Logarithm Sobolev and Shannon’s Inequalities Associated with the Deformed Fourier Transform and Applications

Abstract

By using the symmetry of the Dunkl Laplacian operator, we prove a sharp Shannon-type inequality and a logarithmic Sobolev inequality for the Dunkl transform. Combining these inequalities, we obtain a new, short proof for Heisenberg-type uncertainty principles in the Dunkl setting. Moreover, by combining Nash’s inequality, Carlson’s inequality and Sobolev’s embedding theorems for the Dunkl transform, we prove new uncertainty inequalities involving the $L^{\infty }$-norm. Finally, we obtain a logarithmic Sobolev inequality in $L^p$-spaces, from which we derive an $L^p$-Heisenberg-type uncertainty inequality and an $L^p$-Nash-type inequality for the Dunkl transform.

1. Introduction

Let $T_j$, $j=1,\dots ,d$ be the the Dunkl operators (see [1]) associated to arbitrary finite reflection group G and non-negative multiplicity function k. These are differential-difference operators generalizing the usual partial derivatives. The Dunkl kernel $\mathcal{K}$ on ${\mathbb{R}}^d\times {\mathbb{R}}^d$ associated with G and k, introduced by C. F. Dunkl in [2], generalizes the usual exponential function (to which it reduces in the case $k=0$). This kernel is especially of interest as it gives rise to an integral transformation on ${\mathbb{R}}^d$ associated with G, k and to the weight measure $\mathrm{d}{\mu }_k\left(x\right)=w_k\left(x\right)\mathrm{d}x$ (where $w_k$ is the weight function given in (23)). It is called the Dunkl transform ${\mathcal{F}}_D$, and it is defined on $L_k^1\left({\mathbb{R}}^d\right)\cap L_k^2\left({\mathbb{R}}^d\right)$ by:

$${\mathcal{F}}_D\left(f\right)\left(\xi \right):={\mathbf{c}}_k{\int }_{{\mathbb{R}}^d}\mathcal{K}(-i\xi ,x)f\left(x\right)\mathrm{d}{\mu }_k\left(x\right),\xi \in {\mathbb{R}}^d,$$

(1)

where ${\mathbf{c}}_k$ is the Mehta-type constant given in (24), and $L_k^p\left({\mathbb{R}}^d\right)=L^p({\mathbb{R}}^d,{\mu }_k)$ and $1\le p<\infty$ are the Banach spaces consisting of measurable functions f on ${\mathbb{R}}^d$ equipped with the norms:

$${∥f∥}_{p,{\mu }_k}={\left({\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^p\mathrm{d}{\mu }_k\left(x\right)\right)}^{1/p}.$$

(2)

The Dunkl transform ${\mathcal{F}}_D$ extends uniquely to an isometric isomorphism on $L_k^2\left({\mathbb{R}}^d\right)$. In particular, if $k=0$, then the Dunkl operators are the usual partial derivatives and the Dunkl kernel is the usual exponential function so that the Dunkl transform is exactly the usual Fourier transform $\mathcal{F}$ defined by,

$$\mathcal{F}\left(f\right)\left(\xi \right)=\widehat{f}\left(\xi \right)={\left(2\pi \right)}^{-d/2}{\int }_{{\mathbb{R}}^d}f\left(x\right)e^{-i⟨x,\xi ⟩}\mathrm{d}x,\xi \in {\mathbb{R}}^d,$$

(3)

This is why the Dunkl transform can be considered one of the deformed Fourier transforms (see e.g., [3] for other deformed Fourier transforms). Moreover, if $f\left(x\right)=\tilde{f}\left(\right|x\left|\right)$ is a radial function on ${\mathbb{R}}^d$, then

$${\mathcal{F}}_D\left(f\right)\left(\xi \right)={\mathcal{H}}_{\gamma +d/2-1}\left(\tilde{f}\right)\left(\right|\xi \left|\right),\xi \in {\mathbb{R}}^d,$$

(4)

where ${\mathcal{H}}_{\alpha }$, $\alpha \ge -1/2$, is the Hankel transform (also known as the Fourier-Bessel transform) defined by (see [4]),

$${\mathcal{H}}_{\alpha }\left(\tilde{f}\right)\left(\lambda \right)={\int }_0^{\infty }\tilde{f}\left(x\right)j_{\alpha }\left(x\lambda \right)\frac{x^{2\alpha +1}\mathrm{d}x}{2^{\alpha }\mathrm{\Gamma }(\alpha +1)},\lambda \ge 0,$$

(5)

where $j_{\alpha }$ is the spherical Bessel function (see [5]), and $\mathrm{\Gamma }$ is the gamma function.

In order to present our result, let $L_{k,s}^p\left({\mathbb{R}}^d\right)=\left\{f\in L_k^p\left({\mathbb{R}}^d\right):{\left|x\right|}^{s}f\in L_k^p\left({\mathbb{R}}^d\right)\right\}$, $s>0$ and $1\le p<\infty$ be the weighted Lebesgue spaces, and let ${\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)=\{f\in L_k^2\left({\mathbb{R}}^d\right):$${\left|\xi \right|}^{\beta }{\mathcal{F}}_D\left(f\right)\in L_k^2\left({\mathbb{R}}^d\right)\}$, $\beta >0$ be the Dunkl–Sobolev space on ${\mathbb{R}}^d$. Then we prove the following new Heisenberg-type uncertainty inequalities for the Dunkl transform.

Theorem 1.

Let$s,\beta >0$.

1.

If$s,\beta >\gamma +d/2$, then there exists a constant$C(s,k,\beta ,d)$such that for every nonzero function$f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge C(s,k,\beta ,d){\left(\frac{{∥f∥}_{1,{\mu }_k}{∥f∥}_{\infty }}{{∥f∥}_{2,{\mu }_k}^2}\right)}^{\frac{s\beta }{\gamma +d/2}}{∥f∥}_{2,{\mu }_k}^{s+\beta }.$$

(6)

2.

If$\beta >\gamma +d/2$, then there exists a constant$C(s,k,\beta ,d)$such that for every nonzero function$f\in L_{k,s}^1\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge C(s,k,\beta ,d){\left(\frac{{∥f∥}_{1,{\mu }_k}{∥f∥}_{\infty }}{{∥f∥}_{2,{\mu }_k}^2}\right)}^{\frac{s\beta }{\gamma +d/2}}{∥f∥}_{1,{\mu }_k}^{\beta }{∥f∥}_{2,{\mu }_k}^s.$$

(7)

The proof of these inequalities is based on Nash’s inequality [6] (Proposition 3.2) and Carlson’s inequality [6] (Proposition 3.3) for the Dunkl transform, combined with the following new inequalities:

1.

For all $s>\beta +\frac{2\gamma +d}{p^{\prime }}$, $\beta \ge 0$ and $f\in L_{k,s}^p\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^{\beta }{f\parallel }_{1,{\mu }_k}\le {C(s,p,k,\beta ,d)\parallel f\parallel }_{p,{\mu }_k}^{1-\frac{\beta p^{\prime }+2\gamma +d}{s{p}^{\prime }}}{\parallel {\left|x\right|}^{s}f\parallel }_{p,{\mu }_k}^{\frac{\beta p^{\prime }+2\gamma +d}{s{p}^{\prime }}}.$$

(8)

2.

For all $\beta >\frac{2\gamma +d}{2}$ and $f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$${\parallel f\parallel }_{\infty }\le {c(k,\beta ,d)\parallel f\parallel }_{2,{\mu }_k}^{1-\frac{2\gamma +d}{2\beta }}{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^{\frac{2\gamma +d}{2\beta }}.$$

(9)

Theorem 1 is a variation of the following uncertainty inequalities (see [6,7]):

1.

There exists a constant $c(s,k,\beta ,d)$ such that for all $f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge c(s,k,\beta ,d){\parallel f\parallel }_{2,{\mu }_k}^{s+\beta },s,\beta >0.$$

(10)

2.

There exists a constant $c(s,k,\beta ,d)$ such that for all $f\in L_{k,s}^1\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge c(s,k,\beta ,d){∥f∥}_{1,{\mu }_k}^{\beta }{∥f∥}_{2,{\mu }_k}^s,s,\beta >0.$$

(11)

Heisenberg-type uncertainty inequalities (6) and (7) are related to Inequality (11), since in both cases the proof is based on Nash’s inequality and Carlson’s inequality for the Dunkl transform. Unfortunately, these three inequalities are not sharp, and the best constants and extremal functions remained unknown until now. However, the proof of Heisenberg-type uncertainty inequality (10) can be obtained from either the Faris-type local uncertainty principle [7] (Theorem A) or from the Benedicks–Amrein–Berthier uncertainty principle [7] (Theorem B). Moreover, the sharp constant in (10) is well known only for the special case $s=\beta =1$, and it is equal to $\gamma +d/2$; equality in (10) occurs if and only if $f\left(x\right)=c{e}^{-{\lambda \left|x\right|}^2}$ for some $c\in \mathbb{C}$, and $\lambda >0$ (see [8,9,10]).

Our second result will be the following sharp Shannon-type and logarithmic Sobolev inequalities for the Dunkl transform.

Theorem 2.

Let$s,\beta >0$.

1.

There exists an optimal constant$c(k,s,d)$such that for all nonzero$f\in L_{k,s}^1\left({\mathbb{R}}^d\right)$,

$$-{\int }_{{\mathbb{R}}^d}\frac{\left|f\right(x\left)\right|}{{\parallel f\parallel }_{1,{\mu }_k}}ln\left(\frac{\left|f\right(x\left)\right|}{{\parallel f\parallel }_{1,{\mu }_k}}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{s}ln\left(c(k,s,d)\frac{\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}}{{\parallel f\parallel }_{1,{\mu }_k}}\right).$$

(12)

2.

If$2\gamma +d>2$, then there exists a positive constant${\mathcal{S}}_{k,d}$such that for all nonzero f belongs to${\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$,

$${\int }_{{\mathbb{R}}^d}\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}ln\left({\mathcal{S}}_{k,d}\frac{\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).$$

(13)

3.

There exists a positive constant$A_{k,\beta }\left(d\right)$such that for all nonzero$f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$${\int }_{{\mathbb{R}}^d}\frac{{\left|f\left(x\right)\right|}^2}{{∥f∥}_{2,{\mu }_k}^2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{∥f∥}_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2\beta }ln\left(A_{k,\beta }\left(d\right)\frac{{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).$$

(14)

Shannon-type inequality (12) is sharp, and the constant ${\mathcal{S}}_{k,d}$ that appears in (13) is the optimal constant of the following Sobolev-type inequality for the Dunkl transform (see [11]): For all $f\in {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$,

$${\parallel f\parallel }_{2^{*},{\mu }_k}^2\le {\mathcal{S}}_{k,d}{\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k}^2,2\gamma +d>2,2^{*}=\frac{2(2\gamma +d)}{2\gamma +d-2}.$$

(15)

More generally (see [11] (Theorem 1.1)), if $1\le p<2\gamma +d$, and if $p^{*}=\frac{p(2\gamma +d)}{2\gamma +d-p}$, then there exists a constant $C_k(d,p)$ such that for all f in the Dunkl–Sobolev space

$$W_k^{1,p}\left({\mathbb{R}}^d\right)=\{f\in L_k^p\left({\mathbb{R}}^d\right):{\nabla }_{k}f\in L_k^p\left({\mathbb{R}}^d\right)\},$$

we have

$${\parallel f\parallel }_{p^{*},{\mu }_k}\le C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}.$$

(16)

Using the sharp Shannon’s inequality (12) and the logarithmic Sobolev inequality (13) we derive for $d\gg 1$, the well known sharp Heisenberg’s uncertainty inequality for the Dunkl transform (see [9]) is:

$${\parallel \left|x\right|f\parallel }_{2,{\mu }_k}\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}\ge \frac{2\gamma +d}{2}{\parallel f\parallel }_{2,{\mu }_k}^2.$$

(17)

Moreover, from Inequalities (12) and (14), we give another, short proof of Inequality (10).

Next, from the Sobolev-type inequality (16), we prove the following $L_k^p$-logarithmic Sobolev inequality.

Theorem 3.

If$1\le p<2\gamma +d$, then for all$f\in W_k^{1,p}\left({\mathbb{R}}^d\right)$such that${\parallel f\parallel }_{p,{\mu }_k}=1$, we have

$${\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{p}ln\left({\left|f\left(x\right)\right|}^p\right)\mathrm{d}{\mu }_k\left(x\right)\le (2\gamma +d)ln\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right),$$

(18)

where$C_k(d,p)$is the constant of the Sobolev inequality (16).

Consequently, we obtain the following $L_k^p$-Heisenberg-type uncertainty inequality and $L_k^p$-Nash-type inequality for the Dunkl transform.

Theorem 4.

1.

If$1\le p<2\gamma +d$, then there exists a constant$C(k,p)$such that for all$f\in L_{k,1}^p\left({\mathbb{R}}^d\right)\cap W_k^{1,p}\left({\mathbb{R}}^d\right)$,

$${\parallel \left|x\right|f\parallel }_{p,{\mu }_k}{\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\ge C(k,p){∥f∥}_{p,{\mu }_k}^2.$$

(19)

2.

If$1\le q<p<2\gamma +d$, then for all$f\in L_k^q\left({\mathbb{R}}^d\right)\cap W_k^{1,p}\left({\mathbb{R}}^d\right)$,

$${∥f∥}_{p,{\mu }_k}^{1+\frac{pq}{(p-q)(2\gamma +d)}}\le C_k(d,p){∥f∥}_{q,{\mu }_k}^{\frac{pq}{(p-q)(2\gamma +d)}}{\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k},$$

(20)

where$C_k(d,p)$is the constant of the Sobolev inequality (16).

The remainder of this paper is arranged as follows: in the next section we recall some useful results associated to the Dunkl transform. In Section 3, we prove a sharp Shannon-type inequality, and in Section 4, we prove some new logarithmic Sobolev inequalities. Section 4 is devoted to the study of a Nash-type inequality. Connecting the previous results, we give an application of the uncertainty principle, showing some new Heisenberg-type uncertainty inequalities.

2. Preliminaries

2.1. Notation

Let us denote by $⟨·,·⟩$ the scalar product and by $|·|$ the Euclidean norm on ${\mathbb{R}}^d$.

We denote by $L^{\infty }\left({\mathbb{R}}^d\right)$ the space of essentially bounded functions on ${\mathbb{R}}^d$, equipped with the standard essential supremum norm ${\parallel ·\parallel }_{\infty }$ defined by: For all $f\in L^{\infty }\left({\mathbb{R}}^d\right)$,

$${\parallel f\parallel }_{\infty }=\mathrm{ess}\underset{x\in {\mathbb{R}}^d}{sup}\left|f\left(x\right)\right|.$$

(21)

We denote by ${\mathcal{C}}_c^{\infty }\left({\mathbb{R}}^d\right)$ the set of compactly supported, infinitely differentiable functions on ${\mathbb{R}}^d$, and by ${\mathcal{C}}_0\left({\mathbb{R}}^d\right)$ we denote the space of continuous functions f on ${\mathbb{R}}^d$, vanishing at infinity.

We denote by $\mathcal{S}\left({\mathbb{R}}^d\right)$ the Schwartz space, constituted by the infinitely differentiable functions on ${\mathbb{R}}^d$, rapidly decreasing together with all their derivatives.

2.2. The Dunkl Transform

Let us fix some notation and present some necessary material on the Dunkl theory, which can be found in [2,12,13]. Let G be a finite reflection group on ${\mathbb{R}}^d$, associated with a root system R, and $R_{+}$ will be the positive subsystem of R. We denote by k a non-negative multiplicity function defined on R, with the property that k is G-invariant. We denote by $\left|G\right|$ the order of the reflection group associated to the root system R. We associate with k the index

$$\gamma :=\gamma \left(k\right)=\sum _{\xi \in R_{+}}k\left(\xi \right)\ge 0,$$

(22)

and the weight function $w_k$ defined by

$$w_k\left(x\right)=\prod _{\xi \in R_{+}}{\left|⟨\xi ,x⟩\right|}^{2k\left(\xi \right)}.$$

(23)

Further, we introduce the Mehta-type constant ${\mathbf{c}}_k$ by

$${\mathbf{c}}_k={\left({\int }_{{\mathbb{R}}^d}{e}^{-\frac{{\left|x\right|}^2}{2}}\mathrm{d}{\mu }_k\left(x\right)\right)}^{-1}.$$

(24)

Moreover,

$${\displaystyle {\int }_{{\mathbb{S}}^{d-1}}{w}_k\left(x\right)\mathrm{d}\sigma \left(x\right)=\frac{{\mathbf{c}}_k^{-1}}{2^{\gamma +d/2-1}\mathrm{\Gamma }(\gamma +d/2)}=d_k,}$$

(25)

where $\mathrm{d}\sigma$ is the Lebesgue measure on the unit sphere ${\mathbb{S}}^{d-1}$ of ${\mathbb{R}}^d$.

If $f\in L_k^1\left({\mathbb{R}}^d\right)$ is a radial function, that is $f\left(x\right)=\tilde{f}\left(\right|x\left|\right)$, then function $\tilde{f}$ defined on ${\mathbb{R}}_{+}=[0,\infty )$ is integrable with respect to the measure $r^{2\gamma +d-1}\mathrm{d}r$, and we have,

$$\begin{array}{ccc}\hfill {\int }_{{\mathbb{R}}^d}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)& =& {\int }_0^{\infty }\left({\int }_{{\mathbb{S}}^{d-1}}{w}_k\left(ry\right)\mathrm{d}\sigma \left(y\right)\right)\tilde{f}\left(r\right)r^{d-1}\mathrm{d}r\hfill \\ & =& {\displaystyle d_k{\int }_0^{\infty }\tilde{f}\left(r\right)r^{2\gamma +d-1}\mathrm{d}r.}\hfill \end{array}$$

(26)

Introduced by Dunkl in [1], the Dunkl operators $T_j$, $1\le j\le d$ on ${\mathbb{R}}^d$ associated with the reflection group G and the multiplicity function k are the first-order differential-difference operators given by

$$T_{j}f\left(x\right)=\frac{\partial f}{\partial x_j}+\sum _{\xi \in R_{+}}k\left(\xi \right){\xi }_j\frac{f\left(x\right)-f\left({\sigma }_{\xi }\left(x\right)\right)}{⟨\xi ,x⟩},x\in {\mathbb{R}}^d,$$

(27)

where f is an infinitely differentiable function on ${\mathbb{R}}^d$, ${\xi }_j=⟨\xi ,e_j⟩$, $(e_1,\dots ,e_d)$ being the canonical basis of ${\mathbb{R}}^d$, and ${\sigma }_{\xi }$ denotes the reflection with respect to the hyperplane orthogonal to $\xi$.

We will denote by ${\nabla }_k=\left(T_1,\dots ,T_d\right)$ the Dunkl gradient. Note that for $k=0$, the Dunkl operators reduce to partial derivatives, and ${\nabla }_0=\nabla$ is the usual gradient.

The Dunkl kernel $\mathcal{K}$ on ${\mathbb{R}}^d\times {\mathbb{R}}^d$ has been introduced by Dunkl in [2]; that is, for $\xi \in {\mathbb{R}}^d$, the function $x\mapsto \mathcal{K}(x,\xi )$ can be viewed as the solution on ${\mathbb{R}}^d$ of the following initial problem:

$$T_{j}u(x,\xi )={\xi }_{j}u(x,\xi ),1\le j\le d;u(0,\xi )=1.$$

(28)

Therefore, for all $\lambda \in \mathbb{C}$, $z,z^{\prime }\in {\mathbb{C}}^d$ and $x,\xi \in {\mathbb{R}}^d$, we have,

$$\mathcal{K}(\lambda z,z^{\prime })=\mathcal{K}(z,\lambda z^{\prime }),\overline{\mathcal{K}(-i\xi ,x)}=\mathcal{K}(i\xi ,x),\left|\mathcal{K}(-i\xi ,x)\right|\le 1.$$

(29)

The Dunkl transform ${\mathcal{F}}_D$ associated with G and k is defined for an integrable function f by:

$${\mathcal{F}}_D\left(f\right)\left(\xi \right):=c_k{\int }_{{\mathbb{R}}^d}\mathcal{K}(-i\xi ,x)f\left(x\right)\mathrm{d}{\mu }_k\left(x\right),\xi \in {\mathbb{R}}^d.$$

(30)

Then, by (29) we have,

$${∥{\mathcal{F}}_D\left(f\right)∥}_{\infty }\le {\mathbf{c}}_k{∥f∥}_{1,{\mu }_k}.$$

(31)

It is well known that if $f\in L_k^1\left({\mathbb{R}}^d\right)$, then ${\mathcal{F}}_D\left(f\right)$ is a bounded continuous function on ${\mathbb{R}}^d$, and according to the Riemann–Lebesgue lemma for the Dunkl transform: ${\mathcal{F}}_D\left(f\right)\left(\xi \right)\to 0$ as $\left|\xi \right|\to \infty$ (i.e., ${\mathcal{F}}_D\left(f\right)\in {\mathcal{C}}_0\left({\mathbb{R}}^d\right)$. If, in addition, ${\mathcal{F}}_D\left(f\right)$ is in $L_k^1\left({\mathbb{R}}^d\right)$, then the inverse Dunkl transform is defined for almost every $x\in {\mathbb{R}}^d$ by:

$$f\left(x\right)={\mathbf{c}}_k{\int }_{{\mathbb{R}}^d}{\mathcal{F}}_D\left(f\right)\left(\xi \right)\mathcal{K}(ix,\xi )\mathrm{d}{\mu }_k\left(\xi \right).$$

(32)

The Dunkl transform is an isomorphism from $\mathcal{S}\left({\mathbb{R}}^d\right)$ into itself. In particular, the Dunkl transform can be extended to an isometric isomorphism from $L_k^2\left({\mathbb{R}}^d\right)$ onto itself, satisfying the following Plancherel formula: For all $f\in L_k^2\left({\mathbb{R}}^d\right)$, we have

$${∥{\mathcal{F}}_D\left(f\right)∥}_{2,{\mu }_k}={∥f∥}_{2,{\mu }_k}.$$

(33)

By using the Riesz–Thorin theorem and Inequalities (31) and (33), we have

$$\parallel {\mathcal{F}}_D{\left(f\right)\parallel }_{p^{\prime },{\mu }_k}\le {\mathbf{c}}_k^{2/p-1}{\parallel f\parallel }_{p,{\mu }_k},f\in L_k^p\left({\mathbb{R}}^d\right)1\le p\le 2,$$

(34)

where $p^{\prime }=\frac{p}{p-1}$ is the conjugate index of p. Then the Dunkl transform can be extended to a bounded linear operator from $L_k^p\left({\mathbb{R}}^d\right)$ to $L_k^{p^{\prime }}\left({\mathbb{R}}^d\right)$ if and only if $1\le p\le 2$.

Using the property that, for $f\in \mathcal{S}\left({\mathbb{R}}^d\right)$ and for $j=1,\dots ,d$,

$${\mathcal{F}}_D\left(T_{j}f\right)\left(\xi \right)=i{\xi }_j{\mathcal{F}}_D\left(f\right)\left(\xi \right),\xi \in {\mathbb{R}}^d,$$

(35)

we have, by Plancherel Formula (33),

$${∥{\nabla }_{k}f∥}_{2,{\mu }_k}={∥\left|\xi \right|{\mathcal{F}}_D\left(f\right)∥}_{2,{\mu }_k}.$$

(36)

Thus, Equality (35) can be extended to any function f in the Dunkl–Sobolev space $\{f\in L_k^2\left({\mathbb{R}}^d\right):|\xi |{\mathcal{F}}_D\left(f\right)\in L_k^2\left({\mathbb{R}}^d\right)\}$ (cf. [14]).

Finally, we define the dilation operator ${\delta }_{\lambda }^k$, $\lambda >0$ by

$${\delta }_{\lambda }^{k}f\left(x\right)={\lambda }^{\gamma +d/2}f\left(\lambda x\right),x\in {\mathbb{R}}^d.$$

(37)

Then for all $f\in L_k^2\left({\mathbb{R}}^d\right)$, we have

$$\parallel {\delta }_{\lambda }^k{f\parallel }_{2,{\mu }_k}={\parallel f\parallel }_{2,{\mu }_k},{\mathcal{F}}_D\left({\delta }_{\lambda }^{k}f\right)={\delta }_{1/\lambda }^k{\mathcal{F}}_D\left(f\right).$$

(38)

3. Sharp Shannon-Type Inequality

Following Shannon (see [15]), the entropy of a probability density function $\rho$ on ${\mathbb{R}}^d$ is defined by

$$\mathbf{E}\left(\rho \right)=-{\int }_{{\mathbb{R}}^d}\rho \left(x\right)ln\left(\rho \left(x\right)\right)\mathrm{d}{\mu }_k\left(x\right).$$

(39)

In this section, we continue the study of the sharp Shannon-type inequality in the Dunkl setting started in [16]. First, we consider the weighted Lebesgue spaces $L_{k,s}^p\left({\mathbb{R}}^d\right)$, $s>0$ and $1\le p<\infty$ by

$$L_{k,s}^p\left({\mathbb{R}}^d\right)=\left\{f\in L_k^p\left({\mathbb{R}}^d\right):{⟨x⟩}^{s}f\in L_k^p\left({\mathbb{R}}^d\right)\right\},$$

(40)

where $⟨x⟩={(1+|x|}^2{)}^{\frac{1}{2}}$, for $x\in {\mathbb{R}}^d$ (cf. [16]). Notice that if $f\in L_{k,s}^p\left({\mathbb{R}}^d\right)$, we have:

$${\parallel f\parallel }_{p,{\mu }_k}^p+{\parallel \left|x\right|}^s{f\parallel }_{p,{\mu }_k}^p\le C{∥{⟨x⟩}^{s}f∥}_{p,{\mu }_k}^p<\infty .$$

(41)

Then f and ${\left|x\right|}^{s}f$ belong to $L_k^p\left({\mathbb{R}}^d\right)$. On the other hand, if $f\in L_k^p\left({\mathbb{R}}^d\right)$ and ${\left|x\right|}^{s}f$ are in $L_k^p\left({\mathbb{R}}^d\right)$, then

$${∥{⟨x⟩}^{s}f∥}_{p,{\mu }_k}^p={\int }_{{\mathbb{R}}^d}{(1+|x|}^2{)}^{\frac{sp}{2}}{\left|f\left(x\right)\right|}^p\mathrm{d}{\mu }_k\left(x\right)\le 2^{\frac{sp}{2}}\left({\parallel f\parallel }_{p,{\mu }_k}^p+{\parallel \left|x\right|}^s{f\parallel }_{p,{\mu }_k}^p\right)<\infty .$$

(42)

Hence,

$$L_{k,s}^p\left({\mathbb{R}}^d\right)=\left\{f\in L_k^p\left({\mathbb{R}}^d\right):{\left|x\right|}^{s}f\in L_k^p\left({\mathbb{R}}^d\right)\right\}.$$

(43)

Moreover, we have the following result.

Lemma 1.

Let $\beta \ge 0$ and let $s>\beta +\frac{2\gamma +d}{p^{\prime }}$. Then there exists a positive constant $C(s,p,k,\beta ,d)$ such that for all $f\in L_{k,s}^p\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^{\beta }{f\parallel }_{1,{\mu }_k}\le {C(s,p,k,\beta ,d)\parallel f\parallel }_{p,{\mu }_k}^{1-\frac{\beta p^{\prime }+2\gamma +d}{s{p}^{\prime }}}{\parallel {\left|x\right|}^{s}f\parallel }_{p,{\mu }_k}^{\frac{\beta p^{\prime }+2\gamma +d}{s{p}^{\prime }}}.$$

(44)

Proof. 

Let $r>0$ and let ${\chi }_r={\mathbf{1}}_{B_r}$, where $B_r=\{x\in {\mathbb{R}}^d:\left|x\right|<r\}$ is the open ball of ${\mathbb{R}}^d$ of radius r. Then by Hölder’s inequality and (26),

$$\begin{array}{ccc}\hfill {\parallel \left|x\right|}^{\beta }{f\parallel }_{1,{\mu }_k}& =& \parallel {\left|x\right|}^{\beta }{\chi }_r{f\parallel }_{1,{\mu }_k}+{\parallel {\left|x\right|}^{\beta }(1-{\chi }_r)f\parallel }_{1,{\mu }_k}\hfill \\ & \le & {\left({\int }_{B_r}{\left|x\right|}^{\beta p^{\prime }}\mathrm{d}{\mu }_k\left(x\right)\right)}^{1/p^{\prime }}{\parallel f\parallel }_{p,{\mu }_k}+{\left({\int }_{B_r^c}{\left|x\right|}^{p^{\prime }(\beta -s)}\mathrm{d}{\mu }_k\left(x\right)\right)}^{1/p^{\prime }}{\parallel {\left|x\right|}^{s}f\parallel }_{p,{\mu }_k}\hfill \\ & \le & {\left(d_k{\int }_0^r{t}^{2\gamma +d+\beta p^{\prime }-1}\mathrm{d}t\right)}^{1/p^{\prime }}{\parallel f\parallel }_{p,{\mu }_k}+{\left(d_k{\int }_r^{\infty }{t}^{2\gamma +d+p^{\prime }(\beta -s)-1}\mathrm{d}t\right)}^{1/p^{\prime }}{\parallel {\left|x\right|}^{s}f\parallel }_{p,{\mu }_k}\hfill \\ & =& {\left(\frac{d_k}{\beta p^{\prime }+2\gamma +d}\right)}^{1/p^{\prime }}{r}^{\beta +\frac{2\gamma +d}{p^{\prime }}}{\parallel f\parallel }_{p,{\mu }_k}+{\left(\frac{d_k}{p^{\prime }(s-\beta )-(2\gamma +d)}\right)}^{1/p^{\prime }}{r}^{\beta -s+\frac{2\gamma +d}{p^{\prime }}}{\parallel {\left|x\right|}^{s}f\parallel }_{p,{\mu }_k}.\hfill \end{array}$$

Minimizing the right-hand side of the last inequality by

$$r={\left(\frac{p^{\prime }(s-\beta )-(2\gamma +d)}{\beta p^{\prime }+2\gamma +d}\right)}^{\frac{1}{sp}}{\left(\parallel x^s{f\parallel }_{p,{\mu }_k}{\parallel f\parallel }_{p,{\mu }_k}^{-1}\right)}^{1/s},$$

we obtain (44), with

$$\begin{array}{ccc}\hfill C(s,p,k,\beta ,d)& =& d_k^{1/p^{\prime }}{\left(\frac{p^{\prime }(s-\beta )-(2\gamma +d)}{\beta p^{\prime }+2\gamma +d}\right)}^{\frac{\beta p^{\prime }+2\gamma +d}{sp{p}^{\prime }}}\hfill \\ & \times & \left({(\beta p^{\prime }+2\gamma +d)}^{-\frac{1}{p^{\prime }}}+s{p}^{\prime }{\left(p^{\prime }(s-\beta )-(2\gamma +d)\right)}^{-1-\frac{1}{p^{\prime }}}\right).\hfill \end{array}$$

This completes the proof. □

Remark 1.

From the previous lemma, if $s>\frac{2\gamma +d}{p^{\prime }}$, then for all $f\in L_{k,s}^p\left({\mathbb{R}}^d\right)$,

$${\parallel f\parallel }_{1,{\mu }_k}\le {C(s,p,k,0,d)\parallel f\parallel }_{p,{\mu }_k}^{1-\frac{2\gamma +d}{s{p}^{\prime }}}{\parallel {\left|x\right|}^{s}f\parallel }_{p,{\mu }_k}^{\frac{2\gamma +d}{s{p}^{\prime }}}.$$

(45)

This means that

$$L_{k,s}^p\left({\mathbb{R}}^d\right)\hookrightarrow L_k^1\left({\mathbb{R}}^d\right)\mathrm{if}s>\frac{2\gamma +d}{p^{\prime }}.$$

(46)

In particular, we have $L_{k,s}^2\left({\mathbb{R}}^d\right)\hookrightarrow L_k^1\left({\mathbb{R}}^d\right)$ if $s>\frac{2\gamma +d}{2}$ and

$${\parallel f\parallel }_{1,{\mu }_k}\le {c(s,k,d)\parallel f\parallel }_{2,{\mu }_k}^{1-\frac{2\gamma +d}{2s}}{\parallel {\left|x\right|}^{s}f\parallel }_{2,{\mu }_k}^{\frac{2\gamma +d}{2s}}.$$

(47)

In the following, we recall the sharp Beckner-type logarithmic inequality proved in [16] (Theorem 6.1).

Theorem 5.

Let$s>1$. Then for all nonzero$f\in L_{k,s}^1\left({\mathbb{R}}^d\right)$,

$$-{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\frac{\left|f\right(x\left)\right|}{{\parallel f\parallel }_{1,{\mu }_k}}\right)\mathrm{d}{\mu }_k\left(x\right)\le (2\gamma +d){\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(C_{k,s}\left(d\right)(1+{\left|x\right|}^s)\right)\mathrm{d}{\mu }_k\left(x\right),$$

(48)

where

$$C_{k,s}\left(d\right)={\left(\frac{\mathrm{\Gamma }\left(\frac{2\gamma +d}{s}\right)\mathrm{\Gamma }\left(\frac{2\gamma +d}{s^{\prime }}\right)}{s{\mathbf{c}}_k{2}^{\gamma +d/2-1}\mathrm{\Gamma }(\gamma +d/2)\mathrm{\Gamma }(2\gamma +d)}\right)}^{\frac{1}{2\gamma +d}}$$

(49)

is the sharp constant, and equality holds if and only if

$$f\left(x\right)={\left(C_{k,s}\left(d\right)(1+{\left|x\right|}^s)\right)}^{-(2\gamma +d)}.$$

(50)

Consequently, by following the same process as in [17], we derive the following Shannon-type inequality in the Dunkl setting.

Corollary 1.

Let $s>1$. Then for all nonzero $f\in L_{k,s}^1\left({\mathbb{R}}^d\right)$,

$$\mathbf{E}\left(\frac{\left|f\right(x\left)\right|}{{\parallel f\parallel }_{1,{\mu }_k}}\right)\le \frac{2\gamma +d}{s}ln\left(C(k,s,d)\frac{\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}}{{\parallel f\parallel }_{1,{\mu }_k}}\right),$$

(51)

where

$$C(k,s,d)=s^s{(s-1)}^{1-s}{\left(\frac{\mathrm{\Gamma }\left(\frac{2\gamma +d}{s}\right)\mathrm{\Gamma }\left(\frac{2\gamma +d}{s^{\prime }}\right)}{s{\mathbf{c}}_k{2}^{\gamma +d/2-1}\mathrm{\Gamma }(\gamma +d/2)\mathrm{\Gamma }(2\gamma +d)}\right)}^{\frac{s}{2\gamma +d}}.$$

(52)

Proof. 

Let $f\in L_{k,s}^1\left({\mathbb{R}}^d\right)$ such that ${\parallel f\parallel }_{1,{\mu }_k}=1$. Then by Inequality (48) and Jensen’s inequality,

$$\begin{array}{ccc}\hfill -{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right)& \le & (2\gamma +d){\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(C_{k,s}\left(d\right)(1+{\left|x\right|}^s)\right)\mathrm{d}{\mu }_k\left(x\right)\hfill \\ & =& (2\gamma +d){\int }_{{\mathbb{R}}^d}ln\left(C_{k,s}\left(d\right)(1+{\left|x\right|}^s)\right)\mathrm{d}{\nu }_k\left(x\right)\hfill \\ & \le & (2\gamma +d)ln\left({\int }_{{\mathbb{R}}^d}{C}_{k,s}\left(d\right)(1+{\left|x\right|}^s)\mathrm{d}{\nu }_k\left(x\right)\right)\hfill \\ & =& (2\gamma +d)ln\left(C_{k,s}\left(d\right)\left(1+\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}\right)\right).\hfill \end{array}$$

(53)

Now let $f_{\lambda }$, $\lambda >0$, the function is defined by:

$$f_{\lambda }\left(x\right)={\lambda }^{2\gamma +d}f\left(\lambda x\right),x\in {\mathbb{R}}^d.$$

(54)

Then $\parallel f_{\lambda }{\parallel }_{1,{\mu }_k}={\parallel f\parallel }_{1,{\mu }_k}=1$, and by replacing f with $f_{\lambda }$ in Inequality (53), we obtain

$$-{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right)\le (2\gamma +d)ln\left(\lambda +{\lambda }^{1-s}{\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}\right)+(2\gamma +d)ln\left(C_{k,s}\left(d\right)\right).$$

(55)

Minimizing the right-hand side of the last inequality with $\lambda ={(s-1)}^{1/s}{\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}^{1/s}$, we get

$$-{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{s}ln\left(s^s{\left(C_{k,s}\left(d\right)\right)}^s{(s-1)}^{1-s}{\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}\right).$$

(56)

Finally, if f is any nonzero function in $L_{k,s}^1\left({\mathbb{R}}^d\right)$, then by replacing f by $f/{\parallel f\parallel }_{1,{\mu }_k}$ in (56), we obtain (51). □

Moreover, we have the following improvement:

Theorem 6.

Let$s>0$. Then for any nonzero$f\in L_{k,s}^1\left({\mathbb{R}}^d\right)$, we have

$$\mathbf{E}\left(\frac{\left|f\right(x\left)\right|}{{\parallel f\parallel }_{1,{\mu }_k}}\right)\le \frac{2\gamma +d}{s}ln\left(c(k,s,d)\frac{\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}}{{\parallel f\parallel }_{1,{\mu }_k}}\right),$$

(57)

where

$$c(k,s,d)=\frac{se}{2\gamma +d}{\left(\frac{\mathrm{\Gamma }\left(\frac{2\gamma +d}{s}\right)}{s{\mathbf{c}}_k{2}^{\gamma +d/2-1}\mathrm{\Gamma }(\gamma +d/2)}\right)}^{\frac{s}{2\gamma +d}}$$

(58)

is the sharp constant, and it is attained by$f\left(x\right)=exp\left(-c_{s,k}\left(d\right){\left|x\right|}^s\right)$up to dilation, where$c_{s,k}\left(d\right)$is given by

$$c_{s,k}\left(d\right)={\left({\int }_{{\mathbb{R}}^d}{e}^{-{\left|x\right|}^s}\mathrm{d}{\mu }_k\left(x\right)\right)}^{\frac{s}{2\gamma +d}}={\left(s^{-1}{d}_k\mathrm{\Gamma }\left(\frac{2\gamma +d}{s}\right)\right)}^{\frac{s}{2\gamma +d}}.$$

(59)

Proof. 

Let $f\in L_{k,s}^1\left({\mathbb{R}}^d\right)$ be a nonzero function such that ${\parallel f\parallel }_{1,{\mu }_k}=1$, and let function $F_s\left(x\right)=e^{-c_{s,k}\left(d\right){\left|x\right|}^s}$ satisfy $\parallel F_s{\parallel }_{1,{\mu }_k}=1$. Then by Jensen’s inequality,

$$exp\left({\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\frac{F_s\left(x\right)}{\left|f\right(x\left)\right|}\right)\mathrm{d}{\mu }_k\left(x\right)\right)\le {\parallel F_s\parallel }_{1,{\mu }_k}=1.$$

Therefore,

$$-{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right)\le -{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(F_s\left(x\right)\right)\mathrm{d}{\mu }_k\left(x\right)=c_{s,k}\left(d\right){\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}.$$

(60)

Now replacing f by $f_{\lambda }$ (which is defined in (54)) in Inequality (60), we obtain:

$$-\frac{s}{2\gamma +d}{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{s{c}_{s,k}\left(d\right)}{2\gamma +d}{\lambda }^{-s}{\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}+sln\left(\lambda \right).$$

(61)

By optimizing (61) with $\lambda ={\left(\frac{s{c}_{s,k}\left(d\right){\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}}{2\gamma +d}\right)}^{1/s}$, we have

$$-\frac{s}{2\gamma +d}{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right)\le 1+ln\left(\frac{s{c}_{s,k}\left(d\right){\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}}{2\gamma +d}\right)=ln\left(\frac{se{c}_{s,k}\left(d\right){\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}}{2\gamma +d}\right).$$

Hence,

$$-{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{s}ln\left(c(k,s,d)\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}\right),$$

(62)

where $c(k,s,d)=\frac{se{c}_{s,k}\left(d\right)}{2\gamma +d}$.

Now, since $F_s\left(x\right)$ gives the equality of (60), then $F_{s,{\lambda }_0}\left(x\right)={\lambda }_0^{2\gamma +d}{F}_s\left({\lambda }_{0}x\right)$ with

$${\lambda }_0^s=\frac{s{c}_{s,k}\left(d\right){\parallel {\left|x\right|}^s{F}_s\parallel }_{1,{\mu }_k}}{2\gamma +d},$$

attains the equality in (62). Moreover, since

$${\int }_{{\mathbb{R}}^d}{e}^{-t{c}_{s,k}{\left|x\right|}^s}\mathrm{d}{\mu }_k\left(x\right)=t^{-\frac{2\gamma +d}{s}},$$

(63)

then by taking $t=1$ in the following equality

$$\frac{\mathrm{d}}{\mathrm{d}t}{\int }_{{\mathbb{R}}^d}{e}^{-t{c}_{s,k}{\left|x\right|}^s}\mathrm{d}{\mu }_k\left(x\right)=-c_{s,k}\left(d\right){\int }_{{\mathbb{R}}^d}{\left|x\right|}^s{e}^{-t{c}_{s,k}{\left|x\right|}^s}\mathrm{d}{\mu }_k\left(x\right)=-\frac{2\gamma +d}{s}{t}^{-\frac{2\gamma +d}{s}-1}$$

(64)

we obtain

$$\parallel {\left|x\right|}^s{F}_s{\parallel }_{1,{\mu }_k}=\frac{2\gamma +d}{s{c}_{s,k}\left(d\right)},$$

(65)

and this yields ${\lambda }_0^s=1$.

Finally, if f is any nonzero function in $L_{k,s}^1\left({\mathbb{R}}^d\right)$, then by replacing f by $f/{\parallel f\parallel }_{1,{\mu }_k}$ in (62), we obtain Inequality (57). □

Remark 2.

The constant $C(k,s,d)$ in the Shannon-type inequality (51) coincides with the sharp constant $c(k,s,d)$ of the Shannon-type inequality (57) for large enough $d\gg 1$. In fact, from Stirling’s approximation

$$\mathrm{\Gamma }\left(z\right)\approx \sqrt{2\pi }{e}^{-z}{z}^{z-1/2},\mathrm{for}z\gg 1$$

(66)

we have $C(k,s,d)\approx c(k,s,d)$ for $d\gg 1$. Moreover, the optimal Shannon-type inequality (57) can be written as

$$-{\int }_{{\mathbb{R}}^d}\left|f\left(x\right)\right|ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{s}{\parallel f\parallel }_{1,{\mu }_k}ln\left(c(k,s,d)\frac{\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}}{{\parallel f\parallel }_{1,{\mu }_k}^{1+\frac{s}{2\gamma +d}}}\right).$$

(67)

Now, if $f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\setminus \left\{0\right\}$, then ${\left|f\right|}^2\in L_{k,2s}^1\left({\mathbb{R}}^d\right)$. Thus by (57), we have for all $s>0$ and all $f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\setminus \left\{0\right\}$

$$\mathbf{E}\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\le \frac{2\gamma +d}{2s}ln\left(c(k,2s,d)\frac{\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).$$

(68)

Moreover, from Theorem 5, we have for all $s>1/2$ and all $f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\setminus \left\{0\right\}$

$$\mathbf{E}\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\le (2\gamma +d){\int }_{{\mathbb{R}}^d}\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}ln\left(C_{k,2s}\left(d\right)(1+{\left|x\right|}^{2s})\right)\mathrm{d}{\mu }_k\left(x\right).$$

(69)

As a corollary of Theorem 6, we obtain the following:

Corollary 2.

Let $2\gamma +d\ge 2$ and $s>0$. Then for any non-negative function $f\in L_{k,s}^1\left({\mathbb{R}}^d\right)$

$${\int }_{\{f\le 1\}}f\left(x\right)ln{\left(\frac{f\left(x\right)}{{\parallel f\parallel }_{1_{*},k}}\right)}^{-1}\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{s}{\parallel f\parallel }_{1_{*},k}ln\left(\frac{{\displaystyle c\left(k,s,d\right){\int }_{\{f\le 1\}}{\left|x\right|}^{s}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}}{{\parallel f\parallel }_{1_{*},k}}\right),$$

(70)

where ${\displaystyle {\parallel f\parallel }_{1_{*},k}\equiv {\int }_{\{f\le 1\}}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}$, and the constant $c(k,s,d)$ on the right-hand side is the best possible and is given by (58).

4. Nash-Type and Logarithmic Sobolev-Type Inequalities and Applications

For $\beta >0$, we consider the Dunkl–Sobolev space on ${\mathbb{R}}^d$ (cf. [14]) defined by

$${\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)=\left\{f\in L_k^2\left({\mathbb{R}}^d\right):{⟨\xi ⟩}^{\beta }{\mathcal{F}}_D\left(f\right)\in L_k^2\left({\mathbb{R}}^d\right)\right\},$$

(71)

or equivalently (by Plancherel’s Formula (33))

$${\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)=\left\{f\in L_k^2\left({\mathbb{R}}^d\right):{\left|\xi \right|}^{\beta }{\mathcal{F}}_D\left(f\right)\in L_k^2\left({\mathbb{R}}^d\right)\right\}.$$

(72)

Noting that if $f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$, then ${\mathcal{F}}_D\left(f\right)$ and ${\left|\xi \right|}^{\beta }{\mathcal{F}}_D\left(f\right)$ are in $L_k^2\left({\mathbb{R}}^d\right)$. Therefore, from (47), ${\mathcal{F}}_D\left(f\right)$ is in $L_k^1\left({\mathbb{R}}^d\right)$ if $\beta >\gamma +d/2$. Thus, it follows from the Riemann–Lebesgue lemma that f is uniformly continuous on ${\mathbb{R}}^d$, vanishing at infinity. If, in addition $f\in L_k^1\left({\mathbb{R}}^d\right)$ (e.g., if ${\left|x\right|}^{\beta }f\in L_k^2\left({\mathbb{R}}^d\right)$), then by the inverse Dunkl transform Theorem (32), f is almost everywhere equal to a function in ${\mathcal{C}}_0\left({\mathbb{R}}^d\right)$. Moreover, we have the following Sobolev’s embedding theorem:

Proposition 1

(Sobolev’s embedding theorem). If $\beta >\gamma +d/2$, then for all $f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$${\parallel f\parallel }_{\infty }^2\le \left(\frac{{\mathbf{c}}_k\mathrm{\Gamma }(\beta -\gamma -d/2)}{2^{\gamma -\beta +d/2}\mathrm{\Gamma }\left(\beta \right)}\right)\left({\parallel f\parallel }_{2,{\mu }_k}^2+{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2\right).$$

(73)

Moreover, there exists a constant $c(k,\beta ,d)$ such that for all $f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$${\parallel f\parallel }_{\infty }\le {c(k,\beta ,d)\parallel f\parallel }_{2,{\mu }_k}^{1-\frac{2\gamma +d}{2\beta }}{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^{\frac{2\gamma +d}{2\beta }},$$

(74)

where

$$c(k,\beta ,d)={\left(\frac{{\mathbf{c}}_k\beta \mathrm{\Gamma }(\beta -\gamma -d/2)}{2^{\gamma -\beta +d/2}(\beta -\gamma -d/2)\mathrm{\Gamma }\left(\beta \right)}{\left(\frac{\beta }{\gamma +d/2}-1\right)}^{\frac{2\gamma +d}{2\beta }}\right)}^{1/2}.$$

(75)

Proof. 

Let $f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$. Then by the inversion formula for the Dunkl transform, we have

$$f\left(x\right)={\mathbf{c}}_k{\int }_{{\mathbb{R}}^d}{\mathcal{F}}_D\left(f\right)\left(\xi \right)\mathcal{K}(ix,\xi )\mathrm{d}{\mu }_k\left(\xi \right).$$

(76)

Therefore, by the Cauchy–Schwarz inequality and by (33),

$$\begin{array}{ccc}\hfill \left|f\right(x\left)\right|& \le & {\mathbf{c}}_k{\left({\int }_{{\mathbb{R}}^d}{(1+|\xi |}^2{)}^{-\beta }\mathrm{d}{\mu }_k\left(\xi \right)\right)}^{1/2}{\left({\int }_{{\mathbb{R}}^d}|{\mathcal{F}}_D{\left(f\right)\left(\xi \right)|}^2{(1+|\xi |}^2{)}^{\beta }\mathrm{d}{\mu }_k\left(\xi \right)\right)}^{1/2}\hfill \\ & \le & {\mathbf{c}}_k{2}^{\beta /2}{\left(d_k{\int }_0^{\infty }\frac{r^{2\gamma +d-1}}{{(1+r^2)}^{\beta }}\mathrm{d}r\right)}^{1/2}{\left({\parallel f\parallel }_{2,{\mu }_k}^2+{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2\right)}^{1/2}\hfill \\ & =& {\left(\frac{{\mathbf{c}}_k\mathrm{\Gamma }(\beta -\gamma -d/2)}{2^{\gamma -\beta +d/2}\mathrm{\Gamma }\left(\beta \right)}\right)}^{1/2}{\left({\parallel f\parallel }_{2,{\mu }_k}^2+{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2\right)}^{1/2}.\hfill \end{array}$$

Thus

$${\parallel f\parallel }_{\infty }\le {\left(\frac{{\mathbf{c}}_k\mathrm{\Gamma }(\beta -\gamma -d/2)}{2^{\gamma -\beta +d/2}\mathrm{\Gamma }\left(\beta \right)}\right)}^{1/2}{\left({\parallel f\parallel }_{2,{\mu }_k}^2+{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2\right)}^{1/2}.$$

Now replacing f by ${\delta }_{\lambda }^{k}f$ in the last inequality, we obtain

$${\parallel f\parallel }_{\infty }\le {\left(\frac{{\mathbf{c}}_k\mathrm{\Gamma }(\beta -\gamma -d/2)}{2^{\gamma -\beta +d/2}\mathrm{\Gamma }\left(\beta \right)}\right)}^{1/2}{\left({\lambda }^{-2\gamma -d}{\parallel f\parallel }_{2,{\mu }_k}^2+{\lambda }^{2\beta -2\gamma -d}{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2\right)}^{1/2}.$$

Minimizing the right-hand side with

$$\lambda ={\left(\frac{{(2\gamma +d)\parallel f\parallel }_{2,{\mu }_k}^2}{{(2\beta -2\gamma -d)\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}\right)}^{\frac{1}{2\beta }},$$

we obtain the desired result. □

Remark 3.

We recall that the inhomogeneous Dunkl–Sobolev spaces ${\mathbb{H}}_k^{\beta ,p}\left({\mathbb{R}}^d\right)$ endowed with the norm

$${\parallel f\parallel }_{{\mathbb{H}}_k^{\beta ,p}}={∥{\mathcal{F}}_D^{-1}\left({⟨\xi ⟩}^{\beta }{\mathcal{F}}_D\left(f\right)\left(\xi \right)\right)∥}_{p,{\mu }_k}$$

(77)

and the homogeneous Dunkl–Sobolev spaces $H_k^{\beta ,p}\left({\mathbb{R}}^d\right)$ defined in a similar way by replacing $⟨·⟩$ with $|·|$ in (77) have been defined and studied in [18]. The fact that ${\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)=H_k^{\beta ,2}\left({\mathbb{R}}^d\right)$ is a simple consequence of Plancherel’s identity.

When $1\le p\le 2$, Sobolev’s embedding theorem in the Dunkl setting follows from Hölder’s inequality and the Hausdorff–Young inequality (34). Indeed, if $\beta >\frac{2\gamma +d}{p}$, then

$$\begin{array}{ccc}\hfill \left|f\right(x\left)\right|& \le & {\mathbf{c}}_k{\int }_{{\mathbb{R}}^d}|{\mathcal{F}}_D{\left(f\right)\left(\xi \right)|}^2\mathrm{d}{\mu }_k\left(\xi \right)\le {\mathbf{c}}_k{\left({\int }_{{\mathbb{R}}^d}{⟨\xi ⟩}^{-p\beta }\mathrm{d}{\mu }_k\left(\xi \right)\right)}^{1/p}{\parallel {⟨\xi ⟩}^{\beta }{\mathcal{F}}_D\left(f\right)\parallel }_{p^{\prime },{\mu }_k}\hfill \\ & \le & {\mathbf{c}}_k^{2/p}{\left({\int }_{{\mathbb{R}}^d}{⟨\xi ⟩}^{-p\beta }\mathrm{d}{\mu }_k\left(\xi \right)\right)}^{1/p}{\parallel f\parallel }_{{\mathbb{H}}_k^{\beta ,p}}={\left(\frac{{\mathbf{c}}_k\mathrm{\Gamma }\left(\frac{1}{2}(p\beta -2\gamma -d)\right)}{2^{\gamma +d/2}\mathrm{\Gamma }\left(\frac{p\beta }{2}\right)}\right)}^{1/p}{\parallel f\parallel }_{{\mathbb{H}}_k^{\beta ,p}}.\hfill \end{array}$$

Hence, for $\beta >\frac{2\gamma +d}{p}$ and $1\le p\le 2$, we have

$${\parallel f\parallel }_{\infty }\le {\left(\frac{{\mathbf{c}}_k\mathrm{\Gamma }\left(\frac{1}{2}(p\beta -2\gamma -d)\right)}{2^{\gamma +d/2}\mathrm{\Gamma }\left(\frac{p\beta }{2}\right)}\right)}^{1/p}{\parallel f\parallel }_{{\mathbb{H}}_k^{\beta ,p}}.$$

(78)

In particular, for $p=1$ and $\beta >2\gamma +d$,

$${\parallel f\parallel }_{\infty }\le \frac{{\mathbf{c}}_k\mathrm{\Gamma }\left(\frac{1}{2}(\beta -2\gamma -d)\right)}{2^{\gamma +d/2}\mathrm{\Gamma }\left(\frac{\beta }{2}\right)}{\parallel f\parallel }_{{\mathbb{H}}_k^{\beta ,1}}.$$

(79)

4.1. New Heisenberg-Type Uncertainty Inequalities

In this section, we revisit and prove new Heisenberg-type uncertainty principles for the Dunkl transform. The first result is the following well-known uncertainty inequality (see [8,9,10]).

Theorem 7.

For every $f\in L_k^2\left({\mathbb{R}}^d\right)$,

$${\parallel \left|x\right|f\parallel }_{2,{\mu }_k}\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}\ge \frac{2\gamma +d}{2}{\parallel f\parallel }_{2,{\mu }_k}^2,$$

(80)

with equality if and only if $f\left(x\right)=c{e}^{-\lambda {\left|x\right|}^2/2}$ for some $c\in \mathbb{C}$ and $\lambda >0$.

It is also well known that by using a dilation argument the last inequality is equivalent to the following sharp additive uncertainty inequality (see e.g., [9] (Theorem 1.1)):

$${\parallel \left|x\right|f\parallel }_{2,{\mu }_k}^2+\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2\ge (2\gamma +d){\parallel f\parallel }_{2,{\mu }_k}^2,$$

(81)

with equality if and only if $f\left(x\right)=c{e}^{-{\left|x\right|}^2/2}$ for some $c\in \mathbb{C}$.

More generally, we recall the following results (see [6,7]):

Theorem 8.

Let$s,\beta >0$.

1.

There exists a constant$c_1(s,k,\beta ,d)$such that for all$f\in L_k^2\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge c_1(s,k,\beta ,d){\parallel f\parallel }_{2,{\mu }_k}^{s+\beta }.$$

(82)

2.

There exists a constant$c_2(s,k,\beta ,d)$such that for all$f\in L_k^1\left({\mathbb{R}}^d\right)\cap L_k^2\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge c_2(s,k,\beta ,d){∥f∥}_{1,{\mu }_k}^{\beta }{∥f∥}_{2,{\mu }_k}^s.$$

(83)

Notice that if the time dispersion $\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}$ (or $\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}$) or the frequency dispersion ${\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}$ are not finite, then the last inequalities are trivial. Hence we may assume that $f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$ in Inequalities (80)–(82), and $f\in L_{k,s}^1\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$ in Inequality (83).

The Proof of Inequality (82) can be obtained from either the Faris-type local uncertainty inequalities [7] (Theorem A) or from the Benedicks–Amrein–Berthier uncertainty principle [7] (Theorem B). The Proof of Inequality (83) can be obtained by combining the following Nash-type inequality [6] (Proposition 3.2) and Carlson-type inequality [6] (Proposition 3.3) in the Dunkl setting:

Proposition 2.

Let $s,\beta >0$.

1.

A Carlson-type inequality: There exists a positive constant$C_1=C(s,k,d)$such that for all$f\in L_{k,s}^1\left({\mathbb{R}}^d\right)\cap L_k^2\left({\mathbb{R}}^d\right)$,

$${∥f∥}_{1,{\mu }_k}^{1+\frac{s}{\gamma +d/2}}\le C_1{∥f∥}_{2,{\mu }_k}^{\frac{s}{\gamma +d/2}}{\parallel {\left|x\right|}^{s}f\parallel }_{1,{\mu }_k}.$$

(84)

2.

A Nash-type inequality: There exists a positive constant$C_2=C(k,\beta ,d)$such that for all$f\in L_k^1\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$${∥f∥}_{2,{\mu }_k}^{1+\frac{\beta }{\gamma +d/2}}\le C_2{∥f∥}_{1,{\mu }_k}^{\frac{\beta }{\gamma +d/2}}{∥{\left|\xi \right|}^{\beta }{\mathcal{F}}_D\left(f\right)∥}_{2,{\mu }_k}.$$

(85)

In [6], we have proved that the Nash-type inequality (85) is a key tool in proving uncertainty inequalities for the Dunkl transform. Moreover, from Lemma 1 and the Nash-type inequality (85), we can give another proof of the Heisenberg-type uncertainty inequality (82) (only for $s>\gamma +d/2$).

Corollary 3.

Let$s>\gamma +d/2$and$\beta >0$. Then

1.

There exists a positive constant$C_3=C(s,k,d)$such that for all$f\in L_{k,s}^2\left({\mathbb{R}}^d\right)$, the time-dispersion satisfies

$$\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}\ge C_3{\parallel f\parallel }_{1,{\mu }_k}^{\frac{s}{\gamma +d/2}}{\parallel f\parallel }_{2,{\mu }_k}^{1-\frac{s}{\gamma +d/2}}.$$

(86)

2.

There exists a positive constant$C_4=C(k,\beta ,d)$such that for all$f\in L_k^1\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$, the frequency-dispersion satisfies

$${∥{\left|\xi \right|}^{\beta }{\mathcal{F}}_D\left(f\right)∥}_{2,{\mu }_k}\ge C_4{∥f∥}_{1,{\mu }_k}^{\frac{-\beta }{\gamma +d/2}}{∥f∥}_{2,{\mu }_k}^{1+\frac{\beta }{\gamma +d/2}}.$$

(87)

3.

There exists a positive constant$C(s,k,\beta ,d)$such that for all$f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$

$$\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge C(s,k,\beta ,d){∥f∥}_{2,{\mu }_k}^{s+\beta }.$$

(88)

Proof. 

Let $f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$ be a nonzero function. Then from Lemma 1, the function f belongs in $L_k^1\left({\mathbb{R}}^d\right)$, and

$${\parallel f\parallel }_{1,{\mu }_k}\le {C(s,k,d)\parallel f\parallel }_{2,{\mu }_k}^{1-\frac{\gamma +d/2}{s}}{\parallel {\left|x\right|}^{s}f\parallel }_{2,{\mu }_k}^{\frac{2\gamma +d}{2s}}.$$

(89)

Then we derive (86). Moreover, from Nash’s inequality (85), we have (87). Combining (86) and (87), we obtain (88). □

Remark 4.

The last corollary is valid only for $s>\gamma +d/2$ (not for all $s>0$), but it is stronger than Inequality (82), since here Inequalities (86) and (87) give separate lower bounds for the values of the time-dispersion $\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}$ and the frequency-dispersion ${∥{\left|\xi \right|}^{\beta }{\mathcal{F}}_D\left(f\right)∥}_{2,{\mu }_k}$, which give more information than the lower bound of the product between them in Heisenberg’s inequality (82).

Moreover, we have the following new uncertainty inequalities involving $L^1$ and $L^{\infty }$ norms.

Theorem 9.

Let$s,\beta >0$.

1.

If$s,\beta >\frac{2\gamma +d}{2}$, then there exists a positive constant$C(s,k,\beta ,d)$such that for every nonzero$f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge C(s,k,\beta ,d){\left(\frac{{∥f∥}_{1,{\mu }_k}{∥f∥}_{\infty }}{{∥f∥}_{2,{\mu }_k}^2}\right)}^{\frac{s\beta }{\gamma +d/2}}{∥f∥}_{2,{\mu }_k}^{s+\beta }.$$

(90)

2.

If$\beta >\frac{2\gamma +d}{2}$, then there exists a positive constant$C(s,k,\beta ,d)$such that for every nonzero$f\in L_{k,s}^1\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$$\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge C(s,k,\beta ,d){\left(\frac{{∥f∥}_{1,{\mu }_k}{∥f∥}_{\infty }}{{∥f∥}_{2,{\mu }_k}^2}\right)}^{\frac{s\beta }{\gamma +d/2}}{∥f∥}_{1,{\mu }_k}^{\beta }{∥f∥}_{2,{\mu }_k}^s.$$

(91)

Proof. 

From Inequality (74), we have

$${\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge {\left(c(k,\beta ,d)\right)}^{-\frac{s\beta }{\gamma +d/2}}{\parallel f\parallel }_{\infty }^{\frac{s\beta }{\gamma +d/2}}{\parallel f\parallel }_{2,{\mu }_k}^{s-\frac{s\beta }{\gamma +d/2}},$$

(92)

and from (84), we have

$$\parallel {\left|x\right|}^s{f\parallel }_{1,{\mu }_k}^{\beta }\ge C_1^{-\beta }{\parallel f\parallel }_2^{\frac{-s\beta }{\gamma +d/2}}{\parallel f\parallel }_{1,{\mu }_k}^{\beta +\frac{s\beta }{\gamma +d/2}}.$$

(93)

Then Inequality (90) follows from (86) and (92), and Inequality (91) follows from (92) and (93). □

4.2. Logarithmic Sobolev Inequalities and the Uncertainty Principles

First recall the Sobolev-type inequality for the Dunkl transform, recently proved in [11] (Theorem 1.1, Theorem 6.1).

Theorem 10.

Suppose that $2\gamma +d>2$, and let $2^{*}=\frac{2(2\gamma +d)}{2\gamma +d-2}$. Then for all $f\in {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$,

$${\parallel f\parallel }_{2^{*},{\mu }_k}\le {\mathcal{S}}_{k,d}{\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k},$$

(94)

where ${\mathcal{S}}_{k,d}$ is the sharp constant given by

$${\mathcal{S}}_{k,d}:={\left(\frac{2}{(2\gamma +d)(2\gamma +d-2)}\right)}^{1/2}{\left({\mathbf{c}}_k\left|G\right|\frac{\mathrm{\Gamma }(2\gamma +d)}{\mathrm{\Gamma }\left(\frac{2\gamma +d}{2}\right)}\right)}^{\frac{1}{2\gamma +d}}.$$

(95)

Then we have the following logarithmic Sobolev inequality.

Corollary 4.

Let $2\gamma +d>2$. Then for all nonzero $f\in {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$,

$${\int }_{{\mathbb{R}}^d}\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}ln\left({\mathcal{S}}_{k,d}^2\frac{\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right),$$

(96)

where ${\mathcal{S}}_{k,d}$ is the constant given in (95).

Proof. 

Let $f\in {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$ such that ${\parallel f\parallel }_{2,{\mu }_k}=1$. Then by (94) and Jensen’s inequality, we have

$$\begin{array}{ccc}\hfill \frac{2}{2\gamma +d-2}{\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln\left({\left|f\left(x\right)\right|}^2\right)\mathrm{d}{\mu }_k\left(x\right)& \le & ln\left({\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2^{*}}\mathrm{d}{\mu }_k\left(x\right)\right)\hfill \\ & \le & \frac{2\gamma +d}{2\gamma +d-2}ln\left({\mathcal{S}}_{k,d}^2{\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k}^2\right).\hfill \end{array}$$

Thus, by (36)

$${\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln\left({\left|f\left(x\right)\right|}^2\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}ln\left({\mathcal{S}}_{k,d}^2\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2\right).$$

(97)

Finally, if f is any nonzero function in $f\in {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$, then by replacing f by $f/{\parallel f\parallel }_{2,{\mu }_k}$ in (97), we obtain (96). □

Now, using the asymptote of the sharp Shannon’s inequality (57) and the sharp Sobolev inequality (94), we derive the sharp Heisenberg’s uncertainty inequality (80).

Corollary 5.

Let $2\gamma +d>2$. Then there exists a positive constant $C(k,d)$ such that for all f belonging to $L_{k,1}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$, we have

$${\parallel \left|x\right|f\parallel }_{2,{\mu }_k}\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}\ge C(k,d){∥f∥}_{2,{\mu }_k}^2,$$

(98)

where

$$C(k,d)={\left(\frac{{(2\gamma +d)}^2(\frac{2\gamma +d}{2}-1)}{e}\right)}^{1/2}{\left(\frac{\mathrm{\Gamma }\left(\frac{2\gamma +d}{2}\right)}{\left|G\right|\mathrm{\Gamma }(2\gamma +d)}\right)}^{\frac{1}{2\gamma +d}}.$$

(99)

Moreover, for $d\gg 1$, we have the sharp inequality

$${\parallel \left|x\right|f\parallel }_{2,{\mu }_k}\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}\ge \frac{2\gamma +d}{2}{∥f∥}_{2,{\mu }_k}^2.$$

(100)

Proof. 

Let $f\in L_{k,1}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$ be a nonzero function. Then, from (68), for $s=1$ we have

$$\mathbf{E}\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\le \frac{2\gamma +d}{2}ln\left(c(k,2,d)\frac{{\parallel \left|x\right|f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).$$

(101)

Moreover, by (96), we have

$$\mathbf{E}\left(\frac{{\left|f\left(x\right)\right|}^2}{{∥f∥}_{2,{\mu }_k}^2}\right)\ge -(\gamma +d/2)ln\left({\mathcal{S}}_{k,d}^2\frac{\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).$$

(102)

Thus

$$ln\left(c(k,2,d)\frac{{\parallel \left|x\right|f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)+ln\left({\mathcal{S}}_{k,d}^2\frac{\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\ge 0.$$

(103)

Hence,

$$ln\left({\mathcal{S}}_{k,d}\sqrt{c(k,2,d)}\frac{{\parallel \left|x\right|f\parallel }_{2,{\mu }_k}\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\ge 0,$$

(104)

which implies that

$${\parallel \left|x\right|f\parallel }_{2,{\mu }_k}\parallel \left|\xi \right|{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}\ge C(k,d){\parallel f\parallel }_{2,{\mu }_k}^2,$$

(105)

where $C(k,d)={\left({\mathcal{S}}_{k,d}\sqrt{c(k,2,d)}\right)}^{-1}$. Now using the sharp constants (58) and (95), we have

$${\left(C(k,d)\right)}^{-2}=\frac{e}{4{(\gamma +d/2)}^2(\gamma +d/2-1)}{\left(\left|G\right|\frac{\mathrm{\Gamma }(2\gamma +d)}{\mathrm{\Gamma }\left(\frac{2\gamma +d}{2}\right)}\right)}^{\frac{2}{2\gamma +d}}.$$

(106)

Then by Stirling’s approximation

$$\mathrm{\Gamma }\left(z\right)\approx \sqrt{2\pi }{e}^{-z}{z}^{z-1/2},\mathrm{for}z\gg 1,$$

(107)

we have for $d\gg 1$

$${\left(\left|G\right|\frac{\mathrm{\Gamma }(2\gamma +d)}{\mathrm{\Gamma }\left(\frac{2\gamma +d}{2}\right)}\right)}^{\frac{2}{2\gamma +d}}\approx (\frac{2(2\gamma +d)}{e}.$$

(108)

Thus for $d\gg 1$,

$$C(k,d)\approx \frac{2\gamma +d}{2},$$

(109)

which is the sharp constant in Heisenberg’s inequality (80). □

Now from [16] (Theorem 6.2) we have the following Beckner-type inequality for the Dunkl transform, although not with the sharp constant c.

Theorem 11.

For any nonzero $f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$, we have

$${\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln\left(\frac{\left|f\right(x\left)\right|}{{∥f∥}_{2,{\mu }_k}}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}{\int }_{{\mathbb{R}}^d}|{\mathcal{F}}_D{\left(f\right)\left(\xi \right)|}^{2}ln\left(\right|\xi \left|\right)\mathrm{d}{\mu }_k\left(\xi \right)-c{∥f∥}_{2,{\mu }_k}^2.$$

(110)

This inequality is stronger than Inequality (96) since it implies the following logarithmic Sobolev inequality.

Corollary 6.

Let $\beta >0$. Then there exists a constant $A_{k,\beta }\left(d\right)$ such that for all nonzero $f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$,

$${\int }_{{\mathbb{R}}^d}\frac{{\left|f\left(x\right)\right|}^2}{{∥f∥}_{2,{\mu }_k}^2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{∥f∥}_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2\beta }ln\left(A_{k,\beta }\left(d\right)\frac{{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).$$

(111)

Proof. 

For any nonzero function $f\in {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$, set the measure $\mathrm{d}{\nu }_k$ by

$$\mathrm{d}{\nu }_k\left(\xi \right)=\frac{|{\mathcal{F}}_D{\left(f\right)\left(\xi \right)|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\mathrm{d}{\mu }_k\left(\xi \right).$$

(112)

Then, by Jensen’s inequality and Plancherel’s Frmula (33),

$$\begin{array}{ccc}\hfill {\int }_{{\mathbb{R}}^d}|{\mathcal{F}}_D{\left(f\right)\left(\xi \right)|}^{2}ln\left(\right|\xi \left|\right)\mathrm{d}{\mu }_k\left(\xi \right)& =& \frac{1}{2\beta }{\parallel f\parallel }_{2,{\mu }_k}^2{\int }_{{\mathbb{R}}^d}{ln\left(\right|\xi |}^{2\beta })\mathrm{d}{\nu }_k\left(\xi \right)\hfill \\ & \le & \frac{1}{2\beta }{\parallel f\parallel }_{2,{\mu }_k}^{2}ln\left({\int }_{{\mathbb{R}}^d}{\left|\xi \right|}^{2\beta }\mathrm{d}{\nu }_k\left(\xi \right)\right)\hfill \\ & =& \frac{1}{2\beta }{\parallel f\parallel }_{2,{\mu }_k}^{2}ln\left(\frac{{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).\hfill \end{array}$$

(113)

Thus by Inequality (110), we obtain the desired result, with $A_{k,\beta }\left(d\right)=e^{-\frac{4\beta c}{2\gamma +d}}$. □

The last result allows us to give a new, short proof for Heisenberg’s Inequality (82).

Corollary 7.

Let $s,\beta >0$. Then there exists a positive constant $\mathcal{C}(s,k,\beta ,d)$ such that for every $f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$, we have

$$\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}^{\beta }{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^s\ge \mathcal{C}(s,k,\beta ,d){∥f∥}_{2,{\mu }_k}^{s+\beta }.$$

(114)

Proof. 

Let $f\in L_{k,s}^2\left({\mathbb{R}}^d\right)\cap {\mathbb{H}}_k^{\beta ,2}\left({\mathbb{R}}^d\right)$ be a nonzero function. Then from (68) and (111), we have

$$ln\left({\left(c(k,2s,d)\frac{\parallel {\left|x\right|}^s{f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)}^{1/s}{\left(A_{k,\beta }\left(d\right)\frac{{\parallel \left|\xi \right|}^{\beta }{\mathcal{F}}_D{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)}^{1/\beta }\right)\ge 0.$$

(115)

This implies (114) with $\mathcal{C}(s,k,\beta ,d)={\left({\left(c(k,2s,d)\right)}^{\beta }{\left(A_{k,\beta }\left(d\right)\right)}^s\right)}^{-1/2}$. □

Following we give another formulation for the logarithmic Sobolev inequality (96).

Proposition 3.

Let $2\gamma +d>2.$ For any $a>0$ and $f\in {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right)$, the following inequality holds

$${\displaystyle {\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)+\frac{2\gamma +d}{2}{(2lna+1)\parallel f\parallel }_{2,{\mu }_k}^2\le \frac{K(k,d)a^2}{2\gamma +d-2}{\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k}^2,}$$

(116)

where

$$K(k,d):=\frac{(2\gamma +d)(2\gamma +d-2){\mathcal{S}}_{k,d}^2}{2}={\left({\mathbf{c}}_k\left|G\right|\frac{\mathrm{\Gamma }(2\gamma +d)}{\mathrm{\Gamma }(\gamma +d/2)}\right)}^{\frac{2}{2\gamma +d}}.$$

(117)

Proof. 

Involving (96) and Plancherel’s Formula (33), we have

$${\int }_{{\mathbb{R}}^d}\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}ln\left({\mathcal{S}}_{k,d}^2\frac{\parallel \left|\xi \right|{\nabla }_k{\left(f\right)\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).$$

(118)

Moreover, if we introduce a parameter $a>0,$ we have

$$\begin{array}{ccc}\hfill \frac{2\gamma +d}{2}ln\left({\mathcal{S}}_{k,d}^2\frac{\parallel {\nabla }_k{f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)& =& \frac{2\gamma +d}{2}ln\left(\frac{K(k,d)a^2}{2\gamma +d-2}\frac{\parallel {\nabla }_k{f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)+\frac{2\gamma +d}{2}ln\left(\frac{2}{(2\gamma +d)a^2}\right)\hfill \\ & \le & \frac{2\gamma +d}{2}ln\frac{2\gamma +d}{2e}+exp\left(ln\left(\frac{K(k,d)a^2}{2\gamma +d-2}\frac{\parallel {\nabla }_k{f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\right)\hfill \\ & +& \frac{2\gamma +d}{2}ln\left(\frac{2}{(2\gamma +d)a^2}\right)\hfill \\ & =& \frac{K(k,d)a^2}{2\gamma +d-2}\frac{\parallel {\nabla }_k{f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}-\frac{2\gamma +d}{2}(2lna+1).\hfill \end{array}$$

(119)

Thus combining the relations (118) and (119), we derive the result. □

Remark 5.

1.

One can choose$a\ge \frac{1}{\sqrt{e}}$and obtain the following inequality

$${\displaystyle {\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{K(k,d)}{e(2\gamma +d-2)}{\parallel \nabla f\parallel }_{2,{\mu }_k}^2.}$$

Involving the inequalities (108) and (117), we derive that for $d\gg 1$

$${\displaystyle {\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le 2{\parallel \nabla f\parallel }_{2,{\mu }_k}^2,}$$

(120)

and then the inequality does not depend on the dimension.

2.

We can derive (96) from (116). Indeed, we optimize the parameter $a>0$ appearing in Proposition 3, and then the equivalent form of (116) is the inequality (96). More precisely, we set

$${\displaystyle {\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{K(k,d)a^2}{2\gamma +d-2}\parallel {\nabla }_k{f\parallel }_{2,{\mu }_k}^2-\frac{2\gamma +d}{2}(2lna+1){\parallel f\parallel }_{2,{\mu }_k}^2\equiv F\left(a\right)}$$

and optimize the right-hand side with the parameter$a>0.$Since

$$F^{\prime }\left(a\right)=\frac{2K(k,d)}{2\gamma +d-2}\parallel {\nabla }_k{f\parallel }_{2,{\mu }_k}^{2}a-\frac{2\gamma +d}{a}{\parallel f\parallel }_{2,{\mu }_k}^2=0,$$

and the minimum of$F\left(a\right)$is realized by$a=a_0$with

$$a_0^2=\frac{(2\gamma +d)(2\gamma +d-2)}{2K(k,d)}\frac{{\parallel f\parallel }_{2,{\mu }_k}^2}{\parallel {\nabla }_k{f\parallel }_{2,{\mu }_k}^2},$$

then the minimum of the right-hand side is

$${\int }_{{\mathbb{R}}^d}\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}ln\left({\mathcal{S}}_{k,d}^2\frac{\parallel {\nabla }_k{f\parallel }_{2,{\mu }_k}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right).$$

Now we will study the logarithmic Sobolev inequalities on the space ${\mathbb{H}}_{k,G}^{1,2}\left({\mathbb{R}}^d\right)$ defined by

$${\mathbb{H}}_{k,G}^{1,2}\left({\mathbb{R}}^d\right):=\left\{f\in {\mathbb{H}}_k^{1,2}\left({\mathbb{R}}^d\right):f\circ {\sigma }_{\xi }=f\mathrm{for}\mathrm{all}:\xi \in R_{+}\right\}.$$

The sharp Sobolev inequality implies the following sharp logarithmic Sobolev inequality.

Theorem 12.

Let $2\gamma +d>2.$ For any $f\in {\mathbb{H}}_{k,G}^{1,2}\left({\mathbb{R}}^d\right)$, the following inequality holds

$${\displaystyle {\int }_{{\mathbb{R}}^d}\frac{{\left|f\left(x\right)\right|}^2}{{\left|\right|f\left|\right|}_{2,{\mu }_k}^2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\left|\right|f\left|\right|}_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}ln\left({\displaystyle \frac{4({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}}{e(2\gamma +d)}{\int }_{{\mathbb{R}}^d}{\left|{\nabla }_{k}f\left(x\right)\right|}^2\mathrm{d}{\mu }_k\left(x\right)}\right).}$$

(121)

The constant in the right-hand side is the best possible, and it is attained on the closure of a Weyl chamber by $G_t\left(x\right)\equiv \frac{{\mathbf{c}}_k}{{\left(2t\right)}^{\frac{2\gamma +d}{2}}}{e}^{-\frac{{\left|x\right|}^2}{4t}}$ for any $t>0$.

Proof. 

Let $f\in {\mathbb{H}}_{k,G}^{1,2}\left({\mathbb{R}}^d\right)$, and for simplicity we assume that ${\parallel f\parallel }_{2,{\mu }_k}=1.$ By (96), we have

$$\begin{array}{ccc}\hfill {\displaystyle {\int }_{{\mathbb{R}}^d}{\left|f\right|}^{2}ln{\left|f\right|}^2\mathrm{d}{\mu }_k\left(x\right)}& \le & \frac{2\gamma +d}{2}ln\left({\displaystyle {\mathcal{S}}_{k,d}^2{\int }_{{\mathbb{R}}^d}{\left|{\nabla }_{k}f\right|}^2\mathrm{d}{\mu }_k\left(x\right)}\right),\hfill \end{array}$$

(122)

where ${\mathcal{S}}_{k,d}$ is the best possible constant for the Dunkl–Sobolev inequality given by (95). Plugging $g\left(\tilde{x}\right)=f\left(x_1\right)f\left(x_2\right)\cdots f\left(x_p\right)$ for $\tilde{x}\in {\mathbb{R}}^{pd}$ with $\tilde{x}=(x_1,x_2,\dots ,x_p)$ and $x_i\in {\mathbb{R}}^d$ into (122) and noticing ${\parallel g\parallel }_{L_k^2\left({\mathbb{R}}^{pd}\right)}^2={\parallel f\parallel }_{2,{\mu }_k}^{2p}=1$ and $\parallel {\nabla }_k{g\parallel }_{L_k^2\left({\mathbb{R}}^{pd}\right)}^2=p{\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k}^2,$ we see that

$$\begin{array}{ccc}& & {\displaystyle p{\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln{\left|f\left(x\right)\right|}^2\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & =& {\displaystyle {\int }_{{\mathbb{R}}^{pd}}|g\left(\tilde{x}\right){|}^{2}ln{\left|g\left(\tilde{x}\right)\right|}^2\mathrm{d}{\mu }_k\left(\tilde{x}\right)}\hfill \\ & \le & \frac{p(2\gamma +d)}{2}ln\left({\displaystyle \frac{2}{p(2\gamma +d)\left(p\right(2\gamma +d)-2)}{\left(({\mathbf{c}}_k{\left|G\right|)}^p\frac{\mathrm{\Gamma }\left(\right(2\gamma +d\left)p\right)}{\mathrm{\Gamma }\left(\right(\gamma +d/2\left)p\right)}\right)}^{\frac{2}{p(2\gamma +d)}}{\int }_{{\mathbb{R}}^{pd}}{\left|{\nabla }_{k}g\left(\tilde{x}\right)\right|}^2\mathrm{d}{\mu }_k\left(\tilde{x}\right)}\right)\hfill \\ & \le & \frac{p(2\gamma +d)}{2}ln\left({\displaystyle \frac{2}{p(2\gamma +d)\left(p\right(2\gamma +d)-2)}{\left(({\mathbf{c}}_k{\left|G\right|)}^p\frac{\mathrm{\Gamma }\left(\right(2\gamma +d\left)p\right)}{\mathrm{\Gamma }\left(p\right(2\gamma +d)/2)}\right)}^{\frac{2}{p(2\gamma +d)}}{\int }_{{\mathbb{R}}^d}p{\left|{\nabla }_{k}f\left(x\right)\right|}^2\mathrm{d}{\mu }_k\left(x\right)}\right).\hfill \end{array}$$

By using the Stirling formula: $\mathrm{\Gamma }\left(x\right)\simeq \sqrt{2\pi x}{e}^{-x}{x}^x$, as $x\to \infty$, we obtain

$$\begin{array}{ccc}& & {\displaystyle {\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln{\left|f\left(x\right)\right|}^2\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & \frac{2\gamma +d}{2}ln\left(\frac{2}{(2\gamma +d)\left(\right(2\gamma +d)p-2)}{\left(({\mathbf{c}}_k{\left|G\right|)}^p\frac{\mathrm{\Gamma }\left(\right(2\gamma +d\left)p\right)}{\mathrm{\Gamma }\left(\right(2\gamma +d)p/2)}\right)}^{\frac{2}{(2\gamma +d)p}}{\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k}^2\right)\hfill \\ & \simeq & \frac{2\gamma +d}{2}ln\left(\frac{2({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}}{(2\gamma +d)(p(2\gamma +d)-2)2^{\frac{2}{p(2\gamma +d)}}}{\left(\frac{\sqrt{2\pi p(2\gamma +d)}{(p(2\gamma +d)/e)}^{p(2\gamma +d)}}{\sqrt{\pi p(2\gamma +d)}{(p(2\gamma +d)/2e)}^{p(2\gamma +d)/2}}\right)}^{\frac{2}{p(2\gamma +d)}}{\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k}^2\right)\hfill \\ & =& \frac{2\gamma +d}{2}ln\left(\frac{2({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}}{(2\gamma +d)(p(2\gamma +d)-2)2^{\frac{2}{p(2\gamma +d)}}}\left(\frac{{\left(2\pi p(2\gamma +d)\right)}^{\frac{1}{p(2\gamma +d)}}{(p(2\gamma +d)/e)}^2}{{\left(\pi p(2\gamma +d)\right)}^{\frac{1}{p(2\gamma +d)}}(p(2\gamma +d)/2e)}\right){\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k}^2\right)\hfill \\ & =& \frac{2\gamma +d}{2}ln\left({\displaystyle \frac{4({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}}{e(2\gamma +d-2/p)2^{\frac{1}{p(2\gamma +d)}}}{\int }_{{\mathbb{R}}^d}{\left|{\nabla }_{k}f\left(x\right)\right|}^2\mathrm{d}{\mu }_k\left(x\right)}\right).\hfill \end{array}$$

Letting $p\to \infty$, we obtain the sharp inequality (121).

To see the optimality, we take $f^2\left(x\right)=G_t\left(x\right)$, and by simple calculations we derive the result. □

Remark 6.

Proceeding as in Proposition 3 and Remark 5, we prove that (121) is equivalent to

$${\displaystyle {\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{2}ln\left(\frac{{\left|f\left(x\right)\right|}^2}{{\parallel f\parallel }_{2,{\mu }_k}^2}\right)\mathrm{d}{\mu }_k\left(x\right)+{(2\gamma +d)(lna+1)\parallel f\parallel }_{2,{\mu }_k}^2\le 2({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}{a}^2{\parallel {\nabla }_{k}f\parallel }_{2,{\mu }_k}^2.}$$

(123)

The equality is attained on the closure of a Weyl chamber by $f\left(x\right)=exp\left(-\frac{{\left({\mathbf{c}}_k\right)}^{\frac{-2}{2\gamma +d}}}{4a^2}{\left|x\right|}^2\right)$.

Corollary 8.

Let $2\gamma +d>2.$ For any non-negative $f\in L_k^1\left({\mathbb{R}}^d\right)$ with $f^{\frac{1}{2}}\in {\mathbb{H}}_{k,G}^{1,2}\left({\mathbb{R}}^d\right),$ it holds that

$${\displaystyle {\int }_{{\mathbb{R}}^d}f\left(x\right)ln\left(f\left(x\right)\right)d{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}{\parallel f\parallel }_{1,{\mu }_k}ln\left({\displaystyle \frac{({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}}{{e(2\gamma +d)\parallel f\parallel }_{1,d{\mu }_k}^{1-\frac{2}{2\gamma +d}}}{\int }_{{\mathbb{R}}^d}f\left(x\right)|{\nabla }_{k}ln\left(f\left(x\right)\right){|}^{2}d{\mu }_k\left(x\right)}\right).}$$

The constant in the right-hand side is the best possible, and it is attained on closure of a Weyl chamber by $f\left(x\right)=G_t\left(x\right)$ with $t>0.$

Remark 7.

The resulting inequality can be seen by the following normalized form:

$${\displaystyle {\int }_{{\mathbb{R}}^d}f\left(x\right)ln\left(f\left(x\right)\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{2}ln\left({\displaystyle \frac{({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}}{e(2\gamma +d)}{\int }_{{\mathbb{R}}^d}f\left(x\right)|{\nabla }_{k}ln\left(f\left(x\right)\right){|}^{2}d{\mu }_k\left(x\right)}\right)}$$

(124)

for any non-negative f with $f^{\frac{1}{2}}\in {\mathbb{H}}_{k,G}^{1,2}\left({\mathbb{R}}^d\right),$ and ${\parallel f\parallel }_{1,{\mu }_k}=1.$

We end this section by involving the logarithmic Sobolev inequality on the space with the Gaussian measure.

Theorem 13

(The Gaussian logarithmic Sobolev inequality). Let $2\gamma +d>2$ and $t>$0. For any non-negative function g and any G invariant and satisfying $\parallel g{G}_t{\parallel }_{1,{\mu }_k}=1$, we have

$${\displaystyle {\int }_{{\mathbb{R}}^d}g\left(x\right)ln\left(g\left(x\right)\right)\mathrm{d}{\gamma }_{k,t}\left(x\right)\le t{\int }_{{\mathbb{R}}^d}g\left(x\right){|\nabla ln\left(g\left(x\right)\right)|}^2\mathrm{d}{\gamma }_{k,t}\left(x\right),}$$

(125)

where

$$\mathrm{d}{\gamma }_{k,t}\left(x\right):=G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right).$$

(126)

Proof. 

Substituting $g\left(x\right)G_t\left(x\right)$ into $f\left(x\right)$ in (124), and since

$$ln{G}_t\left(x\right)=-\frac{{\left|x\right|}^2}{4t}-\frac{2\gamma +d}{2}ln\left(2t{\left({\mathbf{c}}_k\right)}^{\frac{-2}{2\gamma +d}}\right),$$

one can see that

$\begin{array}{ccc}\hfill & \hfill & {\displaystyle {\int }_{{\mathbb{R}}^d}g\left(x\right)ln\left(g\left(x\right)\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)-\frac{1}{4t}{\int }_{{\mathbb{R}}^d}{\left|x\right|}^{2}g\left(x\right)G_t\left(x\right)d{\mu }_k\left(x\right)-\frac{2\gamma +d}{2}ln\left(2t{\left({\mathbf{c}}_k\right)}^{\frac{-2}{2\gamma +d}}\right)}\hfill \\ \hfill & \le \hfill & \frac{2\gamma +d}{2}ln\left({\displaystyle \frac{({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}}{e(2\gamma +d)}{\int }_{{\mathbb{R}}^d}g\left(x\right){\left|{\nabla }_{k}ln\left(g\left(x\right)\right)-\frac{x}{2t}\right|}^2{G}_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)\hfill \\ \hfill & \le \hfill & \frac{2\gamma +d}{2}ln\left({\displaystyle \frac{({\mathbf{c}}_k{\left|G\right|)}^{\frac{2}{2\gamma +d}}}{e(2\gamma +d)}{\int }_{{\mathbb{R}}^d}g\left(x\right)\left[|{\nabla }_{k}ln\left(g\left(x\right)\right){|}^2-⟨x/t,{\nabla }_{k}ln\left(g\left(x\right)\right)⟩+\frac{{\left|x\right|}^2}{4t^2}\right]G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right).\hfill \end{array}$

Since

$$\begin{array}{c}ln\left({\displaystyle -\frac{2t{\left(\right|G\left|\right)}^{\frac{2}{2\gamma +d}}}{(2\gamma +d)e}{\int }_{{\mathbb{R}}^d}⟨x/tg\left(x\right),{\nabla }_{k}ln\left(g\left(x\right)\right)⟩G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)\hfill \\ =ln\left({\displaystyle -\frac{2{\left(\right|G\left|\right)}^{\frac{2}{2\gamma +d}}}{(2\gamma +d)e}\frac{1}{2t}{\int }_{{\mathbb{R}}^d}{\left|x\right|}^{2}g\left(x\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)+\frac{{2\left|G\right|}^{\frac{2}{2\gamma +d}}}{e}}\right)\hfill \end{array}$$

(127)

and

$$\begin{array}{ccc}& & {\displaystyle \frac{1}{4t}{\int }_{{\mathbb{R}}^d}{\left|x\right|}^{2}g\left(x\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & =& {\displaystyle \frac{1}{2}{\int }_{{\mathbb{R}}^d}x{\nabla }_{k}ln\left(g\left(x\right)\right)g\left(x\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)+\frac{2\gamma +d}{2}}\hfill \\ & \le & {\displaystyle \frac{t}{2}{\int }_{{\mathbb{R}}^d}|{\nabla }_{k}ln\left(g\left(x\right)\right){|}^{2}g\left(x\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)+\frac{1}{8t}{\int }_{{\mathbb{R}}^d}{\left|x\right|}^{2}g\left(x\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)+\frac{2\gamma +d}{2},}\hfill \end{array}$$

we derive that

$${\displaystyle \frac{1}{4t}{\int }_{{\mathbb{R}}^d}{\left|x\right|}^{2}g\left(x\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)\le t{\int }_{{\mathbb{R}}^d}|{\nabla }_{k}ln\left(g\left(x\right)\right){|}^{2}g\left(x\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)+2\gamma +d.}$$

(128)

From (127) and (128), we have

$$\begin{array}{ccc}& & {\displaystyle {\int }_{{\mathbb{R}}^d}g\left(x\right)ln\left(g\left(x\right)\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & {\displaystyle \frac{2\gamma +d}{2}ln(\frac{2t{\left(\right|G\left|\right)}^{\frac{2}{2\gamma +d}}}{(2\gamma +d)e}{\int }_{{\mathbb{R}}^d}g\left(x\right){|{\nabla }_{k}ln\left(g\left(x\right)\right)|}^2{G}_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & +& {\displaystyle \frac{2t{\left(\right|G\left|\right)}^{\frac{2}{2\gamma +d}}}{(2\gamma +d)e}\left(-\frac{1}{2t^2}+\frac{1}{4t^2}\right){\int }_{{\mathbb{R}}^d}{\left|x\right|}^{2}g\left(x\right)G_t\left(x\right)d{\mu }_k\left(x\right)+\frac{2}{e})}\hfill \\ & +& {\displaystyle t{\int }_{{\mathbb{R}}^d}g\left(x\right){|{\nabla }_{k}ln\left(g\left(x\right)\right)|}^2{G}_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)+2\gamma +d}\hfill \\ & \le & {\displaystyle \frac{2\gamma +d}{2}ln(\frac{2t{\left(\right|G\left|\right)}^{\frac{2}{2\gamma +d}}}{(2\gamma +d)e}{\int }_{{\mathbb{R}}^d}g\left(x\right){|{\nabla }_{k}ln\left(g\left(x\right)\right)|}^2{G}_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & -& {\displaystyle \frac{{\left(\right|G\left|\right)}^{\frac{2}{2\gamma +d}}}{(2\gamma +d)e}\frac{1}{2t}{\int }_{{\mathbb{R}}^d}{\left|x\right|}^{2}g\left(x\right)G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)+\frac{2}{e})}\hfill \\ & +& {\displaystyle t{\int }_{{\mathbb{R}}^d}g\left(x\right){|{\nabla }_{k}ln\left(g\left(x\right)\right)|}^2{G}_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)+2\gamma +d.}\hfill \end{array}$$

Letting $\mathrm{d}{\gamma }_{k,t}\left(x\right)=G_t\left(x\right)\mathrm{d}{\mu }_k\left(x\right)$, one can see that

$$\begin{array}{ccc}{\displaystyle {\int }_{{\mathbb{R}}^d}g\left(x\right)ln\left(g\left(x\right)\right)\mathrm{d}{\gamma }_{k,t}\left(x\right)}\hfill & \le \hfill & \frac{2\gamma +d}{2}ln\left({\displaystyle \frac{{2te\left|G\right|}^{\frac{2}{2\gamma +d}}}{2\gamma +d}{\int }_{{\mathbb{R}}^d}g\left(x\right){|{\nabla }_{k}ln\left(g\left(x\right)\right)|}^2\mathrm{d}{\gamma }_{k,t}\left(x\right)+\frac{{2\left|G\right|}^{\frac{2}{2\gamma +d}}}{e}}\right)\hfill \\ & +\hfill & {\displaystyle t{\int }_{{\mathbb{R}}^d}g\left(x\right){|{\nabla }_{k}ln\left(g\left(x\right)\right)|}^2\mathrm{d}{\gamma }_{k,t}\left(x\right).}\hfill \end{array}$$

After applying the $L^1$-invariant scaling $g_{\lambda }\left(x\right)\equiv {\lambda }^{2\gamma +d}g\left(\lambda x\right)$, we have

$$\begin{array}{ccc}& & {\displaystyle {\int }_{{\mathbb{R}}^d}g\left(x\right)ln\left(g\left(x\right)\right)d{\gamma }_{k,t}\left(x\right)+(2\gamma +d)ln\lambda {\int }_{{\mathbb{R}}^d}g\left(x\right)d{\gamma }_{k,t}\left(x\right)}\hfill \\ & \le & {\displaystyle \frac{2\gamma +d}{2}ln\left(\frac{{2te\left|G\right|}^{\frac{2}{2\gamma +d}}}{2\gamma +d}{\int }_{{\mathbb{R}}^d}g\left(x\right){|{\nabla }_{k}ln\left(g\left(x\right)\right)|}^{2}d{\gamma }_{k,t}\left(x\right)+\frac{{2\left|G\right|}^{\frac{2}{2\gamma +d}}}{e}\right)+t{\int }_{{\mathbb{R}}^d}g\left(x\right){|\nabla ln\left(g\left(x\right)\right)|}^{2}d{\gamma }_{k,t}\left(x\right).}\hfill \end{array}$$

Since $\parallel g{G}_{\mu }{\parallel }_{1,{\mu }_k}=1,$ choosing

$${\displaystyle {\lambda }^2=\frac{{2te\left|G\right|}^{\frac{2}{2\gamma +d}}}{2\gamma +d}{\int }_{{\mathbb{R}}^d}g\left(x\right){|\nabla ln\left(g\left(x\right)\right)|}^{2}d{\gamma }_{k,t}\left(x\right)+\frac{{2\left|G\right|}^{\frac{2}{2\gamma +d}}}{e},}$$

we obtain the desired result. □

5. $L_k^p$-Nash-Type and Uncertainty Inequalities

Let $1\le p<2\gamma +d$, and let $p^{*}=\frac{p(2\gamma +d)}{2\gamma +d-p}$. Then from [11] (Theorem 1.1)), there exists a constant $C_k(d,p)$ such that for all $f\in {\mathcal{C}}_c^{\infty }\left({\mathbb{R}}^d\right)$:

$${\parallel f\parallel }_{p^{*},{\mu }_k}\le C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}.$$

(129)

This inequality extends naturally to the Dunkl–Sobolev space (see also [19,20])

$$W_k^{1,p}\left({\mathbb{R}}^d\right)=\{f\in L_k^p\left({\mathbb{R}}^d\right):{\nabla }_{k}f\in L_k^p\left({\mathbb{R}}^d\right)\}.$$

Then we have the following $L_k^p$-logarithmic Sobolev inequality.

Theorem 14.

Let $1\le p<2\gamma +d$. Then for all $f\in W_k^{1,p}\left({\mathbb{R}}^d\right)$ such that ${\parallel f\parallel }_{p,{\mu }_k}=1$, we have

$${\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{p}ln\left({\left|f\left(x\right)\right|}^p\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{p}ln\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}^p\right),$$

(130)

where $C_k(d,p)$ is the constant of the Sobolev inequality (129).

Proof. 

This proof is inspired by [21]. Let s be a real number such that $p<s<p^{*}=\frac{p(2\gamma +d)}{2\gamma +d-p}$, and let $f\in {\mathcal{C}}_c^{\infty }\left({\mathbb{R}}^d\right)$ be a function such that ${\parallel f\parallel }_{p,{\mu }_k}=1$, by which ${\left|f\left(x\right)\right|}^p\mathrm{d}{\mu }_k\left(x\right)$ is a probability measure. Then by Hölder’s inequality, we have

$${\parallel f\parallel }_{s,{\mu }_k}\le {\parallel f\parallel }_{p^{*},{\mu }_k}^{\theta }{\parallel f\parallel }_{p,{\mu }_k}^{1-\theta },$$

(131)

where

$$\frac{1}{s}=\frac{\theta }{p^{*}}+\frac{1-\theta }{p}.$$

Then $\theta =\frac{p^{*}(s-p)}{s(p^{*}-p)}$, and since ${\parallel f\parallel }_{p,{\mu }_k}=1$, we have

$${\parallel f\parallel }_{s,{\mu }_k}^{\frac{s}{s-p}}\le {\parallel f\parallel }_{p^{*},{\mu }_k}^{\frac{p^{*}}{p^{*}-p}}.$$

(132)

Thus by the Sobolev inequality (129), we obtain

$${\parallel f\parallel }_{s,{\mu }_k}^{\frac{s}{s-p}}\le {\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right)}^{\frac{2\gamma +d}{p}}.$$

(133)

Now, since (see [22]),

$$\underset{s\to p}{lim}{\parallel f\parallel }_{s,{\mu }_k}^{\frac{s}{s-p}}=\underset{s\to p}{lim}{\left({\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{s-p}{\left|f\left(x\right)\right|}^p\mathrm{d}{\mu }_k\left(x\right)\right)}^{\frac{1}{s-p}}=exp\left({\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{p}ln\left(\right|f\left(x\right)\left|\right)\mathrm{d}{\mu }_k\left(x\right)\right),$$

(134)

then by taking the limit of both sides in (133) as s approaches p, we obtain

$$exp\left({\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{p}ln\left(\right|f\left(x\right)\left|\right)\mathrm{d}{\mu }_k\left(x\right)\right)\le {\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right)}^{\frac{2\gamma +d}{p}}.$$

(135)

Hence,

$${\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{p}ln\left(\right|f\left(x\right)\left|\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{p}ln\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right).$$

(136)

The density argument completes the proof. □

Remark 8.

Instead of the limit in (133), we can obtain the result by Jensen’s inequality:

$$\begin{array}{ccc}\hfill \frac{2\gamma +d}{p}ln\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right)& \ge & \frac{1}{s-p}ln\left({\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{s-p}{\left|f\left(x\right)\right|}^p\mathrm{d}{\mu }_k\left(x\right)\right)\hfill \\ & \ge & \frac{1}{s-p}{\int }_{{\mathbb{R}}^d}ln\left({\left|f\left(x\right)\right|}^{s-p}\right){\left|f\left(x\right)\right|}^p\mathrm{d}{\mu }_k\left(x\right)\hfill \\ & =& {\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{p}ln\left(\left|f\right(x\left)\right|\right)\mathrm{d}{\mu }_k\left(x\right).\hfill \end{array}$$

Hence, we derive the following $L_k^p$-Heisenberg uncertainty inequality for the Dunkl transform.

Corollary 9.

Let $1\le p<2\gamma +d$. Then for all $f\in L_{k,1}^p\left({\mathbb{R}}^d\right)\cap W_k^{1,p}\left({\mathbb{R}}^d\right)$, we have

$${\parallel \left|x\right|f\parallel }_{p,{\mu }_k}{\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\ge C(k,p,d){∥f∥}_{p,{\mu }_k}^2,$$

(137)

where

$$C(k,p,d)=\frac{2\gamma +d}{pe{C}_k(p,d)}{\left(\frac{p{\mathbf{c}}_k{2}^{\gamma +d/2-1}\mathrm{\Gamma }(\gamma +d/2)}{\mathrm{\Gamma }\left(\frac{2\gamma +d}{p}\right)}\right)}^{\frac{p}{2\gamma +d}}.$$

(138)

Proof. 

Let $f\in L_{k,1}^p\left({\mathbb{R}}^d\right)\cap W_k^{1,p}\left({\mathbb{R}}^d\right)$ be a nonzero function such that ${\parallel f\parallel }_{p,{\mu }_k}=1$. Then, from (130) and (57), for $f\to {\left|f\right|}^p$ and $s=p$, we have

$$-(2\gamma +d)ln\left({c(k,p,d)\parallel \left|x\right|f\parallel }_{p,{\mu }_k}\right)\le {\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{p}ln\left({\left|f\left(x\right)\right|}^p\right)\mathrm{d}{\mu }_k\left(x\right)\le (2\gamma +d)ln\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right).$$

Thus

$$ln\left({c(k,p,d)\parallel \left|x\right|f\parallel }_{p,{\mu }_k}\right)+ln\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right)\ge 0.$$

(139)

Therefore,

$$ln\left(c(k,p,d)C_k{(d,p)\parallel \left|x\right|f\parallel }_{p,{\mu }_k}{\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right)\ge 0,$$

(140)

which implies that

$${\parallel \left|x\right|f\parallel }_{p,{\mu }_k}{\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\ge {\left(c(k,p,d)C_k(d,p)\right)}^{-1}.$$

(141)

Replacing f with $f/{\parallel f\parallel }_{p,{\mu }_k}$ in the last inequality, we obtain the result. □

It’s well know that Sobolev’s inequality and Nash’s inequality are equivalent (see e.g., [11,23,24]). The Nash-type inequality (85) (for $\beta =1$ and with the assumption $2\gamma +d>2$) can be derived from the Sobolev-type inequality (94) by using only Hölder’s inequality (see [11] (Theorem 4.2)). Now we will use the $L_k^p$-logarithmic Sobolev inequality (130) and the techniques of Beckner in [24] to derive an $L_k^p$-Nash-type inequality for the Dunkl transform.

Corollary 10.

Let $1\le q<p<2\gamma +d$. Then for all $f\in L_k^q\left({\mathbb{R}}^d\right)\cap W_k^{1,p}\left({\mathbb{R}}^d\right)$,

$${∥f∥}_{p,{\mu }_k}^{1+\frac{pq}{(p-q)(2\gamma +d)}}\le C_k(d,p){∥f∥}_{q,{\mu }_k}^{\frac{pq}{(p-q)(2\gamma +d)}}{\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k},$$

(142)

where $C_k(d,p)$ is the constant of the Sobolev inequality (129).

Proof. 

Let $1\le q<p<2\gamma +d$, and let $f\in L_k^q\left({\mathbb{R}}^d\right)\cap W_k^{1,p}\left({\mathbb{R}}^d\right)$ such that ${∥f∥}_{p,{\mu }_k}=1$. Then, by Jensen’s inequality and the $L_k^p$-logarithmic Sobolev inequality (130),

$$\begin{array}{ccc}\hfill -ln\left({∥f∥}_{q,{\mu }_k}^q\right)& \le & -{\int }_{{\mathbb{R}}^d}ln\left(\frac{1}{{\left|f\left(x\right)\right|}^{p-q}}\right){\left|f\left(x\right)\right|}^p\mathrm{d}{\mu }_k\left(x\right)\hfill \\ & =& \frac{p-q}{p}{\int }_{{\mathbb{R}}^d}{\left|f\left(x\right)\right|}^{p}ln\left({\left|f\left(x\right)\right|}^p\right)\mathrm{d}{\mu }_k\left(x\right)\hfill \\ & \le & \frac{(p-q)(2\gamma +d)}{p}ln\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right).\hfill \end{array}$$

Then

$${∥f∥}_{q,{\mu }_k}{\left(C_k(d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}\right)}^{\frac{(p-q)(2\gamma +d)}{qp}}\ge 1.$$

(143)

Now replacing f with $f/{∥f∥}_{p,{\mu }_k}$ in the last inequality, we obtain

$${∥f∥}_{p,{\mu }_k}^{1+\frac{pq}{(p-q)(2\gamma +d)}}\le C_k(d,p){∥f∥}_{q,{\mu }_k}^{\frac{pq}{(p-q)(2\gamma +d)}}{\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k}$$

(144)

as desired. □

Remark 9.

If instead of the $L_k^p$-logarithmic Sobolev inequality (130) we use the logarithmic Sobolev inequality (96) in the last proof, we obtain, with the assumption $2\gamma +d>2$,

$${∥f∥}_{2,{\mu }_k}^{1+\frac{2}{2\gamma +d}}\le {\mathcal{S}}_{k,d}{∥f∥}_{1,{\mu }_k}^{\frac{2}{2\gamma +d}}{∥\left|\xi \right|{\mathcal{F}}_D\left(f\right)∥}_{2,{\mu }_k},$$

(145)

where ${\mathcal{S}}_{k,d}$ is the sharp constant in Sobolev’s inequality (96) given by (95). Therefore the optimal constant in the Nash-type inequality (145) should be $\le {\mathcal{S}}_{k,d}\approx \frac{2}{e(\gamma +d/2)}$ for $d\gg 1$.

Theorem 15.

Let $1<p<2\gamma +d$. For any non-negative function $f\in L_{k,p^{\prime }}^1\left({\mathbb{R}}^d\right)$ such that $f^{\frac{1}{p}}$ belongs to $W_k^{1,p}\left({\mathbb{R}}^d\right)$, we have

$${\parallel f\parallel }_{1,{\mu }_k}\le \frac{1}{2\gamma +d}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)d{\mu }_k\left(x\right)}\right)}^{1/p^{\prime }}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{\left|{\nabla }_{k}f\left(x\right)\right|}^p{f}^{1-p}\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{1/p},$$

(146)

where ${(2\gamma +d)}^{-1}$ is the best possible constant, and it is attained by $f\left(x\right)=e^{-\tilde{c}\left(k,d,p^{\prime }\right){\left|x\right|}^{p^{\prime }}}$ with

$$\tilde{c}\left(k,d,p^{\prime }\right)={\left(\frac{d_k\mathrm{\Gamma }\left(\frac{2\gamma +d}{p^{\prime }}\right)}{p^{\prime }}\right)}^{\frac{p^{\prime }}{2\gamma +d}}.$$

(147)

Proof. 

Involving the fact that ${\mathrm{div}}_k\left(x\right)=2\gamma +d,$ Green’s identity in the Dunkl setting (cf. [1,14]) and Hölder’s inequality, we get

$$\begin{array}{ccc}\hfill {\displaystyle {\int }_{{\mathbb{R}}^d}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}& =& {\displaystyle \frac{1}{2\gamma +d}{\int }_{{\mathbb{R}}^d}\left({\mathrm{div}}_k\left(x\right)\right)f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)=-\frac{1}{2\gamma +d}{\int }_{{\mathbb{R}}^d}⟨x,{\nabla }_{k}f\left(x\right)⟩\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & {\displaystyle \frac{1}{2\gamma +d}{\int }_{{\mathbb{R}}^d}\left|x\right|f{\left(x\right)}^{1/p^{\prime }}f{\left(x\right)}^{1/p}\frac{|{\nabla }_{k}f\left(x\right)|}{f\left(x\right)}d{\mu }_k\left(x\right)}\hfill \\ & \le & {\displaystyle \frac{1}{2\gamma +d}{\left({\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)\right)}^{1/p^{\prime }}{\left({\int }_{{\mathbb{R}}^d}{\left|{\nabla }_{k}f\left(x\right)\right|}^p{f}^{1-p}\left(x\right)\mathrm{d}{\mu }_k\left(x\right)\right)}^{1/p}.}\hfill \end{array}$$

To see the optimality, we substitute $f\left(x\right)=e^{-\tilde{c}\left(k,d,p^{\prime }\right){\left|x\right|}^{p^{\prime }}}$ in (146), involve the identity (65), use the fact that ${\parallel f\parallel }_{1,{\mu }_k}=1$, and, after standard calculations, we derive the result. □

Remark 10.

Let $W_{k,G}^{1,p}\left({\mathbb{R}}^d\right):=\left\{u\in W_k^{1,p}\left({\mathbb{R}}^d\right):u\circ {\sigma }_{\xi }=u,\mathrm{for}\mathrm{all}\xi \in R\right\}$. If we assume that $f\in W_{k,G}^{1,p}\left({\mathbb{R}}^d\right)$, we derive that the inequality (80) is a direct consequence of (146) by setting $p=p^{\prime }=2$ and ${\left|f\right|}^2$ into f.

We illustrate a different method using the logarithmic Sobolev inequality and the generalized version of the Shannon inequality in $L^p$. Firstly, we recall the following results from [11].

Theorem 16.

Assume that $k>0$. Let $1\le p<2\gamma +d$. Then for any $f\in W_{k,G}^{1,p}\left({\mathbb{R}}^d\right)$, we have

$${\parallel f\parallel }_{p^{*},{\mu }_k}\le C(k,d,p){\parallel {\nabla }_{k}f\parallel }_{p,{\mu }_k},$$

(148)

where

$$C(k,d,p)=\left\{\begin{array}{ccc}{(2\gamma +d)}^{\frac{-1}{p}}{\left(\frac{p-1}{2\gamma +d-p}\right)}^{\frac{1}{p^{\prime }}}{\left[\frac{p^{\prime }\left|G\right|\mathrm{\Gamma }\left(2\gamma +d\right)}{d_k\mathrm{\Gamma }\left(\frac{2\gamma +d}{p}\right)\mathrm{\Gamma }\left(\frac{2\gamma +d}{p^{\prime }}\right)}\right]}^{\frac{1}{2\gamma +d}},\hfill & \mathrm{for}\hfill & 1<p<2\gamma +d\hfill \\ {(2\gamma +d)}^{-1+\frac{1}{2\gamma +d}}{\left[\frac{\left|G\right|\mathrm{\Gamma }\left(\frac{2\gamma +d}{2}\right)}{d_k}\right]}^{\frac{1}{2\gamma +d}},\hfill & \mathrm{for}\hfill & p=1.\hfill \end{array}\right.$$

(149)

Moreover, the constant $C(k,d,p)$ is sharp, and equality is achieved if and only if f is supported on the closure of a Weyl chamber, where it takes the form ${(a+b|x|}^{p^{\prime }}{)}^{1-\frac{2\gamma +d}{p}}$, where $a,b>0$.

Remark 11.

Assume that $k>0$, and let $1\le p<2\gamma +d$. We have (see [11])

$$\underset{f\in W_{k,rad}^{1,p}\left({\mathbb{R}}^d\right)}{sup}\frac{{\parallel f\parallel }_{p^{*},{\mu }_k}}{\parallel {\nabla }_k{f\parallel }_{p,{\mu }_k}}:={\left|G\right|}^{\frac{-1}{2\gamma +d}}C(k,d,p),$$

(150)

where $C(k,d,p)$ is the constant defined by (149), and $W_{k,rad}^{1,p}\left({\mathbb{R}}^d\right):=\left\{f\in W_k^{1,p}\left({\mathbb{R}}^d\right):f\mathrm{is}\mathrm{radial}\right\}$.

We proceed as in the proof of Theorem 14, and, involving (148), we get the following inequality:

$${\displaystyle {\int }_{{\mathbb{R}}^d}f{\left(x\right)}^{p}ln\left(\frac{{\left(f\left(x\right)\right)}^p}{{\parallel f\parallel }_p^p}\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{p}{\parallel f\parallel }_{p,{\mu }_k}^{p}ln\left({\displaystyle \frac{C\left(k,d,p\right)}{{\parallel f\parallel }_{p,{\mu }_k}^p}{\int }_{{\mathbb{R}}^d}{\left|{\nabla }_{k}f\left(x\right)\right|}^p\mathrm{d}{\mu }_k\left(x\right)}\right).}$$

(151)

Corollary 11.

Let $1<p<2\gamma +d.$ For any $f^{\frac{1}{p}}\in W_{k,G}^{1,p}\left({\mathbb{R}}^d\right)$ with $f\ge 0,$

$${\displaystyle {\int }_{{\mathbb{R}}^d}f\left(x\right)ln\left(\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}\right)d{\mu }_k\left(x\right)\le \frac{2\gamma +d}{p}{\parallel f\parallel }_{1,{\mu }_k}ln\left({\displaystyle \frac{C\left(k,d,p\right)}{{\parallel f\parallel }_{1,{\mu }_k}}{\int }_{{\mathbb{R}}^d}f\left(x\right){|{\nabla }_{k}lnf\left(x\right)|}^p\mathrm{d}{\mu }_k\left(x\right)}\right).}$$

(152)

Equivalently,

$${\displaystyle {\int }_{{\mathbb{R}}^d}f\left(x\right)lnf\left(x\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{p}{\parallel f\parallel }_{1,{\mu }_k}ln\left({\displaystyle \frac{C\left(k,d,p\right)}{{\parallel f\parallel }_{1,{\mu }_k}^{1-\frac{p}{2\gamma +d}}}{\int }_{{\mathbb{R}}^d}f\left(x\right){|{\nabla }_{k}lnf\left(x\right)|}^{p}dx}\right),}$$

(153)

where $C\left(k,d,p\right)$ is the constant given by (149).

Theorem 17.

Let $1<p<2\gamma +d$. For any non-negative function $f\in L_{k,p^{\prime }}^1\left({\mathbb{R}}^d\right)$ such that $f^{\frac{1}{p}}\in W_{k,G}^{1,p}\left({\mathbb{R}}^d\right)$, we have

$$\begin{array}{ccc}\hfill -\frac{2\gamma +d}{p^{\prime }}ln\left(\frac{{\displaystyle c(k,p^{\prime },d){\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}}{{\parallel f\parallel }_{1,{\mu }_k}}\right)& \le & {\displaystyle {\int }_{{\mathbb{R}}^d}\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}ln\left(\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & \frac{2\gamma +d}{p}ln\left({\displaystyle \frac{C\left(k,d,p\right)}{{\parallel f\parallel }_{1,{\mu }_k}}{\int }_{{\mathbb{R}}^d}{\left|{\nabla }_{k}f\left(x\right)\right|}^p{f}^{1-p}\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right),\hfill \end{array}$$

where $C\left(k,d,p\right)$ and $c(k,p^{\prime },d)$ are given in (149) and (58), respectively. In particular,

$${\parallel f\parallel }_{1,{\mu }_k}\le c{(k,p^{\prime },d)}^{\frac{2\gamma +d}{p^{\prime }}}C{\left(k,d,p\right)}^{\frac{1}{p}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{1}{p^{\prime }}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{|{\nabla }_{k}lnf\left(x\right)|}^{p}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{1}{p}}.$$

(154)

Proof. 

From Theorem 6 with $s=p^{\prime }$, we have

$$\begin{array}{ccc}\hfill {\displaystyle {\int }_{{\mathbb{R}}^d}\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}ln{\left(f\left(x\right)\right)}^{-1}\mathrm{d}{\mu }_k\left(x\right)}& \le & \frac{2\gamma +d}{p^{\prime }}ln\left(\frac{{\displaystyle c(k,p^{\prime },d){\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}}{{\parallel f\parallel }_1^{1+\frac{p^{\prime }}{2\gamma +d}}}\right)\hfill \\ & =& ln\left(\frac{c{(k,p^{\prime },d)}^{\frac{2\gamma +d}{p^{\prime }}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p^{\prime }}}}{{\parallel f\parallel }_1^{1+\frac{p^{\prime }}{2\gamma +d}}}\right),\hfill \end{array}$$

(155)

where

$${\left(c(k,p^{\prime },d)\right)}^{\frac{2\gamma +d}{p^{\prime }}}={\left(\frac{p^{\prime }e}{2\gamma +d}\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\parallel e^{-{\left|x\right|}^{p^{\prime }}}\parallel }_{1,{\mu }_k}.$$

(156)

From the sharp version of the logarithmic Sobolev inequality (152) with the best constant (149),

$$\begin{array}{ccc}\hfill & & {\displaystyle {\int }_{{\mathbb{R}}^d}\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}ln\left(f\left(x\right)\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & \frac{2\gamma +d}{p}ln\left({\displaystyle \frac{C\left(k,d,p\right)}{{\parallel f\parallel }_{1,{\mu }_k}^{1-\frac{p}{2\gamma +d}}}{\int }_{{\mathbb{R}}^d}{|\nabla lnf\left(x\right)|}^{p}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)\hfill \\ & =& ln\left(\frac{{\left(C\left(k,d,p\right)\right)}^{\frac{2\gamma +d}{p}}}{{\parallel f\parallel }_{1,{\mu }_k}^{\frac{2\gamma +d}{p}-1}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{|\nabla lnf\left(x\right)|}^{p}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p}}\right).\hfill \end{array}$$

(157)

Combining (155) with (156) and (157), we immediately obtain that for any $1\le p<2\gamma +d,$

$$\begin{array}{ccc}\hfill & & -ln\left(\frac{{\displaystyle {\left(c(k,p^{\prime },d)\right)}^{\frac{2\gamma +d}{p^{\prime }}}({\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right){)}^{\frac{2\gamma +d}{p^{\prime }}}}}{{\parallel f\parallel }_1^{\frac{p^{\prime }}{2\gamma +d}}}\right)\hfill \\ & \le & {\displaystyle {\int }_{{\mathbb{R}}^d}\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}ln\left(\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & ln\left(\frac{{\left(C\left(k,d,p\right)\right)}^{\frac{2\gamma +d}{p}}}{{\parallel f\parallel }_{1,{\mu }_k}^{\frac{2\gamma +d}{p}-1}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{|\nabla lnf\left(x\right)|}^{p}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p}}\right).\hfill \end{array}$$

Namely,

$$0\le {\parallel f\parallel }_{1,{\mu }_k}ln\left(\frac{{\left(c(k,p^{\prime },d)\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\left(C\left(k,d,p\right)\right)}^{\frac{2\gamma +d}{p}}}{{\parallel f\parallel }_{1,{\mu }_k}^{\frac{2\gamma +d}{p^{\prime }}+\frac{2\gamma +d}{p}}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{|\nabla lnf\left(x\right)|}^{p}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p}}\right),$$

or equivalently

$${\parallel f\parallel }_{1,{\mu }_k}^{2\gamma +d}\le {\left(c(k,p^{\prime },d)\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\left(C\left(k,d,p\right)\right)}^{\frac{2\gamma +d}{p}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{|\nabla lnf\left(x\right)|}^{p}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p}}.$$

The proof is complete. □

Let p and q be two positive reals (not necessarily Hölder conjugate exponents). We proceed as in Theorem 15 to derive a generalized version of the uncertainty principle.

Proposition 4.

Let $2\gamma +d\ge 2$ and $1<p,q<2\gamma +d.$ Then for any nonnegative $f\in W_{k,G}^{1,p}\left({\mathbb{R}}^d\right)$ with $f\in L_{k,q}^1\left({\mathbb{R}}^d\right)$, it holds

$${\parallel f\parallel }_{1,{\mu }_k}^{\frac{1}{p}+\frac{1}{q}}\le {\left(c(k,q,d)\right)}^{\frac{1}{q}}{\left(C\left(k,d,p\right)\right)}^{\frac{1}{p}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{\left|x\right|}^{q}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{1}{q}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{|\nabla lnf\left(x\right)|}^{p}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{1}{p}},$$

where $C\left(k,d,p\right)$ and $c(k,q,d)$ are given in (149) and (58), respectively.

Theorem 18

(Generalized uncertainty principle). Let $2\gamma +d\ge 2$ and $1<p<2\gamma +d.$ For any non-negative function $f\in L_{k,p^{\prime }}^1\left({\mathbb{R}}^d\right)$ with $f\in W_{k,G}^{1,p}\left({\mathbb{R}}^d\right)$, it holds

$$\begin{array}{c}\hfill {\parallel f\parallel }_{1_{*},k}^{\frac{2\gamma +d}{p^{\prime }}+1}{\parallel f\parallel }_{1^{*},k}^{\frac{2\gamma +d}{p}-1}exp\left({\displaystyle \frac{1}{2\gamma +d}{\int }_{{\mathbb{R}}^d}\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}ln\left|f\left(x\right)\right|\mathrm{d}{\mu }_k\left(x\right)}\right)\le \\ \hfill \frac{1}{2\gamma +d}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{1}{p^{\prime }}}{\left({\displaystyle {\int }_{{\mathbb{R}}^d}f\left(x\right){|\nabla lnf\left(x\right)|}^p\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{1}{p}},\end{array}$$

where ${\parallel f\parallel }_{1_{*},k}$ and ${\parallel f\parallel }_{1^{*},k}$ denote ${\parallel f\parallel }_{L_k^1\left(\{f\le 1\}\right)}$ and ${\parallel f\parallel }_{L_k^1\left(\{f\le 1\}\right)},$ respectively.

Proof. 

Replacing $f\left(x\right)\to f\left(x\right){\chi }_{\{f\le 1\}}\left(x\right)$ in (70) and using ${\parallel f\parallel }_{L^1\left(\{f\le 1\}\right)}={\parallel f\parallel }_{1_{*},k}$, we get

$$\begin{array}{ccc}\hfill & & {\displaystyle {\int }_{\{f\le 1\}}f\left(x\right)ln\left(f{\left(x\right)}^{-1}\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & \frac{2\gamma +d}{p^{\prime }}{\parallel f\parallel }_{1_{*},k}ln\left(\frac{{\displaystyle c(k,p^{\prime },d){\int }_{\{f\le 1\}}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}}{{\parallel f\parallel }_{1_{*},k}^{1+\frac{p^{\prime }}{2\gamma +d}}}\right)\hfill \\ & =& {\parallel f\parallel }_{1_{*},k}ln\left(\frac{{\left(c(k,p^{\prime },d)\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\left({\displaystyle {\int }_{\{f\le 1\}}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p^{\prime }}}}{{\parallel f\parallel }_{1_{*},k}^{1+\frac{2\gamma +d}{p^{\prime }}}}\right).\hfill \end{array}$$

(158)

Similarly substituting $f\left(x\right)\to f\left(x\right){\chi }_{\{f\ge 1\}}\left(x\right)$ in (153) and using ${\parallel f\parallel }_{L^1\left(\{f\ge 1\}\right)}={\parallel f\parallel }_{1^{*},k}$, we derive

$${\displaystyle {\int }_{\{f\ge 1\}}f\left(x\right)ln\left(f\left(x\right)\right)\mathrm{d}{\mu }_k\left(x\right)\le \frac{2\gamma +d}{p}{\parallel f\parallel }_{1^{*},k}ln\left({\displaystyle \frac{C\left(k,d,p\right)}{{\parallel f\parallel }_{1^{*},k}^{1-\frac{p}{2\gamma +d}}}{\int }_{\{f\ge 1\}}f\left(x\right){|\nabla ln{\chi }_{f\ge 1}f\left(x\right)|}^p\mathrm{d}{\mu }_k\left(x\right)}\right).}$$

Thus,

$${\displaystyle {\int }_{\{f\ge 1\}}f\left(x\right)ln\left(\frac{f\left(x\right)}{{\parallel f\parallel }_{1^{*},k}}\right)\mathrm{d}{\mu }_k\left(x\right)\le {\parallel f\parallel }_{1^{*},k}ln{\left(\frac{C\left(k,d,p\right)}{{\parallel f\parallel }_{1^{*},k}}\left({\displaystyle {\int }_{\{f\ge 1\}}f\left(x\right){|\nabla ln{\chi }_{f\ge 1}f\left(x\right)|}^p\mathrm{d}{\mu }_k\left(x\right)}\right)\right)}^{\frac{2\gamma +d}{p}}.}$$

(159)

Combining inequalities (158) and (159),

$$\begin{array}{ccc}\hfill & & {\displaystyle {\int }_{\{f\le 1\}}f\left(x\right)ln\left(f{\left(x\right)}^{-1}\right)\mathrm{d}{\mu }_k\left(x\right)+{\int }_{\{f\ge 1\}}f\left(x\right)ln\left(f\left(x\right)\right)\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & {\parallel f\parallel }_{1*}ln{\left(\frac{{\displaystyle c(k,p^{\prime },d){\int }_{\{f\le 1\}}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}}{{\parallel f\parallel }_{1_{*},k}^{1+\frac{p^{\prime }}{2\gamma +d}}}\right)}^{\frac{2\gamma +d}{p^{\prime }}}\hfill \\ & +& {\parallel f\parallel }_{1^{*},k}ln{\left(\frac{{\displaystyle C\left(k,d,p\right){\int }_{\{f\ge 1\}}f\left(x\right){|\nabla ln{\chi }_{f\ge 1}f\left(x\right)|}^p\mathrm{d}{\mu }_k\left(x\right)}}{{\parallel f\parallel }_{1^{*},k}^{1-\frac{p}{2\gamma +d}}}\right)}^{\frac{2\gamma +d}{p}}.\hfill \end{array}$$

(160)

Since ${\parallel f\parallel }_{1_{*},k}{,\parallel f\parallel }_{1^{*},k}\le {\parallel f\parallel }_{1,{\mu }_k},$ it follows from (160) that

$$\begin{array}{ccc}& & {\displaystyle {\int }_{{\mathbb{R}}^d}\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}|lnf\left(x\right)|\mathrm{d}{\mu }_k\left(x\right)}\hfill \\ & \le & ln\left(\frac{{\left(c(k,p^{\prime },d)\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\left(C\left(k,d,p\right)\right)}^{\frac{2\gamma +d}{p}}}{{\parallel f\parallel }_{1_{*},k}^{\frac{2\gamma +d}{p^{\prime }}+1}{\parallel f\parallel }_{1^{*},k}^{\frac{2\gamma +d}{p}-1}}{\left({\left|x\right|}^{p^{\prime }}f\left(x\right)\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\left({\displaystyle {\int }_{\{f\ge 1\}}f\left(x\right){|\nabla ln{\chi }_{f\ge 1}f\left(x\right)|}^{p}dx}\right)}^{\frac{2\gamma +d}{p}}\right),\hfill \end{array}$$

and thus we have

$$\begin{array}{c}{\parallel f\parallel }_{1_{*},k}^{\frac{2\gamma +d}{p^{\prime }}+1}{\parallel f\parallel }_{1^{*},k}^{\frac{2\gamma +d}{p}-1}exp\left({\displaystyle {\int }_{{\mathbb{R}}^d}\frac{f\left(x\right)}{{\parallel f\parallel }_{1,{\mu }_k}}|lnf\left(x\right)|\mathrm{d}{\mu }_k\left(x\right)}\right)\le \hfill \\ {\left(c(k,p^{\prime },d)\right)}^{\frac{2\gamma +d}{p^{\prime }}}{\left(C\left(k,d,p\right)\right)}^{\frac{2\gamma +d}{p}}{\left({\displaystyle {\int }_{\{f\le 1\}}{\left|x\right|}^{p^{\prime }}f\left(x\right)\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p^{\prime }}}\hfill \\ \times {\left({\displaystyle {\int }_{\{f\ge 1\}}f\left(x\right){|\nabla ln{\chi }_{f\ge 1}f\left(x\right)|}^p\mathrm{d}{\mu }_k\left(x\right)}\right)}^{\frac{2\gamma +d}{p}}.\hfill \end{array}$$

The proof is complete. □

6. Conclusions

We have obtained new Heisenberg-type uncertainty principles, new Shannon’s inequalities and new logarithmic Sobolev inequalities for functions in certain weighted Lebesgue spaces in the Dunkl setting. Moreover we have given another, short proof of some known results.