Abstract
This paper is perhaps the first attempt at a study of the Hardy space \(H^1\) in the rational Dunkl setting. Following Uchiyama’s approach, we characterize \(H^1\) atomically and by means of the heat maximal operator. We also obtain a Fourier multiplier theorem for \(H^1\). These results are proved here in the one-dimensional case and in the product case.
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1 Introduction
Dunkl theory is a far reaching generalization of Euclidean Fourier analysis which includes most special functions related to root systems, such as spherical functions on Riemannian symmetric spaces. It started in the late 1980s with Dunkl’s seminal article [7] and developed extensively afterwards. We refer to the lecture notes [18] for the rational Dunkl theory, to the lecture notes [15] for the trigonometric Dunkl theory, and to the books [4, 11] for the generalized quantum theories.
The theory of classical real Hardy spaces in \({\mathbb {R}}^n\) originates from the study of holomorphic functions of one variable in the upper half-plane. We refer the reader to the original works of Stein and Weiss [22], Burkholder et al. [3], and Fefferman and Stein [9]. An important contribution to this theory lies in the atomic decomposition introduced by Coifman [5] and extended to spaces of homogeneous type by Coifman and Weiss [6] (see also [12]). More information can be found in the book [21] and references therein.
This paper deals with the real Hardy space \(H^1\) in the rational Dunkl setting, where the underlying space is of homogeneous type in the sense of Coifman and Weiss. In such a setting, the theory of Hardy spaces goes back to the 1970s [6, 12]. Here we follow Uchiyama’s approach [25], and we characterize the Hardy space \(H^1\) in two ways, by means of the heat maximal operator and atomically. The first characterization, which requires precise heat kernel estimates, has led us to a seemingly new observation, namely that the heat kernel has a rather slow decay in certain directions and is in particular not Gaussian in the present setting (see Remark 2.4). The second characterization is used to prove a Fourier multiplier theorem for \(H^1\).
Throughout the paper, we shall restrict our considerations to the one-dimensional case and to the product case. This restriction is due to our present lack of knowledge in general about the behavior of the Dunkl kernel on the one hand and about generalized translations on the other hand.
Let us introduce some notation and state our main results. On \(\mathbb {R}^n\) we consider the Dunkl operators
associated with the reflections
and the multiplicities \(k_j\!\ge 0\). Their joint eigenfunctions constitute the Dunkl kernel
where
(see for instance [18, p. 107, Example 2.1]). Here \({{}_1}{F_{1}}(a;b;z)\) is the confluent hypergeometric function which is also known as the Kummer function and denoted by \(M(a,b,z)\). Notice that \(\mathbf{E}(\mathbf{x},\mathbf{y})=e^{\langle \mathbf{x},\mathbf{y}\rangle }\) if all multiplicities \(k_j\) vanish.
Let us first define the Hardy space \(H^1\) by means of the heat maximal operator. The Dunkl Laplacian
is the infinitesimal generator of the heat semigroup
which acts by linear self-adjoint operators on \(L^2(\mathbb {R}^n,\mathrm{d}\varvec{\mu })\) and by linear contractions on \(L^p(\mathbb {R}^n,\mathrm{d}\varvec{\mu })\), for every \(1\le p\le \infty \), where
The heat semigroup consists of integral operators
associated with the heat kernel
see, e.g., [17], where
is the homogeneous dimension and
From this point of view, the Hardy space \(H^1_{\text {max}}\) consists of all functions \(f\!\in \!L^1(\mathbb {R}^n,\mathrm{d}\varvec{\mu })\) whose maximal heat transform
belongs to \(L^1(\mathbb {R}^n,\mathrm{d}\varvec{\mu })\) and the norm is given by
Let us turn next to the atomic definition of the Hardy space \(H^1\). Notice that \(\mathbb {R}^n\), equipped with the Euclidean distance \(d(\mathbf{x},\mathbf{y})=|\mathbf{x}\!-\!\mathbf{y}|\) and with the measure \(\varvec{\mu }\), is a space of homogeneous type in the sense of Coifman and Weiss (see Appendix 1). Recall that an atom is a measurable function \(a:\mathbb {R}^n\!\rightarrow \mathbb {C}\) such that
-
\(\;a\) is supported in a ball \(\mathbf{B}\),
-
\(\;\Vert a\Vert _{L^\infty }\!\lesssim \varvec{\mu }(\mathbf{B})^{-1}\),
-
\(\;\displaystyle \int _{\mathbb {R}^n}\mathrm{d}\varvec{\mu }(\mathbf{x})\,a(\mathbf{x})=0\).
By definition, the atomic Hardy space \(H^1_{\text {atom}}\) consists of all functions \(f\!\in \!L^1(\mathbb {R}^n,\mathrm{d}\varvec{\mu })\) that can be written as \(f=\sum _{\ell }\lambda _{\ell }\,a_{\ell }\), where the \(a_\ell \)’s are atoms, the \(\lambda _{\ell }\)’s are complex numbers such that \(\sum _{\ell }|\lambda _{\ell }|\!<\!+\infty \), and the norm is given by
where the infimum is taken over all atomic decompositions of \(f\).
Our first main result is the following theorem.
Theorem 1.8
The spaces \(H^1_{\mathrm{max}}\) and \(H^1_{\mathrm{atom}}\) coincide and their norms are equivalent, i.e., there exists a constant \(C\!>\!0\) such that
The Fourier transform in the Dunkl setting is given by
It is an isometric isomorphism of \(L^2(\mathbb {R}^n,\mathrm{d}\varvec{\mu })\) onto itself and the inversion formula reads
Notice that, if all multiplicities \(k_j\) vanish, then (1.9) boils down to the classical Fourier transform
Our second main result is the following Hörmander type multiplier theorem (see [10] for the original multiplier theorem on \(L^p\) spaces).
Theorem 1.10
Let \(\chi =\chi (\varvec{\xi })\) be a smooth radial function on \(\mathbb {R}^n\) such that
If a function \(m=m(\varvec{\xi })\) on \(\mathbb {R}^n\) satisfies
for some \(\epsilon \!>\!0\), then the multiplier operator
is bounded on the Hardy space \(H^1\) and
Here \(W_{2}^{\sigma }(\mathbb {R}^n)\) denotes the classical \(L^2\) Sobolev space on \(\mathbb {R}^n\), whose norm is given by
Notice that the multiplier \(m\) is continuous and bounded, as \(\frac{\mathbf{N}}{2}+\epsilon >\frac{n}{2}\).
Our paper is organized as follows. Section 2 is devoted to the heat kernel in dimension \(1\). There we analyze its behavior thoroughly, and we remove a small part of it, in order to get Gaussian estimates similar to those in the Euclidean setting. These results are extended to the product case in Sect. 3. Section 4 is devoted to the proof of Theorem 1.8 and Sect. 5 to the proof of Theorem 1.10. Section consists of three appendices. Appendix 1 contains information about the measure of balls, which is used throughout the paper. Appendices 2 and 3 are devoted to so-called folklore results in connection with Uchiyama’s theorem, which have been used for instance in [8].
This paper results from two independent research projects, which were carried out by the first and third authors and by the second and fourth authors, respectively, and which have been merged into a joint article.
2 Heat Kernel Estimates in Dimension \(1\)
Consider first the one-dimensional Dunkl kernel \(E(x,y)=E_k(x,y)\). As the case \(k=0\) is trivial, we may assume that \(k>0\).
Lemma 2.1
-
(a)
\(E(x,y)\) is a holomorphic function of \((x,y)\!\in \!\mathbb {C}^2\).
-
(b)
\(E(x,y)>0\) for every \(x,y\!\in \!\mathbb {R}\).
-
(c)
\(E(x,y)\) has the following symmetry and rescaling properties :
$$\begin{aligned} {\left\{ \begin{array}{ll} \,E(x,y)=E(y,x) \qquad \forall \;x,y\!\in \!\mathbb {C},\\ \,E(\lambda x,y)=E(x,\lambda y) \qquad \forall \;\lambda ,x,y\!\in \!\mathbb {C}.\\ \end{array}\right. } \end{aligned}$$ -
(d)
For every \(y\!\in \!\mathbb {C}\), \(x\mapsto E(x,y)\) is an eigenfunction of the Dunkl operator
$$\begin{aligned} Df(x)=f^{\prime }(x)+\frac{k}{x}\,\{f(x)\!-\!f(-x)\} \end{aligned}$$and of the Dunkl Laplacian
$$\begin{aligned} Lf(x)=D^{2}f(x) =f^{\prime \prime }(x) +\frac{2k}{x}f^{\prime }(x) -\frac{k}{x^2}\,\{f(x)\!-\!f(-x)\}\,. \end{aligned}$$More precisely,
$$\begin{aligned} D_xE(x,y)=y\,E(x,y) \quad \text {and}\quad L_xE(x,y)=y^{2}E(x,y)\,. \end{aligned}$$ -
(e)
As \(xy\rightarrow 0\),
$$\begin{aligned} E(x,y)=1+\mathrm{O}\left( |xy|\right) \,. \end{aligned}$$ -
(f)
As \(xy\rightarrow +\infty \),
$$\begin{aligned} E(x,y)=\frac{2^{k}\,\Gamma (k+\frac{1}{2})}{\sqrt{\pi }}\, e^{xy}\,(xy)^{-k}\, \left\{ 1-\frac{k^2}{2}\frac{1}{xy}+\mathrm{O}\left( \frac{1}{x^2y^2}\right) \right\} \!. \end{aligned}$$ -
(g)
As \(xy\rightarrow -\infty \),
$$\begin{aligned} E(x,y)=\frac{2^{k-1}k\,\Gamma (k+\frac{1}{2})}{\sqrt{\pi }}\, e^{-xy}\,(-xy)^{-k-1}\, \left\{ 1+\frac{k^2-1}{2}\frac{1}{xy}+\mathrm{O}\left( \frac{1}{x^2y^2}\right) \right\} \!. \end{aligned}$$
Proof
The first four properties are known to hold in general. In dimension \(1\), they can also be deduced from the explicit expression (1.3), as can (e). As already observed in [20, Section 2] (see also [18, Example 5.1]), the asymptotics of \(E(x,y)\) at infinity follow from the asymptotics of the confluent hypergeometric function which read, let us say for \(0<a<b\),
as \(z\rightarrow +\infty \) and
as \(z\rightarrow -\infty \) (see for instance [1], (13.5.1)] or [14], (13.7.2)]). \(\square \)
Consider next the one-dimensional heat kernel
where \(c_{k}=2^{2k+1}\Gamma \left( k\!+\!\frac{1}{2}\right) \).
Proposition 2.3
-
(a)
\(h_{t}(x,y)\) is a \(C^\infty \) function of \((t,x,y)\!\in \!(0,+\infty )\!\times \mathbb {R}^2\).
-
(b)
\(h_{t}(x,y)>0\) for every \(t\!>\!0\) and \(x,y\!\in \!\mathbb {R}\).
-
(c)
\(h_{t}(x,y)\) has the following symmetry and rescaling properties :
$$\begin{aligned} {\left\{ \begin{array}{ll} \;h_{t}(x,y)=h_{t}(y,x) \qquad \forall \;x,y\!\in \!\mathbb {R},\\ \;h_{\lambda ^2t}(\lambda x,\lambda y) =|\lambda |^{-2k-1}\,h_{t}(x,y) \qquad \forall \;\lambda \!\in \!\mathbb {R}^*,\,\forall \;t\!>\!0,\,\forall \;x,y\!\in \!\mathbb {R}.\\ \end{array}\right. } \end{aligned}$$ -
(d)
\(h_{t}(x,y)\) satisfies the heat equation
$$\begin{aligned} {\left\{ \begin{array}{ll} \;\partial _{t}h_{t}(x,y)=L_yh_{t}(x,y),\\ \;\lim _{\,t\searrow 0}h_{t}(x,y)\,|y|^{2k}dy=\delta _x(y).\\ \end{array}\right. } \end{aligned}$$ -
(e)
The heat kernel has the following global behavior :
$$\begin{aligned} h_{t}(x,y)\,\asymp \,{\left\{ \begin{array}{ll} \;t^{-k-\frac{1}{2}}\, e^{-\frac{x^2+y^2}{4t}} &{}\text {if}\;|xy|\!\le t,\\ \;t^{-\frac{1}{2}}\,(xy)^{-k}\, e^{-\frac{(x-y)^2}{4t}} &{}\text {if }\;xy\ge t,\\ \;t^{\frac{1}{2}}\,(-xy)^{-k-1}\, e^{-\frac{(x+y)^2}{4t}} &{}\text {if }\,-xy\ge t,\\ \end{array}\right. } \end{aligned}$$and the following asymptotics :
$$\begin{aligned} h_{t}(x,y)\!=\!{\left\{ \begin{array}{ll} \;c_{k}^{-1}\,t^{-k-\frac{1}{2}}\, e^{-\frac{x^2+y^2}{4t}}\, \left\{ 1+\mathrm{O}\left( \frac{|xy|}{t}\right) \right\} &{}\text {if }\; \frac{xy}{t}\!\rightarrow 0,\\ \;\frac{1}{2\sqrt{\pi }}\;e^{-\frac{(x-y)^2}{4t}}\, t^{-\frac{1}{2}}\,(xy)^{-k}\, \left\{ 1-k^2\frac{t}{xy}+\mathrm{O}\left( \frac{t^2}{x^2y^2}\right) \right\} &{}\text {if}\; \frac{xy}{t}\!\rightarrow \!+\infty ,\\ \;\frac{k}{2\sqrt{\pi }}\;e^{-\frac{(x+y)^2}{4t}}\, t^{\frac{1}{2}}\,(-xy)^{-k-1}\, \left\{ 1+\mathrm{O}\left( -\frac{t}{xy}\right) \right\} &{}\text {if} \; \frac{xy}{t}\!\rightarrow \!-\infty . \end{array}\right. } \end{aligned}$$ -
(f)
The following gradient estimates hold for the heat kernel :
$$\begin{aligned}&\left| \frac{\partial }{\partial y}h_{t}(x,y)\right| \\&\lesssim \,{\left\{ \begin{array}{ll} \;t^{-k-\frac{3}{2}}\left( |x|\!+\!|y|\right) \, e^{-\frac{x^2+y^2}{4t}} &{}\text {if }\; |xy|\!\le t,\\ \,\left\{ t^{-\frac{3}{2}}|x\!-\!y| +t^{-\frac{1}{2}}|y|^{-1}\right\} \, (xy)^{-k}\,e^{-\frac{(x-y)^2}{4t}} &{}\text {if }\; xy\ge t,\\ \,\left\{ t^{-\frac{1}{2}}|x\!+\!y| +t^{\frac{1}{2}}(|x|^{-1}\!+|y|^{-1})\right\} \, (-xy)^{-k-1}\, e^{-\frac{(x+y)^2}{4t}} &{}\text {if} \, -xy\ge t.\\ \end{array}\right. } \end{aligned}$$
Proof
The first five properties follow from the expression (2.2) and from Lemma 2.1. Let us turn to the proof of (f). By differentiating (2.2) with respect to \(y\) and by using the well-known formula
(see for instance [1], (13.4.8)] or [14], (13.3.15)]), we get
We conclude by using again the behavior of the confluent hypergeometric function. \(\square \)
Remark 2.4
It follows from Proposition 2.3(e) and Appendix 1 that
for every \(t\!>\!0\) and \(x\!\in \!\mathbb {R}\). Observe in particular that the heat kernel has no global Gaussian behavior and decays rather slowly in certain directions. This phenomenon is even more striking in the product case (3.1), where
if \(t\!>\!0\), \(\mathbf{x}\!\in \!\mathbb {R}^n\), and \(\mathbf{y}=(-x_1,x_2,\ldots ,x_n)\).
Let us introduce a variant of the heat kernel with a Gaussian behavior. Given two smooth bump functions \(\chi _1\) and \(\chi _2\) on \(\mathbb {R}\) such that
consider the smooth cutoff function
and the truncated heat kernel
Remark 2.5
The truncated heat kernel \(H_t(x,y)\) inherits the following properties of the heat kernel \(h_{t}(x,y)\) :
-
(a)
Smoothness : \(H_{t}(x,y)\) is a \(C^\infty \) function of \((t,x,y)\!\in \!(0,+\infty )\!\times \mathbb {R}^2\).
-
(b)
Nonnegativity : \(H_{t}(x,y)\ge 0\) for every \(t\!>\!0\) and \(x,y\!\in \!\mathbb {R}\).
-
(c)
Rescaling : \(H_{\lambda ^2t}(\lambda x,\lambda y)=|\lambda |^{-2k-1}H_t(x,y)\) for every \(\lambda \!\in \!\mathbb {R}^*\), \(t\!>\!0,\) and \(x,y\!\in \!\mathbb {R}\).
-
(d)
Approximation of identity : \(\lim _{\,t\searrow 0}H_{t}(x,y)\,|y|^{2k}\mathrm{d}y=\delta _x(y)\) for every \(x,y\!\in \!\mathbb {R}\).
Theorem 2.6
The following estimates hold for the truncated heat kernel \(H_t(x,y)\).
-
(a)
On-diagonal estimate :
$$\begin{aligned} H_t(x,x)\asymp \mu \big (B\big (x,\sqrt{t\,}\big )\big )^{-1} \qquad \forall \;t\!>\!0,\,\forall \;x\!\in \!\mathbb {R}. \end{aligned}$$ -
(b)
Off-diagonal Gaussian estimate :
$$\begin{aligned} 0\le H_t(x,y)\lesssim \mu \big (B\big (x,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(x-y)^2}{c\,t}} \qquad \forall \;t\!>\!0,\,\forall \;x,y\!\in \!\mathbb {R}. \end{aligned}$$ -
(c)
Gradient estimate :
$$\begin{aligned} \left| \frac{\partial }{\partial y}H_t(x,y)\right| \lesssim t^{-\frac{1}{2}}\,\mu \big (B\big (x,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(x-y)^2}{c\,t}} \qquad \forall \;t\!>\!0,\,\forall \;x,y\!\in \!\mathbb {R}. \end{aligned}$$ -
(d)
Lipschitz estimates :
$$\begin{aligned}\textstyle |H_t(x,y)-H_t(x,y^{\prime })| \lesssim \mu \big (B\big (x,\sqrt{t\,}\big )\big )^{-1}\, \frac{|y-y^{\prime }|}{\sqrt{t\,}} \qquad \forall \;t\!>\!0,\,\forall \;x,y,y^{\prime }\in \!\mathbb {R}, \end{aligned}$$with the following improvement, if \(|y\!-\!y^{\prime }|\le \frac{1}{2}\,|x\!-\!y|\) :
$$\begin{aligned}\textstyle |H_t(x,y)-H_t(x,y^{\prime })| \lesssim \mu \big (B\big (x,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(x-y)^2}{c\,t}}\, \frac{|y-y^{\prime }|}{\sqrt{t\,}}. \end{aligned}$$
Here \(c\) denotes some positive constant and the ball measure has the following behavior, according to Appendix 1 :
Proof
As far as (a), (b), (c) are concerned, the case \(x=0\) follows immediately from the previous heat kernel estimates. Thus we may assume that \(x\ne 0\) and reduce furthermore to \(x=1\) by rescaling.
(a) is immediate:
Let us next prove (b) (Fig. 1).
-
Case 1. Assume that \(|y|\!\le t\).
\(\circ \) Subcase 1.1. Assume that \(t\) is bounded above, say \(t\le \!\frac{1}{2}\). Then
$$\begin{aligned} H_t(1,y)\le h_{t}(1,y)\asymp t^{-k-\frac{1}{2}}\,e^{-\frac{1+y^2}{4t}}\! =t^{-\frac{1}{2}}\,e^{-\frac{(1-y)^2}{8t}}\, t^{-k}\,e^{-\frac{1+y^2}{8t}}\,e^{-\frac{y}{4t}} \end{aligned}$$is bounded above by
$$\begin{aligned} \mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1} \,e^{-\frac{(1-y)^2}{8t}} \end{aligned}$$as \(t^{\frac{1}{2}}\asymp \mu (B(1,\sqrt{t\,}))\), \(t^{-k}\lesssim e^{\frac{1}{8t}}\le e^{\frac{1+y^2}{8t}}\), and \(e^{\frac{y}{4t}}\asymp 1\). \(\circ \) Subcase 1.2. Assume that \(t\) is bounded below, say \(t\ge \!\frac{1}{2}\). Then
$$\begin{aligned} H_t(1,y) \le h_{t}(1,y) \asymp t^{-k-\frac{1}{2}}\,e^{-\frac{1+y^2}{4t}}\! =t^{-k-\frac{1}{2}}\,e^{-\frac{(1-y)^2}{4t}}\,e^{-\frac{y}{2t}} \end{aligned}$$with \(t^{k+\frac{1}{2}}\asymp \mu (B(1,\sqrt{t\,}))\) and \(e^{\frac{y}{2t}}\asymp 1\).
-
Case 2. Assume that \(y\) is close to \(-x=\!-1\), say \(-2\le y\le \!-\frac{1}{2}\).
\(\circ \) Subcase 2.1. If \(t\le \!\frac{1}{2}(\le \!-y)\), then
$$\begin{aligned} H_t(1,y)=0\,. \end{aligned}$$\(\circ \) Subcase 2.2. If \(t\) is bounded below, say \(t\ge \!\frac{1}{2}\), we argue as in Subcase 1.2.
-
Case 3. Assume that \(y\ge t\).
\(\circ \) Subcase 3.1. Assume that t is bounded below, say \((y\!\ge )t\!\ge \!\frac{1}{2}\). Then
$$\begin{aligned} H_t(1,y)\le h_{t}(1,y)\asymp t^{-\frac{1}{2}}\,y^{-k}\,e^{-\frac{(1-y)^2}{4t}}\le t^{-k-\frac{1}{2}}\,e^{-\frac{(1-y)^2}{4t}} \end{aligned}$$with \(t^{k+\frac{1}{2}}\asymp \mu (B(1,\sqrt{t\,}))\).
\(\circ \) Subcase 3.2. Assume that \(y\ge \!\frac{1}{2}\!\ge t\). Then
$$\begin{aligned} H_t(1,y)\le h_{t}(1,y)\asymp t^{-\frac{1}{2}}\,y^{-k}\,e^{-\frac{(1-y)^2}{4t}}\lesssim t^{-\frac{1}{2}}\,e^{-\frac{(1-y)^2}{4t}} \end{aligned}$$with \(t^{\frac{1}{2}}\asymp \mu (B(1,\sqrt{t\,}))\).
\(\circ \) Subcase 3.3. Assume that \(t\le y\le \!\frac{1}{2}\). Then
$$\begin{aligned} H_t(1,y)\le h_{t}(1,y)\asymp t^{-\frac{1}{2}}\,y^{-k}\,e^{-\frac{(1-y)^2}{4t}} \end{aligned}$$is bounded above by
$$\begin{aligned} \mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1} \,e^{-\frac{(1-y)^2}{8t}} \end{aligned}$$as \(t^{\frac{1}{2}}\asymp \mu (B(1,\sqrt{t\,}))\) and \(y^{-k}\le t^{-k}\lesssim e^{\frac{1}{32\,t}}\le e^{\frac{(1-y)^2}{8t}}\).
-
Case 4. Assume that \(y\le \!-t\,(<\!0)\) and that \(y\) stays away from \(-1\), say \(y\!\notin \!\left( -2, -\frac{1}{2}\right) \). Notice that \((1\!+\!y)^2\!\ge \!\frac{(1-\,y)^2}{9}\) if and only if \(y\!\notin \!\left( -2, -\frac{1}{2}\right) \).
\(\circ \) Subcase 4.1. Assume that \(2\le t\le \!-y\). Then
$$\begin{aligned} H_t(1,y)\le h_{t}(1,y)\asymp t^{\frac{1}{2}}\,(-y)^{-k-1}\,e^{-\frac{(1+y)^2}{4t}}\! \le t^{-k-\frac{1}{2}}\,e^{-\frac{(1-y)^2}{36\,t}} \end{aligned}$$with \(t^{k+\frac{1}{2}}\asymp \mu (B(1,\sqrt{t\,}))\).
\(\circ \) Subcase 4.2. Assume that \(t\le 2\le \!-y\). Then
$$\begin{aligned} H_t(1,y)\le h_{t}(1,y)\asymp t^{\frac{1}{2}}\,(-y)^{-k-1}\,e^{-\frac{(1+y)^2}{4t}}\! \lesssim t^{-\frac{1}{2}}\,e^{-\frac{(1-y)^2}{36\,t}} \end{aligned}$$with \(t^{\frac{1}{2}}\asymp \mu (B(1,\sqrt{t\,}))\).
\(\circ \) Subcase 4.3. Assume that \(t\le \!-y\le \frac{1}{2}\). Then
$$\begin{aligned} H_t(1,y)\le h_{t}(1,y)\asymp t^{\frac{1}{2}}\,(-y)^{-k-1}\,e^{-\frac{(1+y)^2}{4t}}\! \le t^{-k-\frac{1}{2}}\,e^{-\frac{(1+y)^2}{8t}}\,e^{-\frac{(1-y)^2}{72\,t}} \end{aligned}$$is bounded above by
$$\begin{aligned} \mu (B(1,\sqrt{t\,}))^{-1}\,e^{-\frac{(1-y)^2}{72\,t}} \end{aligned}$$as \(t^{\frac{1}{2}}\asymp \mu (B(1,\sqrt{t\,}))\) and \(t^{-k}\lesssim e^{\frac{1}{32\,t}}\le e^{\frac{(1+y)^2}{8t}}\).
The proof of (c) follows the same pattern. To begin with, observe that the derivative
of the cut-off is bounded and vanishes unless \(y\in \!\left( -3,-2\right) \cup \left( -\frac{1}{2},0\right) \) and \(t\le \!1\). According to the subcases 1.1, 4.2, and 4.3 above, the contribution of \(\frac{\partial }{\partial y}\left\{ 1-\chi _{t}(1,y)\right\} \, h_{t}(1,y)\) to \(\frac{\partial }{\partial y}H_t(1,y)\) is bounded by
Thus it remains to estimate the contribution of \(\{1-\chi _{t}(1,y)\} \,\frac{\partial }{\partial y}h_{t}(1,y)\).
-
Case 1. Assume that \(|y|\!\le t\).
\(\circ \) Subcase 1.1. Assume that \(t\le \!\frac{1}{2}\). Then
$$\begin{aligned} \{1-\chi _{t}(1,y)\}\, \left| \frac{\partial }{\partial y}h_{t}(1,y)\right|&\lesssim t^{-k-\frac{3}{2}}(1\!+\!|y|)\,e^{-\frac{1+y^2}{4t}}\!\\&\lesssim \overbrace{ t^{-k-\frac{1}{2}}e^{-\frac{1+y^2}{8t}}e^{-\frac{y}{4t}} }^{\text {bounded}}t^{-1}e^{-\frac{(1-y)^2}{8t}}\\&\lesssim t^{-\frac{1}{2}}\mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(1-y)^2}{8t}}. \end{aligned}$$\(\circ \) Subcase 1.2. Assume that \(t\ge \!\frac{1}{2}\). Then
$$\begin{aligned} \{1-\chi _{t}(1,y)\}\, \left| \frac{\partial }{\partial y}h_{t}(1,y)\right|&\lesssim t^{-k-\frac{3}{2}}(1\!+|y|)\, e^{-\frac{1+y^2}{4t}}\\&\!\lesssim t^{-k-1}e^{-\frac{(1-y)^2}{8t}} \overbrace{ \left( \frac{1+y^2}{t}\right) ^{\frac{1}{2}}e^{-\frac{1+y^2}{8t}}e^{-\frac{y}{4t}} }^{\text {bounded}}\\&\lesssim t^{-\frac{1}{2}}\mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(1-y)^2}{8t}}. \end{aligned}$$ -
Case 2. Assume that \(-2\le y\le \!-\frac{1}{2}\).
\(\circ \) Subcase 2.1. If \(t\le \!\frac{1}{2}(\le \!-y)\), then
$$\begin{aligned} \{1-\chi _{t}(1,y)\}\, \frac{\partial }{\partial y}h_{t}(1,y)=0\,. \end{aligned}$$\(\circ \) Subcase 2.2. If \(t\) is bounded below, say \(t\ge \!\frac{1}{2}\), we argue as in Subcase 1.2.
-
Case 3. Assume that \(y\ge t\).
\(\circ \) Subcase 3.1. Assume that \((y\!\ge )t\!\ge \!\frac{1}{2}\). Then
$$\begin{aligned} \{1-\chi _{t}(1,y)\}\, \left| \frac{\partial }{\partial y}h_{t}(1,y)\right|&\lesssim \left\{ t^{-\frac{3}{2}}|1-\!y|+t^{-\frac{1}{2}}y^{-1}\right\} \,y^{-k}\,e^{-\frac{(1-y)^2}{4t}}\\&\lesssim t^{-k-1}\,e^{-\frac{(1-y)^2}{8t}} \underbrace{\left\{ 1+\frac{|1-y|}{\sqrt{t\,}}\,e^{-\frac{(1-y)^2}{8t}} \right\} }_{\text {bounded}}\\&\lesssim t^{-\frac{1}{2}}\mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(1-y)^2}{8t}}. \end{aligned}$$\(\circ \) Subcase 3.2. Assume that \(y\ge \!\frac{1}{2}\!\ge t\). Then
$$\begin{aligned} \{1-\chi _{t}(1,y)\}\, \left| \frac{\partial }{\partial y}h_{t}(1,y)\right|&\lesssim \left\{ t^{-\frac{3}{2}}|1-\!y|+t^{-\frac{1}{2}}y^{-1}\right\} \,y^{-k}\,e^{-\frac{(1-y)^2}{4t}}\\&\lesssim t^{-1}\,e^{-\frac{(1-y)^2}{8t}} \underbrace{\left\{ \sqrt{t\,}\! +\frac{|1-y|}{\sqrt{t\,}}\,e^{-\frac{(1-y)^2}{8t}} \right\} }_{\text {bounded}}\\&\lesssim t^{-\frac{1}{2}}\mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(1-y)^2}{8t}}. \end{aligned}$$\(\circ \) Subcase 3.3. Assume that \(t\le y\le \!\frac{1}{2}\). Then
$$\begin{aligned} \{1-\chi _{t}(1,y)\}\, \left| \frac{\partial }{\partial y}h_{t}(1,y)\right|&\lesssim \left\{ t^{-\frac{3}{2}}|1-\!y|+t^{-\frac{1}{2}}y^{-1}\right\} \,y^{-k}\,e^{-\frac{(1-y)^2}{4t}}\\&\lesssim t^{-1}\,e^{-\frac{(1-y)^2}{8t}} \underbrace{ t^{-k-\frac{1}{2}}\,e^{-\frac{1}{32t}} }_{\text {bounded}}\\&\lesssim t^{-\frac{1}{2}}\mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(1-y)^2}{8t}}. \end{aligned}$$ -
Case 4. Assume that \(y\le \!-t\,(<\!0)\) and that \(y\!\notin \!\left( -2, -\frac{1}{2}\right) \). Recall that \((1\!+\!y)^2\!\ge \!\frac{(1-\,y)^2}{9}\) if and only if \(y\!\notin \!\left( -2, -\frac{1}{2}\right) \).
\(\circ \) Subcase 4.1. Assume that \(2\le t\le \!-y\). Then
$$\begin{aligned} \{1-\chi _{t}(1,y)\}\, \left| \frac{\partial }{\partial y}h_{t}(1,y)\right|&\lesssim \left\{ t^{-\frac{1}{2}}|1+\!y|+t^{\frac{1}{2}}\frac{1+|y|}{|y|}\right\} \, |y|^{-k-1}\,e^{-\frac{(1+y)^2}{4t}}\\&\lesssim t^{-k-1}\,e^{-\frac{(1+y)^2}{8t}} \underbrace{\frac{|1+y|}{\sqrt{t\,}}\, e^{-\frac{(1+y)^2}{8t}}}_{\text {bounded}}\\&\lesssim t^{-\frac{1}{2}}\mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(1-y)^2}{72\,t}}. \end{aligned}$$\(\circ \) Subcase 4.2. Assume that \(t\le 2\le \!-y\). Then
$$\begin{aligned} \{1-\chi _{t}(1,y)\}\, \left| \frac{\partial }{\partial y}h_{t}(1,y)\right|&\lesssim \left\{ t^{-\frac{1}{2}}|1+\!y|+t^{\frac{1}{2}}\frac{1+|y|}{|y|}\right\} \, |y|^{-k-1}\,e^{-\frac{(1+y)^2}{4t}}\\&\lesssim t^{-1}\,e^{-\frac{(1+y)^2}{8t}} \underbrace{\left\{ \frac{|1+y|}{\sqrt{t\,}}\,e^{-\frac{(1+y)^2}{8t}}\! +\sqrt{t}\right\} }_{\text {bounded}}\\&\lesssim t^{-\frac{1}{2}}\mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(1-y)^2}{72\,t}}. \end{aligned}$$\(\circ \) Subcase 4.3. Assume that \(t\le \!-y\le \frac{1}{2}\). Then
$$\begin{aligned} \chi _{t}(1,y)\, \left| \frac{\partial }{\partial y}h_{t}(1,y)\right|&\lesssim \left\{ t^{-\frac{1}{2}}|1+\!y|+t^{\frac{1}{2}}\frac{1+|y|}{|y|}\right\} \, |y|^{-k-1}\,e^{-\frac{(1+y)^2}{4t}}\\&\lesssim t^{-1}\,e^{-\frac{(1+y)^2}{8t}} \underbrace{ t^{-k-\frac{1}{2}}\,e^{-\frac{1}{32t}} }_{\text {bounded}}\\&\lesssim t^{-\frac{1}{2}}\mu \big (B\big (1,\sqrt{t\,}\big )\big )^{-1}\, e^{-\frac{(1-y)^2}{72\,t}}. \end{aligned}$$
Finally, (d) is an immediate consequence of (c). For every \(y^{\prime \prime }\in \![\,y,y^{\prime }]\), we have indeed
Moreover, if \(|y\!-\!y^{\prime }|\!\le \!\frac{1}{2}|x\!-\!y|\), then \(|x\!-\!y^{\prime \prime }|\ge |x\!-\!y|-|y\!-\!y^{\prime \prime }|\ge |x\!-\!y|-|y\!-\!y^{\prime }|\ge \frac{1}{2}|x\!-\!y|\), hence
\(\square \)
Remark 2.7
Contrarily to \(h_{t}(x,y)\), \(H_t(x,y)\) is not symmetric in the space variables \(x,y\). Nevertheless, according to the following result, we may replace \(\mu (B(x,\sqrt{t\,}))\) by \(\mu (B(y,\sqrt{t\,}))\) in the estimates (b), (c) and in the second estimate (d).
Lemma 2.8
For every \(\varepsilon \!>\!0\), there exists \(C\!>\!0\) such that
Proof
By rescaling (see Appendix 1), we can reduce to the case \(t=\!1\). The estimate
is obvious if \(x\) and \(y\) are bounded or if \(|x|/|y|\) is bounded from above. In the remaining case, let us say, when \(|x|\!\ge \!1\!+2|y|\), we have \(|x|\le |x\!-\!y|+|y|\le |x\!-\!y|+\frac{1}{2}\left| x\right| \), hence \(|x|\le 2\,|x\!-\!y|\). Furthermore, as \(|x\!-\!y|\ge |x|-|y|\ge 1\), we have \(|x|\le 2\,(x\!-\!y)^2\). Thus
\(\square \)
The next proposition, which will be used in the proof of Theorem 1.8, shows that the truncated heat kernel \(H_t(x,y)\) captures the main features of the heat kernel \(h_{t}(x,y)\).
Proposition 2.9
The maximal operator
associated with the error
is bounded from \(L^1(\mathbb {R},\mathrm{d}\mu )\) into itself.
Proof
It suffices to check that
The case \(y=0\) is trivial, as \(\chi _{t}(x,0)\) and hence \(Q_t(x,0)\) vanish, for every \(t\!>\!0\) and \(x\!\in \!\mathbb {R}\). Consider next the case \(y\!\in \!\mathbb {R}^*\), which reduces to \(y=1\) by rescaling. Then \(\chi _{t}(x,1)\) and \(Q_t(x,1)\) vanish, unless \(t\!<\!9\) and \(-3\!<\!x\!<\!-\frac{1}{3}\), and in this range (see Proposition 2.3)
is bounded. Hence
\(\square \)
3 Heat Kernel Estimates in the Product Case
According to (1.5) and (1.2), the heat kernel in the product case splits up into one-dimensional heat kernels:
By expanding
we get
Here
and \({\mathbf{P}}_{t}(\mathbf{x},{\mathbf{y}})\) is the sum of all possible products
where each factor \(p_{\,t}^{(j)}(x_j,y_j)\) is equal to \(H_{\,t}^{(j)}(x_j,y_j)\) or \(Q_{\,t}^{(j)}(x_j,y_j)\), and at least one factor \(p_{\,t}^{(j)}(x_j,y_j)\) is equal to \(Q_{\,t}^{(j)}(x_j,y_j)\). Notice the rescaling property
and similarly for the other product kernels. The following estimates follow from the one-dimensional case (see Theorem 2.6 and Remark 2.7).
Theorem 3.2
-
(a)
On-diagonal estimate :
$$\begin{aligned} {\mathbf{H}}_{t}(\mathbf{x},\mathbf{x})\asymp \varvec{\mu }\big (\mathbf{B}\big (\mathbf{x},\sqrt{t\,}\big )\big )^{-1}, \qquad \forall \;t\!>\!0,\,\forall \;\mathbf{x}\!\in \!\mathbb {R}^n. \end{aligned}$$ -
(b)
Off-diagonal Gaussian estimate :
$$\begin{aligned} 0\le {\mathbf{H}}_{t}(\mathbf{x},{\mathbf{y}}) \lesssim \varvec{\max }\left\{ \varvec{\mu }\big (\mathbf{B}\big (\mathbf{x},\sqrt{t\,}\big )\big ), \varvec{\mu }\big (\mathbf{B}\big ({\mathbf{y}},\sqrt{t\,}\big )\big )\right\} ^{-1}\, e^{-\frac{|\mathbf{x}-{\mathbf{y}}|^2}{c\,t}} \end{aligned}$$for every \(t\!>\!0\) and for every \(\mathbf{x},\mathbf{y}\!\in \!\mathbb {R}^n\).
-
(c)
Gradient estimate :
$$\begin{aligned} |\nabla _{\mathbf{y}} {\mathbf{H}}_{t}(\mathbf{x},{\mathbf{y}})| \lesssim t^{-\frac{1}{2}}\, \varvec{\max }\left\{ \varvec{\mu }\big (\mathbf{B}\big (\mathbf{x},\sqrt{t\,}\big )\big ), \varvec{\mu }\big (\mathbf{B}\big ({\mathbf{y}},\sqrt{t\,}\big )\big )\right\} ^{-1}\, e^{-\frac{|\mathbf{x}-{\mathbf{y}}|^2}{c\,t}} \end{aligned}$$for every \(t\!>\!0\) and \(\mathbf{x},\mathbf{y}\!\in \!\mathbb {R}^n\).
-
(d)
Lipschitz estimates :
$$\begin{aligned} |{\mathbf{H}}_{t}(\mathbf{x},{\mathbf{y}})-{\mathbf{H}}_{t}(\mathbf{x},{\mathbf{y}}^{\prime })| \lesssim \varvec{\mu }\big (\mathbf{B}\big (\mathbf{x},\sqrt{t\,}\big )\big )^{-1}\, \frac{|{\mathbf{y}}-{\mathbf{y}}^{\varvec{\prime }}|}{\sqrt{t\,}}\,, \end{aligned}$$for every \(t\!>\!0\) and \(\mathbf{x},\mathbf{y},\mathbf{y}^{\varvec{\prime }}\in \!\mathbb {R}^n\), with the following improvement, if \(|{\mathbf{y}}\!-\!{\mathbf{y}}^{\varvec{\prime }}|\le \frac{1}{2}\,|\mathbf{x}\!-\!{\mathbf{y}}|\) :
$$\begin{aligned} |{\mathbf{H}}_{t}(\mathbf{x},{\mathbf{y}})-{\mathbf{H}}_{t}(\mathbf{x},{\mathbf{y}}^{\varvec{\prime }})| \lesssim \varvec{\max }\left\{ \varvec{\mu }\big (\mathbf{B}\big (\mathbf{x},\sqrt{t\,}\big )\big ), \varvec{\mu }\big (\mathbf{B}\big ({\mathbf{y}},\sqrt{t\,}\big )\big )\right\} ^{-1} e^{-\frac{|\mathbf{x}-{\mathbf{y}}|^2}{c\,t}} \frac{|{\mathbf{y}}-{\mathbf{y}}^{\varvec{\prime }}|}{\sqrt{t\,}}. \end{aligned}$$
Let us turn to the analog of Proposition 2.9 in the product case.
Proposition 3.3
The maximal operator
is bounded from \(L^1(\mathbb {R}^n,\mathrm{d}{\varvec{\mu }})\) into itself.
Proof
We will show again that
but the proof will be more involved in the product case than in the one-dimensional case. Let us begin with some observations. First of all, by using the symmetries
and by interchanging variables, we may reduce to products of the form
where \(1\le n^{\prime }\!\le n\) and \(0\le y_1\!\le {\cdots }\le y_{n^{\prime }}\). Next we may assume that, for every \(1\le j\le n^{\prime }\),
because otherwise \(\chi _{t}(x_j,y_j)\) and hence \(Q_{\,t}^{(j)}(x_j,y_j)\) vanish. By rescaling, we may reduce to the case \(y_1\!=\!1\). Consequently, \(t\) is bounded by \(x_1^{2}\!<9\,y_1^{2}\!=9\) and each factor \(Q_{\,t}^{(j)}(x_j,y_j)\) is bounded by
Thus, on the one hand, the integral
is bounded, uniformly in \({\mathbf{y}}^{\prime }\). On the other hand, let us prove the uniform boundedness of
when \(n^{\prime \prime }\!=n-n^{\prime }\!>\!0\). For this purpose, let us deduce from the Gaussian estimate
that
Assume first that \(\left| \mathbf{x}^{\prime \prime } -{\mathbf{y}}^{\prime \prime }\right| \ge \!\sqrt{t}\) with \(0\!<\!t\!<\!9\). Then, by using (6.2),
Assume next that \(0<|\mathbf{x}^{\prime \prime } -{\mathbf{y}}^{\prime \prime }| \le \!\sqrt{t}\,(\le 3)\). Then, by using (6.2) again,
Now that we have estimated \(t^{\frac{n^\prime }{2}}{\mathbf{H}}_{\,t}^{\prime \prime } \left( \mathbf{x}^{\prime \prime }, {\mathbf{y}}^{\prime \prime }\right) \), let us split up the integral
according to the decomposition \(\mathbb {R}^{n^{\prime \prime }}\backslash \!\{0\}\!=\!\bigsqcup _{\,j\in \mathbb {Z}}\underbrace{ \left\{ \mathbf{x}^{\prime \prime }\!\in \!\mathbb {R}^{n^{\prime \prime }}| \,\,2^{\,j-\frac{1}{2}}\!\!\le \!|\mathbf{x}^{\prime \prime }\!-\!\mathbf{y}^{\prime \prime }| \!<\!2^{\,j+\frac{1}{2}}\right\} }_{\Omega _j}\). Let us show that
If \(j\ge 0\), we have indeed
and, if \(j\le 0\),
By summing up over \(j\!\in \!\mathbb {Z}\), we obtain the uniform boundedness of \({\mathbf{I}}^{\prime \prime } \left( {\mathbf{y}}^{\prime \prime }\right) \). \(\square \)
4 Proof of Theorem 1.8
Theorem 1.8 relies on the following result due to Uchiyama [25].
Theorem 4.1
Assume that a set \(X\) is equipped with
-
a quasi-distance \(\widetilde{d}\), i.e., a distance except that the triangular inequality is replaced by the weaker condition
$$\begin{aligned} \widetilde{d}(x,y)\le A\,\{\widetilde{d}(x,z)+\widetilde{d}(z,y)\} \qquad \forall \;x,y,z\!\in \!X, \end{aligned}$$ -
a measure \(\mu \) whose values on quasi-balls satisfy
$$\begin{aligned} \frac{r}{A}\le \mu \left( \widetilde{B}(x,r)\right) \le r \qquad \forall \;x\!\in \!X,\,\forall \;r\!>\!0\,, \end{aligned}$$ -
a continuous kernel \(K_r(x,y)\!\ge \!0\) such that, for every \(r\!>\!0\) and \(x,y,y^{\prime }\in \!X\),
-
\(\circ \) \(K_r(x,x)\ge \frac{1}{Ar}\) ,
-
\(\circ \) \(K_r(x,y)\le r^{-1}\left( 1+\frac{\widetilde{d}(x,y)}{r} \right) ^{-1-\delta }\) ,
-
\(\circ \) \(\left| K_r(x,y)-K_r(x,y^{\prime })\right| \le r^{-1}\left( 1+\frac{\widetilde{d}(x,y)}{r}\right) ^{-1-2\delta }\left( \frac{\widetilde{d}(y,y^{\prime })}{r}\right) ^{\delta }\) when \(\widetilde{d}(y,y^{\prime })\! \le \!\frac{r+\widetilde{d}(x,y)}{4A}\) .
Here \(A\!\ge \!1\) and \(\delta \!>\!0\). Let us introduce the following definitions :
-
an atom is a measurable function \(a:X\!\rightarrow \mathbb {C}\) such that
$$\begin{aligned} a \; is\; supported \; in \; a \; quasi{\text{- }}ball \,\widetilde{B} , \,\Vert a\Vert _{L^\infty (X,\, \mathrm{d}\mu )}\lesssim \mu (\widetilde{B})^{-1} and \, \displaystyle \int _Xd\mu a=0, \end{aligned}$$ -
the atomic Hardy space \(H^1_\mathrm{atom}(X,\mu ,\widetilde{d}\,)\) consists of all functions \(f\!\in \!L^1(X,\mathrm{d}\mu )\) which can be written as
$$\begin{aligned} f=\sum \nolimits _{\ell }\lambda _{\ell }\,a_{\ell }\,, \end{aligned}$$(4.2)where the \(a_{\ell }\)’s are atoms and the \(\lambda _{\ell }\)’s are complex numbers such that \(\sum _{\ell }|\lambda _{\ell }|\!<\!+\infty \),
-
the atomic norm is given by
$$\begin{aligned} \Vert f\Vert _{H^1_\mathrm{atom}(X,\mu ,\widetilde{d}\,)} =\,\inf \;\sum \nolimits _{\ell }|\lambda _{\ell }|\,, \end{aligned}$$(4.3)where the infimum is taken over all representations (4.2),
-
the maximal Hardy space \(H^1_\mathrm{max}(X,\mu ,K_r)\) consists of all functions \(f\!\in \!L^1(X,\mathrm{d}\mu )\) such that
$$\begin{aligned} K_*f(x)=\sup \nolimits _{\,r>0}\, \left| {\displaystyle \int _X}\mathrm{d}\mu (y)\,K_r(x,y)\,f(y)\,\right| \end{aligned}$$belongs to \(L^1(X,\mathrm{d}\mu )\).
Then \(H^1_\mathrm{atom}(X,\mu ,\widetilde{d}\,)\) coincides with \(H^1_\mathrm{max}(X,\mu ,K_r)\), and (4.3) is comparable to the maximal norm \(\Vert K_*f\Vert _{L^1(X,\, d\mu )}\).
We now adapt Uchiyama’s Theorem to our setting. For \(X\!=\mathbb {R}^n\), equipped with the Euclidean distance \(d(\mathbf{x},\mathbf{y})=|\mathbf{x}-\mathbf{y}|\) and the measure (1.4), set
where the infimum is taken over all closed balls \(\mathbf{B}\) containing \(\mathbf{x}\) and \(\mathbf{y}\), and
where \(t=t(\mathbf{x},r)\) is defined by \(\varvec{\mu }\left( \mathbf{B}\left( \mathbf{x},\sqrt{t}\right) \right) =r\). In Appendices 2 and 3, we check that \(\left( X,\varvec{\mu }, \widetilde{d}, K_r\right) \) satisfy the assumptions of Uchiyama’s theorem with \(\delta =\frac{1}{\mathbf{N}}\). Actually the conditions in Theorem 4.1 are obtained up to constants, and they can be achieved by considering suitable multiples of \(\varvec{\mu }\) and \(K_r(\mathbf{x},\mathbf{y})\). Thus the conclusion of Uchiyama’s theorem holds for the quasi-distance \(\widetilde{d}\) and for the maximal operator \(K_*\).
On the one hand, \(d\) and \(\widetilde{d}\) define the same atomic Hardy space \(H^1_\mathrm{atom}\), with equivalent norms, as balls and quasi-balls are comparable. Let us elaborate. For every \(\mathbf{x},\mathbf{y}\!\in \!\mathbb {R}^n\) and \(t\!>\!0\), we have
where \(r=\varvec{\mu }\left( \mathbf{B}\left( \mathbf{x},\sqrt{t\,}\right) \right) \). The first implication is an immediate consequence of the definition of \(\widetilde{d}\) and the second one is obtained by combining Lemma 6.4(b) in Appendix 2 with (6.2) in Appendix 1. Hence, there exists a constant \(c\!>\!0\) such that
and these sets have comparable measures, according to Appendix 1.
On the other hand, the maximal operators \(K_*\) and \(\mathbf{H}_{*}\) coincide, and they define the same maximal Hardy space \(H^1_\mathrm{max}\), with equivalent norms, as the heat maximal operator \(\mathbf{h}_{*}\) (see (1.7)), according to Propositions 2.9 and 3.3. Indeed, for every \(f\!\in \!L^1(\mathbb {R}^n,d\varvec{\mu })\), the integrals
differ at most by a multiple of \(\Vert f\Vert _{L^1({\mathbb {R}}^n,\, d\varvec{\mu })}\), which is itself controlled by either integral above, as \(\mathbf{h}_{t}(\mathbf{x},\mathbf{y})\,\mathrm{d}\varvec{\mu }(\mathbf{y})\) and \(\mathbf{H}_{t}(\mathbf{x},\mathbf{y})\,\mathrm{d}\varvec{\mu }(\mathbf{y})\) are approximations of the identity.
In conclusion, the atomic Hardy space \(H^1_\mathrm{atom}\) associated with Euclidean balls coincides with the Hardy space \(H^1_\mathrm{max}\) defined by the heat maximal operator \(\mathbf{h}_{*}\). \(\square \)
5 Proof of Theorem 1.10
The proof of Theorem 1.10 requires some weighted estimates in Dunkl analysis, which are well-known in the Euclidean setting corresponding to \(\mathbf{k}=\mathbf{0}\). Let us first prove a weak analog of the Euclidean estimate
Lemma 5.1
For every \(\ell \!\in \!\mathbb {N}\) and \(r\!>\!0\), there is a constant \(C=C_{\ell ,r}\!>\!0\) such that
for every \(f\!\in \!C^\ell (\mathbb {R}^n)\) with \(\mathrm{supp }f\!\subset \mathbf{B}(0,r)\).
Proof
By using the Riemann–Lebesgue lemma for the Fourier transform (1.9), we get
The last expression is bounded by \(\Vert f\Vert _{C^\ell }\) as, by induction on \(\ell \), \(\mathrm{supp }( D_j^{\ell }f)\!\subset \!B(0,r)\) and \(\Vert D_j^{\ell }f\Vert _{L^\infty }\! \lesssim \Vert f\Vert _{C^\ell }.\) \(\square \)
Corollary 5.2
For every \(\ell \!\in \!\mathbb {N}\), \(r\!>\!0\), and \(\epsilon \!>\!0\), there is a constant \(C=C_{\ell ,r,\epsilon }\!>\!0\) such that
for every \(f\!\in \!W_{2}^{\ell +n/2+\epsilon }(\mathbb {R}^n)\) with \(\mathrm{supp }f\!\subset \mathbf{B}(0,r)\), where \(\mathbf{N}\) denotes the homogeneous dimension (1.6).
Proof
This result is deduced from Lemma 5.1, by using on the left-hand side the finiteness of the integral
and on the right-hand side the Euclidean Sobolev embedding theorem. \(\square \)
Proposition 5.3
For every \(\sigma \!>\!0\), \(r\!>\!0\), and \(\epsilon \!>\!0\), there is a constant \(C=C_{\sigma ,r,\epsilon }\!>\!0\) such that
for every \(f\!\in \!W_{2}^{\sigma +\epsilon }(\mathbb {R}^n)\) with \(\mathrm{supp }f\!\subset \mathbf{B}(0,r)\).
Proof
Let \(\chi \!\in \!C_c^\infty (\mathbb {R}^n)\). Following an argument due to Mauceri and Meda [13], this result is obtained by interpolation between the \(L^2\) estimate
which is deduced from Plancherel’s formula, and the following estimate for \(\ell \!\in \!\mathbb {N}\) large, which is deduced from Corollary 5.2:
\(\square \)
By using the Cauchy–Schwarz inequality, we deduce finally the following result.
Corollary 5.4
For every \(\sigma \!>\!0\), \(r\!>\!0\), and \(\epsilon \!>\!0\), there is a constant \(C=C_{\sigma ,r,\epsilon }\!>\!0\) such that
for every \(f\!\in \! W_{2}^{\sigma +\mathbf{N}/2+\epsilon }(\mathbb {R}^n)\) with \(\mathrm{supp }f\!\subset \mathbf{B}(0,r)\).
Let us next prove analogs in the Dunkl setting of the Euclidean estimates
and
Recall that Dunkl translations are defined via the Fourier transform (1.9) by
(see [17, 19, 23, 24]) and have an explicit integral representation
in dimension \(1\) (see [2, 16, 23]) and hence in the product case. Specifically,
where
and
Thus \(\varvec{\nu }_{\mathbf{x},\mathbf{y}}\) is a signed measure which is supported in the product
of the one-dimensional sets
and which is generically given by
Moreover, it is known (see [2, 16, 23]) that
Lemma 5.5
For every \(\delta \ge 0\), \(L^1((1\!+\!|\mathbf{x}|)^\delta \mathrm{d}\varvec{\mu }(\mathbf{x}))\) is an algebra with respect to the Dunkl convolution product
Proof
By using the symmetries
we have
We conclude by estimating
\(\square \)
Lemma 5.6
For every \(\delta \!>\!0\), there is a constant \(C\!>\!0\) such that, for every \(\mathbf{y}\in \mathbb {R}^n\) and \(r\!>\!0\),
where
is the orbit of the cube \(\mathbf Q(\mathbf{y},r)=\prod _{j=1}^nB(y_j,r)\) under the group generated by the reflections (1.1).
Proof
As \(\mathbb {R}^n\!\backslash \!\mathcal {O}(\mathbf{y},r)\) is contained in the union of the sets
we have
As
when \(\mathbf{x}\!\in \!A_j\) and \(\mathbf{z}\!\in \!{\mathcal {I}}_{\mathbf{x},\mathbf{y}}\), the latter expression is bounded above by
We conclude by using the uniform estimate
\(\square \)
Let us turn to the proof of Theorem 1.10, which consists in estimating
for every atom \(a\) in the Hardy space \(H^1\!=H^1_\mathrm{atom}\). By rescaling, it suffices to prove (5.7) for any atom \(a\) associated with a unit ball \(\mathbf{B}(\mathbf{z},1)\). As \(\mathbf{h}_{*}\) and \({\mathcal {T}}_{m}\) are bounded on \(L^2(\mathbb {R}^n,d\varvec{\mu })\), we have
Thus it remains to show that
For this purpose, let us introduce a dyadic partition of unity on the Dunkl transform side. More precisely, given a smooth radial function \(\psi \) on \(\mathbb {R}^n\) such that
let us split up
Set
Then \(e^{-t|\varvec{\xi }|^2}m(\varvec{\xi })={\displaystyle \sum \nolimits _{\ell \in \mathbb {Z}}}m_{t,\ell }(2^{-\ell }\varvec{\xi })\). Consider the convolution kernel
of the multiplier operator \(\mathcal T_{m_{t,\ell } (2^{-\ell }. )}\).
Lemma 5.9
-
(a)
On the one hand, for every \(0\!\le \!\delta \!<\!\epsilon \), we have
$$\begin{aligned} \int _{\mathbb {R}^n\backslash \mathcal {O}(\mathbf{z},2)} \mathrm{d}\varvec{\mu }(\mathbf{x})\, \sup _{t>0}|F_{t,\ell }(\mathbf{x},\mathbf{y})|\, \lesssim \,M\,2^{-\delta \ell }, \quad \forall \;\ell \!\in \!\mathbb {Z}, \,\forall \,\mathbf{z}\in \!\mathbb {R}^n, \;\forall \;\mathbf{y}\!\in \!\mathcal {O}(\mathbf{z},1). \end{aligned}$$ -
(b)
On the other hand,
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\mathbf{x})\,\sup _{t>0}|F_{t,\ell }(\mathbf{x},\mathbf{y})-F_{t,\ell }(\mathbf{x},\mathbf{y}^{\prime })|\, \lesssim \,M\,2^{\ell }|\mathbf{y}\!-\!\mathbf{y}^{\prime }|\, , \quad \forall \;\ell \!\in \!\mathbb {Z},\; \forall \;\mathbf{y},\mathbf{y}^{\prime }\in \!\mathbb {R}^n. \end{aligned}$$
Proof
On the one hand, as
Lemma 5.1 yields the estimate
which holds for any \(d\!\in \!\mathbb {N}\) and which is uniform in \(t\!>\!0\) and \(\ell \!\in \!\mathbb {Z}\). On the other hand, Corollary 5.4 yields the estimate
which holds uniformly in \(\ell \!\in \!\mathbb {Z}\). By resuming the proof of Lemma 5.5, we deduce that
We reach our first conclusion by rescaling and by using Lemma 5.6:
Let us turn to the proof of (b). This time we factorize
and accordingly
On the one hand, by resuming the proof of (5.10), we get
On the other hand, \(\mathbf{h}(\mathbf{x},\mathbf{y})=(\tau _{-\mathbf{y}}h)(\mathbf{x})\) is the heat kernel at time \(t=1\), which satisfies
according to the next lemma. After rescaling, we reach our second conclusion:
\(\square \)
Lemma 5.11
The following gradient estimate holds for the heat kernel:
Proof
We can reduce to the one-dimensional case and moreover to \(t=\!1\) by rescaling. It follows from our gradient estimates for the heat kernel in dimension \(1\) (see Proposition 2.3) that
-
Case 1: Assume that \(|y|\le 2\). Then \(|\partial _{y}h_1(x,y)|\lesssim e^{-x^2/16}\), hence
$$\begin{aligned} \int _{-\infty }^{+\infty }\mathrm{d}x\,|x|^{2k}\, \left| \frac{\partial }{\partial y}h_1(x,y)\right| \,\lesssim \,1\,. \end{aligned}$$ -
Case 2: Assume that \(|y|\ge 2\). Then \(|x|/|y|\le 1+\frac{1}{2}\left| |x|-|y|\right| \), hence
$$\begin{aligned}&|x|^{2k}\,\left| \frac{\partial }{\partial y}h_1(x,y)\right| \lesssim \left( \frac{|x|}{|y|}\right) ^{k}\, e^{-\frac{1}{8}(|x|-|y|)^2}\\&\qquad \lesssim \left( 1+\left| |x|-|y|\right| \right) ^{k}\, e^{-\frac{1}{8}(|x|-|y|)^2} \lesssim e^{-\frac{1}{16}(|x|-|y|)^2} \end{aligned}$$and
$$\begin{aligned} \int _{-\infty }^{+\infty }\mathrm{d}x\,|x|^{2k}\, \left| \frac{\partial }{\partial y}h_1(x,y)\right| \, \lesssim \int _{\,0}^{+\infty }\mathrm{d}x\,e^{-\frac{1}{16}(x-|y|)^2} \lesssim \int _{-\infty }^{+\infty }\mathrm{d}z\,e^{-\frac{1}{16}z^2} \lesssim \,1\,. \end{aligned}$$
Conclusion of proof of Theorem 1.10
Let us split up and estimate
Then (5.8) follows from Lemma 5.9. \(\square \)
Example 5.1
The Riesz transforms \({\mathcal {R}}_j=D_j(-\mathbf L)^{-1\slash 2}\) in the Dunkl setting correspond to the multipliers \(\xi _j/|\varvec{\xi }|\), up to a constant. Hence, by Theorem 1.10, they are bounded operators on the Hardy space \(H^1\).
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Acknowledgments
JPA and JD were partially supported by the European Commission (European program ToK HANAP 2005–2009, Polish doctoral program SSDNM 2010–2013) and by a French–Polish cooperation program (PHC Polonium 22529RF 2010–2011). JPA, NBS, NH partially supported by a French–Tunisian cooperation program (PHC Utique / CMCU 10G1503 2010–2013). JD was supported by the Polish National Science Centre (Narodowe Centrum Nauki, Grant DEC-2012/05/B/ST1/00672) and by the University of Orléans. The third author wishes to thank Paweł Głowacki and Jacek Zienkiewicz for conversations on the subject of the paper.
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Appendices
Appendices
1.1 Appendix 1: Measure of Balls
Recall that \(k_1,\dots ,k_n\ge 0\) and \(\mathbf{N}\!=n+\sum \nolimits _{j=1}^{\,n}2k_j\). On \(\mathbb {R}^n\), equipped with the Euclidean distance, the product measure \(\mathrm{d}\varvec{\mu }(\mathbf{x})\) (see (1.4)) has the following rescaling properties:
and
Moreover,
Hence
In particular, \(\varvec{\mu }\) is doubling, i.e.,
Let us prove (6.1) and (6.2). In dimension \(n=1\), we have
On the one hand, if \(r\le \frac{|x|}{2}\), we deduce that
On the other hand, if \(|x|\le 2r\), we estimate from above
and from below
Thus \(\mu (B(x,r))\!\asymp \!(|x|\!+\!r)^{2k}r\) in all cases and
The product case follows from the one-dimensional case, since the ball \(\mathbf{B}(\mathbf{x},r)\) and the cube
have comparable measures. More precisely, we have
with
1.2 Appendix 2: Distances
The following result, which is used in Sect. 4, is certainly known among specialists. We include nevertheless a proof for lack of reference and for the reader’s convenience.
Lemma 6.4
Let \((X,d,\mu )\) be a metric measure space such that balls have finite positive measure and satisfy the doubling property, i.e.,
Set
where the infimum is taken over all closed balls \(B\) containing \(x\) and \(y\). Then
-
(a)
\(\,\widetilde{d}\) is a quasi-distance,
-
(b)
\(\,\widetilde{d}(x,y)\asymp \mu (B(x,d(x,y)))\) \(\forall \;x,y\!\in \!X\),
Moreover, if the measure \(\mu \) has no atoms and \(\mu (X)=+\infty \), then
-
(c)
\(\,\mu (\widetilde{B}(x,r))\asymp r\), for every \(x\!\in \!X\) and \(r\!>\!0\), where \(\widetilde{B}(x,r)\) denotes the closed quasi-ball with center \(x\) and radius \(r\).
Proof
Let us first prove (b). Set \(R=d(x,y)\). On the one hand, we have \(\widetilde{d}(x,y)\le \mu (B(x,R))\), as \(x\) and \(y\) belong to \(B(x,R)\). On the other hand, if \(x\) and \(y\) belong to a ball \(B=B(z,r)\), then \(R\le 2\,r\), hence \(B(x,R)\subset B(z,3r)\) and \(\mu (B(x,R))\le \mu (B(z,3r))\asymp \mu (B(z,r))\). By taking the infimum over all balls \(B\) containing both \(x\) and \(y\), we conclude that \(\mu (B(x,R))\lesssim \widetilde{d}(x,y)\). Let us next prove (a). For every \(x,y,z\!\in \!X\), we have \(d(x,y)\le d(x,z)+d(z,y)\). Assume that \(r=d(x,z)\ge d(z,y)\). Then \(x,y\!\in \!B(z,r)\). By using (b), we deduce that
Let us finally prove (c). Given \(x\!\in \!X\), notice that \(\mu (B(x,r))\) is an increasing càdlàg function of \(r\!\in \!(0,+\infty )\) such that
Here we have used our additional assumptions. Let \(x\!\in \!X\) and \(r\!>\!0\). On the one hand, for every \(y\!\in \!\widetilde{B}(x,r)\), we have \(\mu (B(x,d(x,y))\asymp \widetilde{d}(x,y)\le r\). Hence
Let \(y\!\in \!\widetilde{B}(x,r)\) such that \(d(x,y)\ge \frac{R}{2}\). Then \(\widetilde{B}(x,r)\subset B(x,R)\subset B(x,2\,d(x,y))\). Hence
On the other hand,
As \(\mu (B(x,T/2))\), we have \(\widetilde{d}(x,y)\!<\!r\) for every \(y\!\in \!B(x,T/2)\), hence \(B(x,T/2)\!\subset \!\widetilde{B}(x,r)\). Consequently,
\(\square \)
1.3 Appendix 3: Kernel bounds
Recall from Sect. 4 that the kernels \(K_r(\mathbf{x},\mathbf{y})\) and \(\mathbf{H}_{t}(\mathbf{x},\mathbf{y})\) are related by (4.4). In this appendix, we check that the Gaussian estimates of \(\mathbf{H}_{t}(\mathbf{x},\mathbf{y})\) in Theorem 3.2 imply the estimates of \(K_r(\mathbf{x},\mathbf{y})\) required in Uchiyama’s theorem (Theorem 4.1). This result is certainly well known among specialists. We include nevertheless a proof for lack of reference and for the reader’s convenience.
According to Appendices 1 and 2, we may consider the quasi-distance \(\widetilde{d}\) on \(\mathbb {R}^n\) associated with the Euclidean distance \(d(\mathbf{x},\mathbf{y})=|\mathbf{x}-\mathbf{y}|\) and the product measure (1.4). The on-diagonal lower estimate
is an immediate consequence of Theorem 3.2(a). For every \(\delta \!>\!0\), the upper estimate
follows from Theorem 3.2(b), more precisely by combining
with
The main problem consists in checking the following Lipschitz estimate.
Lemma 6.7
There exists \(C_3\!>\!0\), and, for every \(\delta \!>\!0\), there exists \(C_4\!>\!0\) such that
if \(\widetilde{d}(\mathbf{y},\mathbf{y}^{\prime })\le C_3\max \,\{r,\widetilde{d}(\mathbf{x},\mathbf{y})\}\).
Proof
Let us begin with some observations. First of all, (6.8) follows from (6.5), as long as \(\widetilde{d}(\mathbf{y},\mathbf{y}^{\prime })\asymp r\). In this case, we have indeed
Next, notice that
This follows indeed from (6.2) and the estimates
Similarly, we have
In particular, there exists \(C_3\!>\!0\) such that
Let us turn to the proof of (6.8) and assume first that \(\widetilde{d}(\mathbf{x},\mathbf{y})\ge r\). In this case, \(|\mathbf{x}-\mathbf{y}|\gtrsim \!\sqrt{t\,}\) and \(\widetilde{d}(\mathbf{y},\mathbf{y}^{\prime }) \le C_3\,\widetilde{d}(\mathbf{x},\mathbf{y})\), hence \(|\mathbf{y}-\mathbf{y}^{\prime }| \le \frac{1}{2}|\mathbf{x}-\mathbf{y}|\). Thus, according to Theorem 3.2(d),
is bounded above by
After substituting \(r=\varvec{\mu }\left( \mathbf{B}\left( \mathbf{x},\sqrt{t}\right) \right) \) and estimating
as in (6.6), it remains to show that
If \(|\mathbf{y}-\mathbf{y}^{\prime }|\le \!\sqrt{t\,}\), then
with
and
If \(|\mathbf{y}-\mathbf{y}^{\prime }|\ge \!\sqrt{t\,}\), we argue similarly, estimating this time
and
Assume next that \(\widetilde{d}(\mathbf{x},\mathbf{y})\le r\). Then \(|\mathbf{x}-\mathbf{y}|\lesssim \!\sqrt{t\,}\), \(\widetilde{d}(\mathbf{y},\mathbf{y}^{\prime })\le C_3\,r\) and (6.8) amounts to
According to Theorem 3.2(d),
As
we have
As \(\frac{\widetilde{d}(\mathbf{y},\mathbf{y}^{\prime })}{r} \le C_3\) and \(\frac{\varvec{\mu }(\mathbf{B}(\mathbf{y},|\mathbf{y}-\mathbf{y}^{\prime }|))}{\varvec{\mu }(\mathbf{B}(\mathbf{y},\sqrt{t}))}\) is bounded below by a power of \(\frac{|\mathbf{y}-\mathbf{y}^{\prime }|}{\sqrt{t\,}}\), we deduce first that \(|\mathbf{y}-\mathbf{y}^{\prime }| \lesssim \sqrt{t}\) and next that
This concludes the proof of Lemma 6.7. \(\square \)
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Anker, JP., Ben Salem, N., Dziubański, J. et al. The Hardy Space \(H^1\) in the Rational Dunkl Setting. Constr Approx 42, 93–128 (2015). https://doi.org/10.1007/s00365-014-9254-2
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DOI: https://doi.org/10.1007/s00365-014-9254-2