\lambda }(f_0)(r)$, where $f_0\in \mathcal {S}(\mathbb {R}_+)$. Taking into account (2.3) and (3.2), we note that for $\lambda =\lambda _k$ the operator $R^t$ is a restriction of $T^t$ on radial functions; that is, for $f_0\in L^{p}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$,

$$\begin{aligned} T^tf_0(|x|)=R^tf_0(r),\quad r=|x|. \end{aligned}$$

We also mention the following useful properties of the generalized translation operator $T^t$.

Lemma 3.3

Let $t\in \mathbb {R}$.

• (1) If $f\in L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, then $\int _{\mathbb {R}^d}T^tf\,\mathrm{d}\mu _k=\int _{\mathbb {R}^d}f\,\mathrm{d}\mu _k$.

• (2) Let $r>0$, $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $1\le p<\infty$. If ${\text {supp}}f\subset B_r$, then ${\text {supp}}T^tf\subset B_{r+|t|}$. If ${\text {supp}}f\subset \mathbb {R}^d\setminus B_r$, $r>|t|$, then ${\text {supp}}T^tf\subset \mathbb {R}^d\setminus B_{r-|t|}$.

Proof

Due to the $L^p$-boundedness of $T^t$ and the density of $\mathcal {S}(\mathbb {R}^d)$ in $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, we can assume that $f\in \mathcal {S}(\mathbb {R}^d)$.

• (1) Let $s>0$. By integral representation of $j_{\lambda _k}(z)$ (see, e.g., [2, Sect. 7.12]) we have

  $$\begin{aligned} T^t(e^{-s|\cdot |^2})(x)&=R^t(e^{-s(\cdot )^2})(|x|)= c_{\lambda _k}\int _{0}^{\pi }e^{-s(|x|^2+t^2-2|x|t\cos \varphi )}\sin ^{2\lambda _k} \varphi \,\mathrm{d}\varphi \\&=e^{-s(|x|^2+t^2)}c_{\lambda _k}\int _{0}^{\pi }e^{2s|x|t\cos \varphi } \sin ^{2\lambda _k}\varphi \,\mathrm{d}\varphi \\&=e^{-s(|x|^2+t^2)}j_{\lambda _k}(2is|x|t), \end{aligned}$$

  and, in particular,

  $$\begin{aligned} T^t(e^{-s|\cdot |^2})(x)\le e^{-s(|x|^2+t^2)}e^{2s|x|t}=e^{-s(|x|-t)^2}\le 1. \end{aligned}$$

  Using the self-adjointness of $T^t$, we obtain

  $$\begin{aligned} \int _{\mathbb {R}^d}T^tf(x)\,e^{-s|x|^2}\,\mathrm{d}\mu _k(x)=\int _{\mathbb {R}^d}f(x)T^t(e^{-s|\cdot |^2})(x)\,\mathrm{d}\mu _k(x). \end{aligned}$$

  Since for any $t\in \mathbb {R}$, $x\in \mathbb {R}^d$,

  $$\begin{aligned} \lim \limits _{s\rightarrow 0}e^{-s|x|^2}=\lim \limits _{s\rightarrow 0}T^t(e^{-s|\cdot |^2})(x)=1, \end{aligned}$$

  by Lebesgue’s dominated convergence theorem we derive (1).

• (2) If ${\text {supp}}f\subset B_r$ and $|x|>r+|t|$, then, in light of (3.4) and (3.3), for $z\in {\text {supp}}\sigma _{x,t}^k$ and $g\in G$, we have that

  $$\begin{aligned} |z|\ge |gx|-|z-gx|=|x|-|z-gx|>r \end{aligned}$$

  and $f(z)=0$, which yields $T^tf(x)=0$.

If ${\text {supp}}f\subset \mathbb {R}^d\setminus B_r$, $|x|<r-|t|$, then, for $z\in {\text {supp}}\sigma _{x,t}^k$ and $g\in G$, we similarly obtain $|z|\le |gx|+|z-gx|=|x|+|z-gx|<r$, $f(z)=0$, and $T^tf(x)=0$. $\square$

Let g be a radial function, $g(y)=g_0(|y|)$, where $g_0(t)$ is defined on $\mathbb {R}_+$. Note that by virtue of (2.2),

$$\begin{aligned} \Vert g\Vert _{p,\mathrm{d}\mu _k}=\Vert g_0\Vert _{p,\mathrm{d}\nu _{\lambda _k}},\quad \mathcal {F}_k(g)(y)=\mathcal {H}_{\lambda _k}(g_0)(|y|). \end{aligned}$$

(3.9)

By means of operators $T^t$ and $\tau ^y$, define two convolution operators:

$$\begin{aligned} (f*_{ \lambda _k}g_0)(x)= & {} \int _0^{\infty }T^tf(x)g_0(t)\,\mathrm{d}\nu _{\lambda _k}(t), \end{aligned}$$

(3.10)

$$\begin{aligned} (f*_{ k}g)(x)= & {} \int _{\mathbb {R}^d}f(y)\tau ^{x}g(-y)\,\mathrm{d}\mu _{k}(y). \end{aligned}$$

(3.11)

Note that operator (3.10) was defined in [48], while (3.11) was investigated in [48, 53].

Thangavelu and Xu [48] proved that if $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $1\le p\le \infty$, and $g\in L^{1}_\mathrm {rad}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, then

$$\begin{aligned} \Vert (f*_{ k}g)\Vert _{p,\mathrm{d}\mu _k}\le \Vert f\Vert _{p,\mathrm{d}\mu _k}\Vert g\Vert _{1,\mathrm{d}\mu _k}, \end{aligned}$$

(3.12)

and if $1\le p\le 2$, $g\in L^{p}_\mathrm {rad}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, then, for any $y\in \mathbb {R}^d$,

$$\begin{aligned} \Vert \tau ^{y}g\Vert _{p,\mathrm{d}\mu _k}\le \Vert g\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(3.13)

Lemma 3.4

If $f\in \mathcal {A}_k$, $g_0\in L^{1}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$, $g(y)=g_0(|y|)$, then, for any $x, y\in \mathbb {R}^d$,

$$\begin{aligned} (f*_{ \lambda _k}g_0)(x)= & {} (f*_{ k}g)(x)=\int _{\mathbb {R}^d}\tau ^{-y}f(x)g(y)\,\mathrm{d}\mu _k(y), \end{aligned}$$

(3.14)

$$\begin{aligned} \mathcal {F}_k(f*_{ \lambda _k}g_0)(y)= & {} \mathcal {F}_k(f*_{ k}g)(y)=\mathcal {F}_k(f)(y)\mathcal {F}_k(g)(y). \end{aligned}$$

(3.15)

Proof

Using (3.2) and (3.9), we get

$$\begin{aligned} (f*_{ \lambda _k}g_0)(x)&=\int _0^{\infty }T^tf(x)g_0(t)\,\mathrm{d}\nu _{\lambda _k}(t)\\&=\int _0^{\infty }\int _{\mathbb {R}^d}j_{\lambda _k}(t|y|)e_k(x, y)\mathcal {F}_k(f)(y)\,\mathrm{d}\mu _k(y)g_0(t)\,\mathrm{d}\nu _{\lambda _k}(t)\\&=\int _{\mathbb {R}^d}e_k(x, y)\mathcal {F}_k(f)(y)\mathcal {F}_k(g)(y)\,\mathrm{d}\mu _k(y), \end{aligned}$$

which gives

$$\begin{aligned} \mathcal {F}_k(f*_{ \lambda _k}g_0)(y)=\mathcal {F}_k(f)(y)\mathcal {F}_k(g)(y). \end{aligned}$$

If $g\in \mathcal {A}_k$, then, by (3.1),

$$\begin{aligned} (f*_{ k}g)(x)&=\int _{\mathbb {R}^d}f(y)\tau ^{x}g(-y)\,\mathrm{d}\mu _{k}(y)\\&=\int _{\mathbb {R}^d}f(y)\int _{\mathbb {R}^d}e_k(-y, z)e_k(x, z)\mathcal {F}_k(g)(z)\,\mathrm{d}\mu _k(z)\,\mathrm{d}\mu _{k}(y)\\&=\int _{\mathbb {R}^d}e_k(x, z)\mathcal {F}_k(f)(z)\mathcal {F}_k(g)(z)\,\mathrm{d}\mu _k(z), \end{aligned}$$

and hence the first equality in (3.14) and the second equality in (3.15) are valid for $g\in \mathcal {A}_k$.

Assuming that $g_0\in L^{1}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda })$, $(g_n)_0\in \mathcal {S}(\mathbb {R}_+)$, $g_{n}\rightarrow g$ in $L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, and taking into account (3.8)–(3.11) and (3.13), we arrive at

$$\begin{aligned} \left| (f*_{ \lambda _k}g_0)(x)-(f*_{ k}g)(x)\right|&\le \left| (f*_{ \lambda _k}(g_0-(g_n)_0))(x)\right| +\left| (f*_{ k}(g-g_n))(x)\right| \\&\le 2\Vert f\Vert _{\infty }\,\Vert g-g_n\Vert _{1, \mathrm{d}\mu _k}. \end{aligned}$$

Thus, the first equality in (3.14) holds.

Finally, using (3.1), we get

$$\begin{aligned} \int _{\mathbb {R}^d}\tau ^{-y}f(x)g(y)\,\mathrm{d}\mu _{k}(y)&=\int _{\mathbb {R}^d}g(y)\int _{\mathbb {R}^d}e_k(-y, z)e_k(x, z)\mathcal {F}_k(f)(z)\,\mathrm{d}\mu _k(z)\,\mathrm{d}\mu _{k}(y)\\&=\int _{\mathbb {R}^d}e_k(x, z)\mathcal {F}_k(f)(z)\mathcal {F}_k(g)(z)\,\mathrm{d}\mu _k(z), \end{aligned}$$

and the second part in (3.14) is valid. $\square$

Let $y\in \mathbb {R}^d$ be given. Rösler [42] proved that the operator $\tau ^y$ is positive on $C^{\infty }_\mathrm {rad}(\mathbb {R}^d)$, i.e., $\tau ^y\ge 0$, and moreover, for any $x\in \mathbb {R}^d$,

$$\begin{aligned} \tau ^yf(x)=\int _{\mathbb {R}^d}f(z)\,\mathrm{d}\rho _{x,y}^k(z), \end{aligned}$$

(3.16)

where $\rho _{x,y}^k$ is a radial probability Borel measure such that ${\text {supp}}\rho _{x,y}^k\subset B_{|x|+|y|}$.

Theorem 3.5

If $1\le p\le \infty$, then, for any $x\in \mathbb {R}^d$ and $f\in \mathcal {S}(\mathbb {R}^d)$,

$$\begin{aligned} \Vert T^tf(x)\Vert _{p,\mathrm{d}\nu _{\lambda _k}}= \left( \int _{\mathbb {R}_{+}}|T^tf(x)|^{p}\,\mathrm{d}\nu _{\lambda }(t)\right) ^{1/p}\le \Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(3.17)

Proof

Let $x\in \mathbb {R}^d$ be given. Let an operator $B^x$ be defined on $\mathcal {S}(\mathbb {R}^d)$ as follows (cf. (3.2) and (3.3)): for $f\in \mathcal {S}(\mathbb {R}^d)$,

$$\begin{aligned} B^xf(t)=T^tf(x)=\int _{\mathbb {R}^d}j_{\lambda _k}(t|y|)e_k(x, y)\mathcal {F}_k(f)(y)\,\mathrm{d}\mu _k(y)=\int _{\mathbb {R}^d}f(z)\,\mathrm{d}\sigma _{x,t}^k(z). \end{aligned}$$

Let $p=2$. We have

$$\begin{aligned} T^tf(x)=\int _{0}^{\infty }j_{\lambda _k}(tr)\int _{\mathbb {S}^{d-1}}e_k(x, ry')\mathcal {F}_k(f)(ry')\,\mathrm{d}\sigma _k(y')\,\mathrm{d}\nu _{\lambda _k}(r) \end{aligned}$$

and

$$\begin{aligned} \mathcal {H}_{\lambda _k}(T^tf(x))(r)=\int _{\mathbb {S}^{d-1}}e_k(x, ry')\mathcal {F}_k(f)(ry')\,\mathrm{d}\sigma _k(y'). \end{aligned}$$

This, Hölder’s inequality, and the fact that the operators $\mathcal {H}_{\lambda _k}$ and $\mathcal {F}_{k}$ are unitary imply

$$\begin{aligned} \Vert T^tf(x)\Vert _{2,\mathrm{d}\nu _{\lambda _k}}^2&=\Vert \mathcal {H}_{\lambda _k}(T^tf(x)) (r)\Vert _{2,\mathrm{d}\nu _{\lambda _k}}^2\\&=\int _{0}^{\infty }\left| \int _{\mathbb {S}^{d-1}}e_k(x, ry')\mathcal {F}_k(f)(ry')\,\mathrm{d}\sigma _k(y') \right| ^2\,\mathrm{d}\nu _{\lambda _k}(r)\\&\le \int _{0}^{\infty }\int _{\mathbb {S}^{d-1}}|\mathcal {F}_k(f)(ry')|^2\, \mathrm{d}\sigma _k(y')\,\mathrm{d}\nu _{\lambda _k}(r)\\&=\Vert \mathcal {F}_k(f)\Vert _{2,\mathrm{d}\mu _k}^2=\Vert f\Vert _{2,\mathrm{d}\mu _k}^2, \end{aligned}$$

which yields inequality (3.17) for $p=2$. Moreover, $B^x$ can be extended to the space $L^{2}(\mathbb {R}_+,\mathrm{d}\nu _{\lambda _k})$ with preservation of norm, and, moreover, this extension coincides with (3.2).

Let $p=1$. By (3.14) and (3.16), we obtain

$$\begin{aligned} \Vert T^tf(x)\Vert _{1,\mathrm{d}\nu _{\lambda _k}}&=\sup \left\{ \int _0^{\infty }T^tf(x)g_0(t)\, \mathrm{d}\nu _{\lambda _k}(t):g_0\in \mathcal {S}(\mathbb {R}_+),\ \Vert g_0\Vert _{\infty }\le 1\right\} \\&=\sup \left\{ \int _{\mathbb {R}^d}f(y)\tau ^{x}g(-y)\,\mathrm{d}\mu _{k}(y):g\in \mathcal {S}_\mathrm {rad}(\mathbb {R}^d),\ \Vert g\Vert _{\infty }\le 1\right\} \\&\le \Vert f\Vert _{1,\mathrm{d}\mu _k}\sup \left\{ \Vert \tau ^{x}g(-y)\Vert _{\infty }:g\in \mathcal {S}_\mathrm {rad}(\mathbb {R}^d),\ \Vert g\Vert _{\infty }\le 1\right\} \\&\le \Vert f\Vert _{1,\mathrm{d}\mu _k}, \end{aligned}$$

which is the desired inequality (3.17) for $p=1$. Moreover, $B^x$ can be extended to $L^{1}(\mathbb {R}_+,\mathrm{d}\nu _{\lambda _k})$ with preservation of norm such that the extension coincides with (3.2) on $L^{1}(\mathbb {R}_+,\mathrm{d}\nu _{\lambda _k})\cap L^{2}(\mathbb {R}_+,\mathrm{d}\nu _{\lambda _k})$.

By the Riesz–Thorin interpolation theorem we obtain (3.17) for $1<p<2$.

If $2<p<\infty$, $1/p+1/p'=1$, then by (3.14) and (3.13),

$$\begin{aligned} \Vert T^tf(x)\Vert _{p,\mathrm{d}\nu _{\lambda _k}}&=\sup \left\{ \int _0^{\infty }T^tf(x) g_0(t)\,\mathrm{d}\nu _{\lambda _k}(t):g_0\in \mathcal {S}(\mathbb {R}_+),\ \Vert g_0\Vert _{p',\mathrm{d}\nu _{\lambda _k}}\le 1\right\} \\&=\sup \left\{ \int _{\mathbb {R}^d}f(y)\tau ^{x}g(-y)\,\mathrm{d}\mu _{k}(y):g\in \mathcal {S}_\mathrm {rad}(\mathbb {R}^d),\ \Vert g\Vert _{p',\mathrm{d}\mu _k}\le 1\right\} \\&\le \Vert f\Vert _{p,\mathrm{d}\mu _k}\sup \{\Vert \tau ^{x}g(-y)\Vert _{p',\mathrm{d}\mu _k}:g\in \mathcal {S}_\mathrm {rad}(\mathbb {R}^d),\ \Vert g\Vert _{p',\mathrm{d}\mu _k}\le 1\}\\&\le \Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

Finally, for $p=\infty$, (3.17) follows from representation (3.3). $\square$

We are now in a position to prove the Young inequality for the convolutions (3.10) and (3.11).

Theorem 3.6

Let $1\le p, q\le \infty$, $\frac{1}{p}+\frac{1}{q}\ge 1$, and $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}-1$. We have that, for any $f\in \mathcal {S}(\mathbb {R}^d)$, $g_0\in \mathcal {S}(\mathbb {R}_+)$, and $g\in \mathcal {S}_\mathrm {rad}(\mathbb {R}^d)$,

$$\begin{aligned} \Vert (f*_{ \lambda _k}g_0)\Vert _{r,\mathrm{d}\nu _{\lambda _k}}\le & {} \Vert f\Vert _{p,\mathrm{d}\mu _k}\Vert g_0\Vert _{q,\mathrm{d}\nu _{\lambda _k}}, \end{aligned}$$

(3.18)

$$\begin{aligned} \Vert (f*_{ k}g)\Vert _{r,\mathrm{d}\mu _k}\le & {} \Vert f\Vert _{p,\mathrm{d}\mu _k}\Vert g\Vert _{q,\mathrm{d}\mu _k}. \end{aligned}$$

(3.19)

Proof

Since for $g(y)=g_0(|y|)$ we have

$$\begin{aligned} \Vert (f*_{ \lambda _k}g_0)\Vert _{r,\mathrm{d}\nu _{\lambda _k}}=\Vert (f*_{ k}g)\Vert _{r,\mathrm{d}\mu _k},\quad \Vert g_0\Vert _{q,\mathrm{d}\nu _{\lambda _k}}=\Vert g\Vert _{q,\mathrm{d}\mu _k}, \end{aligned}$$

it is enough to show inequality (3.18). The proof is straightforward using Hölder’s inequality and estimates (3.6) and (3.17). For the sake of completeness, we give it here. Let $\frac{1}{\mu }=\frac{1}{p}-\frac{1}{r}$ and $\frac{1}{\nu }=\frac{1}{q}-\frac{1}{r}$, then $\frac{1}{\mu }\ge 0$, $\frac{1}{\nu }\ge 0$, and $\frac{1}{r}+\frac{1}{\mu }+\frac{1}{\nu }=1$. In virtue of (3.17), we have

$$\begin{aligned}&\left| \int _0^{\infty }T^tf(x)g_0(t)\,\mathrm{d}\nu _{\lambda _k}(t)\right| \le \left( \int _0^{\infty }|T^tf(x)|^p|g_0(t)|^q\,\mathrm{d}\nu _{\lambda _k}(t)\right) ^{1/r}\\&\quad \quad \times \left( \int _0^{\infty }|T^tf(x)|^p\,\mathrm{d}\nu _{\lambda _k}(t)\right) ^{1/\mu } \left( \int _0^{\infty }|g_0(t)|^q\,\mathrm{d}\nu _{\lambda _k}(t)\right) ^{1/\nu }\\&\quad \le \left( \int _0^{\infty }|T^tf(x)|^p|g_0(t)|^q\,\mathrm{d}\nu _{\lambda _k}(t)\right) ^{1/r} \Vert f\Vert _{p,\mathrm{d}\mu _k}^{p/\mu }\Vert g_0\Vert _{q,\mathrm{d}\nu _{\lambda _k}}^{q/\nu }. \end{aligned}$$

Using (3.6), this gives

$$\begin{aligned}&\Vert (f*_{ \lambda _k}g_0)\Vert _{r,\mathrm{d}\nu _{\lambda _k}}\le \left( \int _{\mathbb {R}^d}\int _0^{\infty }|T^tf(x)|^p|g_0(t)|^q\,\mathrm{d}\nu _{\lambda _k}(t)\,\mathrm{d}\mu _k(x)\right) ^{1/r}\\&\quad \times \Vert f\Vert _{p,\mathrm{d}\mu _k}^{p/\mu }\Vert g_0\Vert _{q,\mathrm{d}\nu _{\lambda _k}}^{q/\nu }\le \Vert f\Vert _{p,\mathrm{d}\mu _k}\Vert g_0\Vert _{q,\mathrm{d}\nu _{\lambda _k}}. \end{aligned}$$

$\square$

Theorem 3.7

Let $1\le p\le \infty$ and $g\in \mathcal {S}_\mathrm {rad}(\mathbb {R}^d)$. We have that, for any $y\in \mathbb {R}^d$,

$$\begin{aligned} \Vert \tau ^{y}g\Vert _{p,\mathrm{d}\mu _k}\le \Vert g\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(3.20)

Remark 3.8

Since $\mathcal {S}(\mathbb {R}^d)$ is dense in $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $1\le p<\infty$, the operator $\tau ^{y}$ can be defined on $L^{p}_\mathrm {rad}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ so that (3.20) holds.

Proof

In the case $1\le p\le 2$, this result was proved in [48]. The case $p=\infty$ follows from (3.16).

Let $2<p<\infty$. Since $\mathcal {F}_k(g)$ is a radial function and

$$\begin{aligned} \tau ^{y}g(-x)&=\int _{\mathbb {R}^d}e_k(y, z)e_k(-x, z)\mathcal {F}_k(g)(z)\,\mathrm{d}\mu _k(z)\\&=\int _{\mathbb {R}^d}e_k(-y, z)e_k(x, z)\mathcal {F}_k(g)(z)\,\mathrm{d}\mu _k(z)=\tau ^{-y}g(x), \end{aligned}$$

using (3.19) for $r=\infty$, $q=p$, we obtain

$$\begin{aligned} \Vert \tau ^{-y}g\Vert _{p,\mathrm{d}\mu _k}&=\sup \left\{ \int _{\mathbb {R}^d}\tau ^{-y}g(x)f(x)\,\mathrm{d}\mu _{k}(x):f\in \mathcal {S}(\mathbb {R}^d),\ \Vert f\Vert _{p',\mathrm{d}\mu _k}\le 1\right\} \\&\le \sup \{\Vert (f*_{ k}g)(y)\Vert _{\infty ,\mathrm{d}\mu _k}:f\in \mathcal {S}(\mathbb {R}^d),\ \Vert f\Vert _{p',\mathrm{d}\mu _k}\le 1\}\le \Vert g\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

$\square$

Now we give an analogue of Lemma 3.4 for the case when $f\in L^{p}$.

Lemma 3.9

Let $1\le p\le \infty$, $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})\cap C_b(\mathbb {R}^d)\cap C^{\infty }(\mathbb {R}^d)$, $g_0\in \mathcal {S}(\mathbb {R}_{+})$, and $g(y)=g_0(|y|)$. Then, for any $x\in \mathbb {R}^d$,

$$\begin{aligned} (f*_{ \lambda _k}g_0)(x)=(f*_{ k}g)(x)\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})\cap C_b(\mathbb {R}^d)\cap C^{\infty }(\mathbb {R}^d), \end{aligned}$$

(3.21)

and, in the sense of tempered distributions,

$$\begin{aligned} \mathcal {F}_k(f*_{ \lambda _k}g_0)=\mathcal {F}_k(f*_{ k}g) =\mathcal {F}_k(f)\mathcal {F}_k(g). \end{aligned}$$

(3.22)

Proof

First, in light of (3.6) and (3.18), we note that the convolution (3.10) belongs to $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$. Moreover, (3.3) implies that it is in $C_b(\mathbb {R}^d)$.

Taking into account that $g\in \mathcal {S}(\mathbb {R}^d)$ and $(-\Delta _k)^re_k(\cdot , z)=|z|^{2r}e_k(\cdot , z),$ we have

$$\begin{aligned} (-\Delta _k)^r(f*_{ k}g)(x)=\int _{\mathbb {R}^d}f(y)\int _{\mathbb {R}^d}e_k(x, z)e_k(-y, z)|z|^{2r}\mathcal {F}_k(g)(z)\,\mathrm{d}\mu _{k}(z)\,\mathrm{d}\mu _{k}(y). \end{aligned}$$

Let us show that the integral converges uniformly in x. We have

$$\begin{aligned} \int _{\mathbb {R}^d}e_k(x, z)e_k(-y, z)|z|^{2r}\mathcal {F}_k(g)(z)\,\mathrm{d}\mu _{k}(z)=\tau ^xG(-y), \end{aligned}$$

where $G\in \mathcal {S}_\mathrm {rad}(\mathbb {R}^d)$ is such that $\mathcal {F}_k(G)(z)=|z|^{2r}\mathcal {F}_k(g)(z)$. Using Hölder’s inequality and (3.20), we get

$$\begin{aligned}&\left| \int _{\mathbb {R}^d}f(y)\int _{\mathbb {R}^d}e_k(x, z)e_k(-y, z)|z|^{2r}\mathcal {F}_k(g)(z)\,\mathrm{d}\mu _{k}(z)\,\mathrm{d}\mu _{k}(y)\right| \\&\quad =\left| \int _{\mathbb {R}^d}f(y)\tau ^xG(-y)\,\mathrm{d}\mu _{k}(y)\right| \le \Vert f\Vert _{p, \mathrm{d}\mu _{k}}\Vert \tau ^xG\Vert _{p', \mathrm{d}\mu _{k}}\le \Vert f\Vert _{p, \mathrm{d}\mu _{k}}\Vert G\Vert _{p', \mathrm{d}\mu _{k}}. \end{aligned}$$

Thus, convolution (3.11) belongs to $C^{\infty }(\mathbb {R}^d)$.

By Lemma 3.4, the equality in (3.21) holds for any function $f\in \mathcal {S}(\mathbb {R}^d)$. If $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $f_n\in \mathcal {S}(\mathbb {R}^d)$, and $f_n\rightarrow f$ in $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, then Minkowski’s inequality and (3.6) give

$$\begin{aligned} \Vert ((f-f_n)*_{ \lambda _k}g_0)\Vert _{p, \mathrm{d}\mu _{k}}\le \Vert f-f_n\Vert _{p, \mathrm{d}\mu _{k}}\,\Vert g_0\Vert _{1, \mathrm{d}\nu _{\lambda _{k}}}, \end{aligned}$$

(3.23)

while Hölder’s inequality and (3.20) imply

$$\begin{aligned} |((f-f_n)*_{ k}g)(x)|\le \Vert f-f_n\Vert _{p, \mathrm{d}\mu _{k}}\,\Vert g\Vert _{p', \mathrm{d}\mu _{k}}. \end{aligned}$$

By (3.23), there is a subsequence $\{n_k\}$ such that $(f_{n_k}*_{ \lambda _k}g_0)(x)\rightarrow (f*_{ \lambda _k}g_0)(x)$ a.e., therefore the relation $(f*_{ \lambda _k}g_0)(x)=(f*_{ k}g)(x)$ holds almost everywhere. Since both convolutions are continuous, it holds everywhere.

To prove the second equation of the lemma, we first remark that Lemma 3.4 implies that (3.22) holds pointwise for any $f\in \mathcal {S}(\mathbb {R}^d)$. In the general case, since $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $\, (f*_{ \lambda _k}g_0)\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, and $\mathcal {F}_k(g)\in \mathcal {S}(\mathbb {R}^d)$, the left- and right-hand sides of (3.22) are tempered distributions. Recall that the Dunkl transform of tempered distribution is defined by

$$\begin{aligned} \langle \mathcal {F}_k(f),\varphi \rangle =\langle f,\mathcal {F}_k(\varphi )\rangle ,\quad f\in \mathcal {S}'(\mathbb {R}^d),\quad \varphi \in \mathcal {S}(\mathbb {R}^d). \end{aligned}$$

Let $f_n\in \mathcal {S}(\mathbb {R}^d)$ and $f_n\rightarrow f$ in $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $\varphi \in \mathcal {S}(\mathbb {R}^d)$. Then

$$\begin{aligned} \langle \mathcal {F}_k((f-f_n)*_{ \lambda _k}g_0),\varphi \rangle= & {} \langle ((f-f_n)*_{ \lambda _k}g_0),\mathcal {F}_k(\varphi )\rangle ,\\ \langle \mathcal {F}_k(g)\mathcal {F}_k(f-f_n),\varphi \rangle= & {} \langle (f-f_n),\mathcal {F}_k(\mathcal {F}_k(g)\varphi )\rangle \end{aligned}$$

and

$$\begin{aligned} |\langle \mathcal {F}_k((f-f_n)*_{ \lambda _k}g_0),\varphi \rangle |\le & {} \Vert f-f_n\Vert _{p, \mathrm{d}\mu _{k}}\,\Vert g_0\Vert _{1, \mathrm{d}\nu _{\lambda _{k}}}\,\Vert \mathcal {F}_k(\varphi )\Vert _{p', \mathrm{d}\mu _{k}},\\ |\langle \mathcal {F}_k(g)\mathcal {F}_k(f-f_n),\varphi \rangle |\le & {} \Vert f-f_n\Vert _{p, \mathrm{d}\mu _{k}}\,\Vert \mathcal {F}_k(\mathcal {F}_k(g)\varphi )\Vert _{p', \mathrm{d}\mu _{k}}. \end{aligned}$$

Thus, the proof of (3.22) is now complete. $\square$

4 Boundedness of the Riesz Potential

Recall that $\lambda _k=d/2-1+\sum _{a\in R_+}k(a)$. For $0<\alpha <2\lambda _k+2$, the weighted Riesz potential $I_{\alpha }^kf$ is defined on $\mathcal {S}(\mathbb {R}^d)$ (see [49]) by

$$\begin{aligned} I_{\alpha }^kf(x)=\left( d_k^{\alpha }\right) ^{-1}\int _{\mathbb {R}^d}\tau ^{-y}f(x) \frac{1}{|y|^{2\lambda _k+2-\alpha }}\,\mathrm{d}\mu _k(y), \end{aligned}$$

where $d_k^{\alpha }=2^{-\lambda _k-1+\alpha }\Gamma (\alpha /2)/\Gamma (\lambda _k+1-\alpha /2)$. We have, in the sense of tempered distributions,

$$\begin{aligned} \mathcal {F}_k(I_{\alpha }^kf)(y)=|y|^{-\alpha }\mathcal {F}_k(f)(y). \end{aligned}$$

Using (2.2) and (3.2), we obtain

$$\begin{aligned} I_{\alpha }^kf(x)=\left( d_k^{\alpha }\right) ^{-1}\int _{0}^{\infty }T^{t}f(x) \frac{1}{t^{2\lambda _k+2-\alpha }}\,\mathrm{d}\nu _{\lambda _k}(t). \end{aligned}$$

(4.1)

To estimate the $L^p$-norm of this operator, we use the maximal function defined for $f\in \mathcal {S}(\mathbb {R}^d)$ as follows [48]:

$$\begin{aligned} M_kf(x)=\sup _{r>0}\frac{|(f*_{ k}\chi _{B_r})(x)|}{\int _{B_r}\,\mathrm{d}\mu _k}, \end{aligned}$$

where $\chi _{B_r}$ is the characteristic function of the Euclidean ball $B_r$ of radius r centered at 0.

Using (2.2), (3.2), and (3.14), we get

$$\begin{aligned} M_kf(x) = \sup _{r>0}\frac{\left| \int _{0}^{r}T^{t}f(x)\,\mathrm{d}\nu _{\lambda _k}(t)\right| }{\int _{0}^{r}\, \mathrm{d}\nu _{\lambda _k}}. \end{aligned}$$

It is proved in [48] that the maximal function is bounded on $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $1<p\le \infty$,

$$\begin{aligned} \Vert M_kf\Vert _{p, \mathrm{d}\mu _k}\lesssim \Vert f\Vert _{p, \mathrm{d}\mu _k}, \end{aligned}$$

(4.2)

and it is of weak type (1, 1); that is,

$$\begin{aligned} \int _{\{x:M_kf(x)>a\}}\,\mathrm{d}\mu _k\lesssim \frac{\Vert f\Vert _{1,\mathrm{d}\mu _{k}}}{a},\quad a>0. \end{aligned}$$

(4.3)

Theorem 4.1

If $1<p<q<\infty$, $0<\alpha <2\lambda _k+2$, $\frac{1}{p}-\frac{1}{q}=\frac{\alpha }{2\lambda _k+2}$, then

$$\begin{aligned} \Vert I_{\alpha }^kf\Vert _{q,\mathrm{d}\mu _k} \lesssim \Vert f\Vert _{p,\mathrm{d}\mu _k},\quad f\in \mathcal {S}(\mathbb {R}^d). \end{aligned}$$

(4.4)

The mapping $f\mapsto I_{\alpha }^kf$ is of weak type (1, q); that is,

$$\begin{aligned} \int _{\{x:|I_{\alpha }^kf(x)|>a\}}\,\mathrm{d}\mu _k\lesssim \left( \frac{\Vert f\Vert _{1,\mathrm{d}\mu _{k}}}{a}\right) ^q. \end{aligned}$$

(4.5)

Remark 4.2

In the case $k\equiv 0$, inequality (4.4) was proved by Soboleff [44] and Thorin [50] and the weighted inequality was studied by Stein and Weiss [46]. For the reflection group $G=\mathbb {Z}_2^d$, Theorem 4.1 was proved in [49]. The general case was obtained in [21]. We give another simple proof based on the $L^p$-boundedness of $T^t$ given in Theorem 3.5 and follow the proof given in [49] for $G=\mathbb {Z}_2^d$.

Remark 4.3

In Theorem 4.1, dealing with (4.4), we may assume that $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $1<p<\infty$, while proving (4.5), we may assume that $f\in L^{1}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$.

Proof

Let $R>0$ be fixed. We write (4.1) as sum of two terms,

$$\begin{aligned} I_{\alpha }^kf(x)&=\left( d_k^{\alpha }\right) ^{-1} \int _{0}^{R}T^{t}f(x)\frac{1}{t^{2\lambda _k+2-\alpha }}\,\mathrm{d}\nu _{\lambda _k}(t)\nonumber \\&\quad +\left( d_k^{\alpha }\right) ^{-1}\int _{R}^{\infty }T^{t}f(x)\frac{1}{t^{2\lambda _k+2-\alpha }}\,\mathrm{d}\nu _{\lambda _k}(t) =J_1+J_2. \end{aligned}$$

(4.6)

Integrating $J_1$ by parts, we obtain

$$\begin{aligned} d_k^{\alpha }J_1&=\int _{0}^{R}t^{-(2\lambda _k+2-\alpha )}\,d \left( \int _{0}^{t}T^{s}f(x)\,\mathrm{d}\nu _{\lambda _k}(s)\right) \nonumber \\&=R^{\alpha }\cdot R^{-(2\lambda _k+2)}\int _{0}^{R}T^{s}f(x)\,\mathrm{d}\nu _{\lambda _k}(s)\nonumber \\&\quad +(2\lambda _k+2-\alpha )\int _{0}^{R}t^{-(2\lambda _k+2)}\int _{0}^{t}T^{s}f(x)\,\mathrm{d}\nu _{\lambda _k}(s)\, t^{\alpha -1}\,\mathrm{d}t. \end{aligned}$$

(4.7)

Here we have used that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0+0}\varepsilon ^{\alpha }\cdot \varepsilon ^{-(2\lambda _k+2)}\int _{0}^{\varepsilon }T^{s}f(x)\,\mathrm{d}\nu _{\lambda _k}(s)=0, \end{aligned}$$

since

$$\begin{aligned} \varepsilon ^{\alpha }\cdot \varepsilon ^{-(2\lambda _k+2)} \left| \int _{0}^{\varepsilon }T^{s}f(x)\,\mathrm{d}\nu _{\lambda _k}(s)\right| \lesssim \varepsilon ^{\alpha }\sup _{\varepsilon >0}\frac{\left| \int _{0}^{\varepsilon }T^{t}f(x)\,\mathrm{d}\nu _{\lambda _k}(t)\right| }{\int _{0}^{\varepsilon }\,\mathrm{d}\nu _{\lambda _k}} =\varepsilon ^{\alpha }M_kf(x). \end{aligned}$$

In light of (4.7), we have

$$\begin{aligned} |J_1|\lesssim R^{\alpha }M_kf(x)+\int _{0}^{R}M_kf(x)t^{\alpha -1}\,\mathrm{d}t\lesssim R^{\alpha }M_kf(x). \end{aligned}$$

(4.8)

To estimate $J_2$, we use Hölder’s inequality, the relation $\frac{1}{p}-\frac{1}{q}=\frac{\alpha }{2\lambda _k+2}$, and (3.17):

$$\begin{aligned} |J_2|&\le \left( d_k^{\alpha }\right) ^{-1} \left( \int _{R}^{\infty }t^{-(2\lambda _k+2-\alpha )p'}\,\mathrm{d}\nu _{\lambda _k}(t)\right) ^{1/p'}\Vert T^tf(x)\Vert _{p, \mathrm{d}\nu _{\lambda _k}}\\&\lesssim R^{-(2\lambda _k+2)q}\Vert f\Vert _{p, \mathrm{d}\mu _k}. \end{aligned}$$

This, (4.6) and (4.8) yield

$$\begin{aligned} |I_{\alpha }^kf(x)|\lesssim R^{\alpha }M_kf(x)+R^{-(2\lambda _k+2)q}\Vert f\Vert _{p, \mathrm{d}\mu _k}, \end{aligned}$$

for any $R>0$. Choosing $R=\left( M_kf(x)/\Vert f\Vert _{p, \mathrm{d}\mu _k}\right) ^{-q/(2\lambda _k+2)}$ implies the inequality

$$\begin{aligned} |I_{\alpha }^kf(x)|\lesssim (M_kf(x))^{p/q}(\Vert f\Vert _{p, \mathrm{d}\mu _k})^{1-p/q} \end{aligned}$$

(4.9)

for any $1\le p<q$. Integrating (4.9) and using (4.2), we have

$$\begin{aligned} \Vert I_{\alpha }^kf\Vert _{q,\mathrm{d}\mu _k}\lesssim \Vert M_kf\Vert _{p, \mathrm{d}\mu _k}^{p/q}\Vert f\Vert _{p, \mathrm{d}\mu _k}^{1-p/q}\lesssim \Vert f\Vert _{p, \mathrm{d}\mu _k},\quad p>1. \end{aligned}$$

Finally, we use inequality (4.3) for the maximal function and inequality (4.9) with $p=1$ to obtain

$$\begin{aligned} \int _{\{x:|I_{\alpha }^kf(x)|>a\}}\,\mathrm{d}\mu _k\le \int _{\{x:(M_kf(x))^{1/q}(\Vert f\Vert _{1, \mathrm{d}\mu _k})^{1-1/q}\gtrsim a\}}\,\mathrm{d}\mu _k\lesssim \left( \frac{\Vert f\Vert _{1,\mathrm{d}\mu _{k}}}{a}\right) ^q. \end{aligned}$$

$\square$

5 Entire Functions of Exponential Type and Plancherel–Polya–Boas-Type Inequalities

Let $\mathbb {C}^d$ be the complex Euclidean space of d dimensions. Let also $z=(z_1,\dots ,z_d)\in \mathbb {C}^d$, $\mathrm {Im}\,z=(\mathrm {Im}\,z_1,\dots ,\mathrm {Im}\,z_d)$, and $\sigma >0$.

In this section, we define several classes of entire functions of exponential type and study their interrelations. Moreover, we prove the Plancherel–Polya–Boas-type estimates and the Paley–Wiener-type theorems. These classes will be used later to study the approximation of functions on $\mathbb {R}^d$ by entire functions of exponential type.

First, we define two classes of entire functions: $B_{p, k}^\sigma$ and $\widetilde{B}_{p, k}^\sigma$. We say that a function $f\in B_{p, k}^\sigma$ if $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ is such that its analytic continuation to $\mathbb {C}^d$ satisfies

$$\begin{aligned} |f(z)|\le c_{\varepsilon }e^{(\sigma +\varepsilon )|z|},\quad \forall \varepsilon >0,\ \forall z\in \mathbb {C}. \end{aligned}$$

The smallest $\sigma =\sigma _{f}$ in this inequality is called a spherical type of f. In other words, the class $B_{p, k}^\sigma$ is the collection of all entire functions of spherical type at most $\sigma$.

We say that a function $f\in \widetilde{B}_{p, k}^\sigma$ if $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ is such that its analytic continuation to $\mathbb {C}^d$ satisfies

$$\begin{aligned} |f(z)|\le c_{f}e^{\sigma |\mathrm {Im}\,z|},\quad \forall z\in \mathbb {C}^d. \end{aligned}$$

Historically, functions from $\widetilde{B}_{p, k}^\sigma$ were basic objects in the Dunkl harmonic analysis. It is clear that $\widetilde{B}_{p, k}^\sigma \subset B_{p, k}^\sigma$. Moreover, if $k\equiv 0$, then both classes coincide (see, e.g., [29]). Indeed, if $f\in B_{p, 0}^\sigma$, $1\le p<\infty$, then Nikol’skii’s inequality [31, 3.3.5]

$$\begin{aligned} \Vert f\Vert _{\infty }\le 2^\mathrm{d}\sigma ^{d/p}\Vert f\Vert _{p,\mathrm{d}\mu _0} \end{aligned}$$

and the inequality [31, 3.2.6]

$$\begin{aligned} \Vert f(\cdot +iy)\Vert _{\infty }\le e^{\sigma |y|}\Vert f\Vert _{\infty },\quad y\in \mathbb {R}^{d}, \end{aligned}$$

imply that, for $z=x+iy\in \mathbb {C}^{d}$,

$$\begin{aligned} |f(z)|\le 2^\mathrm{d}\sigma ^{d/p} \Vert f\Vert _{p,\mathrm{d}\mu _0}e^{\sigma |\mathrm {Im}\,z|}, \end{aligned}$$

i.e., $f\in \widetilde{B}_{p, 0}^\sigma$.

In fact, the classes $B_{p, k}^\sigma$ and $\widetilde{B}_{p, k}^\sigma$ coincide in the weighted case ($k\ne 0$) as well. To see that, it is enough to show that functions from $B_{p, k}^\sigma$ are bounded on $\mathbb {R}^d$.

Theorem 5.1

If $0<p<\infty$, then $B_{p, k}^\sigma =\widetilde{B}_{p, k}^\sigma$.

We will actually prove the more general statement. Let $m\in \mathbb {Z}_+,$$\alpha ^1,\dots ,\alpha ^m\in \mathbb {R}^d\setminus \{0\}$, $k_0\ge 0,$$k_1,\dots ,k_m>0$, and

$$\begin{aligned} v(x)=|x|^{k_0}\prod _{j=1}^m|\langle \alpha ^j,x\rangle |^{k_j} \end{aligned}$$

(5.1)

be the power weight. The Dunkl weight is a particular case of such weighted functions. The weighted function (5.1) arises in the study of the generalized Fourier transform (see, e.g., [3]).

Let $L^{p,v}(\mathbb {R}^d)$, $0<p<\infty$, be the space of complex-valued Lebesgue measurable functions f for which

$$\begin{aligned} \Vert f\Vert _{p,v}=\left( \int _{\mathbb {R}^d}|f(x)|^pv(x)\,\mathrm{d}x\right) ^{1/p}<\infty . \end{aligned}$$

Let $\varvec{\upsigma }=(\sigma _1,\dots ,\sigma _d)$, $\sigma _1,\dots ,\sigma _d>0$.

Again, let us define three anisotropic classes of entire functions: $B^{\varvec{\upsigma }}$, $B_{p,v}^{\varvec{\upsigma }}$, and $\widetilde{B}_{p,v}^{\varvec{\upsigma }}$.

We say that a function f defined on $\mathbb {R}^d$ belongs to $B^{\varvec{\upsigma }}$ if its analytic continuation to $\mathbb {C}^d$ satisfies

$$\begin{aligned} |f(z)|\le c_{\varepsilon }e^{(\sigma _1+\varepsilon )|z_1|+\dots +(\sigma _d+\varepsilon )|z_d|},\quad \forall \varepsilon >0,\ \forall z\in \mathbb {C}^d. \end{aligned}$$

We say that a function $f\in B_{p,v}^{\varvec{\upsigma }}$ if $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ is such that its analytic continuation to $\mathbb {C}^d$ belongs to $B^{\varvec{\upsigma }}$.

We say that a function $f\in \widetilde{B}_{p,v}^{\varvec{\upsigma }}$ if $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ is such that its analytic continuation to $\mathbb {C}^d$ satisfies

$$\begin{aligned} |f(z)|\le c_fe^{\sigma _1|\mathrm {Im}\,z_1|+\dots +\sigma _d|\mathrm {Im}\,z_d|},\quad \forall z\in \mathbb {C}^d. \end{aligned}$$

We will use the notation $L^p(\mathbb {R}^d)$, $\Vert \cdot \,\Vert _{p}$, $B_p^{\varvec{\upsigma }}$, and $\widetilde{B}_p^{\varvec{\upsigma }}$ in the case of the unit weight, i.e., $v\equiv 1$.

Theorem 5.2

If $0<p<\infty$, then

• (1) $B_{p,v}^{\varvec{\upsigma }}\subset B_{p}^{\varvec{\upsigma }}$,

• (2) $B_{p,v}^{\varvec{\upsigma }}=\widetilde{B}_{p,v}^{\varvec{\upsigma }}$,

• (3) $B_{p,v}^{\sigma }=\widetilde{B}_{p,v}^{\sigma }$.

Remark 5.3

• (i) Part (3) of Theorem 5.2 implies Theorem 5.1.

• (ii) Note that in some particular cases ($k_0=0$ and $p\ge 1$) a similar result was discussed in [23].

Parts (2) and (3) of Theorem 5.2 follow from (1). Indeed, the embedding in (1) implies that $B_{p,v}^{\varvec{\upsigma }}\subset B_{p}^{\varvec{\upsigma }}\subset B_{\infty }^{\varvec{\upsigma }}$. Hence, a function $f\in B_{p,v}^{\varvec{\upsigma }}$ is bounded on $\mathbb {R}^{d}$ and then $f\in \widetilde{B}_{p,v}^{\varvec{\upsigma }}$, which gives (2). Further, $B_{p,v}^{\sigma }\subset B_{p,v}^{\varvec{\upsigma }}$ holds, where $\varvec{\upsigma }=(\sigma ,\dots ,\sigma )\in \mathbb {R}_{+}^{d}$ since $|z|\le |z_{1}|+\dots +|z_{d}|$. Hence, similar to the above, we have $B_{p,v}^{\sigma }\subset B_{\infty }^{\varvec{\upsigma }}$ and (3) follows. Thus, to prove Theorem 5.2, it is sufficient to verify part (1).

The main difficulty to prove Theorem 5.2 is that the weight v(x) vanishes. In order to overcome this problem, we will first prove two-sided estimates of the $L^p$ norm of entire functions in terms of the weighted $l_p$ norm, $\left( \sum _{n} v(\lambda ^{(n)})|f(\lambda ^{(n)})|^p\right) ^{1/p}$, $0<p<\infty$, where v does not vanish at $\{\lambda ^{(n)}\}\subset \mathbb {R}^d$.

Such estimates are of their own interest. They generalize the Plancherel–Polya inequality [33, 6, Chapt. 6, 6.7.15]

$$\begin{aligned} \sum _{k\in \mathbb {Z}}|f(\lambda _k)|^p\le c(\delta ,\sigma ,p)\int _{-\infty }^{\infty }|f(x)|^p\,\mathrm{d}x,\quad 0<p<\infty , \end{aligned}$$

where $\lambda _k$ is an increasing sequence such that $\lambda _{k+1}-\lambda _k\ge \delta >0$, and f is an entire function of exponential type at most $\sigma$; and the Boas inequality [5, 6, Chapt. 10, 10.6.8],

$$\begin{aligned} \int _{-\infty }^{\infty }|f(x)|^p\,\mathrm{d}x\le C(\delta ,L,\sigma ,p)\sum _{k\in \mathbb {Z}}|f(\lambda _k)|^p,\quad 0<p<\infty , \end{aligned}$$

(5.2)

where, additionally, $\left| \lambda _k-\frac{\pi }{\sigma }\,k\right| \le L$ and the type of f is $<\sigma$.

We write $\varvec{\upsigma }'=(\sigma _1',\dots ,\sigma _d')<\varvec{\upsigma }=(\sigma _1,\dots ,\sigma _d)$ if $\sigma _1'<\sigma _1,\dots ,\sigma _d'<\sigma _d$. Let $n=(n_1,\dots ,n_d)\in \mathbb {Z}^d$ and $\lambda ^{(n)}:\mathbb {Z}^d\rightarrow \mathbb {R}^d$. In what follows, we consider the sequences of the following type:

$$\begin{aligned} \lambda ^{(n)}=(\lambda _1(n_1), \lambda _2(n_1, n_2),\dots , \lambda _d(n_1,\dots , n_d)), \end{aligned}$$

(5.3)

where $\lambda _{i}^{(n)}=\lambda _i(n_1,\dots ,n_i)$ are sequences increasing with respect to $n_i$, $i=1,\dots ,d$ for fixed $n_1,\dots ,n_{i-1}$.

Definition 5.4

We say that the sequence $\lambda ^{(n)}$ satisfies the separation condition$\Omega _\mathrm {sep}[\delta ]$, $\delta >0$, if, for any $n\in \mathbb {Z}^d$,

$$\begin{aligned} \lambda _i(n_1,\dots ,n_{i-1},n_i+1)-\lambda _i(n_1,\dots ,n_{i-1},n_i)\ge \delta ,\quad i=1,\dots ,d. \end{aligned}$$

Note that if the sequence $\lambda ^{(n)}$ satisfies the separation condition $\Omega _\mathrm {sep}[\delta ]$, then it also satisfies the condition $\inf _{n\ne m}|\lambda ^{(n)}-\lambda ^{(m)}|> 0$.

Definition 5.5

We say that the sequence $\lambda ^{(n)}$ satisfies the close-lattice condition$\Omega _\mathrm {lat}[\mathbf {a},L]$, $\mathbf {a}=(a_1,\dots ,a_d)>0$, $L>0$, if, for any $n\in \mathbb {Z}^d$,

$$\begin{aligned} \left| \lambda _i(n_1,\dots ,n_i)-\frac{\pi n_i}{a_i}\right| \le L,\quad i=1,\dots ,d. \end{aligned}$$

We start with the Plancherel–Polya-type inequality.

Theorem 5.6

Assume that $\lambda ^{(n)}$ satisfies the condition $\inf _{n\ne m}|\lambda ^{(n)}-\lambda ^{(m)}|> 0$. Then for $f\in B^{\varvec{\upsigma }}_p$, $0<p<\infty$, we have

$$\begin{aligned} \sum _{n\in \mathbb {Z}^d}|f(\lambda ^{(n)})|^p\lesssim \int _{\mathbb {R}^d}|f(x)|^p\,\mathrm{d}x. \end{aligned}$$

Proof

For simplicity, we prove this result for $d=2$. The proof in the general case is similar.

The function $|f(z)|^p$ is plurisubharmonic, and therefore for any $x=(x_1, x_2)\in \mathbb {R}^2$, one has [38]

$$\begin{aligned} |f(x_1,x_2)|^p\le \frac{1}{(2\pi )^2}\int _0^{2\pi }\int _0^{2\pi }|f(x_1+\rho _1e^{i\theta _1},\, x_2+\rho _2e^{i\theta _2}|^p\,\mathrm{d}\theta _1 \mathrm{d}\theta _2, \end{aligned}$$

where $\rho _1,\,\rho _2>0$. Following [31, 3.2.5], for $\delta >0$ and $\xi +i\eta =(\xi _1+i\eta _1, \xi _2+i\eta _2)$, we obtain that

$$\begin{aligned} |f(x_1,x_2)|^p\le \frac{1}{(\pi \delta ^2)^2}\int _{-\delta }^{\delta }\int _{-\delta }^{\delta } \int _{x_1-\delta }^{x_1+\delta }\int _{x_2-\delta }^{x_2+\delta }|f(\xi +i\eta )|^p\,\mathrm{d}\xi _1\,\mathrm{d}\xi _2\,\mathrm{d}\eta _1\,\mathrm{d}\eta _2. \end{aligned}$$

(5.4)

The separation condition implies that for some $\delta >0$, the boxes $[\lambda _1^{(n)}-\delta , \lambda _1^{(n)}+\delta ]\times [\lambda _2^{(n)}-\delta , \lambda _2^{(n)}+\delta ]$ do not overlap for any n.

Since

$$\begin{aligned} f(x+iy)=\sum _{k\in \mathbb {Z}_+^2}\frac{f^{(k)}(x)}{k!}\,(iy)^{k}, \end{aligned}$$

where $f^{(k)}$ is a partial derivative f of order $k=(k_1,k_2)$, $k!=k_1!\,k_2!$, and $(iy)^k=(iy_1)^{k_{1}}(iy_{2})^{k_{2}}$, by applying Bernstein’s inequality (see [31, 3.2.2 and 3.3.5] and [37]), we derive that

$$\begin{aligned} \Vert f(\cdot +iy)\Vert _{p}\lesssim e^{\sigma _1|y_1|+\sigma _2|y_2|}\Vert f\Vert _{p}. \end{aligned}$$

Using this and (5.4), we derive that

$$\begin{aligned} \sum _{n\in \mathbb {Z}^2}|f(\lambda ^{(n)})|^p&\le \frac{1}{(\pi \delta ^2)^2}\int _{-\delta }^{\delta }\int _{-\delta }^{\delta } \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }|f(\xi +i\eta )|^p\,\mathrm{d}\xi _1\,\mathrm{d}\xi _2\,\mathrm{d}\eta _1\,\mathrm{d}\eta _2\\&\lesssim \int _{-\delta }^{\delta }\int _{-\delta }^{\delta }e^{p(\sigma _1|\eta _1|+\sigma _2|\eta _d|)}\,\mathrm{d}\eta _1\,\mathrm{d}\eta _2 \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }|f(\xi )|^p\,\mathrm{d}\xi _1\,\mathrm{d}\xi _2\\&\lesssim \int _{\mathbb {R}^2}|f(x)|^p\,\mathrm{d}x. \end{aligned}$$

$\square$

Theorem 5.7

Let the sequence $\lambda ^{(n)}$ of form (5.3) satisfy the conditions $\Omega _\mathrm {sep}[\delta ]$ and $\Omega _\mathrm {lat}[\varvec{\upsigma },L]$. Assume that $f\in B^{\varvec{\upsigma }'}$, $\varvec{\upsigma }'<\varvec{\upsigma }$, is such that $\sum _{n\in \mathbb {Z}^d}|f(\lambda ^{(n)})|^p<\infty$, $0<p<\infty$. Then $f\in L^p(\mathbb {R}^d)$ and

$$\begin{aligned} \int _{\mathbb {R}^d}|f(x)|^p\,\mathrm{d}x\lesssim \sum _{n\in \mathbb {Z}^d}|f(\lambda ^{(n)})|^p. \end{aligned}$$

Remark 5.8

For $p\ge 1$, a similar two-sided Plancherel–Polya–Boas-type inequality was obtained from [32].

Proof

For simplicity, we consider the case $d=2$. Integrating $|f(x_1,x_2)|^p$ at $x_1$ and applying inequality (5.2), we get, for any $x_2$,

$$\begin{aligned} \int _{-\infty }^{\infty }|f(x_1,x_2)|^p\,\mathrm{d}x_1\lesssim \sum _{n_1\in \mathbb {Z}} |f(\beta _1(n_1),x_{2})|^p. \end{aligned}$$

Since by (5.2), for any $n_1$,

$$\begin{aligned} \int _{-\infty }^{\infty }|f(\beta _1(n_1),x_2)|^p\,\mathrm{d}x_2\lesssim \sum _{n_2\in \mathbb {Z}} |f(\beta _1(n_1),\beta _2(n_1, n_2))|^p, \end{aligned}$$

we then have

$$\begin{aligned} \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }|f(x_1,x_2)|^p\,\mathrm{d}x_1\,\mathrm{d}x_2&\lesssim \sum _{n_1\in \mathbb {Z}}\int _{-\infty }^{\infty }|f(\beta _1(n_1),x_2)|^p\,\mathrm{d}x_2\\&\lesssim \sum _{n_1\in \mathbb {Z}}\sum _{n_2\in \mathbb {Z}} |f(\beta _1(n_1),\beta _2(n_1, n_2))|^p<\infty . \end{aligned}$$

$\square$

Using Theorems 5.6 and 5.7 we arrive at the following statement:

Theorem 5.9

Let the sequence $\{\lambda ^{(n)}\}$ of form (5.3) satisfy the conditions $\Omega _\mathrm {sep}[\delta ]$ and $\Omega _\mathrm {lat}[\varvec{\upsigma },L]$. If $f\in B^{\varvec{\upsigma }'}$, $\varvec{\upsigma }'<\varvec{\upsigma }$, then, for $0<p<\infty$,

$$\begin{aligned} \sum _{n\in \mathbb {Z}^d}|f(\lambda ^{(n)})|^p \lesssim \int _{\mathbb {R}^d}|f(x)|^p\,\mathrm{d}x\lesssim \sum _{n\in \mathbb {Z}^d}|f(\lambda ^{(n)})|^p. \end{aligned}$$

We will need the weighted version of the Plancherel–Polya–Boas equivalence. We start with three auxiliary lemmas.

Lemma 5.10

[18] If $\gamma \ge -1/2$, then there exists an even entire function $\omega _\gamma (z)$, $z\in \mathbb {C}$, of exponential type 2 such that, uniformly in $x\in \mathbb {R}_+$,

$$\begin{aligned} \omega _\gamma (x)\asymp {\left\{ \begin{array}{ll}x^{2k+2},&{} 0 \le x\le 1,\\ x^{2\gamma +1},&{} x\ge 1,\end{array}\right. } \end{aligned}$$

where $k=[\gamma +1/2]$ and [a] is the integral part of a. In particular, we can take

$$\begin{aligned} \omega (z)=z^{2k+2}j_{k-\gamma }(z+i)j_{k-\gamma }(z-i). \end{aligned}$$

Lemma 5.11

Let $m\in \mathbb {N}$, $j=1,\dots ,m$, $b^j=(b_1^j,\dots ,b_d^j)\in \mathbb {R}^d\setminus \{0\}$, and either $|b_i^j|\ge 1$, or $b_i^j=0$, $i=1,\dots ,d$. Then there exists a sequence $\{\rho ^{(n)}\}\subset \mathbb {Z}^d\setminus \{0\}$ of the form (5.3) such that, for any $j=1,\dots ,m$ and $i=1,\dots ,d$,

$$\begin{aligned} |\rho _i(n_1,\dots ,n_i)-n_i|\le & {} m, \end{aligned}$$

(5.5)

$$\begin{aligned} |\langle b^j, \rho ^{(n)}\rangle |\ge & {} 1/2. \end{aligned}$$

(5.6)

Proof

To construct a desired sequence

$$\begin{aligned} \rho ^{(n)}=(\rho _1(n_1), \rho _2(n_1, n_2),\dots ,\rho _d(n_1,\dots ,n_d))\in \mathbb {Z}^d, \end{aligned}$$

we will use the following simple remark. If we throw out m points from $\mathbb {Z}$, then the rest can be numbered such that the obtained sequence will be increasing and (5.5) holds.

Let $J_1=\{j:b_1^j\ne 0,\, b_2^j=\dots =b_d^j=0\}$. If $J_1=\varnothing$, then we set $\rho _1(n_1)=n_1$. If $J_1\ne \varnothing$, then $\rho _1(n_1)$ is an increasing sequence formed from $\mathbb {Z}\setminus \{0\}$. In both cases (5.5) is valid, and, moreover, for $j\in J_1$ and any $\rho _2(n_1,n_2),\dots ,\rho _d(n_1,\dots ,n_d)$, one has (5.6) since $|\langle b^j, \rho ^{(n)}\rangle |=|b_1^j\rho _1(n_1)|\ge 1$.

Let $J_2=\{j:b_2^j\ne 0,\, b_3^j=\dots =b_d^j=0\}$, $n_1\in \mathbb {Z}$. If $J_2=\varnothing$, then we set $\rho _2(n_1, n_2)=n_2$. Let $J_2\ne \varnothing$. If $j\in J_2$ and $b_1^j\rho _1(n_1)+b_2^jt_j=0$, then $t_j=l_j+\varepsilon _j$, $l_j\in \mathbb {Z}$, $|\varepsilon _j|\le 1/2$. Here $l_j$ is the nearest integer to $t_j$. Note that if $\rho _2\ne l_j$, then $|b_1^j\rho _1(n_1)+b_2^j\rho _2|=|b_2^j(\rho _2-l_j-\varepsilon _j)|\ge 1/2$.

Let $\rho _2(n_1, n_2)$ be an increasing sequence at $n_2$ formed from $\mathbb {Z}\backslash \{l_j:j\in J_2\}$. For this sequence (5.5) holds and, for $j\in J_2$ and any $\rho _3(n_1,n_2,n_3),\dots ,\rho _d(n_1,\dots ,n_d)$, one has

$$\begin{aligned} |\langle b^j, \rho ^{(n)}\rangle |=|b_1^j\rho _1(n_1)+b_2^j\rho _2(n_1, n_2)|\ge 1/2; \end{aligned}$$

that is, (5.6) holds as well.

Assume that we have constructed the sets $J_1,\dots ,J_{d-1}$, and the sequence $(\rho _1(n_1), \rho _2(n_1, n_2),\dots ,\rho _{d-1}(n_1,\dots ,n_{d-1}))\in \mathbb {Z}^{d-1}$.

Let $J_d=\{j:b_d^j\ne 0\}$, $(n_1,\dots ,n_{d-1})\in \mathbb {Z}^{d-1}$. If $J_d=\varnothing$, then we set $\rho _d(n_1,\dots ,n_{d-1},n_d)=n_d$. Assume now that $J_d\ne \varnothing$. If $j\in J_d$ and

$$\begin{aligned} b_1^j\rho _1(n_1)+\cdots +b_{d-1}^j\rho _{d-1}(n_1,\dots ,n_{d-1})+b_d^jt_j=0, \end{aligned}$$

then $t_j=l_j+\varepsilon _j$, $|\varepsilon _j|\le 1/2$. Note that if $\rho _d\ne l_j$, then

$$\begin{aligned} |b_1^j\rho _1(n_1)+\cdots +b_{d-1}^j\rho _{d-1}(n_1,\dots ,n_{d-1})+b_d^j\rho _d|=|b_d^j(\rho _d-l_j-\varepsilon _j)|\ge 1/2. \end{aligned}$$

Let $\rho _d(n_1,\dots ,n_d)$ be an increasing sequence in $n_d$ formed from $\mathbb {Z}\setminus \{l_j:j\in J_d\}$, $\rho ^{(n)}=(\rho _1(n_1), \rho _2(n_1, n_2),\dots ,\rho _d(n_1,\dots ,n_d))$. For the sequence $\rho _d(n_1,\dots ,n_d)$, inequality (5.5) holds, and, for $j\in J_d$, one has $|\langle b^j, \rho ^{(n)}\rangle |\ge 1/2$.

Thus, we construct the desired sequence since, for any $j\in \{1,\dots ,m\}$ and some $i\in \{1,\dots ,d\}$, $b^j\in J_i$ holds. $\square$

An important ingredient of the proof of Theorem 5.2 is the following corollary of Lemma 5.11:

Lemma 5.12

If $\mathbf {a}>0$, $\alpha ^1,\dots ,\alpha ^m\in \mathbb {R}^d\setminus \{0\}$, then there exists a sequence $\lambda ^{(n)}$ of the form (5.3) such that for some $\delta ,\,L>0$ the conditions $\Omega _\mathrm {sep}[\delta ]$, $\Omega _\mathrm {lat}[\mathbf {a}, L]$, and $\xi _{j}(\lambda ^{(n)})\ge \delta$, $j=0,1,\dots ,m$, $n\in \mathbb {Z}^d$, hold, where

$$\begin{aligned} \xi _0(x)=|x|,\quad \xi _j(x)=|\langle \alpha ^j,x\rangle |,\ j=1,\dots ,m. \end{aligned}$$

(5.7)

Indeed, for $m\ge 1$, it is enough to define

$$\begin{aligned} \lambda ^{(n)}&=(\lambda _1(n_1), \lambda _2(n_1, n_2),\dots , \lambda _d(n_1,\dots , n_d))\nonumber \\&:=\left( \frac{\pi \rho _1(n_1)}{a_1}, \frac{\pi \rho _2(n_1, n_2)}{a_2},\dots , \frac{\pi \rho _d(n_1,\dots , n_d)}{a_d}\right) , \end{aligned}$$

(5.8)

where $\rho ^{(n)}$ is the sequence defined in Lemma 5.11. For $m=0$ in (5.8), we can take $\{\rho ^{(n)}\}=\mathbb {Z}^{d}\setminus \{0\}$.

We are now in a position to state the Plancherel–Polya–Boas inequalities with weights.

Theorem 5.13

Let $f\in B^{\varvec{\upsigma }}$ and $\lambda ^{(n)}$ be the sequence satisfying all conditions of Lemma 5.12 with some $\mathbf {a}>\varvec{\upsigma }$. Then, for $0<p<\infty$,

$$\begin{aligned} \sum _{n\in \mathbb {Z}^d}v(\lambda ^{(n)})|f(\lambda ^{(n)})|^p \lesssim \int _{\mathbb {R}^d}|f(x)|^pv(x)\,\mathrm{d}x\lesssim \sum _{n\in \mathbb {Z}^d}v(\lambda ^{(n)})|f(\lambda ^{(n)})|^p. \end{aligned}$$

Proof

Recall that $v(x)=\prod _{j=0}^mv_j(x)$, where $v_j(x)=\xi _j^{k_j}(x)$, $j=0,1,\dots ,m$ (see (5.1) and (5.7)).

By Lemma 5.10, we construct an entire function of exponential type

$$\begin{aligned} w(z)=\prod _{j=0}^m w_j(z), \end{aligned}$$

where $w_0(z)=\omega _{\gamma _{0}}(|z|)$, $w_j(z)=\omega _{\gamma _j}(\langle \alpha ^j, z\rangle )$, $j=1,\dots ,m$, and

$$\begin{aligned} \gamma _j=\frac{k_j}{2p}-\frac{1}{2},\quad j=0,1,\dots ,m. \end{aligned}$$

For $j=0,1,\dots ,m$, we have $w_j\in B^{2\varvec{\upmu }^j}$, where

$$\begin{aligned} \varvec{\upmu }^0=(1,\dots ,1)\in \mathbb {R}^{d},\quad \varvec{\upmu }^j= \left( \left| \alpha _1^j\right| ,\dots ,\left| \alpha _d^j\right| \right) ,\ j=1,\dots ,m, \end{aligned}$$

and $w\in B^{2\varvec{\upmu }}$, $\varvec{\upmu }=\sum _{j=0}^m\varvec{\upmu }^j$. Moreover, for any $j=0,1,\dots ,m$,

$$\begin{aligned} \begin{aligned} w_j^p(x)&\lesssim v_j(x),\quad x\in \mathbb {R}^d,\\ w_j^p(x)&\gtrsim v_j(x)\gtrsim 1,\quad \text {for}\quad \xi _j(x)\ge \delta >0. \end{aligned} \end{aligned}$$

(5.9)

Let $f\in B^{\varvec{\upsigma }}_{p,v}$, $0<p<\infty$, $\varvec{\upsigma }<\mathbf {a}$, and $\lambda ^{(n)}$ be the sequence satisfying all conditions of Lemma 5.12. Then, for some $s>0$ such that $\varvec{\upsigma }+2s\varvec{\upmu }<\mathbf {a}$, we have that $f(x)w(sx)\in B^{\varvec{\upsigma }+2s\varvec{\upmu }}$.

Using Theorem 5.6 and properties (5.9), we derive

$$\begin{aligned} \sum _{n\in \mathbb {Z}^d}v(\lambda ^{(n)})|f(\lambda ^{(n)})|^p&\lesssim \sum _{n\in \mathbb {Z}^d}|f(\lambda ^{(n)})w(\lambda ^{(n)})|^p\\&\lesssim \int _{\mathbb {R}^d}|f(x)w(x)|^p\,\mathrm{d}x\lesssim \int _{\mathbb {R}^d}|f(x)|^pv(x)\,\mathrm{d}x. \end{aligned}$$

Let $\delta >0$, $J\subset J_{m}:=\{0,1,\dots ,m\}$ or $J=\varnothing$,

$$\begin{aligned} E_{\delta }(J)=\{x\in \mathbb {R}^d:\xi _{j}(x)\ge \delta ,\ j\in J\ \text {and}\ \xi _{j}(x)\le \delta ,\ j\in J_{m}\setminus J\}. \end{aligned}$$

Since $f(x)\prod _{j\in J}w_j(sx)\in B^{\varvec{\upsigma }+2s\varvec{\upmu }}$, using Theorems 5.6 and 5.7 and properties (5.9) for $\delta$ from Lemma 5.12, we obtain

$$\begin{aligned} \int _{\mathbb {R}^d}|f(x)|^pv(x)\,\mathrm{d}x&=\sum _{J}\int _{E_{\delta }(J)}| f(x)|^pv(x)\,\mathrm{d}x\lesssim \sum _{J}\int _{E_{\delta }(J)}|f(x)|^p\prod _{j\in J}v_j(sx)\,\mathrm{d}x\\&\lesssim \sum _{J}\int _{E_{\delta }(J)}\left| f(x)\prod _{j\in J}w_j(sx)\right| ^p\,\mathrm{d}x\lesssim \sum _{J}\int _{\mathbb {R}^d}\left| f(x)\prod _{j\in J}w_j(sx)\right| ^p\,\mathrm{d}x\\&\lesssim \sum _{n}|f(\lambda ^{(n)})|^p\sum _{J}\prod _{j\in J}w_j^p(s\lambda ^{(n)}) \lesssim \sum _{n}|f(\lambda ^{(n)})w(s\lambda ^{(n)})|^p\\&\lesssim \sum _{n}|f(\lambda ^{(n)})|^pv(s\lambda ^{(n)}) \lesssim \sum _{n}|f(\lambda ^{(n)})|^pv(\lambda ^{(n)}), \end{aligned}$$

where we have assumed that $\prod _{j\in \varnothing }=1$. $\square$

Proof of Theorem 5.2

Recall that it is enough to show that $B_{p,v}^{\varvec{\upsigma }}\subset B_{p}^{\varvec{\upsigma }}$, and the latter follows from $B^{\varvec{\upsigma }}_{p,v}\subset L^{p}(\mathbb {R}^d)$.

Let $f\in B^{\varvec{\upsigma }}_{p,v}$, $0<p<\infty$, $\mathbf {a}>\varvec{\upsigma }$, and $\lambda ^{(n)}$ be the sequence satisfying all conditions of Lemma 5.12. Using Theorem 5.6 and properties (5.9) as in Theorem 5.13, we have

$$\begin{aligned} \int _{\mathbb {R}^d}|f(x)|^p\,\mathrm{d}x&\lesssim \sum _{n\in \mathbb {Z}^d}|f(\lambda ^{(n)})|^p\lesssim \sum _{n\in \mathbb {Z}^d}|w(\lambda ^{(n)})f(\lambda ^{(n)})|^p\\&\lesssim \int _{\mathbb {R}^d}|f(x)w(x)|^p\,\mathrm{d}x\lesssim \int _{\mathbb {R}^d}|f(x)|^pv(x)\,\mathrm{d}x. \end{aligned}$$

$\square$

By the Paley–Wiener theorem for tempered distributions (see [25, 53]) and Theorem 5.1, we arrive at the following result.

Theorem 5.14

A function $f\in B_{p, k}^\sigma$, $1\le p<\infty$, if and only if $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})\cap C_b(\mathbb {R}^d)$ and ${\text {supp}}\mathcal {F}_k(f)\subset B_\sigma$.

The Dunkl transform $\mathcal {F}_k(f)$ in Theorem 5.14 is understood as a function for $1\le p\le 2$ and as a tempered distribution for $p>2$.

We conclude this section by presenting the concept of the best approximation. Let

$$\begin{aligned} E_{\sigma }(f)_{p,\mathrm{d}\mu _k}=\inf \{\Vert f-g\Vert _{p,\mathrm{d}\mu _k}:g\in B_{p, k}^\sigma \} \end{aligned}$$

be the best approximation of a function $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ by entire functions of spherical exponential type $\sigma$. We show that the best approximation is achieved.

Theorem 5.15

For any $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, $1\le p\le \infty$, there exists a function $g^{*}\in B_{p, k}^\sigma$ such that $E_{\sigma }(f)_{p,\mathrm{d}\mu _k}=\Vert f-g^{*}\Vert _{p,\mathrm{d}\mu _k}$.

Proof

The proof is standard. Let $g_n$ be a sequence from $B_{p, k}^\sigma$ such that $\Vert f-g_n\Vert _{p,\mathrm{d}\mu _k}\rightarrow E_{\sigma }(f)_{p,\mathrm{d}\mu _k}$. Since it is bounded in $L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$, it is also bounded in $C_b(\mathbb {R}^d)$. A compactness theorem for entire functions [31, 3.3.6] implies that there exist a subsequence $g_{n_k}$ and an entire function $g^{*}$ of exponential type at most $\sigma$ such that

$$\begin{aligned} \lim _{k\rightarrow \infty }g_{n_k}(x)=g^{*}(x),\quad x\in \mathbb {R}^d, \end{aligned}$$

and, moreover, convergence is uniform on compact sets. Therefore, for any $R>0$,

$$\begin{aligned} \Vert g^{*}\chi _{B_R}\Vert _{p,\mathrm{d}\mu _k}=\lim _{k\rightarrow \infty }\Vert g_{n_k} \chi _{B_R}\Vert _{p,\mathrm{d}\mu _k}\le M. \end{aligned}$$

Letting $R\rightarrow \infty$, we have that $g^{*}\in B_{p, k}^\sigma$. In light of

$$\begin{aligned} \Vert (f-g^{*})\chi _{B_R}\Vert _{p,\mathrm{d}\mu _k}&=\lim _{k\rightarrow \infty }\Vert (f-g_{n_k}) \chi _{B_R}\Vert _{p,\mathrm{d}\mu _k}\\&\le \lim _{k\rightarrow \infty }\Vert f-g_{n_k}\Vert _{p,\mathrm{d}\mu _k}=E_{\sigma }(f)_{p,\mathrm{d}\mu _k}, \end{aligned}$$

we have $\Vert f-g^{*}\Vert _{p,\mathrm{d}\mu _k}\le E_{\sigma }(f)_{p,\mathrm{d}\mu _k}$. $\square$

6 Jackson’s Inequality and Equivalence of Modulus of Smoothness and K-Functional

6.1 Smoothness Characteristics and K-Functional

We define the r-th power of the Dunkl Laplacian as a tempered distribution:

$$\begin{aligned} \langle (-\Delta _k)^rf,\varphi \rangle =\langle f,(-\Delta _k)^r\varphi \rangle ,\quad f\in \mathcal {S}'(\mathbb {R}^d),\quad \varphi \in \mathcal {S}(\mathbb {R}^d),\quad r\in \mathbb {N}. \end{aligned}$$

The Dunkl Laplacian can also be written in terms of the Dunkl transform

$$\begin{aligned} (-\Delta _k)^rf=\mathcal {F}_k^{-1}(|\cdot |^{2r}\mathcal {F}_k(f)). \end{aligned}$$

(6.1)

Let $W^{2r}_{p, k}$ be the Sobolev space, that is,

$$\begin{aligned} W^{2r}_{p, k}=\{f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k}):(-\Delta _k)^rf\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})\} \end{aligned}$$

equipped with the Banach norm

$$\begin{aligned} \Vert f\Vert _{W^{2r}_{p,k}}=\Vert f\Vert _{p,\mathrm{d}\mu _k}+\Vert (-\Delta _k)^rf\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

Note that $(-\Delta _k)^rf\in \mathcal {S}(\mathbb {R}^d)$ whenever $f\in \mathcal {S}(\mathbb {R}^d)$ and $\mathcal {S}(\mathbb {R}^d)$ is dense in $W^{2r}_{p, k}$. Indeed, if $f\in W^{2r}_{p, k}$, defining

$$\begin{aligned} \Phi (x)=e^{-|x|^2/2},\quad \Phi _{\varepsilon }(x)=\frac{1}{\varepsilon ^{2\lambda _k+2}}\, \Phi \left( \frac{x}{\varepsilon }\right) , \end{aligned}$$

we obtain that $(f*_{ k}\Phi _{\varepsilon })\in \mathcal {S}(\mathbb {R}^d)$ and (see [48])

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\Vert f-(f*_{ k}\Phi _{\varepsilon })\Vert _{p,\mathrm{d}\mu _k}=\lim _{\varepsilon \rightarrow 0}\Vert (-\Delta _k)^rf-((-\Delta _k)^rf*_{ k}\Phi _{\varepsilon })\Vert _{p,\mathrm{d}\mu _k}=0. \end{aligned}$$

Define the K-functional for the couple $(L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k}), W^{2r}_{p, k})$ as follows:

$$\begin{aligned} K_{2r}(t, f)_{p,\mathrm{d}\mu _k}=\inf \{\Vert f-g\Vert _{p,\mathrm{d}\mu _k}+t^{2r}\Vert (-\Delta _k)^rg\Vert _{p,\mathrm{d}\mu _k}:g\in W^{2r}_{p, k}\}. \end{aligned}$$

Note that for any $f_1,f_2\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ and $g\in W^{2r}_{p, k}$, we have

$$\begin{aligned}&\Vert f_1-g\Vert _{p,\mathrm{d}\mu _k}+t^{2r}\Vert (-\Delta _k)^rg\Vert _{p,\mathrm{d}\mu _k}\\&\quad \le \Vert f_2-g\Vert _{p,\mathrm{d}\mu _k}+t^{2r}\Vert (-\Delta _k)^rg\Vert _{p,\mathrm{d}\mu _k}+\Vert f_1-f_2\Vert _{p,\mathrm{d}\mu _k}, \end{aligned}$$

and hence,

$$\begin{aligned} |K_{2r}(t, f_1)_{p,\mathrm{d}\mu _k}-K_{2r}(t, f_2)_{p,\mathrm{d}\mu _k}|\le \Vert f_1-f_2\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.2)

If $f\in W^{2r}_{p, k}$, then $K_{2r}(t, f)_{p,\mathrm{d}\mu _k}\le t^{2r}\Vert (-\Delta _k)^rf\Vert _{p,\mathrm{d}\mu _k}$ and $\lim _{t\rightarrow 0}K_{2r}(t, f)_{p,\mathrm{d}\mu _k}=0$. This and (6.2) imply that, for any $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$,

$$\begin{aligned} \lim _{t\rightarrow 0}K_{2r}(t, f)_{p,\mathrm{d}\mu _k}=0. \end{aligned}$$

(6.3)

Another important property of the K-functional is

$$\begin{aligned} K_{2r}(\lambda t, f)_{p,\mathrm{d}\mu _k}\le \max \{1,\,\lambda ^{2r}\}K_{2r}(t, f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.4)

Let I be an identical operator and $m\in \mathbb {N}$. Consider the following three differences:

$$\begin{aligned} \varDelta _t^mf(x)=(I-T^t)^mf(x)= & {} \sum _{s=0}^m(-1)^s\left( {\begin{array}{c}m\\ s\end{array}}\right) (T^{t})^sf(x), \end{aligned}$$

(6.5)

$$\begin{aligned} {^{*}}\varDelta _t^mf(x)= & {} \sum _{s=0}^m(-1)^s\left( {\begin{array}{c}m\\ s\end{array}}\right) T^{st}f(x), \end{aligned}$$

(6.6)

$$\begin{aligned} {^{**}}\varDelta _t^mf(x)= & {} \left( {\begin{array}{c}2m\\ m\end{array}}\right) ^{-1}\sum _{s=-m}^m(-1)^s \left( {\begin{array}{c}2m\\ m-s\end{array}}\right) T^{st}f(x). \end{aligned}$$

(6.7)

Differences (6.5) and (6.6) coincide with the classical difference for the translation operator $T^{t}f(x)=f(x+t)$ and correspond to the usual definition of the modulus of smoothness of order m. Difference (6.7) can be seen as follows. Define $\mu _s=(-1)^s\left( {\begin{array}{c}m\\ s\end{array}}\right)$, $s\in \mathbb {Z}$. Then the convolution $\mu *\mu$ is given by

$$\begin{aligned} \nu _s:=(\mu *\mu )_s=\sum _{l\in \mathbb {Z}}\mu _l\mu _{s+l}=(-1)^s\left( {\begin{array}{c}2m\\ m-s\end{array}}\right) . \end{aligned}$$

Note that $\nu _s\ne 0$ if $|s|\le m$. Moreover, if $k\equiv 0$, then

$$\begin{aligned} \frac{1}{\nu _0}\sum _{s=-m}^m\nu _sT^{st}f(x)=f(x)+\frac{2}{\nu _0} \sum _{s=1}^m\nu _sS^{st}f(x)=f(x)-V_{m, t}f(x), \end{aligned}$$

where the operator $S^{t}$ was given in (3.5) and the averages

$$\begin{aligned} V_{m, t}f(x)=\frac{-2}{\nu _0}\sum _{s=1}^m\nu _sS^{st}f(x) \end{aligned}$$

were defined by Dai and Ditzian in [9].

Definition 6.1

The moduli of smoothness of a function $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$ are defined by

$$\begin{aligned} \omega _m(\delta , f)_{p,\mathrm{d}\mu _k}= & {} \sup _{0<t\le \delta }\Vert \varDelta _t^mf(x)\Vert _{p,\mathrm{d}\mu _k}, \end{aligned}$$

(6.8)

$$\begin{aligned} {^{*}}\omega _m(\delta , f)_{p,\mathrm{d}\mu _k}= & {} \sup _{0<t\le \delta }\Vert {^{*}}\varDelta _t^mf(x)\Vert _{p,\mathrm{d}\mu _k}, \end{aligned}$$

(6.9)

$$\begin{aligned} {^{**}}\omega _m(\delta , f)_{p,\mathrm{d}\mu _k}= & {} \sup _{0<t\le \delta }\Vert {^{**}}\varDelta _t^mf(x)\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.10)

Let us mention some basic properties of these moduli of smoothness. Define by $\Omega _m(\delta , f)_{p,\mathrm{d}\mu _k}$ any of the three moduli in Definition 6.1. Using the triangle inequality, estimate (3.6) reveals

$$\begin{aligned} \Omega _m(\delta , f_1+f_2)_{p,\mathrm{d}\mu _k}\le \Omega _m(\delta , f_1)_{p,\mathrm{d}\mu _k}+\Omega _m(\delta , f_2)_{p,\mathrm{d}\mu _k} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Omega _m(\delta , f)_{p,\mathrm{d}\mu _k}&\lesssim \Vert f\Vert _{p,\mathrm{d}\mu _k},\\ |\Omega _m(\delta , f_1)_{p,\mathrm{d}\mu _k}-\Omega _m(\delta , f_2)_{p,\mathrm{d}\mu _k}|&\lesssim \Vert f_1-f_2\Vert _{p,\mathrm{d}\mu _k}. \end{aligned} \end{aligned}$$

(6.11)

If $f\in \mathcal {S}(\mathbb {R}^d)$, then, by (3.2),

$$\begin{aligned} \begin{aligned} \mathcal {F}_k(\varDelta _t^mf)(y)&=j_{\lambda _{k},m}(t|y|)\mathcal {F}_k(f)(y),\\ \mathcal {F}_k({^{*}}\varDelta _t^rf)(y)&=j_{\lambda _{k},m}^*(t|y|)\mathcal {F}_k(f)(y),\\ \mathcal {F}_k({^{**}}\varDelta _t^mf)(y)&=j^{**}_{\lambda _{k},m}(t|y|)\mathcal {F}_k(f)(y), \end{aligned} \end{aligned}$$

(6.12)

where $\lambda _k=d/2-1+\sum _{a\in R_+}k(a)>-1/2$,

$$\begin{aligned} j_{\lambda _{k},m}(t)&=\sum _{s=0}^m(-1)^s\left( {\begin{array}{c}m\\ s\end{array}}\right) \left( j_{\lambda _{k}}(t)\right) ^s=(1-j_{\lambda _{k}}(t))^m,\\ j_{\lambda _{k},m}^*(t)&=\sum _{s=0}^m(-1)^s\left( {\begin{array}{c}m\\ s\end{array}}\right) j_{\lambda _{k}}(st), \end{aligned}$$

and

$$\begin{aligned} j^{**}_{\lambda _{k},m}(t)&=\left( {\begin{array}{c}2m\\ m\end{array}}\right) ^{-1}\sum _{s=-m}^m(-1)^s \left( {\begin{array}{c}2m\\ m-s\end{array}}\right) j_{\lambda _{k}}(st)\nonumber \\&=1+2\left( {\begin{array}{c}2m\\ m\end{array}}\right) ^{-1}\sum _{s=1}^m(-1)^s\left( {\begin{array}{c}2m\\ m-s\end{array}}\right) j_{\lambda _{k}}(st). \end{aligned}$$

(6.13)

These formulas alow us to prove the following remark, which will be important further in Theorem 6.6.

Remark 6.2

The functions $j_{\lambda _{k},m}(t)$ and $j^{**}_{\lambda _{k},m}(t)$ have zero of order 2m at the origin, while the function $j^{*}_{\lambda _{k},m}(t)$ has zero of order $m+1$ if m is odd and of order m if m is even.

Indeed, first we study $j_{\lambda _{k},m}(t)=(1-j_{\lambda _{k}}(t))^{m}$. Since, for any t,

$$\begin{aligned} j_{\lambda }(t)=\sum _{k=0}^{\infty }\frac{(-1)^{k} \Gamma (\lambda +1)(t/2)^{2k}}{k!\,\Gamma (k+\lambda +1)}, \end{aligned}$$

(6.14)

we get $j_{\lambda _{k},m}(t)\asymp t^{2m}$ as $t\rightarrow 0$. Second, since

$$\begin{aligned} \sum _{s=0}^m(-1)^s\left( {\begin{array}{c}m\\ s\end{array}}\right) s^{2k}=0, \quad 0\le 2k\le m-1, \end{aligned}$$

(see [36, Sect. 4.2]), using (6.14), we obtain that $j_{\lambda _{k},m}^*(t)\asymp t^{2[(m+1)/2]}$. Finally, taking into account

$$\begin{aligned} \begin{aligned}&\sum _{s=1}^m(-1)^s\left( {\begin{array}{c}2m\\ m-s\end{array}}\right) =-\frac{1}{2}\left( {\begin{array}{c}2m\\ m\end{array}}\right) ,\\&\sum _{s=1}^m(-1)^s\left( {\begin{array}{c}2m\\ m-s\end{array}}\right) s^{2k}=0,\quad k=1,\dots ,m-1, \end{aligned} \end{aligned}$$

(see [36, Sect. 4.2]) and using again (6.14), we arrive at $j^{**}_{\lambda _{k},m}(t)\asymp t^{2m}$. Some of these properties were known (see [9, 34, 35]).

Remark 6.3

In the paper [9], the authors obtained that $j^{**}_{\lambda _{k},m}(t)>0$ for $t>0$.

6.2 Main Results

First we state the Jackson-type inequality.

Theorem 6.4

Let $\sigma > 0$, $1\le p \le \infty$, $r\in \mathbb {Z}_+$, $m\in \mathbb {N}$. We have, for any $f\in W_{p, k}^{2r}$,

$$\begin{aligned} E_{\sigma }(f)_{p,\mathrm{d}\mu _k} \lesssim \frac{1}{\sigma ^{2r}}\,\Omega _m\left( \frac{1}{\sigma }, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}, \end{aligned}$$

(6.15)

where $\Omega _m$ is any of the three moduli of smoothness (6.8)–(6.10).

Remark 6.5

• (i) For radial functions, inequality (6.15) is the Jackson inequality in $L^{p}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$. In this case it was obtained in [34, 35] for moduli (6.8) and (6.9). For $k\equiv 0$ and the modulus of smoothness (6.10), inequality (6.15) was obtained by Dai and Ditzian [9], see also the paper [10].

• (ii) From the proof of Theorem 6.4, we will see that inequality (6.15) for moduli (6.8) and (6.10) can be equivalently written as

  $$\begin{aligned} E_{\sigma }(f)_{p,\mathrm{d}\mu _k}&\lesssim \frac{1}{\sigma ^{2r}} \left\| \varDelta _{1/\sigma }^m ((-\Delta _k)^rf) \right\| _{p,\mathrm{d}\mu _k},\\ E_{\sigma }(f)_{p,\mathrm{d}\mu _k}&\lesssim \frac{1}{\sigma ^{2r}} \left\| {^{**}}\varDelta _{1/\sigma }^m ((-\Delta _k)^rf) \right\| _{p,\mathrm{d}\mu _k}. \end{aligned}$$

The next theorem provides an equivalence between moduli of smoothness and the K-functional.

Theorem 6.6

If $\delta > 0$, $1\le p\le \infty$, $r\in \mathbb {N}$, then for any $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$,

$$\begin{aligned} \begin{aligned} K_{2r}(\delta , f)_{p,\mathrm{d}\mu _k}&\asymp \omega _r(\delta , f)_{p,\mathrm{d}\mu _k}\asymp {^{**}}\omega _r(\delta , f)_{p,\mathrm{d}\mu _k}\\&\asymp {^{*}}\omega _{2r-1}(\delta , f)_{p,\mathrm{d}\mu _k}\asymp {^{*}}\omega _{2r}(\delta , f)_{p,\mathrm{d}\mu _k}. \end{aligned} \end{aligned}$$

(6.16)

Remark 6.7

If $k\equiv 0$, the equivalence between the classical modulus of smoothness and the K-functional is well known [8, 26], while the equivalence between modulus (6.10) and the K-functional was shown in [9]. For radial functions, a partial result of (6.16), more precisely, an equivalence of the K-functional and moduli of smoothness (6.8) and (6.9), was proved in [34, 35].

Remark 6.8

One can continue equivalence (6.16) as follows (see also Remark 6.16):

$$\begin{aligned} \ldots \asymp \Vert \varDelta _\delta ^r f \Vert _{p,\mathrm{d}\mu _k} \asymp \Vert {^{**}}\varDelta _\delta ^r f \Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

We give the proof for the difference (6.7) and the modulus of smoothness (6.10). We partially follow the proofs in [27, 34, 35], which are different from those given in [9]. For moduli of smoothness (6.8) and (6.9), the proofs are similar and will be omitted here (see also [34, 35]). The proof makes use of radial multipliers and is based on boundedness of the translation operator $T^t$. Note that by (6.2) and (6.11), the K-functional and moduli of smoothness depend continuously on a function. Moreover, the best approximations also depend continuously on a function, and therefore one can assume that functions belong to the Schwartz space.

6.3 Properties of the de la Vallée Poussin Type Operators

Let $\eta \in \mathcal {S}_\text {rad}(\mathbb {R}^d)$ be such that $\eta (x)=1$ if $|x|\le 1$, $\eta (x)>0$ if $|x|<2$, and $\eta (x)=0$ if $|x|\ge 2$. We write

$$\begin{aligned} \eta _r(x)=\frac{1-\eta (x)}{|x|^{2r}},\quad \widehat{\eta }_{k,r}(y)=\mathcal {F}_{k}(\eta _r)(y), \end{aligned}$$

where $\mathcal {F}_{k}(\eta _r)$ is a tempered distribution. If $t=|x|$, $\eta _0(t)=\eta (x)$, and $\eta _{r 0}(t)=\eta _r(x)$, then $\mathcal {F}_{k}(\eta _r)(y)=\mathcal {H}_{\lambda _k}(\eta _{r 0})(|y|)$.

Lemma 6.9

We have $\widehat{\eta }_{k,r}\in L^{1}(\mathbb {R}^d, \mathrm{d}\mu _k)$, where $r>0$.

Proof

It is sufficient to prove that $\mathcal {H}_{\lambda _k}(\eta _{r 0})\in L^{1}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$. In the case $r\ge 1$, this was proved in [35, (4.25)]. We give the proof for any $r>0$.

Letting $u_j(t)=(1+t^2)^{-j}$ and taking into account that

$$\begin{aligned} \frac{1}{t^{2r}}=\frac{1}{(1+t^2)^r}\left( 1-\frac{1}{1+t^2}\right) ^{-r}= \sum _{j=0}^{\infty }\left( {\begin{array}{c}j+r-1\\ j\end{array}}\right) \frac{1}{(1+t^2)^{j+r}},\quad t\ne 0, \end{aligned}$$

we obtain, for any $M\in \mathbb {N}$ and $t\ge 0$,

$$\begin{aligned} \eta _{r 0}(t)&=\sum _{j=0}^{\infty }\left( {\begin{array}{c}j+r-1\\ j\end{array}}\right) (1-\eta _0(t))u_{j+r}(t)\\&=\sum _{j=0}^{M-1}\left( {\begin{array}{c}j+r-1\\ j\end{array}}\right) u_{j+r}(t)- \eta _0(t)\sum _{j=0}^{M-1}\left( {\begin{array}{c}j+r-1\\ j\end{array}}\right) u_{j+r}(t)\\&\quad + \sum _{j=M}^{\infty }\left( {\begin{array}{c}j+r-1\\ j\end{array}}\right) (1-\eta _0(t))u_{j+r}(t) =:\psi _1(t)+\psi _2(t)+\psi _3(t). \end{aligned}$$

For any $r>0$, we have $\mathcal {H}_{\lambda _k}(u_r)\in L^{1}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$ (see [35, Lemma 3.2], [47, Chapt 5, 5.3.1], [31, Chapt 8, 8.1]); therefore $\mathcal {H}_{\lambda _k}(\psi _1)\in L^{1}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$. Because of $\psi _2\in \mathcal {S}(\mathbb {R}_+)$, $\mathcal {H}_{\lambda _k}(\psi _2)\in L^{1}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$. Thus, we are left to show that, for sufficiently large M, $\mathcal {H}_{\lambda _k}(\psi _3)\in L^{1}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$.

Let $M+r> \lambda _k+1$, $t\ge 1$. Since $\frac{\Gamma (j+r)}{\Gamma (j+1)}\lesssim j^{r-1}$, we have

$$\begin{aligned} |\psi _3(t)|\le \frac{M^{r-1}}{(1+t^2)^{M+r}}\sum _{j=0}^{\infty }\frac{(1+j)^{r-1}}{2^{j}}\lesssim \frac{1}{(1+t^2)^{M+r}}\lesssim \frac{1}{t^{2M+2r}}, \end{aligned}$$

and

$$\begin{aligned} \int _0^{\infty }|\psi _3(t)|\,\mathrm{d}\nu _{\lambda _k}(t)\lesssim \int _1^{\infty }t^{-(2M+2r-2\lambda _k-1)}\,\mathrm{d}t<\infty . \end{aligned}$$

Thus, $\psi _3\in L^{1}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda _k})$, $\mathcal {H}_{\lambda _k}(\psi _3)\in C(\mathbb {R}_{+})$, and $\mathcal {H}_{\lambda _k}(\psi _3)\in L^{1}([0,2],\mathrm{d}\nu _{\lambda _k})$.

Recall that the Bessel differential operator is defined by

$$\begin{aligned} \mathcal {B}_{\lambda _k}=\frac{d^2}{\mathrm{d}t^2}+\frac{(2\lambda _k+1)}{t}\frac{\mathrm{d}}{\mathrm{d}t}. \end{aligned}$$

Using $\psi _3\in C^{\infty }(\mathbb {R}_{+})$, we have, for any $s\in \mathbb {N}$, $\mathcal {B}_{\lambda _k}^s\psi _3\in L^{1}([0,2],\mathrm{d}\nu _{\lambda _k})$.

If $t\ge 2$, then $(1-\eta _0(t))u_{j+r}(t)=u_{j+r}(t)$ and

$$\begin{aligned} \mathcal {B}_{\lambda _k}u_{j+r}(t)=4(j+r)(j+r-\lambda _k)u_{j+r+1} (t)-4(j+r)(j+r+1)u_{j+r+2}(t). \end{aligned}$$

This gives

$$\begin{aligned} |\mathcal {B}_{\lambda _k}u_{j+r}(t)|\le 2^3(j+r+\lambda _k+1)^2u_{j+r+1}(t). \end{aligned}$$

By induction on s,

$$\begin{aligned} |\mathcal {B}_{\lambda _k}^su_{j+r}(t)|\le 2^{3s}(j+r+2s+\lambda _k-1)^{2s}u_{j+r+s}(t), \end{aligned}$$

and then, for $t\ge 2$,

$$\begin{aligned} |\mathcal {B}_{\lambda _k}^s\psi _3(t)|\lesssim \frac{1}{(1+t^2)^{M+r+s}}\sum _{j=0}^{\infty }\frac{(1+j)^{r+2s-1}}{5^{j}}\lesssim \frac{1}{(1+t^2)^{M+r+s}}\lesssim \frac{1}{t^{2M+2r+2s}}, \end{aligned}$$

and $\mathcal {B}_{\lambda _k}^s\psi _3\in L^{1}([2,\infty ),\mathrm{d}\nu _{\lambda _k})$. Thus, we have $\mathcal {B}_{\lambda _k}^s\psi _3\in L^{1}(\mathbb {R}_+,\mathrm{d}\nu _{\lambda _k})$ for any s. Choosing $s>\lambda _k+1$ and using the inequality

$$\begin{aligned} |\mathcal {H}_{\lambda _k}(\psi _3)(\tau )|\le \frac{1}{\tau ^{2s}} \int _{0}^{\infty }|\mathcal {B}_{\lambda _k}^s\psi _3(t)|\,\mathrm{d}\nu _{\lambda _k}(t)\lesssim \frac{1}{\tau ^{2s}}, \end{aligned}$$

we arrive at $\mathcal {H}_{\lambda _k}(\psi _3)\in L^{1}([2,\infty ),\mathrm{d}\nu _{\lambda _k})$. Finally, we obtain that $\mathcal {H}_{\lambda _k}(\psi _3)\in L^{1}(\mathbb {R}_+,\mathrm{d}\nu _{\lambda _k})$. $\square$

For $m, r\in \mathbb {N}$ and $m\ge r$, we set

$$\begin{aligned} g_{m, r}^*(y):= & {} |y|^{-2r}j^{**}_{\lambda _{k},m}(|y|),\quad g_{m,r}(x):=\mathcal {F}_{k}(g_{m, r}^*)(x),\\ g_{m,r}^t(x):= & {} t^{2r-2\lambda _k-2}g_{m,r}\left( \frac{x}{t}\right) . \end{aligned}$$

Since

$$\begin{aligned} g_{m,r}^*(y)= & {} j^{**}_{\lambda _{k},m}(|y|)\eta _r(y)+\frac{j^{**}_{\lambda _{k},m} (|y|)}{|y|^{2r}}\,\eta (y),\\&\frac{j^{**}_{\lambda _{k},m}(|y|)}{|y|^{2r}}\in C^{\infty }(\mathbb {R}^d),\quad \frac{j^{**}_{\lambda _{k},m}(|y|)}{|y|^{2r}}\,\eta (y)\in \mathcal {S}(\mathbb {R}^d), \end{aligned}$$

and

$$\begin{aligned} \mathcal {F}_{k}(j^{**}_{\lambda _{k},m}\eta _r)(x)=\left( {\begin{array}{c}2m\\ m\end{array}}\right) ^{-1} \sum _{s=-m}^m(-1)^s\left( {\begin{array}{c}2m\\ m-s\end{array}}\right) T^{s}\widehat{\eta }_{\lambda _k,r}(x), \end{aligned}$$

boundedness of the operator $T^s$ in $L^{1}(\mathbb {R}^d,\mathrm{d}\mu _k)$ and Lemma 6.9 imply that

$$\begin{aligned} \begin{aligned}&g_{m,r},\,g_{m,r}^t\in L^{1}(\mathbb {R}^d,\mathrm{d}\mu _{k}),\quad \Vert g_{m,r}^t\Vert _{1,\mathrm{d}\mu _{k}}=t^{2r}\Vert g_{m,r}\Vert _{1,\mathrm{d}\mu _{k}},\\&\mathcal {F}_{k}^{-1}(g_{m,r}^t)(y)= \mathcal {F}_{k}(g_{m,r}^t)(y)=t^{2r}g_{m,r}^*(ty)=|y|^{-2r}j^{**}_{\lambda _{k},m}(t|y|). \end{aligned} \end{aligned}$$

(6.17)

Lemma 6.10

Let $m, r \in \mathbb {N}$, $m \ge r$, $1\le p\le \infty$, and $f\in \mathcal {S}(\mathbb {R}^d)$. We have

$$\begin{aligned} {^{**}}\varDelta _t^mf = (-\Delta _k)^rf*_{ k}g_{m,r}^t \end{aligned}$$

(6.18)

and

$$\begin{aligned} \Vert {^{**}}\varDelta _t^mf\Vert _{p,\mathrm{d}\mu _k}\lesssim t^{2r}\Vert (-\Delta _k)^rf\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.19)

Proof

Combining (3.15), (6.1), (6.12), and (6.17), we obtain that

$$\begin{aligned} \mathcal {F}_k({^{**}}\varDelta _t^mf)(y)&=j^{**}_{\lambda _{k},m}(t|y|)\mathcal {F}_k(f)(y) =|y|^{2r}\mathcal {F}_k(f)(y)\frac{j^{**}_{\lambda _{k},m}(t|y|)}{|y|^{2r}}\\&=\mathcal {F}_k((-\Delta _k)^rf)(y) \mathcal {F}_k(g_{m,r}^t)(y). \end{aligned}$$

Then (6.18) follows from (3.11) and Lemma 3.4. Inequality (6.19) follows from (6.17), (6.18), Lemma 3.4, and (3.12). Note that a constant in (6.19) can be taken as $\Vert g_{m,r}\Vert _{1,\mathrm{d}\mu _{k}}$. $\square$

Remark 6.11

Since $\mathcal {S}(\mathbb {R}^d)$ is dense in $W_{p, k}^{2r}$, in light of (6.11), inequality (6.19) holds for any function from $W_{p, k}^{2r}$.

Let $f\in \mathcal {S}(\mathbb {R}^d)$. We set $\theta (x)=\mathcal {F}_k(\eta )(x)$ and $\theta _\sigma (x) = \theta (x/\sigma )$. Then $\theta$, $\theta _\sigma \in \mathcal {S}(\mathbb {R}^d)$. The de la Vallée Poussin type operator is given by $P_\sigma (f)= f*_{ k}\theta _\sigma$. By Lemma 3.4,

$$\begin{aligned} \mathcal {F}_k(P_\sigma (f))(y)=\eta (y/\sigma )\mathcal {F}_k (f)(y). \end{aligned}$$

Lemma 6.12

If $\sigma >0$, $1 \le p\le \infty$, $f\in \mathcal {S}(\mathbb {R}^d)$, then

• (1) $P_\sigma (f)\in B_{p, k}^{2\sigma }$ and $P_\sigma (g)=g$ for any $g\in B_{p, k}^\sigma$;

• (2) $\Vert P_\sigma (f)\Vert _{p,\mathrm{d}\mu _{k}} \lesssim \Vert f\Vert _{p,\mathrm{d}\mu _{k}}$;

• (3) $\Vert f-P_\sigma (f)\Vert _{p,\mathrm{d}\mu _{k}} \lesssim E_{\sigma } (f)_{p,\mathrm{d}\mu _{k}}$.

Remark 6.13

Property (3) in this lemma means that $P_\sigma (f)$ is the near best approximant of f in $L^p(\mathbb {R}^d,\mathrm{d}\mu _{k})$.

Proof

(1) We observe that ${\text {supp}}\eta (\cdot /\sigma )\subset B_{2\sigma }$ and then ${\text {supp}}\mathcal {F}_k(P_\sigma (f))\subset B_{2\sigma }$. Theorem 5.14 yields $P_\sigma (f)\in B_{p, k}^{2\sigma }$. If $g\in B_{p,k}^\sigma$, then by Theorem 5.14, ${\text {supp}}\mathcal {F}_k(g)\subset B_{\sigma }$ and $\mathcal {F}_k(P_\sigma (g))(y)=\eta (y/\sigma )\mathcal {F}_k(g)(y)=\mathcal {F}_k (g)(y).$ Hence, $P_\sigma (g)=g$.

(2) In light of (3.12),

$$\begin{aligned} \Vert P_\sigma (f)\Vert _{p,\mathrm{d}\mu _{k}}&=\Vert f*_{ k}\theta _\sigma \Vert _{p,\mathrm{d}\mu _{k}}\le \Vert \theta _\sigma \Vert _{1,\mathrm{d}\mu _{k}} \Vert f\Vert _{p,\mathrm{d}\mu _{k}}\\&=\Vert \theta \Vert _{1,\mathrm{d}\mu _{k}} \Vert f\Vert _{p,\mathrm{d}\mu _{k}}\lesssim \Vert f\Vert _{p,\mathrm{d}\mu _{k}}. \end{aligned}$$

(3) Using Theorem 5.15, there exists an entire function $g^*\in B_{p, k}^\sigma$ such that $\Vert f-g^*\Vert _{p,\mathrm{d}\mu _{k}}=E_\sigma (f)_{p,\mathrm{d}\mu _{k}}$. Then using $P_\sigma (g^*)=g^*$ implies

$$\begin{aligned} \Vert f- P_\sigma (f)\Vert _{p,\mathrm{d}\mu _{k}}&= \Vert f-g^*+P_\sigma (g^*-f)\Vert _{p,\mathrm{d}\mu _{k}}\\&\le \Vert f-g^*\Vert _{p,\mathrm{d}\mu _{k}}+\Vert P_\sigma (f-g^*)\Vert _{p,\mathrm{d}\mu _{k}} \lesssim E_\sigma (f)_{p,\mathrm{d}\mu _{k}}. \end{aligned}$$

$\square$

In the proof of the next lemma we will use the estimate

$$\begin{aligned} |j^{(n)}_{\lambda }(t)|\lesssim (|t|+1)^{-(\lambda +1/2)},\quad t\in \mathbb {R},\ \lambda \ge -1/2,\ n\in \mathbb {Z}_+, \end{aligned}$$

(6.20)

which follows, by induction on n, from the known properties of the Bessel function [2, Chap. 7]

$$\begin{aligned} |j_{\lambda }(t)|\lesssim (|t|+1)^{-(\lambda +1/2)},\quad j'_{\lambda }(t)=-\frac{t}{2(\lambda +1)}j_{\lambda +1}(t). \end{aligned}$$

Lemma 6.14

If $\sigma >0$, $1 \le p\le \infty$, $m\in \mathbb {N}$, $r\in \mathbb {Z}_+$, $f\in \mathcal {S}(\mathbb {R}^d)$, then

$$\begin{aligned} \Vert f-P_{\sigma /2}(f)\Vert _{p,\mathrm{d}\mu _{k}} \lesssim \sigma ^{-2r}\Vert {^{**}}\varDelta _{a/\sigma }^m((-\Delta _k)^rf)\Vert _{p,\mathrm{d}\mu _k} \end{aligned}$$

(6.21)

for some $a=a(\lambda _k, m)>0$.

Proof

We have

$$\begin{aligned} \mathcal {F}_k(f-P_{\sigma /2}(f))(y)&=(1-\eta (2y/\sigma ))\mathcal {F}_kf(y)\nonumber \\&=\sigma ^{-2r}\frac{1-\eta (2y/\sigma )}{(|y|/\sigma )^{2r} j^{**}_{\lambda _{k},m}(a|y|/\sigma )}\mathcal {F}_k({^{**}}\varDelta _{a/\sigma }^m ((-\Delta _k)^r f))(y)\nonumber \\&=\sigma ^{-2r} \varphi (y/\sigma )\mathcal {F}_k({^{**}}\varDelta _{a/\sigma }^m ((-\Delta _k)^r f))(y), \end{aligned}$$

(6.22)

where

$$\begin{aligned} \varphi (y)= \frac{1-\eta (2y)}{|y|^{2r} j^{**}_{\lambda _{k},m}(a|y|)}, \quad {^{**}}\varDelta _{a/\sigma }^m ((-\Delta _k)^r f)\in \mathcal {S}(\mathbb {R}^d). \end{aligned}$$

(6.23)

Setting $j^{**}_{\lambda _{k},m}(t)= 1- \tau _0(t)$, in light of (6.13) and (6.20), we observe that $j^{**}_{\lambda _{k},m}(t)\rightarrow 1$ as $t\rightarrow \infty$. Then we can choose $a>0$ such that $|\tau _0(t)| \le 1/2$ for $|t| \ge a/2$. For such $a=a(\lambda _k, m)$, we have that $\varphi (y)=0$ for $|y|\le 1/2$, $\varphi (y)>0$ for $|y|> 1/2$, and $\varphi \in C^{\infty }(\mathbb {R}^d)$. Moreover, the derivatives $\varphi ^{(k)}(y)$ grow at infinity not faster than $|y|^{a_k}$, which yields $\varphi \in \mathcal {S}'(\mathbb {R}^d)$.

We will use the following decomposition:

$$\begin{aligned} \varphi (y)=\varphi _1(|y|)+\varphi _2(|y|), \end{aligned}$$

where

$$\begin{aligned} \varphi _1(|y|)=2^{2r}\eta _{r}(2y)\left( \frac{1}{1- \tau _0(a|y|)}-S_N(\tau _0(a|y|)\right) \end{aligned}$$

and

$$\begin{aligned} \varphi _2(|y|)=2^{2r}\eta _{r}(2y)S_N(\tau _0(a|y|)), \quad \eta _{r}(y)=\frac{1-\eta (y)}{|y|^{2r}}, \quad S_N(t)=\sum _{j=0}^{N-1}t^j. \end{aligned}$$

First, we show that $\mathcal {F}_k(\varphi _1(|\cdot |))\in L^{1}(\mathbb {R}^d, \mathrm{d}\mu _k)$. Since for a radial function we have

$$\begin{aligned} \Delta _k\varphi _1(|y|)=\varphi _1''(|y|)+\frac{2\lambda _k+1}{|y|}\,\varphi _1'(|y|) \end{aligned}$$

and, for $|t|\le 1/2$,

$$\begin{aligned} (1-t)^{-1}-\sum _{j=0}^{N-1}t^j=(1-t)^{-1}-S_N(t)=\frac{t^N}{1-t}, \end{aligned}$$

then, by (6.13) and (6.20), we obtain

$$\begin{aligned} \Delta _k^s\varphi _1(|y|)=O(|y|^{-2r-N(\lambda _k+1/2)}),\quad |y|\ge 1/2,\ s\in \mathbb {Z}_+. \end{aligned}$$

Hence, for a fixed $N\ge 2+2/(2\lambda _k+1)$, we have $\Delta _k^s\varphi _1(|y|)\in L^{1}(\mathbb {R}^d, \mathrm{d}\mu _k)$, where $s\in \mathbb {Z}_+$. Applying (6.1) we derive that

$$\begin{aligned} |\mathcal {F}_k(\varphi _1(|\cdot |))(x)|= \frac{|\mathcal {F}_k((-\Delta _k)^s \varphi _1(|\cdot |))(x)|}{|x|^{2s}}\le \frac{\Vert (-\Delta _k)^s \varphi _1(|\cdot |)\Vert _{1,\mathrm{d}\mu _k}}{|x|^{2s}}. \end{aligned}$$

Setting $s>\lambda _k+1$ yields $\mathcal {F}_k(\varphi _1(|\cdot |))\in L^{1}(\mathbb {R}^d, \mathrm{d}\mu _k)$.

Second, let us show that $\mathcal {F}_k(\varphi _2(|\cdot |))\in L^{1}(\mathbb {R}^d, \mathrm{d}\mu _k)$ for $r\in \mathbb {N}$. Let

$$\begin{aligned} \tau _0(t)= & {} \sum _{s=1}^m\nu _sj_{\lambda _{k}}(st),\quad \psi _r(x)=2^{2r}\mathcal {F}_k(\eta _{r}(2\cdot ))(x),\\ A^af(x)= & {} \sum _{s=1}^m\nu _sT^{as}f(x),\quad B^af(x)=\sum _{j=0}^{N-1}(A^a)^jf(x). \end{aligned}$$

Boundedness of the operator $T^t$ in $L^{p}(\mathbb {R}^d, \mathrm{d}\mu _k)$ implies

$$\begin{aligned} \Vert A^a\Vert _{p\rightarrow p}=\sup \{\Vert Af\Vert _{p,\mathrm{d}\mu _k}:\Vert f\Vert _{p,\mathrm{d}\mu _k}\le 1\}\le \sum _{s=1}^m|\nu _s| \end{aligned}$$

and

$$\begin{aligned} \Vert B^a\Vert _{p\rightarrow p}\le \sum _{j=0}^{N-1}(\Vert A\Vert _{p\rightarrow p})^j\le N\left( 1+\sum _{s=1}^m|\nu _s|\right) ^{N-1},\quad 1\le p<\infty . \end{aligned}$$

(6.24)

Then for $p=1$, taking into account Lemma 6.9, we have

$$\begin{aligned} \Vert \mathcal {F}_k(\varphi _2(|\cdot |))\Vert _{1,\mathrm{d}\mu _k}=\bigl \Vert B^a\psi _r\bigr \Vert _{1,\mathrm{d}\mu _k}\le N\left( 1+\sum _{s=1}^m|\nu _s|\right) ^{N-1}\Vert \psi _r\Vert _{1,\mathrm{d}\mu _k}<\infty . \end{aligned}$$

Thus, $\mathcal {F}_k(\varphi ) \in L^{1}(\mathbb {R}^d, \mathrm{d}\mu _k)$. Combining Lemma 3.4, relations (3.12), (6.22), (6.23), and the formula $\Vert \mathcal {F}_k(\varphi (\cdot /\sigma ))\Vert _{1,\mathrm{d}\mu _k}=\Vert \mathcal {F}_k(\varphi )\Vert _{1,\mathrm{d}\mu _k}$, we obtain inequality (6.21) for $r\in \mathbb {N}$.

Now let $r=0$. Define the operators $A_1$ and $A_2$ as follows:

$$\begin{aligned} \mathcal {F}_k(A_1g)(y)=\varphi _1(|y|/\sigma )\mathcal {F}_k(g)(y) \end{aligned}$$

and

$$\begin{aligned} \mathcal {F}_k(A_2g)(y)=\varphi _2(|y|/\sigma )\mathcal {F}_k(g)(y),\quad \varphi _2(|y|)=(1-\eta (2y))S_N(\tau _0(a|y|)). \end{aligned}$$

Since $\mathcal {F}_k(\varphi _1(|\cdot |))\in L^{1}(\mathbb {R}^d, \mathrm{d}\mu _k)$,

$$\begin{aligned} \begin{aligned} \Vert A_1g\Vert _{p,\mathrm{d}\mu _k}&\le \Vert \mathcal {F}_k(\varphi _1(|y|))\Vert _{1,\mathrm{d}\mu _k}\,\Vert g\Vert _{p,\mathrm{d}\mu _k}\lesssim \Vert g\Vert _{p,\mathrm{d}\mu _k},\\ 1&\le p\le \infty ,\ g\in \mathcal {S}(\mathbb {R}^d). \end{aligned} \end{aligned}$$

(6.25)

We are left to show that

$$\begin{aligned} \Vert A_2g\Vert _{p,\mathrm{d}\mu _k}\lesssim \Vert g\Vert _{p,\mathrm{d}\mu _k},\quad 1\le p\le \infty ,\ g\in \mathcal {S}(\mathbb {R}^d). \end{aligned}$$

We have

$$\begin{aligned} \mathcal {F}_k(A_2g)(y)&=(1-\eta (2y/\sigma ))S_N(\tau _0(a|y|/\sigma ))\mathcal {F}_k(g)(y)\\&=(1-\eta (2y/\sigma ))\mathcal {F}_k(B^{a/\sigma }g)(y)\\&=\mathcal {F}_k(B^{a/\sigma }g-P_{\sigma /2}(B^{a/\sigma }g))(y). \end{aligned}$$

Since $B^{a/\sigma }g\in \mathcal {S}(\mathbb {R}^d)$, using Lemma 6.12 and inequality (6.24), we get

$$\begin{aligned} \Vert A_2g\Vert _{p,\mathrm{d}\mu _k}\le \Vert B^{a/\sigma }g\Vert _{p,\mathrm{d}\mu _k}\le N\left( 1+\sum _{s=1}^m|\nu _s|\right) ^{N-1}\Vert g\Vert _{p,\mathrm{d}\mu _k}\lesssim \Vert g\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.26)

Using (6.25) and (6.26) with $g= {^{**}}\varDelta _{a/\sigma }^m f$, we finally obtain (6.21) for $r=0$. $\square$

Lemma 6.15

If $\sigma >0$, $1 \le p\le \infty$, $m\in \mathbb {N}$, $f\in \mathcal {S}(\mathbb {R}^d)$, then

$$\begin{aligned} \Vert ((-\Delta _k)^{m}P_\sigma (f)\Vert _{p,\mathrm{d}\mu _k} \lesssim \sigma ^{2m} \Vert {^{**}}\varDelta _{a/(2\sigma )}^{m} f\Vert _{p,\mathrm{d}\mu _k}, \end{aligned}$$

(6.27)

where $a=a(\lambda _k, m)>0$ is given in Lemma 6.14.

Proof

We have

$$\begin{aligned} \mathcal {F}_k(((-\Delta _k)^{m}P_\sigma (f))(y)&=|y|^{2m}\eta (y/\sigma )\mathcal {F}_k(f)(y)\\&=\sigma ^{2m}\varphi (y/\sigma )j^{**}_{\lambda _{k},m}(a/(2\sigma ))\mathcal {F}_k(f)(y)\\&=\sigma ^{2m}\varphi (y/\sigma )\mathcal {F}_k({^{**}}\varDelta _{a/(2\sigma )}^{m} f)(y), \end{aligned}$$

where

$$\begin{aligned} \varphi (y)=\frac{|y|^{2m}\eta (y)}{j^{**}_{\lambda _{k},m}(a|y|/2)}. \end{aligned}$$

Since $j^{**}_{\lambda _{k},m}(a|y|/2)/|y|^{2m}>0$ for $|y|>0$, we observe that $\varphi \in \mathcal {S}(\mathbb {R}^d)$ and $\mathcal {F}_k(\varphi )\in L^{1}(\mathbb {R}^d, \mathrm{d}\mu _k)$. Then estimate (6.27) follows from Lemma 3.4, Young’s inequality (3.12), and $\Vert \mathcal {F}_k(\varphi (\cdot /\sigma ))\Vert _{1,\mathrm{d}\mu _k}=\Vert \mathcal {F}_k(\varphi )\Vert _{1,\mathrm{d}\mu _k}$. $\square$

6.4 Proofs of Theorems 6.4 and 6.6

Proof of Theorem 6.6

In connection with Lemma 6.10 and Remark 6.11, observe that, for $f\in \mathcal {S}(\mathbb {R}^d)$ and $g\in W_{p, k}^{2r}$,

$$\begin{aligned} \Vert {^{**}}\varDelta _{\delta }^rf\Vert _{p,\mathrm{d}\mu _k}&\le {^{**}}\omega _r(\delta , f)_{p,\mathrm{d}\mu _k} \le {^{**}}\omega _r(\delta , f-g)_{p,\mathrm{d}\mu _k} +{^{**}}\omega _r(\delta , g)_{p,\mathrm{d}\mu _k}\\&\lesssim \Vert f-g\Vert _{p,\mathrm{d}\mu _k}+\delta ^{2r}\Vert (-\Delta _k)^{r}g\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

Then

$$\begin{aligned} \Vert {^{**}}\varDelta _{\delta }^rf\Vert _{p,\mathrm{d}\mu _k}\le {^{**}}\omega _r(\delta , f)_{p,\mathrm{d}\mu _k} \lesssim K_{2r}(\delta , f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.28)

On the other hand, $P_\sigma (f)\in W_{p, k}^{2r}$ and

$$\begin{aligned} K_{2r}(\delta , f)_{p,\mathrm{d}\mu _k} \le \Vert f-P_\sigma (f)\Vert _{p,\mathrm{d}\mu _k}+ \delta ^{2r}\Vert (-\Delta _k)^{r}P_\sigma (f) \Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.29)

In light of Lemma 6.14,

$$\begin{aligned} \Vert f-P_{\sigma }(f)\Vert _{p,\mathrm{d}\mu _{k}} \lesssim \Vert {^{**}}\varDelta _{a/(2\sigma )}^rf\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

Further, Lemma 6.15 yields

$$\begin{aligned} \Vert ((-\Delta _k)^{r}P_\sigma (f)\Vert _{p,\mathrm{d}\mu _k} \lesssim \sigma ^{2r} \Vert {^{**}}\varDelta _{a/(2\sigma )}^{r} f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.30)

Setting $\sigma =a/(2\delta )$, from (6.29)–(6.30) we arrive at

$$\begin{aligned} K_{2r}(\delta , f)_{p,\mathrm{d}\mu _k}\lesssim \Vert {^{**}}\varDelta _{\delta }^{r} f\Vert _{p,\mathrm{d}\mu _k}\lesssim {^{**}}\omega _r(\delta , f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.31)

Proof of Theorem 6.4

Using property (6.4) and inequalities (6.28) and (6.31), we obtain

$$\begin{aligned} E_\sigma (f)_{p,\mathrm{d}\mu _k}&\le \Vert f-P_{\sigma /2}(f)\Vert _{p,\mathrm{d}\mu _{k}} \lesssim \sigma ^{-2r}\Vert {^{**}}\varDelta _{a/\sigma }^m((-\Delta _k)^rf)\Vert _{p,\mathrm{d}\mu _k}\nonumber \\&\lesssim \frac{1}{\sigma ^{2r}}\,K_{2m}\left( \frac{a}{\sigma }, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}\lesssim \frac{1}{\sigma ^{2r}}\,K_{2m}\left( \frac{1}{\sigma }, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}\nonumber \\&\lesssim \frac{1}{\sigma ^{2r}}\,\Vert {^{**}}\varDelta _{1/\sigma }^m((-\Delta _k)^rf)\Vert _{p,\mathrm{d}\mu _k} \lesssim \frac{1}{\sigma ^{2r}}\,{^{**}}\omega _m\left( \frac{1}{\sigma }, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(6.32)

Remark 6.16

The proofs of estimates (6.31) and (6.32) for the difference (6.7) are based on the fact that the parameter a in Lemmas 6.14 and 6.15 is the same. It is possible due to the fact that $j^{**}_{\lambda _{k},m}(t)>0$ for $t>0$, see Remark 6.3. This estimate is valid for the difference (6.5) as well, since $j_{\lambda _{k},m}(t)=(1-j_{\lambda _{k}}(t))^m>0$ for $t>0$.

Therefore, the moduli of smoothness (6.8) and (6.10) in inequalities (6.15) and (6.16) can be replaced by the norms of the corresponding differences (6.5) and (6.7). For the modulus of smoothness (6.9), this observation is not valid since $j^{*}_{\lambda _{k},m}(t)$ does not keep its sign.

Remark 6.17

Properties (6.3) and (6.4) of the K-functional and the equivalence (6.16) imply the following properties of moduli of smoothness:

$$\begin{aligned}&(1) \ \lim _{\delta \rightarrow 0+0}\omega _m(\delta , f)_{p,\mathrm{d}\mu _k}=\lim _{\delta \rightarrow 0+0} {^{*}}\omega _m(\delta , f)_{p,\mathrm{d}\mu _k}=\lim _{\delta \rightarrow 0+0} {^{**}}\omega _m(\delta , f)_{p,\mathrm{d}\mu _k}=0;\\&(2) \ \omega _m(\lambda \delta , f)_{p,\mathrm{d}\mu _k}\lesssim \max \{1, \lambda ^{2m}\}\,\omega _m(\delta , f)_{p,\mathrm{d}\mu _k};\\&(3) \ {^{*}}\omega _{l}(\lambda \delta , f)_{p,\mathrm{d}\mu _k}\lesssim \max \{1, \lambda ^{2m}\}\,{^{*}}\omega _l(\delta , f)_{p,\mathrm{d}\mu _k},\quad l=2m-1,\,2m;\\&(4) \ {^{**}}\omega _m(\lambda \delta , f)_{p,\mathrm{d}\mu _k}\lesssim \max \{1, \lambda ^{2m}\}\,{^{**}}\omega _m(\delta , f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

7 Some Inequalities for Entire Functions

In this section, we study weighted analogues of the inequalities for entire functions. In particular, we obtain Nikolskii’s inequality ([31], see Theorem 7.1 below), Bernstein’s inequality ([31], Theorem 7.3), Nikolskii–Stechkin’s inequality ([30, 45], Theorem 7.5), and Boas-type inequality ([4], Theorem 7.7).

Theorem 7.1

If $\sigma >0$, $0<p\le q\le \infty$, $f\in B_{p, k}^\sigma$, then

$$\begin{aligned} \Vert f\Vert _{q,\mathrm{d}\mu _k} \lesssim \sigma ^{(2\lambda _k+2)(1/p-1/q)}\Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(7.1)

Remark 7.2

Observe that the obtained Nikolskii inequality is sharp, i.e., we actually have

$$\begin{aligned} \sup _{f\in B_{p, k}^\sigma , f\ne 0}\frac{\Vert f\Vert _{q,\mathrm{d}\mu _k}}{\Vert f\Vert _{p,\mathrm{d}\mu _k}}\asymp \sigma ^{(2\lambda _k+2)(1/p-1/q)}, \end{aligned}$$

and an extremizer can be taken as

$$\begin{aligned} f_{\sigma ,m}(x)=\frac{\sin ^{2m}(\theta |x|)}{|x|^{2m}},\quad \theta =\frac{\sigma }{2m}, \end{aligned}$$

for sufficiently large $m\in \mathbb {N}$.

Proof

Let $f\in B_{p, k}^\sigma$, $p\ge 1$, $q=\infty$. By Theorem 5.14, we have ${\text {supp}}\mathcal {F}_k(f)\subset B_\sigma$, and then

$$\begin{aligned} \mathcal {F}_k(f)(y)=\eta (y/\sigma )\mathcal {F}_k(f)(y),\quad \eta (y)=\eta _0(|y|). \end{aligned}$$

(7.2)

Lemma 3.9 implies

$$\begin{aligned} f(x)=(f*_{ \lambda _k}\mathcal {H}_{\lambda _k}(\eta _0(\cdot /\sigma )))(x)= \int _0^{\infty }T^tf(x)\mathcal {H}_{\lambda _k}(\eta _0(\cdot /\sigma ))(t)\,\mathrm{d} \nu _{\lambda _k}(t). \end{aligned}$$

Taking into account that

$$\begin{aligned} \mathcal {H}_{\lambda _k}(\eta _0(\cdot /\sigma ))(t)= & {} \sigma ^{2\lambda _k+2} \mathcal {H}_{\lambda _k}(\eta _0)(\sigma t),\\ \Vert \mathcal {H}_{\lambda _k}(\eta _0)(\sigma t)\Vert _{p',\mathrm{d}\mu _k}= & {} \sigma ^{-\frac{2\lambda _k+2}{p'}}\Vert \mathcal {H}_{\lambda _k} (\eta _0)(t)\Vert _{p',\mathrm{d}\mu _k}, \end{aligned}$$

Hölder’s inequality and Theorem 3.5 yield

$$\begin{aligned} |f(x)|&\le \sigma ^{2\lambda _k+2}\Vert T^tf(x)\Vert _{p,\mathrm{d}\nu _{\lambda _k}}\Vert \mathcal {H}_{\lambda _k} (\eta _0)(\sigma t)\Vert _{p',\mathrm{d}\mu _k}\\&\le \sigma ^{(2\lambda _k+2)/p}\Vert \mathcal {H}_{\lambda _k}(\eta _0)(t) \Vert _{p',\mathrm{d}\mu _k}\Vert f\Vert _{p,\mathrm{d}\mu _k} \lesssim \sigma ^{(2\lambda _k+2)/p}\Vert f\Vert _{p,\mathrm{d}\mu _k}, \end{aligned}$$

i.e., (7.1) holds.

Let $f\in B_{p, k}^\sigma$, $0<p<1$, $q=\infty$. By Theorem 5.1, f is bounded and $f\in B_{1, k}^\sigma$. We have

$$\begin{aligned} \Vert f\Vert _{1,\mathrm{d}\mu _k} = \Vert |f|^{1-p} |f|^p\Vert _{1,\mathrm{d}\mu _k} \le \Vert |f|^{1-p}\Vert _{\infty } \Vert |f|^p\Vert _{1,\mathrm{d}\mu _k} = \Vert f\Vert _\infty ^{1-p} \Vert f\Vert _{p,\mathrm{d}\mu _k}^p. \end{aligned}$$

Using (7.1) with $p=1$ and $q=\infty$,

$$\begin{aligned} \Vert f\Vert _{1,\mathrm{d}\mu _k}\lesssim \sigma ^{2\lambda _k+2}\Vert f\Vert _{1,\mathrm{d}\mu _k} \Vert f\Vert _\infty ^{-p} \Vert f\Vert _{p,\mathrm{d}\mu _k}^p, \end{aligned}$$

which gives

$$\begin{aligned} \Vert f\Vert _\infty \lesssim \sigma ^{(2\lambda _k+2)/p}\Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

Thus, the proof of (7.1) for $q=\infty$ is complete.

If $0<p\le q<\infty$, we obtain

$$\begin{aligned} \Vert f\Vert _{q,\mathrm{d}\mu _k}&= \Vert |f|^{1-p/q}|f|^{p/q}\Vert _{q,\mathrm{d}\mu _k} \le \Vert f\Vert _\infty ^{1-p/q} \Vert f\Vert _{p,\mathrm{d}\mu _k}^{p/q}\\&\le \sigma ^{(2\lambda _k+2)(1-p/q)/p} \Vert f\Vert _{p,\mathrm{d}\mu _k}^{1-p/q} \Vert f\Vert _{p,\mathrm{d}\mu _k}^{p/q} = \sigma ^{(2\lambda _k+2)(1/p-1/q)}\Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

$\square$

Theorem 7.3

If $\sigma >0$, $r\in \mathbb {N}$, $1 \le p\le \infty$, $f\in B_{p, k}^\sigma$, then

$$\begin{aligned} \Vert (-\Delta _k)^rf\Vert _{p,\mathrm{d}\mu _k} \lesssim \sigma ^{2r}\Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(7.3)

Proof

It is enough to consider the case $r=1$. As in the previous theorem, we use (7.2) to obtain

$$\begin{aligned} \mathcal {F}_k((-\Delta _k)f)(y)=|y|^2\eta (y/\sigma )\mathcal {F}_k(f)(y)= \sigma ^2\varphi _0(|y|/\sigma )\mathcal {F}_k(f)(y), \end{aligned}$$

where $\varphi _0(t)=t^2\eta _0(t)\in \mathcal {S}(\mathbb {R}_+)$. Combining Lemma 3.9, inequality (3.12), and $\Vert \mathcal {F}_k(\varphi _0(|\cdot |/\sigma ))\Vert _{1,\mathrm{d}\mu _k}=\Vert \mathcal {F}_k (\varphi _0(|\cdot |))\Vert _{1,\mathrm{d}\mu _k}$, we arrive at

$$\begin{aligned} \Vert (-\Delta _k)f\Vert _{p,\mathrm{d}\mu _k}\le \sigma ^2 \Vert \mathcal {F}_k(\varphi _0(|\cdot |))\Vert _{1,\mathrm{d}\mu _k}\Vert f\Vert _{p,\mathrm{d}\mu _k}\lesssim \sigma ^2 \Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

$\square$

The next result follows from Lemma 6.10, Remark 6.11, and Theorem 7.3.

Corollary 7.4

If $\sigma ,\,\delta >0$, $m\in \mathbb {N}$, $1 \le p\le \infty$, $f\in B_{p, k}^\sigma$, then

$$\begin{aligned} \omega _m(\delta , f)_{p,\mathrm{d}\mu _k}&\lesssim (\sigma \delta )^{2m}\Vert f\Vert _{p,\mathrm{d}\mu _k},\\ {^{*}}\omega _{l}(\delta , f)_{p,\mathrm{d}\mu _k}&\lesssim (\sigma \delta )^{2m}\Vert f\Vert _{p,\mathrm{d}\mu _k},\quad l=2m-1,\,2m,\\ {^{**}}\omega _m(\delta , f)_{p,\mathrm{d}\mu _k}&\lesssim (\sigma \delta )^{2m}\Vert f\Vert _{p,\mathrm{d}\mu _k}, \end{aligned}$$

where constants do not depend on $\sigma , \delta ,$ and f.

Theorem 7.5

If $\sigma >0$, $m\in \mathbb {N}$, $1 \le p\le \infty$, $0<t\le 1/(2\sigma )$, $f\in B_{p, k}^\sigma$, then

$$\begin{aligned} \Vert (-\Delta _k)^mf\Vert _{p,\mathrm{d}\mu _k} \lesssim t^{-2m}\Vert {^{**}}\varDelta _{t}^{m} f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(7.4)

Remark 7.6

By Remark 6.8, this inequality can be equivalently written as

$$\begin{aligned} \Vert (-\Delta _k)^mf\Vert _{p,\mathrm{d}\mu _k} \lesssim t^{-2m} K_{2m}(t, f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Proof

We have

$$\begin{aligned} \mathcal {F}_k((-\Delta _k)^mf)(y)=\frac{|y|^{2m}\eta (y/\sigma )}{j^{**}_{\lambda _{k},m}(t|y|)}\, j^{**}_{\lambda _{k},m}(t|y|)\mathcal {F}_k(f)(y). \end{aligned}$$

Since for $0<t\le 1/(2\sigma )$,

$$\begin{aligned} \eta (y/\sigma )=\eta (y/\sigma )\eta (ty), \end{aligned}$$

we obtain that

$$\begin{aligned} \mathcal {F}_k((-\Delta _k)^mf)(y)=t^{-2m}\eta (y/\sigma )\varphi (ty) j^{**}_{\lambda _{k},m}(t|y|)\mathcal {F}_k(f)(y), \end{aligned}$$

where

$$\begin{aligned} \varphi (y)=\frac{|y|^{2m}\eta (y)}{j^{**}_{\lambda _{k},m}(|y|)}\in \mathcal {S}(\mathbb {R}^d). \end{aligned}$$

Using

$$\begin{aligned} j^{**}_{\lambda _{k},m}(t|\cdot |)\mathcal {F}_k(f)=\mathcal {F}_k({^{**}}\varDelta _{t}^{m}f),\quad {^{**}}\varDelta _{t}^{m}f\in L^{p}(\mathbb {R}^d, \mathrm{d}\mu _k), \end{aligned}$$

and

$$\begin{aligned} \Vert \mathcal {F}_k(\eta (\cdot /\sigma ))\Vert _{1,\mathrm{d}\mu _k}=\Vert \mathcal {F}_k(\eta )\Vert _{1,\mathrm{d}\mu _k},\quad \Vert \mathcal {F}_k(\varphi (t\cdot ))\Vert _{1,\mathrm{d}\mu _k}=\Vert \mathcal {F}_k(\varphi )\Vert _{1,\mathrm{d}\mu _k}, \end{aligned}$$

and combining Lemma 3.9 and inequality (3.12), we have

$$\begin{aligned} \Vert (-\Delta _k)^mf\Vert _{p,\mathrm{d}\mu _k}&\le t^{-2m}\Vert \mathcal {F}_k(\eta (\cdot /\sigma ))\Vert _{1,\mathrm{d}\mu _k}\Vert \mathcal {F}_k(\varphi (t\cdot ))\Vert _{1,\mathrm{d}\mu _k} \Vert {^{**}}\varDelta _{t}^{m} f\Vert _{p,\mathrm{d}\mu _k}\\&=t^{-2m}\Vert \mathcal {F}_k(\eta )\Vert _{1,\mathrm{d}\mu _k}\Vert \mathcal {F}_k(\varphi )\Vert _{1,\mathrm{d}\mu _k} \Vert {^{**}}\varDelta _{t}^{m} f\Vert _{p,\mathrm{d}\mu _k}\\&\lesssim t^{-2m}\Vert {^{**}}\varDelta _{t}^{m} f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

$\square$

Theorem 7.7

If $\sigma >0$, $m\in \mathbb {N}$, $1 \le p\le \infty$, $0<\delta \le t\le 1/(2\sigma )$, $f\in B_{p, k}^\sigma$, then

$$\begin{aligned} \delta ^{-2m}\Vert {^{**}}\varDelta _{\delta }^{m} f\Vert _{p,\mathrm{d}\mu _k} \lesssim t^{-2m}\Vert {^{**}}\varDelta _{t}^{m} f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(7.5)

Remark 7.8

Using Remark 6.8, Theorem 7.5, and taking into account that $\delta ^{-2m} K_{2m}(\delta , f)_{p,\mathrm{d}\mu _k}$ is decreasing in $\delta$ (see (6.4)), inequality (7.5) can be equivalently written as

$$\begin{aligned}&\Vert (-\Delta _k)^mf\Vert _{p,\mathrm{d}\mu _k} \asymp \delta ^{-2m}\Vert {^{**}}\varDelta _{\delta }^{m} f\Vert _{p,\mathrm{d}\mu _k} \asymp t^{-2m}\Vert {^{**}}\varDelta _{t}^{m} f\Vert _{p,\mathrm{d}\mu _k},\\&\Vert (-\Delta _k)^mf\Vert _{p,\mathrm{d}\mu _k} \asymp \delta ^{-2m} K_{2m}(\delta , f)_{p,\mathrm{d}\mu _k} \asymp t^{-2m}K_{2m}(t, f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Proof

We have

$$\begin{aligned} \mathcal {F}_k({^{**}}\varDelta _{\delta }^{m} f)(y)&=j^{**}_{\lambda _{k},m}(\delta |y|) \mathcal {F}_k(f)(y)\\&=\eta (y/\sigma )\frac{j^{**}_{\lambda _{k},m}(\delta |y|) \eta (ty)}{j^{**}_{\lambda _{k},m}(t|y|)}\mathcal {F}_k({^{**}}\varDelta _{t}^{m} f)(y)\\&=\theta ^{2m}\eta (y/\sigma )\varphi _{\theta }(ty)\mathcal {F}_k({^{**}}\varDelta _{t}^{m} f)(y), \end{aligned}$$

where $\theta =(\delta /t)^{2m}\in (0, 1]$,

$$\begin{aligned} \varphi _{\theta }(y)=\frac{\psi (\theta y)\eta (y)}{\psi (y)}\in \mathcal {S}(\mathbb {R}^d),\quad \psi (y)=\frac{j^{**}_{\lambda _{k},m}(|y|)}{|y|^{2m}}\in C^{\infty }(\mathbb {R}^d). \end{aligned}$$

Using Lemma 3.9 and estimate (3.12), we arrive at inequality (7.5):

$$\begin{aligned} \Vert {^{**}}\varDelta _{\delta }^{m} f\Vert _{p,\mathrm{d}\mu _k}&\le \theta ^{2m}\Vert \mathcal {F}_k(\eta )\Vert _{1,\mathrm{d}\mu _k}\max _{0\le \theta \le 1}\Vert \mathcal {F}_k(\varphi _{\theta })\Vert _{1,\mathrm{d}\mu _k}\Vert {^{**}}\varDelta _{t}^{m} f\Vert _{p,\mathrm{d}\mu _k}\\&\lesssim \left( \frac{\delta }{t}\right) ^{2m}\Vert {^{**}}\varDelta _{t}^{m} f\Vert _{p,\mathrm{d}\mu _k}, \end{aligned}$$

provided that the function $n(\theta )=\Vert \mathcal {F}_k(\varphi _{\theta })\Vert _{1,\mathrm{d}\mu _k}$ is continuous on [0, 1]. Let us prove this.

Set $\varphi _{\theta }(y)=\varphi _{\theta 0}(|y|)$, $r=|y|$, $\rho =|x|$. Then

$$\begin{aligned} n(\theta )&=\int _{\mathbb {R}^d}\left| \int _{\mathbb {R}^d}\varphi _{\theta }(y)e_k(x,y)\, \mathrm{d}\mu _k(y)\right| \,\mathrm{d}\mu _k(x)\\&=\int _{0}^{\infty }\left| \int _{0}^2\varphi _{\theta 0}(r)j_{\lambda _k}(\rho r)\,\mathrm{d}\nu _{\lambda _k}(r)\right| \,\mathrm{d}\nu _{\lambda _k}(\rho )\\&=b_{\lambda _k}^2\int _{0}^{\infty }\left| \int _{0}^2\varphi _{\theta 0}(r)j_{\lambda _k}(\rho r)r^{2\lambda _k+1}\,\mathrm{d}r\right| \rho ^{2\lambda _k+1}\,\mathrm{d}\rho . \end{aligned}$$

The inner integral continuously depends on $\theta$. Let us show that the outer integral converges uniformly in $\theta \in [0,1]$. Since [2, Sect. 7.2]

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}r}\left( j_{\lambda _k+1}(\rho r)r^{2\lambda _k+2}\right) =(2\lambda _k+2)j_{\lambda _k}(\rho r)r^{2\lambda _k+1}, \end{aligned}$$

integrating by parts implies

$$\begin{aligned}&\int _{0}^2\varphi _{\theta 0}(r)j_{\lambda _k}(\rho r)r^{2\lambda _k+1}\,\mathrm{d}r=\int _{0}^2\varphi _{\theta 0}(r)\,d \left( \int _{0}^{r}j_{\lambda _k}(\rho \tau )\tau ^{2\lambda _k+1}\right) \\&\quad =\frac{1}{2\lambda _k+2}\int _{0}^2\varphi _{\theta 0}(r)\,d\left( j_{\lambda _k+1}(\rho r)r^{2\lambda _k+2}\right) \\&\quad =-\frac{1}{2\lambda _k+2}\int _{0}^2\frac{\varphi _{\theta 0}(r)}{r}j_{\lambda _k+1}(\rho r)r^{2\lambda _k+3}\,\mathrm{d}r=\dots \\&\quad =(-1)^s\left( \prod _{j=1}^s(2\lambda _k+2s)\right) ^{-1} \int _{0}^2 \varphi _{\theta 0}^{[s]}(r)j_{\lambda _k+s}(\rho r)r^{2\lambda _k+2s+1}\,\mathrm{d}r, \end{aligned}$$

where

$$\begin{aligned} \varphi _{\theta 0}^{[s]}(r):=\frac{\mathrm{d}}{\mathrm{d}r}\left( \frac{\varphi _{\theta 0}^{[s-1]}(r)}{r}\right) . \end{aligned}$$

This and (6.20) give

$$\begin{aligned} \left| \int _{0}^2\varphi _{\theta 0}(r)j_{\lambda _k}(\rho r)r^{2\lambda _k+1}\,\mathrm{d}r\right| \le \frac{c_1(\lambda _k, m, s)}{(\rho +1)^{\lambda _k+s+1/2}} \end{aligned}$$

and, for $s>\lambda _k+3/2$,

$$\begin{aligned} n(\theta )\le c_2(\lambda _k, m, s)\int _{0}^{\infty }(1+\rho )^{-(s-\lambda _k-1/2)}\,\mathrm{d}\rho \le c_3(\lambda _k, m, s), \end{aligned}$$

completing the proof. $\square$

Remark 7.9

Combining (7.1) and (7.3), the following Bernstein–Nikolskii inequality is valid:

$$\begin{aligned} \Vert (-\Delta _k)^rf\Vert _{q,\mathrm{d}\mu _k}\lesssim \sigma ^{2r+(2\lambda _k+2)(1/p-1/q)}\Vert f\Vert _{p,\mathrm{d}\mu _k},\quad 1\le p\le q\le \infty . \end{aligned}$$

Remark 7.10

For radial functions, Nikolskii inequality (7.1), Bernstein (7.3), Nikolskii–Stechkin (7.4), and Boas inequality (7.5) follow from corresponding estimates in the space $L^{p}(\mathbb {R}_{+},\mathrm{d}\nu _{\lambda })$ proved in [34].

8 Realization of K-Functionals and Moduli of Smoothness

In the nonweighted case ($k\equiv 0$) the equivalence between the classical modulus of smoothness and the K-functional between $L^p$ and the Sobolev space $W_p^{r}$ is well known [8, 26]: $1\le p \le \infty$, for any integerrone has

$$\begin{aligned} \omega _r(t,f)_{{L^p(\mathbb {R})}} \asymp {K}_{r}(f,t)_p,\quad 1\le p \le \infty , \end{aligned}$$

where

$$\begin{aligned} {K}_{r}(f,t)_p:=\inf _{g\in \dot{W}_p^{r}} \left( \Vert f-g\Vert _p+ t^r \Vert g\Vert _{ {\dot{W}_p^{r}}}\right) . \end{aligned}$$

Starting from the paper [13] (see also [17, Lemma 1.1] for the fractional case), the following equivalence between the modulus of smoothness and the realization of the K-functional is widely used in approximation theory:

$$\begin{aligned} \omega _r(t,f)_{{L^p(\mathbb {R})} } \asymp \mathcal {R}_{r}(t,f)_p=\inf _{g}\left\{ \Vert f-g\Vert _p+t^{r}\Vert g^{(r)}\Vert _p\right\} , \end{aligned}$$

where g is an entire function of exponential type 1 / t.

Let the realization of the K-functional $K_{2r}(t, f)_{p,\mathrm{d}\mu _k}$ be given as follows:

$$\begin{aligned} \mathcal {R}_{2r}(t, f)_{p,\mathrm{d}\mu _k}=\inf \left\{ \Vert f-g\Vert _{p,\mathrm{d}\mu _k}+t^{2r}\Vert (-\Delta _k)^rg\Vert _{p,\mathrm{d}\mu _k}:g\in B_{p, k}^{1/t}\right\} \end{aligned}$$

and

$$\begin{aligned} \mathcal {R}^*_{2r}(t, f)_{p,\mathrm{d}\mu _k}=\Vert f-g^*\Vert _{p,\mathrm{d}\mu _k}+t^{2r}\Vert (-\Delta _k)^rg^*\Vert _{p,\mathrm{d}\mu _k}, \end{aligned}$$

where $g^*\in B_{p, k}^{1/t}$ is a near best approximant.

Theorem 8.1

If $t> 0$, $1\le p\le \infty$, $r\in \mathbb {N}$, then for any $f\in L^{p}(\mathbb {R}^{d},\mathrm{d}\mu _{k})$,

$$\begin{aligned} \begin{aligned} \mathcal {R}_{2r}(t, f)_{p,\mathrm{d}\mu _k}&\asymp \mathcal {R}^*_{2r}(t, f)_{p,\mathrm{d}\mu _k} \asymp K_{2r}(t, f)_{p,\mathrm{d}\mu _k}\asymp \omega _r(t, f)_{p,\mathrm{d}\mu _k}\\&\asymp {^{**}}\omega _r(t, f)_{p,\mathrm{d}\mu _k}\asymp {^{*}}\omega _{2r-1}(t, f)_{p,\mathrm{d}\mu _k}\asymp {^{*}}\omega _{2r}(t, f)_{p,\mathrm{d}\mu _k}. \end{aligned} \end{aligned}$$

Proof

By Theorem 6.6,

$$\begin{aligned} \begin{aligned} \omega _r(t, f)_{p,\mathrm{d}\mu _k}&\asymp {^{**}}\omega _r(t, f)_{p,\mathrm{d}\mu _k} \asymp {^{*}}\omega _{2r-1}(t, f)_{p,\mathrm{d}\mu _k}\asymp {^{*}}\omega _{2r}(t, f)_{p,\mathrm{d}\mu _k} \\&\asymp K_{2r}(t, f)_{p,\mathrm{d}\mu _k}\le \mathcal {R}_{2r}(t, f)_{p,\mathrm{d}\mu _k}\le \mathcal {R}^*_{2r}(t, f)_{p,\mathrm{d}\mu _k}, \end{aligned} \end{aligned}$$

where we have used the fact that $B_{p, k}^{1/t}\subset W_{p, k}^{2r}$, which follows from Theorem 7.3.

Therefore, it is enough to show that

$$\begin{aligned} \mathcal {R}^*_{2r}(t, f)_{p,\mathrm{d}\mu _k} \le C\omega _r(t, f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Indeed, for $g^*$ being the best approximant (or near best approximant), the Jackson inequality given in Theorem 6.4 implies that

$$\begin{aligned} \Vert f-g^{*}\Vert _{p,\mathrm{d}\mu _k}\lesssim E_{1/t}(f)_{p,\mathrm{d}\mu _k}\lesssim \omega _r(t, f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

(8.1)

Using the first inequality in Theorem 7.5 and taking into account (8.1), we have

$$\begin{aligned} \Vert (-\Delta _k)^rg^*\Vert _{p,\mathrm{d}\mu _k}&\lesssim t^{-2r}\Vert \varDelta _{t/2}^{r} g^*\Vert _{p,\mathrm{d}\mu _k}\\&\lesssim t^{-2r}\Vert \varDelta _{t/2}^{r} (g^*-f)\Vert _{p,\mathrm{d}\mu _k}+t^{-2r}\Vert \varDelta _{t/2}^{r} f\Vert _{p,\mathrm{d}\mu _k}\\&\lesssim t^{-2r}\Vert g^*-f\Vert _{p,\mathrm{d}\mu _k}+t^{-2r}\omega _r(t/2, f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Using again (8.1), we arrive at

$$\begin{aligned} \Vert f-g^*\Vert _{p,\mathrm{d}\mu _k}+t^{2r}\Vert (-\Delta _k)^rg^*\Vert _{p,\mathrm{d}\mu _k}\lesssim \omega _r(t, f)_{p,\mathrm{d}\mu _k}, \end{aligned}$$

completing the proof.

The next result answers the following important question (see, e.g., [22, 51]): when does the relation

$$\begin{aligned} \omega _m\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k} \asymp E_n(f)_{p,\mathrm{d}\mu _k} \end{aligned}$$

(8.2)

(or similar relations with concepts in Theorem 8.2) hold?

Theorem 8.2

Let $1\le p\le \infty$ and $m\in \mathbb {N}$. We have that (8.2) is valid if and only if

$$\begin{aligned} \omega _m\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k} \asymp \omega _{m+1}\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}. \end{aligned}$$

(8.3)

Proof

We prove only the nontrivial part that (8.3) implies (8.2). Since, by (6.4), we have $\omega _m(nt, f)_{p,\mathrm{d}\mu _k} \lesssim n^{2m} \omega _m(t, f)_{p,\mathrm{d}\mu _k}$, relation (8.3) implies that

$$\begin{aligned} \omega _{m+1}(nt, f)_{p,\mathrm{d}\mu _k} \lesssim n^{2m} \omega _{m+1}(t, f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

(8.4)

This and Jackson’s inequality give

$$\begin{aligned}&\frac{1}{n^{2(m+1)}} \sum _{j=0}^n (j+1)^{2(m+1)-1} E_j(f)_{p,\mathrm{d}\mu _k}\\&\quad \lesssim \frac{1}{n^{2(m+1)}} \sum _{j=0}^n (j+1)^{2(m+1)-1} \omega _{m+1}\left( \frac{1}{j+1}, f\right) _{p,\mathrm{d}\mu _k}\\&\quad \lesssim \omega _{m+1}\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}. \end{aligned}$$

Moreover, Theorem 9.1 below implies

$$\begin{aligned} \omega _{m+1}\left( \frac{1}{l n}, f\right) _{p,\mathrm{d}\mu _k}&\lesssim \frac{1}{(ln)^{2(m+1)}} \sum _{j=0}^{ln} (j+1)^{2(m+1)-1} E_j(f)_{p,\mathrm{d}\mu _k}\\&\lesssim \frac{1}{l^{2(m+1)}}\,\omega _{m+1}\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}\\&\quad +\frac{1}{(ln)^{2(m+1)}} \sum _{j=n+1}^{ln} (j+1)^{2(m+1)-1} E_j(f)_{p,\mathrm{d}\mu _k}, \end{aligned}$$

or, in other words,

$$\begin{aligned}&\frac{1}{n^{2(m+1)}} \sum _{j=n+1}^{ln} (j+1)^{2(m+1)-1} E_j(f)_{p,\mathrm{d}\mu _k}\\&\quad \gtrsim C l^{2(m+1)} \omega _{m+1}\left( \frac{1}{l n}, f\right) _{p,\mathrm{d}\mu _k}- \omega _{m+1}\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}. \end{aligned}$$

Using again (8.4), we obtain

$$\begin{aligned} \frac{1}{n^{2(m+1)}} \sum _{j=n+1}^{ln} (j+1)^{2(m+1)-1} E_j(f)_{p,\mathrm{d}\mu _k}\gtrsim ( C l^{2}-1) \omega _{m+1}\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}. \end{aligned}$$

Taking into account monotonicity of $E_j(f)_{p,\mathrm{d}\mu _k}$ and choosing l sufficiently large, we arrive at (8.2). $\square$

9 Inverse Theorems of Approximation Theory

Theorem 9.1

Let $m, n \in \mathbb {N}$, $1\le p\le \infty$, $f \in L^{p}(\mathbb {R}^d, \mathrm{d}\mu _k)$. We have

$$\begin{aligned} K_{2m}\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k} \lesssim \frac{1}{n^{2m}} \sum _{j=0}^n (j+1)^{2m-1} E_j(f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

(9.1)

Remark 9.2

By Remark 6.8, $K_{2m}\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}$ in this inequality can be equivalently replaced by $\omega _m\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}$, ${^{**}}\omega _m\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}$, and ${^{*}}\omega _l\left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}$, $l=2m-1,2m$.

Proof

Let us prove (9.1) for $\omega _m \left( \frac{1}{n}, f\right) _{p,\mathrm{d}\mu _k}$. By Theorem 5.15, for any $\sigma >0$, there exists $f_\sigma \in B_{p, k}^\sigma$ such that

$$\begin{aligned} \Vert f-f_\sigma \Vert _{p,\mathrm{d}\mu _k}=E_\sigma (f)_{p,\mathrm{d}\mu _k},\quad E_0(f)_{p,\mathrm{d}\mu _k}=\Vert f\Vert _{p,\mathrm{d}\mu _k}. \end{aligned}$$

For any $s\in \mathbb {Z}_+$,

$$\begin{aligned} \omega _m(1/n, f)_{p,\mathrm{d}\mu _k}&\le \omega _m(1/n, f-f_{2^{s+1}})_{p,\mathrm{d}\mu _k}+ \omega _m(1/n, f_{2^{s+1}})_{p,\mathrm{d}\mu _k}\\&\lesssim E_{2^{s+1}}(f)_{p,\mathrm{d}\mu _k}+\omega _m(1/n, f_{2^{s+1}})_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Using Lemma 6.10,

$$\begin{aligned}&\omega _m(1/n, f_{2^{s+1}})_{p,\mathrm{d}\mu _k}\lesssim {n^{-2m}}\Vert (-\Delta _k)^m f_{2^{s+1}}\Vert _{p,\mathrm{d}\mu _k}\\&\quad \lesssim \frac{1}{n^{2m}} \left( \Vert (-\Delta _k)^mf_1\Vert _{p,\mathrm{d}\mu _k}+\sum _{j=0}^s \Vert (-\Delta _k)^mf_{2^{j+1}}- (-\Delta _k)^mf_{2^j}\Vert _{p,\mathrm{d}\mu _k}\right) . \end{aligned}$$

Then Bernstein inequality (7.3) implies that

$$\begin{aligned} \Vert (-\Delta _k)^mf_{2^{j+1}}-(-\Delta _k)^mf_{2^j}\Vert _{p,\mathrm{d}\mu _k}&\lesssim 2^{2m(j+1)}\Vert f_{2^{j+1}}-f_{2^j}\Vert _{p,\mathrm{d}\mu _k}\\&\lesssim 2^{2m(j+1)}E_{2^j}(f)_{p,\mathrm{d}\mu _k},\\ \Vert (-\Delta _k)^mf_1\Vert _{p,\mathrm{d}\mu _k}&\lesssim E_0(f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Thus,

$$\begin{aligned} \omega _m(1/n, f_{2^{s+1}})_{p,\mathrm{d}\mu _k}\lesssim \frac{1}{n^{2m}}\left( E_0(f)_{p,\mathrm{d}\mu _k}+ \sum _{j=0}^s2^{2m(j+1)}E_{2^j}(f)_{p,\mathrm{d}\mu _k}\right) . \end{aligned}$$

Taking into account that

$$\begin{aligned} \sum _{l=2^{j-1}+1}^{2^j} l^{2m-1}E_l(f)_{p,\mathrm{d}\mu _k} \ge 2^{2m(j-1)}E_{2^j}(f)_{p,\mathrm{d}\mu _k}, \end{aligned}$$

(9.2)

we have

$$\begin{aligned}&\omega _m(1/n, f_{2^{s+1}})_{p,\mathrm{d}\mu _k}\lesssim \frac{1}{n^{2m}}\left( E_0(f)_{p,\mathrm{d}\mu _k}+2^{2m} E_1(f)_{p,\mathrm{d}\mu _k}\right. \\&\quad \left. +\sum _{j=1}^s 2^{4m} \sum _{l=2^{j-1}+1}^{2^j} l^{2m-1} E_l(f)_{p,\mathrm{d}\mu _k}\right) \lesssim \frac{1}{n^{2m}}\sum _{j=0}^{2^s}(j+1)^{2m-1}E_j(f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Choosing s such that $2^s\le n<2^{s+1}$ implies (9.1). $\square$

Theorem 9.1 and Jackson’s inequality imply the following Marchaud inequality.

Corollary 9.3

Let $m \in \mathbb {N}$, $1\le p\le \infty$, $f \in L^{p}(\mathbb {R}^d, \mathrm{d}\mu _k)$. We have

$$\begin{aligned} K_{2m}(\delta , f)_{p,\mathrm{d}\mu _k} \lesssim \delta ^{2m} \left( \Vert f\Vert _{p,\mathrm{d}\mu _k}+ \int _\delta ^1 {t^{-2m}} {K_{2m+2}(t, f)_{p,\mathrm{d}\mu _k}}\,\frac{\mathrm{d}t}{t} \right) . \end{aligned}$$

Theorem 9.4

Let $1\le p\le \infty$, $f \in L^{p}(\mathbb {R}^d, \mathrm{d}\mu _k)$, and $r\in \mathbb {N}$ be such that $\sum _{j=1}^{\infty } j^{2r-1}E_j(f)_{p,\mathrm{d}\mu _k} <\infty .$ Then $f \in W^{2r}_{p, k}$, and, for any $m,n\in \mathbb {N}$, we have

$$\begin{aligned} K_{2m}\left( \frac{1}{n}, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}\lesssim & {} \frac{1}{n^{2r}} \sum _{j=0}^n (j+1)^{2k+2r-1}E_j(f)_{p,\mathrm{d}\mu _k} \nonumber \\&+\sum _{j=n+1}^\infty j^{2r-1} E_j(f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

(9.3)

Remark 9.5

We can replace $K_{2m}\left( \frac{1}{n}, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}$ by any of moduli $\omega _m\left( \frac{1}{n}, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}$, ${^{*}}\omega _l\left( \frac{1}{n}, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k},$ and ${^{**}}\omega _m \left( \frac{1}{n}, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}$, $l=2m-1,2m$.

Proof

Let us prove (9.3) for $\omega _m \left( \frac{1}{n}, (-\Delta _k)^rf\right) _{p,\mathrm{d}\mu _k}$. Consider

$$\begin{aligned} (-\Delta _k)^r f_1+\sum _{j=0}^{\infty }\left( (-\Delta _k)^rf_{2^{j+1}}- (-\Delta _k)^rf_{2^{j}}\right) . \end{aligned}$$

(9.4)

By Bernstein’s inequality (7.3),

$$\begin{aligned} \Vert (-\Delta _k)^rf_{2^{j+1}}-(-\Delta _k)^rf_{2^j}\Vert _{p,\mathrm{d}\mu _k}\lesssim 2^{(j+1)r}E_{2^j}(f)_{p,\mathrm{d}\mu _k}\lesssim \sum _{l=2^{j-1}+1}^{2^j}l^{r-1}E_l(f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Therefore, series (9.4) converges to a function $g\in L^{p}(\mathbb {R}^d, \mathrm{d}\mu _k)$. Let us show that $g=(-\Delta _k)^rf$, i.e., $f\in W^{2r}_{p, k}$. Set

$$\begin{aligned} S_N=(-\Delta _k)^rf_1+\sum _{j=0}^N \left( (-\Delta _k)^rf_{2^{j+1}}-(-\Delta _k)^rf_{2^j}\right) . \end{aligned}$$

Then

$$\begin{aligned} \langle \mathcal {F}_k(g), \varphi \rangle&= \langle g, \mathcal {F}_k(\varphi )\rangle =\lim _{N\rightarrow \infty } \langle S_N, \mathcal {F}_k(\varphi )\rangle \\&= \lim _{N\rightarrow \infty }\langle \mathcal {F}_k(S_N), \varphi \rangle =\lim _{N\rightarrow \infty }\langle |y|^{2r}\mathcal {F}_k(f_{2^{N+1}}), \varphi \rangle = \langle |y|^{2r} \mathcal {F}_k(f), \varphi \rangle , \end{aligned}$$

where $\varphi \in \mathcal {S}(\mathbb {R}^d)$. Hence, $\mathcal {F}_k(g)(y) =|y|^{2r}\mathcal {F}_k(f)(y)$ and $g=(-\Delta _k)^rf$.

To obtain (9.3), we write

$$\begin{aligned} \omega _m(1/n, (-\Delta _k)^r f)_{p,\mathrm{d}\mu _k}\le \omega _m(1/n, (-\Delta _k)^r f-S_N)_{p,\mathrm{d}\mu _k}+\omega _m(1/n, S_N)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

The first term is estimated as follows

$$\begin{aligned} \omega _m(1/n, (-\Delta _k)^r f&-S_N)_{p,\mathrm{d}\mu _k} \lesssim \Vert (-\Delta _k)^r f-S_N\Vert _{p,\mathrm{d}\mu _k}\\&\lesssim \sum _{j=N+1}^{\infty }2^{2r(j+1)}E_{2^j}(f)_{p,\mathrm{d}\mu _k} \lesssim \sum _{l=2^N+1}^{\infty } l^{2r-1}E_l(f)_{p,\mathrm{d}\mu _k}. \end{aligned}$$

Moreover, by Corollary 7.4,

$$\begin{aligned} \omega _m(1/n, S_N)_{p,\mathrm{d}\mu _k}&\le \omega _m(1/n, (-\Delta _k)^r f_1)_{p,\mathrm{d}\mu _k}\\&\quad +\sum _{j=0}^{N}\omega _m\left( 1/n, (-\Delta _k)^r f_{2^{j+1}}-(-\Delta _k)^r f_{2^j}\right) _{p,\mathrm{d}\mu _k}\\&\lesssim \frac{1}{n^{2r}}\left( E_0(f)_{p,\mathrm{d}\mu _k}+ \sum _{j=0}^{N} 2^{2(m+r)(j+1)}E_{2^j}(f)_{p,\mathrm{d}\mu _k}\right) . \end{aligned}$$

Using (9.2) and choosing N such that $2^N\le n<2^{N+1}$ completes the proof of (9.3). $\square$